Category: Class 12

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1

Exercise 3.1 Class 12 Maths State Board Question 1.
If the sides of a cubic box are increased by 1, 2, 3 units respectively to form a cuboid, then the volume is increased by 52 cubic units. Find the volume of the cuboid.
Solution:
Let the side of the cube be ‘x’
Sides of cuboid are (x + 1) (x + 2) (x + 3)
∴ Volume of cuboid = x³ + 52
⇒ (x + 1) (x + 2) (x + 3) = x³ + 52
⇒ (x² + 3x + 2)(x + 3) = x³ + 52
⇒ x³ + 3x² + 3x² + 9x + 2x + 6 – x³ – 52 = 0
⇒ 6x² + 11x – 46 = 0 (÷2)
⇒ (x – 2) (6x + 23) = 0
⇒ x – 2 = 0 or 6x + 23 = 0
⇒ x = 2 or x = \(-\frac{23}{6}\) (not possible)
∴ x = 2
Volume of cube = 2³ = 8
Volume of cuboid = 52 + 8 = 60 cubic units

12th Maths Exercise 3.1 Question 2.
Construct a cubic equation with roots
(i) 1, 2 and 3
(ii) 1, 1 and -2
(iii) 2, \(\frac { 1 }{ 2 }\) and 1
Solution:
(i) Given roots are α = 1, β = 2, γ = 3
The cubic equation is
x³ – x² (α + β + γ) + x (αβ + βγ + γα) – αβγ = 0
⇒ x³ – x² (1 + 2 + 3) + x (2 + 6 + 3) – (1) (2) (3) = 0
⇒ x³ – 6x² + 11x – 6 = 0
(ii) α = 1, β = 1, γ = -2
The cubic equation is
x³ – x² (α + β + γ) + x (αβ + βγ + γα) – αβγ = 0
⇒ x³ – x² (1 + 1 – 2) + x (1 – 2 – 2) – (1) (1) (-2) = 0
⇒ x³ – 0x² – 3x + 2 = 0
⇒ x³ – 3x + 2 = 0
(iii) α = 2, β = \(\frac { 1 }{ 2 }\), γ = 1
The cubic equation is
x³ – x² (α + β + γ) + x (αβ + βγ + γα) – αβγ = 0

2x³ – 7x² + 7x – 2 = 0

12th Maths Theory Of Equations Question 3.
If α, β and γ are the roots of the cubic equation x³ + 2x² + 3x + 4 = 0, form a cubic equation whose roots are
(i) 2α, 2β, 2γ
(ii) \(\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\)
(iii) -α, -β, -γ
Solution:
(i) Given that α, β, γ are the roots of x³ + 2x² + 3x + 4 = 0
Compare with x³ + bx² + cx + d = 0
b = 2, c = 3, d = 4
α + β + γ = -6 = -2
αβ + βγ + γα = c = 3
αβγ = -d = -4
Given roots are 2α, 2β, 2γ
2α + 2β + 2γ = 2 (α + β + γ)
= 2 (-2)
= -4
(2α) (2β) + (2β) (2γ) + (2γ) (2α) = (4αβ + 4βγ + 4γα)
= 4(αβ + βγ + γα)
= 4(3)
= 12
(2α) (2β) (2γ) = 8(αβγ)
= 8(-4)
= -32
The equation is
x³ – x² (2α + 2β + 2γ) + x (4αβ + 4βγ + 4γα) – 8 (αβγ) = 0
⇒ x³ – x² (-4) + x (12) – (-32) = 0
⇒ x³ + 4x² + 12x + 32 = 0
(ii) The given roots are \(\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\)
The cubic equation is

4x³ + 3x² + 2x + 1 = 0 (Multiply by 4)
(iii) The given roots are -α, -β, -γ
The cubic equation is
x³ – x² (-α – β – γ) + x (αβ + βγ + γα) + (αβγ) = 0
⇒ x³ + x² (α + β + γ) + x (αβ + βγ + γα) + (αβγ) = 0
⇒ x³ + x² (-2) + x (3) – 4 = 0
⇒ x³ – 2x² + 3x – 4 = 0

12th Maths Chapter 3 Exercise 3.1 Question 4.
Solve the equation 3x³ – 16x² + 23x – 6 = 0 if the product of two roots is 1.
Solution:
The given equation is 3x³ – 16x² + 23x – 6 = 0
⇒ \(x^{3}-\frac{16}{3} x^{2}+\frac{23}{3} x-2=0\) (÷3)
Let the roots be α, β, γ
α + β + γ = -b = \(\frac{16}{3}\) …….. (1)
αβ + βγ + γα = c = \(\frac{23}{3}\) …….. (2)
αβγ = -d = 2 ……. (3)
Given that αβ = 1
from (3), γ = 2
Substitute \(\beta=\frac{1}{\alpha}\), γ = 2 in (1)
⇒ \(\alpha+\frac{1}{\alpha}+2=\frac{16}{3}\)
\(\Rightarrow \frac{\alpha^{2}+1}{\alpha}=\frac{16}{3}-2\)
⇒ \(\frac{\alpha^{2}+1}{\alpha}=\frac{10}{3}\)
⇒ 3α² + 3 = 10α
⇒ 3α² – 10α + 3 = 0
⇒ (3α – 1) (α – 3) = 0
⇒ α = \(\frac{1}{3}\), 3
α = \(\frac{1}{3}\), β = 3 (or) when a = 3, β = \(\frac{1}{3}\)
∴ The roots are 3, \(\frac{1}{3}\), 2
(or) when γ = 2, by synthetic division method.

The factors are (x – 2) (x – 3) (3x – 1)
∴ The roots are 2, 3, \(\frac{1}{3}\)

12th Maths 3.1 Question 5.
Find the sum of squares of roots of the equation 2x⁴ – 8x³ + 6x² – 3 = 0.
Solution:
The given equation is 2x⁴ – 8x³ + 6x² – 3 = 0.
(÷ 2) ⇒ x⁴ – 4x³ + 3x² – \(\frac{3}{2}\) = 0
Let the roots be α, β, γ, δ
α + β + γ + δ = -b = 4
(αβ + βγ + γδ + αδ + αγ + βδ) = c = 3
αβγ + βγδ + γδα = -d = 0
αβγδ = \(\frac{-3}{2}\)
To Find α² + β² + γ² + δ² = (α + β + γ + δ)² – 2 (αβ + βγ + γδ + αδ + αγ + βδ)
= (4)² – 2(3)
= 16 – 6
= 10

12 Maths Exercise 3.1 Question 6.
Solve the equation x³ – 9x² + 14x + 24 = 0 if it is given that two of its roots are in the ratio 3 : 2.
Solution:
The given equation is x³ – 9x² +14x + 24 = 0.
Since the two roots are in the ratio 3 : 2.
The roots are α, 3λ, 2λ
α + 3λ + 2λ = -b = 9
⇒ α + 5λ = 9 …… (1)
(α) (3λ) (2λ) = -24
6λ²α = -24
⇒ λ²α = -4 …… (2)
(1) ⇒ α = 9 – 5λ
(2) ⇒ λ² (9 – 5λ) = -4
9λ² – 5λ³ + 4 = 0
5λ³ – 9λ² – 4 = 0

(λ – 2) (5λ² + λ + 2) = 0
λ = 2, 5λ² + λ + 2 = 0 has only Imaginary roots Δ < 0
when λ = 2, α = 9 – 5 (2) = 9 – 10 = -1
The roots are α, 3λ, 2λ i.e., -1, 6, 4

Samacheer Kalvi Guru 12th Maths Question 7.
If α, β and γ are the roots of the polynomial equation ax³ + bx² + cx + d= 0, find the value of \(\Sigma \frac{\alpha}{\beta \gamma}\) in terms of the coefficients.
Solution:
The given equation is ax³ + bx² + cx + d = 0.
÷a ⇒ \(x^{3}+\frac{b}{a} x^{2}+\frac{c}{a} x+\frac{d}{a}=0\)
Let the roots be α, β, γ
α + β + γ = \(-\frac{b}{a}\)
αβ + βγ + γα = \(\frac{c}{a}\)
αβγ = \(-\frac{d}{a}\)
To find:

Samacheer Kalvi 12 Maths Solutions Question 8.
If α, β, γ and δ are the roots of the polynomial equation 2x⁴ + 5x³ – 7x² + 8 = 0, find a quadratic equation with integer coefficients whose roots are α + β + γ + δ and αβγδ.
Solution:
The given equation is 2x⁴ + 5x³ – 7x² + 8 = 0.
÷ 2 ⇒ \(x^{4}+\frac{5}{2} x^{3}-\frac{7}{2} x^{2}+4=0\)
Let the roots be α, β, γ, δ
α + β + γ + δ = \(-\frac{5}{2}\)
αβγδ = -4
To form the quadratic equation with the given roots α + β + γ + δ, αβγδ.
x² – x(S.O.R) + P.O.R = 0
\(x^{2}-x\left(\frac{-5}{2}-4\right)+\left(\frac{-5}{2}\right)(-4)=0\)
\(\Rightarrow x^{2}-x\left(\frac{-13}{2}\right)+10=0\)
2x² + 13x + 20 = 0

Samacheer Kalvi 12th Maths Solutions Question 9.
If p and q are the roots of the equation lx² + nx + n = 0, show that \(\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0\)
Solution:
The given equation is lx² + nx + n = 0.
p + q = \(-\frac{n}{l}\), pq = \(\frac{n}{l}\)

Samacheer Kalvi.Guru 12th Maths Question 10.
If the equations x² + px + q = 0 and x² + p’x + q’ = 0 have a common root, show that it must be equal to \(\frac{p q^{\prime}-p^{\prime} q}{q-q^{\prime}}\) or \(\frac{q-q^{\prime}}{p^{\prime}-p}\)
Solution:
If α is the common root, then.
α² + pα + q = 0 ……. (1)
α² + p’α + q’ = 0 ……… (2)
Subtracting α (p – p’) = q’ – q
\(\alpha=\frac{q^{\prime}-q}{p-p^{\prime}}=\frac{q-q^{\prime}}{p^{\prime}-p}\) …….. (3)
Eliminating α from (1) & (2)

Samacheer Kalvi Class 12 Maths Solutions Question 11.
Formulate into a mathematical problem to find a number such that when its cube root is added to it, the result is 6.
Solution:
Let the number be x.
Given that \(\sqrt[3]{x}+x=6\)
\(\Rightarrow \sqrt[3]{x}=6-x\)
Cubing on both sides
x = (6 – x)³
⇒ x = 216 – 3 (6)² (x) + 3(6) (x)² – x³
⇒ x = 216 – 108x + 18x² – x³
⇒ x³ – 18x² + 109x – 216 = 0

Samacheerkalvi.Guru 12th Maths Question 12.
A 12 metre tall tree was broken into two parts. It was found that the height of the part which was left standing was the cube root of the length of the part that was cut away. Formulate this into a mathematical problem to find the height of the part which was cut away.
Solution:
Let the two parts be x and (12 – x)
Given that \(x=\sqrt[3]{12-x}\)
Cubing on both side,
x³ = 12 – x
⇒ x³ + x – 12 = 0

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1 Additional Problems

Samacheer Kalvi 12th Maths Example Sums Question 1.
Construct a cubic equation with roots 2, 3, 4.
Solution:
Given roots are 2, 3, 4
Take α = 1; β = 3; γ = 4
The required cubic polynomial is
x³ – (α + β + γ) x² + (αβ + βγ + γα) x – αβγ = 0
x³ – (1 + 3 + 4)x² + (3 + 12 + 4) x – 12 = 0
x³ – 8x² + 19 x – 12 = 0

Samacheer Kalvi 12th Maths Question 2.
If α, β, γ are the roots of the cubic equation x³ – 6x² + 11x – 6 = 0. From a cubic equation whose roots are 2α, 2β, 2γ.
Solution:
Given that α, β, γ are the roots of x³ – 6x² + 11x – 6 = 0 … α, β, γ
α + β + γ = 6 …(1)
αβ + βγ + γα = 11 …(2)
αβγ = 6 …(3)
Form a cubic equation whose roots are 2α, 2β, 2γ.
∴ 2α + 2β + 2γ = 2(α + β + γ) = 2(6) = 12

The required cubic equation is
x³ – (2α + 2β + 2γ)x² + (4αβ + 4βγ + 4γα) x – (2α) (2β) (2γ) = 0
x³ – 12x² + 44x – 48 = 0

Samacheer Kalvi Maths 12th Question 3.
If the roots of x⁴ + 5x³ – 30x² – 40x + 64 = 0 are in G.P; then find the roots.
Solution:

Samacheer Kalvi 12th Maths Guide Question 4.
Determine the value of k such that the equation (2k – 5)x² – 4x – 15 = 0 and (3k – 8)x² – 5x – 21 = 0 may have a common root.
Solution:
If α be the common root, the two equations.
(2k – 5) α² – 4α – 15 = 0
(3k – 8) α² – 5α – 21 = 0 These are the linear equation is α² and α.
By cross multiplication rule

12th Maths Samacheer Kalvi Question 5.
If α, β, γ are the roots of the equation x³ + px² + qx +1 = 0. Find the value of the following in terms of coefficients.

Solution:

Samacheer Kalvi Guru 12 Maths Question 6.

Solution:

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1

12th Maths 6th Chapter Solutions Question 1.
Prove by vector method that if a line is drawn from the centre of a circle of a circle to the midpoint of a chord, then the line is perpendicular to the chord.
Solution:
Let ‘C’ be the mid point of the chord AB
Take ‘O’ on the centre of the circle.
Since, OA = OB (Radii)

12th Maths Exercise 6.1 Question 2.
Prove by vector method that the median to the base of an isosceles triangle is perpendicular to the base.
Solution:
Let OAB be an isosceles triangle with OA = OB
Let OC be the median to the base AB
C is the midpoint of AB
Take O as origin.

12th Maths 6th Chapter Question 3.
Prove by vector method that an angle in a semi-circle is a right angle.
Solution:
Let AB be the diameter of the circle with centre ‘O’
Let P be any point on the semi-circle.


This gives ∠APB = 90°. Hence the result.

12th Maths Chapter 6 Exercise 6.1 Question 4.
Prove by vector method that the diagonals of a rhombus bisect each other at right angles.
Solution:
Let ABCD be a rhombus

12 Maths Samacheer Kalvi Solutions Question 5.
Using vector method, prove that if the diagonals of a parallelogram are equal, then it is a rectangle.
Solution:
Let ABCD be a parallelogram
To prove ABCD be a rectangle provided the diagonals are equal.

\(\overrightarrow{\mathrm{AB}}\) ⊥^r to \(\overrightarrow{\mathrm{BC}}\)
⇒ ABCD is a rectangle.

12th Maths Chapter 6 Question 6.
Prove by vector method that the area of the quadrilateral ABCD having diagonals AC and BD is \(\frac{1}{2}|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}|\)
Solution:
Vector area of quadrilateral ABCD = {Vector area of ∆ABC} + {Vector area of ∆ ACD}

Class 12 Maths Chapter 6 Exercise 6.1 Question 7.
Prove by vector method that the parallelogram on the same base and between the same parallels are equal in area.
Solution:
Let ABCD and ABC’D’ be two parallelogram between the parallels with same base
To prove: Area of ABCD = Area of ABC’D’

Samacheer Kalvi Guru 12th Maths Question 8.
If G is the centroid of a AABC, prove that.
(area of ∆GAB) = (area of ∆GBC) = (area of ∆GAC) = \(\frac{1}{3}\) [area of ∆ABC]
Solution:


Similarly we can prove
Area of ∆GBC = Area of ∆GAC = \(\frac{1}{3}\) [Area of ∆ABC]

Samacheer Kalvi 12 Maths Solutions Question 9.
Using vector method, prove that cos(α – β) = cos α cos β + sin α sin β.
Solution:

From (1) and (2), we get
cos(α + β) = cos α cos β + sin α sin β

Samacheer Kalvi 12th Maths Guide Question 10.
Prove by vector method that sin(α + β) = sin α cos β + cos α sin β.
Solution:
Take two points A and B on the unit circle with centre as origin ‘O’, so \(|\overrightarrow{\mathrm{OA}}|=|\overrightarrow{\mathrm{OB}}|\) = 1


From (1) & (2), we get
sin (α + β) = sin α cos β + cos α sin β

Std 12 Maths Solutions Samacheer Kalvi Question 11.
A particle acted on by constant forces \(8 \vec{i}+2 \vec{j}-6 \vec{k}\) and \(\overrightarrow{6 i}+2 \vec{j}-2 \vec{k}\) is displaced from the point (1, 2, 3) to the point (5, 4, 1). Find the total work done by the forces.
Solution:

From (1) & (2), we get
Work done by the force = \(\overrightarrow{\mathrm{F}} \cdot \vec{d}\) = 56 + 8 + 16 = 80 units.

Samacheer Kalvi Class 12 Maths Solutions Question 12.
Forces of magnitude \(5 \sqrt{2}\) and \(10 \sqrt{2}\) units acting in the directions \((3 \vec{i}+4 \vec{j}+5 \vec{k})\) and \((10 \vec{i}+6 \vec{j}-8 \vec{k})\), respectively, act on a particle which is displaced from the point with position vector \((4 \vec{i}-3 \vec{j}-2 \vec{k})\) to the point with position vector \((\overrightarrow{6 i}+\vec{j}-3 \vec{k})\). Find the work done by the forces.
Solution:

12 Maths Solutions Samacheer Kalvi Question 13.
Find the magnitude and direction cosines of the torque of a force represented by \(3 \vec{i}+4 \vec{j}-5 \vec{k}\) about the point with position vector \(2 \vec{i}-3 \vec{j}+4 \vec{k}\) acting through a point whose position vector is \(\overrightarrow{4 i}+2 \vec{j}-3 \vec{k}\).
Solution:

Samacheer Kalvi Guru 12 Maths Question 14.
Find the torque of the resultant of the three forces represented by \(-3 \vec{i}+6 \vec{j}-3 \vec{k}\), \(\overrightarrow{4 i}-10 \vec{j}+12 \vec{k}\) and \(\overrightarrow{4 i}+7 \vec{j}\) acting at the point with position vector \(8 \vec{i}-\overrightarrow{6} \vec{j}-4 \vec{k}\), about the point with position vector \(18 \vec{i}+3 \vec{j}-9 \vec{k}\)
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 Additional Problems

12th Maths Solutions Samacheer Kalvi Question 1.
The work done by the force \(\overrightarrow{\mathrm{F}}=a \vec{i}+\vec{j}+\vec{k}\) in moving the point of application from (1, 1, 1) to (2, 2, 2) along a straight line is given to be 5 units. Find the value of a.
Solution:

Samacheer Kalvi.Guru 12th Maths Question 2.
If the position vectors of three points A, B and C are respectively and \(7(\vec{i}+\vec{k})\). Find \(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}\). Interpret the result geometrically.
Solution:

12 Maths Solutions Pdf Samacheer Kalvi Question 3.
A force given by and \(3 \vec{i}+2 \vec{j}-4 \vec{k}\) is applied at the point (1, – 1, 2). Find the moment of the force about the point (2, – 1, 3).
Solution:

Samacheer Kalvi 12 Maths Guide Question 4.
Show that the area of a parallelogram having diagonals is \(5 \sqrt{3}\)
Solution:

Students can Download Accountancy Chapter 4 Goodwill in Partnership Accounts Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 4 Goodwill in Partnership Accounts

Samacheer Kalvi 12th Accountancy Goodwill in Partnership Accounts Text Book Back Questions and Answers

I. Choose the Correct Answer

12th Accountancy 4th Chapter Solutions Question 1.
Which of the following statements is true?
(a) Goodwill is an intangible asset
(b) Goodwill is a current asset
(c) Goodwill is a fictitious asset
(d) Goodwill
Answer:
(a) Goodwill is an intangible asset

12th Accountancy Chapter 4 Question 2.
Super profit is the difference between ………………..
(a) Capital employed and average profit
(b) Assets and liabilities
(c) Average profit and normal profit
(d) Current year’s profit and average profit
Answer:
(c) Average profit and normal profit

12th Accountancy 4th Chapter Question 3.
The average rate of return of similar concerns is considered as ………………..
(a) Average profit
(b) Normal rate of return
(c) Expected rate of return
(d) None of these
Answer:
(b) Normal rate of return

Class 12 Accountancy Chapter 4 Solutions Question 4.
Which of the following is true?
(a) Super profit = Total profit / number of years
(b) Super profit = Weighted profit / number of years
(c) Super profit = Average profit – Normal profit
(d) Super profit = Average profit x Years of purchase
Answer:
(c) Super profit = Average profit – Normal profit

Chapter 4 Accountancy Class 12 Solutions Question 5.
Identify the incorrect pair ………………..
(a) Goodwill under Average profit method – Average profit x Number of years of purchase
(b) Goodwill under Super profit method – Super profit x Number of years of purchase
(c) Goodwill under Annuity method – Average profit x Present value of annuity factor
(d) Goodwill under Weighted average profit method – Weighted average profit x Number of years of purchase
Answer:
(c) Goodwill under Annuity method – Average profit x Present value of annuity factor

Class 12th Accounts Chapter 4 Solutions Question 6.
When the average profit is ₹ 25,000 and the normal profit is ₹ 15,000, super profit is ………………..
(a) ₹ 25,000
(b) ₹ 5,000
(c) ₹ 10,000
(d) ₹ 15,000
Answer:
(c) ₹ 10,000

Accountancy Class 12 Chapter 4 Solutions Question 7.
Book profit of 2017 is ₹ 35,000; non – recurring income included in the profit is ₹ 1,000 and abnormal loss charged in the year 2017 was ₹ 2,000, then the adjusted profit is ………………..
(a) ₹ 36,000
(b) ₹ 35,000
(c) ₹ 38,000
(d) ₹ 34,000
Answer:
(a) ₹ 36,000

Accountancy 4th Chapter Question 8.
The total capitalised value of a business is ₹ 1,00,000; assets are ₹ 1,50,000 and liabilities are ₹ 80,000. The value of goodwill as per the capitalisation method will be ………………..
(a) ₹ 40,000
(b) ₹ 70,000
(c) ₹ 1,00,000
(d) ₹ 30,000
Answer:
(d) ₹ 30,000

II. Very Short Answer Questions

Accountancy Chapter 4 Class 12 Question 1.
What is goodwill?
Answer:
Goodwill is the good name or reputation of the business which brings benefit to the business. It enables the business to earn more profit. It is the present value of a firm’s future excess earnings.

Accountancy Chapter 4 Question 2.
What is acquired goodwill?
Answer:
Goodwill acquired by making payment in cash or kind is called acquired or purchased goodwill. When a firm purchases an existing business, the price paid for purchase of such business may exceed the net assets (Assets – Liabilities) of the business acquired.

Chapter 4 Accounts Class 12 Question 3.
What is super profit?
Answer:
Super profit is the excess of average profit over the normal profit of a business.
Super profit = Average profit – Normal profit.
Average profit is calculated by dividing the total of adjusted actual profit of certain number of years by the total number of such years. Normal profit is the profit earned by the similar business firms under normal conditions.
Normal profit = Capital employed x Normal rate of return Capital employed = Fixed assets + Current assets – Current liabilities

Samacheer Kalvi Guru 12th Accountancy Question 4.
What is normal rate of return?
Answer:
It is the rate at which profit is earned by similar business entities in the industry under normal circumstances.

Question 5.
State any two circumstances under which goodwill of a partnership firm is valued?
Answer:

1. When there is a change in the profit sharing ratio.
2. When a new partner is admitted into a firm.
3. When an existing partner retires from the firm or when a partner dies.
4. When a partnership firm is dissolved.

III. Short Answer Questions

Question 1.
State any six factors determining goodwill.
Answer:

1. Profitability of the firm.
2. Favourable location of the business enterprise.
3. Good quality of goods or services offered.
4. Tenure of the business enterprise.
5. Efficiency of management.
6. Degree of competition.
7. Other factors.

Question 2.
How is goodwill calculated under the super profits method?
Answer:
1. Purchase of super profit method: Goodwill is calculated by multiplying the super profit by a certain number of years of purchase.
Goodwill = super profit x No. of years of purchase

2. Annuity method: Value of goodwill is calculated by multiplying the super profit with the present value of annuity.
Goodwill = Super profit x Present value annuity factor

3. Capitalisation of super profit method: Goodwill = x 100

Question 3.
How is the value of goodwill calculated under the capitalisation method?
Answer:
Capitalisation method:
Under Capitalisation method, goodwill is the excess of capitalised value of average profit of the business over the actual capital employed in the business.
Goodwill = Total capitalised value of the business – Actual capital employed
The total capitalised value of the business is calculated by capitalising the average profits on the basis of the normal rate of return.
Capitalised value of the business = x 100
Actual capital employed = Fixed assets (excluding goodwill) + Current assets – Current liabilities

Question 4.
Compute average profit from the following information.
Answer:
Calculation of Average profit:
2016 – ₹ 8,000; 2017 – ₹ 10,000; 2018 – ₹ 9,000

Valuation of goodwill = ₹ 9,000

Question 5.
Calculate the value of goodwill at 2 years purchase of average profit when average profit is ₹ 15,000.
Answer:
Goodwill: ₹ 30,000

IV. Exercises

Simple average profit method:
Question 1.
The following are the profits of a firm in the last five years:

Answer:

Valuation of goodwill = Average profit x No. of years purchase
= ₹ 12,000 x 2 years
= ₹ 24,000

Question 2.
From the following information, calculate the value of goodwill on the basis of 3 years purchase of average profits of last four years.

Answer:
Calculation of goodwill:

Valuation of goodwill = Average profit x No. of years purchase = ₹ 4,000 x 3
= ₹ 12,000

Question 3.
From the following information relating to a partnership firm, find out the value of its goodwill based on 3 years purchase of average profits of the last 4 years:

1. Profits of the years 2015, 2016, 2017 and 2018 are ₹ 10, 000, ₹ 12, 500, ₹ 12, 000 and ₹ 11, 500, respectively.
2. The business was looked after by a partner and his fair remuneration amounts to ₹ 1, 500 per year. This amount was not considered in the calculation of the above profits.

Answer:
Valuation of goodwill
Calculation of average profit Year Profit

(-) Remuneration = ₹ 1,500
Net average profit = ₹ 10,000
Valuation of goodwill = Average profit x No. of years purchase
= ₹ 10,000 x 3
= ₹ 30,000

Question 4.
From the following information relating to Sridevi enterprises, calculate the value of goodwill on the basis of 4 years purchase of the average profits of 3 years.

1. Profits for the years ending 31^st December 2016, 2017 and 2018 were ₹ 1,75,000, ₹ 1,50,000 and ₹ 2,00,000, respectively.
2. A non – recurring income of ₹ 45,000 is included in the profits of the year 2016.
3. The closing stock of the year 2017 was overvalued by ₹ 30,000.

Answer:
Calculation of adjusted profit

Average profit = <
Average profit = \(\frac { 4, 80, 000 }{ 3 }\) = ₹ 1,60,000
Goodwill = Average profit x No. of year purchase = ₹ 1,60,000 x 4
= ₹ 64,00,000

Question 5.
The following particulars are available in respect of the business carried on by a partnership firm:

1. Profits earned: 2016: ₹ 25,000; 2017: ₹ 23,000 and 2018: ₹ 26,000.
2. Profit of 2016 includes a non – recurring income of ₹ 2,500.
3. Profit of 2017 is reduced by ₹ 3,500 due to stock destroyed by fire.
4. The stock was not insured. But, it is decided to insure the stock in future. The insurance premium is estimated to be ₹ 250 per annum.

You are required to calculate the value of goodwill of the firm on the basis of 2 years purchase of average profits of the last three years.
Answer:
Calculation of adjusted profit

Total profit = 22,250 + 26, 250 + 25, 750 = ₹ 74, 250

Valuation of goodwill = Average profit x No.of year purchase
= ₹ 24, 750 x 2 years = ₹ 49, 500

Weighted average profit method:
Question 6.
Find out the value of goodwill at three years purchase of weighted average profit of last four years.

Answer:
Calculation of weighted average profit

Goodwill = Weighted average profit x No. of years purchase
= ₹ 15,400 x 3 = ₹ 46,200

Purchase of super profit method:
Question 7.
From the following details, calculate the value of goodwill at 2 years purchase of super profit:

1. Total assets of a firm are ₹ 5,00,000
2. The liabilities of the firm are ₹ 2,00,000
3. Normal rate of return in this class of business is 12.5 %.
4. Average profit of the firm is ₹ 60,000.

Answer:
Capital employed = fixed assets + current assets – current liabilities
= 5, 00, 000 – 2, 00, 000 = 3, 00, 000
Normal profit = Capital employed x Normal rate of return
= 3,00,000 x \(\frac { 12.5 }{ 100 }\) = 3, 75, 000
Super profit = Average profit – Normal profit
= 60, 000 – 37, 500 = 22, 500
Goodwill = Super profit x Number of years of purchase
= ₹ 22,500 x 2
= ₹ 45,000

Question 8.
A partnership firm earned net profits during the last three years as follows:
2016: ₹ 20,000; 2017: ₹ 17,000 and 2018: ₹ 23,000
The capital investment of the firm throughout the above mentioned period has been ₹ 80,000. Having regard to the risk involved, 15% is considered to be a fair return on capital employed in the business. Calculate the value of goodwill on the basis of 2 years purchase of super profit.
Answer:
Calculation of average profit Year Profit

Normal profit = Capital employed x Normal rate of return
= 80,000 x \(\frac { 15 }{ 100 }\) = 12,000
Super profit = 8,000
Valuation of goodwill = Super profit x No. of years purchase
= ₹ 8,000 x 2
= ₹ 16,000

Annuity method:
Question 9.
From the following information, calculate the value of goodwill under annuity method:

1. Average profit – ₹ 14,000
2. Normal profit – ₹ 4,000
3. Normal rate of return – 15%
4. Years of purchase of goodwill – 5

Present value of ₹ 1 for 5 years at 15% per annum as per the annuity table is 3.352
Answer:
Super profit = Average profit – Normal profit
= 14, 000 – ₹ 4, 000 = ₹ 10, 000
Goodwill = Super profit x Present value of annuity factor
= ₹ 10,000 x 3.352 = ₹ 33, 520

Capitalisation of super profit method:
Question 10.
Find out the value of goodwill by capitalising super profits:

1. Normal Rate of Return 10%
2. Profits for the last four years are ₹ 30,000, ₹ 40,000, ₹ 50,000 and ₹ 45,000.
3. Anon – recurring income of ₹ 3,000 is included in the above mentioned profit of ₹ 30,000.
4. Average capital employed is ₹ 3,00,000.

Answer:

Average profit = = \(\frac { 1, 62, 000 }{ 4 }\)
= ₹ 40,500
Normal profit = Capital employed x Normal rate of return
= 3,00,00 x \(\frac { 10 }{ 100 }\) = 30,000
Super profit = 10,500
Capitalisation super profit method = x 100
= \(\frac { 10, 500 }{ 10 }\) = ₹ 1, 05, 000

Capitalisation method:
Question 11.
From the following information, find out the value of goodwill by capitalisation method :

1. Average profit ₹ 20, 000
2. Normal rate of return 10%
3. Tangible assets of the firm ₹ 2, 20, 000
4. Liabilities of the firm ₹ 70, 000

Answer:
Capital Assets – Liabilities
= 2, 20, 000 – 70, 000 = ₹ 1, 50, 000
Capitalised value of business = x 100
= \(\frac { 20, 000 }{ 10 }\) x 100 = ₹ 2, 00, 000
Value of goodwill = Total capitalised average profit – Capital employed
= 2, 00, 000 – 1, 50, 000
= ₹ 50,000

Samacheer Kalvi 12th Accountancy Goodwill in Partnership Accounts Additional Questions and Answers

I. Choose the correct answer

Question 1.
Goodwill is valued under ………………
(a) Average profit method
(b) Super profit method
(c) Capitalisation method
(d) All of these
Answer:
(d) All of these

Question 2.
Average profit method can be further divided ………………
(a) Simple average profit
(b) Weighted average profit
(c) Both (a) and (b)
(d) None of these
Answer:
(c) Both (a) and (b)

Question 3.
Super profit is
(a) Average profit – Normal profit
(b) Normal profit – Average profit
(c) Both (a) and (b)
(d) None of these
Answer:
(a) Average profit – Normal profit

Question 4.
Kinds of goodwill ………………
(a) Purchased goodwill
(b) Self – generated goodwill
(c) None of these
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 5.
Nature of goodwill is ………………
(a) Reputation of business firm
(b) Profit of firm
(c) total asset of firm
(d) None of these
Answer:
(a) Reputation of business firm

Question 6.
Factor of goodwill is ………………
(a) Location of the business
(b) Efficient management
(c) Nature of the goods
(d) All of these
Answer:
(b) Efficient management

Question 7.
Arrange the following in ascending or descending order:
1. Super profit = Weighted average profit x No. of years purchase
2. Capitalistion method = x 100
3. Weighted Average method = Average profit – Normal profit
4. Capitalisation super profit = Total profit/No. of years purchase
(a) 1, 2, 3, 4
(b) 2, 4, 1, 2
(c) 3, 4, 1, 2
(d) 4, 3, 2, 1
Answer:
(c) 3, 4, 1, 2

Question 8.
Goodwill of the firm on the basis of 2 years purchase of average profit of the last 3 years is ₹ 25,000. Find average profit ………………
(a) ₹ 50,000
(b) ₹ 25,000
(c) ₹ 10,000
(d) ₹ 12,500
Answer:
(a) ₹ 50,000

Question 9.
Calculate the value of goodwill at 3 years purchase when capital employed is ₹ 2,50,000. Average profit ₹ 30,000 and normal rate of return is 10% ………………
(a) ₹ 3,000
(b) ₹ 25,000
(c) ₹ 30,000
(d) ₹ 15,000
Answer:
(d) ₹ 15,000

Question 10.
Adjusted profit = Actual profit
(a) + past expenses not required in the future
(b) – past revenue not likely to be eached in the future
(c) + additional income expected in the future
(d) All of these
Answer:
(d) All of these

II. Fill in the blanks:

Question 11.
Goodwill brought in cash by new partner is divided among the old partner by debiting is ________ A/c and crediting ________ A/c.
Answer:
Goodwill; capital.

Question 12.
Calculation of goodwill under simple average profit method is ________
Answer:
Total profit =

Question 13.
Super profit method normal profit is ________
Answer:
Normal profit = Capital employed x Normal rate of return

Question 14.
Under average profit method ________
Answer:
x 100

Question 15.
Super profit = ________
Answer:
Average profit – Normal profit

III. Short answer questions

Question 1.
What are the nature of goodwill?
Answer:
The nature of the goodwill can be described as follows:

1. Goodwill is an tangible fixed asset. It is tangible because it has no physical existence. It cannot be seen or touched.
2. It has a definite value depending on the profitability of the business enterprise.
3. It cannot be separated from the business.
4. It helps in earning more profit and attracts more customers.
5. It can be purchased or sold only when the business is purchased or sold in full or in part.

Question 2.
What is average profit method?
Answer:
Under this method, goodwill is calculated as certain years of purchase of average profits of the past years. The number of years of purchase is generally determined on the basis of the average period a new business will take in order to bring it to the current state of the existing business.

Question 3.
What is meant by simple average profit method?
Answer:
Goodwill is calculated by multiplying the average profit by a certain number of years of purchase. Simple average profit is calculated by adding the adjusted profits of certain number of years by dividing the total number of such years.
Goodwill = Average profit x Number of years purchase Total profit.
Average profit =

Question 4.
Weighted average profit method.
Answer:
Goodwill is calculated by multiplying the weighted average profit by a certain number of years of purchase.
Goodwill = Weighted average profit x Number of years purchase
In this method, weights are assigned to each year’s profit. Weighted profit is ascertained by multiplying the weights assigned with the respective year’s profit.
Weighted average profit =

Question 5.
What is meant by annuity method of valuation of goodwill?
Answer:
Value of goodwill is calculated by multiplying the super profit with the present value of annuity.
Goodwill = Super profit x Present value annuity factor
Present value annuity factor is the present value of annuity of rupee one at a given time. It can be found out from annuity table or by using formula.

Question 6.
What is annuity factor?
Answer:
Annuity refers to series of uniform cash flows at regular intervals. The table value gives the present value of annuity of rupee one received at the end of every year for a specified number of years.

where, i = interest rate ; n = estimated number of years

Question 7.
What is capitalisation of super profit method?
Answer:
Under this method, value of goodwill is calculated by capitalising the super profit at normal rate of return, that is, goodwill is the capitalised value of super profit.
Goodwill = x 100

IV. Exercise

Question 1.
Goodwill is to be valued at three years purchase of five years average profits. The profit for the last five years were 2010 – ₹ 4,200; 2011 — ₹ 4,500; 2012 – ₹ 4,700; 2013 – ₹ 4,600; and 2014 – ₹ 5,000.
Solution:
Calculation of average profit:

Calculation of goodwill = Average profit x No. of years purchase
= ₹ 4,600 x 3 = ₹ 13,800

Question 2.
A firm’s profit for the last 5 years were ₹ 20,000, ₹ 30,000, ₹ 40,000. ₹ 50,000, and ₹ 60,000. Calculated the value of firm’s goodwill on the basis of three years purchase of weighted average profit after using weight of 1, 2, 3, 4, 5 respectively.
Solution:

Goodwill = Weighted avg. profit x Number of years purchase = 46,667 x 3 = ₹ 1,40,000 (appr.)

Question 3.
A business has earned average profit ₹ 1,00,000 during the last few years and the normal rate of return in similar business is 10%. Find out the value of goodwill.

1. Capitalisation of super profit method.
2. Super profit method of the goodwill is valued at 3 years purchase of super profit. The assets of the business were ₹ 10,00,000 and liabilities of ₹ 1,80,000.

Solution:
1. Capital employed = Assets – libilities
= 10,00,000 – 1,80,000
= ₹ 8,20,000
Normal profit = Capital employed x
8,20,000 = 1 x \(\frac { 10 }{ 100 }\) = 82,000
Super profit = Average profit – Normal profit
= ₹ 1,00,000 – ₹ 82,000 = ₹ 18,000

2. Capitalisation method Super profit
Goodwill = x 100
= \(\frac { 18, 000 }{ 10 }\) = ₹ 1, 80,000
as per super profit method Goodwill = super profit x No. of years purchase
= ₹ 18000 x 3 = ₹ 54,000

Question 4.
The average profit earned by the firm is Z 80,000 which includes under valuation of stock an average basis the capital invested in the business and normal rate of return is 8%. Calculate the goodwill of the firm on the basis of times the super profit.
Solution:
Average profit – ₹ 80,000
(A) Under value of stock – ₹ 8,000
Actual Average profit – ₹ 88,000
Normal profit = Capital investment x Normal rate of return
= ₹ 88,000 – ₹ 64,000
= ₹ 24,000
Goodwill = Super profit x 7
= ₹ 24,000 x 7
= ₹ 1,68,000

Question 5.
Capital investment is ₹ 5,00,000; firms profit ₹ 1,50,000 Assuming that normal rate of return is 20%. Calculate the goodwill

1. Capitalisation method.
2. Super profit method if the goodwill is valued @ 2 years purchase

Solution:
1. Capitalisation of average profit method.
Goodwill = x 100
= \(\frac { 1,50,000 }{ 20 }\) = ₹ 7,50,000
Goodwill = Total capital employed – capital employed
= ₹ 7,50,000 – ₹ 5,00,000 = ₹ 2,50,000

2. Super profit method
Normal profit = Capital employed x
Normal profit = ₹ 5,00,000 x \(\frac { 20 }{ 100 }\) = ₹ 1,00,000
Super profit = Average profit – Normal profit
= ₹ 1,50,000 – 1,00,000
= ₹ 50,000
Value of Goodwill = Super profit x No. of years purchase
= ₹ 50,000 x 2
= ₹ 1,00,000

Question 6.
Calculate the value of goodwill of the firm of 2 partners
(a) At the 3 years purchase of average profits
(b) At 3 years purchase of super profits
(c) On the basis of capitalisation of super profits
(d) On the basis of capitalisation of average profits

(i) Average capital employed ₹ 7,00,000
(ii) Net trading results of the firm 2014 – ₹ 1,47,600 2015 – Loss ₹ 1,48,100. Profit for 2016 – ₹ 4,48,700
(iii) Rate of interest on capital @ 18%
(iv) Remunuration ₹ 500/- per month
Solution:
Calculation of Average profit and super profit
Total profit = ₹ 1,47,600 – ₹ 1,48,100 + ₹ 4,48,700 = ₹ 4,48,200
Average profit = = ₹ 1,49,400
Less: Remunration of 2 partner (2 x 500 x 1/2) \(\frac { 12,000 }{ 1,37,400 }\)
Less: Normal profit = Capital employed x
= 7,00,000 x \(\frac { 18 }{ 100 }\) = ₹ 1,26,000
Super profit = ₹ 11,400

(a) Average profit – Average profit x No. of year purchase
= ₹ 1,37,400 x 3 = ₹ 4,12,200

(b) Super profit – Super profit x No. of year purchase
= ₹ 11,400 x 3 = ₹ 34,200

(c) Capitalisation of super profit – x 100
= \(\frac { 11,400 }{ 18 }\) x 100
= ₹ 63,333

(d) Capitalisation of average profit = x 100
= \(\frac { 1,37,400 }{ 100 }\) x 100
= ₹ 7,63,333
(-) Actual capital employed = ₹ 7,00,000
Goodwill = ₹ 63,333

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Samacheer Kalvi 12th Chemistry Metallurgy TextBook Evalution

I. Choose the correct answer.

12th Chemistry Chapter 1 Book Back Answers Question 1.
Bauxite has the composition ………………
(a) Al₂O₃
(b) Al₂O₃.nH₂O
(c) Fe₂O₃.2H₂O
(d) None of these
Answer:
(b) Al₂O₃.nH₂O

Metallurgy Class 12 State Board Question 2.
Roasting of sulphide ore gives the gas (A). (A) is a colourless gas. Aqueous solution of (A) is acidic. The gas (A) is ………………
(a) CO₂
(b) SO₃
(c) SO₂
(d) H,S
Answer:
(c) SO₂

Metallurgy Class 12 Important Questions State Board Question 3.
Which one of the following reaction represents calcination?
(a) 2Zn + O₂ → 2ZnO
(b) 2ZnS + 3O₂ → 2ZnO + 2SO₂
(c) MgCO₃ → MgO + CO₂
(d) Both (a) and (c)
Answer:
(c) MgCO₃ → MgO + CO₂

Samacheer Kalvi Guru 12th Chemistry Question 4.
The metal oxide which cannot be reduced to metal by carbon is ………………
(a) PbO
(b) Al₂O₃
(C) ZnO
(d) FeO
Answer:
(b) Al₂O₃

12th Chemistry Samacheer Kalvi Question 5.
Which of the metal is extracted by Hall-Herold process?
(a) Al
(b) Ni
(c) Cu
(d) Zn
Answer:
(a) Al

12th Chemistry Metallurgy Book Back Answers Question 6.
Which of the following statements, about the advantage of roasting of sulphide ore before reduction is not true?
(a) ∆G_f° of sulphide is greater than those for CS₂ and H₂S.
(b) ∆G_r° is negative for roasting of sulphide ore to oxide.
(c) Roasting of the sulphide to its oxide is thermodynamically feasible.
(d) Carbon and hydrogen are suitable reducing agents for metal sulphides.
Answer:
(d) Carbon and hydrogen are suitable reducing agents for metal sulphides.

12th Chemistry Metallurgy Pdf Question 7.
Match items in column -1 with the items of column – II and assign the correct code:


Answer:
(c) A – (iv), B – (ii), C – (iii), D – (i)

Chemistry Class 12 Samacheer Kalvi Question 8.
Wolframite ore is separated from tinstone by the process of ………………
(a) Smelting
(b) Calcination
(c) Roasting
(d) Electromagnetic separation
Answer:
(d) Electromagnetic separation

12th Chemistry Metallurgy Question 9.
Which one of the following is not feasible?
(a) Zn(s) + Cu²⁺(aq) → Cu(s) + Zn²⁺(aq)
(b) Cu(s) + Zn²⁺+(aq) → Zn(s) + Cu²⁺(aq)
(c) Cu(s) + 2Ag⁺(aq) → Ag(s) + Cu²⁺(aq)
(d) Fe(s) + Cu²⁺(aq) → Cu(s) + Fe²⁺(aq)
Answer:
(b) Cu(s) + Zn²⁺+(aq) → Zn(s) + Cu²⁺(aq)

Samacheer Kalvi 12 Chemistry Solutions Question 10.
Electrochemical process is used to extract ………………
(a) Iron
(b) Lead
(c) Sodium
(d) Silver
Answer:
(c) Sodium

12 Chemistry Samacheer Kalvi Question 11.
Flux is a substance which is used to convert ………………
(a) Mineral into silicate
(b) Infusible impurities to soluble impurities
(c) Soluble impurities to infusible impurities
(d) All of these
Answer:
(b) Infusible impurities to soluble impurities

12th Chemistry Metallurgy Important Questions Question 12.
Which one of the following ores is best concentrated by froth – floatation method?
(a) Magnetite
(b) Hematite
(c) Galena
(d) Cassiterite
Answer:
(c) Galena

Samacheer Kalvi 12th Chemistry Solutions Chapter 1 Question 13.
In the extraction of aluminium from alumina by electrolysis, cryolite is added to ………………
(a) Lower the melting point of alumina
(b) Remove impurities from alumina
(c) Decrease the electrical conductivity
(d) Increase the rate of reduction
Answer:
(a) Lower the melting point of alumina

Chemistry Metallurgy Pdf Download Question 14.
Zinc is obtained from ZnO by ………………
(a) Carbon reduction
(b) Reduction using silver
(c) Electrochemical process
(d) Acid leaching
Answer:
(a) Carbon reduction

12th Samacheer Kalvi Chemistry Question 15.
Cupellation is a process used for the refining of ………………
(a) Silver
(b) Lead
(c) Copper
(d) Iron
Answer:
(a) Silver

Samacheer Kalvi 12th Chemistry Question 16.
Extraction of gold and silver involves leaching with cyanide ion. Silver is later recovered by ………………
(a) Distillation
(b) Zone refining
(c) Displacement with zinc
(d) liquation
Answer:
(c) Displacement with zinc

Samacheer Kalvi Guru 12 Chemistry Question 17.
Considering Ellingham diagram, which of the following metals can be used to reduce alumina?
(a) Fe
(b) Cu
(c) Mg
(d) Zn
Answer:
(c) Mg

Samacheer Kalvi 12th Chemistry Book Solutions Question 18.
The following set of reactions are used in refining Zirconium

This method is known as
(a) Liquation
(b) Van Arkel process
(c) Zone refining
(d) Monds process
Answer:
(b) Van Arkel process

12th Chemistry Solution Book Samacheer Kalvi Question 19.
Which of the following is used for concentrating ore in metallurgy?
(a) Leaching
(b) Roasting
(c) Froth floatation
(d) Both (a) and (c)
Answer:
(d) Both (a) and (c)

Class 12 Chemistry Chapter 1 Notes Question 20.
The incorrect statement among the following is ………………
(a) Nickel is refined by Monds process
(b) Titanium is refined by Van Arkels process
(c) ZinC blende is concentrated by froth floatation
(d) In the metallurgy of gold, the metal is leached with dilute sodium chloride solution
Answer:
(d) In the metallurgy of gold, the metal is leached with dilute sodium chloride solution

12th Chemistry 1 Chapter Question 21.
In the electrolytic refining of copper, which one of the following is used as anode?
(a) Pure copper
(b) Impure copper
(c) Carbon rod
(d) Platinum electrode
Answer:
(b) Impure copper

Metallurgy Class 12 Important Questions Pdf Question 22.
Which of the following plot gives Ellingham diagram?
(a) ∆S Vs T
(b) ∆G° Vs T
(c) ∆G° Vs
(d) ∆G° Vs T
Answer:
(b) ∆G° Vs T

Samacheer Kalvi 12th Chemistry Solutions Question 23.
In the Ellingham diagram, for the formation of carbon monoxide
(a) \(\left( \frac { \triangle { S }^{ 0 } }{ \triangle T } \right) \)is negative
(b) \(\left( \frac { \triangle { G }^{ 0 } }{ \triangle T } \right) \)is positive
(c) \(\left( \frac { \triangle { G }^{ 0 } }{ \triangle T } \right) \)is negative
(d) initially \(\left( \frac { \triangle T }{ \triangle { G }^{ 0 } } \right) \)is positive, after 700°C, \(\left( \frac { \triangle { G }^{ 0 } }{ \triangle T } \right) \)is negative
Answer:
(c) \(\left( \frac { \triangle { G }^{ 0 } }{ \triangle T } \right) \)is negative

Question 24.
Which of the following reduction is not thermodynamically feasible?
(a) Cr₂O₃ → Al₂O₃ + 2Cr
(b) Al₂O₃ → Cr₂O₃ + 2Al
(c) 3TiO₂ + 4Al → 2Al₂O₃ + 2Al
(d) none of these
Answer:
(b) Al₂O₃ → Cr₂O₃ + 2Al

Question 25.
Which of the following is not truc with respect to Ellingham diagram?
(a) Free energy changes follow a straight line. Deviation occurs when there is a phase change.
(b) The graph for the formation of CO₂ is a straight line almost parallel to free energy axis.
(c) Negative slope of CO shows that it becomes more stable with increase in temperature.
(d) Positive slope of metal oxides shows that their stabilities decrease with increase in temperature.
Answer:
(b) The graph for the fonnation of CO₂ is a straight line almost parallel to free energy axis.

II. Answer the following questions:

Question 1.
What is the difference between minerals and ores?
Answer:
Minerals:

1. Minerals contain a low percentage of metal.
2. Metal cannot be extracted easily from minerals.
3. Clay Al₂O₃. SiO₂. 2H₂O is the mineral of aluminium.

Ores:

1. Ores contain a large percentage of metal.
2. Ores can be used for the extraction of metals on a large scale readily and economically.
3. Bauxite Al₂O₃. 2H₂O is the ore of aluminium.

Question 2.
What are the various steps involved in extraction of pure metals from their ores?
Answer:
The extraction of pure metals from the concentrated ores is carried out in two steps:

1. Conversion of the ore into oxides of the metal of interest.
2. Reduction of the metal oxides to elemental metals.

Question 3.
What is the role of Limestone in the extraction of Iron from its oxide Fe₂O₃?
Answer:
In the extraction of iron, a basic flux limestone is used. Limestone decomposes to form CaO which reacts with silica gangue present in the iron ore is acidic in nature to form calcium silicate (slag).

Question 4.
Which type of ores can be concentrated by froth floatation method? Give two examples for such ores.
Answer:
Sulphide ores can be concentrated by froth floatation method, e.g.,

1. Copper pyrites (CuFeS₂H₂)
2. Zinc blende (ZnS)
3. Galena (PbS)

Question 5.
Out of coke and CO, which is better reducing agent for the reduction of ZnO? Why?
Answer:
Coke (C) is a better reducing agent for the reduction of ZnO. Because, when we use coke, the reduction can be easily carried out at 673 K. Thus Carbon (Coke) reduces zinc oxide more easily than carbon monoxide (CO). From the Ellingham diagrams, it is quite clear that the reduction of zinc oxide is more favourable using coke ∆G for the formation of carbon monoxide from carbon is more negative).

Question 6.
Describe a method for refining nickel.
Answer:
The impure nickel is heated in a stream of carbon monoxide at around 350K. The nickel reacts with the CO to form a highly volatile nickel tetracarbonyl. The solid impurities are left behind.
Ni (s) + 4 CO (g) → Ni(CO)₄(g)

On heating the nickel tetracarbonyl around 460 K, the complex decomposes to give pure metal.
Ni(CO)₄ (g) → Ni (s) + 4 CO (g)

Question 7.
Explain zone refining process with an example using the Ellingham diagram given below.
Answer:
1. Zone Refining method is based on the principles of fractional crystallisation. When an impure metal is melted and allowed to solidify, the impurities will prefer to be in the molten region, i.e. impurities are more soluble in the melt than in the solid state metal.

2. In this process, the impure metal is taken in the form of a rod. One end of the rod is heated using a mobile induction heater which results in melting of the metal on that portion of the rod.

3. When the heater is slowly moved to the other end the pure metal crystallises while the impurities will move on to the adjacent molten zone formed due to the movement of the heater. As the heater moves further away, the molten zone containing impurities also moves along with it.

4. The process is repeated several times by moving the heater in the same direction again and again to achieve the desired purity level.

5. This process is carried out in an inert gas atmosphere to prevent the oxidation of metals.

6. Elements such as germanium (Ge), silicon (Si) and galium (Ga) that are used as semiconductor are refined using this process.

Question 8.
1. Predict the conditions under which
(a) Aluminium might be expected to reduce magnesia.
(b) Magnesium could reduce alumina.

2. Carbon monoxide is more effective reducing agent than carbon below 983 K but, above this temperature, the reverse is true -Explain.

3. it is possible to reduce Fe₂O₃ by coke at a temperature around 1200 K.
Answer:
1. The conditions under which:
(a) Ellingham diagram is used to predict thermodynamic feasibility of reduction of oxides of one metal by another metal. Any metal can reduce the oxides of other metals that are located above it in the Ellingham diagram. In the Ellingham diagram, for the formation of magnesia (magnesium oxide) occupy lower position than aluminium oxide. Therefore aluminium cannot be used to reduce the oxides of magnesium (magnesia). Above 1623K, A1 can reduce MgO to Mg, so that ArG° becomes negative and the process becomes thermodynamically feasible.

(b)

• \(\left( \frac { 4 }{ 3 } \right) \)Al + O₂ → \(\left( \frac { 2 }{ 3 } \right) \)Al₂O₃
  
• 2Mg + O₂ → 2MgO

At the point of intersection of the Al₂O₃ and MgO curves in Ellingham diagram. ∆G°
becomes zero for the reaction:

\(\left( \frac { 2 }{ 3 } \right) \)Al₂O₃ → 2MgO + \(\left( \frac { 4 }{ 3 } \right) \)Al
Below that point magnesium can reduce alumina.

2. From the Ellingham diagram, we find that at 983 K, the curves intersect.

The value of ∆G° for change of C to CO₂ is less than the value of ∆G° for change of CO to CO₂. Therefore, coke (C) is a better reducing agent than CO at 983K or above temperature. However below this temperature (e.g. at 673K), CO is more effective reducing agent than C.

3. Yes, it is possible to reduce Fe₂O₃ by coke at a temperature around 1200 K. In the Ellingham diagram, carbon line cuts across the lines of many metal oxides and hence it can reduce all those metal oxides at sufficiently high temperature. Ellingham diagram for the formation of Fe₂O₃ and CO intersects around 1000 K.

Below this temperature, the carbon line lies above the iron line which indicates that Fe₂O₃ is more stable than CO and hence at this temperature range the reduction is not thermodynamically feasible. However above 1000 K carbon line lies below the iron line and hence we can use coke as a reducing agent around 1200 K. Around 1200 K, coke is better reducing agent because above 1000 K, Gibb’s free energy for the formation of Fe₂O₃ is more than the formation of CO₂ from C.

Question 9.
Give the uses of zinc.
Answer:
Applications of Zinc (Zn):

1. Metallic zinc is used in galvanising metals such as iron and steel structures to protect them from rusting and corrosion.
2. Zinc is also used to produce die-castings in the automobile, electrical and hardware industries.
3. Zinc oxide is used in the manufacture of many products such as paints, rubber, cosmetics, pharmaceuticals, plastics, inks, batteries, textiles arid electrical equipment. Zinc sulphide is used in making luminous paints, fluorescent lights and x-ray screens.
4. Brass an alloy of zinc is used in water valves and communication equipment as it is highly resistant to corrosion.

Question 10.
Explain the electrometallurgy of aluminium.
Answer:
Electrochemical extraction of aluminium Hall-Herold process:
In this method, electrolysis is carried out in an iron tank lined with carbon, which acts as a cathode. The carbon blocks immersed in the electrolyte acts as a anode. A 20% solution of alumina, obtained from the . bauxite ore is mixed with molten cyrolite and is taken in the electrolysis chamber. About 10%, calcium chloride is also added to the solution. Here calcium chloride helps to lower the melting point of the mixture. The fused mixture is maintained at a temperature of above 1270 K. The chemical reactions involved in this process are as follows:

Ionisation of alumina:
Al₂O₃ → 2Al₃ + 3OO^2-

Reaction at cathode:
2Al³⁺ (melt) + 3e^– → Al_(l)

Reaction at anode:
2O^2- (melt) → O₂ + 3e^–

Since carbon acts as anode the following reaction also takes place on it.

• C (s) + O^2- (melt) → CO + 2e^–
• C (s) + 2O^2- (melt) → CO₂ + 4e^–

Due to the above two reactions, anodes are slowly consumed during the electrolysis. The pure aluminium is formed at the cathode and settles at the bottom. The net electrolysis reaction can be written as follows:
4Al³⁺ (melt) + 6O^2- (melt) + 3C_(s) → 4A_(l) + 3CO_2(g)

Question 11.
Explain the following terms with suitable examples.

1. Gangue
2. Slag

Answer:
1. Gangue:
The impurities associated with the minerals are known as Gangue or Matrix.

2. Slag:
A compound formed when gangue is combined with flux is called slag.
Flux + Gangue → Slag

For example, the oxide of iron can be reduced by carbon monoxide as follows:
Fe₂O₃ + 3CO → 2Fe + 3CO₂

In this extraction a basic flux, limestone is used.
Since the silica gangue present in the ore is acidic in nature, the limestone combines with it to form Calcium silicate (Slag).

Question 12.
Give the basic requirement for vapour phase refining.
Answer:
The two requirements for vapour phase refining are:

1. The metal should form a volatile compound with a suitable reagent.
2. The volatile compound is decomposed to give the pure metal.

Question 13.
Describe the role of the following in the process mentioned.

1. Silica in the extraction of copper.
2. Cryolite in the extraction of aluminium.
3. Iodine in the refining of Zirconium.
4. Sodium cyanide in froth floatation.

Answer:
1. The role of silica in the extraction of copper is to remove the iron oxide obtained during the process of roasting as slag. If the sulphide ore of copper contains iron, the silica (SiO₂) is added as flux before roasting. Then, FeO combines with silica to form iron silicate, FeSiO₃ (Slag).

2. Cryolite reduces the melting point of Al₂O₃ and increases its electrical conductivity. Aluminium is produced by the electrolytic reduction of fused alumina in the electrolytic cell. Alumina is not an electrolyte. So it is made as an electrolyte by dissolving it in the fused cryolite. The function of cryolite is to lower the fusion temperature.

3. Zirconium crude metal is heated with iodine in an evacuated vapour to separate from impurities and this decomposes at 1800 K to give a pure zirconium metal and iodine. Initially iodine is heated with zirconium to form a volatile compound.

4. Sulphide ores which are concentrated by the froth floatation process. Depressants are used to prevent certain type of particles from forming the froth. NaCN act as a depressant to separate ZnS from PbS.

Question 14.
Explain the principle of electrolytic refining with an example.
Answer:
The crude metal is refined by electrolysis. It is carried out in an electrolytic cell containing aqueous solution of the salts of the metal of interest. The rods of impure metal are used as anode and thin strips of pure metal are used as cathode.

The metal of interest dissolves from the anode, pass into the solution while the same amount of metal ions from the solution will be deposited at the cathode. During electrolysis, the less electropositive impurities in the anode, settle down at the bottom and are removed as anode mud. Let us understand this process by considering electrolytic refining of silver as an example.

Cathode:
Pure silver

Anode:
Impure silver rods

Electrolyte:
Acidified aqueous solution of silver nitrate. When a current is passed through the electrodes the following reactions will take place Reaction at anode.
2Ag (s) → Ag⁺(aq) + 1 e^–

Reaction at cathode:
Ag⁺ (aq) + 1 e^– → Ag (s)

During electrolysis, at the anode the silver atoms lose electrons and enter the solution. The positively charged silver cations migrate towards the cathode and get discharged by gaining electrons and deposited on the cathode. Other metals such as copper, zinc etc.,can also be refined by this process in a similar manner.

Question 15.
The selection of reducing agent depends on the thermodynamic factor. Explain with an example.
Answer:
From the Ellingham diagram, it is clear that metals for which the standard free energy of formation (∆_fG⁰) of their oxides is more negative can reduce the metal oxides for which the standard free energy of formation (∆_fG⁰) of oxides is less negative.

Thermodynamic factor has a major role in selecting the reducing agent for a particular reaction. Only that reagent will be preferred which will lead to decrease in the free energy (AG°) at a certain specific temperature.
E.g – Carbon reduce ZnO to Zn but not CO.

• ZnO + C → Zn + CO …………..(1)
• ZnO + CO → Zn + CO₂ ………………(2)

In the first case, there is increase in the magnitude of ∆S° while in the second case, it almost remains the same. In other words, ∆G° will have more negative value in the first case, when C is the reducing agent then in the second case when CO acts as the reducing agent. Therefore, C is a better reducing agent.

Question 16.
Give the limitations of Ellingham diagram.
Answer:
Limitations of Ellingham diagram:
1. Ellingham diagram is constructed based only on thermodynamic considerations. It gives information about the thermodynamic feasibility of a reaction. It does not tell anything about the rate of the reaction. Moreover, it does not give any idea about the possibility of other reactions that might be taking place.

2. The interpretation of ∆G is based on the assumption that the reactants are in equilibrium with the product which is not always true.

Question 17.
Write a short note on electrochemical principles of metallurgy.
Answer:
Electrochemical principles also find applications in metallurgical process. The reduction of oxides of active metals such as sodium, potassium etc., by carbon is thermodynamically not feasible. Such metals are extracted from their ores by using electrochemical methods. In this technique, the metal salts are taken in a fused form or in solution form. The metal ion present can be reduced by treating it with some suitable reducing agent or by electrolysis. Gibbs free energy change for the electrolysis process is given by the following expression
∆G° = -nFE°

Where n is number of electrons involved in the reduction process, F is the Faraday and E° is the electrode potential of the redox couple. If E° is positive then the ∆G is negative and the reduction is spontaneous and hence a redox reaction is planned in such a way that the e.m.f of the net redox reaction is positive. When a more reactive metal is added to the solution containing the relatively less reactive metal ions, the more reactive metal will go into the solution. For example,

• Cu (s) + 2Ag⁺(s) → Cu²⁺ (aq) + 2Ag (s)
• Cu²⁺ (aq) + Zn (s) → Cu (s) + Zn⁺(aq)

Evaluate Yourself

Question 1.
Write the equation for the extraction of silver by leaching with sodium cyanide and show that the leaching process is a redox reaction.
Answer:
The crushed ore of argentite (Ag₂S) is leached with sodium cyanide solution. This reaction forms sodium argento cyanide
Na[Ag(CN)₂]

Step 1:
Ag₂S + 4NaCN \(\rightleftharpoons\) 2Na[Ag(CN)₂] + Na₂S
The solution of sodium argento cyanide combines with zinc dust and forms sodium tetra cyano zincate and precipitated silver.

Step 2:
Zn + 2Na[Ag(CN)₂] → Na₂[Ag(CN)₄] + 2 Ag↓
In the step 2, redox reaction take place.

Question 2.
Magnesite (Magnesium carbonate) is calcined to obtain magnesia, which is used to make refractory bricks. Write the decomposition reaction.
Answer:
Magnesite is a carbonate of magnesium. Magnesite when heated at 800°C to 1000°C at the CO₂ content in it is driven off. The residue so obtained is known as calcined magnesite.

Question 3.
Using Ellingham diagram indicate the lowest temperature at which ZnO can be reduced to zinc metal by carbon. Write the overall reduction reaction at this temperature.
Answer:
Ellingham diagram shows variation in standard Gibbs free energy change with temperature for the formation of oxide. The Ellingham diagram shows straight line upward slope with formation of oxide, but in case of ZnO there is sudden change. Ellingham diagram helps in the selecting suitable reducing agent.

By seeing the Ellingham diagram, the free energy formation (∆_fG°) of CO from C becomes lower temperatures above 1120 K while that of CO₂ from C becomes lower above 1323 K than ∆_fG° of ZnO. As ∆_fG° of C0₂ from CO is always higher than that of ZnO. So C can reduce ZnO to Zn but not CO. Thus carbon is better reducing agent than CO for ZnO.

Question 4.
Metallic sodium is extracted by the electrolysis of brine (aq. NaCl). After electrolysis the electrolytic solution becomes basic in nature. Write the possible electrode reactions.
Answer:
Brine is a solution of sodium chloride (molten state):
The process of electrolysis involves using an electric current to bring about a chemical change and make new chemicals. In the electrolysis of brine, sodium ions migrate to the cathode, where electrons enter the melt and are reduced to sodium metal.
Na⁺ + e^– → Na (at cathode)

Chloride ions migrate the other way toward the anode. They give up their electrons to the anode and are oxidised to chlorine gas.
Cl ^– → \(\frac { 1 }{ 2 } \)Cl₂ + e^– (at anode)

Overall reaction:
2NaCl → 2Na(s) + Cl₂ (g)

For aqueous solution of NaCl:

• H₂O + 2e^– → H₂\(\uparrow \)+ 20H^– (at cathode)
• Cl^– → \(\frac { 1 }{ 2 } \)Cl₂ + e^– (at anode)

Overall reaction:
NaCl (aq) + H₂O(1) → Na⁺(aq) + OH^–(aq) + H₂(g) + \(\frac { 1 }{ 2 } \)Cl₂(g)
After electrolysis the electrolytic solution becomes basic in nature. [Due to formation of hydroxide (OH^–) ion].

Samacheer Kalvi 12th Chemistry Metallurgy Additional Questions

Samacheer Kalvi 12th Chemistry Metallurgy 1 Mark Questions and Answers

I. Choose the best answer:

Question 1.
Chemical formula of bauxite is ……………
(a) Fe₂O₃
(b) Fe₂O₃. FeO
(c) Al₂O₃ . H₂O
(d) Al₂O₃.2H₂O
Answer:
(d) Al₂O₃.2H₂O

Question 2.
An example of an oxide ore is ……………
(a) malachite
(b) bauxite
(c) galena
(d) zinc blende
Answer:
(b) bauxite

Question 3.
Froth floatation process involves the ……………
(a) treatment of the ore with water and pine oil
(b) washing of the ore with a steam of water
(c) owing off the ore over a conveyor belt rolling over magnetic roller
(d) melting of ore
Answer:
(a) treatment of the ore with water and pine oil

Question 4.
In the froth floatation process for the purification of ores the particles float because ……………
(a) they are light
(b) their surface is not easily wetted by water
(c) they bear electrostatic charge
(d) they are insoluble
Answer:
(b) their surface is not easily wetted by water

Question 5.
In a metallurgical process, an acid flux is used for removing ……………
(a) Slag
(b) basic flux
(c) acidic gangue
(d) basic gangue
Answer:
(d) basic gangue

Question 6.
The process of the removal of impurities from a crude metal is called ……………
(a) concentration
(b) calcination
(c) refining
(d) roasting
Answer:
(c) refining

Question 7.
Which of the following metal is obtained by self reduction method?
(a) Fe
(b) Cu
(c) Ag
(d) Mg
Answer:
(b) Cu

Question 8.
Which one of the following ore is best concentrated by froth floatation method?
(a) Magnetite
(b) Malachite
(c) Galena
(d) Haematite
Answer:
(c) Galena

Question 9.
Heating mixture of Cu₂O and Cu₂S will give ……………
(a) Cu + SO₂
(b) Cu + SO₃
(c) CuO + CuS
(d) Cu₂SO₃
Answer:
(a) Cu + SO₂

Question 10.
Which of the following pairs of metals is purified by Van-Arkel method?
(a) Ag and Au
(b) Ni and Fe
(c) Ga and In
(d) Zr and Ti
Answer:
(d) Zr and Ti

Question 11.
Aluminium is extracted from alumina (Al₂O₃) by electrolysis of a molten mixture of ……………
(a) (Al₂O₃) + KF + Na₃AlF₆
(b) (Al₂O₃) + HF + NaAlF₄
(c) (Al₂O₃) + CaF₂ + NaAlF₄
(d) (Al₂O₃) + Na₃AlF₆ + CaF₂
Answer:
(d) (Al₂O₃) + Na₃AlF₆ + CaF₂

Question 12.
The ore which contains both copper and iron ……………
(a) Cuprite
(b) Haematite
(c) Copper pyrite
(d) Malachite
Answer:
(c) Copper pyrite

Question 13.
Match the extraction processes listed in Column-I with metals listed in Column-II.

(a) A – 2, B – 1, C – 4, D – 3
(b) A – 3, B – 4, C – 1, D – 2
(c) A – 3, B – 1, C – 2, D – 4
(d) A – 4, B – 2, C – l, D – 3
Answer:
(c) A – 3, B – 1, C – 2, D – 4

Question 14.
Chief ore of aluminium is ……………
(a) bauxite
(b) clay
(c) haematite
(d) magnetite
Answer:
(a) bauxite

Question 15.
Which one of the following metal having least chemical reactive?
(a) Na
(b) Mg
(c) Al
(d) Au
Answer:
(d) Au

Question 16.
Pick out the more reactive metal.
(a) Cu
(b) Ag
(c) Au
(d) Na
Answer:
(d) Na

Question 17.
Consider the following statements.
(i) All ores are minerals but all minerals are not ores.
(ii) Bauxite is an ore of aluminium while clay is not.
(iii) Extraction of aluminium form clay is profitable one.

Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i), (ii) and (iii)
Answer:
(c) (iii) only

Question 18.
Match the List-I and List-II correctly using the code given below.

Answer:
(c)  3, 4, 2, 1

Question 19.
The impurities associated with ores is ……………
(a) flux
(b) slag
(c) gangue
(d) metal
Answer:
(c) gangue

Question 20.
Oxide ores are concentrated by ……………
(a) handpicking
(b) hydraulic washing
(c) froth floatation process
(d) Magnetic separation process
Answer:
(b) hydraulic washing

Question 21.
Haematite and tin stone are purified by ……………
(a) gravity separation process
(b) magnetic separation process
(c) froth floatation process
(d) handpicking
Answer:
(a) gravity separation process

Question 22.
Froth floatation process is applicable for ……………
(a) oxide ores
(b) carbonate ores
(c) chloride ores
(d) sulphide ores
Answer:
(d) sulphide ores

Question 23.
Copper pyrite and zinc blende are purified by ……………
(a) gravity separation process
(b) magnetic separation process
(c) froth floatation process
(c) handpicking
Answer:
(c) froth floatation process

Question 24.
Frothing agent used in froth floatation process is ……………
(a) pine oil
(b) olive oil
(c) mustard oil
(d) neem oil
Answer:
(a) pine oil

Question 25.
Depressing agents used to separate ZnS from PbS is ……………
(a) NaCN
(b) NaCl
(c) NaNO₃
(d) NaNO₂
Answer:
(a) NaCN

Question 26.
Leaching is also called as ……………
(a) hand picking
(b) Electrolysis
(c) Chemical process
(d) magnetic separation process
Answer:
(c) Chemical process

Question 27.
In the leaching process, the metal present in the ore is converted into ……………
(a) soluble salt
(b) soluble complex
(c) insoluble complex
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 28.
Gold ore is concentrated by ……………
(a) cyanide leaching
(b) alkali leaching
(c) acid leaching
(d) hand picking
Answer:
(a) cyanide leaching

Question 29.
Bauxite is purified by ……………
(a) cyanide leaching
(b) alkali leaching
(c) acid leaching
(d) hand picking
Answer:
(b) alkali leaching

Question 30.
Which type of leaching process convert insoluble sulphide ores into soluble sulphates?
(a) cyanide leaching
(b) alkali leaching
(c) acid leaching
(d) hand picking
Answer:
(c) acid leaching

Question 31.
A chemist involves mining process and he got tow ores mixed together. If one is tinstone and another one is chromite, which type of process will be used to separate that two ores?
(a) Leaching process
(b) Froth floatation process
(c) Zone refining process
(d) Magnetic separation process
Answer:
(d) Magnetic separation process

Question 32.
The process in which the concentrated ore is strongly heated in the absence of air is called as ……………
(a) Roasting
(b) Calcination
(c) Smelting
(d) Leaching
Answer:
(b) Calcination

Question 33.
A chemical substance that forms an easily fusible slag with gangue is called as ……………
(a) flux
(b) pure metal
(c) ore
(d) impure metal
Answer:
(a) flux

Question 34.
Blistered copper is ……………
(a) 98% pure copper
(b) 96% pure copper
(c) 97% pure copper
(d) 88% pure copper
Answer:
(a) 98% pure copper

Question 35.
Ignition mixture used in aluminothermic process is ……………
(a) Mg + BaO₂
(b) MgO + BaO
(c) Al₂O₃ + Mg
(d) Al₂O₃ + BaO₂
Answer:
(a) Mg + BaO₂

Question 36.
For spontaneous reaction, the change in free energy should be ……………
(a) positive
(b) negative
(c) zero
(d) neutral
Answer:
(b) negative

Question 37.
The change in Gibbs free energy for a reaction is expressed by ……………
(a) ∆G = ∆H + T∆S
(b) ∆G = ∆H – TS
(c) G = H – TS
(d) ∆G = ∆H – T∆S
Answer:
(d) ∆G = ∆H – T∆S

Question 38.
Relationship between ∆G° and K_p is ……………
(a) ∆G° = – RT In K_p
(b) ∆G° = – R In K_p
(c) ∆G° = – T In K_p
(d) ∆G° = RT In K_p
Answer:
(a) ∆G° = – RT In K_p

Question 39.
Consider the following statements.
(i) Ellingham drawn on a plot by considering the temperature in the x-axis and the standard free energy change for the formation of metal oxide in y-axis.
(ii) The resultant plot is straight line.
(ii) In the Ellingham diagram, ∆H as slope and ∆S as y-intercept.

Which of the above statement(s) is / are not correct?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (ii) only
(d) (iii) only
Answer:
(d) (iii) only

Question 40.
Which of the following oxides is unstable j at moderate temperature?
(a) Al₂O₃
(b) Cr₂O₃
(c) MgO
(d) HgO
Answer:
(d) HgO

Question 41.
The oxides will decompose on heating even in the absence of a reducing agent is ……………
(a) Ag₂O
(b) HgO
(c) MgO
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 42.
Consider the following statements.
(i) Ellingham diagram gives information about the thermodynamic feasibility of a reaction.
(ii) It explains the rate of the reaction.
(iii) Below 1000 K temperature, FeO is more stable than CO.

Which of the above statements (s) is / are not correct?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) only
(d) (ii) only
Answer:
(d) (ii) only

Question 43.
Gibbs free energy change for the electrolysis process is expressed by ……………
(a) ∆G° = – nFE°
(b) ∆G° = nF
(c) ∆G° = – nE°
(d) ∆G° = nFE°
Answer:
(a) ∆G° = – nFE°

Question 44.
The technique used to refining zinc and mercury is ……………
(a) Liquation
(b) Distillation
(c) Zone refining
(d) Van-Arkel method
Answer:
(b) Distillation

Question 45.
Which of the following is not purified by zone refining process?
(a) Ge
(b) Si
(c) Ga
(d) Al
Answer:
(d) Al

Question 46.
Nickel is purified by ……………
(a) Mond process
(b) Van-Arkel method
(c) Zone refining
(d) Electrolytic refining
Answer:
(a) Mond process

Question 47.
Titanium is purified by ……………
(a) Mond process
(b) Van-Arkel method
(c) Zone refining
(d) Electrolytic refining
Answer:
(b) Van-Arkel method

Question 48.
The metal used for galvanisation of iron is ……………
(a) Al
(b) Zn
(c) Cu
(d) Au
Answer:
(b) Zn

Question 49.
Which metal alloy is used in design of aeroplane parts?
(a) Al
(b) Zn
(c) Cu
(d) Au
Answer:
(a) Al

Question 50.
Which metal is used for making coins and ornaments along with gold and other metals?
(a) Zn
(b) Al
(c) Cu
(d) Fe
Answer:
(c) Cu

II. Fill in the blanks:

1. Minerals that contains a high percentage of metal from which it can be extracted conveniently and economically are called ……………
2. …………… helps us to select a suitable reducing agent and appropriate temperature range for reduction.
3. If E° is positive, then the ∆G is ……………
4. Chemical formula of cuprite is ……………
5. Gravity separation is also called as ……………
6. …………… ore is purified by gravity separation process.
7. In froth floatation process …………… acts as a collector.
8. Sodium cyanide is added to depresses the floatation property of ZnS by forming a layer of ……………
9. The process of gold reduced to its elemental state is called ……………
10. Leaching process is a …………… reaction.
11. Magnesite is calcined to give ……………
12. …………… a chemical substance that forms an easily fusible slag with gangue.
13. Flux + Gangue → ……………
14. In the extraction of iron …………… is removed as slag.
15. In the extraction of copper …………… is removed as slag.
16. Cr₂O₃ can be reduced by …………… an process.
17. …………… is used as a reducing agent for the reduction of chromic oxide.
18. …………… is purified by zone refining.
19. In the mond process, impure nickel is heated with …………… compound.
20. …………… filament is used to decompose titanium tetraoxide.
21. …………… is the most abundant metal.
22. …………… is used in packing materials for food items.
23. …………… is used in the manufacture of paints, rubber and cosmetics.
24. …………… is the first metal used by the humans.
25. …………… is one of the expensive and precious metals.
26. …………… is used for increasing the efficiency of solar cells and also used as catalysts.
27. …………… ore is concentrated by froth floatation process.
28. Zinc blende is concentrated by …………… process.
29. Gold ore is leached by adding of ……………

Answer:

1. Ores
2. Ellingham diagram
3. negative
4. Cu₂O
5. Hydraulic washing
6. oxide
7. Sodium ethyl xanthate
8. Na₂[Zn(CN)₄] – Sodium negative
9. Cementation
10. Redox
11. Magnesia
12. Flux
13. slag
14. Calcium silicate (CaSiO₃)
15. Ferrous silicate (FeSiO₃)
16. Aluminothermic process
17. Aluminium
18. semi-conductior
19. carbon monoxide
20. Tungsten
21. Aluminium
22. Aluminium foils
23. Aluminium
24. Zinc oxide
25. Copper
26. Gold
27. Gold nanoparticle
28. Sulphide
29. Froth floatation
30. NaCN

III. Match the following:

Question 1.
(i) Haematite – (a) Fe₃O₄
(ii) Siderite – (b) Fe₃O₃
(iii) Iron pyrite – (c) Fe₃O₃
(iv) Magnetite – (d) FeS₂
Answer:
(i) – (b)
(ii) – (c)
(iii) – (d)
(iv) – (a)

Question 2.
(i) Copper glance – (a) CuCO₃. Cu(OH)₂
(ii) Malachite – (b) Cu₂S
(iii) Copper pyrite – (c) 2CuCO₃. Cu(OH)₂
(iv) Azurite – (d) CuFeS₂
Answer:
(i) – (b)
(ii) – (a)
(iii) – (d)
(iv) – (c)

Question 3.
(i) Zinc blende – (a) Al₂0₃.2H₂O
(ii) Bauxite – (b) ZnC0₃
(iii) Zincite – (c) ZnS
(iv) Calamine – (d) ZnO
Answer:
(i) – (c)
(ii) – (a)
(iii) – (d)
(iv) – (b)

Question 4.
(i) Tin stone – (a) AgCl
(ii) Argentite – (b) Ag₃SbS₃
(iii) Ruby silver – (c) Ag₂S
(iv) Horn silver – (d) SnO₂
Answer:
(i) – (d)
(ii) – (c)
(iii) – (b)
(iv) – (a)

Question 5.
(i) Oxide ore – (a) Zinc blende
(ii) Carbonate ore – (b) Horn silver
(Hi) Sulphide ore – (c) Haematite
(iv) Chloride ore – (d) Calamine
Answer:
(i) – (c)
(ii) – (d)
(iii) – (a)
(iv) – (b)

Question 6.
(i) Tin stone – (a) Magnetic separation process
(ii) Copper pyrite – (b) Leaching process
(Hi) Bauxite – (c) Froth floatation process
(iv) Chromite – (d) Hydraulic washing process
Answer:
(i) – (d)
(ii) – (c)
(iii) – (b)
(iv) – (a)

Question 7.
(i) Aluminium – (a) cosmetics
(ii) Zinc oxide – (b) gas pipelines
(iii) Iron – (c) making coins
(iv) Copper – (d) bicycle chains
Answer:
(i) – (b)
(ii) – (a)
(iii) – (d)
(iv) – (c)

Question 8.
(i) Gold nanopaticle – (a)Cast iron
(ii) Cast iron – (b) Cooking vessels
(iii) Metallic zinc – (c) Solar cells
(iv) Aluminium – (d) Pump stoves
Answer:
(i) – (c)
(ii) – (d)
(iii) – (a)
(iv) – (b)

IV. Assertion and reasons:

Note:
In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(a) A and R are correct, R explains A
(b) A and R are correct, R does not explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct

Question 1.
Assertion(A) – Clay is an ore of aluminium while bauxite is not.
Reason (R) – Aluminium can be economically extracted from bauxite not from clay.
Answer:
(d) A is wrong but R is correct

Question 2.
Assertion(A) – Haematite and tin stone are concentrated by hydraulic washing process.
Reason (R) – Oxide ores are heavy and have high specific gravity.
Answer:
(a) A and R are correct, R explains A .

Question 3.
Assertion(A) – Sulphide ores are concentrated by froth floatation process.
Reason (R) – Sulphide ores are preferentially wetted by oil can be separated from gangue.
Answer:
(a) A and R are correct, R explains A

Question 4.
Assertion(A) – A suitable reducing agent is selected based on the thermodynamic consideration.
Reason (R) – The reduction of metal oxide with a given reducing agent can occur if the free energy change for the coupled reaction is positive.
Answer:
(c) A is correct but R is wrong

Question 5.
Assertion(A) – Aluminium can be used as a reducing agent for the reduction of chromic oxide.
Reason (R) – In the Ellingham diagram, formation of chromium oxide lies above that of the aluminium, therefore Al₂O₃ is more stable than Cr₂O₃.
Answer:
(a) A and R are correct, R explains A

Question 6.
Assertion(A) – Zone refining is based on the principles of fractional crystallisation,
Reason (R) – This process is carried out in an inert gas temperature.
Answer:
(b) A and R are correct, R does not explain A

Question 7.
Assertion(A) – Aluminium is used in the j design of chemical reactors.
Reason (R) – Aluminium has high resistance to corrosion.
Answer:
(a) A and R are correct, R explains A

V. Find the odd one out and give the reasons:

Question 1.
(a) Nickel
(b) Silicon
(c) Germanium
(d) Galium
Answer:
(a) Nickel
Reason:
Si, Ge and Ga are purified by I zone refining while Ni is not.

Question 2.
(a) Copper
(b) Zinc
(c) Silver
(d) Galium
Answer:
(d) Galium
Reason:
Cu, Zn and Ag are purified by electrolytic refining while Ga is not.

Question 3.
(a) ∆G = -ve
(b) ∆H = -ve
(c) ∆S = +ve
(d) ∆S = -ve
Answer:
(d) ∆S = -ve

Reason:
∆G=-ve, ∆H=-ve, ∆S=+ve are the conditions for spontaneous reaction, ∆S = -ve is condition for non-spontaneous process.

Question 4.
(a) Haematite
(b) Siderite
(c) Limonite
(d) Azunte
Answer:
(a) Azurite

Reason:
Azurite is ore of copper but others arc ore of iron.

Question 5.
(a) Copper glance
(b) Zinc blende
(c) Argentite
(d) Magnetite
Answer:
(d) Magnetite

Reason:
Magnetite is oxide ore but others are sulphide ores.

Question 6.
(a) Silver
(b) Gold
(c) Sodium
(d) Platinum
Answer:
(c) Sodium

Reason:
Sodium is more reactive element than Ag, Au and Pt.

VI. Find out the correct pair:

Question 1.
(a) Metallurgy – Extraction of metals
(b) Clay – Ore of Al
(c) Na – Native element
(d) Ore – Gangue
Answer:
(a) Metallurgy Extraction of metals

Question 2.
(a) Bauxite – Iron
(b) Siderite – Aluminium
(c) Malachite – Copper
(d) Argentite – Gold
Answer:
(c) Malachite – Copper

Question 3.
(a) Limonite – Sulphide ore
(b) Cuprite – Oxide ore
(c) Calamine – Suiphide ore
(d) Horn silver – Oxide ore
Answer:
(b) Cuprite – Oxide ore

Question 4.
(a) Oxide ore – Froth floatation process
(b) Suiphide ore – Gravity separation process
(c) Gold ore – Leaching method
(d) Oxide ore – Magnetic separation process
Answer:
(c) Gold ore – Leaching method

Question 5.
(a) Aluminium – Galvanising metals
(b) Zinc – Cooking vessels
(c) Iron – Cutlery
(d) Copper – Dental fillings
Answer:
(c) Iron -Cutlery

VII. Find out the incorrect pair:

Question 1.
(a) Copper – Least reactive
(b) Clay – Mineral of Al
(c) Bauxite – Mineral of Al
(d) Gangue – Impurity
Answer:
(c) Bauxite – Mineral of Al

Question 2.
(a) Aluminium – Corundum
(b) Limonite – Iron
(c) Galena – Lead
(d) Tin – Siderite
Answer:
(d) Tin – Siderite

Question 3.
(a) Magnetite – FeO₄
(b) Malachite – CuCO₃ .Cu(OH)₂
(c) Horn silver – Ag₂S
(d) Stefinite – Ag₂SbS₄
Answer:
(c) Horn silver – Ag₂S

Question 4.
(a) Tin stone – Oxide ore
(b) Copper pyrite – Oxide ore
(c) Zincite – Oxide ore
(d) Bauxite – Oxide ore
Answer:
(b) Copper pyrite Oxide ore

Question 5.
(a) Haematitc – Gravity separation process
(b) Copper pyrite – froth floatation process
(c) Bauxite – Leaching process
(a) pyrolusite – Magnetic separation process
Answer:
(b) Copper pyrite – froth floatation process

Question 6.
(a) Gold ore – Cyanide leaching
(b) Nickel ore – Ammonia leaching
(c) Aluminium ore – Alkali leaching
(d) Silver ore – Acid leaching
Answer:
(a) Silver ore – Acid leaching

Question 7.
(a) Aluminium – Design of aeroplane
(b) Zinc oxide – Cosmetics
(c) Zinc suiphide – X-ray screens
(d) Iron – Artifici al limb joints
Answer:
(d) Iron – Artificial limb joints

Samacheer Kalvi 12th Chemistry Metallurgy 2 Mark Questions and Answers

VIII. Answer the following:

Question 1.
Define Metallurgy.
Answer:
The various steps involved in the extraction of metals from their ores as well as refining of crude metals are collectively known as metallurgy.

Question 2.
What are minerals?
Answer:
A naturally occurring substance obtained by mining which contains the metal in free state or in the form of compounds like oxides, sulphides etc… is called as mineral. Bauxite and clay are minerals of Aluminium.

Question 3.
What are ores?
Answer:
A mineral from which the metal can be extracted easily and economically are called ores. Bauxite is an ore of Aluminium.

Question 4.
What are all the steps involved in metallurgical process?
Answer:
The extraction of a metal from its ore consists the following metallurgical process.

1. Concentration of the ore
2. Extraction of crude metal
3. Refining of crude metal

Question 5.
What is Gangue?
Answer:
Generally, the ores are associated with nonmetallic impurities, rocky materials and siliceous matter which arc collectively known as gangue.

Question 6.
Give the depressing agents used in the froth floatation process and why we use depressing agents in that process?
Answer:
Generally sodium cyanide and sodium carbonate are used as an depressing agents for froth floatation process. These reagents are used to selectively prevent other metal sulphides from coming to the froth. For example, when impurities such as ZnS is present in galena (PbS), sodium cyanide (NaCN) is added to depresses the flotation property of ZnS by forming a layer of zinc complex Na₂[Zn(CN)₄] on the surface of zinc sulphide.

Question 7.
What are leaching process?
Answer:
This method is based on the solubility of the ore in a suitable solvent and the reactions in aqueous solution. In this method, the crushed ore is allowed to dissolve in a suitable solvent, the metal present in the ore is converted to its soluble salt or complex while the gangue remains insoluble. This process is also called chemical method.

Question 8.
Explain Cyanide leaching.
Answer:
In the concentration of gold ore, the crashed ore of gold is leached with aerated dilute solution of sodium cyanide. Gold is converted into a soluble cyanide complex. The gangue, aluminosilicate remains insoluble.
4Au (s) + 8CN^– (aq) + O₂ (g) + 2H₂O (1) → 4[Au(CN)₂]^–(aq) + 4OH^– (aq)

Question 9.
What is Cementation?
Answer:
Gold can be recovered by reacting the deoxygenated leached solution with zinc. In this process the gold is reduced to its elemental state (zero oxidation sate) and the process is called
Zn (s)+ 2[Au(CN)₂]^– (aq) → [Zn(CN)₄]^2-(aq)+ 2Au (s)

Question 10.
What is Ammonia leaching.
Answer:
When a crashed ore containing nickel, copper and cobalt is treated with aqueous ammonia . under suitable pressure, ammonia selectively leaches these metals by forming their soluble complexes viz. [Ni(NH₃)₆]²⁺, [Cu(NH₃)₄]²⁺, and [Co(NH₃)₅H₂O]³⁺ respectively from the ore leaving behind the gangue, iron(III) oxides/hydroxides and aluminosilicate.

Question 11.
What is Acid leaching?
Answer:
Leaching of sulphide ores such as ZnS, PbS etc,, can be done by treating them with hot aqueous sulphuric acid.
2ZnS(s) + 2H₂SO₄(aq) + O₂ (g) → 2ZnSO₄ (aq) + 2S (s) + H₂O
In this process the insoluble sulphide is converted into soluble sulphate and elemental sulphur.

Question 12.
What is roasting?
Answer:
The process of heating an ore (generally, sulphide are) is strongly below its melting point in the presence of an excess of air is called roasting.

Question 13.
Define Calcination.
Answer:
It is the process in which the concentrated ore is strongly heated in the absence of air. This method can also be carried out with a limited supply of air.

Question 14.
How will you manage sulphur dioxide produced during roasting process?
Answer:
The sulphur dioxide produced during roasting process is harmful to the environment. In modem metallurgical factories, this by product is trapped and converted into sulphuric acid to avoid air pollution.

Question 15.
What is smelting?
Answer:
Smelting is a process of reducing the roasted metallic oxide to metal in molten condition. In this process, impurities are removed by the addition of flux as slag.

Question 16.
Explain Auto reduction.
Answer:
Simple roasting of some of the ores give the crude metal. In such cases, the use of reducing agents is not necessary. For example, mercury is obtained by roasting of its ore cinnabar (HgS)
HgS _(s) + O_2(g) → Hg _(l) + SO₂ \(\uparrow \)

Question 17.
What is Ellingham diagram?
Answer:
The graphical representation of variation of the standard Gibbs free energy of reaction for the formation of various metal oxides with temperature is called Ellingham diagram.

Question 18.
CO is more stable at higher temperature. Why?
Answer:
In the Ellingham diagram, the formation of carbon monoxide is a straight line with negative slope. In this case ∆S is positive as 2 moles of CO gas is formed by the consumption of one mole of oxygen. Hence, CO is more stable at higher temperature.

Question 19.
Ag_2(g)O and HgO are unstable at moderate temperature and they will decompose on heating even in the absence of a reducing agent. Why?
Answer:
Ellingham diagram for the formation of Ag_2(g)O and HgO is at upper part of the diagram and their decomposition temperatures are 600 and 700 K respectively. It indicates that these oxides are unstable at moderate temperatures and will decompose on heating even in the absence of a reducing agent.

Question 20.
What is refining process?
Answer:
Generally the metal extracted from its ore contains some impurities such as unreacted oxide ore, other metals, nonmetal etc. Removal of such impurities associated with the isolated crude metal is called refining process.

Question 21.
List out the common refining methods.
Answer:

1. Distillation
2. Liquation
3. Electrolytic refining
4. Zone refining
5. Vapour phase method

Question 22.
Explain Distillation process with suitable example.
Answer:
This method is employed for low boiling volatile metals like zinc (boiling point 1180 K) and mercury (630 K). In this method, the impure metal is heated to evaporate and the vapours are condensed to get pure metal.

Question 23.
Write the applications of copper.
Answer:

1. Copper is the first metal used by the humans and extended use of its alloy bronze resulted in a new era, Bronze age.
2. Copper is used for making coins and ornaments along with gold and other metals.
3. Copper and its alloys are used for making wires, water pipes and other electrical parts.

Question 24.
Why aluminium cannot be extracted by reducing alumina with carbon?
Answer:
Alumina (Al₂O₃) cannot be reduced by using carbon because aluminum has more affinity for oxygen than carbon. Therefore, aluminium cannot be extracted by reducing alumina with carbon.

Question 25.
Write the names of two ores of copper.
Answer:
Main ores of copper are:

1. Copper pyrites – CuFeS₂
2. Copper glance – Cu₂S
3. Malachite – CuCO₃. Cu(OH)₂

Question 26.
Explain the role of carbon monoxide in the purification of nickel?
Answer:
During the purification of Nickel by Mond’s process, carbon monoxide (CO) is used to convert impure nickel to nickel carbonyl. Nickel carbonyl is an unstable compound. Heating to higher temperature decomposes it to give pure nickel.

Question 27.
ZnO can be reduced to the metal by heating with carbon but not Cr₂O₃. Justify your answer.
Answer:
Carbon has more affinity for oxygen than zinc, whereas chromium has higher affinity for oxygen than zinc. Hence ZnO can be reduced to the metal by heating with carbon but not Cr₂O₃.

Question 28.
Name the method used for the refining of

1. Nickel
2. Zirconium

Answer:

1. Mond’s process
2. Van Arkel’s method

Question 29.
Give one example for

1. Acidic flux
2. Basic flux

Answer:

1. Acidic flux: SiO₂
2. Basic flux: CaCO₃, MgCO₃

Question 30.
Why is the reduction of a metal oxide easier if the metal formed is in liquid state at the temperature of reduction?
Answer:
The entropy is higher if the metal is in liquid state than when it is in solid state. The value of entropy change (∆S) of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus, the value of ∆G° becomes more on negative side and the reduction becomes easier.

Samacheer Kalvi 12th Chemistry Metallurgy 3 Mark Questions and Answers

IX. Answer the following questions:

Question 1.
All ores are minerals, but all minerals are not ores. Explain.
Answer:
A naturally occurring substance obtained by mining which contains the metal in free state or in the form of compounds is called a mineral. In most of the minerals, the metal of interest is present only in small amounts and some of them contains a reasonable percentage of metal. Such minerals that contains a high percentage of metal, from which it can be extracted conveniently and economically are called ores. Hence all ores are minerals but all minerals are not ores.

Question 2.
Bauxite is an ore of aluminium while china clay is not. Why?
Answer:
Bauxite and china clay, both are minerals of aluminium. However, aluminium can be commercially extracted from bauxite while extraction from china clay is not a profitable one. Hence the mineral, bauxite is an ore of aluminium while china clay is not.

Question 3.
Explain Gravity separation process or Hydraulic washing process?
[OR]
How will you concentrate oxide ores?
[OR]
Explain the suitable method to concentrate hematite and tinstone ores.
Answer:
In this method, the ore having high specific gravity is separated from the gangue that has low specific gravity by simply washing with running water. Ore is crushed to a finely powdered form and treated with rapidly flowing current of water. During this process the lighter gangue particles are washed away by the running water. This method is generally applied to concentrate the native ore such as gold and oxide ores such as hematite (Fe₂O₃), tin stone (SnO₂) etc.

Question 4.
Write a notes on alkali leaching process?
Answer:
In this method, the ore is treated with aqueous alkali to form a soluble complex. For example, bauxite, an important ore of aluminium is heated with a solution of sodium hydroxide or sodium carbonate in the temperature range 470 – 520 K at 35 atm to form soluble sodium meta- aluminate leaving behind the impurities, iron oxide and titanium oxide.
Al₂O₃ (S) + 2NaOH (aq) + 3H₂O (l) → 2Na[Al(OH)₄] (aq)

The hot solution is decanted, cooled, and diluted. This solution is neutralised by passing CO₂ gas, to the form hydrated Al₂O₂ precipitate.
2Na[Al(OH)₄](aq) + CO₂(g) → Al₂O₃. xH₂O(s) + 2NaHCO₃ (aq)
The precipitate is filtered off and heated around 1670 K to get pure alumina Al₂O₃.

Question 5.
Complete the following reaction.

1. ZnS + O₂ \(\underrightarrow { \Delta } \) ?
2. P₄+ O₂ \(\underrightarrow { \Delta } \)?
3. CaCO₃ \(\underrightarrow { \Delta } \) ?

Answer:

1. 2ZnS + 3O₂ \(\underrightarrow { \Delta } \) 2ZnO + 2SO₂ \(\uparrow \)
2. P₄+ 5O₂ \(\underrightarrow { \Delta } \) P₄O₁₀ \(\uparrow \)
3. CaCO₃ \(\underrightarrow { \Delta } \) CaO + C0₂ \(\uparrow \)

Question 6.
Explain Aluminothermic process.
Answer:
Metallic oxides such as Cr₂O₃ can be reduced by an aluminothermic process. In this process, the metal oxide is mixed with aluminium powder and placed in a fire clay crucible. To initiate the reduction process, an ignition mixture (usually magneisium and barium peroxide) is used.
BaO₂ + Mg → BaO + MgO

During the above reaction a large amount of heat is evolved (temperature upto 2400°C, is generated and the reaction enthalpy is – 852 kJ mol ) which facilitates the reduction of Cr₂O₃ by aluminium power.
Cr₂O₃ + 2Al \(\underrightarrow { \Delta } \) 2Cr + Al₂O₃

Question 7.
Why aluminium can be used as a reducing agent for the reduction of chronic oxide?
Answer:
Ellingham diagram is used to predict thermodynamic feasibility of reduction of oxides of one metal by another metal. Any metal can reduce the oxides of other metals that are located above it in the diagram. For example, in the Ellingham diagram, for the fonnation of chromium oxide lies above that of the aluminium, meaning that Al₂O₃, is more stable than Cr₂O₃. Hence aluminium can be used as a reducing agent for the reduction of chromic oxide. However, it cannot be used to reduce the oxides of magnesium and calcium which occupy lower position than aluminium oxide.

Question 8.
Write notes on liquation.
Answer:
Liquation:
This method, is employed to remove the impurities with high melting points from metals having relatively low melting points such a stin(Sb;mp=904K), lead(Pb;mp=600K), mercury(Hg; mp = 234 K), and bismuth (Bi; mp = 545K). In this process, the crude metal is heated to form fusible liquid and allowed to flow on a sloping surface.

The impure metal is placed on sloping hearth of a reverberatory furnace and it is heated just above the melting point of the metal in the absence of air, the molten pure metal flows down and the impurities are left behind. The molten metal is collected and solidified.

Question 9.
Write the application of Iron (Fe).
Answer:
1. Iron is one of the most useful metals and its alloys are used everywhere including bridges, electricity pylons, bicycle chains, cutting tools and rifle barrels.

2. Cast iron is used to make pipes, valves and pumps stoves etc.

3. Magnets can be made of iron and its alloys and compounds.

4. An important alloy of iron is stainless steel, and it is very resistant to corrosion. It is used in architecture, bearings, cutlery, surgical instruments and jewellery. Nickel steel is used for making cables, automobiles and aeroplane parts. Chrome steels are used for manufacturing cutting tools and curshing machines

Question 10.
Mention the uses of Gold (Au).
Answer:
1. Gold, one of the expensive and precious metals. It is used for coinage, and has been used as standard for monetary systems in some countries.

2. It is used extensively in jewellery in its alloy form with copper. It is also used in electroplating to cover other metals with a thin layer of gold which are used in watches, artificial limb joints, cheap jewellery, dental fillings and electrical connectors.

3. Gold nanoparticles are also used for increasing the efficiency of solar cells and also used an catalysts.

Question 11.
The extraction of Au by leaching with NaCN involves both oxidation and reduction. Justify giving equation.
Answer:

• 4Au (s) + 8CN^–(aq) + 2H₂O₂(aq) + O₂(g) → 4[Au(CN)₂]^– (aq) + 4OH^–(aq)
• 2[Au(CN)₂]^–(aq) + Zn(s) → 2Au(s)+ [Zn(CN)₄]^-2(aq)

In the first reaction Au changes into Au⁺, i.e. its oxidation takes place. In the second reaction:
Au⁺ → Au°
(i.e.) reduction takes place.

Question 12.
Account for the following facts:

1. The reduction of a metal oxide is easier if the metal formed is in liquid state at the temperature of reduction.
2. The reduction of Cr₂O₃ with Al is thermodynamically feasible, yet it does not occur at room temperature.
3. Pine oil is used in froth floatation method.

Answer:

1. In liquid state entropy is higher than the solid form, this makes ∆_fG more negative.
2. By increasing the temperature, fraction of activated molecules increases, which helps in crossing over the energy barrier.
3. Pine oil enhances the non-wetting property of the ore particles and also acts as the froth collector.

Question 13.
Write the chemical reactions for purification of Zirconium by Van Arkel’s process.
Answer:

Samacheer Kalvi 12th Chemistry Metallurgy 5 Mark Questions and Answers

X. Answer the following questions:

Question 1.
Explain froth floatation process.
(Or)
How will you concentrate sulphide ores?
(Or)
Explain-the concentration of copper pyrites and galena ores.
Answer:
Froth flotation:
This method is commonly used to concentrate sulphide ores such as galena (PbS), zinc blende (ZnS) etc. In this method, the metallic ore particles which are preferentially wetted by oil can be separated from gangue. In this method, the crushed ore is suspended in water and mixed with frothing agent such as pine oil, eucalyptus oil etc. A small quantity of sodium ethyl xanthate which acts as a collector is also added. A froth is generated by blowing air through this mixture.

The collector molecules attach to the ore particle and make them water repellent. As a result, ore particles, wetted by the oil, rise to the surface along with the froth. The froth is skimmed off and dried to recover the concentrated ore. The gangue particles that are preferentially wetted by water settle at the bottom.

When a sulphide ore of a metal of interest contains other metal sulphides as impurities, depressing agents such as sodium cyanide, sodium carbonate etc are used to selectively prevent other metal sulphides from coming to the froth. For example, when impurities such as ZnS is present in galena (PbS), sodium cyanide (NaCN) is added to depresses the flotation property of ZnS by forming a layer of zinc complex Na₂[Zn(CN)₄] on the surface of zinc sulphide.

Question 2.
Discuss the magnetic separation process.
(Or)
How will you separate magnetic ores from non-magnetic ores?
Answer:
Magnetic separation:
This method is applicable to ferromagnetic ores and it is based on the difference in the magnetic properties of the ore and the impurities. For example tin stone can be separated from the wolframite impurities which is magnetic. Similarly, ores such as chromite, pyrolusite having magnetic property can be removed from the non magnetic siliceous impurities.

The crushed ore is poured on to an electromagnetic separator consisting of a belt moving over two rollers of which one is magnetic. The magnetic part of the ore is attracted towards the magnet and falls as a heap close to the magnetic region while the nonmagnetic part falls away from it as shown in the figure.

Question 3.
Write a note on thermodynamic principle of metallurgy.
Answer:
Thermodynamic principle of metallurgy:
The extraction of metals from their oxides can be carried out by using different reducing agents. For example, consider the reduction of a metal
oxide M_xO_y.
\(\frac { 2 }{ y } \)M_xO_y(s) → \(\frac { 2x }{ y } \)M(s) + O₂(g) ……………..(1)

The above reduction may be carried out with carbon. In this case, the reducing agent carbon may be oxidised to either CO or CO₂.

• C + O₂ → CO_2(g) ……….(2)
• 2C + O₂ → 2CO_(g) …………(3)

If carbon monoxide is used as a reducing agent, it is oxidised to CO₂ as follows,
2CO + O₂ → 2CO_2(g)  ……………(4)

A suitable reducing agent is selected based on the thermodynamic considerations. We know that for a spontaneous reaction, the change in free energy (AG) should be negative. Therefore, thermodynamically, the reduction of metal oxide [equation (1)] with a given reducing agent [Equation (2), (3) or (4)] can occur if the free energy change for the coupled reaction. [Equations (1) & (2), (1) & (3) or (1) & (4)] is negative. Hence, the reducing agent is selected in such a way that it provides a large negative AG value for the coupled reaction.

Ellingham diagram:
The change in Gibbs free energy (∆G) for a reaction is given by the expression.
∆G = ∆H – T∆S ……….(1)
where, ∆H is the enthalpy change , T the temperature in kelvin and ∆S the entropy change. For an equilibrium process, ∆G° can be calculated using the equilibrium constant by the following expression ∆G° = – RT lnKp

Harold Ellingham used the above relationship to calculate the ∆G° values at various temperatures for the reduction of metal oxides by treating the reduction as an equilibrium process. He has drawn a plot by considering the temperature in the x-axis and the standard free energy change for the formation of metal oxide in y-axis. The resultant plot is a straight line with ∆S as slope and ∆H as y-intercept. The graphical representation of variation of the standard Gibbs free energy of reaction for the formation of various metal oxides with temperature is called Ellingham diagram.

Question 4.
Explain the observations from the Ellingham diagram.
Answer:
1. For most of the metal oxide formation, the slope is positive. It can be explained as follows. Oxygen gas is consumed during the formation of metal oxides which results in the decrease in randomness. Hence, ∆S becomes negative and it makes the term, T∆S positive in the straight line equation.

2. The graph for the formation of carbon monoxide is a straight line with negative slope. In this case ∆S is positive as 2 moles of CO gas is formed by the consumption of one mole of oxygen gas. It indicates that CO is more stable at higher temperature.

3. As the temperature increases, generally ∆G value for the formation of the metal oxide become less negative and becomes zero at a particular temperature. Below this temperature, ∆G is negative and the oxide is stable and above this temperature ∆G is positive. This general trend suggests that metal oxides become less stable at higher temperature and their decomposition becomes easier.

4. There is a sudden change in the slope at a particular temperature for some metal oxides like MgO, HgO. This is due to the phase transition (melting or evaporation).

Question 5.
Discuss.the applications of the Ellingham diagram:
Answer:
Ellingham diagram helps us to select a suitable reducing agent and appropriate temperature range for reduction. The reduction of a metal oxide to its metal can be considered as a competition between the element used for reduction and the metal to combine with oxygen.

If the metal oxide is more stable, then oxygen remains with the metal and if the oxide of element used for reduction is more stable, then the oxygen from the metal oxide combines with elements used for the reduction. From the Ellingham diagram, we can infer the relative stability of different metal oxides at a given temperature.

1. Ellingham diagram for the formation of Ag₂O and HgO is at upper part of the diagram and their decomposition temperatures are 600 and 700 K respectively. It indicates that these oxides are unstable at moderate temperatures and will decompose on heating even in the absence of a reducing agent.

2. Ellingham diagram is used to predict thermodynamic feasibility of reduction of oxides of one metal by another metal. Any metal can reduce the oxides of other metals that are located above it in the diagram. For example, in the Ellingham diagram, for the formation of chromium oxide lies above that of the aluminium, meaning that Al₂O₃ is more stable than Cr₂O₃. Hence aluminium can be used as a reducing agent for the reduction of chromic oxide. However, it cannot be used to reduce the oxides of magnesium and calcium which occupy lower position than aluminium oxide.

3. The carbon line cuts across the lines of many metal oxides and hence it can reduce all those metal oxides at sufficiently high temperature. Let us analyse the thermodynamically favourable conditions for the reduction of iron oxide by carbon. Ellingham diagram for the formation of FeO and CO intersects around 1000 K.

Below this temperature the carbon line lies above the iron line which indicates that FeO is more stable than CO and hence at this temperature range, the reduction is not thermodynamically feasible. However, above 1000 K carbon line lies below the iron line and hence, we can use coke as reducing agent above this temperature. The following free energy calculation also confirm that the reduction is thermodynamically favoured.

From the Ellingham Diagram at 1500 K:

• 2Fe_(s) + O_2(g) → 2FeO_(g)
• 2C_(s) + O_2(g) → 2CO_(g)
  
• ∆G₁ = – 350 kJ mol^-1 …………(5)
• ∆G₂ = – 480 kJ mol^-1 ………….(6)

Reverse the reaction (1)

• 2FeO_(s) → 2Fe_(s) + O_2(g)
• – ∆G₁= +350 kJ mol^-1 ………. (7)

Now couple the reactions (2) and (3)

• 2FeO_(s) + 2C 2Fe (l,s) + 2CO_(g)
• ∆G₃ = -130 kJ mol^-1 ……………. (8)

The standard free energy change for the reduction of one mole of FeO is:
∆G₃/2 = -65 kJ mol^-1.

Question 6.
Explain the method to purify Titanium metal.
[OR]
Explain Van-Arkel method for refining Titanium.
[OR]
How will you purify metals by using iodine?
Answer:
This method is based on the thermal decomposition of metal compounds which lead to the formation of pure metals. Titanium and zirconium can be purified using this method. For example, the impure titanium metal is heated in an evacuated vessel with iodine at a temperature of 550 K to form the volatile titanium tetra-iodide.(TiI₄). The impurities are left behind, as they do not react with iodine.
Ti_(s) + 2I_2(s) → TiI₄ (vapour)

The volatile titanium tetraiodide vapour is passed over a tungsten filament at a temperature around 1800 K. The titanium tetraiodide is decomposed and pure titanium is deposited on the filament. The iodine is reused.
Til₄ (vapour) → Ti_(s)+ 2I_2(s)

Question 7.
Mention the applications of Aluminium.
Answer:
Aluminium is the most abundant metal and is a good conductor of electricity and heat. It also resists corrosion. The following are some of its applications.

1. Many heat exchangers/sinks and our day to day cooking vessels are made of aluminium.
2. It is used as wraps (aluminium foils) and is used in packing materials for food items,
3. Aluminium is not very strong, However, its alloys with copper, manganese, magnesium and silicon are light weight and strong and they are used in design of aeroplanes and other forms of transport.
4. As Aluminium shows high resistance to corrosion, it is used in the design of chemical reactors, medical equipments,refrigeration units and gas pipelines.
5. Aluminium is a good electrical conductor and cheap, hence used in electrical overhead electric cables with steel core for strength.

Question 8.
Complete the following reactions,

1. Cr₂O₃ + A1 \(\underrightarrow { \Delta } \) ?
2. B₂O₂ + Na \(\underrightarrow { \Delta } \) ?
3. ThO₂ + Ca \(\underrightarrow { \Delta } \) ?
4. Mn₃O₄ + C \(\underrightarrow { \Delta } \) ?
5. Ag₂O + H₂ \(\underrightarrow { \Delta } \) ?

Answer:

1. Cr₂O₃ + 2Al \(\underrightarrow { \Delta } \) 2Cr +Al₂O₃
2. B₂O₃ + 6Na \(\underrightarrow { \Delta } \) 2B +3Na₂O
3. ThO₂ + 2Ca \(\underrightarrow { 1250K }\) Th + 2CaO
4. Mn₃O₄ + 4C \(\underrightarrow { \Delta } \) 3Mn + 4CO
5. Ag₂O + H₂ \(\underrightarrow { \Delta } \) 2Ag + H₂O

Common Errors and its Rectifications

Common Errors:

1. In balancing inorganic reactions, they may struggle.
2. Ores formula they may forget.
3. In metallurgy, the steps are very important.

Rectifications:

1. Students are advised to balance first the negative charged atoms like oxygen, chlorine in the equation.
2. Simple way is to remember one oxide ore (or) one sulphide ore from which the metal is extracted.
3. Students should practice to write steps headings first along with explanation.

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Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

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Samacheer Kalvi 12th Physics Electrostatics Textual Evaluation Solved

Samacheer Kalvi 12th Physics Electrostatics Multiple Choice Questions

Physics Class 12 Samacheer Kalvi Question 1.
Two identical point charges of magnitude -q are fixed as shown in the figure below. A third charge +q is placed midway between the two charges at the point P. Suppose this charge +q is displaced a small distance from the point P in the directions indicated by the arrows, in which direction(s) will +q be stable with respect to the displacement?

(a) A₁ and A₂
(b) B₁ and B₂
(c) both directions
(d) No stable
Answer:
(b) B₁ and B₂

Samacheer Kalvi 12th Physics Chapter 1 Question 2.
Which charge configuration produces a uniform electric field?
(a) point charge
(b) infinite uniform line charge
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer:
(c) uniformly charged infinite plane

Samacheer Kalvi Guru 12th Physics Question 3.
What is the ratio of the charges \(\left|\frac{q_{1}}{q_{2}}\right|\) for the following electric field line pattern?

(a) \(\frac { 1 }{ 5 }\)
(b) \(\frac { 25 }{ 11 }\)
(c) 5
(d) \(\frac { 12 }{ 25}\)
Answer:
(d) \(\frac { 12 }{ 25}\)

Samacheer Kalvi 12th Physics Question 4.
An electric dipole is placed at an alignment angle of 30° with an electric field of 2 x 10⁵ N C^-1. It experiences a torque equal to 8 N m. The charge on the dipole if the dipole length is 1 cm is-
(a) 4 mC
(b) 8 mC
(c) 5 mC
(d) 1 mC
Answer:
(b) 8 mC

12 Physics Samacheer Kalvi Question 5.
Four Gaussian surfaces are given below with charges inside each Gaussian surface. Rank the electric flux through each Gaussian surface in increasing order-

(a) D < C < B < A
(b) A < B = C < D
(c) C < A = B < D
(d)D > C > B > A
Answer:
(a) D < C < B < A

Class 12 Physics Samacheer Kalvi Question 6.
The total electric flux for the following closed surface which is kept inside water-

(a) \(\frac { 80q }{{ ε }_{0}}\)
(b) \(\frac { q }{{ 40ε }_{0}}\)
(c) \(\frac { q }{{ 80ε }_{0}}\)
(d) \(\frac { q }{{ 40ε }_{0}}\)
Answer:
(b) \(\frac { q }{{ 40ε }_{0}}\)

12th Physics 1st Chapter Question 7.
Two identical conducting balls having positive charges q₁ and q₂ are separated by a center to center distance r. If they are made to touch each other and then separated to the same distance, the force between them will be- (NSEP 04-05)
(a) less than before
(b) same as before
(c) more than before
(d) zero
Answer:
(c) more than before

12th Physics Samacheer Kalvi Question 8.
Rank the electrostatic potential energies for the given system of charges in increasing order

(a) 1 = 4 < 2 < 3
(b) 2 = 4 < 3 < 1
(c) 2 = 3 < 1 < 4
(d) 3 < 1 < 2 < 4
Answer:
(a) 1 = 4 < 2 < 3

Samacheerkalvi.Guru 12th Physics Question 9.
An electric field \(\vec { E } \) = 10x\(\hat{i} \) exists in a certain region of space. Then the potential difference V = V₀ – V_A, Where V₀ is the potential at the origin and V_A is the potential at x = 2 m is-
(a) 10 J
(b) -20 J
(c) + 20 J
(d) – 10 J
Answer:
(a) 10 J

Electrostatics Notes Class 12 State Board Question 10.
A thin conducting spherical shell of radius R has a charge Q which is uniformly distributed on its surface. The correct plot for electrostatic potential due to this spherical shell is-

Answer:

12th Physics Lesson 1-Electrostatics Question 11.
Two points A and B are maintained at a potential of 7 V and -4 V respectively. The work done in moving 50 electrons from A to B is-
(a) 8.80 x 10^-17 J
(b) -8.80 x 10^-17 J
(c) 4.40 x 10^-17 J
(d) 5.80 x 10^-17 J
Answer:
(a) 8.80 x 10^-17 J

Electrostatics Class 12 Questions And Answers Pdf Question 12.
If voltage applied on a capacitor is increased from V to 2V, choose the correct conclusion.
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains same, Q doubled
(d) Both Q and C remain same
Answer:
(c) C remains same, Q doubled

Electrostatic Problems And Solutions Pdf Question 13.
A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
(a) Capacitance
(b) Charge
(c) Voltage
(d) Energy density
Answer:
(d) Energy density

Samacheer Kalvi Physics Question 14.
Three capacitors are connected in triangle as shown in the figure. The equivalent capacitance between the points A and C is

(a) 1 μF
(b) 2 μF
(c) 3 μF
(d) \(\frac { 1 }{ 4 }\) μF
Answer:
(b) 2 μF

Physics Solution Class 12 Samacheer Kalvi Question 15.
Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 x 10^-2 C and 5 x 10^-2 C respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is (AIIPMT 2012)
(a) 3 x 10^-2 C
(b) 4 x 10^-2 C
(c) 1 x 10^-2 C
(d) 2 x 10^-2 C
Answer:
(a) 3 x 10^-2 C

Samacheer Kalvi 12th Physics Electrostatics Short Answer Questions

Samacheer Kalvi Class 12 Physics Solutions Question 1.
What is meant by quantisation of charges?
Answer:
The charge q on any object is equal to an integral multiple of this fundamental unit of charge e.
q = ne
Here n is any integer (0, ±1, ±2, ±3, ±4). This is called quantisation of electric charge.

Samacheer Kalvi Guru Physics Question 2.
Write down Coulomb’s law in vector form and mention what each term represents.
Answer:
The force on a charge q₁ exerted by a point charge q₁ is given by
\(\vec { F } \)₁₂ = \(\frac { 1 }{{ 4πε }{0}}\) \(\frac {{ q }_{1}{ q }_{2}}{{ r }^{2}}\) \(\hat{r} \)₂₁
Here \(\hat{r} \)₂₁ is the unit vector from charge q₁ to q₁.
But \(\hat{r} \)₂₁ = –\(\hat{r} \)₁₂,

Therefore, the electrostatic force obeys Newton’s third law.

Samacheer Kalvi 12th Physics Solutions Question 3.
What are the differences between Coulomb force and gravitational force?
Answer:

• The gravitational force between two masses is always attractive but Coulomb force between two charges can be attractive or repulsive, depending on the nature of charges.
• The value of the gravitational constant G = 6.626 x 10^-11 N m² kg^-2. The value of the constant k in Coulomb law is k = 9 x 10⁹ N m² C².
• The gravitational force between two masses is independent of the medium. The electrostatic force between the two charges depends on nature of the medium in which the two charges are kept at rest.
• The gravitational force between two point masses is the same whether two masses are at rest or in motion. If the charges are in motion, yet another force (Lorentz force) comes into play in addition to coulomb force.

Samacheer Kalvi.Guru 12th Physics Question 4.
Write a short note on superposition principle.
Answer:
According to this superposition principle, the total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.
\({ \vec { F } }_{ 1 }^{ tot }\) = \(\vec { F } \)₁₂ + \(\vec { F } \)₁₃ + \(\vec { F } \)₁₄ + \(\vec { F } \)_1n

Samacheer Kalvi 12 Physics Question 5.
Define ‘Electric field’.
Answer:
The electric field at the point P at a distance r from the point charge q is the force experienced
by a unit charge and is given by
\(\vec { E } \) = \(\frac { \vec { F } }{ { q }_{ 0 } } \)
The electric field is a vector quantity and its SI unit is Newton per Coulomb (NC^-1).

Question 6.
What is mean by ‘Electric field lines’?
Answer:
Electric field vectors are visualized by the concept of electric field lines. They form a set of continuous lines which are the visual representation of the electric field in some region of space.

Question 7.
The electric field lines never intersect. Justify.
Answer:
As a consequence, if some charge is placed in the intersection point, then it has to move in two different directions at the same time, which is physically impossible. Hence, electric field lines do not intersect.

Question 8.
Define ‘Electric dipole’
Answer:
Two equal and opposite charges separated by a small distance constitute an electric dipole.

Question 9.
What is the general definition of electric dipole moment?
Answer:
The electric dipole moment vector lies along the line joining two charges and is directed from -q to + q. The SI unit of dipole moment is coulomb meter (Cm).
\(\vec { P } \) = qa\(\hat{i} \) -qa(\(\hat{-i} \)) = 2 qa\(\hat{i} \)

Question 10.
Define “electrostatic potential”.
Answer:
The electric potential at a point P is equal to the work done by an external force to bring a unit positive charge with constant velocity from infinity to the point P in the region of the external
electric field \(\vec { E } \).

Question 11.
What is an equipotential surface?
Answer:
An equipotential surface is a surface on which all the points are at the same potential.

Question 12.
What are the properties of an equipotential surface?
Answer:
Properties of equipotential surfaces
(i) The work done to move a charge q between any two points A and B,
W = q (V_B – V_A). If the points A and B lie on the same equipotential surface, work done is zero because V_A = V_B.

(ii) The electric field is normal to an equipotential surface. If it is not normal, then there is a component of the field parallel to the surface. Then work must be done to move a charge between two points on the same surface. This is a contradiction. Therefore the electric field must always be normal to equipotential surface.

Question 13.
Give the relation between electric field and electric potential.
Answer:
Consider a positive charge q kept fixed at the origin. To move a unit positive charge by a small distance dx in the electric field E, the work done is given by dW = -E dx. The minus sign implies that work is done against the electric field. This work done is equal to electric potential difference. Therefore,
dW = dV.
(or) dV = -Edx
Hence E = \(\frac { dV }{ dx }\)
The electric field is the negative gradient of the electric potential.

Question 14.
Define electrostatic potential energy?
Answer:
The potential energy of a system of point charges may be defined as the amount of work done in assembling the charges at their locations by bringing them in from infinity.

Question 15.
Define ‘electric flux’.
Answer:
The number of electric field lines crossing a given area kept normal to the electric field lines is called electric flux. Its unit is N m² C^-1. Electric flux is a scalar quantity.

Question 16.
What is meant by electrostatic energy density?
Answer:
The energy stored per unit volume of space is defined as energy density u_E = \(\frac { U }{ Volume }\)
From equation u_E = \(\frac { 1 }{ 2 }\) \(\frac{\left(\varepsilon_{0} A\right)}{d}\) (Ed)² = \(\frac { 1 }{ 2 }\) ε₀ (Ad) E² or u_E = \(\frac { 1 }{ 2 }\) ε₀E²

Question 17.
Write a short note on ‘electrostatic shielding’.
Answer:
Consider a cavity inside the conductor. Whatever the charges at the surfaces and whatever the electrical disturbances outside, the electric field inside the cavity is zero. A sensitive electrical instrument which is to be protected from external electrical disturbance is kept inside this cavity. This is called electrostatic shielding.

Question 18.
What is Polarisation?
Answer:
Polarisation \(\vec { P } \) is defined as the total dipole moment per unit volume of the dielectric.
\(\vec { P } \) = X_e \(\vec { P } \)_ext

Question 19.
What is dielectric strength?
Answer:
The maximum electric field the dielectric can withstand before it breakdowns is called dielectric strength.

Question 20.
Define ‘capacitance’. Give its unit.
Answer:
The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between the conductors.
C = \(\frac { q }{ V }\) or Q ∝ V.
The SI unit of capacitance is coulomb per volt or farad (F).

Question 21.
What is corona discharge?
Answer:
The electric field near the edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge. This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

Samacheer Kalvi 12th Physics Electrostatics Long Answer Questions

Question 1.
Discuss the basic properties of electric charges.
Answer:
Basic properties of charges
(i) Electric charge:
Most objects in the universe are made up of atoms, which in turn are made up of protons, neutrons and electrons. These particles have mass, an inherent property of particles. Similarly, the electric charge is another intrinsic and fundamental property of particles. The SI unit of charge is coulomb.

(ii) Conservation of charges:
Benjamin Franklin argued that when one object is rubbed with another object, charges get transferred from one to the other. Before rubbing, both objects are electrically neutral and rubbing simply transfers the charges from one object to the other. (For example, when a glass rod is rubbed against silk cloth, some negative charge are transferred from glass to silk. As a result, the glass rod is positively charged and silk cloth becomes negatively charged).

From these observations, he concluded that charges are neither created or nor destroyed but can only be transferred from one object to other. This is called conservation of total charges and is one of the fundamental conservation laws in physics. It is stated more generally in the following way. The total electric charge in the universe is constant and charge can neither be created nor be destroyed. In any physical process, the net change in charge will always be zero.

(iii) Quantisation of charges:
The charge q on any object is equal to an integral multiple of this fundamental unit of charge e.
q = ne
Here n is any integer (0, ±1, ±2, ±3, ± ….). This is called quantisation of electric charge. Robert Millikan in his famous experiment found that the value of e = 1.6 x 10^-19C. The charge of an electron is -1.6 x 10^-19 C and the charge of the proton is +1.6 x 10^-19C. When a glass rod is rubbed with silk cloth, the number of charges transferred is usually very large, typically of the order of 10¹⁰. So the charge quantisation is not appreciable at the macroscopic level. Hence the charges are treated to be continuous (not discrete). But at the microscopic level, quantisation of charge plays a vital role.

Question 2.
Explain in detail Coulomb’s law and its various aspects.
Answer:
Consider two point charges q₁ and q₂ at rest in vacuum, and separated by a distance of r. According to Coulomb, the force on the point charge q₂ exerted by another point charge q₁ is
\(\vec { F } \) ₂₁ = K\(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)₁₂,
where [/latex] \(\hat{r} \)₁₂ is the unit vector directed from charge q₁ to charge q₂ and k is the proportionality constant.

Important aspects of Coulomb’s law:
(i) Coulomb’s law states that the electrostatic force is directly proportional to the product of the magnitude of the two point charges and is inversely proportional to the square of the distance between the two point charges.

(ii) The force on the charge q₂exerted by the charge q₁ always lies along the line joining the two charges. \(\hat{r} \)₂₁is the unit vector pointing from charge q₁ to q₂ Likewise, the force on the charge q₁ exerted by q₂ is along – (i.e., in the direction opposite to \(\hat{r} \)₂₁).

(iii) In SI units, k = \(\frac { 1 }{{ 4πε }_{0}}\) and its value is 9 x 10⁹ Nm²C^-2. Here e0 is the permittivity of free space or vacuum and the value of ε₀ = \(\frac { 1 }{{ 4πε }_{0}}\) = 8.85 x 10^-12 C² N^-1 m^-2

(iv) The magnitude of the electrostatic force between two charges each of one coulomb and separated by a distance of 1 m is calculated as follows:
[F] = \(\frac{9 \times 10^{9} \times 1 \times 1}{1^{2}}\) = 9 x 10⁹N. This is a huge quantity, almost equivalent to the weight of one million ton. We never come across 1 coulomb of charge in practice. Most of the electrical phenomena in day-to-day life involve electrical charges of the order of pC (micro coulomb) or nC (nano coulomb).

(v) In SI units, Coulomb’s law in vacuum takes the form \(\vec { F } \) ₂₁ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)₁₂. sin Since ε > ε₀, the force between two point charges in a medium other than vacuum is always less than that in vacuum. We define the relative permittivity for a given medium as ε = \(\frac { ε }{{ ε }_{0}}\) .For vacuum or air, ε_r = 1 and for all other media ε_r > 1

(vi) Coulomb’s law has same structure as Newton’s law of gravitation. Both are inversely proportional to the square of the distance between the particles. The electrostatic force is directly proportional to the product of the magnitude of two point charges and gravitational force is directly proportional to the product of two masses.

(vii) The force on a charge q₁ exerted by a point charge q₂ is given by \(\vec { F } \)₁₂ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)₁₂ Here \(\hat{r} \)₂₁ is sthe unit vector from charge q₂ to q₁.

Therefore, the electrostatic force obeys Newton’s third law.

(viii) The expression for Coulomb force is true only for point charges. But the point charge is an ideal concept. However we can apply Coulomb’s law for two charged objects whose sizes are very much smaller than the distance between them. In fact, Coulomb discovered his law by considering the charged spheres in the torsion balance as point charges. The distance between the two charged spheres is much greater than the radii of the spheres.

Question 3.
Define ‘Electric field’ and discuss its various aspects.
Answer:
The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by

Here \(\hat{r} \) is the unit vector pointing from q to the point of interest P. The electric field is a vector quantity and its SI unit is Newton per Coulomb (NC^-1).

Important aspects of Electric field:
(i) If the charge q is positive then the electric field points away from the source charge and if q is negative, the electric field points towards the source charge q.

(ii) If the electric field at a point P is \(\vec { E } \), then the force experienced by the test charge qo placed at the point P is \(\vec { F } \) = q₀ \(\vec { E } \). This is Coulomb’s law in terms of electric field. This is shown in the below Figure.

(iii) The equation implies that the electric field is independent of the test charge q0 and it depends only on the source charge q.

(iv) Since the electric field is a vector quantity, at every point in space, this field has unique direction and magnitude as shown in Figures (a) and (b). From equation, we can infer that as distance increases, the electric field decreases in magnitude. Note that in Figures (a) and (b) the length of the electric field vector is shown for three different points. The strength or magnitude of the electric field at point P is stronger than at the point Q and R because the point P is closer to the source charge.

(v) In the definition of electric field, it is assumed that the test charge q₀ is taken sufficiently small, so that bringing this test charge will not move the source charge. In other words, the test charge is made sufficiently small such that it will not modify the electric field of the source charge.

(vi) The expression is valid only for point charges. For continuous and finite size charge distributions, integration techniques must be used. However, this expression can be used as an approximation for a finite-sized charge if the test point is very far away from the finite sized source charge.

(vii) There are two kinds of the electric field: uniform (constant) electric field and non-uniform electric field. Uniform electric field will have the same direction and constant magnitude at all points in space. Non-uniform electric field will have different directions or different magnitudes or both at different points in space. The electric field created by a point charge is basically a non-uniform electric field. This non-uniformity arises, both in direction and magnitude, with the direction being radially outward (or inward) and the magnitude changes as distance increases.

Question 4.
How do we determine the electric field due to a continuous charge distribution? Explain. Electric field due to continuous charge distribution.
Answer:
The electric charge is quantized microscopically. The expressions of Coulomb’s Law, superposition principle force and electric field are applicable to only point charges. While dealing with the electric field due to a charged sphere or a charged wire etc., it is very difficult to look at individual charges in these charged bodies. Therefore, it is assumed that charge is distributed continuously on the charged bodies and the discrete nature of charges is not considered here. The electric field due to such continuous charge distributions is found by invoking the method of calculus.

Consider the following charged object of irregular shape. The entire charged object is divided into a large number of charge elements ∆q₁, ∆q₂, ∆q₃ ……..∆q_n,…… and each charge element Δq is taken as a point charge.
The electric field at a point P due to a charged object is approximately given by the sum of the fields at P due to all such charge elements.

Here ∆ qi is the i^th charge element, r_ip is the distance of the point P frome the i^th charge element, r_ip is the unit vector from ith charge element to the pont P.
However the equation is only an approximation. To incorporate the continuous distribution of charge, we take the limit ∆q → 0(= dq). In this limit, the summation in the equation becomes an integration and takes the following form

Here r is the distance of the point P from the infinitesimal charge dq and \(\hat{r} \) is the unit vector from dq to point P. Even though the electric field for a continuous charge distribution is difficult to evaluate, the force experienced by some test charge q in this electric field is still given by \(\vec { F } \) = q\(\vec { E } \).

(a) Line charge distribution: If the charge Q is uniformly distributed along the wire of length L, then linear charge density (charge per unit length) is λ = \(\frac { Q }{ L }\). Its unit is colomb per meter (Cm^-1). The charge present in the infinitestimal length dl is dq = λdl.

The electric field due to the line of total charge Q is given by

(b) Surface charge distribution: If the charge Q is uniformly distributed on a surface of area A, then surface charge density (charge per unit area) is σ = \(\frac { Q }{ A }\). Its unit is coulomb per square meter (C m^-2). The charge present in the infinitesimal area dA is dq = σdA. The electric field due to a of total charge Q is given by

(c) Volume charge distribution: If the charge Q is uniformly distributed in a volume V, then volume charge density (charge per unit volume) is given by ρ = \(\frac { Q }{ V }\). Its unit is coulomb per cubic meter (Cm^-3) The charge present in the infinitesimal volume element dV is dq = ρdV. The electric field due to a volume of total charge Q is given by

Question 5.
Calculate the electric field due to a dipole on its axial line and equatorial plane.
Case (I) :
Electric field due to an electric dipole at points on the axial line. Consider an electric dipole placed on the x-ax is as shown in figure. A point C is located at a distance of r from the midpoint O of the dipole along the axial line. Axial line

The electric field at a point C due to +q is

Since the electric dipole moment vector \(\vec { P } \) is from -q to +q and is directed along BC, the above equation is rewritten as

where \(\hat{p} \) is the electric dipole moment unit vector from -q to +q. The electric field at a point C due to -q is

Since +q is located closer to the point C than -q, \(\vec { E } \) _. \(\vec { E } \) + us stronger than \(\vec { E } \) _. Therefore, the length of the E + vector is drawn large than that of \(\vec { E } \) _vector.
The total electric field at point C is calculated using the superposition principle of the electric field.

Note that the total electric field is along \(\vec { E } \)₊, since +q is closer to C than -q.

The direction of \(\vec { E } \)_tot is shown in Figure
If the point C is very far away from the dipole then (r >> a). Under this limit the term(r² – a²)² ≈ r⁴ Substituting this into equation, we get

If the point C is chosen on the left side of the dipole, the total electric field is still in the

Case (II) :
Electric field due to an electric dipole at a point on the equatorial plane
Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure. Since the point C is equi-distant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of E+ is along BC and the direction of E is along CA. E+ and E_ are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it.

The perpendicular components \(\left|\vec{E}_{+}\right|\) sin θ and \(\left|\vec{E}_{-}\right|\) sin θ are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the paralle component of \(\vec { E } \)₊ and \(\vec { E } \)_– and its direction is along \(\hat{-p} \).


The magnitudes \(\vec { E } \)₊ and \(\vec { E } \)_– are the same and are given by

By substituting equation (1) into equation (2), we get

At very large distances (r >> a), the equation becomes
\(\vec { E } \)_tot \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{ r }^{3}}\) (r >>) …… (4)

Question 6.
Derive an expression for the torque experienced by a dipole due to a uniform electric field.
Answer:
Torque experienced by an electric dipole in the uniform electric field:
Consider an electric dipole of dipole moment \(\vec { p } \) placed in a uniform electric field E whose field lines are equally spaced and point in the same direction. The charge +q will experience a force q\(\vec { E } \) in the direction of the field and charge -q will experience a force -q\(\vec { E } \) in a direction opposite to the field.

Since the external field \(\vec { E } \) is uniform, the total force acting on the dipole is zero. These two forces acting at different points will constitute a couple and the dipole experience a torque. This torque tends to rotate the dipole. (Note that electric field lines of a uniform field are equally spaced and point in the same direction). The total torque on the dipole about the point O
\(\vec { τ } \) = \(\overrightarrow{\mathrm{OA}}\) × (-q\(\vec { E } \)) + \(\overrightarrow{\mathrm{OB}}\) × q\(\vec { E } \)
Using right-hand corkscrew rule, it is found that total torque is perpendicular to the plane of the paper and is directed into it.

The magnitude of the total torque
\(\vec { τ } \) = \(|\overrightarrow{\mathrm{OA}}|\)(-q\(\vec { E } \)) sin θ + \(|\overrightarrow{\mathrm{OB}}|\) \(|q \overrightarrow{\mathrm{E}}|\) sin θ
where θ is the angle made by \(\vec { P } \) with \(\vec { E } \). Since p = 2aq, the torque is written in terms of the vector product as
\(\vec { τ } \) = \(\vec { p } \) x \(\vec { E } \)
The magnitude of this torque is τ = pE sin θ and is maximum Torque on dipole
when θ =90°.
This torque tends to rotate the dipole and align it with the electric field \(\vec { E } \). Once \(\vec { E } \) is aligned with \(\vec { E } \), the total torque on the dipole becomes zero.

Question 7.
Derive an expression for electrostatic potential due to a point charge.
Answer:
Electric potential due to a point charge:
Consider a positive charge q kept fixed at the origin. Let P be a point at distance r from the charge q.
The electric potential at the point P is


Electric field due to positive point charge q is

The infinitesimal displacement vector, d\(\vec { r } \) = dr\(\hat{r} \) and using \(\hat{r} \) . \(\hat{r} \) = 1, we have

After the integration,

Hence the electric potential due to a point charge q at a distance r is
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{ r }\) …… (2)
Important points (If asked in exam)
(i) If the source charge q is positive, V > 0. If q is negative, then V is negative and equal to
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{ r }\)
(ii) The description of motion of objects using the concept of potential or potential energy is simpler than that using the concept of field.
(i) From expression (2), it is clear that the potential due to positive charge decreases as the distance increases, but for a negative charge the potential increases as the distance is increased. At infinity (r = ∞) electrostatic potential is zero (V = 0).
(iv) The electric potential at a point P due to a collection of charges q₁,q₂,q₃… q_n is equal to sum of the electric potentials due to individual charges.

Where r₁, r₂,r₃,…..r_n are the distances of q₁,q₂,q₃… q_n
respectively from P

Question 8.
Derive an expression for electrostatic potential due to an electric dipole.
Answer:
Electrostatic potential at a point due to an electric dipole:
Consider two equal and opposite charges separated by a small distance 2a. The point P is located at a distance r from the midpoint of the dipole. Let 0 be the angle between the line OP and dipole axis AB.
Let r₁ be the distance of point P from +q and r₁ be the distance of point P from -q.
Potential at P due to charge +q = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }_{1}}\)
Potential at P due to charge -q = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }_{2}}\)
Total Potential at the point P,
V = \(\frac { 1 }{{ 4πε }_{0}}\)q \(\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)\) ….. (1)
Suppose if the point P is far away from the dipole, such that r >> a, then equation can be expressed in terms of r. By the cosine law for triangle BOP,

\(r_{1}^{2}\) = r² + a² – 2ra cos θ = r² \(\left(1+\frac{a^{2}}{r^{2}}-\frac{2 a}{r} \cos \theta\right)\)
Since the point P is very far from dipole, then r >> a. As a result the term \(\frac {{ a }^{ 2 }}{{ r }^{ 2 }}\) is very small and can be neglected. Therefore

since \(\frac { a }{ r }\) << 1, we can use binominal theorem and retain the terms up to first order
\(\frac { 1 }{{ r_{1}} }\) = \(\left(1+\frac{a}{r} \cos \theta\right)\) ……. (2)
Similarly applying the cosine law for triangle AOP,
\(r_{2}^{2}\) = r² + a² – 2ra cos (180 – θ)
Since cos (180 – θ) = cos θ we get
\(r_{2}^{2}\) = r² + a² + 2ra cos θ
Neglecting the term \(\frac {{ a }^{ 2 }}{{ r }^{ 2 }}\) (because r >> a)
\(r_{2}^{2}\) = r² \(\left(1+\frac{2 a \cos \theta}{r}\right)\) (or) r₂ = r \(\left(1+\frac{2 a \cos \theta}{r}\right)^{\frac{1}{2}}\)
Using Binomial theorem, we get
\(\frac { 1 }{{ r_{2}} }\) = \(\frac { 1 }{ r }\) \(\left(1-a \frac{\cos \theta}{r}\right)\)
Substituting equations (3) and (2) in equation (1)

But the electric dipole moment p = 2qa and we get,
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\left(\frac{p \cos \theta}{r^{2}}\right)\)
Now we can write p cos θ = \(\vec { P } \), \(\hat{r} \) where \(\hat{r} \) is the unit vector from the point O to point P. Hence the electric potential at a point P due to an electric dipole is given by
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{\vec{p} \cdot \hat{r}}{r^{2}}\) (r >> a) ….. (4)
Equation (4) is valid for distances very large compared to the size of the dipole. But for a point dipole, the equation (4) is valid for any distance.
Special cases:

Case (I):
If the point P lies on the axial line of the dipole on the side of +q, then θ = 0. Then the electric potential becomes
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{r}^{ 2 }}\)

Case (II):
If the point P lies on the axial line of the dipole on the side of -q, then θ = 180°, then
V = – \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{r}^{ 2 }}\)

Case (III):
If the point P lies on the equatorial line of the dipole, then θ = 90°. Hence, V = 0.

Question 9.
Obtain an expression for potential energy due to a collection of three point charges which are separated by finite distances.
Answer:
Electrostatic potential energy for collection of point charges:
The electric potential at a point at a distance r from point charge ql is given by
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 1 }}{r}\) …… (1)
This potential V is the work done to bring a unit positive charge from infinity to the point. Now if the charge q₂ is brought from infinity to that point at a distance r from qp the work done is the product of q₂ and the electric potential at that point. Thus we have W = q₂V …… (2)
This work done is stored as the electrostatic potential energy U of a system of charges q₁ and q₂ separated by a distance r. Thus we have
U = q₂ V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r}\) …… (3)
The electrostatic potential energy depends only on the distance between the two point charges. In fact, the expression (3) is derived by assuming that q₁ is fixed and q₂ is brought from infinity. The equation (3) holds true when q₂ is fixed and q₁ is brought from infinity or both q₂and q₂ are simultaneously brought from infinity to a distance r between them.
Three charges are arranged in the following configuration as shown in Figure.

To calculate the total electrostatic potential energy, we use the following procedure. We bring all the charges one by one and arrange them according to the configuration.
(i) Bringing a charge q₁ from infinity to the point A requires no work, because there are no other charges already present in the vicinity of charge q₁

(ii) To bring the second charge q₂ to the point B, work must be done against the electric field created by the charge q₁ So the work done on the charge q₁ is W = q₂V_1B. Here V_1B is the electrostatic potential due to the charge q₁ at point B.
U = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{{r}_{12}}\) ….. (4)
Note that the expression is same when q₂ is brought first and then q₁ later.

(iii) Similarly to bring the charge q3 to the point C, work has to be done against the total electric field due to both charges q₁ and q₂. So the work done to bring the charge q₃ is = q₃ (V_1C + V_2C). Here V_1C is the electrostatic potential due to charge q₁ at point C and V_2C is the electrostatic potential due to charge q₂ at point C. The electrostatic potential is

(iv) Adding equations (4) and (5), the total electrostatic potential energy for the system of three charges q₁,q₂ and q₃ is

Note that this stored potential energy U is equal to the total external work done to assemble the three charges at the given locations. The expression (6) is same if the charges are brought to their positions in any other order. Since the Coulomb force is a conservative force, the electrostatic potential energy is independent of the manner in which the configuration of charges is arrived at.

Question 10.
Derive an expression for electrostatic potential energy of the dipole in a uniform electric field.
Answer:
Electrostatic potential energy of a dipole in a uniform electric field:
Consider a dipole placed a torque when kept in an uniform electric field \(\vec { E } \). A dipole experiences a torque when kept in an uniform electric field \(\vec { E } \). This torque rotates the dipole to align it with the direction of the electric field. To rotate the dipole (at constant angular velocity) from its initial angle θ’ to another angle θ against the torque exerted by the electric field, an equal and opposite external torque must be applied on the dipole.

The work done by the external torque to rotate the dipole from angle θ’ to θ at constant angular velocity is

Since τ_ext is equal and opposite to τ_E = \(\vec { P } \) x \(\vec { E } \), we have
\(\left|\overrightarrow{\mathrm{r}}_{\mathrm{ext}}\right|\) = \(\left|\overrightarrow{\mathrm{r}}_{\mathrm{E}}\right|\)= \(|\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{E}}|\) …. (2)
Substituting equation (2) in equation (1) We get,

This work done is equal to the potential energy difference between the angular positions θ and θ’.
U(θ) – (Uθ’) = AU = -pE cos θ +PE cos θ’.
If the initial angle is = θ’ = 90° and is taken as reference point, then U(θ’) + pE cos θ’ = θ.
The potential energy stored in the system of dipole kept in the uniform electric field is given by El = -pE cos θ = –\(\vec { P } \) . \(\vec { E } \) ….. (3)
In addition to p and E, the potential energy also depends on the orientation θ of the electric dipole with respect to the external electric field.
The potential energy is maximum when the dipole is aligned anti-parallel (θ = π) to the external electric field and minimum when the dipole is aligned parallel (θ = 0) to the external electric field.

Question 11.
Obtain Gauss law from Coulomb’s law.
Answer:
Gauss law: Gauss’s law states that if a charge Q is enclosed by an arbitrary closed surface, then the total electric flux Φ_E through the closed surface is
Φ_E = \(\oint { \vec { E } } \) .d \(\vec { A } \) = \(\frac{\mathrm{Q}_{\mathrm{end}}}{\varepsilon_{0}}\)
A positive point charge Q is surrounded by an imaginary sphere of radius r as shown in figure. We can calculate the total electric flux through the closed surface of the sphere using the equation.

Φ_E = \(\oint { \vec { E } } \) .d \(\vec { A } \) = \(\oint { EdA } \) cos θ …… (1)
The electric field of the point charge is directed radially outward at all points on the surface of the sphere. Therefore, the direction of the area element d \(\vec { A } \) is along the electric field \(\vec { E } \) and θ = 0°.
Φ_E = \(\oint { EdA } \) since cos 0° = 1 ….. (2)
E is uniform on the surface of the sphere,
Φ_E = \(\oint { EdA } \) ….. (3)
Substituting for
\(\oint { dA } \) = 4π² and E = \(\frac { 1 }{{ 4πε }_{0}}\) Q in equation 3, we get
Φ_E = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }^{2}}\) × 4π² = 4π \(\frac { 1 }{{ 4πε }_{0}}\) = \(\frac { q }{{ ε }_{0}}\) ……. (4)
The equation (4) is called as Gauss’s law. The remarkable point about this result is that the equation (4) is equally true for any arbitrary shaped surface which encloses the charge Q.

Question 12.
Obtain the expression for electric field due to an infinitely long charged wire.
Answer:
Electric field due to an infinitely long charged wire:
Consider an infinitely long straight wire having uniform linear charge density λ. Let P be a point located at a perpendicular distance r from the wire. The electric field at the point P can be found using Gauss law. We choose two small charge elements A₁ and A₁ on the wire which are at equal distances from the point P.

The resultant electric field due to these two charge elements points radially away from the charged wire and the magnitude of electric field is same at all points on the circle of radius r. From this property, we can infer that the charged wire possesses a cylindrical symmetry.

Let us choose a cylindrical Gaussian surface of radius r and length L. The total electric flux in this closed surface is

It is seen that for the curved surface, \(\vec { E } \) is parallel to \(\vec { A } \) and \(\vec { E } \).d \(\vec { A } \) = EdA. For the top and bottom surface, \(\vec { E } \) is perpendicular to \(\vec { A } \) and \(\vec { E } \).d\(\vec { A } \) = 0
Substituting these values in the equation (2) and applying Gauss law


Since the magnitude of the electric field for the entire curved surface is constant, E is taken out of the integration and Q_encl is given by Q_encl = λL.

Here,

dA = total area of the curved surface = 2πrL. Substituting this in
equation (4), We get

The electric field due to the infinite charged wire depends on \(\frac { 1 }{ r }\) rather than \(\frac { 1 }{{r}^{ 2 }}\) for a point charge.
Equation (6) indicates that the electric field is always along the perpendicular direction (\(\hat{r} \) ) to wire. In fact, if λ > 0 then E points perpendicular outward (\(\hat{r} \) ) from the wire and if λ < 0, then E points perpendicular inward (- \(\hat{r} \) ).

Question 13.
Obtain the expression for electric field due to an charged infinite plane sheet.
Answer:
Electric field due to charged infinite plane sheet: Consider an infinite plane sheet of charges with uniform surface charge density o. Let P be a point at a distance of r from the sheet. Since the plane is infinitely large, the electric field should be same at all points equidistant from the plane and radially directed at all points. A cylindrical shaped Gaussian surface of length 2r and area A of the flat surfaces is chosen such that the infinite plane sheet passes perpendicularly through the middle part of the Gaussian surface.

Applying Gauss law for this cylindrical surface,

The electric field is perpendicular to the are a element at all points on the curved surface and is parallel to the surface areas at P and P’. Then,

Since the magnitude of the electric field at these two equal surfaces is uniform, E is taken out of the integration and Q_encl is given by Q_encl = σA, we get

The total area of surface either at P or P’

Hence 2EA = \(\frac { σA }{{ ε }_{0}}\) or E = \(\frac { σ }{{ 2ε }_{0}}\) …… (3)
In vector from, E = \(\frac { σ }{{ 2ε }_{0}}\) \(\hat{n} \) ….. (4)
Hence \(\hat{n} \) is the outward unit vector normal to the plane. Note that the electric field due to an infinite plane sheet of charge depends on the surface charge density and is independent of the distance r.

The electric field will be the same at any point farther away from the charged plane. Equation (4) implies that if o > 0 the electric field at any point P is outward perpendicular n to the plane and if σ < 0 the electric field points inward perpendicularly (\(\hat{n} \) ) to the plane. For a finite charged plane sheet, equation (4) is approximately true only in the middle region of the plane and at points far away from both ends.

Question 14.
Obtain the expression for electric field due to an uniformly charged spherical shell.
Answer:
Electric field due to a uniformly charged spherical shell:
Consider a uniformly charged spherical shell of radius R and total charge Q. The electric field at points outside and inside the sphere is found using Gauss law.

Case (a):
At a point outside the shell (r > R): Let us choose a point P outside the shell at a distance r from the center as shown in figure (a). The charge is uniformly distributed on the surface of the sphere (spherical symmetry). Hence the electric field must point radially outward if Q > 0 and point radially inward if Q < 0. So we choose a spherical Gaussian surface of radius r and the total charge enclosed by this Gaussian surface is Q. Applying Gauss law,
\(\oint { \vec { E } } .d\vec { A } \) = \(\frac { Q }{{ ε }_{0}}\) …….(1)
The electric field \(\vec { E } \) and d\(\vec { A } \) point in the same direction (outward normal) at all the points on the Gaussian surface. The magnitude of \(\vec { E } \) is also the same at all points due to the spherical symmetry of the charge distribution.

But

dA = total area of Gaussian surface = 4πr². Substituting this value in equation (2).

The electric field is radially outward if Q > 0 and radially inward if Q < 0. From equation (3), we infer that the electric field at a point outside the shell will be same as if the entire charge Q is concentrated at the center of the spherical shell. (A similar result is observed in gravitation, for gravitational force due to a spherical shell with mass M)

Case (b):
At a point on the surface of the spherical shell (r = R): The electrical field at points on the spherical shell (r = R) is given by
\(\vec { E } \) = \(\frac{\mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\) \(\hat{r} \) …… (4)

Case (c):
At a point inside the spherical shell (r < R): Consider a point P inside the shell at a distance r from the center. A Gaussian sphere of radius r is constructed as shown in the figure (b). Applying Gauss law

Since Gaussian surface encloses no charge, So Q = 0. The equation (5) becomes E = 0 (r < R) …(6)
The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.

Question 15.
Discuss the various properties of conductors in electrostatic equilibrium.
Answer:
Properties of conductors in electrostatic equilibrium:
(i) The electric field is zero everywhere inside the conductor. This is true regardless of whether the conductor is solid or hollow. This is an experimental fact. Suppose the electric field is not zero inside the metal, then there will be a force on the mobile charge carriers due to this electric field.

As a result, there will be a net motion of the mobile charges, which contradicts the conductors being in electrostatic equilibrium. Thus the electric field is zero everywhere inside – the conductor. We can also understand this fact by applying an external uniform electric field on the conductor.

Before applying the external electric field, the free electrons in the conductor are uniformly distributed in the conductor. When an electric field is applied, the free electrons accelerate to the left causing the left plate to be negatively charged and the right plate to be positively charged.

Due to this realignment of free electrons, there will be an internal electric field created inside the conductor which increases until it nullifies the external electric field. Once the external electric field is nullified the conductor is said to be in electrostatic equilibrium. The time taken by a conductor to reach electrostatic equilibrium is in the order of 10^-6s, which can be taken as almost instantaneous.

(ii) There is no net charge inside the conductors. The charges must reside only on the surface of the conductors. We can prove this property using Gauss law. Consider an arbitarily shaped conductor. A Gaussian surface is drawn inside the conductor such that it is very close to the surface of the conductor.

Since the electric field is zero everywhere inside the conductor, the net electric flux is also zero over this Gaussian surface. From Gauss’s law, this implies that there is no net charge inside the conductor. Even if some charge is introduced inside the conductor, it immediately reaches the surface of the conductor.

(iii) The electric field outside the conductor is perpendicular to the surface of the conductor and has a magnitude of \(\frac { σ }{{ ε }{0}}\) where a is the surface charge density at that point. If the electric field has components parallel to the surface of the conductor, then free electrons on the surface of the conductor would experience acceleration. This means that the conductor is not in equilibrium. Therefore at electrostatic equilibrium, the electric field must be perpendicular to the surface of the conductor.

We now prove that the electric field has magnitude \(\frac { σ }{{ ε }{0}}\) just outside the conductor’s surface. Consider a small cylindrical Gaussian surface. One half of this cylinder is embedded inside the conductor. Since electric field is normal to the surface of the conductor, the curved part of the cylinder has zero electric flux. Also inside the conductor, the electric field is zero. Hence the bottom flat part of the Gaussian surface has no electric flux. Therefore the top flat surface alone contributes to the electric flux. The electric field is parallel to the area vector and the total charge inside the surface is σA. By applying Gauss’s law,
EA = \(\frac { σA }{{ ε }{0}}\)
In vector from, \(\vec { E } \) = \(\frac { σ }{{ ε }{0}}\) \(\hat{n} \)
Here n represents the unit vector outward normal to the surface of the conductor. Suppose σ < 0, then electric field points inward perpendicular to the surface.

(iv) The electrostatic potential has the same value on the surface and inside of the conductor. We know that the conductor has no parallel electric component on the surface which means that charges can be moved on the surface without doing any work. This is possible only if the electrostatic potential is constant at all points on the surface and there is no potential difference between any two points on the surface. Since the electric field is zero inside the conductor, the potential is the same as the surface of the conductor. Thus at electrostatic equilibrium, the conductor is always at equipotential.

Question 16.
Explain the process of electrostatic induction.
Answer:
Whenever a charged rod is touched by another conductor, charges start to flow from charged rod to the conductor. This type of charging without
actual contact is called electrostatic induction:
(i) Consider an uncharged (neutral) conducting sphere at rest on an insulating stand. Suppose a negatively charged rod is brought near the conductor without touching it, as shown in figure (a). The negative charge of the rod repels the electrons in the conductor to the opposite side.

Various steps in electrostatic induction
As a result, positive charges are induced near the region of the charged rod while negative charges on the farther side. Before introducing the charged rod, the free electrons were distributed uniformly on the surface
of the conductor and the net charge is zero. Once the charged rod is brought near the conductor, the distribution is no longer uniform with more electrons located on the farther side of the rod and positive charges are located closer to the rod. But the total charge is zero.

(ii) Now the conducting sphere is connected to the ground through a conducting wire. This is called grounding. Since the ground can always receive any amount of electrons, grounding removes the electron from the conducting sphere. Note that positive charges will not flow to the ground because they are attracted by the negative charges of the rod (figure (b)).

(iii) When the grounding wire is removed from the conductor, the positive charges remain near the charged rod (figure (c)).

(iv) Now the charged rod is taken away from the conductor. As soon as the charged rod is removed, the positive charge gets distributed uniformly on the surface of the conductor (figure (d)). By this process, the neutral conducting sphere becomes positively charged.

Question 17.
Explain dielectrics in detail and how an electric field is induced inside a dielectric.
Answer:
Induced Electric field inside the dielectric:
When an external electric field is applied on a conductor, the charges are aligned in such a way that an internal electric field is created which cancels the external electric field. But in the case of a dielectric, which has no free electrons, the external electric field only realigns the charges so that an internal electric field is produced.

The magnitude of the internal electric field is smaller than that of external electric field. Therefore the net electric field inside the dielectric is not zero but is parallel to an external electric field with magnitude less than that of the external electric field. For example, let us consider a rectangular dielectric slab placed between two oppositely charged plates (capacitor) as shown in the figure.

The uniform electric field between the plates Induced electric field lines inside the dielectric acts as an external electric field \(\vec { E } \)_ext which polarizes the dielectric placed between plates. The positive charges are induced on one side surface and negative charges are induced on the other side of surface But inside the dielectric, the net charge is zero even in a small volume. So the dielectric in the external field is equivalent to two oppositely charged sheets with the surface charge densities +σb and -σb. These charges are called bound charges. They are not free to move like free electrons in conductors. This is shown in the figure.

(a) Balloon sticks to the wall
(b) Polarisation of wall due to the electric field created by the balloon
For example, the charged balloon after rubbing sticks onto a wall. The reason is that the negatively charged balloon is brought near the wall, it polarizes opposite charges on the surface of the wall, which attracts the balloon.

Question 18.
Obtain the expression for capacitance for a parallel plate capacitor.
Answer:
Capacitance of a parallel plate capacitor:
Consider a capacitor with two parallel plates each of cross-sectional area A and separated by a distance d. The electric field between two infinite parallel plates is uniform and is given by E = \(\frac { σ }{{ ε }{0}}\) where σ is the surface charge density on the plates σ = \(\frac { Q }{ A }\) .If the separation distance d is very much smaller than the size of the plate (d² << A), then the above result is used even for finite-sized
parallel plate capacitor.

Capacitance of a parallel plate capacitor
The electric field between the plates is
E = \(\frac { Q }{{ Aε }{0}}\) ….. (1)
Since the electric field is unifonn, the electric potential between the plates having separation d is given by
V = Ed = \(\frac { Qd }{{ Aε }{0}}\) ….. (2)
Therefore the capacitance of the capacitor is given by
C = \(\frac { Q }{ V }\) = \(\frac{\mathrm{Q}}{\left(\frac{\mathrm{Q} d}{\mathrm{A} \varepsilon_{0}}\right)}\) = \(\frac{\varepsilon_{0} \mathrm{A}}{d}\) ….. (3)
From equation (3), it is evident that capacitance is directly
proportional to the area of cross section and is inversely proportional to the distance between the plates. This can be understood from the following.
(i) If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.

(ii) If the distance d between the two plates is reduced, the potential difference between the plates (V = Ed) decreases with E constant.

Question 19.
Obtain the expression for energy stored in the parallel plate capacitor.
Answer:
Energy stored in the capacitor:
Capacitor not only stores the charge but also it stores energy. When a battery is connected to the capacitor, electrons of total charge -Q are transferred from one plate to the other plate. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor. To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
dW = VdQ ….. (1)
Where V = \(\frac { Q }{ C }\)
The total work done to charge a capacitor is

This work done is stored as electrostatic potential energy (U_E) in the capacitor.
U_E = \(\frac {{ Q }^{2}}{ 2C }\) = \(\frac { 1 }{ 2 }\) CV² (∴ Q = CV) ….. (3)
where Q = CV is used. This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor. But where is this energy stored in the capacitor? To understand this question, the equation (3) is rewritten as follows using the results
C = \(\frac{\varepsilon_{0} \mathrm{A}}{d}\) and V = Ed
U_E = \(\frac { 1 }{ 2 }\) \(\left(\frac{\varepsilon_{0} \mathrm{A}}{d}\right)\) (Ed)² = \(\frac { 1 }{ 2 }\) ε₀(Ad)² …… (4)
where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined as energy density \(\overline { Volume } \). Frome equation (4) we get
u_E = \(\frac { 1 }{ 2 }\) ε₀E²
From equation (5), we infer that the energy is stored in the electric field existing between the plates of the capacitor. Once the capacitor is allowed to discharge, the energy is retrieved.

Question 20.
Explain in detail the effect of a dielectric placed in a parallel plate capacitor.
Answer:
(i) When the capacitor is disconnected from the battery:
Consider a capacitor with two parallel plates each of cross-sectional area A and are separated by a distance d. The capacitor is charged by a battery of voltage V₀ and the charge stored is Q₀. The capacitance of the capacitor without the dielectric is
C₀ = \(\frac {{ Q }_{0}}{{ V }_{0}}\) ….. (1)
The battery is then disconnected from the capacitor and the dielectric is inserted between the plates. The introduction of dielectric between the plates will decrease the electric field. Experimentally it is found that the modified electric field is given by

(a) Capacitor is charged with a battery
(b) Dielectric is inserted after the battery is disconnected
E = \(\frac {{ E }_{0}}{{ ε }_{r}}\) …… (2)
Here E₀ is the electric field inside the capacitors when there is no dielectric and ε_r is the relative permeability of the dielectric or simply known as the dielectric constant. Since ε_r > 1, the electric field E < E₀. As a result, the electrostatic potential difference between the plates (V = Ed) is also reduced. But at the same time, the charge Q₀ will remain constant once the battery is disconnected. Hence the new potential difference is
V = Ed = \(\frac {{ E }_{0}}{{ ε }_{r}}\)d = \(\frac {{ V }_{0}}{{ ε }_{r}}\) ….. (3)
We know that capacitance is inversely proportional to the potential difference. Therefore as V decreases, C increases. Thus new capacitance in the presence of a dielectric is
C = \(\frac {{ Q }_{0}}{ V }\) = ε_r \(\frac {{ Q }_{0}}{{ V }_{0}}\) = ε_r C₀ …… (4)
Since ε_r > 1, we have C > C₀. Thus insertion of the dielectric constant ε_r increases the capacitance. Using equation,
C = \(\frac { { \varepsilon }_{ 0 }A }{ d } \)
C = \(\frac{\varepsilon_{r} \varepsilon_{o} A}{d}\) = \(\frac { εA }{ d }\) …… (5)
where ε = ε_rε₀ is the permittivity of the dielectric medium. The energy stored in the capacitor before the insertion of a dielectric is given by U₀ = \(\frac { 1 }{ 2 }\) \(\frac{\mathrm{Q}_{0}^{2}}{\mathrm{C}_{0}}\) ….. (6)
After the dielectric is inserted, the charge Q₀ remains constant but the capacitance is increased. As a result, the stored energy is decreased.

Since ε_r> 1 we get U < U₀. There is a decrease in energy because, when the dielectric is inserted, the capacitor spends some energy in pulling the dielectric inside.

(ii) When the battery remains connected to the capacitor: Let us now consider what happens when the battery of voltage V₀ remains connected to the capacitor when the dielectric is inserted into the capacitor.
The potential difference V₀ across the plates remains constant. But it is found experimentally (first shown by Faraday) that when dielectric is inserted, the charge stored in the capacitor is increased by a factor ε_r.

(a) Capacitor is charged through a battery
(b) Dielectric is inserted when the battery is connected.
Q = ε_rQ₀ ….. (1)
Due to this increased charge, the capacitance is also increased. The new capacitance is
C = \(\frac {{ Q }_{0}}{ V }\) = ε_r \(\frac {{ Q }_{0}}{{ V }_{0}}\) = ε_r C₀ …… (2)
However the reason for the increase in capacitance in this case when the battery remains connected is different from the case when the battery is disconnected before introducing the dielectric.
Now, C₀ = \(\frac {{ ε }{_0}A}{ d }\) and, C = \(\frac { εA }{ d }\) …… (3)
U₀ = \(\frac { 1 }{ 2 }\) C₀ \({ V }_{ 0 }^{ 2 }\) ….. (4)
Note that here we have not used the expression
U₀ = \(\frac { 1 }{ 2 }\)\({{ V }_{ 0 }^{ 2 }}{{C}_{0}}\)
because here, both charge and capacitance are changed, whereas in equation 4, V₀ remains constant. After the dielectric is inserted, the capacitance is increased; hence the stored energy is also increased.
U = \(\frac { 1 }{ 2 }\) \({ CV }_{ 0 }^{ 2 }\) = \(\frac { 1 }{ 2 }\) ε_r \({ CV }_{ 0 }^{ 2 }\) = ε_r U₀
Since e_r > 1 we have U > U₀
It may be noted here that since voltage between the capacitor V₀ is constant, the electric field between the plates also remains constant.

Question 21.
Derive the expression for resultant capacitance, when capacitors are connected in series and in parallel.
Answer:
Capacitor in series and parallel:
(i) Capacitor in series:
Consider three capacitors of capacitance C₁, C₂ and C₃ connected in series with a battery of voltage V as shown in the figure (a).
As soon as the battery is connected to the capacitors in series, the electrons of charge -Q are transferred from negative terminal to the right plate of C₃which pushes the electrons of same amount -Q from left plate of C₃ to the right plate of C₂ due to electrostatic induction. Similarly, the left plate of C₂ pushes the charges of Q to the right plate of which induces the positive charge +Q on the left plate of C₁ At the same time, electrons of charge -Q are transferred from left plate of C₁ to positive terminal of the battery.

By these processes, each capacitor stores the same amount of charge Q. The capacitances of the capacitors are in general different, so that the voltage across each capacitor is also different and are denoted as V₁, V₂ and V₃ respectively.
The total voltage across each capacitor must be equal to the voltage of the a battery.
V = V₁ + V₂ + V₃ ….. (1)
Since Q = CV, we have V = \(\frac { Q }{{ C }_{1}}\) + \(\frac { Q }{{ C }_{2}}\) + \(\frac { Q }{{ C }_{3}}\)
Q = \(\left( \frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } \right) \) ….. (2)
If three capacitors in series are considered to form an equivalent single capacitor Cs shown in figure (b), then we have V = \(\frac { Q }{{ C }_{s}}\)
Substituting this expression into equation (2) we get
V = \(\frac { Q }{{ C }_{s}}\) = Q\(\left( \frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } \right) \)
\(\frac { 1 }{{ C }_{s}}\) = \(\frac { 1 }{{ C }_{1}}\) + \(\frac { 1 }{{ C }_{2}}\) + \(\frac { 1 }{{ C }_{3}}\) ….. (3)
Thus, the inverse of the equivalent capacitance C_s of three capacitors connected in series is equal to the sum of the inverses of each capacitance. This equivalent capacitance C_s is always less than the smallest individual capacitance in the series.

(ii) Capacitance in parallel:
Consider three capacitors of capacitance C₁,C₂ and C₃ connected in parallel with a battery of voltage V as shown in figure (a).

Since corresponding sides of the capacitors are connected to the same positive and negative terminals of the battery, the voltage across each capacitor is equal to the battery’s voltage. Since capacitance of the capacitors is different, the charge stored in each capacitor is not the same. Let the charge stored in the three capacitors be Q₁,Q₂, and Q₂ respectively. According to the law of conservation of total charge, the sum of these three charges is equal to the charge Q transferred by the battery,
Q = Q₁ + Q₂ + Q₃ ….. (1)
Now, since Q = CV, we have
Q = C₁V + C₂ V + C₃ V ….. (2)
If these three capacitors are considered to form a single capacitance C_P which stores the total charge Q as shown in the figure (b), then we can write Q = CPV. Substituting this in equation (2), we get
C_p V = C₁ V + C₂ V + C₃ V
C_p = C₁ + C₂ + C₃
Thus, the equivalent capacitance of capacitors connected in parallel is equal to the sum of the individual capacitance. The equivalent capacitance C_p in a parallel connection is always greater than the largest individual capacitance. In a parallel connection, it is equivalent as area of each capacitance adds to give more effective area such that total capacitance increases.

Question 22.
Explain in detail how charges are distributed in a conductor, and the principle behind the lightning conductor.
Answer:
Distribution of charges in a conductor: Consider two conducting spheres A and B of radii r₁ and r₂ respectively connected to each other by a thin conducting wire as shown in the figure. The distance between the spheres is much greater than the radii of either spheres.

If a charge Q is introduced into any one of the spheres, this charge Q is redistributed into both the spheres such that the electrostatic potential is same in both the spheres. They are now uniformly charged and attain electrostatic equilibrium. Let q₁ be the charge residing on the surface of sphere A and q₂ is the charge residing on the surface of sphere B such that Q = q₁ + q₂ The charges are distributed only on the surface and there is no net charge inside the conductor. The electrostatic potential at the surface of the sphere A is given by
V_A = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) …. (1)
The electrostatic potential at the surface of the sphere B is given by
V_B = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) ….. 2)
The surface of the conductor is an equipotential. Since the spheres are connected by the conducting wire, the surfaces of both the spheres together form an equipotential surface. This implies that
V_A = V_B or \(\frac {{ q }_{ 1 }}{{ r }_{ 1 }}\) = \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) ….. (3)
Let us take the charge density on the surface of sphere A is σ₁ and charge density on the surface of sphere B is σ₁. This implies that q₁ = \({ 4\pi r }_{ 1 }^{ 2 }\)σ₁ and q₁ = \({ 4\pi r }_{ 1 }^{ 2 }\)σ₂. Substituting these values into equation (3), we get
σ₁ r₁ = σ₂r₂ ….. (4)
from which we conclude that
σr = constant …. (5)
Thus the surface charge density o is inversely proportional to the radius of the sphere. For a smaller radius, the charge density will be larger and vice versa.

Lightning arrester or lightning conductor:
This is a device used to protect tall buildings from lightning strikes. It works on the principle of action at points or corona discharge. The device consists of a long thick copper rod passing from top of the building to the ground. The upper end of the rod has a sharp spike or a sharp needle.

The lower end of the rod is connected to the copper plate which is buried deep into the ground. When a negatively charged cloud is passing above the building, it induces a positive charge on the spike. Since the induced charge density on thin sharp spke is large, it results in a corona discharge.

This positive charge ionizes the surrounding air which in turn neutralizes the negative charge in the cloud. The negative charge pushed to the spikes passes through the copper rod and is safely diverted to the Earth. The lightning arrester does not stop the •lightning; rather it divers the lightning to the ground safety.

Question 23.
Explain in detail the construction and working of a Van de Graaff generator.
Answer:
Principle: Electrostatic induction and action at points.
Construction:
A large hollow spherical conductor is fixed on the insulating stand. A pulley B is mounted at the center of the hollow sphere and another pulley C is fixed at the bottom. A belt made up of insulating materials like silk or rubber runs over both pulleys. The pulley C is driven continuously by the electric motor.

Two comb shaped metallic conductors E and D are fixed near the pulleys. The comb D is maintained at a positive potential of 104 V by a power supply. The upper comb E is connected to the inner side of the hollow metal sphere.

Working:
Due to the high electric field near comb D, air between the belt and comb D gets ionized. The positive charges are pushed towards the belt and negative charges are attracted towards the comb D. The positive charges stick to the belt and move up. When the positive charges reach the comb E, a large amount of negative and positive charges are induced on either side of comb E due to electrostatic induction. As a result, the positive charges are pushed away from the comb E and they reach the outer surface of the sphere. Since the sphere is a conductor, the positive charges are distributed uniformly on the outer surface of the hollow sphere. At the same time, the negative charges nullify the positive charges in the belt due to corona discharge before it passes over the pulley.

When the belt descends, it has almost no net charge. At the bottom, it again gains a large positive charge. The belt goes up and delivers the positive charges to the outer surface of the sphere. This process continues until the outer surface produces the potential difference of the order of 10⁷ which is the limiting value. We cannot store charges beyond this limit since the extra charge starts leaking to the surroundings due to ionization of air. The leakage of charges can be reduced by enclosing the machine in a gas filled steel chamber at very high pressure. Uses: The high voltage produced in this Van de Graaff generator is used to accelerate positive ions (protons and deuterons) for nuclear disintegrations and other applications.

Samacheer Kalvi 12th Physics Electrostatics Numarical Problems

Question 1.
When two objects are rubbed with each other, approximately a charge of 50 nC can be produced in each object. Calculate the number of electrons that must be transferred to produce this charge.
Solution:
Charge produced in each object q = 50 nC
q = 50 x 10^-9 C
Charge of electron (e) = 1.6 x 10^-9 C
Number of electron transferred, n = \(\frac { q }{ e }\) = \(\frac {{ 50 × 10 }^{-9}}{{ 1.6 × 10 }^{-19}}\)
=31. 25 × 10^-9 × 10¹⁹
n = 31.25 x 10¹⁰ electrons
Ans. n = 31.25 x 10¹⁰ electrons

Question 2.
The total number of electrons in the human body is typically in the order of 10²⁸. Suppose, due to some reason, you and your friend lost 1% of this number of electrons. Calculate the electrostatic force between you and your friend separated at a distance of 1 m. Compare this with your weight. Assume mass of each person is 60 kg and use point charge approximation.
Solution:
Number of electrons in the human body = 10²⁸
Number of electrons in me and my friend after lost of 1% = 10²⁸ x 1%
= 10²⁸ x \(\frac { 1 }{ 100 }\)
n = 10²⁶ electrons
Separation distance d = 1 m
Charge of each person q = 10²⁶ x 1.6 x 10^-19
q = 1.6 x 10⁷ C
Electrostatic force, F = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r^{2}}\) = \(\frac{9 \times 10^{9} \times 1.6 \times 10^{7} \times 1.6 \times 10^{7}}{1^{2}}\)
F = 2.304 x 10²⁴N
Mass of the person, M = 60 kg
Acceleration due to gravity, g = 9.8 ms^-2
Weight (W) = mg
= 60 x 9.8
W = 588 N
Comparison: Electrostatic force is equal to 3.92 x 10²¹ times of weight of the person.

Question 3.
Five identical charges Q are placed equidistant on a semicircle as shown in the figure. Another point charge q is kept at the center of the circle of radius R. Calculate the electrostatic force experienced by the charge q.

Solution:
Force acting on q due to Q₁ and Q₅ are opposite direction, so cancel to each other.
Force acting on q due to Q₃ is F₃ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ qQ }_{3}}{{ R }^{2}}\)
Force acting on q due to Q₂ and Q₄

Resolving in two component method:
(i) Vertical Component:
Q₂ Sin θ and Q₄ Sinθ are equal and opposite direction, so they are cancel to each other.

(ii) Horizontal Component:
Q₂ Sin θ and Q₄ cos θ are equal and same direction, so they can get added.
F₂₄ = F_2q + F_4q = F₂ cos 55° + F₄ cos 45°
F₂₄ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ qQ }_{2}}{{ R }^{2}}\) cos 45° + \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { qQ 4}{{ R }^{2}}\) cos 45°
Resultant net force F

Question 4.
Suppose a charge +q on Earth’s surface and another +q charge is placed on the surface of the Moon, (a) Calculate the value of q required to balance the gravitational attraction between Earth and Moon (b) Suppose the distance between the Moon and Earth is halved, would the charge q change? (Take m_E = 5.9 x 10²⁴ kg, m_M = 7.348 x 10²² kg)
Solution:
Mass of the Earth, M_E = 5.9 x 10²⁴ kg
Mass of the Moon, M_M = 7.348 x 10²² kg
Charge placed on the surface of Earth and Moon = q
(a) Required charge to balance the F_G between Earth and Moon
F_C = F_G (or) \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }^{2}}{{ r }^{2}}\) = \(\frac{\mathrm{G} \mathrm{M}_{\mathrm{E}} \times \mathrm{M}_{\mathrm{M}}}{r^{2}}\)
q² = G × M_E × M_M × 4πε₀ = 320.97 × 10²⁵
q = \(\sqrt { 320.97\times { 10 }^{ 25 } } \) = 5.665 x 10¹³ = 5.67 x 10¹³ C

(b) The distance between Moon and Earth is

so q = 5.67 x 10¹³ C
There is no change.

Question 5.
Draw the free body diagram for the following charges as shown in the figure (a), (b) and (c).

Solution:

Question 6.
Consider an electron travelling with a speed V_Φ and entering into a uniform electric field \(\vec { E } \) which is perpendicular to \(\overrightarrow{\mathrm{V}_{0}}\) as shown in the Figure. Ignoring gravity, obtain the electron’s acceleration, velocity and position as functions of time.
Solution:

Speed of an electron = V₀
Uniform electric field = \(\vec { E } \)
(а) Electron’s acceleration:
Force on electron due to uniform electric field, F = Ee
Downward acceleration of electron due to electric field, a = \(\frac { F }{ m }\) = – \(\frac { eE }{ M }\)
Vector from, \(\vec { a } \) = – \(\frac { eE }{ M }\) \(\hat{j} \)

(b) Electron’s velocity:
Speed of electron in horizontal direction, u = V₀ From the equation of motion, V = u + at
V = V₀ \(\frac { eE }{ M }\) t
Vector from \(\vec { V } \) = V₀ \(\hat{j} \) – \(\frac { eE }{ M }\) t \(\hat{j} \)

(c) Electron’s position:
Position of electron, s = r
From equation of motion, r = V₀ t + \(\frac { 1 }{ 2 }\) \(\left(-\frac{e \mathrm{E}}{\mathrm{M}}\right)\) t²
r = V₀ t + \(\frac { 1 }{ 2 }\) \(\frac { eE }{ M }\) t² \(\hat{j} \)
Vector from,
\(\vec { r } \) = V₀ t \(\hat{j} \) \(\frac { 1 }{ 2 }\) \(\frac { eE }{ M }\) t² \(\hat{j} \)

Question 7.
A closed triangular box is kept in an electric field of magnitude E = 2 × 10³ N C^-1 as shown in the figure.

Calculate the electric flux through the (a) vertical rectangular surface (b) slanted surface and (c) entire surface.
Answer:
Electric field of magnitude E = 2 × 10³ NC^-1
(a) Vertical rectangular surface:
Rectangular area A= 5 × 10^-2 × 15 × 10^-2
A= 75 × 10^-24 m²
θ =180°
⇒ cos 180° = -1

Electric flux, Φ_v.s = EA cos θ
= 2 × 10³ × 75 × 10^-4 × cos 180°
= -150 × 10^-1
Φ_v.s = -15 Nm² C^-1

(b) Slanted surface:
cos θ = cos 60° = 0.5
sin θ = sin 30° = \(\frac { Opposite }{ hyp }\)

hyp = \(\frac {{ 5 × 10 }^{2}}{ 0.5 }\)
hyp = 0.1m
Area of slanted surface A₂ = (0.1 × 15 × 10^-2)
A₂ = 0.015 M²
Electric flux, Φ_v.s = EA = cos θ
= 2 × 10³ × 0.015 × cos 60°
= 2 × 10³ × 0.015 × 10³
= 0.015 × 10³
Φ_v.s = 15 Nm² C^-1
Horizontal surface
θ = 90° ; cos 90° = 0
Electric flux, Φ_H.S = E. A₃ Cos 90° = 0

(c) Entire surface:
Φ_Total = Φ_V.S + Φ_S.S + Φ_H.S = -15 + 15 + 0
Φ_Total = 0

Question 8.
The electrostatic potential is given as a function of x in figure (a) and (b). Calculate the corresponding electric fields in regions A, B, C and D. Plot the electric field as a function of x for the figure (b).

Answer:
The relation between electric field and potential
E = – \(\frac { dv }{ dx }\)

(a) Region A :
dv = -3V ; dx = 0.2 m
Electric field, E_A = \(\frac { (-3) }{ 0.2 }\) = 15 V m^-1
Region B:
dv = 0V ; dx = 0.2 m
Electric field, E_B = \(\frac { 0 }{ 0.2 }\) = 0
Region C:
dv = 2V ; dx = 0.2 m
Electric field, E_C = \(\frac { -2 }{ 0.2 }\) = 10 V m^-1
Region D:
dv = -6V ; dx = 0.2 m
Electric field, E_D = \(-\left(\frac{-6}{0.2}\right)\) = 10 V m^-1 = 30 V m^-1

Electric field, E_A = 15 V m^-1
Electric field, E_B = 0
Electric field, E_C = \(\frac { (-3) }{ 0.2 }\) = 10 V m^-1
Electric field, E_D = 30 V m^-1

(b)

Question 9.
A spark plug in a bike or a car is used to ignite the air-fuel mixture in the engine. It consists of two electrodes separated by a gap of around 0.6 mm gap as shown in the figure.

To create the spark, an electric field of magnitude 3 x 10⁶Vm^-1 is required, (a) What potential difference must be applied to produce the spark? (b) If the gap is increased, does the potential difference increase, decrease or remains the same? (c) find the potential difference if the gap is 1 mm.
Answer:
Separation gap between two electrodes, d = 0.6 mm
d = 0.6 × 10^-3 m
Magnetude of electric field Electric field = E = 3 × 10⁶ V m^-1
Electric field E = \(\frac { V }{ d }\)
(a) Applied potential difference, V = E . d
= 3 × 10⁶ × 0.6 10^-13 = 1.8 × 10³
V = 1800 V

(b) From equation, V = E . d
If the gap (distance) between the electrodes increased, the potential difference also increases.

(c) Gap between the electrodes, d = 1mm = 1 x 10^-3 m
Potential difference, V = E.d
= 3 × 10⁶ × 1 × 10^-3 = 3 × 103
V = 3000 V

Question 10.
A point charge of +10 μC is placed at a distance of 20 cm from another identical point charge of +10 μC. A point charge of -2 μC is moved from point a to b as shown in the figure. Calculate the
change in potential energy of the system? Interpret your result.

q₁ = 10μC = 10 x 10^-6 C
q₂ = 2μC = -2 x 10^-6 C
distance, r = 5cm = 5 x 10^-2 m
Answer:
Change in potential energy,

= -36 × 1 × 10⁹ × 10^-12 × 10² = -36 × 10^-1
∆ U = -3.6 J

Negative sign implies that to move the charge -2pC no external work is required. System spends its stored energy to move the charge from point a to point b.
Ans:
∆ U = -3.6 J, negative sign implies that to move the charge -2μC no external work is required. System spends its stored energy to move the charge from point a to point b.

Question 11.
Calculate the resultant capacitances for each of the following combinations of capacitors.

Answer:

Parallel combination of capacitor 1 and 2
C_p = C₀ + C₀ = 2C₀
Series combination of capacitor C_p and 3
\(\frac { 1 }{{ C }_{S}}\) = \(\frac { 1 }{{ C }_{p}}\) + \(\frac { 1 }{{ C }_{3}}\) = \(\frac { 1 }{{ 2C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = (or) \(\frac { 1 }{{ C }_{S}}\) = \(\frac { 3 }{ 2 }\) C₀  (or)C_S = \(\frac { 2 }{ 3 }\) C₀ 

\(\frac { 1 }{ { C }_{ { S }_{ 1 } } } \) = \(\frac { 1 }{{ C }_{1}}\) + \(\frac { 1 }{{ C }_{2}}\) = \(\frac { 1 }{{ C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = \(\frac { 1 }{{ C }_{0}}\) (or)
\(\frac { 1 }{ { C }_{ { S }_{ 1 } } } \) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 1 } }\) = \(\frac {{ C }_{0}}{ 2 }\)
Similarly 3 and 4 are series combination
\(\frac { 1 }{ { C }_{ { S }_{ 2 } } } \) = \(\frac { 1 }{{ C }_{3}}\) + \(\frac { 1 }{{ C }_{4}}\) = \(\frac { 1 }{{ C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }{0}}{ 2 }\)

\({ C }_{ { S }_{ 1 } }\) and \({ C }_{ { S }_{ 2 } }\) are in parallel combination
C_p = \({ C }_{ { S }_{ 1 } }\) + \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }_{0}}{ 2 }\) + \(\frac {{ C }_{0}}{ 2 }\) (or) C_p = \(\frac {{ 2C }_{0}}{ 2 }\) C_p = C₀

(c) Capacitor 1, 2 and 3 are in parallel combination
C_p = C₀ + C₀ + C₀ = 3C₀
C_p = 3C₀

(d) Capacitar C₁ and C₂ are in combination

Similarly C₃ and C₄ are in series combination


\({ C }_{ { S }_{ 1 } }\) and \({ C }_{ { S }_{ 2 } }\) are in parallel combination across RS:

(e) Capacitor 1 and 2 are series combination

Similarly 3 and 4 are series combination
\(\frac { 1 }{ { C }_{ { S }_{ 2 } } } \) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }_{0}}{ 2 }\)
Three capacitors are in parallel combination

Question 12.
An electron and a proton are allowed to fall through the separation between the plates of a parallel plate capacitor of voltage 5 V and separation distance h = 1 mm as shown in the figure.

(a) Calculate the time of flight for both electron and proton
(b) Suppose if a neutron is allowed to fall, what is the time of flight?
(c) Among the three, which one will reach the bottom first?
(Take m_p = 1.6 x 10^-27 kg, m_e= 9.1 x 10^-31 kg and g = 10 m s^-2)
Answer:
Potential difference between the parallel plates V = 5 V
Separation distance, h = 1 mm =1 x 10^-3 m
Mass of proton, mp = 1.6 x 10^-27 kg
Mass of proton, m =9.1 x 10^-31 kg
Charge of an a proton (or) electron, e— 1.6 x 10^-19 C
[u = 0; s = h]
From equation of motion, S = ut + \(\frac { 1 }{ 2 }\) at²
From equation of motion, h = \(\frac { 1 }{ 2 }\) at²
t = \(\sqrt { \frac { 2h }{ a } } \)
Acceleration of an electron due to electric field, a = \(\frac { F }{ m }\) = \(\frac { eE }{ m }\)
[E = \(\frac { V }{ d }\)]

(a) Time of flight for both electron and proton,

t_p = 63 ns……. (2)

(b) time of flight of neutron t_n = \(\sqrt { \frac { 2h }{ g } } \) = \(\sqrt{\frac{2 \times 1 \times 10^{3}}{10}}\) = \(\sqrt{0.2 \times 10^{-3}}\)
t_n = 0.0141 s = 14.1 x 10^-3 s
t_n = 14.1 x 10^-3 ms ……. (3)
(c) Compairision of values 1,2 and 3. The electron will reach the bottom first.

Question 13.
During a thunder storm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 x 10⁶ Vm^-1 ), lightning will occur.

(a) If the bottom part of the cloud is 1000 m above the ground, determine the electric potential difference that exists between the cloud and ground.
(b) In a typical lightning phenomenon, around 25C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?
Answer:
(a) Electric field between the cloud and ground,
V = E.d
V= 3 x 10⁶ x 1000 = 3 x 10⁹V
(a) Electrons transfered from cloud to ground,
q = 25 C
Electron static potential energy,
U = \(\frac { 1 }{ 2 }\) CV²
[C = \(\frac { q }{ V }\)]
= \(\frac { 1 }{ 2 }\) qV = \(\frac { 1 }{ 2 }\) x 25 x 3 x 10⁹
U = 37.5 x 10⁹ J

Question 14.
For the given capacitor configuration
(a) Find the charges on each capacitor
(b) potential difference across them
(c) energy stored in each capacitor.

Answer:
Capacitor b and c in parallel combination
C_p = C_b + C_c = (6 + 2) μF = 8 μF
Capacitor a, c_p and d are in series combination, so the resulatant copacitance
\(\frac { 1 }{{ C }_{s}}\) = \(\frac { 1 }{{ C }_{a}}\) + \(\frac { 1 }{{ C }_{cp}}\) + \(\frac { 1 }{{ C }_{d}}\) = \(\frac { 1 }{ 8 }\) + \(\frac { 1 }{ 8 }\) + \(\frac { 1 }{ 8 }\) = \(\frac { 3 }{ 8 }\)
C_s  = \(\frac { 8 }{ 3 }\) μF

(a) Charge on each capacitor,
Charge on capacitor a, Q_a = C_s V = \(\frac { 8 }{ 3 }\) x 9
Q_a = 24 μC
Charge on capacitor, d, Q_d = C_s V = \(\frac { 8 }{ 3 }\) x 9
Q_d = 24 μC
Capacitor b and c in parallel
Charge on capacitor, b, Q_b = \(\frac { 6 }{ 3 }\) x 9 = 18
Q_b = 18 μC
Charge on capacitor, c, Q_c = \(\frac { 2 }{ 3 }\) x 9 = 6
Q_c = 6 μC

(b) Potential difference across each capacitor, V = \(\frac { q }{ C }\)
Capacitor C_a, V_a = \(\frac{ { q }_{a}}{{ C }_{a}}\) = \(\frac {{ 24 × 10 }^{6}}{{ 8 × 10 }^{6}}\) = 3 V
Capacitor C_b, V_b = \(\frac{ { q }_{b}}{{ C }_{b}}\) = \(\frac {{ 18 × 10 }^{6}}{{ 6 × 10 }^{6}}\) = 3 V
Capacitor C_c, V_c = \(\frac{ { q }_{c}}{{ C }_{c}}\) = \(\frac {{ 6 × 10 }^{6}}{{ 2 × 10 }^{6}}\) = 3 V
Capacitor C_d, V_d = \(\frac{ { q }_{d}}{{ C }_{d}}\) = \(\frac {{ 24 × 10 }^{6}}{{ 8 × 10 }^{6}}\) = 3 V

(c) Energy stores in a capacitor, U = \(\frac { 1 }{ 2 }\) CV²
Energy in capacitor C_a, U_a = \(\frac { 1 }{ 2 }\) C_a \({ V }_{ a }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 8 x 10^-6 x (3)²
U_a = 36 μJ
Capacitor C_b, U_b = \(\frac { 1 }{ 2 }\) C_b \({ V }_{ b }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 6 x 10^-6 x (3)²
U_a = 27 μJ
C_c, U_c = \(\frac { 1 }{ 2 }\) C_c \({ V }_{ c }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 2 x 10^-6 x (3)²
U_a = 9 μJ
C_d, U_d = \(\frac { 1 }{ 2 }\) C_d \({ V }_{ d }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 8 x 10^-6 x (3)²
U_a = 36 μJ

Question 15.
Capacitors P and Q have identical cross sectional areas A and separation d. The space between the capacitors is filled with a dielectric of dielectric constant er as shown in the figure. Calculate the capacitance of capacitors P and Q.

Answer:
Cross-sectional area of parallel plate capacitor = A
Each area of different medium between parallel plate capacitor = \(\frac { A }{ 2 }\)
Separation distance = d
Capacitance of parallel plate capacitor, C = \(\frac { εA }{ d }\)
Air medium of dielectric constant, ε_r = 1
dielectric medium of dielectric constant = ε_r

Case 1:
Capacitance of air filled capacitor

Capacitance of dielectric filled capacitor

Capacitance of parallel plate capacitor

Case 2:
Each distance of different medium between the parallel plate capacitor = \(\frac { d }{ 2 }\)
Capacitance of dielectric filled capacitor

Capacitance of air filled capacitor,

Capacitance of parallel plate capacitor,

Samacheer Kalvi 12th Physics Electrostatics Additional Questions Solved

I. Multiple Choice Questions

Question 1.
When a solid body is negatively charged by friction, it means that the body has
(a) acquired excess of electrons
(b) lost some, problems
(c) acquired some electrons and lost a lesser number of protons
(d) lost some positive ions
Answer:
(a) acquired excess of electrons

Question 2.
A force of 0.01 N is exerted on a charge of 1.2 x 10^-5 G at a certain point. The electric field at that point is
(a) 5.3 x 10⁴ NC^-1
(b) 8.3 x 10^-4 NC^-1
(c) 5.3 x 10² NC^-1
(d) 8.3 x 10⁴ NC^-1
Answer:
(d) 8.3 x 10⁴ NC^-1
Hint:
E = \(\frac { F }{ q }\) = \(\frac { 0.01 }{{ 1.2 × 10 }^{-5}}\) = 8.3 x 10² NC^-1

Question 3.
The electric field intensity at a point 20 cm away from a charge of 2 x 10 5 C is
(a) 4.5 x 10⁶ NC^-1
(b) 3.5 x 10⁵ NC^-1
(c) 3.5 x 10⁶ NC^-1
(d) 4.5 x 10⁵ NC^-1
Answer:
(a) 4.5 x 10⁶ NC^-1
Hint:
E = \(\frac{q}{4 \pi \varepsilon_{0} r^{2}}\) = \(\frac{9 \times 10^{9} \times 2 \times 10^{-5}}{(0.2)^{2}}\) = 4.5 x 10⁶ NC^-1

Question 4.
How many electrons will have a charge of one coulomb?
(a) 6.25 x 10¹⁸
(b) 6.25 x 10¹⁹
(c) 1.6 x 10¹⁸
(d) 1.6 x 10¹⁹
Answer:
(a) 6.25 x 10¹⁸
Hint:
Number of electron, n = \(\frac { q }{ e }\) = \(\frac { 1 }{{ 1.6 × 10 }^{-19}}\) = 6.25 × 10¹⁸

Question 5.
The ratio of the force between two charges in air and that in a medium of dielectric constant K is
(a) K : 1
(b) 1 : K
(c) K² : 1
(d) 1 : K²
Answer:
(a) K : 1

Question 6.
The work done in moving a positive charge on an equipotential surface is
(a) finite and positive
(b) infinite
(c) finite and negative
(d) zero
Answer:
(d) zero

Question 7.
If a charge is moved against the coulomb force of an electric field.
(a) work is done by the electric field
(b) energy is used from some outside source
(c) the strength of the field is decreased
(d) the energy of the system is decreased
Answer:
(b) energy is used from some outside source

Question 8.
No current flows between two charged bodies when connected
(a) if they have the same capacitance
(b) if they have same quantity of charge
(c) if they have the same potential
(d) if they have the same charge density
Answer:
(c) if they have the same potential

Question 9.
Electric field lines about a negative point charge are
(a) circular, anticlockwise
(b) circular, clockwise
(c) radial, inwards
(d) radial, outwards
Answer:
(c) radial, inwards

Question 10.
Two plates are 1 cm apart and the potential difference between them is 10 V. The electric field between the plates is
(a) 10 NC^-1
(b) 250 NC^-1
(c) 500 N^-1
(d) 1000 NC^-1
Answer:
(d) 1000 NC^-1
Hint:
E = \(\frac { V }{ d }\) = \(\frac { 10 }{{ 1 × 10 }^{-2}}\) = 8.3 x 10² NC^-1

Question 11.
At a large distance (r), the electric field due to a dipole varies as
(a) \(\frac { 1 }{ r }\)
(b) \(\frac { 1 }{{ r }^{2}}\)
(c) \(\frac { 1 }{{ r }^{3}}\)
(d) \(\frac { 1 }{{ r }^{4}}\)
Answer:
(c) \(\frac { 1 }{{ r }^{3}}\)

Question 12.
Two thin infinite parallel plates have uniform charge densities +c and -σ. The electric field in the space between then is
(a) \(\frac { σ }{{ 2ε }_{0}}\)
(b) \(\frac { σ }{{ ε }_{0}}\)
(c) \(\frac { 2σ }{{ 2ε }_{0}}\)
(d) Zero
Answer:
(b) \(\frac { σ }{{ ε }_{0}}\)

Question 13.
Two isolated, charged conducting spheres of radii R₁, and R₂ produce the same electric field near their surfaces. The ratio of electric potentials on their surfaces is-
(a) \(\frac {{ R }_{1}}{{ R }_{2}}\)
(b) \(\frac {{ R }_{2}}{{ R }_{1}}\)
(c) \(\frac { { R }_{ 1 }^{ 2 } }{ { R }_{ 2 }^{ 2 } } \)
(d) \(\frac { { R }_{ 2 }^{ 2 } }{ { R }_{ 1 }^{ 2 } } \)
Answer:
(b) \(\frac {{ R }_{2}}{{ R }_{1}}\)

Question 14.
A 100 μF capacitor is to have an energy content of 50 J in order to operator a flash lamp. The voltage required to charge the capacitor is
(a) 500 V
(b) 1000 V
(c) 1500 V
(d) 2000 V
Answer:
(b) 1000 V
Hint:

Question 15.
A 1 μF capacitor is placed in parallel with a 2 μF capacitor across a 100 V supply. The total charge on the system is
(a) \(\frac { 100 }{ 3 }\) μC
(b) 100 μC
(c) 150 μC
(d) 300 μC
Answer:
(d) 300 μC
Hint:
Equivalent capacitor = 1 + 2 = 3 μF
Total charge, q = CV = 3 x 100 = 300 μF

Question 16.
A parallel plate capacitor of capacitance 100 μF is charged to 500 V. The plate separation is then reduce to half its original value. Then the potential on the capacitor becomes
(a) 250 V
(b) 500 V
(c) 1000V
(d) 2000 V
Answer:
(a) 250 V
Hint:
Here, C’ = 2C, since the charge remains the same.
q = C’V’ = CV ⇒ V = \(\frac { CV }{ 2C }\) = \(\frac { 500 }{ 2 }\) = 250 V

Question 17.
A point charge q is placed at the midpoint of a cube of side L. The electric flux emerging from the cube is ‘
(a) \(\frac { q }{{ ε }_{0}}\)
(b) \(\frac { q }{{ 6Lε }_{0}}\)
(c) \(\frac { 6Lq }{{ ε }_{0}}\)
(d) zero
Answer:
(a) \(\frac { q }{{ ε }_{0}}\)

Question 18.
The capacitor C of a spherical conductor of radius R is proportional to
(a) R²
(b) R
(c) R^-1
(d) R⁰
Answer:
(b) R

Question 19.
Energy of a capacitor of capacitance C, when subjected to a potential V, is given by
(a) \(\frac { 1 }{ 2 }\) CV²
(b) \(\frac { 1 }{ 2 }\) C²V
(c) \(\frac { 1 }{ 2 }\) CV
(d) \(\frac { 1 }{ 2 }\) \(\frac { C }{ V }\)
Answer:
(a) \(\frac { 1 }{ 2 }\) CV²

Question 20.
The electric field due to a dipole at a distance r from its centre is proportional to
(a) \(\frac { 1 }{{ r }^{3/2}}\)
(b) \(\frac { 1 }{{ r }^{3}}\)
(c) \(\frac { 1 }{ r }\)
(d) \(\frac { 1 }{{ r }^{3}}\)
Answer:
(b) \(\frac { 1 }{{ r }^{3}}\)

Question 21.
A point charge q is rotating around a charge Q in a circle of radius r. The workdone on it by the coulomb force is
(a) 2πrq
(b) 2πQq
(c) \(\frac { Q }{{ 2ε }^{0}r}\)
(d) zero
Answer:
(d) zero

Question 22.
The workdone in rotating an electric dipole of moment P in an electric field E through an angle 0 from the direction of the field is
(a) pE (1 – cos θ)
(b) 2pE
(c) zero
(d) -pE cos θ
Answer:
(a) pE (1 – cos θ)
Hint:
W = pE(cos θ₀ – cos θ)
[θ₀ = cos 0, cos 0 = 1]
W = pE(1 – cos θ)

Question 23.
Capacitance of a parallel plate capacitor can be increased by
(a) increasing the distance between the plates
(b) increasing the thickness of the plates
(c) decreasing the thickness of the plates
(d) decreasing the distance between the plates
Answer:
(d) decreasing the distance between the plates

Question 24.
Two charges are placed in vacuum at a distance d apart. The force between them is F. If a medium of dielectric constant 2 is introduced between them, the force will now be
(a) 4F
(b) 2F
(c) F/2
(d) F/4
Answer:
(d) F/4

Question 25.
An electric charge is placed at the centre of a cube of side a. The electric flux through one of its faces will be
(a) \(\frac { q }{{ 6ε }^{0}}\)
(b) \(\frac { q }{ { ε }_{ 0 }{ a }^{ 2 } } \)
(c) \(\frac { q }{ { 4πε }_{ 0 }{ a }^{ 2 } } \)
(a) \(\frac { q }{{ ε }^{0}}\)
Answer:
(a) \(\frac { q }{{ 6ε }^{0}}\)
Hint:
According to Gauss’s law, the electric flux through the cube is \(\frac { q }{{ ε }^{0}}\). Since there are six faces, the flux through one face is \(\frac { q }{{ 6ε }^{0}}\).

Question 26.
The electric field in the region between two concentric charged spherical shells-
(a) is zero
(b) increases with distance from centre
(c) is constant
(d) decreases with distance from centre
Answer:
(d) decreases with distance from centre

Question 27.
A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 80 V. The potential at the centre of the sphere is-
(a) 800 V
(b) zero
(c) 8 V
(d) 80 V
Answer:
(d) 80 V

Question 28.
A 4 μF capacitor is charged to 400 V and then its plates are joined through a resistance of 1 K Ω. The heat produced in the resistance is-
(a) 0.16 J
(b) 0.32 J
(c) 0.64 J
(d) 1.28 J
Answer:
(b) 0.32 J
Hint:
The energy stored in capacitor is converted into heat
U = H = \(\frac { 1 }{ 2 }\) CV² = \(\frac { 1 }{ 2 }\) x 4 x 10^-6 x (400)² = 0.32 J

Question 29.
The workdone in carrying a charge Q, once round a circle of radius R with a charge Q₂ at the centre is-
(a) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\)
(b) zero
(c) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}}\)
(d) infinite
Answer:
(b) zero
Hint:
The electric field is conservative. Therefore, no work is done in moving a charge around a closed path in a electric field.

Question 30.
Two plates are 2 cm apart. If a potential difference of 10 V is applied between them. The electric field between the plates will be
(a) 20 NC^-1
(b) 500 NC^-1
(c) 5 NC^-1
(d) 250 NC^-1
Answer:
(b) 500 NC^-1
Hint:
\(\frac { V }{ d }\) = \(\frac { 10 }{{ 2 ×10 }^{-2}}\) 500 NC^-1

Question 31.
The capacitance of a parallel plate capacitor does not depend on
(a) area of the plates
(b) metal of the plates
(c) medium between the plates
(d) distance between the plates
Answer:
(b) metal of the plates

Question 32.
A capacitor of 50 μF is charged to 10 volts. Its energy in joules is
(a) 2.5 x 10^-3
(b) 5 x 10^-3
(c) 10 x 10^-4
(d) 2.5 x 10^-4
Answer:
(a) 2.5 x 10^-3
Hint:
U = \(\frac { 1 }{ 2 }\) CV² = \(\frac { 1 }{ 2 }\) x 50 x 10^-6 x (10)² = 2.5 x 10^-3 J

Question 33.
A cube of side b has a charge q at each of its vertices. The electric field due to this charge distribution at the centre of the cube is
(a) \(\frac { q }{{b}^{ 2 }}\)
(b) \(\frac { q }{{2b}^{ 2 }}\)
(c) \(\frac { 32q }{{b}^{ 2 }}\)
(d) zero
Answer:(d) zero
Hint:
There
is an equal charge at diagonally opposite comer. The fields due the these at the centre cancel out. Therefore, the net field at the centre is zero.

Question 34.
Total electric fulx coming out of a unit positive charge put in air is
(a) ε₀
(b) \({ \varepsilon }_{ 0 }^{ -1 }\)
(c) (4πε₀)^-1
(d) 4πε₀
Answer:
(b) \({ \varepsilon }_{ 0 }^{ -1 }\)

Question 35.
Electron volt (eV) is a unit of
(a) energy
(b) potential
(c) current
(d) charge
Answer:
(a) energy

Question 36.
A point Q lies on the perpendicular bisector of an electric dipole of dipole moment P. If the distance of Q from the dipole is r, then the electric field at Q is proportional to-
(a) p^-1 and r^-2
(b) p and r^-2
(c) p and r^-3
(d) p² and r^-3
Answer:
(c) p and r^-3

Question 37.
A hollow insulated conducting sphere is given a positive charge of 10 μC. What will be the electric field at the centre of the sphere is its radius is 2 metres?
(a) zero
(b) 8 μCm^-2
(c) 20 μCm^-2
(d) 5 μCm^-2
Answer:
(d) zero

Question 38.
A particle of charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is-
(a) qE²y
(b) q²Ey
(c) qEy²
(d) qEy
Answer:
(d) qEy
Hint:
Force on the particle = qE
KE = Workdone by the force = F.y = qEy

Question 39.
Dielectric constant of metals is-
(a) 1
(b) greater then 1
(c) zero
(d) infinite
Answer:
(d) infinite

Question 40.
When a positively charged conductor is earth connected
(a) protons flow from the conductor to the earth
(b) electrons flow from the earth to the conductor
(c) electrons flow from the conductor to the earth
(d) no charge flow occurs
Answer:
(b) electrons flow from the earth to the conductor

Question 41.
The SI unit of electric flux is
(a) volt metre²
(b) newton per coulomb
(c) volt metre
(d) joule per coulomb
Answer:
(c) volt metre

Question 42.
Twenty seven water drops of the same size are charged to the same potential. If they are combined to form a big drop, the ratio of the potential of the big drop to that of a small drop is-
(a) 3
(b) 6
(c) 9
(d) 27
Answer:
(c) 9
Hint:
V’ = n^2/3 V
⇒ \(\frac { V’ }{ V }\) = (27)^2/3 = 9

Question 43.
A point charge +q is placed at the midpoint of a cube of side l. The electric flux emerging ’ from the cube is-
(a) \(\frac { q }{{ ε }^{0}}\)
(b) \(\frac {{ 6ql }^{2}}{{ ε }^{0}}\)
(c) \(\frac { q }{ { 6l }^{ 2 }{ { ε }^{ 0 } } } \)
(d) \(\frac { { C }^{ 2 }{ V }^{ 2 } }{ 2 } \)
Answer:
(a) \(\frac { q }{{ ε }^{0}}\)

Question 44.
The energy stored in a capacitor of capacitance C, having a potential difference V between the plates, is-
Answer:
(c)

Question 45.
The electric potential at the centre of a charged conductor is-
(a) zero
(b) twice that on the surface
(c) half that on the surface
(d) same as that on the surface
Answer:
(d) same as that on the surface

Question 46.
The energy stored in a capacitor is given by
(a) qV
(b) \(\frac { 1 }{ 2 }\)qV
(c) \(\frac { 1 }{ 2 }\) CV
(d) \(\frac { q }{ 2C }\)
Answer:
(b) \(\frac { 1 }{ 2 }\)qV

Question 47.
The unit of permitivity of free space so is
(a) coulomb/newton-metre
(b) newton-metre²/coulomb²
(c) coulomb²/newton-metre²
(d) coulomb/(newton-metre)²
Answer:
(c) coulomb²/newton-metre²

Question 48.
An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in a uniform electric field E. It its dipole moment is along the direction of the field, the force on it and its potential energy are, respectively.
(a) 2qE and minimum
(b) qE and pE
(c) zero and minimum
(d) qE and maximum
Answer:
(c) zero and minimum
Hint:
Potential energy, U = -pE cos θ
For q = 0°; U = -pE, which is minimum.

Question 49.
An electric dipole of moment \(\vec { P } \) is lying along a uniform electric field \(\vec { E } \) . The workdone in rotating the dipole by 90° is
(a) \(\frac { pE }{ 2 }\)
(b) 2pE
(c) pE
(d) √2pE
Answer:
(c) pE

Question 50.
Aparallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates
(a) does not charge
(b) becomes zero
(c) increases
(d) decreases
Answer:
(c) increases

Question 51.
When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance
(a) increases K times
(b) increases K^-1 times
(c) decreases K times
(d) remains constant
Answer:
(c) decreases K times

Question 52.
A comb run through one’s dry hair attracts small bits of paper. This is due to the fact that
(a) comb is a good conductor
(b) paper is a good conductor
(c) the atoms in the paper gets polarised by the charged comb
(d) the comb posseses magnetic properties
Answer:
(c) the atoms in the paper gets polarised by the charged comb

Question 53.
Which of the following is not a property of equipotential surfaces?
(a) they do not cross each other
(b) they are concentric spheres for uniform electric field
(c) the rate of change of potential with distance on them is zero
(d) they can be imaginary spheres.
Answer:
(b) they are concentric spheres for uniform electric field

Question 54.
A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will be
(a) reduced to half
(b) doubled
(c) becomes 4 times
(d) remains the same
Answer:
(d) remains the same

Question 55.
If the electric field in a region is given by \(\vec { E } \) = 5\(\hat{j} \) + 4\(\hat{j} \) + 9\(\hat{k} \) , then the electric flux through a surface of area 20 units lying in the y-z plane will be-
(a) 20 units
(b) 80 units
(c) 100 units
(d) 180 units
Answer:
(c) 100 units
Hints:
The area vector \(\vec { A } \) = 20\(\hat{j} \); \(\vec { E } \) = (5\(\hat{j} \) + 4\(\hat{j} \) + 9\(\hat{k} \))
Flux (Φ) = \(\vec { E } \) – \(\vec { A } \) = 5 x 20 =100 units

Question 56.
A, B and C are three points in a uniform electric field. The electric potential is-

(a) maximum at A
(b) maximum at B
(c) maximum at B
(d) same at all the three points A, B and C
Answer:
(b) maximum at B
Hint:
The potential decreases in the direction of the field. Therefore V_B > V_C>C_A.

Question 57.
A conducting sphere of radius R is give a charge Q. The electric potential and the electric field at the centre of the sphere are, respectively-
(a) zero, \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R }^{ 2 } } \)
(b) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \)
(c) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \), zero
(d) zero,zero
Answer:
(c) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \), zero.

II. Fill in the blanks

Question 1.
A dipole is placed in a uniform electric field with its axis parallel to the field. It experiences …………………
Answer:
neither a net force nor a torque

Question 2.
The unit of permittivity is…………………
Answer:
C²N^-1m^-2

Question 3.
The branch of physics which deals with static electric charges or charges at rest is …………………
Answer:
electrostatics

Question 4.
The charges in a electrostatics field are analogous to ………………… in a gravitational field.
Answer:
mass

Question 5.
The substances which acquire charges on rubbing are said to be …………………
Answer:
electrified

Question 6.
Electron means …………………
Answer:
amber

Question 7.
A glass rod rubbed with a silk cloth. Glass rod and silk cloth acquires …………………
Answer:
positive and negative charge respectvely .

Question 8.
When ebonite rod is rubbed with fur, ebonite rod and fur acquires …………………
Answer:
negative and positive charge respectively

Question 9.
………………… termed the classification of positive and negative charges.
Answer:
Franklin

Question 10.
Applications such as electrostatic point spraying and powder coating, are based on the property of ………………… between charged bodies.
Answer:
attraction and repulsion

Question 11.
Bodies which allow the charge to pass through them are called …………………
Answer:
conductor

Question 12.
Bodies which do not allow the charge to pass through them are called …………………
Answer:
insulators

Question 13.
The unit of electric charge is …………………
Answer:
coulomb

Question 14.
Total charge in an isolated sysem …………………
Answer:
remains a constant

Question 15.
The force between two charged bodies was studied by …………………
Answer:
coulomb

Question 16.
The unit of permittivity in free space (s0) is …………………
Answer:
C²N^-1m^-2

Question 17.
The value of s, for air or vacuum is …………………
Answer:1

Question 18.
Charges can neither be created nor be destroyed is the statement of law of conservation of …………………
Answer:
charge

Question 19.
The space around the test charge, in which it experiences a force is known as field …………………
Answer:
electric

Question 20.
Electric field at a point is measued in terms of …………………
Answer:
electric field intensity

Question 21.
The unit of electric field in tensity is …………………
Answer:
NC^-1.

Question 22.
The lines of force are far apart, when electric field E is …………………
Answer:
small

Question 23.
The lines of force are close together, when electric field E is …………………
Answer:
large

Question 24.
Electric dipole moment …………………
Answer:
P = 2qd

Question 25.
Torque experienced by electric dipole is …………………
Answer:
x = PE sin θ

Question 26.
An electric dipole placed in a non-uniform electric field at an angle 0 experiences …………………
Answer:
both torque and force

Question 27.
When thee dipole is aligned parallel to the field, its electric potential energy is …………………
Answer:
u = -PE

Question 28.
Change of potential with distance is known as …………………
Answer:
potential distance

Question 29.
The number of electric lines of force crossing through the given area is …………………
Answer:
electric flux

Question 30.
The process of isolating a certain region of space from external field is called …………………
Answer:
electrostatic shielding

Question 31.
Capacitor is a device to store …………………
Answer:
charge

Question 32.
The charge density in maximum at …………………
Answer:
pointed

Question 33.
The principle made use of in lightning arrestor is …………………
Answer:
action of points

Question 34.
Van de Graaff generator producers large electrostatic potential difference of the order of …………………
Answer:
10⁷ V

III. Match the following

Question 1.

Answer:
(i) → (d)
(ii) → (a)
(iii) → (b)
(iv) → (c)

Question 2.

Answer:
(i) → (c)
(ii) → (d)
(iii) → (a)
(iv) → (b)

Question 3.

Answer:
(i) → (b)
(ii) → (d)
(iii) → (a)
(iv) → (c)

Question 4.

Answer:
(i) → (b)
(ii) → (d)
(iii) → (a)
(iv) → (c)

IV. Assertion and reason type

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but the reason is not correct explanation of the assertion.
(c) If assertion is true but reason is false.
(d) If the assertion and reason both are false.
(e) If assertion is false but reason is true.

Question 1.
Assertion: Electric lines of force cross each other.
Reason: Electric field at a point supermpose to give one resultant electric field.
Answer:
(e) Both assertion and reason are true but the reason is not correct explanation of the assertion.
Explanation: If electric lines of forces cross each other, then the electric field at the point of intersection will have two direction simultaneously which is not possible physically.

Question 2.
Assertion: Charge is quantized.
Reason: Charge, which is less than 1 C is not possible.
Answer:
(c) If assertion is true but reason is false.
Explanation: Q = ±ne and charge lesser than 1 C is possible.

Question 3.
Assertion:
A point charge is brought in an electric field. The field at a nearby point will increase, whatever be the nature of the charge.
Reason: The electric field is independent of the nature of charge.
(d) If the assertion and reason both are false.
Explanation: Electric field at the nearby-point will be resultant of existing field and field due to the charge brought. It may increase or decrease if the charge is positive or negative depending on the position of the point with respect to the charge brought.

Question 4.
Assertion: The tyre’s of aircraft’s are slightly conducting.
Reason: If a conductor is connected to ground, the extra charge induced on conductor will flow to ground.
Answer:
(b) Both assertion and reason are true but the reason is not correct explanation of the assertion.
Explanation: During take off and landing, the friction between treys and the run way may cause electrification of treys. Due to conducting to a ground and election sparking is avoided.

Question 5.
Assertion: The lightening conductor at the top of a high building has sharp ends.
Reason: The surface density of charge at sharp points is very high, resulting in setting up of electric wind.
Answer:
(a) Both assertion and reason are true and the reason is the correct explanation of the assertion.

Samacheer Kalvi 12th Physics Electrostatics Short Answer Questions

Question 1.
What is meant by triboelectric charging?
Answer:
Charging the objects through rubbing is called triboelectric charging.

Question 2.
What is meant by conservation of total charges?
Answer:
The total electric charge in the universe is constant and charge can neither be created nor be destroyed. In any physical process, the net change in charge will always be zero.

Question 3.
State Gauss’s Law?
Answer:
Definition:
Gauss’s law states that if a charge Q is enclosed by an arbitrary closed surface, then the total electric flux OE through the closed surface is
Φ_E = \(\oint { \vec { E } } \) .d\(\vec { A } \) = \(\frac {{ q }_{encl}}{{ ε }_{0}}\)

Question 4.
What is meant by electrostatic shielding?
During lightning accompanied by a thunderstorm, it is always safer to sit inside a bus than in open ground or under a tree. The metal body of the bus provides electrostatic shielding, since the electric field inside is zero. During lightning, the charges flow through the body of the conductor to the ground with no effect on the person inside that bus.

Question 5.
What is meant by dielectric?
Answer:
A dielectric is a non-conducting material and has no free electrons. The electrons in a dielectric are bound within the atoms. Ebonite, glass and mica are some examples of dielectrics.

Question 6.
What are non-polar molecules? Give examples.
A non-polar molecule is one in which centers of positive and negative charges coincide. As a result, it has no permanent dipole moment. Examples of non-polar molecules are hydrogen (H₂), oxygen (O₂), and carbon dioxide (CO₂) etc.

Question 7.
What are polar molecules? Give examples.
Answer:
In polar molecules, the centers of the positive and negative charges are separated even in the absence of an external electric field. They have a permanent dipole moment.
The net dipole moment is zero in the absence of an external electric field. Examples of polar molecules are H₂O, N₂O, HCl, NH₃.

Question 8.
What is a capacitors?
Answer:
Capacitor is a device used to store electric charge and electrical energy. Capacitors are widely used in many electronic circuits and have applications in many areas of science and technology.

Samacheer Kalvi 12th Physics Electrostatics Long Answer Questions

Question 1.
Derive an expression for electric field due to the system of point charges?
Answer:
Electric field due to the system of point charges:
Suppose a number of point charges are distributed in space. To find the electric field at some point P due to this collection of point charges, superposition principle is used. The electric field at an arbitrary point due to a collection of point charges is simply equal to the vector sum of the electric fields created by the individual point charges. This is called superposition of electric fields.
Consider a collection of point charges q₁, q₂, q₃,…., q_n located at various points in space. The ‘ total electric field at some point P due to all these n charges is given by

Here r_1p, r_2p, r_3p,…., r_np, are the distance of the charges ₁, q₂, q₃,…., q_n from the point respectively. Also \(\hat{r} \)_1p + \(\hat{r} \)_2p + \(\hat{r} \)_3p,…., \(\hat{r} \)_np are the corresponding unit vectors directed from q₁, q₂, q₃,…., q_n tpo P.

Equation (2) can be re-written as,

For example in figure, the resultant electric field due to three point charges q₁, q₂, q₃ at point P is shown. Note that the relative lengths of the electric field vectors for the charges depend on relative distantes of the charges to the point P.

Question 2.
Derive an expression for electric flux of rectangular area placed in uniform electric field.
Answer:
(i) Electric flux for uniform Electric field:
Consider a uniform electric field in a region of space. Let uschoose an area A normal to the electric field lines as shown in figure (a). The electric flux for this case is
Φ_E = EA ….. (1)
Suppose the same area A is kept parallel to the uniform electric field, then no electric field lines pierce through the area A, as shown in figure (b). The electric flux for this case is zero.
Φ_E = 0 ….. (2)
If the area is inclined at an angle θ with the field, then the component of the electric field perpendicular to the area alone contributes to the electric flux. The electric field component parallel to the surface area will not contribute to the electric flux. This is shown in figure (c). For this case, the electric flux
ΦE = (E cosθ) A …(3)
Further, θ is also the angle between the electric field and the direction normal to the area. Hence in general, for uniform electric field, the electric flux is defined as
Φ_E= \(\vec { E } \).\(\vec { A } \) = EA cos θ …(4)

Here, note that \(\vec { A } \) is the area vector \(\vec { A } \) = A\(\hat{n} \). Its magnitude is simply the area A and the direction is along the unit vector h perpendicular to the area. Using this definition for flux, Φ_E= \(\vec { E } \).\(\vec { A } \), equations (2) and (3) can be obtained as special cases.
In figure (a), θ = 0° so Φ_E= \(\vec { E } \).\(\vec { A } \) = EA
In figure (b), θ = 90° so Φ_E= \(\vec { E } \).\(\vec { A } \) = 0

(ii) Electric flux in a non uniform electric field and an arbitrarily shaped area: Suppose the electric field’is not uniform and the area A is not flat, then the entire area is divided
into n small area segments ∆\(\vec { A } \)₁ ∆\(\vec { A } \)₂, ∆\(\vec { A } \)₃,…..∆\(\vec { A } \)_n, such that each area element is almost flat and the electric field over each area element is considered to be uniform.
The electric flux for the entire area A is approximately written as

By taking the limit ∆\(\vec { A } \)₁ → 0 (for all i) the summation in equation (5) becomes integration. The total electric flux for the entire area is given by
Φ_E = ∫\(\vec { E } \).d\(\vec { A } \) ….. (6)
From Equation (6), it is clear that the electric flux for a given surface depends on both the electric field pattern on the surface area and orientation of the surface with respect to the electric field.

(iii) Electric flux for closed surfaces: In the previous section, the electric flux for any arbitrary curved surface is discussed. Suppose a closed surface is present in the region of the non-uniform electric field as shown in figure (a).

The total electric flux over this closed surface is written as
Φ_E = \(\oint { \vec { E } } \).d\(\vec { A } \) …… (7)
Note the difference between equations (6) and (7). The integration in equation (7) is a closed surface integration and for each areal element, the outward normal is the direction of d\(\vec { A } \) as shown in the figure (b).
The total electric flux over a closed surface can be negative,
positive or zero. In the figure (b), it is shown that in one area element, the angle between d\(\vec { A } \) and \(\vec { E } \) is less than 90°, then the electric flux is positive and in another areal element, the angle between dA and E is greater than 90°, then the electric flux is negative. In general, the electric flux is negative if the electric field lines enter the closed surface and positive if the electric field lines leave the closed surface.

Samacheer Kalvi 12th Physics Electrostatics Numerical Problems

Question 1.
Electrons are caused to fall through a potential difference of 1500 volts. If they were initially at rest. Then calculate their final speed.
Solution:
The electrical potential energy is converted into kinetic energy. If v is the final speed then

Question 2.
Small mercury drops of the same size are charged to the same potential V. If n such drops coalesce to form a single large drop, then calculate its potential.
Solution:
Let r be the radius of a small drop and R that of the large drop. Then, since the volume remains conserved,
\(\frac { 1 }{ 2 }\) πR² = \(\frac { 4 }{ 3 }\) πR³n
⇒ R³ = r³n
R = r³(n)^1/3
Further, since the total charge remains conserved, we have, using Q = CV
C_large V = n C_small v
Where V is the potential of the large drop.
4πε₀ RV = n (4πε₀r)v
V = \(\frac { nrv }{ R }\) = \(\frac { nrv }{{ r(n) }^{1/3}}\)
V = vn^2/3

Question 3.
Two particles having charges Q₁ and Q₂ when kept at a certain distance, exert a force F on each other. If the distance between the two particles is reduced to half and the charge on each particle is doubled. Find the force between the particles.
Solution:
F = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{r^{2}}\)
If the distance is educed by half and two particles of charges are doubled.

Question 4.
Two charged spheres, separated by a distance d, exert a force F on each other. If they are immersed in a liquid of dielectric constant 2, then what is the force.
Solution:
Force between the charges (vacuum)

Force between the charges (medium)

Question 5.
Find the force of attraction between the plates of a parallel plate capacitor.
Solution:
Let d be the distance between the plates. Then the capacitor is
C = \(\frac { { \varepsilon }_{ 0 }A }{ d } \)
Energy stored in a capacitor,

Energy magnitude of the force is,

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Students can Download English Lesson 3 In Celebration of Being Alive Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 12th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th English Solutions Prose Chapter 3 In Celebration of Being Alive

12th English In Celebration Of Being Alive Paragraph Warm Up

There are several physically-challenged people who have lived successful and meaningful lives. Here are a few personalities who have fought great odds and lived a life of blazing achievements. Let’s share wdiat we know about each of them and complete the table below.

Answer:

Name of the Personality	Nature of Challenge	Field of achievement
e.g. Beethovan	Hearing impairment	Music
Demosthenes	Speech impaired	Oration
Helen Keller	Vision and multiple disorders	Writing, Public Service
Mariyappan Thangavelu	Physically handicapped	High Jump
Mozart	Hearing impairment	Music
John Milton	Hearing impairment	Poetry
Sudha Chandran	Hearing impairment	Dancing

Samacheer Kalvi 12th English In Celebration of Being Alive Textual Questions

1. Answer the following questions in one or two sentences based on your understanding of the lesson.

In Celebration Of Being Alive Summary Question (a)
What thoughts troubled Dr. Christiaan Barnard as he neared the end of his career as a heart surgeon?
Answer:
Towards the end of his career, Dr. Christiaan Barnard was troubled by the suffering of people and especially of young children. He could not accept the fact that 12 million children are unlikely to reach the age of one and about 6 million children die annually before reaching the age of five.

In Celebration Of Being Alive Paragraph Question (b)
What were Dr. Barnard’s feelings when he was hospitalized after an accident?
Answer:
He experienced not only agony but also anger after they had met with an accident. He had eleven broken ribs and perforated lungs. His wife had a badly fractured shoulder. He could not understand why they should undergo pain when they had other important things to do in life during that time.

In Celebration Of Being Alive Question (c)
When and where did the accident occur?
Answer:
He and his wife were crossing the road after a lovely meal. A car hit him and knocked him ’ into his wife. His wife was thrown into the other lane and was struck by a car coming in the opposite direction.

In Celebration Of Being Alive Questions And Answers Question (d)
How did the hospitalization of Dr. Barnard and his wife affect their routine?
Answer:
As a heart surgeon, he had to operate many heart patients. He was helpless as he had perforated lungs and broken ribs. His wife could not take care of the baby. Thus the routine life of both Dr. Barnard and his wife was affected adversely.

12th English Unit 3 Prose Question (e)
How was Dr. Barnard’s attitude to suffering different from that of his father’s?
Answer:
Dr. Barnard’s father accepted suffering as God’s will. He also believed that suffering ennobles humans. But Dr. Barnard found no meaning in the agony and suffering of patients and especially of the young children.

In Celebration Of Being Alive Book Back Answers Question (f)
How was the unattended trolley put to use?
Answer:
The unattended trolley was used as a car for a race (Grand prix of Cape Town’s Red cross childrens’ hospital) within the hospital.

In Celebration Of Being Alive Summary In English Question (g)
What roles did the duo take up?
Answer:
The unattended trolley was commandeered by a crew of two boys. One served as the driver and the other as a mechanic.

12th English 3rd Prose Question (h)
Why did the choice of roles prove to be easy for them?
Answer:
The mechanic provided motor power by galloping along behind the trolley with his head down. The driver steered the trolley by scraping his foot on the floor. The choice of the roles was easy because the mechanic was totally blind and the driver had only one arm.

Question (i)
Who encouraged them and how?
Answer:
They were encouraged by laughter and shouts of encouragement from the rest of the patients.

Question (j)
What does Dr. Barnard compare this entertainment to?
Answer:
Dr. Barnard compared the trolley race as much better entertainment than anything anyone puts on at the Indianapolis 500 car race.

Question (k)
What happened in the grand finale?
Answer:
There was a grand finale of scattered plates and silverware before the nurse and ward sister caught up with them.

Question (l)
How does Dr. Barnard know the boy who played the trolley’s driver?
Answer:
Dr. Barnard knew the trolley’s driver better. He had successfully closed a hole in his heart a few years back.

Question (m)
What was the profound lesson that Dr. Barnard learnt from the boys?
Answer:
The boys had taught Dr. Barnard the lesson in getting on with the business of living. The business of living is the celebration of being alive.

Additional Questions

Question (a)
What did the Grand Prix of the hospital teach the author?
Answer:
The author found out that his perception of human suffering was incomplete. The boys taught him how the experience of suffering helps a person value life. This thought gave the author a solace.

2. Answer the following questions in three or four sentences.

Question (а)
Detail the statistics Dr. Barnard has provided in his speech.
Answer:
Of the 125 million children bom that year, 12 million are unlikely to reach the age of one. Another six million will die before the age of five. Among the rest, many will end up as mental or physical cripples.

Question (b)
What happened when the doctor couple were crossing the street?
Answer:
After a nice meal, Dr. Barnard and his wife were crossing the street. A car hit the doctor. He dashed against his wife who was thrown on the other side of the road. She was hit by another car from the opposite side.

Question (c)
What injuries did they sustain in the accident?
Answer:
Dr. Barnard had eleven broken ribs. A lung was profoundly perforated. His wife had a badly fractured shoulder.

Question (d)
Dr. Barnard couldn’t find any nobility in suffering. Why?
Answer:
As a doctor, he does not find any nobility in suffering. There is nothing noble in a patient’s thrashing around in a sweat-soaked bed, mind clouded in agony. He was against his dad’s faith that suffering ennobles human beings.

Question (e)
Why does Dr. Barnard find suffering of children heartbreaking?
Answer:
He has always found suffering of young children heart-breaking. Especially because they have total faith in doctors. They believe doctors will help. They don’t complain even after undergoing a mutilating surgery.

Question (f)
How did the boy who played the mechanic lose his eyesight?
Answer:
The mechanic was seven years old. One night, when his drunk father tortured his mom, she threw a lantern at him. The lantern broke over the child’s head and shoulders. He suffered third degree bums on the upper part of his body and lost both of his eyes.

Question (g)
Why does Dr. Barnard describe the blind boy as a ‘walking horror’?
Answer:
At the time of the grand prix, the boy was a walking horror. His face was disfigured. A long flap of skin was hanging from the side of his neck to his body. As the wound healed around his neck, his lower jaw became gripped in a mass of fibrous tissue. The only way he could open his mouth was to raise his head.

Question (h)
What were the problems the trolley driver suffered from?
Answer:
The trolley’s driver had a malignant tumour of the bone. A few days before the race, his shoulder and arm were amputated. There was no hope of recovery.

Additional Questions

Question (a)
How does suffering ennoble a person?
Answer:
One does not become a better person because one has suffered. One becomes a better person because one has undergone suffering. One can’t appreciate light in the absence of darkness.

3. Answer the following in a paragraph of 100 – 150 words each.

Question (a)
Give an account of the medical problems for which the two boys were hospitalized.
Answer:
The seven years old mechanic suffered third degree bums on the upper part of his body. He had lost both his eyes. He was literally a walking horror. He was disfigured. A long flap of skin w as hanging from the side of his neck to his body. As the wound healed around his neck, his lower jaw became gripped in a mass of fibrous tissue. The trolley driver had a malignant tumour of the bone. A few days before the race, his shoulder and arm were amputated. There was little hope of his recovery. If two adults had similar ailments, they would have got dejected with life. But the boys were just happy celebrating the joy of being alive.

Question (b)
“These two children had given me a profound lesson …” Elucidate.
Answer:
The author had self-pity and was cursing the accident that had caused him and his wife great pain and inconveniences in the daily routine. But the two little boys, one almost scarred to death with both eyes gone and the other with amputated arm and no hope of recovery wrere together celebrating the joy of being alive. They minded the business of living ignoring pain, surgery and the sickly environment. Dr. Barnard leamt the lesson from the children that the business of living is joy in the real sense of the w ord. It was not just something for pleasure, amusement or recreation. The business of living is the celebration of being alive.

Question (c)
Describe the ‘Grand Prix’ at Cape Town’s Red Cross Children’s Hospital.
Answer:
The author describes the event as “the Grand Prix of Cape Town’s Red Cross Childrens’ Hospital”. A nurse had left a breakfast trolley unattended. Very soon this trolley was commandeered by a daring crew of two, a driver and a mechanic. The mechanic provided motor power by galloping along behind the trolley head down. While the driver, seated on the lower deck held on with one hand and steered it by scrapping his foot on the floor. The choice of roles wag easy because the mechanic was totally blind and the driver had only one arm. It was better than Indianapolis 500 car race. Patients shouted and cheered the boys. There was a grand finale of scattered plates and silverware before the nurse and ward sister took control of the situation.

Question (d)
How did a casual incident in a hospital help Dr. Barnard perceive a new dimension of life?
Answer:
Initially Dr. Barnard was grumbling. He wondered why on earth he and his wife should have been subjected to agony and inconvenience. He couldn’t take his dad’s view that suffering ennobles human beings. But the little boys taught him a profound lesson of life. One should get on with the business of living irrespective of whatever misfortune strikes one. You don’t become a better person because you suffered, your suffering does not ennoble you. But you become a better person because you have experienced suffering. It is not what you have lost is important. What is important is what you have left. We can appreciate light better once we have experienced darkness. Similarly, we can appreciate warmth only after experiencing cold.

Question (e)
Life is unjust and cruel to certain people. Do they all resign themselves to their fate? Can you think of some who have fought their disabilities heroically and remained a stellar example for others? (for e.g. the astrophysicist Stephen Hawking, a paraplegic). Give an account of one such person and his/her struggle to live a fruitful life.
Answer:
Alexis Leon lives in Kakkanad, Kerala. He passed B.Tech from the university of Kerala with first rank. Then he did his M.Tech. He met w ith an accident in 1993 which left him paralysed from chest down and confined him to a wheel chair for the rest of his life. After a brief spell of dejection owing to a suspended marriage, he made up his mind to write books. He has written 50 books for Engineering graduates.

His notable w’ork is ‘Internet for Everyone’ and ‘A Guide to Software Configuration Management’. He is also a mentor at International Mentoring Network Association. He offers software consultancy to international IT companies. He travels across the world and delivers lectures to graduates and Corporates. His indomitable will has made him strong. He has become a light house for many aspirants in the software industry.

Additional Questions

(a) What did Dr. Christiaan Barnard know about the mechanic and his family?
Answer:
The mechanic was seven years old. Both of his parents were drunk while quarreling, his mom threw a lantern at his father. It missed him but hit the boy. He suffered severe third degree bums on the upper part of his body. He had lost both his eyes in the process. He got a disfigured face. He was a walking horror with a long flap of skin hanging from the side off his next to his body.

Vocabulary

1. More and more, as I near the end of my career as a heart surgeon, my thoughts have turned to the consideration of why people should suffer. Suffering seems so cruelly prevalent in the world today. Do you know that of the 125 million children born this year, 12 million are unlikely to reach the age of one and another six million will die before the age of five? And, of the rest, many will end up as mental or physical cripples.

2. My gloomy thoughts probably stem from an accident I had a few years ago. One minute I was crossing the street with my wife after a lovely meal together, and the next minute a car had hit me and knocked me into my wife. She was thrown into the other lane and struck by a car coming from the opposite direction.

3. During the next few days in the hospital, I experienced not only agony and fear but also anger. I could not understand why my wife and I had to suffer. I had eleven broken ribs and a perforated lung. My wife had a badly fractured shoulder. Over and over, I asked myself, why should this happen to us? I had work to do, after all; there were patients waiting for me to operate on them. My wife had a young
baby who needed her care

4. My father, had he still been alive, would have said: “My son, it’s God’s will. That’s the way God tests you. Suffering ennobles you – makes you a better person.”

5. But, as a doctor, I see nothing noble in a patient’s thrashing around in a sweat-soaked bed, mind clouded in agony. Nor can I see any nobility in the crying of a lonely child in a ward at night.

6. In those days, they didn’t have sophisticated heart surgery. I have always found the suffering of children particularly heartbreaking–especially because of their total trust in doctors and nurses. They believe you are going to help them. If you can’t they accept their fate. They go through mutilating surgery, and afterwards they don’t complain.

7. One morning, several years ago, I witnessed what I call the Grand Prix of Cape Town’s Red Cross Children’s Hospital. It opened my eyes to the fact that I was missing something in all my thinking about suffering – something basic that was full of solace for me.

8. What happened there that morning was that a nurse had left a breakfast trolley unattended. And very soon this trolley was commandeered by an intrepid crew of two – a driver and a mechanic. The mechanic provided motor power by galloping along behind the trolley with his head down, while the driver, seated on the mower deck, held on with one hand and steered by scraping his foot on the floor. The choice of roles was easy because the mechanic was totally blind and the driver had only one arm.

9. They put on quite a show that day. Judging by the laughter and shouts of encouragement from the rest of the patients, it was a much better entertainment than anything anyone puts on at the Indianapolis 500 car race. There was a grand finale of scattered plates and silverware before the nurse and ward sister caught up with them, scolded them and put them back to bed.

10. Let me tell you about these two. The mechanic was all of seven years old. One night, when his mother and father were drunk, his mother threw a lantern at his father, missed and the lantern broke over the child’s head and shoulders. He suffered severe third-degree burns on the upper part of his body, and lost both his eyes.

At the time of the Grand Prix, he was a walking horror, with a disfigured face and long flap of skin hanging from the side of his neck to his body. As the wound healed around his neck, his lower jaw became gripped in a mass of fibrous tissue. The only way this little boy could open his mouth was to raise his head. When I stopped by to see him after the race, he said, “You know, we won.” And he was laughing.

11. The trolley’s driver I knew better. A few years earlier, I had successfully closed a hole in his heart. He had returned to the hospital because he had a malignant tumour of the bone. A few days before the race, his shoulder and arm were amputated. There was little hope of his recovery. After the Grand Prix, he proudly informed me that the trolley’s wheels were not properly oiled, but he was a good driver, and he had full confidence in the mechanic.

12. Suddenly, I realized that these two children had given me a profound lesson in getting on with the business of living. Because the business of living is joy in the real sense of the word, not just something for pleasure, amusement, recreation. The business of living is the celebration of being alive.

13. I had been looking at suffering from the wrong end. You don’t become a better person because you are suffering; but you become a better person because you have experienced suffering. We can’t appreciate light if we haven’t known darkness. Nor can we appreciate warmth if we have never suffered cold. These children showed me that it’s not what you’ve lost that’s important. What is important is what you have left

(a) Go through the lesson and spot the words which mean the same as the following.

1. profession (para 1)
2. sorrowful (para 2)
3. decency (para 5)
4. destiny (para 6)
5. hijacked (para 8)
6. motivation (para 9)
7. serious (para 10)
8. significant (para 13)

Answer:

1. career
2. gloomy
3. nobility
4. fate
5. commandeered
6. encouragement
7. severe
8. important

(b) Go through the lesson and spot the words opposite to the meaning of the following.

1. rare (para 1)
2. primitive (para 6)
3. fiction (para 7)
4. fearful (para 8)
5. benign (para 11)
6. diffidence (para 11)
7. boredom (para 12)
8. criticize (para 13)

Answers

1. prevalent
2. sophisticated
3. fact
4. intrepid
5. malignant
6. confidence
7. recreation
8. appreciate

(c) Frame illustrative sentences to distinguish the meaning of the words in the following clusters.

Question 1.
career – carrier – courier
Answer:
A J. Cronin started his career as a doctor but eventually became a writer.
The curry spilled over as the tiffin carrier lid was not tightly closed.
The courier office is closed on Sundays.

Question 2.
patients – patience – patents
Answer:
I saw a large number of patients standing in queue in front of the Government Hospital.
The patience of the gardener got paid well when the garden bloomed.
No other scientist in the world has got so many patents as Edison did.

Question 3.
accident – incident – incidence
Answer:
The accident took place in front of the hospital.
The incident of the French camp is an interesting poem.
The rising incidence of cross border terror attacks has annoyed India.

Question 4.
scraping – scrapping – scrubbing
Answer:
Scraping of NEET may help rural students to get into Government Medical Colleges.
Anil Ambani’s company suffered through several bankruptcies, resulting in scrapping of many deals.
After scrubbing the stains for a long time, she found that the stain was gone but the cloth had got damaged.

Question 5.
accept – except – expect
Answer:
Don’t accept bribe.
Except Raghu all had paid the fees for NEET coaching.
Those who don’t expect much gain a lot in life.

Question 6.
lesson – lessen – lesion
Answer:
Dr. Barnard leamt a great lesson from the two disabled kids.
This medicine will lesson the pain.
The protruding thorn caused a lesion in his forearm.

Question 7.
severe – sever – sewer
Answer:
Prime Minister of India warned Pakistan of severe consequences of Pulwana terror attack.
It is difficult to sever the relationship on flimsy reasons.
Madras Corporation has invested a lot in laying underground sewer.

Question 8.
raise – rise – rice
Answer:
The philanthropist raised funds for the benefit of relations of those warriors who laid down their lives in Pulwana.
“The rise and fall of Roman empire” is an interesting book.
Many poor people do not use the ration rice but exchange it for other items.

Question 9.
quiet – quite – quit
Answer:
Sheela lives in her quiet cottage in Chengalpattu with her retired husband.
Life in Coimbatore is quite expensive.
Some players, who are jealous of Dhoni’s reputation, want him to quit before the world cup matches.

Question 10.
final – finale – feline
Answer:
You must be relaxed before the final examination.
The performance of the child with autism in super singer in the grand finale was amazing.
I have a young feline pet who is very naughty.

(d) Fill in the blanks with the words given in brackets.

[profound, amusement, confidence, agony, solace, intrepid, disfigured, perforated]

1. Theatrical plays were a main source of ________ before the advent of television.of police.
2. The ________ warriors of the Spartan Army marched into battle against a powerful enemy.
3. The ________ of parents finally came to an end when their lost child was found with the help
4. Social media has brought about a ________ impact on the lives of millennial.
5. The tyres of the car got ________ when the vehicle rolled over the rusted nails scattered on the road.
6. Thomas Alva Edison did not lose his ________ , even after facing a series of experimental’ failures in his quest to discover tungsten.
7. Many victims of the pipeline explosion in an oil refinery were left permanently ________
8. The old lady found ________ in the company of the children in the neighbourhood

Answer:

1. amusement
2. intrepid
3. agony
4. profound
5. perforated
6. confidence
7. disfigured
8. solace

(e) Form a phrase with each of the

1. muscle + pain – muscular pain
2. skeleton + system – skeletal system
3. nerve + disorder – nervous disorder
4. digestion + enzymes – digestive enzyme
5. surgery + instruments – surgical instruments
6. agony + experience – agonising experience
7. glory + victory – glorious victory
8. fancy + idea – fanciful idea
9. emotion + song – emotional song
10. sense + issue – sensitive issue

(f) Fill the empty boxes with suitable words under each word class.

NOUN	VERB	ADJECTIVE	ADVERB
amusement
	appreciate
success
			proudly
		hopeful

Answer:

NOUN	VERB	ADJECTIVE	ADVERB
amusement	amuse	amusing	amusingly
appreciation	appreciate	appreciative	appreciably
success	succeed	successful	successfully
pride	be proud	proud	proudly
hope(n)	hope (v)	hopeful	hopefully

(g) Spot the errors in the following sentences and rewrite them correctly.

Question 1.
My grandfather is well-known in the village for his nobel deeds.
Answer:
My grandfather is well-known in the village for his noble deeds.

Question 2.
I had my evening meals in a restaurant near my office.
Answer:
I had my dinner in a restaurant near my office.

Question 3.
The Boss had full confidence on his Manager for successful completion of the project.
Answer:
The Boss had full confidence in his Manager for successful completion of the project.

Question 4.
After the complicated surgery, the patient hoped of complete recovery.
Answer:
After the complicated surgery, the patient hoped for complete recovery.

Question 5.
The new health care scheme announced by the Government will bring relief to the children suffering with acute tuberculosis.
Answer:
The new health care scheme announced by the Government will bring relief to the children suffering from acute tuberculosis.

Question 6.
In spite of his poverty and setbacks, he was able to launch his dream carrier.
Answer:
In spite of his poverty and setbacks, he was able to launch his dream career.

Listening Activity

Listen to the passage being read out. Based on your understanding, complete the statements given below with appropriate answers.

Boredom
We have all experienced boredom sometime or the other. Boredom occurs when a person is unable to stay attentive. It is something more than an unpleasant feeling. It can make you angry and frustrated and lead to negative physical health consequences.

How boredom affects one physically
A study reveals that when a person is affected by acute boredom his eyelids droop and the face assumes a frown. There is a gradual loss of ability to coordinate movements. These symptoms are accompanied by mental fatigue and a slowing down of thought processes. A bored person at work is likely to make many more errors than one who is not bored. We should never let boredom take charge. There are several easy ways to overcome boredom.

Here are a few practical suggestions:

• Set goals for yourself, work towards them.
• Develop an interest in hobbies and crafts.
• Socialize, stay in the company of cheerful people.
• Take up a charitable cause.
• Exercise regularly.

Coming out of boredom will feel like breaking free from a cold, dark room into the outdoors on a warm, sunny day

Questions:

1. Boredom occurs when a person is unable to
2. ______ and ______ are emotional consequences of boredom.
3. Two physical signs of acute boredom are (a) ______ (b) ______
4. How does boredom affect the quality of work a person does?
5. Mention two ways by which one can overcome boredom. (a) ______ (b) ______

Answers:

1. stay attentive
2. Anger, frustration
3. (a) Drooping eyelids, (b) Frown
4. A bored person commits a lot of errors in his work.
5. (a) Goal setting, (b) stay in the company of cheerful people

Speaking Activity

1. You are rushing to attend to an important work and you witness an accident on your way. Will you go to the rescue of the injured person? Share your views with the class.
Answer:
There is nothing more valuable than a human life. I would suspend whatever important work I have in hand and call 108 and summon the ambulance. I will try to get the contact number . of the next of kin of the injured person if he/she is conscious. I will inform them the incident without causing panic. I will accompany the injured person to the hospital. In our NSS team, we have blood donors whatsapp group. I will find out if blood is required and inform in my blood donor group to ensure blood donation to the injured person. Only after the doctor confirms that the injured person is out of danger, I will leave the place.

I draw my inspiration to help people in distress from the life of Abraham Lincoln. He was a budding lawyer. He had to walk 2 hours to reach the court. On his way, he found a pig sinking in mud and grunting in agony. It was scared that it would die. Abraham Lincoln was on his only best dress. He just got into the mud, lifted the pig and left him safely on the hard surface. As he reached the court everyone laughed at his dirty coat but Abraham Lincoln just smiled and argued well in favour of his innocent client and got him acquitted too.

2. Every person should take up the responsibility to serve the society in his or her own way. Discuss the various ways in which you can serve the society.
Answer:
Elders keep on telling that students should not enter into politics, instead should focus on studies and scoring high marks. I’ve a sincere feeling that life is more than marks. Students can do a lot of service which would certainly make the society a better place. Though the state government has banned the use of single use plastics, it is widely used. Hot sambar curry and tea are packed in small plastic bags. People just use them.

We can generate awareness to all sellers and buyers in places where large number of people gather that by eating hot things from single use plastic bags, we are increasing the chances of getting cancer.The plastic covers thrown away blatantly does not degenerate for hundreds of years, it does not allow percolation of rain water and soon after rain, we suffer from water scarcity. The rain water runs away and joins the oceans.

We can tell the importance of rain water harvesting and make farming practises economically viable. During elections we can campaign for judicious use of voting rights to elect a person who has both the capacity and love for citizens to serve without expecting anything in return. We can also spread hygiene among the masses.

(a) Pair work: Practise the dialogue with another student. Then write a similar dialogue between a student and the class teacher regarding an educational trip.

Teacher : We plan to go on an excursion.
Student A : Where sir?
Teacher : We will discuss and finalise it today.
Student B : Sir, how about Vandaloor Zoo?
Teacher : it is a very hot season now, all the animals will be taking shelter under distant trees
Student A : We can’t see all of them. What about Vedanthangal?
Teacher : No, only after the rains foreign birds visit it.
Student B : What about Birla Planetarium?
Teacher : Fine, that is a good choice. We will go there next week.

(b) Build a conversation for the following situations with a minimum of five exchanges.

1. A passenger and a railway staff regarding the cancellation of the reserved tickets.

Passenger : Sir, I want to cancel a ticket.
Clerk : When was the journey planned?
Passenger : Sir, next Monday, the 23rd of March.
Clerk : Well, still three days are there.
Passenger : This is my ticket and take the filled in cancellation form. Tell me sir, how much will you deduct.
Clerk : Well, we will deduct only the reservation charges and you will get the balance amount. Take the money and count it, it is Rs. 280/ Thank you sir.

2. Two friends about the NSS camp which they are going to attend.

Tarun : I am leaving for NSS camp.
Kavin : How many days?
Tarun : One week.
Kavin : Where are you going?
Tarun : To Madagupatti.
Kavin : That village has no electricity.
Tarun : Yes. That’s why we are going there to help them in small possible ways.
Kavin : What will you do?
Tarun : We will deepen the lakes, clean the streets, remove the unneeded thorny
Kavin : bushes from the school and temple premises.
Tarun : That is a lot of work.
Kavin : We will conduct a medical camp with the help of doctors too.
Tarun : Sounds exciting. Can I join you?
Tarun : No, brother. Wait till you become old enough to become a volunteer.
Kavin : Well, I’m ten years old already.

3. A salesman and a customer at an electronic shop.

Customer : Good evening sir.
Salesman : Good evening. What shall I do for you?
Customer : We bought a Smart TV last week. It is not working properly.
Salesman : May be the problem is with your dish antenna or cable connection.
Customer : No, the fault is only with the TV.
Salesman : Please leave your address and phone number. I will send the TV mechanic in half an hour.
Customer : Thanks.

4. A father and his daughter about the advantages of the habit of newspaper- reading.

Father : Maydhini my dear, you must read newspapers daily.
Maydhini : How does it help dad? Will they ask questions in the examination from them?
Father : No, newspapers help you know about the world.
Maydhini : I’m a small girl. Are there sections in a newspaper which will have things to my taste?
Father : Why not? You read young world in the Hindu. In fact, children write and send things to be published here.
Maydhini : Can I send my puzzles and paintings?
Father : Yes, of course.
Maydhini : Ok dad, I will read newspaper everyday.

(c) Extend the conversation with two more relevant exchanges.

Question 1.
Receptionist : Good evening, sir. Welcome to Chennai.
Traveller : I would like to book a deluxe room in your hotel for 3 days.
Receptionist : (i) ________
Traveller :(ii) ________
Receptionist :(iii) ________
Traveller :(iv) ________
Answers:
(i) How many people intend to stay sir?
(ii) Three members
(iii) For Deluxe A/C room the charge is Rs. 5000/- per day and for Non A/C it is Rs. 3500/-
(iv) If it includes GST, please book Deluxe A/C room. Take my card and book the room.

Question 2.
Student : Good morning, sir. May I come in?
Teacher : Good morning, why are you late today?
Student : (i) ________
Teacher : (ii) ________
Student : (iii) ________
Teacher : (iv) ________
Answers:
(i) There was a traffic jam sir.
(ii) Why?
(iii) A political party had staged a protest blocking the vehicles. I was stranded and helpless.
(iv) Oh! The politicians should keep in mind the difficulties common people face when doing such protests. Okay, go to your seat now.

Reading

Read the passage given below and answer the questions that follow.

Humans have long been fascinated by fiction. We experience excitement in assigning supernatural power to imaginary characters in fictional stories – and so we have Spider man, Batman, He man, Titans and many more. The ‘Cyborg’ was an offshoot of such wild imagination of humans to invest our species with superhuman powers. Today, the Cyborg is no more an imaginary organism. We are living in a world where a sizeable population of humans have merged their bodies with technological implants. The term ‘Cyborg’, short for ‘cybernetic organism’, was coined to describe a man, whose body is implanted with technological devices to supplement and substitute body functions.

Cyborgs include people with cardiac pacemakers, contact lenses, bionic ears and eyes, prosthetics and so on. In other words, a cyborg is partly human and partly machine. The technological innovations in the field of medicine and healthcare augment humans with machines, producing a beta version of the human body. The advent of brain machine interfaces is certain to blur the boundary between humans and machines. Scientists are working hard to find a technique for age reversal too. People do not want to die, so mankind is striving to get to the final frontier, which is development of machines and devices that would accord man immortality.

The needs of humans are not limited. As time passes, food habits change, thinking patterns change, and even appearances change. We are about to travel by driverless, folly automated vehicles. Computers and smart phones have become our masters. The more we depend and merge with technological advancements, the more the humanness in us slowly erodes. Intelligence is sought to be infused into machines and robotics are designed in such a way to give man a virtual human companion.

The field of artificial intelligence is overtaking the human brain and many fear that it could even harm the human race. Despite certain limitations and potential threats, many believe that cyborgs will be the next step in the evolution of mankind. The amalgamation of man and machine is sure to add a new dimension to the life of mankind and this will prove to be the ‘biggest evolution in Biology’ since the emergence of life, four billion years ago.

Question (a)
Account for the popularity’ of characters with supernatural powers.
Answer:
Human beings want to achieve things far above their natural capacity. The superheroes like Spiderman, Batman and He man do great feats on screen. So, people like them.

Question (b)
Who is referred to as a ‘Cyborg’?
Answer:
Cyborg is a man whose body is implanted with technological devices to supplement and substitute body functions.

Question (c)
What is expected to happen with the advent of the brain machine interface?
Answer:
The advent of brain machine interface is certain to blur the boundary between machines and humans.

Question (d)
The needs of humans are not limited. How is this statement elaborated in the passage?
Answer:
As the time passes, food habits change, thinking patterns change, even appearances change. We are about to travel by driverless, folly automated
vehicles.

Question (e)
How can a machine turn into a virtual companion for humans?
Answer:
Intelligence is sought to be infused into machines and robots are designed in such a way to give man a virtual human companion.

Question (f)
Explain the flipside of the rapid technological advancement.
Answer:
The more we depend and merge with technological advancements, the humanness in us slowly erodes.

Question (g)
Identify the word in para 2 which means ‘everlasting life’.
Answer:
Immortality in para 2 means ‘everlasting life’.

Question (h)
Which of the following words is synonymous with ‘amalgamation’?
(a) recreation
(b) integration
(c) exploration
(d) proposition
Answer:
(b) integration

Question (i)
Which of the following options is the antonym of the word ‘advent’?
(a) drawback
(b) dispute
(c) departure
(d) danger
Answer:
(c) departure

Question (J)
Find out the word which is the antonym of ‘natural’ in para 2.
Answer:
Prosthetics

Grammar

Tenses
Task 1
Change the following sentences into Passive Voice.

Question (a)
The Governor inaugurated the exhibition at ten o’ clock.
Answer:
The exhibition was inaugurated by the Governor at ten o’ clock

Question (b)
The crowd expected their leader to arrive early in the morning.
Answer:
The leader was expected to arrive early in the morning.

Question (c)
Who taught her Computer Science?
Answer:
By who was she taught Computer Science.

Question (d)
They unanimously named Ravi the captain of team.
Answer:
Ravi was named the captain of the team unanimously.

Question (e)
The President gave the commander an award.
Answer:
An award was given to the commander by the president.

Question (f)
Do not tell a lie.
Answer:
Let not a lie be told.

Question (g)
Please open the door.
Answer:
Let the door be opened please.

Question (h)
It is time to stop the work.
Answer:
It is time for the work to be stopped.

Question (i)
They say he is a spy.
Answer:
It is said that he is a spy.

Question (j)
One should keep one’s promise.
Answer:
Promise must be kept.

Question (k)
People burn a great deal of w ood in winter.
Answer:
A great deal of wood is burnt in the winter by people.

Question (l)
Where had you kept the book?
Answer:
Where had the book been kept by you?

Question (m)
When did you feel the tremors?
Answer:
When was the tremors felt by you?

Question (n)
How did you do the experiment?
Answer:
How was the experiment done by you?

Question (o)
Whose car did someone park in front of your gate?
Answer:
Whose car was parked in front of your gate?

Task 2
Change the following sentences into Active Voice.

Question (а)
The smuggler has been nabbed by the police.
Answer:
The police have nabbed the smuggler.

Question (b)
By whom were you interviewed?
Answer:
Who interviewed you?

Question (c)
Why were you scolded by your parents?
Answer:
Why did your parents scold you?

Question (d)
Not a word was spoken by the convict in self-defence.
Answer:
The convict spoke not a word in self-defence.

Question (e)
Good news is expected shortly.
Answer:
We expect good news shortly.

Question (f)
The mail has just been received.
Answer:
We have received the mail just now.

Question (g)
Sundari has been taken to hospital by her husband.
Answer:
Sundari’s husband has taken her to the hospital.

Question (h)
Our television is being repaired now.
Answer:
We are repairing our TV now.

Question (i)
Sweets have not been distributed to children by the organisers.
Answer:
The organization have not distributed sweets to children.

Question (j)
Prizes were being given by the chief guest.
Answer:
The chief guest was giving prizes.

Question (k)
Nobody has been seen in the library this week.
Answer:
The librarian has not seen anybody in the library this week.

Question (l)
Nobody would have known the truth if you had not disclosed it.
Answer:
Nobody will know the truth if you have not disclosed it.

Question (m)
You are advised to help the poor and needy.
Answer:
Help the poor and the needy.

Question (n)
You are requested to make a cup of tea for the guest.
Answer:
Please make a cup of tea for the guest.

Interrogations Or Questions

Task 1
Add suitable question tags to the following sentences and punctuate properly.

Question 1.
The children are very happy today.
Answer:
The children are very happy today, aren’t they?

Question 2.
You have not returned my books yet.
Answer:
You have not returned my books yet, have you?

Question 3.
We enjoyed the trip very much.
Answer:
We enjoyed the trip very much, didn’t we?

Question 4.
Let’s clean the shelves this weekend.
Answer:
Let’s clean the shelves this weekend, shall we?

Question 5.
My mother rarely travels by bus.
Answer:
My mother rarely travels by bus, does she?

Question 6.
Somebody must bell the cat.
Answer:
Somebody must bell the cat, mustn’t they?

Question 7.
Anita never comes late to office.
Answer:
Anita never comes late to office, does she?

Question 8.
Iam always the winner.
Answer:
I am always the winner, aren’t I?

Question 9.
Don’t commit this mistake again.
Answer:
Don’t commit this mistake again, will you?

Question 10.
There is a pharmacy near that bus stand.
Answer:
There is a pharmacy near that bus stand, isn’t it?

Question 11.
Bacteria can never survive in extreme weather conditions.
Answer:
Bacteria can never survive in extreme weather conditions, can it?

Question 12.
I am not as smart as you are.
Answer:
I am not as smart as you are, am I?

Question 13.
The boys broke the window pane last evening.
Answer:
The boys broke the window pane last evening, didn’t they?

Question 14.
Leaves wither during autumn.
Answer:
Leaves wither during autumn, don’t they?

Question 15.
You should add a little salt to the buttermilk.
Answer:
You should add a little salt to the buttermilk, shouldn’t you?

Task 2
Correct the error found in the question tag in each of the following.

Question 1.
The evil doers cannot cross the path of truth, can’t they?
Answer:
The evil doers cannot cross the path of truth, can they?

Question 2.
The vegetables in the fridge are still fresh, aren’t it?
Answer:
The vegetables in the fridge are still fresh, aren’t they?

Question 3.
The village head understood the intention of the politician, doesn’t he?
Answer:
The village head understood the intention of the politician, didn’t he?

Question 4.
I claim to be a person of faith and prayer, aren’t I?
Answer:
I claim to be a person of faith and prayer, don’t I?

Question 5.
The employees are seldom allowed to meet their boss, aren’t they?
Answer:
The employees are seldom allowed to meet their boss, are they?

Question 6.
Let’s organize a trip to Goa, can we?
Answer:
Let’s organize a trip to Goa, shall we?

Question 7.
The landlady will charge me for the damage, shan’t she?
Answer:
The landlady will charge me for the damage, won’t she?

Question 8.
Both the sisters have left for Canada, aren’t they?
Answer:
Both the sisters have left for Canada, haven’t they?

Question 9.
That’s definitely not the right thing to do in this situation, isn’t that?
Answer:
That’s definitely not the right thing to do in this situation, isn’t it?

Question 10.
We needn’t apply for a bank loan, do we?
Answer:
We needn’t apply for a bank loan, need we?

Question 11.
The Chief Guest spoke a few words, did he?
Answer:
The Chief Guest spoke a few words, didn’t he?

Question 12.
The rhinoceros has a horn made of keratin, haven’t they?
Answer:
The rhinoceros has a horn made of keratin, hasn’t it?

Writing

Story Writing
Task 1
Expand the following outlines into complete stories and supply a suitable title for each.

Question 1.
Big cotton merchant – owned a factory – many employees – one day a heap of cotton stolen – no clue – merchant’s secretary assured to find out – asked him to host dinner – invite all workers – merchant agreed – middle of feast – secretary suddenly shouted – cotton sticking to hair of thieves – the guilty dusted their heads – tried to clear – caught in the trap – punished.
Answer:
Once upon a time there was a big cotton merchant. He owned a factory. Many employees were working in it. One day a bale of cotton was stolen. The merchant had no clue as to who might have stolen the bale of cotton. The merchant’s secretary assured him that she will find out the thief. She asked him to host a dinner and invite all the workers. Merchant agreed. When the feast was in progress, the secretary shouted suddenly, “There is cotton sticking on to the hair of the thieves. The guilty ones involuntarily dusted their heads to clear it. They were caught unawares. They were sacked from their jobs as a punishment for their theft.

Question 2.
Mr. X, a rich businessman – runs a company – always very busy with office work – one day his son – 10 years old – approaches dad and asks – how much he earns in one hour – father gets furious – boy persuades – father says Rs. 500 – immediately son asks for ’ Rs. 300 – father shouts – wasting money on toys – son leaves to his room crying – father feels bad – thinks might need some stationery – enters boy’s room and gives money – boy becomes happy – takes some crumpled notes – under his pillow – counts everything together – total Rs.500 – gives it to dad – wants to buy – one hour of his time – father realizes his mistake – feels sorry and guilty – hugs son – closes all office files – takes him on a picnic – decides to spend more time w ith near and dear ones.
Answer:
Mr. X is a rich business man. He runs a company. He is always busy with his office work. Like a snail, he carries his office work everywhere. His young son one day asks him, “How much do you earn in an hour?” The father gets furious and refuses to reply. But the son coaxes him to give the answer. The father grudgingly says, “Rs. 500/- an hour.” Immediately the boy asks his dad to give him three hundred rupees. He reprimands his son for trying to waste his hard earned money. The boy cries and goes to the bedroom without dinner. The father feels guilty.

He realises that the boy might really need the money for buying some stationary. He gives him three hundred rupees. The next moment, the boy collected all the crumpled notes from under his pillow. Dad asks, “why did you ask for money when you have so much?” Without replying the boy handed him Rs. 500/- and said, “this is the amount you earn in an hour, I have given you that money. Now will you spend an hour with me?” The father realises his mistake. He suspends his work and takes the boy out on a picnic. Then onwards he regularly spends time with near and dear ones.

Task 2
Continue and complete the following stories and suggest suitable titles for the same.

Question 1.
A rich man had a neighbour who was suffering from acute poverty. The rich man was
proud of his wealth and treated his poor neighbour with disrespect and derision. One day, a fortune-teller told the rich man that all his wealth would be possessed by his neighbour within a month. The rich man became greatly worried and spent sleepless nights. He did not know how to safeguard his wealth round the clock. Suddenly he thought of a plan. He disposed of everything he had and with all that money, he bought a large, precious diamond. He sewed up the diamond in his turban. He proudly said to himself, “Now, there’s no way. My poor neighbour can never secure my wealth. The words of the fortune-teller will prove false.” _______
Answer:
Once, he had to go to a distant town with his merchandise. His poor neighbour’s wife came to borrow the turban on the occasion of attending a marriage function in the town. The rich man’s wife wasn’t aware of the value of the turban. She gave it to him. This poor neighbour wore the turban proudly and attended the wedding. When he was returning, he had a feeling that something hurt like a stone. On reaching home, he took a needle and untied the stitches.

He was amazed at the hidden diamond. He threw the turban into the rich man’s compound and vacated the house at night and ran away with his family. When the rich man returned, he saw the turban lying in his compound. He shouted at his wife for not keeping the turban in its proper place. He hurriedly took the turban and touched at the bottom. He was shocked to find the diamond missing. He realized that the fortune-teller w as proved right.
Title: The Rich man’s turban and the fortune-teller

Question 2.
Four friends decided to go to a restaurant for dinner. They ordered an extra-large pizza with grated cheese and other choice toppings. The next 20 minutes seemed to be too long a time. Their eyes widened and their mouth watered, when the server brought the steaming hot pizza and placed it on the table. They could barely control the drool. Simultaneously, all the four hands pulled at a slice from the plate, their faces beaming with a victorious grin. Silence prevailed as they were absorbed in the taste of their favourite food. They relished every mouthful to the core and savoured the taste of each topping with a smile of approval. Soon, the plate was empty and clean with no trace.

The boys dabbed their mouths and wiped their hands with tissues. Mission accomplished, they leaned back with immense joy and satisfaction not knowing, it would be short-lived. The waiter arrived with the bill. Joseph, who had brought the others to the restaurant for a treat casually slipped his hand into his pocket to get his wallet. He gave a soft shriek accompanied by an expression of dismay and utter disbelief. He exclaimed, “It’s not there! Someone has pinched my wallet! What are we to do now?”______
Answer:
One of the friends sprang up saying, “take it easy. Our government has made things simple for us. See we are going digital, isn’t it. So now consider the problem we encountered to be solved.'”Joseph, still will, asked, “What do you mean?” Simple, let me pay the bill through the Paytm app in my mobile. The money will be transferred instantly. Everyone was overjoyed at the turn of the event. They walked out with their pride being saved. It is important to be safe and to have an alternative option in unwarranting situations.

UTHIRA is an event conducted by the NSS unit of ABC Hr Sec School. Imagine you are a volunteer and help a parent fill in the following registration form. (Invent necessary details)

Question 1.

Answer:

Question 2.

Answer:

In Celebration of Being Alive About The Author

Christiaan Neethling Barnard was bom in rural South Africa in 1922 to poor parents as their fourth child. After the loss of his brother to heart ailment, he resolved to become a doctor. He performed the world’s first successful human heart transplantation in the year 1967. He instantly gained worldwide recognition. He has penned 14 books and 235 scientific articles. Some of his prominent books are, ‘The Best Medicine’ and ‘The Faith’. He has been conferred with 11 honorary doctorates, and 36 International Awards. In his late years, he established the Christiaan Barnard Foundation to serve and promote the cause of underprivileged children throughout the world and died at the age of 78.

In Celebration of Being Alive Summary in English

Introduction
The piece “In celebration of being alive” portrays a new dawn of understanding of human life after he met with an accident.

Is there divinity behind suffering?
Christian Barnard was a Doctor who performed the world’s first human heart transplant operation. He found the suffering of children particularly heart-breaking. During his lifetime, Christian Barnard and his wife met with an accident while they were crossing the road. His eleven ribs were broken and lung was perforated. His wife had a badly fractured shoulder. Both experienced fear and agony in the hospital. He totally disagreed with his father’s view that God tests human beings and suffering ennobles a person.

A life changing event
His brother died of an abnormal heart. This incident brought awareness to Dr. Barnard of the sufferings of little children. Several years ago. Dr. Barnard witnessed an incident at Cape Town’s Red Cross Children’s Hospital. That event made him realize that he was missing something in all his thinking about suffering.

Grand prix in the hospital
While he and his wife were undergoing treatment in the hospital, one day a nurse left a breakfast trolley unattended. Two children who were patients took charge of the trolley. One was blind and the other was crippled. One of them played the role of a driver and the other played that of a mechanic. The blind boy provided motor power, and the crippled sat on the lower deck and steered the trolley. The rest of the patients laughed and gave shouts of encouragement. The nurse and the ward sister finally took control over the situation

Harsh truth about the heroes
The mechanic was a seven-year old boy who was admitted in the hospital due to serious bums on the upper part of his body and lost both of I .eyes.
The driver had a harmful tumor and his shoulder and arm were amputated with little hope of recovery. Both did not lose hope. They enjoyed life despite its stings

Learning life’s lesson
These two children taught Dr. Barnard a profound lesson that the business of living is in the celebration of being alive and not just something for pleasure, amusement and recreation. They made it clear that being alive is more important than the suffering they are experiencing. They made him understand that sufferings does not ennobles humans but celebrating life inspite of the sufferings makes it noble

Conclusion
People try to understand life always with their own experiences. Only when a misfortune strikes, they look at life from a different angle. They may have scars but then they understand what makes life noble.

In Celebration of Being Alive Summary in Tamil

முன்னுரை:
ஆசிரியர் தாம் விபத்துக்குள்ளாகிய பிறகு எழுதிய இந்தக் கட்டுரை மனித வாழ்க்கையின் புது அர்த்தத்தை விவரிக்கக் கூடிய ஒரு விடிவெள்ளியாகும்.

நாம் படும் வேதனைக்குப் பின்னால் தெய்வீக காரணம் உள்ளதா?
கிரிஸ்டியன் பார்னாட் உலகத்திலேயே முதன் முதலாக மனித இதய மாற்று சிகிச்சையை மேற்கொண்ட மருத்துவர் ஆவார். அவர், குழந்தைகள் இதய நோயால் அவதிப்படுவதை கண்டு மனமுடைந்து போனார். அவர்தம் மனைவியுடன் சாலை ஒன்றை கடக்கும் போது இருவரும் விபத்திற்கு உள்ளானார்கள். அவரின் 11 விலா எலும்புகள் முறிந்தன மற்றும் நுரையீரல் சல்லடைத் துளையாகிப் போனது. அவர் மனைவிக்கு தோற்பட்டை முறிந்து போனது. மருத்துவமனையில் பயமும், வேதனையும் அவரைத் தொற்றிக் கொண்டது. ஆசிரியர் தம் தகப்பனாரின் கருத்தான ‘கடவுள் மனிதனை சோதிப்பான், அந்த சோதனை அவனை மேன்மைபடுத்தும்’ என்பதை ஆட்சேபித்தார்.

ஒரு வாழ்க்கை மாறும் நிகழ்வு
அவரின் சகோதரர் அசாதாரண இதயத்துடன் பிறந்ததால் இறந்து போனார். இந்த சம்பவம் குழந்தைகள் படும் துன்பத்தை உணர்த்த வல்லதாக அமைந்தது. பல வருடங்களுக்கு முன்னதாக டாக்டர். பார்னாட் கேப் டவுனின் செஞ்சிலுவை குழந்தைகள் மருத்துவமனையில் ஒரு சம்பவத்தை காண நேரிட்டது. இந்த சம்பவம் அவர் வேதனை என்று தாம் நினைப்பதில் ஏதோ ஒரு குறைபாடு இருப்பதை உணர்ந்தார்.

கிரிஸ்டியன் பார்னாட்டின் மருத்துவமனை அனுபவம்
ஆசிரியரும் அவரது மனைவியும் மருத்துவமனையில் சிகிச்சை மேற்கொண்டிருந்த போது ஒரு சமயம் செவிலிப் பெண் சிற்றுண்டி வண்டியை நிறுத்தி சென்று விட்டார். அங்கு அனுமதிக்கப் பட்டிருந்த இரண்டு நோயாளிகளான சிறுவர்கள் அந்த சிற்றுண்டி வண்டியைக் கையாண்டனர். ஒருவன் குருடன், மற்றவனோ முடவன். அதில் ஒரு சிறுவன் தன்னை ஓட்டுனர் போல் சித்தரித்துக் கொண்டான். மற்றொருவன் இயந்திர தொழில் நிபுணராக வேடம் ஏற்றுக் கொண்டான், கண் இல்லாத சிறுவன் வண்டியை தள்ள வண்டியின் அடித்தளத்தில் அமர்ந்த முடவனான சிறுவன் வண்டியை செலுத்துவது போல் பாவனை செய்தான். இதைக் கண்ட இதர நோயாளிகள் உற்சாகக் கூக்குரல் எழுப்பினார்கள். ஒரு வழியாக செவிலிப் பெண்ணும் மற்றும் வார்ட் சிஸ்ட்டரும் நிலைமையை கட்டுக்குள் கொண்டு வந்தனர்.

இரு கத இயந்திர தெரு வயது சிங்கம் கருகி
இரு கதாநாயகர்களைப் பற்றிய திடுக்கிடும் தகவல்:
இயந்திர தொழில் நிபுணராக தன்னை சித்தரித்துக் கொண்ட 7 வயது சிறுவனின் தீப்புண் காயங்களால் உடலின் மேல் பாகம் கருகி கண் பார்வையை இழந்த நிலையில் மருத்துவமனையில் அனுமதிக்கப்பட்டான். தன்னை ஓட்டுநர் போல் சித்தரித்துக் கொண்டு இருந்த சிறுவனுக்கு தோள்பட்டை புற்றுநோய் பாதித்ததால் கை ஒன்று வெட்டி எடுக்கப்பட்டு உடல்நலம் தேற சிறிது வாய்ப்பே இருந்தது. ஆனால், இருவரும் நம்பிக்கையை இழக்கவில்லை. வாழ்க்கையின் வலியைத் தாண்டிய குதூகலம் அவர்களிடம் காணப்பட்டது.

வாழ்க்கை தந்த பாடம்:
இந்த இரு சிறுவர்களும் டாக்டர். பர்னாட் அவர்களுக்கு கற்றுத் தந்த ஆழ்ந்த உண்மை நாம் | உயிருடன் இருப்பதை கொண்டாடுவதே வாழ்க்கை. | அல்லாமல் சொகுசு. வேடிக்கை மற்றும் பொழுது போக்கு அல்ல என்பதாகும். அவர்கள் தாம் அனுபவிக்கின்ற வலியைக் காட்டிலும் உயிருடன் இருப்பதை |முக்கியத்துவம் வாய்ந்ததாக தெளிவுபடுத்தினர். வேதனைகள் மனிதனை மேம்படுத்துவதில்லை. மாறாக,வேதனைகளை அனுபவித்த பின் வாழ்வதின் சுகம் எளிதில் புலப்படுகிறது.

முடிவுரை:
தம் அனுபவத்தின் வாயிலாகதான் மனிதன் வாழ்க்கையை அறிந்து கொள்கிறான். துன்பப்படும் போது தான் வாழ்க்கையை வேறு கோணத்தில் பார்க்க முயல்கிறான். காயங்கள் அடைந்தபோதிலும் வாழ்க்கையை எது உன்னதப்படுத்துகிறது என்பதை அறிந்து கொள்கிறான்.

In Celebration of Being Alive Glossary

Textual:

Additional:

In Celebration of Being Alive Synonyms

Find out the synonym of the underlined word in each of the following sentences.

Question 1.
My thoughts have turned to the consideration of why people suffer.
(a) imagination
(b) curiosity
(c) absurdity
(d) careful thought
Answer:
(d) careful thought

Question 2.
Suffering seems cruelly prevalent in the world today.
(a) unique
(b) uncommon
(c) common
(d) fair
Answer:
(c) common

Question 3.
Of the rest, many will end up as mental or physical cripples
(a) people with sound bodies
(b) people with rare gifts
(c) people with disabilities
(d) people with perfect health
Answer:
(c) people with disabilities

Question 4.
My gloomy thoughts stem from an accident.
(a) ecstatic
(b) blissful
(c) depressed
(d) lofty
Answer:
(c) depressed

Question 5.
I experienced agony.
(a) relaxation
(b) liberation
(c) amusement
(d) extreme suffering
Answer:
(d) extreme suffering

Question 6.
Dr. Barnard had a perforated lung.
(a) damaged with holes
(b) inflated
(c) shrunk
(d) inebriated
Answer:
(a) damaged with holes

Question 7.
The wound healed around his neck.
(a) cursed
(b) cured
(c) curtailed
(d) contorted
Answer:
(b) cured

Question 8.
He had a malignant tumour of the bone.
(a) harmless
(b) friendly
(c) harm full
(d) helpful
Answer:
(c) harm full

Question 9.
His shoulder and arm were amputated.
(a) attached
(b) surgically cut off
(c) frustrated
(d) transplant
Answer:
(b) surgically cut off

Question 10.
There was little hope for his recovery.
(a) recuperation
(b) discovery
(c) loss
(d) damage
Answer:
(a) recuperation

Question 11.
He had full confidence in the mechanic.
(a) trust
(b) disbelief
(c) doubt
(d) scepticism
Answer:
(a) trust

Question 12.
Two children had given me a profound lesson.
(a) humble
(b) modest
(c) shallow
(d) very great
Answer:
(d) very great

Question 13.
The business of living is joy
(a) amnesia
(b) dyslexia
(c) dementia
(d) delight
Answer:
(d) delight

Question 14.
Life is not something for recreation.
(a) earnestness
(b) meditation
(c) entertainment
(d) distress
Answer:
(c) entertainment

Question 15.
Nor can we appreciate warmth if we have not
(a) heat
(b) wellness
(c) chillness
(d) wetness
Answer:
(a) heat

Question 16.
Suffering ennobles you.
(a) destroys
(b) dignifies
(c) distress
(d) encourages
Answer:
(b) dignifies

Question 17.
I see nothing noble in thrashing around in a sweat-soaked bed
(a) heated
(b) warmed
(c) drenched
(d) smellled
Answer:
(c) drenched

Question 18.
There was a grand finale of scattered plates.
(a) anti-climax
(b) climax
(c) boring end
(d) gloomy denouement
Answer:
(b) climax

Question 19.
They didn’t have sophisticated surgery.
(a) crude
(b) undeveloped
(c) cumbersome
(d) well-advanced
Answer:
(d) well-advanced

Question 20.
Something basic that w as full of solace for me.
(a) pity
(b) neglect
(c) abandon
(d) consolation
Answer:
(d) consolation

In Celebration of Being Alive Antonyms

Find out the antonym of the underlined word in each of the following sentences.

Question 1.
Why people should suffer.
(a) undergo pain
(b) struggle
(c) enjoy
(d) heal
Answer:
(c) enjoy

Question 2.
Suffering is cruelly prevalent.
(a) sarcastically
(b) tortuously
(c) mercilessly
(d) mercifully
Answer:
(d) mercifully

Question 3.
My gloomy thoughts stem from an accident.
(a) sad
(b) murky
(c) happy
(d) vague
Answer:
(c) happy

Question 4.
I experienced agony.
(a) calamity
(b) delight
(c) horror
(d) misery
Answer:
(b) delight

Question 5.
Suffering ennobles you.
(a) exalts
(b) dignifies
(c) praises
(d) humilites
Answer:
(d) humilites

Question 6.
This trolley was commandeered by an intrepid crew of two.
(a) timid
(b) bold
(c) daring
(d) adventurous
Answer:
(a) timid

Question 7.
They did not have sophisticated heart surgery.
(a) advanced
(b) well-developed
(c) backward
(d) cultured
Answer:
(c) backward

Question 8.
He was a walking horror.
(a) disgust
(b) delight
(c) shock
(d) fear
Answer:
(b) delight

Question 9.
They go through mutilating surgery.
(a) crippling
(b) paralysing
(c) maiming
(d) heating
Answer:
(d) heating

Question 10.
There was a grand finale.
(a) end
(b) climax
(c) exciting finish
(d) begining
Answer:
(d) begining

Question 11.
It was solace for me.
(a) anguish
(b) cheer
(c) consolation
(d) reassurance
Answer:
(a) anguish

Question 12.
The trolley was commandeered by the two boys.
(a) hijacked
(b) snatched
(c) usurped
(d) abandoned
Answer:
(d) abandoned

Question 13.
The wound healed around his neck.
(a) cured
(b) worsened
(c) got healthy again
(d) returned to normalcy
Answer:
(b) worsened

Question 14.
He had full confidence.
(a) trust
(b) belief
(c) disabilities
(d) distrust
Answer:
(c) disabilities

Question 15.
Two children had given me profound lesson.
(a) deep
(b) great
(c) intense
(d) superficial/hollow
Answer:
(d) superficial/hollow

Question 16.
The business of living is the celebration of being alive.
(a) commemoration
(b) praise
(c) honour
(d) criricism
Answer:
(d) criricism

Question 17.
Business of living is not just for joy.
(a) bliss
(b) sorrow
(c) indulgence
(d) elation
Answer:
(b) sorrow

Question 18.
We can’t appreciate light if we haven’t known darkness.
(a) criticise
(b) acknowledge
(c) welcome
(d) be obliged
Answer:
(a) criticise

Question 19.
Living is not for just recreation.
(a) dalliance
(b) amusement
(c) drudgery
(d) pastime
Answer:
(c) drudgery

Question 20.
He was laughing.
(a) crying
(b) chuckling
(c) whooping
(d) simpering
Answer:
(a) crying

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.1

12th Maths Exercise 1.1 Answers Question 1.
Find the adjoint of the following:

Solution:

Exercise 1.1 Class 12 Maths State Board Question 2.
Find the inverse (if it exists) of the following:

Solution:
For a matrix A, \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{A}|}(\mathrm{adj} \mathrm{A})\). Where |A| ≠ 0. If |A| = 0 then A is called a singular matrix and so \(\mathrm{A}^{-1}\) does not exist.

12th Maths Exercise 1.1 Question 3.
If F(α) = \(\left[\begin{array}{ccc}{\cos \alpha} & {0} & {\sin \alpha} \\ {0} & {1} & {0} \\ {-\sin \alpha} & {0} & {\cos \alpha}\end{array}\right]\) show that \([\mathrm{F}(\alpha)]^{-1}=\mathrm{F}(-\alpha)\)
Solution:
Let A = F (α)
So \([\mathrm{F}(\alpha)]^{-1}=\mathrm{A}^{-1}\)
Now

12th Maths Chapter 1 Exercise 1.1 Question 4.
If A = \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]\) show that A² – 3A – 7I₂ = O₂. Hence find A^-1.
Solution:
A = \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]\)

To Find A-1
Now we have proved that A² – 3A – 7I₂ = O₂
Post multiply by A^-1 we get
A – 3I – 7A^-1 = O₂

12th Maths 1.1 Exercise Question 5.
If \(\mathbf{A}=\frac{1}{9}\left[\begin{array}{ccc}{-8} & {1} & {4} \\ {4} & {4} & {7} \\ {1} & {-8} & {4}\end{array}\right]\) prove that A^-1 = A^T
Solution:

12 Maths Exercise 1.1 Question 6.
If \(\mathbf{A}=\left[\begin{array}{rr}{8} & {-4} \\ {-5} & {3}\end{array}\right]\), verify that A(adj A) = (adj A)A = |A| I₂
Solution:

12th Maths 1st Chapter Exercise 1.1 Question 7.
If \(\mathbf{A}=\left[\begin{array}{ll}{3} & {2} \\ {7} & {5}\end{array}\right]\), and \(\mathbf{B}=\left[\begin{array}{cc}{-1} & {-3} \\ {5} & {2}\end{array}\right]\) verify that (AB)^-1 = B^-1 A^-1.
Solution:

12th Maths Application Of Matrices And Determinants Question 8.
If adj (A) = \(\left[\begin{array}{ccc}{2} & {-4} & {2} \\ {-3} & {12} & {-7} \\ {-2} & {0} & {2}\end{array}\right]\) find A
Solution:

12th Maths Ex 1.1 Question 9.
If adj(A) = \(\left[\begin{array}{ccc}{0} & {-2} & {0} \\ {6} & {2} & {-6} \\ {-3} & {0} & {6}\end{array}\right]\) find A^-1
Solution:

12th Exercise 1.1 Question 10.
Find adj(adj(A)) if adj A = \(\left[\begin{array}{ccc}{1} & {0} & {1} \\ {0} & {2} & {0} \\ {-1} & {0} & {1}\end{array}\right]\)
Solution:

12th Maths Exercise 1.1 Answers In Tamil Medium Question 11.

Solution:

12th Maths 1st Chapter Question 12.
Find the matrix A for which A \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]=\left[\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right]\)
Solution:
Given A \(\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]=\left[\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right]\)
Let \(\mathrm{B}=\left(\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right) \text { and } \mathrm{C}=\left(\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right)\)
Given AB = C, To find A
Now AB = C
Post multiply by B^-1 on both sides
ABB^-1 = CB^-1 (i.e) A (BB^-1) = CB^-1
⇒ A(I) = CB^-1 (i.e) A = CB^-1
To find B^-1:

12th Maths 1.1 Question 13.
Given \(\mathbf{A}=\left[\begin{array}{cc}{1} & {-1} \\ {2} & {0}\end{array}\right], \mathbf{B}=\left[\begin{array}{cc}{3} & {-2} \\ {1} & {1}\end{array}\right] \text { and } \mathbf{C}\left[\begin{array}{ll}{1} & {1} \\ {2} & {2}\end{array}\right]\), find a matrix X such that AXB = C.
Solution:
A × B = C
Pre multiply by A^-1 and post multiply by B^-1 we get
A^-1 A × BB^-1 = A^-1CB^-1 (i.e) X = A^-1CB^-1

12 Maths Samacheer Kalvi Solutions Question 14.

Solution:

12th Maths Exercise 1.1 5th Sum Question 15.
Decrypt the received encoded message \(\left[\begin{array}{cc}{2} & {-3}\end{array}\right]\left[\begin{array}{ll}{20} & {4}\end{array}\right]\) with the encryption matrix \(\left[\begin{array}{cc}{-1} & {-1} \\ {2} & {1}\end{array}\right]\) and the decryption matrix as its inverse, where the system of codes are described by the numbers 1-26 to the letters A- Z respectively, and the number 0 to a blank space.
Solution:
Let the encoding matrix be \(\left[\begin{array}{cc}{-1} & {-1} \\ {2} & {1}\end{array}\right]\)


So the sequence of decoded matrices is [8 5], [12 16].
Thus the receivers read this message as HELP.

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.1 Additional Problems

12th Maths Chapter 1 Question 1.
Using elementary transformations find the inverse of the following matrix
Solution:

12th Maths Guide Question 2.
Using elementary transformations find the inverse of the matrix
Solution:

12th Maths Solutions Samacheer Kalvi Question 3.
Using elementary transformation find the inverse of the matrix
Solution:

12 Maths Chapter 1 Exercise 1.1 Question 4.
Using elementary transformations find the inverse of the matrix
Solution:

12th Maths Exercise 1.1 Solutions Question 5.
Using elementary transformation, find the inverse of the following matrix
Solution:

Samacheer Kalvi 12th Maths Guide Question 6.

Solution:

Samacheer Kalvi 12 Maths Solutions Question 7.

Solution:

12 Maths Solutions Samacheer Kalvi Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Students can Download English Lesson 1 Two Gentlemen of Verona Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 12th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th English Solutions Prose Chapter 1 Two Gentlemen of Verona

Warm up

Samacheer Kalvi 12th English Two Gentlemen of Verona Textual Questions

1. Answer the following questions in one or two sentences each based on your understanding of the story.

12th English Two Gentlemen Of Verona Paragraph Question (a)
Who did the narrator meet at the outskirts of Verona?
Answer:
The Narrator met Nicola and Jacopo at the outskirts of Verona.

12th English Two Gentlemen Of Verona Questions And Answers Question (b)
Why did the driver not approve of the narrator buying fruits from the boys?
Answer:
The fruit vendors were shabby in their appearance. So, the driver did not approve of the narrator’s idea of buying fruits from them.

Two Gentlemen Of Verona Questions And Answers Question (c)
The boys did not spend much on clothes and food. Why?
Answer:
The boys had to save every penny towards the weekly medical bill of their sister Lucia. So, they did not spend much on the clothes and food.

Two Gentlemen Of Verona Book Back Answers Question (d)
Were the boys saving money to go the States? How do you know?
Answer:
No, the boys did not save the money to go to States. The boys themselves admit that they have other plans.

Two Gentlemen Of Verona Summary 12th English Question (e)
Why did the author avoid going to Lucia’s room?
Answer:
The author did not want to disturb the privacy of the family party and the sense of pride the boys had in supporting their elder sister.

12th English Two Gentlemen Of Verona Book Back Answers Question (f)
What was Lucia suffering from?
Answer:
Lucia was suffering from tuberculosis of the spine.

Two Gentlemen Of Verona Paragraph Pdf Question (g)
What made the boys join the resistance movement against the Germans?
Answer:
The boys hated the Germans who had destroyed their town and killed their father. So, they joined the resistance movement against the Germans.

Two Gentlemen Of Verona Class 12 Questions And Answers Question (h)
What made the boys work so hard?
Answer:
They did not want charity from the hospital. The determination to save the life of their sister made the boys work hard.

Two Gentlemen Of Verona Paragraph Question (i)
Why didn’t the boys disclose their problem to the author?
Answer:
The little boys did not want sympathy but any work which would fetch them money. So, they did not disclose their problem to the author.

12th English Two Gentlemen Of Verona Summary Additional Questions

12th English Two Gentlemen Of Verona Question (a)
What were the two boys selling?
Answer:
The two boys were selling wild strawberries.

The Two Gentlemen Of Verona Question Answer Question (b)
What were the boys doing at night in a deserted square?
Answer:
The boys were resting on the stone pavement in the windy and deserted square beneath the street light. They were waiting for the last bus from Padua to sell the unsold newspapers.

Two Gentlemen Of Verona Question Answers Question (c)
How did the boys impress the narrator?
Answer:
Both the boys were cheerfully ready to work. In their boyish faces, there was a seriousness which was far beyond their years.

The Gentlemen Of Verona Question Answers Question (d)
What did the boys do other than selling fruits?
Answer:
The boys shined shoes, hawked newspapers, conducted tourists round the town and ran errands in addition to selling fruits.

2. Answer the following questions in three or four sentences each.

12th English Two Gentlemen Of Verona In Tamil Question (a)
Describe the appearance of Nicola and Jacopo.
Answer:
One boy had worn Jersey and cut-off khaki pants. The other had a shortened army tunic gathered in loose folds about his skinny frame. They had tangled hair and dark earnest eyes.

Two Gentlemen Of Verona Answers Question (b)
What were the various jobs undertaken by the little boys?
Answer:
The little boys sold wild berries. They shined shoes. They showed visitors through the town to Juliet’s tomb and other places of interest. They even booked a seat for the narrator in a theatre and got American cigars too.

12th English Two Gentlemen Of Verona Story In Tamil Question (c)
How did the narrator help the boys on Sunday?
Answer:
The author took the boys in his car to a tiny village called Poleta,set high upon a hill. He drove the car up to a large red-roofed villa. The boys asked the author to come there after an hour and went inside.

Two Gentlemen Of Verona Class 12 Question (d)
Who took the author to the cubicle?
Answer:
Attained nurse who was very familiar with Nicola and Jacopo took the narrator in. She led him through a cool, tiled vestibule into the hospital, the villa had become. She left him at the door of a little cubicle from where he can watch unseen Nicola and Jacopa chatting with their ailing sister.

Two Gentlemen Of Verona Question Answers Pdf Question (e)
Describe the girl with whom the boys were talking to in the cubicle.
Answer:
Lucia was a girl of about twenty. She was propped up on pillows wearing a pretty lace jacket. Her eyes were soft and tender.

Two Gentlemen Of Verona Synonyms And Antonyms Question (f)
Recount the untold sufferings undergone by the siblings after they were rendered homeless.
Answer:
The boys and their sister were thrown in the streets. Lucia had to give up her singing lessons. They had suffered horribly from near starvation and exposure to the cold winter. Lucia developed tuberculosis of the spine. The boys built a home from the rubbles. They had to admit Lucia in a nursing home. To pay the weekly medical bill, they worked from dawn to midnight doing odd jobs eating very little.

12th Two Gentlemen Of Verona Question (g)
The narrator did not utter a word and preferred to keep the secret to himself. Why? Substantiate the statement with reference to the story.
Answer:
After coming out of the hospital the boys sat beside the narrator. Throughout the 30 km ride, they did not say a word. The narrator did not say a word because he wanted the boys to feel that their sacrifice and sufferings remained a secret. However, the narrator was touched by the devotion of the little boys.

Additional Questions

Two Gentlemen Of Verona Notes Pdf Question (a)
Why was the narrator tempted to ask if the brothers wanted to emigrate to America?
Answer:
The boys earned a lot of money doing odd jobs from dawn to midnight. Yet, they spent little on clothes and food. This tempted the narrator to ask the boys if they had plans to emigrate to America.

Gentlemen Of Verona Question Answers Question (b)
How did the narrator gather information about Nicola and Jacopo without meeting Lucia?
Answer:
The author could not bear to intrude upon the happy family party. Yet, he wanted to know everything about the boys. So, he requested the nurse to tell him all she knew about them. She did share all she knew about Nicola and Jacopo with the narrator.

3. Answer the following in a paragraph of 100-150 words each.

Question (а)
What was the driving force that made the boys do various jobs?
Answer:
In the war-hit Verona everything was difficult. Food had become scarce and dear. The hospital which offered to treat Lucia was forced to charge a weekly fee. Work was scarce. But the boys were determined to save the life of their beloved sister. After the untimely death of the father

in the war, the two little boys Nicola and Jacopo took the responsibility of earning money by doing all odd jobs for defraying the medical expenses for their sister on their tender shoulders. They spent very little on their food and clothes. The driving force behind their selfless sacrifice and hard work from dawn to midnight was the love they had for their elder sister. They wanted her to get well soon and become an accomplished singer in life. Once you set a goal far above your capacity but work hard ceaselessly, you are bound to achieve it.

Question (b)
How was the family affected by the war?
Answer:
Before the war, Nicola and Jacopo’s widowed father was a well-known singer. He was killed in the early part of the war leaving Lucia, his daughter behind to take care of the little boys. Shortly after a bomb destroyed their home. The three children were thrown into the streets. They had always known a comfortable and cultured life. Lucia was training to become a singer. Suddenly they were uprooted from their cozy life. They had suffered from near starvation and exposure to cold winter. The home they rebuilt from the rubbles was not weather-proof. The boys joined the secret services of resistance moment as they hated Germans. When they returned home after the war, they found their sister affected by tuberculosis of the spine. They persuaded a private hospital to take Lucia in. Every week they worked hard to pay for their sister’s medical expenses. To achieve it they had to sacrifice a lot.

Question (c)
Write a character sketch of Nicola and Jacopo.
Answer:
Nicola was 13 and Jacopo only 12. They were brothers. They were tanned, had tangled hair and dark earnest eyes. Though they were just kids, they were serious about their work. They did hundreds of errands for the tourists. They were found doing brisk business shining shoes or selling wild berries. They had the skill to find seats in theater for the tourists and also guide the tourists through many important sites of the city of Verona such as Juliet’s tomb. Jacopo was lively as a squirrel. Nicola’s smile was steady and engaging yet in their innocent faces one could find seriousness far beyond their years. What struck one was the extreme willingness of both the boys to work. Under the scorching Sun, they hawked newspapers. When the narrator enquired what they did with the money they earned as they were not spending it on clothes or food or saving it for emigrating to America, Nicola coloured but he did not reveal the secret family adversity. Both are gentlemen because they did not want any one’s sympathy.

Question (d)
What message is conveyed through the story ‘Two Gentlemen of Verona’?
Answer:
Adversity is a touchstone of virtue. Both the little boys resemble tea leaves. Their best essence comes to limelight when they find themselves in hot waters. But they don’t complain. One does not give up on family relations when misfortune strikes. The bond becomes strong as the bond demands selfless sacrifice. The nobility of human life emerges from the precious lessons one learns from the supreme sacrifices of Nicola and Jacopo. One who shoulders the responsibility of taking care of the loved ones, irrespective of age, is an exemplary gentleman worthy of emulation. The primary motivational force of an individual is to find meaning in life. Both the little boys had a specific purpose for their life. They had made up their minds to do what ever work possible to earn money to save their sister and restore her singing career.

Question (e)
Justify the title of the story ‘Two Gentlemen of Verona’.
Answer:
In the story ‘Two gentlemen of Verona’, the two orphan boys Jacopo and Nicola work immensely and exhaustively. They shine shoes, sell newspapers, run errands and show tourists round the city. They take only figs and black tea as food. They save every penny for the medical treatment of their sister Lucia who is suffering from Tuberculosis of the spine. The narrator himself admits that their willingness to work impressed him. They demonstrate an unusual devotion to their work. The objective of their hard work and purpose behind their half-starved daily routine is not known to many in Verona. Anyone may polish shoes or hawk newspapers but it is the magnanimity of heart and nobleness of purpose that actually determines whether one is a real gentleman or a hypocrite. Hence the title ‘‘Two gentlemen of Verona” is justified.

Question (f)
Adversity brings out the best as well as the worst in people. Elucidate this statement with reference to the story.
Answer:
Whenever confronted by an adversity, there is a psychological reaction among most of us. Either fight or flee from it. Even adults run away from crisis and seek asylum or support of others. We find hordes of people, when living becomes difficult, migrate to other places to survive. In the face of adversity, even educated people turn to evil ways. They worked hard from dawn to midnight. They did any job that came their way. They shined shoes, hawked newspapers, sold wild fruits and took the people round the city. They spent very little on their food and clothes. Every week they cycled to Poleta and paid Lucia’s medical bill systematically. In this story, the nurse of the private hospital reveals the bitter truth about the economy of Verona. Jobs are scarce. There is inflation. It is difficult to buy food with limited money. When the hospital laid the condition that the little boys Nicola and Jacopo should pay the weekly medical bill for their sister Lucia’s treatment for tuberculosis, they did not back off or give up. Their best qualities came out during adversity. Like tea leaves, they gave their best while in hot waters.

Question (g)
Which character do you like the most in the story and why?
Answer:
Of all the characters, I like Nicola who is a 13 year old boy (i.e.) one of the two gentlemen of Verona. Though he is small, he is mature beyond his years. Like John Keets, he hated sympathy and self-pity. He does not want any favour except the opportunity to work. He has an engaging smile. He and his brother Jacopo were hardworking and devoted to their sister Lucia. He has seriousness far beyond his years. Even when kindly enquired by the narrator as to why he was spending little on clothes and food, he doesn’t open up. He colours deeply and grows pale. He even avoided the eyes of the narrator. When his squirrel-like brother requested the narrator to send them in his car to Poleta on Sunday, Nicola doesn’t like it. He glares at his brother in vexation and says, “we couldn’t think of troubling you sir.” He wins the respect of all readers because of his maturity, willingness to work and devotion to his sister.

Vocabulary

(a) Read the following words taken from the story. Give two synonyms and one antonym for each of these words. Use a dictionary, if required.

Answer:

S. No.	Word	Synonyms		Antonym
1.	cautious	careful	alert	reckless
2.	disapprove	express dislike	object to	approve
3.	brisk	swift	fast	sloppy
4.	engaging	charming	agreeable	repulsive
5.	humble	meek	submissive	proud
6.	eager	anxious	impatient	apathetic
7.	resistance	opposition	hostility	acceptance
8.	persuade	convince	induce	dissuade
9.	scarce	scant	insufficient	plentiful
10.	nobility	virtue	goodness	dishonour

(b) Homophones and Confusables

Question 1.
Given below is a list of common confusables. Distinguish the meaning of each pair of words by framing your own sentences.
Answer:
(a) emigrate (leaving homeland) – immigrate (come to live permanently in a foreign country)
Priyanka Chopra does not wish to emigrate to UK where her husband is living. Candidates who wish to immigrate to USA are increasing in number every year.

(b) beside (at the side of) – besides (in addition to)
I remember keeping my watch beside my glasses.
Besides supporting terrorist outfits within its borders, Pakistan claims innocence in the Pulwana attack.

(c) judicial (relating to justice) – judicious (wise/prudent)
The fraudsters who swindled the money of an IT personnel in Navalur through ATM were taken into judicial custody.
Kalpana reacted judiciously when she realised that a mobile thief was standing next to her.

(d) eligible (Adj.- satisfying the appropriate conditions) – illegible (Adj.- not clear enough to be read)
Even at the age of 30, Vinod is an eligible bachelor for the manager post.
Priya’s handwriting is illegible.

(e) conscience (a person’s moral sense of right and wrong) – conscious (being aware of) “Conscience makes cowards of us all”, said Shakespeare.
The lorry driver was conscious of the dangers involved in carrying the explosives in his truck.

(f) industrial (relating to industry) – industrious (diligent/hardworking)
Perungudi is an industrial block.
Man should be as industrious as ants and bees.

(g) eminent (famous) – imminent (something bad likely to happen very soon)
Every eminent scientist has an unseen past in which he must have struggled to prove his theory or inventions to the world.
Environmentalists are apprehensive of the imminent earthquakes in mega multipurpose projects’ catchment areas as they fall under seismicity-prone areas.

(h) illicit (adj- illegal) – elicit (v – to evoke a response)
Two hundred miners perished after consuming illicit liquor in Bihar.
The police personnel tried to elicit the truth about the murder of the school girl in a pump set from the arrested relative and three other culprits.

(i) prescribed (set down rules) – proscribed (prohibited)
The students were advised to follow the guidelines prescribed by the Directorate of Government Examinations.
The tourists were proscribed from taking selfies from the suicide point in Kodaikanal.

(j) affect (to influence something or someone) – effect (result/impact of something)
The athlete did not imagine that a steroid shot might affect his sports career adversely.
The rising number of cybercrimes proves the adverse effect of educated unemployment.

(k) aural (relating to hearing) – oral (verbal)
In primary schools, teachers must assess the aural skills of children.
There must be a fool-proof mechanism to assess the oral skills of students at the secondary level especially in languages.

(l) born (related to birth) – borne (carried by)
Divya was not born with a silver spoon in her mouth.
The entire village paid homage to army personnel, U. Saravanan’s body which was borne in a decorated coffin by military soldiers and homeguards.

Question 2.
Fill in the blanks with suitable homophones or confusables

Answer:

wallet	valet	hoard	horde
fairy	ferry	desert	dessert
medal	meddle	night	knight
wait	weight	sweet	sweat
yoke	yolk	plain	plane
grown	groan	might	mite
earn	yearn	quite	quiet

(c) Give the meanings of the following phrasal verbs and frame sentences using them.

Question 1.
cut off: block the way
Answer:
The village was cut off due to floods.

Question 2.
come upon : to meet or encounter unexpectedly
Answer:
I came upon an old friend in the street today.

Question 3.
put out: extinguish
Answer:
The firemen put out the fire after a stmggle of six hours.

Question 4.
draw up : halt
Answer:
Ram drew up his car on seeing the red signal.

Question 5.
pass out: become unconscious
Answer:
The college student who fell down from the bus passed out and an ambulance was called.

Question 6.
take off: leave the ground
Answer:
The plane took off two hours late from Delhi due to fog.

Question 7.
turn away : move one’s face away
Answer:
Whenever I find able-bodied men and women beg in the trains, I turn away.

Question 8.
stand by : ready to substitute
Answer:
The eleventh player was standing by to play if any player gets hurt or a substitute is needed.

Question 9.
bank on : rely on confidently
Answer:
We can always bank on our English teacher whenever a debate is announced.

Prefixes and Suffixes

(d) Read the list of words formed by adding suffixes.

frequently	satisfaction	willingness
comfortable	resemblance	nobility

Form two derivatives from each of the following words by adding prefixes and suffixes.

Answer:

Word	Prefix	Suffix
patient	impatient	patiently
honour	dishonour	honourable
respect	disrespect	respectable
manage	mismanage	management
fertile	infertile	fertility
different	indifferent	differently
friend	befriend	friendly
obey	disobey	obedience

Listening Activity

Floods are an inevitable natural disaster which can occur in any part of the world. Floods can prove all the more disastrous in localities, where population density is high. Preparation for Disaster Management has become imperative for any city, town or village during monsoons. The Government Department entrusted with Disaster Management makes periodic announcements about the precautions to be taken whenever floods are anticipated.

Now, you are going to listen to the cautionary instructions that are given to the general public living in flood-prone areas. Listen carefully and complete the following sentences. For the attention of the public, here is an announcement from the Department of Disaster Management. As per the warning issued today by the Meteorological Department, there exists a high probability of a widespread heavy downpour from the early hours of Sunday and consequent flooding of low-lying areas. In order to ensure safety of life and property, everyone is hereby warned and advised to take certain precautionary measures:

• First of all, prepare a household flood plan and be ready to respond to the situation.
• Find out the locations of the closest flood shelters available and routes to reach them.
• Maintain an emergency kit comprising water bottles, biscuit packets, medical supplies, a torch light, and a whistle to signal for help.
• Paste or fix a list of emergency telephone numbers on the wall in a visible spot.
• Switch off hazardous items like gas cylinders and disconnect electrical gadgets.
• Secure important personal documents and valuables in a waterproof case and place it in an accessible location.
• Place small pieces of furniture and clothing on tables and cots.
• Shift all the small objects safely to the loft.
• Empty your refrigerators and leave their doors open to avoid damage in case they float.
• Charge your mobile phones as well as your battery banks so as to communicate with friends, relatives and emergency services.
• Place sandbags in the toilet bowls and bathroom drain holes to prevent sewage inflow.
• Prepare and pack food with a long shelf life.
• Finally, listen to the periodic news updates through your portable communication devices and follow the instructions implicitly.

1. The announcement was made by the Department of _________
2. Widespread heavy rains are expected from the early hours of _________
3. The public is asked to find out the locations of _________
4. An emergency kit should contain water bottles, biscuit packets and a _________
5. A list _________ of should be displayed on the wall.
6. Important documents can be secured by keeping them in a _________ case.
7. Damage to refrigerators can be avoided by _________
8. Mobile phones should be charged to enable the marooned to contact their friends, relatives and _________
9.  _________ should be placed in the toilet bowls to prevent sewage inflow.
10. Listen to the _________ and follow the instructions implicitly.

Answer:

1. Disaster Management
2. Sunday
3. closest flood shelters
4. whistle
5. emergency telephone numbers
6. waterproof
7. emptying refrigerators and leaving their doors open
8. emergency services
9. Sandbags
10. periodic news updates

Speaking Activity

Task 1
On the occasion of World Environment Day, you have been asked to deliver a speech during morning assembly on the importance of tree planting. Write the speech in about 100 -150 words.

• Introduction
• Suggested value points
• Pollution control – Medicine – Necessary for wildlife – Cause rainfall
• Conclusion

Respected Principal, chief guest and dignitaries on the dias and off the dias.
Good evening. I am immensely delighted to speak to you on World Environment Day. All of you know that we celebrate Environment Day on 5th June, every year, let me explain why we celebrate World Environment Day. The relationship between humanity and environment is a delicate balance. Since the industrial revolution, the world’s population has increased exponentially. With the population growth, the environment has been profoundly affected. Deforestation, pollution and global climatic changes are amongst the adverse effects the explosive population and technological expansion have introduced. The main objective of Environment Day is to gain greater understanding of the environment. This enhanced awareness, I believe, will help the country reduce carbon foot prints, increase the use of solar power, reduce and recycle non-biodegradable waste and evolve alternative technologies which would reduce our dependence on fossil fuels.

Moreover, tree planting will be done on a mission modeby school children and college students representing NCC and NSS wings. This would spread the forest cover and give shelter to the wild life and also ensure plentiful rainfall sustaining life on this lovely planet. Student community shall disseminate the urgent need to conserve water and other natural resources. I appeal to the younger generation, I mean my friends to be socially responsive and do their part to preserve the environment.

Task 2
Prepare a speech on “The importance of a reading habit” in about 100-150 words using the hints given below together with your own ideas.

• Introduction
• Suggested value points
• Knowledge enrichment – Skill development – Meaningful usage of time – Overall  development
• Conclusion

Most distinguished chief guest, revered Principal, dignitaries on the dias and off the dias, my beloved teachers and friends. This morning 1 wish to share my views on a man-making habit called ‘Reading’.

Books are the quietest and most constant of friends. They are the most accessible and wisest of counselors and most patient of teachers. Some schools do focus on developing a flair for reading by keeping a library period. Children are allowed to borrow books and also share their observations with their peers after the library hour is over. Such sharing introduces children to the wonder world of books. The art of reading is in great part that of acquiring a better understanding of life from one’s encounter with it in a book. Through reading, one exposes oneself to new things, new ways to solve a problem. Reading is a very useful hobby. Wisemen have lauded the importance of reading and the hobby of reading. Kofi Annan said, “Literacy is a bridge from misery to hope.”

William James said, “So it is with children who learn to read fluently and well. They begin to take flight into whole new worlds as effortlessly as young birds take to the sky.” Dear friends let us read books and conquer the world.

Reading

Read the passage given below and make notes.

To match the best cities across the world, the Government of India initiated ‘smart cities’ to drive economic growth and improve the quality of life of people. The agenda under smart city promises to resolve urban sustainability problems. Urban forests provide a range of important ecosystem services that are critical for the sustainability of cities. Urban forestry, which is defined more as ‘Management of Trees’ contributes to the physiological, sociological andeconomic well-being of the society. Mangroves, lakes, grasslands, and forests in and around our cities, act as sponges that absorb the air and noise pollution and they present themselves as our cultural and recreational hotspots. However, these spots are rapidly being reclaimed” and replaced in the name of development. Presence of urban green has shown to increase the economic value of the place.

Urban forests contribute to reduce the cost of building storm water drain systems for municipalities and neutralizing urban heat island effect. Plants not only provide shade but also help in regulating the micro-climate. They help regulate energy budgets, improve air quality, and curtail noise pollution. Trees, herbs, shrubs and grasses arrest sedimentation and prevent other pollutants from entering our water systems. This will give a chance for our urban lakes and rivers to recover and help improve aquatic ecosystems. Biodiversity also gets a boost through the urban forests and helps create corridors connecting the forest areas. High biodiversity areas can also help to build resilient ecosystems. Availability of forests within our urban areas gives an opportunity for children to connect to the natural environment and learn about native species.

Notes

1. Agenda of smart cities
Resolve urban sustainability problems-manage trees-urban forestry-psychological and sociological well-being of society.

2. Main advantages of urban forestry
Mangroves, lakes, grasslands and forests-sponges absorbing noise and air pollutants Recreational hotspots-urban green-additive economic value-threat in the name of development.

3. Impact on climate
Reduce cost of storm water drain system-neutralise urban heat island effect-maintains I microclimate.

4. Enhancing environmental health
Regulate energy budget-air quality improved-sedimentation arrested-pollutants prevented I from entering water systems-urban lakes and rivers revive aquatic ecosystems.

5. Biodiversity
Biodiversity enhanced-resilient ecosystems developed-children get connected to natural ‘ environment.

Grammar

Tenses
Task 1
Tick the correct options and complete the dialogue.

1. A: Hello. What do you watch / are you watching?
2. B: A programme about the Jallian Wala Bagh massacre, which I recorded last night. I study /I’m studying about it this term.
3. A: All that I know / I’ve known about it is that hundreds of people died / had died in it.
4. B: Yes, it was much, much worse than anyone has expected / had expected. It went on / has gone on for hours. Do you want / Have you wanted to watch the programme with me?
5. A: No, thanks. I’ve got to do some veena practice. I’ve just remembered /1 just remembered that we’ve got a concert tomorrow, and I don’t have / haven’t had time to practise my new piece this week.
6. B: OK. I’ve already done /I already did my practice, so I’ve got time to watch TV. See you later.

Answer:

1. are you watching
2. I’m studying
3. I’ve known, had died
4. had expected, went on, Do you want
5. I’ve just remembered, haven’t had
6. I already did

Task 2
Complete the sentences with the correct tense form of the verbs in brackets.

(a) ________ (tell) me exactly what ________ (happen) last night!
(b) Mrs. Mageswari is my Maths teacher. She ________ (teach) me for four years.
(c) I ________ (never /think) of a career in medicine before I spoke to my Biology teacher but now I (seriously/ consider) it.
(d) Oh no! I ________ (forget) to bring my assignment! What am I going to do? This is the second time I (do) this!
(e) I can’t remember what my teacher ________ (say) yesterday about our homework. I (not listen) properly because Hussain (talk) to me at the same time.
(f) Last year we ________ (go) on a school trip to Kanyakumari. We (have) a very interesting time.
(g) At the moment I ________ (think) about what course to pursue next year but I (notmake) a final decision yet.
(h) I ________ (get) up at 7 every morning but this morning I (sleep) for a long time and I ________ (not get) up until 8.

Answer:

(a) Tell, happened
(b) has been teaching
(c) never thought, am seriously considering it
(d) forgot, have done
(e) said, was not listening, was talking
(f) went, had
(g) am thinking, have not made
(h) get, slept, did not get

Task 3
Fill in the blanks with the correct form of the verbs given in the brackets.

(a) Everyone ________ when the earthquake hit the small town, (sleep)
(b) Evangelene ________ her job a couple of years ago. (quit)
(c) Where ________ your last holidays? (you spend)
(d) I think Suresh ________ for Tiruvallur next morning, (leave)
(e) I was angry that I ________ such a mistake (make).
(f) My mother was tired yesterday because she ________ well the night before (not sleep).
(g) Her parents ________ in Coimbatore for two weeks from today (be).
(h) Nothing much ________ when I got to the meeting (happen).
(i) Scientists predict that by 2050, man ________ on Mars, (land)
(J) Sh! Someone ________ to our conversation! (listen)
(k) The plane ________ off in a few minutes, (take)
(l) They ________ about me when I interrupted their conversation, (talk)
(m) Justin and his parents ________ in an apartment right now because they can’t find a cheap house, (live)
(n) Rajini Prem’s family ________ in Chengalpet now. (be).
(o) Yusuf ________ to the movies once in a while (go)
(p) This ________ an easy quiz so far (be).
(q) Our team ________ any games last year, (not win)
(r) We ________ a wonderful film at the cinema last night, (see)
(s) Hurry up! The movie ________ (already begin)

Answer:

(a) was sleeping
(b) had quit
(c) did you spend
(d) left
(e) had made
(f) had not slept
(g) will be
(h) had happened
(i) will land
(j) is listening
(k) will take
(l) were talking
(m) are living
(n) is
(o) goes
(p) has been
(q) did not win
(r) saw
(s) may have already begun

Modal Auxiliaries
Task 1
Fill in the blanks with modal auxiliaries.

(a) The candidates ________ answer five out of ten questions.
(b) How ________ you open my bag?
(c) Tajudeen ________ finish this work by Monday.
(d) ________ I go to school today?
(e) I wish you ________ tell me the truth.
(f) Poonam ________ not catch the bus yesterday.
(g) People who live in glass houses ________ not throw stones.
(h) You ________ not go to the market as I have brought vegetables.
(i) ________ you have taken all this trouble?
(j) You ________ be joking.
(k) I tried to climb up the tree, but ________ not.
(l) Hima Das ran so fast that she ________ win the medal.
(m) You ________ lead a horse to water, but you make it drink.
(n) I ________ like to have a cup of coffee.
(o) My grandfather ________ visit this temple when he was young.

Answer:

(a) should
(b) dare
(c) can
(d) May
(e) could
(J) could
(g) should
(h) need
(i) Should
(j) must
(k) could
(l) could
(m) can, can’t
(n) would
(o) used to

Task 2
Complete the following sentences with modals using the clues given.

(a) You ________ help the needy, (moral obligation)
(b) If I were you, I ________ not behave like that, (conditional sentence)
(c) I ________ never tell a lie. (determination)
(d) My uncle ________ have reached by now. (possibility)
(e) to play hockey when I was a student, (past habit)
(f) You not attend my class, (order)
(h) He come today, (remote possibility)
(i) You rollow the traffic rules, (regulation)
(j) He play the match, (willingness)
(k) You not waste time on it. (necessity)
(l) Had the doctor come in time, he have saved the patient.(probability)

Answer:

(a) ought to
(b) would
(c) will
(d) may
(e) must
(f) used
(g) should
(h) might
(i) should
(j) would
(k) need
(l) would

Reported Speech

Task 1
Read what these people say and rewrite as sentences.

Question 1.
I am very busy.
Answer:
Raja said that he was very busy.

Question 2.
I have completed my work.
Answer:
Satya said that she had completed her work

Question 3.
I don’t like to go out.
Answer:
Johnson said that he didn’t like to go out

Question 4.
I have just come back from Chennai.
Answer:
Akshita said that she had just comeback from Chennai

Question 5.
I am learning English.
Answer:
Jayan said that he was learning English

Question 6.
I bought a pen yesterday.
Answer:
Madhu said that she had bought a pen the previous day

Question 7.
We will go shopping tomorrow.
Answer:
Joseph and Mary said that they would go shopping the next day

Question 8.
We can’t attend the party.
Answer:
Afsar and Ayesha said that they couldn’t attend the party

Question 9.
How are you?
Answer:
Satish

Question 10.
I am line. Thank you.
Answer:
Victor enquired how Victor was replied that he was fine and thanked him

Task 2
Read the following dialogue and complete the report in the space provided.

Question 1.
Priya : Where are you going?
Vijay : I am going to the railway station.
Priya : Why are you going there?
Vijay : I want to receive my uncle who is coming from Bangalore.

Answer:

Priya asked Vijay : (a) where he was going.
Vijay replied : (b) that he was going to the railway station.
Priya further inquired : (c) why he was going there.
Vijay stated that : (d) he wanted to receive his uncle who was coming from Bangalore.

Question 2.
Teacher : Why are you late?
Divya : I missed the bus.
Teacher : You should have reached the bus stop on time.
Divya : My grandmother is ill. So, I had to take her to the doctor.
Teacher : I am sorry. What ails her?
Divya : She has high fever.
The teacher asked Divya why she was late. Divya replied that ___(a)___The teacher told her that ___(b)___ Divya said that her grandmother was ill so she had to take her to the doctor. The teacher felt sorry and further asked her ___(c)___Divya explained that she ___(d) ___ high fever.

Answer:

(a) she had missed the bus
(b) she should have reached the bus stop on time .
(c) what ailed her
(d) had

Task 3
Rewrite the following passage in indirect speech.

Pradeep got out of bed with much excitement. “It is going to be a lovely sunny day,” he remarked to his sister Varshini.
I “Just let me sleep a bit longer, Pradeep,” Varshini begged, “and since you are feeling so
enthusiastic,” she suggested, “Why don’t you go and help mother in cooking?”
“Sure, I will”, said Pradeep. “Can I help you, mum?”, he said to his mother.
“Yes, of course. There are idlis and vadas on the dining table. Have your breakfast.”
“Thank you, mum; I’ll surely help you by eating them.”

‘Pradeep got out of his bed with much excitement and told his sister that it was going to be a lovely Sunday. Varshini requested him to let her sleep a bit longer. She added that as he was feeling so enthusiastic he should go and help mother in cooking. Pradeep agreed readily. He asked his mom if he could help her. Mom agreed and asked him to have breakfast adding that there were idlis and vadas on the dining table. Pradeep thanked mom and said that he would surely help her by eating them.

Writing
Letter Writing

(a) Read the following letter carefully, discuss with your partner and answer the questions.

Sir,
I request you o publish the following letter in your daily to address an important issue which needs immediate redressal.
During week ends motorcyclists are seen racing on the East Coast road and the Old Mahabalipuram road of Chennai. The amateur racers are risking their own lives and of the public as well. They are fearless and irresponsible, not knowing the price of human lives. Risking others and their lives for their pleasure is highly condemnable.

Even school students indulge in such activities and cause fatal accidents. The racers have started occupying the lanes and streets of busy localities. As such, the violation of traffic rules often results in loss of young lives. Such reckless riders who violate traffic rules should be punished severely.

As a responsible citizen of the society, I request the authorities concerned to take appropriate measures to put an end to this menace.
Yours truly,
Srivastsav

Questions:

Question (a)
Who is the sender of the letter?
Answer:
The sender is Srivatsav

Question (b)
Who is the receiver?
Answer:
The editor of a Newspaper (unnamed).

Question (c)
What is the issue?
Answer:
Bike racing on the East Coast road and the Old Mahabalipuram road of Chennai is the issue.

Question (d)
What is the request of the sender?
Answer:
The reckless riders who violate traffic rules must be punished. This is the request of the sender.

Question (e)
Who will take steps after reading it?
Answer:
The authorities concerned will take steps after reading it.

(b) You have chosen Computer Science in the Higher Secondary Course. Write a letter to your friend giving reasons for your choice. Read the clues given in brackets to complete the letter.

(1) _____ (Greeting),
Hope (2) ______ (enquire about his well being). I would like to (3) ______ Firstly (4) ______ (reason for your choice of group). I wish (5) ______ (state your ambition). We come to hear (6) ______ (discuss recent unhealthy happenings in social media and the society). These have made me (7) ______ (demand or need of this profession). I have plans to pursue (8) ______ (higher studies or specialization).
All the best for your CA preparation.
Convey my (9) ______
Yours lovingly,
(your name)

Answer:

1. Dear Ram
2. you are doing fine
3. inform you that I have chosen Computer Science stream in eleventh
4. one has to have sound knowledge of computers and without computer literacy one can’t succeed in life.
5. to become a Computer Engineer
6. cyber bullying and sexploitation
7. understand the need of ethical computer engineers in tackling such issues
8. Computer Science Engineering in Anna University
9. kind regards to your parents

Task
(a) You had been to your Grandma’s house during the summer holidays. You enjoyed your stay in her company. Write a letter to your Grandma stating how much you miss her after returning to your home.
Dear Grandma,

I am fine. I hope your health is fine. My stay in your house was really wonderful. You spent a lot of time with me sharing recipes for many delicious dishes. Everyday you prepared curry with vegetables from your own garden. The food you gave was so tasty that I envied my mom who had spent the best part of her life with you. I just love the stories you shared every evening. Grandma, you had herbal remedies for every kind of ailment. I wish I could come again during next summer and stay with you.
Yours lovingly,
Anu
To
K. Savithri
18, Mettu St.
Chengalpattu – 603001

(b) You are the head of the English department in a renowned institution. You are invited to preside over the inauguration of the English Literary club in your alma mater. Respond to the letter you have received either accepting the invitation or expressing your inability to attend the function.

Dear Vijaya Sarathy,
Received your invitation. I am really glad to accept your invitation to inaugurate the English Literary Club in the school where I had completed my twelfth standard 15 years ago. Don’t worry about sending a car. I will come in my own car. You can fix any working day in the second week of November.
With Warm
Regards Yours sincerely
Anjali
To
Vijaya Sarathy
Secretary, Literary Club
ABC HSS,
XXXX

Inability to attend the function

Dear Sarathy,

Received your invitation, I wish I could attend the inaugural meeting of the literary club. It would be such an honour as I have studied in that school. But I need to attend an international conference in Thailand. I will be away during second and third week of October. 1 would recommend you to invite Dr. Sangeetha, a great scholar who has recently done her Ph.D. in Modem Poetry (i.e.) Pudhukavithai. Though she is young, she is a versatile genius and it would be appropriate if you would come and invite her in person. Good luck for the success of the programme.
With Warm Regards Yours sincerely Anjali Head
Dept, of English
ABC College of Arts and Science Chennai – 78.
To
Vijaya Sarathy
Secretary, Literary Club
ABC HSS,
XXXX

(c) Write a letter to the Headmaster of your school requesting him to help you obtain a duplicate mark sheet of class XII, which you lost while travelling.

From
S. Kavinilavan
18, Gangu Reddy
Egmore Chennai To
The Headmaster
Govt. HSS
Nungambakkam
Respected Sir,
Sub: Loss of Twelfth Marksheet
I completed my twelfth in your esteemed school in the year 2016-2017. My reg. no. is 98781. I lost my twelfth marksheet. I request you to kindly guide me to get a duplicate marksheet from the Directorate of Government Examinations at the earliest.
With warm regards.
Yours sincerely
S. Kavinilavan
To;
The Headmaster
Govt. HSS
Nungambakkam

(d) Write a letter to AZ Company requesting them to replace the defective juicer that you bought recently. Include the following details: the problem, date of purchase, receipt number, model and warranty.

From
K. Manimuthu
18, East Street
Kulamangalam
To,
The Manager
A2 Company
18, Dindigul Road
Madurai
Sir,
Sub: Request to replace defective juicer-Reg.
I bought Preethi juicer mixie on 18.02.19 vide bill no. 1781. On reaching home I tried to make apple juice. The juicer rotates but the cut apples were not crushed. I believe this is a mechanical fault. Could you please replace the faulty Preethi juicer with a new one? Photocopies of warranty card and the bill are enclosed.
Yours sincerely K. Manimuthu
To,
The Manager
A2 Company
Company
18, Dindigul Road
Madurai

(e) You wish to become a pilot. Write a letter to a college enquiring about the details of the pilot training course offered by the college. Include the following details in your enquiry: duration of the course, fee structure, scholarships, hostel facilities and placement details.

From
L. Kathiravan
17, Awaiyar St.
K.K. Nagar
Madurai – 20 To
The Principal
PTC Aviation Academy
Chromepet
Chennai – 600044
Sir,
Warm Greetings.
Sub: Enquiry about Pilot training course-Reg.
I saw your advertisement in the Indian Express newspaper about pilot training course yesterday. I would like to become a commercial pilot. Kindly clarify my doubts by answering the following questions.
1. How long do you conduct the course for the candidates who want to become pilots?
2. What is the fee structure of the course?
3. I have 90+ marks in I Group in my +2 Board examinations. Is there any scholarship or concession in fee structure for meritorious candidates?
4. What is the success rate of placements in your institute for trained pilots?
I shall contact you in person after receiving your information-packed reply.
Yours sincerely
L. Kathiravan
To,
The Principal
PTC Aviation Academy
Chromepet
Chennai – 600044

(f) Write a letter to the manager of Waves Furniture Company ordering furniture for a coaching centre. Include the following details: description of the furniture, number of pieces, mode of payment, time and delivery options.

From
M. Usharani Coordinator
NEET Coaching Centre Government Girl’s Hr. Sec. School Egmore To
The Sales Manager
Waves Furniture Company
Station Road
Nungambakkam
Chennai
Sir,
We need 100 double desks (i.e.) Desk + bench enabling four students to sit comfortably and write at a time. On receipt of the green painted iron double desks, we will make payment through cheque. Delivery can be effected only after personal verification of sample piece tailor-made for our school requirements. You can deliver the pieces before 28th July without fail.
Yours sincerely
S/d Usharani
To,
The Sales Manager
Waves Furniture Company
Station Road
Nungambakkam
Chennai

(g) Write an application for the post of Personal Secretary to the Managing Director of a company. Include the following details: Educational qualification, experience, various other qualifications required for the post.

From
R. Kalaivani, B.Sc.
23, Karameswar Koil St.
Saidapet Chennai – 16 To
The Managing Director Mathew Garments Rameshwaram Road T. Nagar Chennai – 17
Sir,
With reference to your advertisement in The Hindu dated 18.02.19,1 wish to apply for the post of Personal Secretary. I am a Science graduate from Loyola College. I have done my short hand and P.G. Diploma in Human Resource Management. I can take notes and type in the computer at a reasonable speed. I have an experience of working as a part-time Personal Secretary to Shri Ram Chits Company Manager for a period of three years. I have good working knowledge of English and Tamil. I request you to consider my application favourably and send me a call letter. I assure you, I would execute my job to the best satisfaction of my superiors.
With Regards
Yours Sincerely
R. Kalaivani
To,
The Managing Director
Mathew Garments
Rameshwaram Road, T. Nagar
Chennai – 17

(h) Write a letter the Editor of a newspaper about the nuisance created by the roadside vendors blocking the pavements and occupying the parking zone.

From
M. Kalavathi
24, South Car St
Madurai
To,
The Editor
The Hindu
Chennai – 2
Sir,
Sub: Nuisance of vendors in roads and parking spaces
Many foot path vendors are blocking the pavements and occupying parking areas in South Car St. and the nearby parking areas. All those vendors who eke out a living may be allotted shops elsewhere to curtail this menace. Kindly publish this in your newspapers so that Madurai corporation and the traffic police may take the necessary action.
Yours Sincerely
M. Kalavathy
To,
The Editor
The Hindu
Chennai – 2

(i) Write a letter to your relative or friend who is admitted in hospital for treatment of jaundice. Advise him/her not to worry about the illness and be positive. Assure him/her of your psychological and financial help during the crisis.

Dear Mala,
Received the letter informing me of your ailment. Jaundice can be cured by herbs very quickly without serious side effects. Sirukurinjan Thalai is a remarkable herb which would expedite the cure. In case you have already started the treatment, don’t worry, if you need money, I will transfer it to your account by NEFT or I can transfer it from my Paytm App.’ Drink only boiled water. Avoid oily and spicy food items. You must take fruit juices to recover completely. Don’t worry, I’m there for you. Get well soon.
With love KuralMathi
To,
Mala
Meenakshi Nagar
Madurai

Two Gentlemen of Verona About The Author

Archibald Joseph Cronin (1896-1981) was a Scottish novelist, dramatist and physician. A.J. Cronin was a doctor by profession and a writer by accident. While recovering from an illness, he started writing his first novel “Hatter’s Castle” and instantly became popular. Many of his stories have emerged from his medical career and are noted for their narrative skill, deep social conscience and finely drawn characters. Cronin’s books were not only bestsellers but some of them were made into successful films and were adapted for radio and television. His novella Country Doctor was adapted for a long running BBC radio and TV series. Drawing inspiration from an early play of Shakespeare, he wrote “Two Gentlemen of Verona”.

Two Gentlemen of Verona Summary in English

Introduction
‘Two Gentlemen of Verona’ is the story of two young boys, Nicola and Jacopo. The story recounts how the two brothers devoted themselves to a hard and difficult life to earn money for the medical treatment of their sister.She was afflicted with tuberculosis of the spine. The author was impressed by their sincerity and devotion to the cause. They displayed maturity and understanding beyond their years. Their selfless action ‘brought a new nobility to human life and gave promise of a greater hope for human society’.

Nicola and Jacopo – the fruit vendors
Two young boys stopped the narrator on the outskirts of Verona. They were selling wild strawberries. The elder brother Nicola was thirteen while the younger brother Jacopo was nearly twelve. Both were in rags. He was strangely attracted towards them. The earnestness in their eyes touched the narrator. He bought their biggest basket inspite of Luigi’s (his driver) protests.

Doing odd jobs

Next morning, the narrator saw both the brothers bending over shoeshine boxes in the public square. They were doing a brisk business. They not only picked up fruits but also did many other things to earn a living. Often they showed visitors round the town-to Juliet’s tomb and other places of interest. They proved extremely useful to the narrator and his driver Luigi. They could buy a pack of American cigarettes or get seats for the opera. They could be relied upon to satisfy their needs. They shined shoes, sold fruits and hawked newspapers, took tourists round the city and did all odd jobs for others to earn money.

Working till late night

One night, the narrator saw both the brothers resting on the stone pavement. Jacopo’s head was resting upon his elder brother’s shoulders. He was asleep. It was nearly midnight. Nicola told them that they were waiting for the last bus from Padua. They could sell all their papers when the bus came in there. The narrator thought that they must earn quite a lot. They worked so hard and ate so little. They spent nothing on their clothes. The narrator thought that perhaps they were saving up to emigrate to America. But they had other plans.

A rare privilege
The author was to leave Verona on Monday. He asked the brothers if he could do anything for them before leaving. Jacopo told that every Sunday they made a visit to the country to Poleta, 30 kilometers from there. Usually they hired bicycles to go there. He asked the narrator if he could send them there in his car. Luigi had taken the Sunday off but the narrator replied that he would drive them out himself. The following afternoon they drove to a tiny village. The two brothers came out of the car and went towards a red-roofed villa. They promised to be back within an hour.

Unravelled Mystery
The narrator found a grilled slide entrance. He determinedly rang the bell. A pleasant-looking young woman appeared in the white uniform of a trained nurse.She led him into the hospital. She stopped at the door of a little cubicle. She asked him to look through the glass partition. The two brothers were seated at the bedside of a girl of about twenty. She resembled her brothers. She was listening to their chatter. The nurse asked the narrator to go in. Lucia would be pleased to see him. The narrator did not want to intrude upon that happy family party. He begged the nurse to tell him all she knew about those boys.

Family Destroyed by the War
Nicola and Jacopo were quite alone in the world, except their sister Lucia. Their father was a widower. He was a well-known singer and was killed in the early part of the war. A bomb had destroyed their home and they were thrown into the streets. They rebuilt their house using the rubbles. They suffered horribly. They faced starvation and were exposed to the cold winter. They hated the Germans who brought all these troubles on them. The boys served as secret agents and carried messages about movement of German troops to freedom fighters. When the war was over, they came to their beloved sister. They found her suffering from tuberculosis of the spine.

Love for their Sister
The boys did all that they could do for their ill sister. They brought her to the hospital and persuaded them to take her into the hospital. She had made a lot ofj progress in the last twelve months. It I was hoped that one day she would walk and sing again.Lucia’s brothers paid weekly bills regularly. The narrator was shocked to know the story of the two boys and their sister. He hurriedly came out and waited until the boys rejoined him. He drove them back to the city. They sat beside him and spoke nothing. Even he didn’t say a word to them. He knew they would prefer to feel that they had ‘safely kept their secret’. Their devotion, selfless action and sincerity touched his heart deeply. Their devotion to the cause brought a new nobility to human life. They gave a new hope for humanity.

Conclusion
The selfless hardships bravely faced by the two little brothers of Lucia teach us the meaning of life. The story conveys the message that as long as people are willing to make sacrifices for the well-being of others, there is hope for humanity.

Two Gentlemen of Verona Summary in Tamil

முடிவுரை:
லூசியாவின் சகோதரர்களின் தன்னலமற்ற கடின உழைப்பு மனித வாழ்க்கையின் அர்த்தத்தை நமக்கு உணர்த்தியது. இந்தக் கதையானது பிறரின் நல்வாழ்விற்காக தங்களை அர்ப்பணிப்பவர்கள் உள்ள வரையில் மனித குலம் தழைக்கும் என்ற செய்தியை நமக்கு அளிக்கிறது.அவர்கள் தங்கை முதுகுத்தண்டு புற்றுநோயால் | பாதிக்கப்பட்டிருந்தார். அச்சகோதரர்கள் தன் குறிக்கோளின் மேல் கொண்ட விசுவாசத்தைக் கண்டு | கதையாசிரியர் வியந்து போனார். அவர்கள் வயதிற்கு மீறிய பக்குவத்தைக் கொண்டிருந்தனர். தன்னலமற்ற அவர்களின் சேவை மனித குலத்திற்கே உண்டான | பெருந்தன்மையையும், மனித குலத்தின் மேல் அதிக நம்பிக்கையையும்

கனி விற்பவர்களாக நிகோலாவும், ஜக்கோப்போவும்:
நிகோலா மற்றும் ஜகாப்போ என்ற இரு சிறுவர்களும் கதையாசிரியரை வெரோனா நகர்ப்புறவெளியில் வழி மறித்து நிறுத்தினர். அவர்கள் காட்டு வகை மென்மையான செந்நிற கனிகளை (Wild Strawberries) விற்றுக் கொண்டிருந்தனர். பெரியவன் நிகோலாவிற்கு வயது 13, சிறியவன் ஜகாப்போவிற்கு வயது 12. இருவரும் மிகவும் ஏழ்மை கோலம் பூண்டிருந்தனர். கதாசிரியர் வினோதமாக அவர்களின்பால் ஈர்க்கப்பட்டார். அவர்களின் கபடமற்ற கண்கள் கதாசிரியரை கவர்ந்தன. கதையாசிரியரின் காரோட்டி லூகி மறுப்பு தெரிவித்த போதிலும் அவர் சிறுவர்களிடமிருந்து பெரிய பழக்கூடையை வாங்கிக் கொண்டார்.

கையில் கிடைத்த வேலைகளை எல்லாம் செய்தனர்:

அடுத்த நாள் காலைப் பொழுதில் ஒரு பொது இடத்தில் இரு சகோதரர்களும் பொது மக்களின் காலணியை மெருகூட்டிக் கொண்டிருப்பதைக் கண்டார் அவர்கள் சுறுசுறுப்பாக வேலையில் ஈடுபட்டிருந்தனர் அவர்கள் வாழ்வாதாரத்திற்காக பழம் மட்டும் விற்கவில்லை . மேலும், பல பணிகளை புரிந்து வந்தனர் பல சமயங்களில் அவர்கள் சுற்றுலாப் பயணிகளை ஊரைச் சுற்றிக் காட்டும் விதமாக ஜுலியட் கல்லறை மற்றும் இதரப் புகழ் பெற்ற இடங்களுக்கு அழைத்துக் சென்றனர். இருவரும் கதையாசிரியருக்கும், லூகிக்கும் மிகவும் உதவியாக இருந்தனர். அவர்களுக்கு சிகரெட் வாங்குதல், கலை கச்சேரிக்கு அனுமதிச் சீட்டு வாங்குதல் என எல்லா உதவிகளையும் செய்து வந்தனர் நம்பிக்கைக்கு பாத்திரமானவர்களாக, அவர்கள் தேவைகளை நிறைவு செய்பவர்களாகத் திகழ்ந்தனர் காலணி பாலிஷ் செய்தல், பழங்கள் மற்றும் செய்தித் தாள்களை விற்றல், சுற்றுலா பயணிகளுக்கு ஊரை சுற்றிக் காட்டுதல் மற்றும் இதர கிடைத்த பணிகளைட் புரிந்து பொருள் ஈட்டினர்.

நடுநிசி வரை பணி புரிந்தனர்:

ஒரு நாள் இரவு கதையாசிரியரும், லூகியும், சிறுவர்கள் இருவரும் நடைபாதை கல் ஒன்றின் மேல் அமர்ந்து இருப்பதைக் கண்டனர். ஜகாப்போ அண்ணனின் தோள் பட்டையின் மேல் சாய்ந்து உறங்கிக் கொண்டு இருந்தான். அது நள்ளிரவு நேரம். நிக்கோலா கதையாசிரியரிடம் பதுவாவிலிருந்து வரும் கடைசி பேருந்திற்கு காத்திருப்பதாகக் கூறினான். ஊரைப் பேருந்து அடைந்ததும் அவர்கள் செய்தித் தாள்கள் முழுவதையும் விற்றுவிடுவார்கள். கதையாசிரியர் இச்சிறுவர்கள் கணிசமான தொகையை சம்பாதிக்கக் கூடும் என நம்பினார். அவர்கள் கடுமையாக உழைத்தாலும் உணவிற்காக கொஞ்சமே செலவழித்தனர். துணிகளுக்காக எதையும் அவர்கள் செலவழிக்கவில்லை. அவர்கள் அமெரிக்க நாட்டிற்கு குடிபெயர்வதற்கு சேமிக்கிறார்களா என கதையாசிரியர் சிறுவர்களிடம் விசாரித்தார். ஆனால் சிறுவர்களுக்கோ வேறு திட்டங்கள் இருந்தன.

அரிதான சலுகை:
கதையாசிரியர் திங்கட்கிழமை அன்று வெரோனாவை விட்டு கிளம்புவதாக இருந்தது. அவர் கிளம்புவதற்கு முன் அச் சகோதரர்களுக்கு ஏதேனும் உதவி தேவைப்படுகிறதா எனக் கேட்டார். அதற்கு ஜகாப்போ அங்கிருந்து 30 கிலோ மீட்டர் தூரத்தில் உள்ள பொலேட்டாவிற்கு செல்லவேண்டும் என்றான். பொதுவாகவாடகைசைக்கிளில் செல்வதுதான் அவர்கள் வழக்கம். ஜகாப்போ கதையாசிரியரின் மகிழுந்தில் செல்லும் விருப்பத்தை தெரிவித்தான். லூகிக்கு விடுமுறை கொடுத்து விட்டதால் கதையாசிரியர் தாமே இவர்களை தன் மகிழுந்தில் அழைத்துச் செல்வதாகக் கூறினார். அன்றைய மதிய நேரத்தில் அக்குறுகிய கிராமத்திற்கு பயணத்தை மேற்கொண்டனர். சிறுவர்கள் மகிழுந்தை விட்டிறங்கி சிவப்பு சாயம் பூசப்பட்ட பெரிய பங்களாவுக்குள் நுழைந்தனர். ஒரு மணி நேரத்தில் திரும்பி விடுவதாக வாக்களித்தனர்.

அவிழ்ந்த மர்மம்:
கதையாசிரியர் அந்த பங்களாவில் கம்பிகளாலான வாயிற் கதவைக் கண்டார். ஒரு முடிவோடு அழைப்பு மணியை அடித்தார். அங்கு இளமையான வெண்ணிற ஆடை உடுத்திய செவிலிப் பெண் ஒருவர் தோன்றினார்.
அவர் கதையாசிரியரை மருத்துவமனைக்குள் அழைத்துச் சென்றார். ஓர் தடுப்புச்சுவர் அருகில் கதையாசிரியரை நிற்கச் சொன்னார். கண்ணாடி தடுப்பு வழியாகப் பார்க்கச் சொன்னார். இரு சகோதரர்களும் 20 வயது மதிக்கத்தக்க பெண் ஒருவர் அருகில் அமர்ந்திருந்தனர். அவர் இச்சகோதரர்களின் சாயலில் இருந்தாள். அப்பெண் தம்பிகளின் அரட்டையை இரசித்துக் கேட்டுக் கொண்டிருந்தாள். செவிலிப் பெண் அவரை உள்ளே போகச் சொன்னார். அதற்கு கதையாசிரியர். அவர்களின் சந்தோசமான குடும்ப சந்தோசத்தில் தலையிட விரும்பவில்லை என்றார். செவிலிப் பெண்ணிடம் சிறுவர்களைப் பற்றிய விவரங்களைக் கூறுமாறு வருந்தி வேண்டிக் கொண்டார்.

போரால் சிதைந்த குடும்பம்:
லூசியாவைத் தவிர நிகோலா, ஜகாப்போவிற்கு இந்த உலகத்தில் சொந்தம் என்று கூறிக்கொள்ள எவரும் இல்லை. சிறுவர்களின் அம்மா சிறுவயதிலேயே இறந்துவிட்டார். அவர்களின் தகப்பனார் ஒரு பாடகர். அவர் ஆரம்ப போர் காலத்திலேயே மரணமடைந்து விட்டார். அதன்பிறகு நடந்த ஒரு குண்டு வெடிப்பில் அவர்கள் வீடு சின்னாபின்னமாகிக் போனதால் வீதிக்கு தள்ளப்பட்டார்கள். இடிந்த கட்டிட குவியலைக் கொண்டு மீண்டும் ஒரு வீட்டைக் கட்டினர். அவர்கள் பிறரைப் போலவே கடினமாகப் பாதிக்கப்பட்டனர். பசியையும், குளிரையும் எதிர் கொண்டார்கள். இந்நிலைக்கு தம்மை ஆளாக்கிய ஜெர்மானியர்களை சிறுவர்கள் வெறுத்தனர். இரகசிய உளவுத்துறை உளவாளிகளாகப் பணிபுரிந்து தகவல்களை விடுதலை வீரர்களிடம் தெரிவித்தனர். போர் முடிந்ததும் தன் அன்பார்ந்த சகோதரியுடன் வந்து சேர்ந்து கொண்டனர். தங்கள் சகோதரி முதுகெலும்பு புற்றுநோயால் அவதியுறுவதை அறிந்தனர்.

சகோதரியின்பால் கொண்ட அன்பு:
நோய் வாய்ப்பட்ட சகோதரிக்காக தம்மால் இயன்ற அனைத்தையும் செய்தனர். சகோதரியை மருத்தவமனைக்கு அழைத்து வந்து அவரை அனுமதித்துக் கொள்ள மன்றாடினர். 12 மாதத்தில் அவர் நல்ல உடல்நலம் தேறினார். அவர் மீண்டும் நடக்கவும், பாடவும் செய்வார் என்ற நம்பிக்கை உண்டாயிற்று.
லூசியாவின் வாராந்திர மருத்துவச் செலவினை சகோதரர்கள் ஏற்றுக்கொண்டனர். அவர்கள் கதையை அறிந்த கதையாசிரியர் அதிர்ச்சிக்குள்ளானார். அவர் அவசரமாக வெளியே வந்து சகோதரர்களின் வருகைக்கு காத்திருந்தார். அவர்களுடன் மீண்டும் வெரோனா நகரத்திற்கு பயணித்தார். இரு சிறுவர்களும் அவரின் அருகே அமர்ந்து அமைதியாகப் பயணம் செய்தனர். கதையாசிரியரும் ஏதும் பேசிக் கொள்ளவில்லை. அவர்கள் இந்த இரகசியத்தை பேணிக் காக்கவே விரும்புவார்கள் என எண்ணினார். அச்சகோதரர்களின் தியாகம், நம்பிக்கை மற்றும் குறிக்கோள் கதையாசிரியரை உருகச் செய்தது. ஒரு குறிக்கோளுக்காக அவர்கள் தங்களை அர்ப்பணித்துக் கொண்டது மனித குலத்திற்கே உரிதான பெருந்தன்மையை வெளிப்படுத்தியது. மனித குலத்திற்கே அவர்கள் நம்பிக்கையை கொண்டு வந்தனர்.

முடிவுரை:
லூசியாவின் சகோதரர்களின் தன்னலமற்ற கடின உழைப்பு மனித வாழ்க்கையின் அர்த்தத்தை நமக்கு உணர்த்தியது. இந்தக் கதையானது பிறரின் நல்வாழ்விற்காக தங்களை அர்ப்பணிப்பவர்கள் உள்ள வரையில் மனித குலம் தழைக்கும் என்ற செய்தியை நமக்கு அளிக்கிறது.

Two Gentlemen of Verona Glossary

Textual:

Additional:

Two Gentlemen of Verona Synonyms

Question 1.
He disannroved of their shabby appearance.
(a) tidy
(b) clean
(c) crisp
(d) wom-out
Answer:
(d) wom-out

Question 2.
He shrugged his shoulders.
(a) lowered
(b) raised lightly
(c) widened
(d) decreased
Answer:
(b) raised lightly

Question 3.
I was gazing at the two little figures.
(a) blinking
(b) looking intently
(c) gruelling
(d) glistening
Answer:
(b) looking intently

Question 4.
They had tangled hair.
(a) smooth
(b) matted/twisted
(c) straight
(d) Well-groomed
Answer:
(b) matted/twisted

Question 5.
The narrator was attracted to their dark earnest eyes.
(a) comic
(b) serious
(c) innocent
(d) yearning
Answer:
(b) serious

Question 6.
Jacopo barely came up to the door handle of the car.
(a) fully
(b) hardly
(c) simply
(d) complexly
Answer:
(b) hardly

Question 7.
They were doing brisk business in the public square.
(a) indolent
(b) active
(c) lethargic
(d) dull
Answer:
(b) active

Question 8.
We went over to the boys as trade slackened.
(a) got brisk
(b) intensified
(c) quickened up
(d) slowed down
Answer:
(d) slowed down

Question 9.
My interest was again provoked.
(a) annoyed
(b) irritated
(c) aroused
(d) exasperated
Answer:
(c) aroused

Question 10.
They had remarkable demeanour.
(a) meanness
(b) manner
(c) make-up
(d) policy
Answer:
(b) manner

Question 11.
In many ways the boys were artless.
(a) artificial
(b) innocent
(c) pretentious
(d) snobbish
Answer:
(b) innocent

Question 12.
They hawked newspapers.
(a) sold in public places
(b) displacment
(c) showcased
(d) brougth
Answer:
(a) sold in public places

Question 13.
We came upon them in the windy and deserted square,
(a) populated
(b) uninhabited
(c) dense
(d) occupied
Answer:
(b) uninhabited

Question 14.
you must be saving to emigrate to America.
(a) visit another country
(b) desert
(c) settle down
(d) become a citizen in another country
Answer:
(b) desert

Question 15.
There was a pause.
(a) closed
(b) temporary stop
(c) evicted
(d) seized
Answer:
(b) temporary stop

Question 16.
We drove to a tiny village.
(a) stupendous
(b) very small
(c) large
(d) spectacular
Answer:
(b) very small

Question 17.
I imagined to see some humble dwellings.
(a) proud
(b) modest
(c) immodest
(d) grand
Answer:
(b) modest

Question 18.
They disappeared beyond the comer of a wall.
(a) left
(b) came
(c) emerged
(d) vanished
Answer:
(d) vanished

Question 19.
I rang the bell determinedly.
(a) resolutely
(b) gradually
(c) intermitently
(d) cautiously
Answer:
(a) resolutely

Question 20.
A pleasant looking woman with a steel rimmed glass opened the door.
(a) irritated
(b) intrepid
(c) morose
(d) happy
Answer:
(d) happy

Question 21.
With a smile she bade me look through the partition.
(a) requested
(b) ordered
(c) humbled
(d) humiliated
Answer:
(b) ordered

Question 22.
A girl was listening to their chatter.
(a) series of meaningless sounds
(b) chopper
(c) copter
(d) callousness
Answer:
(a) series of meaningless sounds

Question 23.
Her eyes were soft and tender.
(a) harsh
(b) gentle
(c) sly
(d) stubborn
Answer:
(b) gentle

Question 24.
I could not bear to intrude upon their privacy.
(a) help
(b) protect
(c) predate
(d) encroach
Answer:
(d) encroach

Question 25.
They persuaded us to take her in.
(a) dissuaded
(b) convinced
(c) disdained
(d) discouraged
Answer:
(b) convinced

Two Gentlemen of Verona Antonyms

Find out the antonym of the underlined word in each of the following sentences.

Question 1.
The driver disapproved of their shabby appearance.
(a) untidy
(b) dirty
(c) smart
(d) unkempt
Answer:
(c) smart

Question 2.
They had tangled hair.
(a) matted
(b) twisted
(c) organised
(d) mangled
Answer:
(c) organised

Question 3.
The children were artless.
(a) innocent
(b) simple
(c) natural
(d) pretentious
Answer:
(d) pretentious

Question 4.
They were doing brisk business in the public square
(a) quick
(b) intensified
(c) sluggish
(d) feverish
Answer:
(c) sluggish

Question 5.
We waited until trade slackened.
(a) slowed down
(b) intensified
(c) got closed
(d) begen
Answer:
(b) intensified

Question 6.
Their earnestness provoked my interest.
(a) aroused
(b) teased
(c) deterred
(d) preempted
Answer:
(c) deterred

Question 7.
We came upon a windy deserted square.
(a) uninhabited
(b) depeopled
(c) unoccupied
(d) crowded
Answer:
(d) crowded

Question 8.
There was a pause.
(a) period
(b) stop
(c) resumption
(d) end
Answer:
(c) resumption

Question 9.
We drove to a tiny village.
(a) very small
(b) huge
(c) minute
(d) dwarfis
Answer:
(b) huge

Question 10.
I expected to see some humble dwellings.
(a) modest
(b) small
(c) meek
(d) proud
Answer:
(d) proud

Question 11.
They disappeared beyond the comer.
(a) vanished
(b) appeared
(c) ran away
(d) melting
Answer:
(b) appeared

Question 12.
I rang the bell determinedly.
(a) firmly
(b) irresolutely
(c) decidedly
(d) unhesitatingly
Answer:
(b) irresolutely

Question 13.
He saw a pleasant looking nurse.
(a) gleeful
(b) happy
(c) unpleasant
(d) kind
Answer:
(c) unpleasant

Question 14.
Her eyes were soft and tender.
(a) caring
(b) kind
(c) noble
(d) unsympathetic
Answer:
(d) unsympathetic

Question 15.
They persuaded us to take her in.
(a) convinced
(b) satisfied
(c) paified
(d) dissuaded
Answer:
(d) dissuaded