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You can Download Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.5

Multiple Choice Questions
Question 1.
Which of the following is not a measure of dispersion?
(1) Range
(2) Standard deviation
(3) Arithmetic mean
(4) Variance
Solution:
(3) Arithmetic mean

Question 2.
The range of the data 8, 8, 8, 8, 8. . . 8 is _____
(1) 0
(2) 1
(3) 8
(4) 3
Answer:
(1) 0
Hint:
Range = L – S = 8 – 8 = 0

Question 3.
The sum of all deviations of the data from its mean is
(1) Always positive
(2) always negative
(3) zero
(4) non-zero integer
Solution:
(3) zero

Question 4.
The mean of 100 observations is 40 and their standard deviation is 3. The sum of squares of all deviations is ______
(1) 40000
(2) 160900
(3) 160000
(4) 30000
Answer:
(2) 160900
Hint:

Question 5.
Variance of the first 20 natural numbers is
(1) 32.25
(2) 44.25
(3) 33.25
(4) 30
Solution:
(3) 33.25

Question 6.
The standard deviation of a data is 3. If each value is multiplied by 5 then the new variance is ______
(1) 3
(2) 15
(3) 5
(4) 225
Answer:
(4) 225
Hint:
Standard deviation = 3
Each value is multiplied by 5
New standard deviation = 3 × 5 = 15
New variance = 152 = 225

Question 7.
If the standard deviation of x, y, z is p then the standard deviation of 3x + 5, 3y + 5, 3z + 5 is
(1) 3p + 5
(2) 3p
(3) p + 5
(4) 9p + 15
Solution:
(2) 3p

Question 8.
If the mean and coefficient of variation of a data are 4 and 87.5% then the standard deviation is _____
(1) 3.5
(2) 3
(3) 4.5
(4) 2.5
Answer:
(1) 3.5
Hint:

Question 9.
Which of the following is incorrect?
(1) P (A) > 1
(2) 0 ≤ P(A) ≤ 1
(3) P(ϕ) = 0 (4)
(4) P (A) + P(\(\overline{\mathbf{A}}\)) = 1
Solution:
(1) P(A) > 1

Question 10.
The probability a red marble selected at random from a jar containing p red, q blue and r green marbles is

Solution:
(2) \(\frac{p}{p+q+r}\)

Question 11.
A page is selected at random from a book. The probability that the digit at units place of the page number chosen is less than 7 is

Solution:
(2) \(\frac{7}{10}\)

Question 12.
The probability of getting a job for a person is \(\frac{x}{3}\). If the probability of not getting the job is \(\frac{2}{3}\) then the value of x is ____
(1) 2
(2) 1
(3) 3
(4) 1.5
Answer:
(2) 1
Hint:

Question 13.
Kamalam went to play a lucky draw contest. 135 tickets of the lucky draw were sold. If the probability of Kamalam winning is \(\frac{1}{9}\), then the number of tickets bought by Kamalam is
(1) 5
(2) 10
(3) 15
(4) 20
Solution:
(3) 15
Hint:
\(=\frac{1}{9} \times 135=15\)

Question 14.
If a letter is chosen at random from the English alphabets {a, b, ……, z} then the probability that the letter chosen precedes x _______
(1) \(\frac{12}{13}\)
(2) \(\frac{1}{13}\)
(3) \(\frac{23}{26}\)
(4) \(\frac{3}{26}\)
Answer:
(3) \(\frac{23}{26}\)
Hint:

Question 15.
A purse contains 10 notes of ₹ 2000, 15 notes of ₹ 500, and 25 notes of ₹ 200. One note is drawn at random. What is the probability that the note is either a ₹ 500 note or ₹ 200 note?
(1) \(\frac{1}{5}\)
(2) \(\frac{3}{10}\)
(3) \(\frac{2}{3}\)
(4) \(\frac{4}{5}\)
Solution:
(4) \(\frac{4}{5}\)

Students can Download Tamil Nadu 11th Maths Model Question Paper 5 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Maths Model Question Paper 5 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 90

PART – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
Let X= {1,2, 3, 4} and R = {(1, 1), (1, 2), (1, 3), (2, 2), (3, 3) (2, 1), (3, 1), (1, 4), (4, 1)} then R is……………
(a) reflexive
(b) symmetric
(c) transitive
(d) equivalence
Solution:
(b) symmetric

Question 2.
Find a so that the sum and product of the roots of the equation 2x² – (a – 3)x + 3a – 5 = 0 are equal is…………
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2

Question 3.
If π < 2θ < \(\frac{3π}{2}\) then \(\sqrt{2+\sqrt{2+2 \cos 4 \theta}}\) = ………….
(a) -2 cos θ
(b) -2 sin θ
(c) 2 cos θ
(d) 2 sin θ
Solution:
(d) 2 sin θ

Question 4.
If f(θ) = |sin θ| + |cos θ|, θ ∈ R then f(θ) is in the interval……………
(a) [0, 2]
(b) [1, √2]
(c) [ 1, 2]
(d) [0, 1]
Solution:
(b) [1, √2]

Question 5.
Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to…………
(a) 60
(b) 600
(c) 720
(d) 7200
Solution:
(d) 7200

Question 6.
The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is…………
(a) 6
(b) 9
(c) 12
(d) 18
Solution:
(d) 18

Question 7.
The nth term of the sequence \(\frac{1}{3}\), \(\frac{3}{4}\), \(\frac{7}{8}\),\(\frac{15}{16}\)…. is …………..
(a) 2ⁿ – n – 1
(b) 1 – 2^-n
(c) 2^-n + n – 1
(d) 2^n-1
Solution:
(b) 1 – 2^-n

Question 8.
The remainder when 3815 is divided by 13 is…………
(a) 12
(b) 1
(c) 11
(d) 5
Solution:
(a) 12

Question 9.
If the straight line joining the points (2, 3) and (-1, 4) passes through the point (α, β) then…………
(a) α + 2β = 7
(b) 3α + β = 9
(c) α + 3β = 11
(d) 3α + β = 11
Solution:
(c) α + 3β = 11

Question 10.
If the equation of the base opposite to the vertex (2, 3) of an equilateral triangle is x + y = 2 then the length of a side is………….
(a) \(\sqrt{\frac{3}{2}}\)
(b) 6
(c) √6
(d) 3√2
Solution:
(c) √6

Question 11.
If A and B are symmetric matrices of order n where A ≠ B then…………
(a) A + B is skew symmetric
(b) A + B is symmetric
(c) A + B is a diagonal matrix
(d) A + B is a zero matrix
Solution:
(b) A + B is symmetric

Question 12.
If A is a square matrix then which of the following is not symmetric?
(a) A + A^T
(b) AA^T
(c) A^TA
(d) A – A^T
Solution:
(d) A – A^T

Question 13.
\(\lim _{x \rightarrow \infty}\) \(\frac{a^x-b^x}{x}\) = …………….
(a) log ab
(b) log \(\frac{a}{b}\)
(c) log log \(\frac{b}{a}\)
(d) \(\frac{a}{b}\)
Solution:
(b) log \(\frac{a}{b}\)

Question 14.
The function f(x) =  is discontinuous at ………..
(a) x = 0
(b) x = 1
(c) x = -2
(d) x = 2
Solution:
(d) x = 2

Question 15.
The function f(x) = \(\left\{\begin{array}{ll} 2 & x \leq 1 \\ x & x>1 \end{array}\right.\) is not differentiable at………..
(a) x = 0
(b) x = 1
(c) x = -1
(d) x = 2
Solution:
(b) x = 1

Question 16.
The number of points in R in which the function f(x) = |x – 1| + |x – 3| + sin x is not differentiable is…………
(a) 3
(a) 2
(c) l
(d) 4
Solution:
(a) 2

Question 17.
If y = 1 +  + …..∞ then \(\frac{dx}{dy}\) =……..
(a) x
(b) x²
(c) y
(d) y²
Solution:
(d) y²

Question 18.
\(\int \frac{\sqrt{\tan x}}{\sin 2 x}\) dx = …………….
(a) \(\sqrt{tan x}\) + c
(b) 2\(\sqrt{tan x}\) + c
(c) \(\frac{1}{2}\) \(\sqrt{tan x}\) + c
(d) \(\frac{1}{4}\) \(\sqrt{tan x}\) + c
Solution:
(a) \(\sqrt{tan x}\) + c

Question 19.
An urn contains 5 red and 5 black balls. A balls is drawn at random, its colour is noted and is returned to the um. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. The probability that the second ball drawn is red will be ………….
(a) \(\frac{5}{12}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{7}{12}\)
(d) \(\frac{1}{4}\)
Solution:
(b) \(\frac{1}{2}\)

Question 20.
A bag contains 6 green, 2 white, and 7 black balls. If two balls are drawn simultaneously then the probability that both are different colours is……….
(a) \(\frac{68}{105}\)
(b) \(\frac{71}{105}\)
(c) \(\frac{64}{105}\)
(d) \(\frac{73}{105}\)
Solution:
(a) \(\frac{68}{105}\)

PART – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
On the set of natural numbers let R be the relation defined by aRb if 2a + 3b = 30. Write down the relation by listing all the pairs. Check whether it is (i) reflexive, (ii) symmetric, (iii) transitive, (iv) equivalence
Solution:
N = {set of natural numbers};
R ={(3,8), (6, 6), (9, 4), (12, 2)}
(3, 3) ∉ R ⇒ R is not reflexive
(3, 8) ∈ R (8, 3) ∉ R
2a + 3b = 30
3b = 30 – 2a
b = \(\frac{30-2a}{3}\)
⇒ R is not symmetric
(a, b) (b, c) ∉ R ⇒ R is transitive
∴ It is not equivalence relation.

Question 22.
Compute log₉27 – log₂₇9.
Solution:
Let log₉27 = x ⇒ 27 = 9^x ⇒ 3³ = (3²)^x = 3^2x
⇒ 2x = 3 ⇒ x = 3/2
Let log₂₇ 9 = x
9 = 27^x
3² = (3³)^x ⇒ 3² = 3^3x
3x = 2 ⇒ x = 2/3
∴ log₉27 – log₂₇9 = \(\frac{3}{2}\) – \(\frac{2}{3}\) = \(\frac{9-4}{6}\) = \(\frac{5}{6}\)

Question 23.
Write the first 6 terms of the exponential series e^5x
Solution:

Question 24.
Find the points on the line x + y = 5, that lie at a distance 2 units from the line 4x + 3y – 12 = 0.
Solution:
Any point on the line x + y = 5 is x = t, y = 5 – t
The distance from (t, 5 – t) to the line 4x.+ 3y – 12 = 0 is given by 2 units.
∴ \(\frac{4(t)+3(5-t)-12}{\sqrt {4^2+3^2}}\) = 2 ⇒ \(\frac{|t+3|}{5}\) = 2
⇒ t + 3 = ± 10
t = -13, t = 7
∴ The points (-13, 18) and (7, -2).

Question 25.
Find |\(\vec{a}\) × \(\vec{b}\)| where \(\vec{a}\) = 3\(\vec{i}\) + 4\(\vec{j}\) and \(\vec{b}\) = \(\vec{i}\) +\(\vec{j}\) + \(\vec{k}\)
Solution:
\(\vec{a}\) × \(\vec{b}\) = \(\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 0 \\ 1 & 1 & 1 \end{array}\right|\) = \(\hat{i}\)(4 – 0) – \(\hat{j}\)(3 – 0) + \(\hat{k}\){3 – 4) = 4\(\hat{i}\) – 3\(\hat{j}\) – \(\hat{k}\)
|\(\vec{a}\) × \(\vec{b}\)| = |4\(\hat{i}\) – 3\(\hat{j}\) – \(\hat{k}\)| = \(\sqrt{16+9+1}\) = \(\sqrt{26}\)

Question 26.
Evaluate \(\lim _{x \rightarrow 1} \frac{x^{m}-1}{x^{n}-1}\) m and n are integers.
Solution:

Question 27.
Find the derivative of sinx² with respect to x²
Solution:
Here u = sinx² and v = x²
Now we have to find

Question 28.
Evaluate ∫[5x⁴ + 3(2x + 3)⁴ – 6(4 – 3x)⁵]
Solution:
∫[5 x⁴ + 3(2x + 3)⁴ – 6(4 – 3x)⁵] dx.
= 5∫x⁴dx + 3∫ (2x + 3)⁴ dx – 6∫ (4 – 3x)⁵ dx
= 5.\(\frac{x^5}{5}\) + 3.\(\frac{1}{2}\)\(\frac{(2x+3)^5}{5}\) – 6.\(\frac{1}{(-3)}\)\(\frac{(4-3x)^6}{6}\) + c.
= x⁵ + \(\frac{3}{10}\)(2x + 3)⁵ + \(\frac{1}{3}\) (4 – 3x)⁶ + c

Question 29.
Given that P(A) = 0.52, P(B) = 0.43, and P(A ∩ B) = 0.24, find P(A ∪ B)
Solution:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.52 + 0.43 – 0.24
P(A ∪ B) = 0.71

Question 30.
For what value of x, the matrix A = \(\left[\begin{array}{rrr} 0 & 1 & -2 \\ -1 & 0 & x^{3} \\ 2 & -3 & 0 \end{array}\right]\) is skew-symmetric.
Solution:

-x³ = -3 ⇒ x³ = 3 ⇒ x = 3^1/3

PART- III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
Find the largest possible domain for the real valued function given by f(x) =\(\frac{\sqrt{9-x^{2}}}{\sqrt{x^{2}-1}}\)
Solution:
If x < -3 or x > 3, then x² will be greater than 9 and hence 9 – x² will become negative which has no square root in R. So x must lie on the interval [-3, 3].
Also if x ≥ – 1 and x ≤ 1, then x² – 1 will become negative or zero. If it is negative, x² – 1^e has no square root in R. If it is zero, f is not defined. So x must lie outside [-1, 1]. That is, x must lie on (- ∞, -1) ∪ (1, ∞). Combining these two conditions, the largest possible domain for/is [-3, 3] ∩ ((-∞, -1) ∪ (i, ∞)). That is, [-3, -1) ∪ (1, 3].

Question 32.
Solve sin x + sin 5x = sin 3x
Solution:
sin x + sin 5x = sin 3x ⇒ 2 sin 3x cos 2x = sin 3x
sin 3x (2 cos 2x – 1) = 0
Thus, either sin 3x = 0 (or) cos 2x = \(\frac{1}{2}\)
If sin 3x = 0, then 3x = nπ ⇒ x = \(\frac{nπ}{3}\) n ∈ Z ………(i)
If cos 2x = \(\frac{1}{2}\) ⇒ cos 2x = cos \(\frac{π}{3}\)
2x = 2nn ± \(\frac{π}{3}\) ⇒ x = nπ ± \(\frac{π}{6}\), n ∈ Z ….(ii)
From (i) and (ii), we have the general solution x = \(\frac{π}{3}\) (or) x = nπ ± \(\frac{π}{6}\), n ∈ Z

Question 33.
How many triangles can be formed by 15 points in which 7 of them lie on one line and the remaining 8 on another parallel line?
Solution:

7 points lie on one line and the other 8 points parallel on another parallel line.
A triangle is obtained by taking one point from one line and 2 points from the other parallel line which can be done as follows.
⁷C₁ × ⁸C₂ or ⁷C₂ × ⁸C₁
⁷C₁ = 7; ⁷C₂ = \(\frac{7×6}{2×1}\) = 21
⁸C₁ = 8; ⁸C₂= \(\frac{8×7}{2×1}\) = 28
∴ No. of triangles = (7) (28) + (21) (8) = 196 + 168 = 364

Question 34.
Using binomial theorem indicate which of the following two numbers is larger (1.01)^1000000 or 10000
Solution:
(1.01)^1000000 = (1 + 0.01)^1000000
= ^1000000C₀(1)^1000000 + ^1000000C₁(1)⁹⁹⁹⁹⁹⁷(0.01)¹
+ ^1000000C₂(1)⁹⁹⁹⁹⁹⁸(0.01)² + ^1000000C₃(1)⁹⁹⁹⁹⁹⁷(0.01)³ +……….
= 1 (1) + 1000000 × \(\frac{1}{10^2}\) + \(\frac{1000000×999999}{2}\) × \(\frac{1}{10000}\) + …………
= 1 + 10000 + 50 × 999999 +…………
which is > 10000
So (1.01)^1000000 > 10000 (i.e.) (1.01)^1000000 is larger.

Question 35.
If a line joining two points (3, 0) and (5, 2) is rotated about the point (3, 0) in counter clockwise direction through an angle 15°, then find the equation of the line in the new position.
Solution:

Let P (3, 0) and Q (5, 2) be the given points.
Slope of PQ = \(\frac{y_2-y_1}{x_2-x_1}\) = 1
⇒ The angle of inclination of the line PQ = tan^-1(1) = \(\frac{π}{4}\) = 45°
∴ The slope of the line in new position is m = tan (45° + 15°)
⇒ Slope = tan (60°) = (√3)
∴ Equation of the straight line passing through (3, 0) and with the slope √3 is y – 0 = √3 (x – 3)
√3 x – y – 3√3 = 0

Question 36.
Find the area of the triangle whose vertices are A(3, -1, 2), B(l, -1, -3) and C(4, -3,1)
Solution:
A = (3, -1, 2), B = (1, -1, -3) and C = (4, -3, 1)
∴ \(\vec{OA}\) = 3\(\hat{i}\) – \(\hat{j}\) + 2\(\hat{k}\); \(\vec{OB}\) = \(\hat{i}\) – \(\hat{j}\) – 3\(\hat{k}\); and \(\vec{OC}\) = 4\(\hat{i}\) – 3\(\hat{j}\) + \(\hat{k}\)
Area of ΔABC = \(\frac{1}{2}\)|\(\vec{AB}\) × \(\vec{AC}\)| = \(\frac{1}{2}\)|\(\vec{BA}\) × \(\vec{BC}\)| = \(\frac{1}{2}\)|\(\vec{CA}\) x \(\vec{CB}\)|
\(\vec{AB}\) = \(\vec{OB}\) – \(\vec{OA}\) = (\(\hat{i}\) – \(\hat{j}\) – 3\(\hat{k}\)) – (3\(\hat{i}\) – \(\hat{j}\) + 2\(\hat{k}\)) = \(\hat{i}\) – \(\hat{j}\) – 3\(\hat{k}\) – 3\(\hat{i}\) + \(\hat{j}\) – 2\(\hat{k}\)
= -2\(\hat{i}\) – 5\(\hat{k}\)
\(\vec{AC}\) = \(\vec{OC}\) – \(\vec{OA}\) = 4\(\hat{i}\) – 3\(\hat{j}\) + \(\hat{k}\) – 3\(\hat{i}\) + \(\hat{j}\) – 2\(\hat{k}\)
= \(\hat{i}\) – 2\(\hat{j}\) – \(\hat{k}\)

Question 37.
Check if \(\lim _{x \rightarrow-5}\) f(x) exists or not, where f(x) = \(\left\{\begin{array}{l} \frac{|x+5|}{x+5}, \text { for } x \neq-5 \\ 0, \quad \text { for } x=-5 \end{array}\right.\)
Solution:
(i) f(-5^–)
For x < -5, |x + 5| = -(x + 5)
Thus f(-5^–) = \(\lim _{x \rightarrow-5^-}\) \(\frac{-(x-5)}{(x+5)}\) = -1

(ii) f(-5⁺)
For x > -5, |x + 5| = (x + 5)
Thus f(-5⁺) = \(\lim _{x \rightarrow-5^+}\) \(\frac{(x+5)}{(x+5)}\) = 1
∴ f(-5^–) ≠ f(-5⁺). Hence the limit does not exist.

Question 38.
Evaluate \(\frac{\sqrt{x}}{1+\sqrt{x}}\)
Solution:

Question 39.
The probability that a girl, preparing for competitive examination will get a State Government service is 0.12, the probability that she will get a Central Government job is 0.25, and the probability that she will get both is 0.07. Find the probability that
(i) she will get atleast one of the two jobs
(ii) she will get only one of the two jobs.
Solution:
Let I be the event of getting State Government service and C be the event of getting Central Government job.
Given that P(I) = 0.12, P(C) = 0.25, and P(I ∩ C) = 0.07
(i) P (at least one of the two jobs) = P(I or C) = P(I ∪ C)
= P(I) + P(C) – P(I ∩ C)
= 0.12 + 0.25 – 0.07 = 0.30

(if) P(only one of the two jobs) = P[only I or only C].
= P(I ∩ \(\bar{C}\)) + P(\(\bar{I}\) ∩ C)
= (P(I) – P(I ∩ C)} + (P(C) – P(I ∩ C)}
= {0.12 – 0.07} + {0.25 – 0.07}
= 0.23.

Question 40.
Find the derivatives of the following function. \(\sqrt{xy}\) = e^(x-y)
Solution:
\(\sqrt{xy}\) = e^x-y
(i.e.) (xy)^1/2 = e^x-y
Taking log on both sides we get
log (xy)^1/2 = log e^x-y
(i.e.) \(\frac{1}{2}\) (log x + log y) = x – y
⇒ log x + log y = 2x – 2y
differentiating w.r. to x we get

PART – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41 (a).
Find the range of the function \(\frac{1}{2cos x -1}\)
Solution:
The range of cos x is – 1 to 1
-1 < cos x < 1
(× by 2) -2 < 2 cos x < 2
adding -1 throughout
-2 – 1 <2 cos x – 1 < 2 – 1
(i.e.,) -3 < 2 cos x – 1 < 1
so 1 < \(\frac{1}{2cos x -1}\) < \(\frac{-1}{3}\)
The range is outside \(\frac{-1}{3}\) and 1
i.e., range is (-∞, \(\frac{-1}{3}\)] ∪ [1, ∞)

[OR]

(b) Solve \(\frac{(x-2)}{(x+4)}\) ≥ \(\frac{5}{(x+3)}\)
Solution:

x + 4 = 0 ⇒ x = -4; x + 3 = 0 ⇒ x = -3
Plotting the points -4, -3 on number line and taking limits (-∞, -4), (-4, -3), (-3, ∞)

The solution for the inequality \(\frac{(x-2)}{(x+4)}\) ≥ \(\frac{5}{(x+3)}\) are the intervals (-∞, -4) and (-4, -3)

Question 42 (a).
Prove that log 2 + 16 log \(\frac{16}{15}\) + 12 log \(\frac{25}{24}\) + 7 log \(\frac{80}{81}\) = 1
Solution:
LHS = log 2 + 16 [log 16 – log 15] + 12 [log 25 – log 24] + 7 [log 81 – log 80]

= log 2 + 16 [log 2⁴ – log 3 × 5 ] + 12 [log 5² – log 2³ × 3] + 7[log 3⁴ – log 2⁴ × 5]
= log 2 + 16 [41og2 – log3 – log5] + 12 [2 log 5 – 3 log 2 – log 3] + 7 [4 log 3 – 4 log 2 – log 5]
= log 2 + 64 log 2 – 16 log 3 – 16 log 5 + 24 log 5 – 36 log 2 – 12 log 3 + 28 log 3 – 28 log 2 – 7 log 5
= log 2 [1+ 64 – 36 – 28] + log 3 [-16 – 12 + 28] + log 5 [-16 + 24 – 7]
= log 2(1) + log 3(0) +log 5(1)
= log 2 + log 5 = log 2 × 5 = log 10 = 1 = RHS

[OR]

(b) If tan α = \(\frac{1}{3}\) and tan β = \(\frac{1}{7}\) show that 2 α + β = \(\frac{π}{4}\)
Solution:

∴ 2 α + β = 45° = \(\frac{π}{4}\)

Question 43 (a).
Using Binomial theorem, prove that 6ⁿ – 5n always leaves remainder 1 when divided by 25 for all positive integer n.
Solution:
To prove this it is enough to prove, 6ⁿ – 5n = 25k + 1 for some integer k. We first consider the expansion
(1 + x)ⁿ = ⁿC₀ + ⁿC₁ x + ⁿC₂ x² +…+ ⁿC_n-1x^n-1 + ⁿC_n xⁿ, n ∈ N.
Taking x = 5 we get (1 + 5)ⁿ = ⁿC₀ + ⁿC₁ 5 + ⁿC₂ 5² +…+ ⁿC_n-1 5^n-1 + ⁿC_n 5ⁿ. The above equality reduces to 6ⁿ = 1 + 5n + 25(ⁿC₂ + 5 ⁿC₃ +…+ ⁿC_n 5^n-2)
That is,
6ⁿ – 5n = 1 + 25(ⁿC₂ + 5 ⁿC₃ +…+ ⁿC_n 5^n-2) = 1 + 25k, k ∈ N.
Thus 6ⁿ – 5n always, leaves remainder 1 when divided by 25 for all positive integer n.

[OR]

(b) Prove that

Solution:
Taking p = 0, we get |A| = \(\left|\begin{array}{ccc} (q+r)^{2} & 0 & 0 \\ q^{2} & r^{2} & q^{2} \\ r^{2} & r^{2} & q^{2} \end{array}\right|\) = 0
Therefore, (p- 0) is a factor. That is, p is a factor.
Since |A| is in cyclic symmetric form in p, q, r and hence q and r also factors.
Putting p + q + r = 0 ⇒ q + r = -p; r + p = -q; and p + q = -r.
|A| = \(\left|\begin{array}{lll} p^{2} & p^{2} & p^{2} \\ q^{2} & q^{2} & q^{2} \\ r^{2} & r^{2} & r^{2} \end{array}\right|\) = 0 since 3 columns are identical.
Therefore, (p + q + r)² is a factor of |A|.
The degree of the obtained factor pqr(p + q + r)² is 5. The degree of |A| is 6.
Therefore required factor is k (p + q + r)

4(16 – 1) -1 (4 – 1) +1 (1 – 4) = 27k
60 – 3 – 3 = 27 k ⇒ k = 2.
|A| = 2pqr (p + q + r)³

Question 44 (a).
The sum of the distance of a moving point from the points (4, 0) and (-4, 0) is always 10 units. Find the equation of the locus of the moving point.
Solution:
Let point (h, k) be a moving point
Here A = (4, 0) and B = (- 4, 0)
Given PA + PB = 10
⇒ \(\sqrt{(h-4)^{2}+k^{2}}+\sqrt{(h+4)^{2}+k^{2}}\) = 10
⇒ \(\sqrt{(h-4)^{2}+k^{2}}\) = 10 – \(\sqrt{(h+4)^{2}+k^{2}}\)
Squaring both sides (h – 4)² + k² = 100 + (h + 4)² + k² – 20\(\sqrt{(h+4)^{2}+k^{2}}\)
(i.e.) h² + l6 – 8h + k² = 100 + h² + 16 + 8h + k² – 20\(\sqrt{(h+4)^{2}+k^{2}}\)
⇒ -16h – 100 = – 20 \(\sqrt{(h+4)^{2}+k^{2}}\)
(÷ by -4) 4h + 25 = 5 \(\sqrt{(h+4)^{2}+k^{2}}\)
Squaring both sides we get,
(4h + 25)² = 25 [(h + 4)² + k²]
(i.e) 16h² + 25 + 200h = 25 [h² + 8h + 16 + k²]
= 16h² + 625 + 200h – 25h² – 200h – 400 – 25k² = 0
= – 9h² – 25k² + 225 = 0
⇒ 9h² + 25 k² = 225
\(\frac{9h^2}{225}\) + \(\frac{25k^2}{225}\) = 1
(i.e) h²/25 + k²/9 = 1, So the locus is \(\frac{x^2}{25 }\) + \(\frac{y^2}{9}\) = 1

[OR]

(b) Show that the vectors \(\hat{i}\) – 2\(\hat{j}\) + 3\(\hat{k}\), -2\(\hat{i}\) + 3\(\hat{j}\) – 4\(\hat{k}\) and – \(\hat{j}\) + 2\(\hat{k}\) are coplanar.
Solution:
Let the given three vectors be \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\). When we are able to write one vector as a linear combination of the other two vectors, then the given vectors are called coplanar vectors.
Let \(\vec{a}\) = m\(\vec{b}\) + n\(\vec{c}\) where
(i.e.) \(\vec{a}\) = \(\hat{i}\) – 2\(\hat{j}\) + 3\(\hat{k}\)
\(\vec{b}\) = -2\(\hat{i}\) + 3\(\hat{j}\) – 4\(\hat{k}\) and \(\vec{c}\) = –\(\hat{j}\) + 2\(\hat{k}\)
⇒ \(\hat{i}\) – 2\(\hat{j}\) + 3\(\hat{k}\) = m (-2\(\hat{i}\) + 3\(\hat{j}\) – 4\(\hat{k}\)) + n (-\(\hat{j}\) + 2\(\hat{k}\))
Equating the \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) components
(i.e.) 1 = -2m ……… (1)
-2 = 3m – n …………(2)
3 = -4m + 2n …………(3)
Now we have to solve (1) and (2) and substitute the value in (3).
Solving (1) and (2)
(1) ⇒ -2m = 1
∴ m = –\(\frac{1}{2}\)
Substituting m = –\(\frac{1}{2}\) in (2) we get,
3(\(\frac{-1}{2}\)) – n = -2
–\(\frac{3}{2}\) – n = -2
∴ -n = -2 + \(\frac{3}{2}\) = \(\frac{-4+3}{2}\) = –\(\frac{1}{2}\)
n = \(\frac{1}{2}\)
∴ m = –\(\frac{1}{2}\); n = \(\frac{1}{2}\)
Substituting the values of m and n in (3).
LHS = 3
RHS = -4 + 2n = -4(\(\frac{-1}{2}\)) + 2(\(\frac{1}{2}\))
= 2 + 1 = 3
⇒ LHS = RHS
∴ we are able to write one vector as a linear combination of the other two
⇒ the given vectors are coplanar.

Question 45 (a).
A committee of 7 peoples has to be formed from 8 men and 4 women. In how many ways can this be done when the committee consists of
(i) exactly 3 women?
(ii) at least 3 women?
(iii) at most 3 women?
Solution:

We need a committee of 7 people with 3 women and 4 men.
This can be done in (⁴C₃) (⁸C₄) ways
⁴C₁ = ⁴C₁ = 4
⁸C₄ = \(\frac{8×7×6×5}{4×3×2×1}\) = 70
The number of ways = (70) (4) = 280

(ii) Atleast 3 women

So the possible ways are (3W and 4M) or (4W and 3M)
(i.e) (⁴C₃) (⁸C₄) + (⁴C₄) (⁸C₃)
⁴C₃ = ⁴C₁ = 4; ⁴C₄ = 1
⁸C₄ = \(\frac{8×7×6×5}{4×3×2×1}\) = 70
⁸C₃ = \(\frac{8×7×6}{3×2×1}\) = 56
The number of ways (4) (70) + (1) (56) = 280 + 56 = 336

(iii) Atmost 3 women

The possible ways are (0W 8M) or (1W 6M) or (2W 5M) or (3W 4M)

∴ The possible ways are
(1) (8) + (4) (28) + (6) (56) + (4) (70) = 8 + 112 + 336 + 280 = 736 ways

[OR]

(b) Evaluate \(\lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x}\)
Solution:

Question 46 (a).
If y = \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\) show that (1 – x²)y₂ – 3xy₁ – y = 0
Solution:
y = \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\)
⇒ y \(\sqrt{1-x^2}\) = sin^-1 x
differentiating w.r.to x we get

multiplying both sides by \(\sqrt{1-x^2}\) we get
-xy + (1 – x²)y₁ = 1
differentiating both sides again w.r.to x.
– [xy^-1 +y(1)] + (1 – x²) (y₂) + y₁(-2x) = 0
(i.e.) -xy₁ – y + (1 – x²)y₂ – 2xy₁ = 0
(1 – x²)y₂ – 3xy₁ – y = 0

[OR]

(b) If y = cos (m sin^-1 x), prove that (1 – x²) y₃ – 3xy₂ + (m² – 1) y₁ = 0
Solution:
We have y = cos (m sin^-1 x)
y₁ = sin(m sin^-1 x) \(\frac{m}{\sqrt{1-x^2}}\)
\(y_{1}^{2}\) = sin² (m sin^-1 x) \(\frac{m^2}{(1-x^2)}\)
This implies (1 – x²) \(y_{1}^{2}\) = m² sin² (m sin^-1 x) = m² [1 – cos² (m sin^-1 x)]
This is, (1 – x²) \(y_{1}^{2}\) = m² (1 – y²).
Again differentiating,
(1 – x²) 2y₁ \(\frac{dy_1}{dx}\) + \(y_{1}^{2}\)(-2x) = m² (-2y\(\frac{dy}{dx}\))
(1 – x²) 2y₁y₂ – 2x\(y_{1}^{2}\) = – 2m²yy₁
(1 – x²) y₂ – xy₁ = m²y
Once again differentiating
(1 – x²) 2y₁ \(\frac{dy_2}{dx}\) + \(y_{2}\)(-2x) – [x.\(\frac{dy_1}{dx}\) + y₁.1] = -m² (\(\frac{dy}{dx}\))
(1 – x²)y₃ – 2xy₂ – xy₂ – y₁ = -m²y₁
(1 – x²)y₃ – 3xy₂ + (m² – 1)y₁ = 0

Question 47 (a).
Evaluate ∫\(\frac{dx}{\sqrt{9+8x-x^2}}\)
Solution:
Let I = ∫\(\frac{dx}{\sqrt{9+8x-x^2}}\) dx
Consider, 9 + 8x – x²
= -[x² – 8x – 9]
= -[(x – 4)² – 16 – 9]
= -[(x – 4)² – (5)²]
= (5)² – (x – 4)²

[OR]

(b) An advertising executive is studying television viewing habits of married men and women during prime time hours. Based on the past viewing records he has determined that during prime time wives are watching television 60% of the time. It has also been determined that when the wife is watching television, 40% of the time the husband is also watching. When the wife is not watching the television, 30% of the time husband is watching the television. Find the probability that
(i) the husband is watching the television during the prime time of television
(ii) if the husband is watching the television, the wife is also watching the television.
Solution:
P(Wife watching TV) = P(W) = \(\frac{60}{100}\)
P(H/W) = \(\frac{40}{100}\); P(H/W’) = 30/100
(i) P(Husband watching TV) = P(H)
= P(H/W) P(W) + P(H/W’) P(W’)

You can Download Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 7 Mensuration Additional Questions

Question 1.
If the radii of the circular ends of a conical bucket which is 45 cm high are 28 cm and 7 cm, find the capacity of the bucket. (Use \(\pi=\frac{22}{7}\))
Solution:
Clearly bucket forms frustum of a cone such that the radii of its circular ends are r₁ = 28 cm, r₂ = 7 cm, h = 45 cm.
Capacity of the bucket = volume of the frustum

Question 2.
Find the depth of a cylindrical tank of radius 28 m, if its capacity is equal to that of a rectangular tank of size 28 m × 16 m × 11 m.
Solution:
Volume of the cylindrical tank = Volume of the rectangle tank

Question 3.
What is the ratio of the volume of a cylinder, a cone, and a sphere. If each has the same diameter and same height?
Solution:

Question 4.
Find the number of coins, 1.5 cm is diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:

Question 5.
A spherical ball of iron has been melted and made into small balls. If the raidus of each smaller ball is one-fourth of the radius of the original one, how many such balls can be made?
Solution:

Question 6.
A wooden article was made by scooping out a hemisphere from each end of a cylinder as shown in figure. If the height of the cylinder is 10cm and its base is of radius 3.5 cm find the total surface area of the article.

Solution:
Radius of the cylinder be r Height of the cylinder be h Total surface area of the article = CSA of cylinder + CSA of 2 hemispheres = 2πrh + 2πr² = 2πr (h + 2r)

You can Download Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 7 Mensuration Unit Exercise 7

Question 1.
The barrel of a fountain-pen cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used for writing 330 words on an average. How many words can be written using a bottle of ink containing one fifth of a litre?
Solution:

Question 2.
A hemi-spherical tank of radius 1.75 m is full of water. It is connected with a pipe which empties the tank at the rate of 7 litre per second. How much time will it take to empty the tank completely?
Solution:
Suppose the pipe takes x seconds to empty the tank. Then, volume of the water that flows out of the tank in x seconds = Volume of the hemispherical tank.
Volume of the water that flows out of the tank in x seconds.
= Volume of hemispherical shell of radius 175 cm.

Question 3.
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r units.
Solution:
Radius of the base of cone = Radius of the hemisphere = r
Height of the cone = Radius of the hemisphere

Question 4.
An oil funnel of tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion be 8cm and the diameter of the top of the funnel be 18 cm, then find the area of the tin sheet required to make the funnel.
Solution:
Slant height of the frustum

Question 5.
Find the number of coins, 1.5 cm in diameter and 2 mm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:
No. of coins required .

Question 6.
A hollow metallic cylinder whose external radius is 4.3 cm and internal radius is 1.1 cm
and whole length is 4 cm is melted and recast into a solid cylinder of 12 cm long. Find the diameter of solid cylinder.
Solution:
Volume of the solid cylinder = Volume of the hollow cylinder melted.

Question 7.
The slant height of a frustum of a cone is 4 m and the perimeter of circular ends are 18 m and 16 m. Find the cost of painting its curved surface area at ₹ 100 per sq. m.
Solution:

Question 8.
A hemi-spherical hollow bowl has material of volume \(\frac{436 \pi}{3}\) cubic cm. Its external diameter is 14 cm. Find its thickness.
Solution:

Question 9.
The volume of a cone is \(1005 \frac{5}{7}\) cu. cm. The area of its base is \(201 \frac{1}{7}\) sq. cm. Find the slant height of the cone.
Solution:

Question 10.
A metallic sheet in the form of a sector of T a circle of radius 21 cm has central angle of 216°. The sector is made into a cone by bringing the bounding radii together. Find the volume of the cone formed.
Solution:

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.5

Multiple choice questions.
Question 1.
The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is ______
(1) 60π cm²
(2) 68π cm²
(3) 120π cm²
(4) 136π cm²
Answer:
(4) 136π cm²
Hint:

Question 2.
If two solid hemispheres of same base radius r units are joined together along with their bases, then the curved surface area of this new solid is
(1) 4πr² sq. units
(2) 67πr² sq. units
(3) 3πr² sq. units
(4) 8πr² sq. units
Solution:
(1) 47πr² sq. units]

Question 3.
The height of a right circular cone whose radius is 5 cm and slant height is 13 cm will be __________
(1) 12 cm
(2) 10 cm
(3) 13 cm
(4) 5 cm
Answer:
(1) 12 cm
Hint:

Question 4.
If the radius of the base of a right circular cylinder is halved keeping the same height, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is
(1) 1 : 2
(2) 1 : 4
(3) 1 : 6
(4) 1 : 8
Solution:
(2) 1 : 4

Question 5.
The total surface area of a cylinder whose radius is \(\frac{1}{3}\) of its height is

Solution:
(3) \(\frac{8 \pi h^{2}}{9}\) sq. units

Question 6.
In a hollow cylinder, the sum of the external and internal radii is 14 cm and the width is 4 cm. If its height is 20 cm, the volume of the material in it is _______
(1) 560π cm³
(2) 1120π cm³
(3) 56π cm³
(4) 360π cm³
Answer:
(2) 1120π cm³
Hint:
R + r = 14 cm
w = 4 cm
h = 90 cm

Question 7.
If the radius of the base of a cone is tripled and the height is doubled then the volume is
(1) made 6 times
(2) made 18 times
(3) made 12 times
(4) unchanged
Solution:
(2) made 18 times
Hint:

Question 8.
The total surface area of a hemisphere is how many times the square of its radius ______
(1) π
(2) 4π
(3) 3π
(4) 2π
Answer:
(3) 3π
Hint:

Question 9.
A solid sphere of radius x cm is melted and cast into a shape of a solid cone of same radius. The height of the cone is
(1) 3x cm
(2) x cm
(3) 4x cm
(4) 2x cm
Solution:
(3) 4x cm
Hint:

Question 10.
A frustum of a right circular cone is of height 16 cm with radii of its ends as 8 cm and 20 cm. Then, the volume of the frustum is _______
(1) 3328π cm3
(2) 3228π cm3
(3) 3240π cm3
(4) 3340π cm3
Answer:
(1) 3328π cm3Hint:

Question 11.
A shuttlecock used for playing badminton has the shape of the combination of
(1) a cylinder and a sphere
(2) a hemisphere and a cone
(3) a sphere and a cone
(4) frustum of a cone and a hemisphere
Solution:
(4) frustum of a cone and a hemisphere

Question 12.
A spherical ball of radius r₁ units is melted to make 8 new identical balls each of radius r₂ units. Then r₁ : r₂ is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Solution:
Hint:

Question 13.
The volume (in cm³) of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is

Solution:
(1) \(\frac{4}{3} \pi\)

Question 14.
The height and radius of the cone of which the frustum is a part are h₁ units and r₁ units respectively. Height of the frustum is h₂ units and the radius of the smaller base is r₂ units. If h₂: h₁ = 1 : 2 then r₂ : r₁ is
(1) 1 : 3
(2) 1 : 2
(3) 2 : 1
(4) 3 : 1
Solution:
(2) 1 : 2

Question 15.
The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is ____
(1) 1 : 2 : 3
(2) 2 : 1 : 3
(3) 1 : 3 : 2
(4) 3 : 1 : 2
Answer:
(4) 3 : 1 : 2
Hint:

Students can Download Tamil Nadu 11th Physics Model Question Paper 4 English Medium Pdf, Tamil Nadu 11th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Physics Model Question Paper 4 English Medium

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts. An internal choice of questions is provided wherever: applicable
3. All questions of Part I, II, III, and IV are to be attempted separately
4. Question numbers 1 to 15 in Part I are Multiple choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

PART – I

Answer all the questions: [15 × 1 = 15]

Question 1.
A parallax of heavenly body measured from two points diametrically opposite on equator of Earth is 1.0 minute. If the radius of Earth is 6400 km the distance of the body is …………
(a) 8.8 × 10¹⁰m
(b) 4.4 × 10¹⁰m
(c) 0.29 × 10^-10m
(d) 8.6 × 10^-10m
Answer:
(b) 4.4 × 10¹⁰m
Hint:
θ = 1 min = \(\frac{1}{60}\) × \(\frac{π}{1800}\)rad
Diameter of earth, d = 2 × R_E = 2 × 6400 × 10³m
Distance of the heavenly body from the centre of the earth, r = \(\frac{d}{θ}\) = \(\frac{2×6400×10^3}{\frac{π}{60×80}}\)
r = 4.4 × 10¹⁰m

Question 2.
A particle is thrown vertically upwards, its velocity at half of the height is 10 m/s then the maximum height attained by it is (g = 10 m/s²)
(a) 8 m
(b) 20 m
(c) 10 m
(d) 16 m
Answer:
(c) 10 m
Hint: From equation of motion, v² = u² – 2as
O = (10)² + 2(-10) s
∴S = 5 m
u = 10 ms^-1
v = 0 ms^-1
a = -10 ms^-2
Total height = 2 × 5 = 10 m

Question 3.
If the velocity is \(\vec{v}\) = 2\(\vec{i}\) + t²\(\vec{j}\) – 9\(\vec{k}\), then the magnitude, of acceleration at t = 0.5 s is……….
(a) 1 ms^-2
(b) 2 ms^-2
(c) zero
(d) -1 ms^-2
Answer:
(a) 1 ms^-2
Hint:
a = \(\frac{dy}{dt}\) = \(\frac{d}{dt}\) (2\(\vec{i}\) + t²\(\vec{j}\) – 9\(\vec{k}\)) = 2t\(\vec{j}\)
at, t = 0.5 s ⇒ a = 2 (0.5).= 1 ms^-2

Question 4.
A uniform force of (2\(\vec{i}\) + \(\vec{j}\))N acts on a particle of mass 1 kg. The particle displaces from position (3\(\vec{j}\) + \(\vec{k}\)) m to (5\(\vec{i}\) + 3\(\vec{j}\)) m. The work done by the force on the particle is……….
(a) 9 J
(b) 6 J
(c) 10 J
(d) 12 J
Answer:
(c) 10 J
Hint:
\(\vec{F}\) = (2\(\vec{i}\)+ \(\vec{j}\))N; Δ\(\vec{r}\) = \(\vec{j}_2\) – \(\vec{r}_2\) = (5\(\vec{i}\) + 3\(\vec{j}\)) – (3\(\vec{i}\) + \(\vec{k}\)); Δr = 5\(\vec{i}\) – \(\vec{k}\)
Workdone, \(\vec{W}\) = \(\vec{F}\). Δ\(\vec{r}\) = (2\(\vec{i}\) + \(\vec{j}\)) . (5\(\vec{i}\) – \(\vec{k}\))
\(\vec{W}\) = 10\(\vec{i}\) \(\vec{W}\) = 10Nm = 10J

Question 5.
A couple produces……….
(a) pure rotation
(b) pure translation
(c) rotation and translation
(d) no motion
Answer:
(a) pure rotation

Question 6.
What is the shape, when a non-wetting liquid is placed in a capillary tube?
(a) convex upwards
(b) concave upwards
(c) concave downwards
(d) convex downwards
Answer:
(a) convex upwards

Question 7.
An ideal gas heat engine operators in a carnot’s cycle between 227°C and 127°C. It absorbs 6 × 10⁴J at high temperature. The amount of heat converted into work is………
(a) 2.4 × 10⁴J
(b) 4.8 × 10⁴J
(c) 1.2 × 10⁴J
(d) 6 × 10⁴J
Answer:
(c) 1.2 × 10⁴J
Hint:
\(\frac{W}{Q}\) = 1 – \(\frac{T_2}{T_1}\)
W = (1 – \(\frac{273+127}{273+227}\)) × 10⁴ = 1.2 × 10⁴J

Question 8.
Four round objects namely a ring, a disc, a hollow sphere and a solid sphere with same radius R start to roll down an incline at the same time. Find out the order of objects reaching the bottom first?
(a) solid sphere, disc, hollow sphere, ring
(b) ring, disc, hollow sphere, solid sphere
(c) disc, ring, solid sphere, hollow sphere
(d) hollow sphere, disc, ring, solid sphere
Answer:
(a) solid sphere, disc, hollow sphere, ring

Question 9.
Two forces of magnitude F having a resultant of the same magnitude of F, the angle between the two forces is…………
(a) 45°
(b) 60°
(c) 120°
(d) 150°
Answer:
(c) 120°
Hint:
Magnitude of each force is F
∴ The resultant force, F = \(\sqrt{F^2+F^2+2F.Fcosθ}\)
F² = 2F² + 2F² cos θ ⇒ cos θ = –\(\frac{1}{2}\) ⇒ θ = 120°

Question 10.
If v₀ and v denote the sound velocity and the rms velocity of the molecules in a gas, then……..
(a) v₀ = v(\(\frac{3}{r}\))^{\(\frac{1}{2}\)}
(b) v₀ = 0
(c) v₀ = v(\(\frac{r}{3}\))^{\(\frac{1}{2}\)}
(d) v₀ and v are not related
Answer:
(c) v₀ = v(\(\frac{r}{3}\))^{\(\frac{1}{2}\)}

Question 11.
The internal energy of an ideal gas depends on ………..
(a) pressure
(b) volume
(c) temperature
(d) size of molecules
Answer:
(c) temperature

Question 12.
The internal energy of an ideal gas increases during an isothermal process, when the gas is ……….
(a) expanded by adding more molecules to it
(b) expanded by adding more heat to it
(c) expanded against zero pressure
(d) compressed by doing work on it
Answer:
(a) expanded by adding more molecules to it

Question 13.
A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its……….
(a) 3 Hz
(b) 2 Hz
(c) 4 Hz
(d) 1 Hz
Answer:
(d) 1 Hz
Hint:
A = 5 cm = 5 × 10^-2m ; υ_max = 31.4 cm/s = 31.4 × 10^-2 m/s
Maximum speed V_max = 2πη × A
∴ n = \(\frac{V_{max}}{2πA}\) = \(\frac{31.4×10^{-2}}{2π×5×10^{-2}}\) = \(\frac{31.4}{10×3.14}\); n = 1 Hz

Question 14.
A hollow sphere is filled with water. It is hung by a long thread. As the water flows out of a hole at the bottom, the period of oscillation will………
(a) first increase and then decrease
(b) first decrease and then increase
(c) increase continuously
(d) decrease continuously
Answer:
(a) first increase and then decrease

Question 15.
A wave travels in a medium according to the equation of displacement given by y(x, t) = 0.03 sin {π(2t – 0.01 x)} where y and x are in metres and t in seconds. The wave length of the wave is……….
(a) 200 m
(b) 100 m
(c) 20 m
(d) 10 m
Answer:
(a) 200 m
Hint:
λ = \(\frac{2π}{K}\) = \(\frac{π}{0.01π}\) = 200 m

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
Distinguish scalar and vector.
Answer:

Question 17.
Calculate the total number of degrees of freedom possessed by the molecular in one cm³ of H₂ gas at NTP.
Answer:
22400 cm³ of every gas contains 6.02 × 10²³ molecules
∴ Number of molecules in 1 cm² of H₂ gas = \(\frac{6.02×10^{23}}{22400}\) = 0.26875 × 10²⁰
Number of degrees of freedom of a H₂ gas molecule = 5
∴ Total number of degrees of freedom of 0.26875 × 10²⁰ molecules
= 0.26875 × 10²⁰ × 5 = 1.34375 × 10²⁰

Question 18.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolution in 25 s, what is the magnitude and direction of acceleration of the stone?
Answer:
The acceleration will be directed towards the centre of the circular loop
angular velocity, ω = 2πf = 2 × 3.14 × \(\frac{14}{25}\); ω = \(\frac{88}{25}\) rad/s
Centripetal acceleration = rω² = \(\frac{0.8×(88)^2}{(25)^2}\); a_c = 9.91 m/s²

Question 19.
There are two identical balls of same material, one being solid and the other being hollow. How will you distinguish them without weighting?
Answer:
Solid and hollow balls can be differentiated by different methods
(a) by spinning than using equal torques
(b) by determining their moment of inertia ie I_h > I_s
(c) by rolling them down is an inclined plane
ie, when torques are equal angular acceleration of hollow must be smaller than that of solid. Similarly, on rolling, solid ball will reach the bottom before the hollow ball.

Question 20.
In a dark room would you be able to tell whether a given note had been produced by a Piano or a Violin?
Answer:
Yes, in a dark room we can easily identify a sound produced by a Piano or a Violin by using the knowledge of timber or quality of sound. The two sources even though having the same intensity and fundamental frequency will be associated with different number of overtones of different relative intensities. These overtones combine and produce different sounds which enables us to identify them.

Question 21.
Will water at the foot of the waterfall be at a different temperature from that at the top? If yes explain.
Answer:
When water reaches the ground, its gravitational potential energy is converted into kinetic energy which is further converted into heat energy. This raises the temperature of water. So, water at the foot of the water fall is at a higher temperature of water at the top of the waterfall.

Question 22.
Which one among a solid, liquid gas of same mass and at the same temperature has the greatest internal energy. Which one has least and why?
Answer:
A gas has greatest value of internal energy. Being a negative potential energy, potential energy of its molecules is smallest. Internal energy of solid is maximum because negative potential energy of its molecules is maximum.

Question 23.
Is it possible if work is done by the internal force. What will be the change in kinetic energy?
Answer:
Yes. this is possible. If work is done by the internal forces then kinetic energy will be increased. As an example, when a bomb explodes, the combined kinetic energy of all the fragments is greater than the initial energy.

Question 24.
What is red shift and blue shift in Doppler effect.
Answer:
If the spectral lines of the star are found to shift towards red end of the spectrum (called as red shift) then the star is receding away from the Earth. Similarly, if the spectral lines of the star are found to shift towards the blue end of the spectrum (called as blue shift) then the star is approaching Earth.

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
Give the difference between systematic errors and random errors.
Answer:
Systematic errors: Systematic errors are reproducible inaccuracies that are consistently in the same direction. These occur often due to a problem that persists throughout the experiment.

Random errors: Random errors may arise due to random and unpredictable variations in experimental conditions like pressure, temperature, voltage supply etc. Errors may also be due to personal errors by the observer who performs the experiment. Random errors are sometimes called “chance error”. When different readings are obtained by a person every time he repeats the experiment, personal error occurs.

Question 26.
Write down the kinematics equation for the object moving in a straight line with constant acceleration and also for free falling body.
Answer:
(a) the equation of motion of a moving object with constant acceleration is

1. v – u + at
2. s = ut+ \(\frac{1}{2}\) at²
3. v² – u² = 2 as

(b) for free falling body u = 0 and a = g

1. V = gt
2. s = \(\frac{1}{2}\) gt2
3. v² = 2 gs

Question 27.
A nucleus is at rest in the laboratory frame of reference show that if it disintegrates into two smaller nuclei. The products must be emitted in opposite directions.
Answer:
Let m₁, m₂ are be the masses of product nuclei and v₁, v₂ are the velocities of it.
∴ linear momentum after disintegration = m₁v₁ + m₂v₂
Before disintegration nucleus is at rest therefore its linear momentum before disintegration is zero. According to the principle of conservation of linear momentum
m₁v₁ + m₂v₂ = 0
v₂ = –\(\frac{m_1v_1}{m_2}\)

Question 28.
How do you classify the physical quantities on the basis of dimension?
Answer:

1. Dimensional variables: Physical quantities, which possess dimensions and have variable values are called dimensional variables. Examples are length, velocity, and acceleration etc.
2. Dimensionless variables: Physical quantities which have no dimensions,. but have variable values are called dimensionless variables. Examples are specific gravity, strain, refractive index etc.
3. Dimensional Constant: Physical quantities which possess dimensions and have constant values are called dimensional constants. Examples are Gravitational constant, Planck’s constant etc.
4. Dimensionless Constant: Quantities which have constant values and also have no dimensions are called dimensionless constants. Examples are π, e, numbers etc.

Question 29.
Write short notes on the oscillations of liquid column in U-tube.
Answer:
Oscillations of liquid in U-tube.

Consider a U-shaped glass tube which consists of two open arms with uniform cross sectional area A. Let us pour a non-viscous uniform incompressible liquid of density ρ in the U-shaped tube to a height h as shown in the figure. If the liquid and tube are not disturbed then the liquid surface will be in equilibrium position O. It means the pressure as measured at any point on the liquid is the same and also at the surface on the arm (edge of the tube on either side), which balances with the atmospheric pressure. Due to this the level of liquid in each arm will be the same. By blowing air one can provide sufficient force in one arm, and the liquid gets disturbed from equilibrium position O, which means, the pressure at blown arm is higher than the other arm. This creates difference in pressure which will cause the liquid to oscillate for a very short duration of time about the mean or equilibrium position and finally comes to rest, Time period of the oscillation is
T = 2π\(\sqrt{\frac{l}{2g}}\) second

Question 30.
What are the factors affecting the surface tension of a liquid.
Answer:

1. The presence of any contamination or impurities.
2. The presence of dissolved substances.
3. Electrification
4. Temperature

Question 31.
Write a note on Brownian motion.
Answer:
Brownian motion is due to the bombardment of suspended particles by molecules of the surrounding fluid. But during 19th century people did not accept that every matter is made up of small atoms or molecules. In the year 1905, Einstein gave systematic theory of Brownian motion based on kinetic theory and he deduced the average size of molecules.

According to kinetic theory, any particle suspended in a liquid or gas is continuously bombarded from all the directions so that the mean free path is almost negligible. This leads to the motion of the particles in a random and zig-zag manner. But when we put our hand in water it causes no random motion because the mass of our hand is so large that the momentum transferred by the molecular collision is not enough to move our hand.

Factors affecting Brownian Motion:

1. Brownian motion increases with increasing temperature.
2. Brownian motion decreases with bigger particle size, high viscosity and density of the liquid (or) gas.

Question 32.
What is the acceleration of the block and troller system as the figure. If the Co-efficient of kinetic friction between the trolley and the surface is 0.04? Also calculate friction in the string. Take G = 10 ms^-2, mass of the string is negligible.
Answer:

Free body diagram of the block
30 – T = 3a ……..(1)
Free body diagram of the trolley
T – f_k = 20a ……(2)
where f_k = = µ_k N = 0.04 × 20 × 10 = 8 N
Solving (1) & (2), a = 0.96 m/s² and T = 27.2 N

Question 33.
An increase in pressure of 100 kPa causes a certain volume of water to decrease by 0.005% of its original volume.
(a) Calculate the bulk modulus of water?
Answer:
Bulk modulus
B = v|\(\frac{Δp}{Δv}\)| = \(\frac{100×10^3}{0.005×10^{-2}}\)
B = 2000 MPa

(b) Compute the speed of sound (compressional waves) in water?
Answer:
v = \(\sqrt{\frac{B}{ρ}}\) = \(\sqrt{\frac{2000×10_6}{1000}}\)
v = 1414 ms^-1

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
Describe the vertical oscillations of a spring?
Answer:
Vertical oscillations of a spring: Let us consider a massless spring with stiffness constant or force constant k attached to a ceiling as shown in figure. Let the length of the spring before loading mass m be L.

If the block of mass m is attached to the other end of spring, then the spring elongates by a length. Let F, be the restoring force due to stretching of spring. Due to mass m, the gravitational force acts vertically downward. We can draw free-body diagram for this system as shown in figure. When the system is under equilibrium,

F₁ + mg = 0 ……………… (1)
But the spring elongates by small displacement 1, therefore,
F₁ ∝ l ⇒ F₁ = -kl ……………….. (2)
Substituting equation (2) in equation (1) we get
-kl + mg = 0
mg = kl or
\(\frac{m}{k}\) = \(\frac{l}{g}\) ………………….. (3)

Suppose we apply a very small external force on the mass such that the mass further displaces downward by a displacement y, then it will oscillate up and down. Now, the restoring force due to this stretching of spring (total extension of spring is y + 1) is
F₂ ∝ (y + l)
F₂ = -k(y + l) = -ky – kl …………………… (4)

Since, the mass moves up and down with acceleration \(\frac{d^{2} y}{d t^{2}}\), by drawing the free body diagram for this case we get
-ky – kl + mg = m \(\frac{d^{2} y}{d t^{2}}\) ……………………. (5)
The net force acting on the mass due to this stretching is
F = F₂ + mg
F = -ky – kl + mg …………………… (6)
The gravitational force opposes the restoring force. Substituting equation (3) in equation (6), we get
F = – ky- kl + kl = -ky
Applying Newton’s law we get
m \(\frac{d^{2} y}{d t^{2}}\) = -ky
\(\frac{d^{2} y}{d t^{2}}\) = –\(\frac{k}{m}\)y ………………… (7)
The above equation is in the form of simple harmonic differential equation. Therefore, we get the time period as
T = 2π\(\sqrt{m/k}\) second ……………………. (8)
The time period can be rewritten using equation (3)
T = 2π\(\sqrt{m/k}\) = 2πl\(\frac{1}{g}\) second ……………………. (9)
The accleration due to gravity g can be computed by the formula
g = 4π²\((\frac { 1 }{ T } )^{ 2 }\)ms^-2 …………………….. (10)

[OR]

(b) Derive poiseuille’s formula for the volume of a liquid flowing per second through a pipe under streamlined flow?
Answer:
Consider a liquid flowing steadily through a horizontal capillary tube. Let v = (\(\frac{1}{g}\)) be the volume of the liquid flowing out per second through a capillary tube. It depends on (1) coefficient of viscosity (η) of the liquid, (2) radius of the tube (r), and (3) the pressure gradient (\(\frac{P}{l}\)) . Then,
v ∝η^ar^b(\(\frac{P}{l}\))^c
v = kη^ar^b(\(\frac{P}{l}\))^c …………………….. (1)

where, k is a dimensionless constant.
Therefore, [v] = \(\frac { Volume }{ Time } \) = [L³T^-1]; [ \(\frac{dP}{dX}\) ] = \(\frac { Pressure }{ Distance } \) = [ML^-2T^-2]
[η] = [Ml^-1T^-1] and [r] = [L]

Substituting in equation (1)
[L³T^-1] = [ML^-1T^-1]^a[L]^b [ML^-2T^-2]^c
M⁰L³T^-1 = M^a+bL^-a+b-2cT^-a-2c = -1

So, equating the powers of M, L and T on both sides, we get
a + c = 0, – a + b – 2c = 3, and – a – 2c = – 1

We have three unknowns a, b and c. We have three equations, on solving, we get
a = – 1, b = 4 and c = 1

Therefore, equation (1) becomes,
v = kη^-1r⁴(\(\frac{P}{l}\))¹

Experimentally, the value of k is shown to be , we have \(\frac{π}{8}\), we have
v = \(\frac{\pi r^{4} \mathrm{P}}{8 \eta /}\)

The above equation is known as Poiseuille’s equation for the flow of liquid through a narrow tube or a capillary tube. This relation holds good for the fluids whose velocities are lesser than the critical velocity (v_c).

Question 35 (a).
Describe briefly simple harmonic oscillation as a projection of uniform circular motion?
Answer:
Consider a particle of mass m moving with unifonn speed v along the circumference of a circle whose radius is r in anti-clockwise direction (as shown in figure). Let us assume that the origin of the coordinate system coincides with the center O of the circle.

If ω is the angular velocity of the particle and θ the angular displacement of the particle at any instant of time t, then θ = ωt. By projecting the uniform circular motion on its diameter gives a simple harmonic motion.

This means that we can associate a map (or a relationship) between uniform circular (or revolution) motion to vibratory motion. Conversely, any vibratory motion or revolution can be mapped to unifonn circular motion. In other words, these two motions are similar in nature.

Let us first project the position of a particle moving on a circle, on to its vertical diameter or on to a line parallel to vertical diameter as shown in figure. Similarly, we can do it for horizontal axis or a line parallel to horizontal axis.

The projection of uniform circular motion on a diameter of SHM:
As a specific example, consider a spring mass system (or oscillation of pendulum). When the spring moves up and down (or pendulum moves to and fro), the motion of the mass or bob is mapped to points on the circular motion.

Thus, if a particle undergoes uniform circular motion then the projection of the particle on the diameter of the circle (or on a line parallel to the diameter) traces straight line motion which is simple harmonic in nature. The circle is known as reference circle of the simple harmonic motion. The simple harmonic motion can also be defined as the motion of the projection of a particle on any diameter of a circle of reference.

[OR]

(b) State and prove Bernoulli’s theorem for a flow of incompressible non viscous and stream lined flow of fluid?
Answer:
Bernoulli’s theorem:
According to Bernoulli’s theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains a constant. Mathematically,
\(\frac{P}{ρ}\) + \(\frac{1}{2}\)v² + gh – constant
This is known as Bernoulli’s equation.
Proof:
Let us consider a flow of liquid through a pipe AB. Let V be the volume of the liquid when it enters A in a time t. Which is equal to the volume of the liquid leaving B in the same time. Let a_A, v_A and P_A be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at A respectively.

Let the force exerted by the liquid at A is
F_A = P_Aa_A

Distance travelled by the liquid in time t is d = v_At
Therefore, the work done is W = F_Ad = P_Aa_Av_At
But a_Av_At = a_Ad = V, volume of the liquid entering at A.

Thus, the work done is the pressure energy (at A), W = F_Ad = P_AV

Pressure energy per unit volume at
A = \($\frac{\text { Pressure energy }}{\text { Volume }}$\) = \(\frac { P_{ A }V }{ V } \) = P_A

Pressure energy per unit mass at
A = \($\frac{\text { Pressure energy }}{\text { Mass }}$\) = \(\frac { P_{ A }V }{ m } \) = \(\frac { P_{ A } }{ \frac { m }{ V } } \) = \(\frac { P_{ A } }{ \rho } \)

Since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is
EP_A = P_AV = P_AV × (\(\frac{m}{m}\)) = m\(\frac { P_{ A } }{ \rho } \)

Potential energy of the liquid at A,
PE_A = mgh_A

Due to the flow of liquid, the kinetic energy of the liquid at A,
KE_A = \(\frac{1}{2}\)mv^2_A

Therefore, the total energy due to the flow of liquid at A,
E_A = EP_A + KE_A + PE_A
E_A = \(m \frac{P_{A}}{\rho}+\frac{1}{2} m v_{A}^{2}+m g h_{A}\)

Similarly, let a_B, V_B and P_B be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at B. Calculating the total energy at F_B, we get .
\(\mathrm{E}_{\mathrm{B}}=m \frac{\mathrm{P}_{\mathrm{B}}}{\rho}+\frac{1}{2} m v_{\mathrm{B}}^{2}+m g h_{\mathrm{B}}\)
From the law of conservation of energy.
E_A = E_B

Thus, the above equation can be written as
\(\frac { P }{ \rho g } \) + \(\frac{1}{2}\) \(\frac { v^{ 2 } }{ g } \) + h = Constant

The above equation is the consequence of the conservation of energy which is true until there is no loss of energy due to friction. But in practice, some energy is lost due to friction. This arises due to the fact that in a fluid flow, the layers flowing with different velocities exert frictional forces on each other. This loss of energy is generally converted into heat energy. Therefore, Bernoulli’s relation is strictly valid for fluids with zero viscosity or non-viscous liquids. Notice that when the liquid flows through a horizontal pipe, then
h = 0 ⇒ \(\frac { P }{ \rho g } \) + \(\frac{1}{2}\) \(\frac { v^{ 2 } }{ g } \) = Constant

Question 36 (a).
Explain perfect inelastic collision and derive an expression for loss of kinetic energy in perfect inelastic collision?
Answer:
In a perfectly inelastic or completely inelastic collision, the objects stick together permanently after collision such that they move with common velocity. Let the two bodies with masses m₁ and m₂ move with initial velocities u₁ and u₂ respectively before collision. Aft er perfect inelastic collision both the objects move together with a common velocity v as shown in figure.
Since, the linear momentum is conserved during collisions,

m₁u₁ + m₂u₂ = (m₁ + m₂) v

The common velocity can be computed by
v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}\) ………………….. (1)

Loss of kinetic energy in perfect inelastic collision;
In perfectly inelastic collision, the loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, heat, light etc. Let KE_i be the total kinetic energy before collision and KE_f be the total kinetic energy after collision.
Total kinetic energy before collision,
KE_e = \(\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}\) …………………… (2)
Total kinetic energy after collision,
KE_f = \(\frac{1}{2}\left(m_{1}+m_{2}\right) v^{2}\) …………………….. (3)
Then the loss of kinetic energy is Loss of KE, ∆Q = KE_f – KE_i
= \(\frac{1}{2}\left(m_{1}+m_{2}\right) v^{2}-\frac{1}{2} m_{1} u_{1}^{2}-\frac{1}{2} m_{2} u_{2}^{2}\) ………………….. (4)
Substituting equation (1) in equation (4), and on simplifying (expand v by using the algebra (a + b)² = a² + b² + 2ab), we get
Loss of KE, ∆Q = \(\frac{1}{2}\) \(\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)\) (u₁ – u₂)²

[OR]

(b) Derive an expression for maximum height attained, time of flight, horizontal range for a projectile in oblique projection?
Answer:

Maximum height (h_max):
The maximum vertical distance travelled by the projectile during the journey is called maximum height. This is determined as follows:
For the vertical part of the motion
\(v_{y}^{2}=u_{y}^{2}+2 a_{y} s\)
Here, u_y= u sin θ, a = -g, s = h_max, and at the maximum height v = 0

Time of flight (T_f):
The total time taken by the projectile from the point of projection till it hits the horizontal plane is called time of flight. This time of flight is the time taken by the projectile to go from point O to B via point A as shown in figure.
We know that s_y = u_yt + \(\frac{1}{2}\)a_yt²
Here, s_y = y = 0 (net displacement in y-direction is zero), u_y = u sin θ, a_y = -g, t = T_f, Then
0 = u sin θ T_f – \(\frac{1}{2} g \mathrm{T}_{f}^{2}\)
T_f = 2u \(\frac{sin θ}{g}\) …………………….. (2)

Horizontal range (R):
The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R). This is found easily since the horizontal component of initial velocity remains the same. We can write.

Range R = Horizontal component of velocity % time of flight = u cos θ × T_f = \(\frac{u^{2} \sin 2 \theta}{g}\)
The horizontal range directly depends on the initial speed (u) and the sine of angle of projection (θ). It inversely depends on acceleration due to gravity ‘g’.

For a given initial speed u, the maximum possible range is reached when sin 2θ is
maximum, sin 2θ = 1. This implies 2θ = π/2 or θ = π/4
This means that if the particle is projected at 45 degrees with respect to horizontal, it attains maximum range, given by
R_max = \(\frac { u^{ 2 } }{ g } \).

Question 37 (a).
Explain the work-energy theorem in detail and also give three examples?
Answer:

1. If the work done by the force on the body is positive then its kinetic energy increases.
2. If the work done by the force on the body is negative then its kinetic energy decreases.
3. If there is no work done by the force on the body then there is no change in its kinetic energy, which means that the body has moved at constant speed provided its mass remains constant.
4. When a particle moves with constant speed in a circle, there is no change in the kinetic energy of the particle. So according to work energy principle, the work done by centripetal force is zero.

[OR]

(b) (i) Define molar specific heat capacity?
Answer:
Molar specific heat capacity is defined as heat energy required to increase the temperature of one mole of substance by IK or 1°C

(ii) Derive Mayer’s relation for an ideal gas?

Mayer’s relation: Consider p mole of an ideal gas in a container with volume V, pressure P and temperature T.

When the gas is heated at constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU.

If C_V is the molar specific heat capacity at constant volume, from equation.
C_V = \(\frac { 1 }{ \mu } \) \(\frac{dU}{dT}\) …………………… (1)
dU = µC_V dT ………………… (2)

Suppose the gas is heated at constant pressure so that the temperature increases by dT. If ‘Q’ is the heat supplied in this process and ‘dV’ the change in volume of the gas.
Q = pCpdT ……………. (3)

If W is the workdone by the gas in this process, then
W = P dV ………………….. (4)

But from the first law of thermodynamics,
Q = dU + W ………………… (5)

Substituting equations (2), (3) and (4) in (5), we get,
For mole of ideal gas, the equation of state is given by
\(\mu \mathrm{C}_{\mathrm{p}} d \mathrm{T}=\mu \mathrm{C}_{\mathrm{v}} d \mathrm{T}+\mathrm{P} d \mathrm{V}\)

Since the pressure is constant, dP = 0
C_pdT = C_VdT + PdV
∴ C_p = C_V + R (or) C_p – C_V = R …………………… (6)
This relation is called Mayer’s relation It implies that the molar specific heat capacity of an ideal gas at constant pressure is greater than molar specific heat capacity at constant volume.
The relation shows that specific heat at constant pressure (s_p) is always greater than specific heat at constant volume (s_v).

Question 38 (a).
Derive an expression of pressure exerted by the gas on the walls of the container?
Answer:
Expression for pressure exerted by a gas : Consider a monoatomic gas of N molecules each having a mass m inside a cubical container of side l.
The molecules of the gas are in random motion. They collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of energy, but a change in momentum occurs.

The molecules of the gas exert pressure on the walls of the container due to collision on it. During each collision, the molecules impart certain momentum to the wall. Due to transfer of momentum, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. It is essential to determine the total momentum transferred by the molecules in a short interval of time.

A molecule of mass m moving with a velocity \(\vec { v } \) having components (v_x, v_y, v_z) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with same speed and its x-component is reversed. This is shown in the figure. The components of velocity of the molecule after collision are (-v_x, v_y, v_z).

The x-component of momentum of the molecule before collision = mv_x
The x-component of momentum of the molecule after collision = – mv_x
The change in momentum of the molecule in x direction
= Final momentum – initial momentum = – mv_x – mv_x = – 2mv_x
According to law of conservation of linear momentum, the change in momentum of the wall = 2 mv_x
The number of molecules hitting the right side wall in a small interval of time ∆t.

The molecules within the distance of v_x∆t from the right side wall and moving towards the right will hit the wall in the time interval ∆t. The number of molecules that will hit the right side wall in a time interval ∆t is equal to the product of volume (Av_x∆t) and number density of the molecules (n).

Here A is area of the wall and n is number of molecules per \(\frac{N}{V}\) unit volume. We have assumed that the number density is the same throughout the cube.

Not all the n molecules will move to the right, therefore on an average only half of the n molecules move to the right and the other half moves towards left side.

The number of molecules that hit the right side wall in a time interval ∆t
= \(\frac{n}{2}\) Av_x∆t
In the same interval of time ∆t, the total momentum transferred by the molecules
\(\Delta \mathrm{P}=\frac{n}{2} \mathrm{A} v_{x} \Delta t \times 2 m v_{x}=\mathrm{A} v_{x}^{2} m n \Delta t\) ………………….. (2)
From Newton’s second law, the change in momentum in a small interval of time gives rise to force.

The force exerted by the molecules on the wall (in magnitude)
F = \(\frac{∆p}{∆t}\) = nmAv^2_x ……………………. (3)

Pressure, P = force divided by the area of the wall
P = \(\frac{F}{A}\) = nmAv^2_x ……………………….. (4)
p = \(nm\bar{v}_{x}^{2}\)

Since all the molecules are moving completely in random manner, they do not have same . speed. So we can replace the term vnmAv^2_x by the average \(\bar { v } \)^2_x in equation (4).
P = nm\(\bar { v } \)^2_x ……………………. (5)

Since the gas is assumed to move in random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So, \(\bar{v}_{x}^{2}\) = \(\bar{v}_{y}^{2}\) = \(\bar{v}_{z}^{2}\). The mean square speed is written as
\(\bar{v}^{2}\) = \(\bar{v}_{x}^{2}\) + \(\bar{v}_{y}^{2}\) + \(\bar{v}_{z}^{2}\) = 3\(\bar{v}_{x}^{2}\)
\(\bar{v}_{x}^{2}\) = \(\frac{1}{3}\) \(\bar{v}^{2}\)
Using this in equation (5), we get
P = \(\frac{1}{3} n m \bar{v}^{2} \quad \text { or } P=\frac{1}{3} \frac{N}{V} m \bar{v}^{2}\) ………………….. (6)

[OR]

(b) Discuss the simple pendulum in detail?
Answer:
Simple pendulum

A pendulum is a mechanical system which exhibits periodic motion. It has a bob with mass m suspended by a long string (assumed to be massless and inextensible string) and the other end is fixed on a stand. At equilibrium, the pendulum does not oscillate and hangs vertically downward.

Such a position is known as mean position or equilibrium position. When a pendulum is displaced through a small displacement from its equilibrium position and released, the bob of the pendulum executes to and fro motion. Let l be the length of the pendulum which is taken as the distance between the point of suspension and the centre of gravity of the bob. Two forces act on the bob of the pendulum at any displaced position.

• The gravitational force acting on the body (\(\vec { F} \) = m\(\vec { g } \)) which acts vertically downwards.
• The tension in the string T which acts along the string to the point of suspension.

Resolving the gravitational force into its components:

1. Normal component: The component along the string but in opposition to the direction of tension, F_as = mg cos θ.
2. Tangential component: The component perpendicular to the string i.e., along tangential direction of arc of swing, F_ps = mg sin θ.

Therefore, The normal component of the force is, along the string,
\(\mathrm{T}-\mathrm{W}_{a s}=m \frac{v^{2}}{l}\)

Here v is speed of bob
T -mg cos θ = m \(\frac{v^{2}}{l}\)

From the figure, we can observe that the tangential component W_ps of the gravitational force always points towards the equilibrium position i.e., the direction in which it always points opposite to the direction of displacement of the bob from the mean position. Hence, in this case, the tangential force is nothing but the restoring force. Applying Newton’s second law along tangential direction, we have
\(m \frac{d^{2} s}{d t^{2}}+\mathrm{F}_{p s}=0 \Rightarrow m \frac{d^{2} s}{d t^{2}}=-\mathrm{F}_{p s}\)
\(m \frac{d^{2} s}{d t^{2}}=-m g \sin \theta\) …………………. (1)

where, s is the position of bob which is measured along the arc. Expressing arc length in terms of angular displacement i.e.,
s = lθ ………………… (2)
then its acceleration, \(\frac{d^{2} s}{d t^{2}}=l \frac{d^{2} \theta}{d t^{2}}\) …………………. (3)

Substituting equation (3) in equation (1), we get
\(\begin{aligned}
l \frac{d^{2} \theta}{d t^{2}} &=-g \sin \theta \\
\frac{d^{2} \theta}{d t^{2}} &=-\frac{g}{l} \sin \theta
\end{aligned}\) ………………….. (4)

Because of the presence of sin θ in the above differential equation, it is a non-linear differential equation (Here, homogeneous second order). Assume “the small oscillation approximation”, sin θ ~ 0, the above differential equation becomes linear differential equation.
\(\frac{d^{2} \theta}{d t^{2}}=-\frac{g}{l} \theta\) …………………… (5)

This is the well known oscillatory differential equation. Therefore, the angular frequency of this oscillator (natural frequency of this system) is
ω² = \(\frac{g}{l}\) …………………… (6)
∴ ω = \(\sqrt{g/l}\) in rad s^-1 ……………….. (7)

The frequency of oscillation is
f = \(f=\frac{1}{2 \pi} \sqrt{\frac{g}{l}} \text { in } \mathrm{Hz}\) ………………… (8)
and time period of sscillations is
T = 2π\(\sqrt{l/g}\) in second. ……………….. (9)

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 2 Introduction to Algebra Ex 2.3

Miscellaneous Practice Problems

Question 1.
Complete the following pattern.
9 – 1 =
98 – 21 =
987 – 321 =
9876 – 4321 =
98765 – 54321 =
What comes next?
Solution:
9 – 1 = 8
98 – 21 = 77
987 – 321 = 666
9876 – 4321 = 5555
98765 – 54321 = 44444
Next will be 987654 – 654321 = 333333

Question 2.
A piece of wire is ’12s’ cm long. What will be the length of the side, if it is formed as
(i) an equilateral triangle
(ii) a square?
Solution:
(i) 4s
(ii) 3s

Question 3.
Identify the value of the shapes and figures in the table given below the verify their addition horizontally and vertically.

Solution:

Question 4.
The table given below shows the results of the matches played by 8 teams in a Kabaddi championship tournament.

Find the value of all the variables in the table given above.
Solution:
We know that
Total matches played = Matches wont + Matches lost.
5 + k = 8
⇒ k = 3
7 + 2 = a
a = 9
7 = 6 + m
⇒ m = 1
b + 3 = 9
⇒ b = 6
4 + 6 = n
⇒ n = 10
6 + c = 10
⇒ c = 4
x + 4 = 8
⇒ x = 4
3 + 6 = y
⇒ y = 10

Challenging Problem (Text book Page No. 53)

Question 5.
Gopal is 8 years younger to Karnan. If the sum of their ages is 30, how old is Karnan?
Solution:
Let Kaman’s age be x years
Gopal age is x – 8 years
Given sum of their ages is 30. i.e, x + (x – 8) = 30
Now we will form the table to find their sums = 30.

For x = 19 (Karnan’s age)
the sum = 30
Kaman’s age = 19 years.

Question 6.
The rectangles made of identical square blocks with varying lengths but having only two square blocks as width are given below.

(i) How many smallest squares are there in each of the rectangles P, Q, R and S?
(ii) Fill in the boxes.

Solution:
(i) No. of smallest squares in P, Q, R and S are 2, 8, 6 and 10 respectively.
(ii)

Question 7.
Find the variables from the clues given below and solve the crossword puzzle.


Solution:

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• Shapes Worksheets

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.5

Integrate the following functions with respect to x
Question 1.
\(\frac{x^{3}+4 x^{2}-3 x+2}{x^{2}}\)
Solution:

Question 2.
\(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}\)
Solution:

Question 3.
(2x – 5)(36 + 4x)
Solution:

Question 4.
cot² x + tan² x
Solution:

Question 5.
\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\)
Solution:

Question 6.
\(\frac{\cos 2 x}{\sin ^{2} x \cos ^{2} x}\)
Solution:

Question 7.
\(\frac{3+4 \cos x}{\sin ^{2} x}\)
Solution:

Question 8.
\(\frac{\sin ^{2} x}{1+\cos x}\)
Solution:

Question 9.
\(\frac{\sin 4 x}{\sin x}\)
Solution:

Question 10.
cos 3x cos 2x
Solution:

Question 11.
sin² 5x
Solution:

Question 12.
\(\frac{1+\cos 4 x}{\cot x-\tan x}\)
Solution:

Question 13.
e^{x log a}e^x
Solution:

Question 14.
(3x + 4) \(\sqrt{3 x+7}\)
Solution:

Question 15.
\(\frac{8^{1+x}+4^{1-x}}{2^{x}}\)
Solution:

Question 16.
\(\frac{1}{\sqrt{x+3}-\sqrt{x-4}}\)
Solution:

Question 17.
\(\frac{x+1}{(x+2)(x+3)}\)
Solution:

x + 1 = A(x + 3) + B(x + 2) ………….. (i)
Put x = -3 in (i)
-2 = B(-1)
B = 2
Put x = -2 in (i)
-1 = A(1)
A = -1

Question 18.
\(\frac{1}{(x-1)(x+2)^{2}}\)
Solution:

Question 19.
\(\frac{3 x-9}{(x-1)(x+2)\left(x^{2}+1\right)}\)
Solution:

put x = 1 in (i)
3 – 9 = A(3)(2) + 0 + 0
– 6 = 6A
A = -1

put x = -2 in (i)
-6 – 9 = 0 + B(-3) (5) + 0
-15 = – 15B
B = 1

put x = 0 in (i)
0 – 9 = A(2)(1) + B(-1)(1) + D(-1)(2)
-9 = 2A – B – 2D
2D = 2A – B + 9
2D = 2(-1) – 1 + 9
= -2 – 1 + 9
2D = 6
D = 3

Equating x³ coefficients on b/s in (i)
A + B + C = 0
-1 + 1 + C = 0
C = 0

Question 20.
\(\frac{x^{3}}{(x-1)(x-2)}\)
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.5 Additional Problems

Integrate the following
Question 1.
\(\frac{e^{2 x}+e^{-2 x}+2}{e^{x}}\)
Solution:

Question 2.
cos³ 2x – sin 6x
Solution:
cos 3x = 4 cos³ x – 3 cos x
∴ 4 cos³ x = cos 3x + 3 cos x

Question 3.
\(\sqrt{1-\sin 2 x}\)
Solution:

Question 4.
cos 2x sin 4x
Solution:

Question 5.
(e^x – 1)² e^-4x
Solution:

Question 6.
(x + 1) \(\sqrt{x+3}\)
Solution:

Question 7.
(2x + 1)\(\sqrt{2x+3}\)
Solution:

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Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.3

Integrate the following with respect to x.

Question 1.
\(e^{x \log a}+e^{a \log a}-e^{n \log x}\)
Solution:

Question 2.
\(\frac{a^{x}-e^{x \log b}}{e^{x \log a} b^{x}}\)
Solution:

Question 3.
\(\left(e^{x}+1\right)^{2} e^{x}\)
Solution:

Question 4.
\(\frac{e^{3 x}-e^{-3 x}}{e^{x}}\)
Solution:

Question 5.
\(\frac{e^{3 x}+e^{5 x}}{e^{x}+e^{-x}}\)
Solution:

Question 6.
\(\left[1-\frac{1}{x^{2}}\right] e^{\left(x+\frac{1}{x}\right)}\)
Solution:

Question 7.
\(\frac{1}{x(\log x)^{2}}\)
Solution:

Question 8.
If f'(x) = e^x and f(0) = 2, then find f(x).
Solution:
f'(x) = e^x
Integrating both sides of the equation,
∫ f'(x) dx = ∫ e^x dx
⇒ f(x) = e^x + c
given f(0) = 2
2 = e⁰ + c
⇒ c = 1
Thus f(x) = e^x + 1

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Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems

Question 1.
Find the rank of the matrix A = \(\left(\begin{array}{rrrr}
1 & -3 & 4 & 7 \\
9 & 1 & 2 & 0
\end{array}\right)\)
Solution:
A = \(\left(\begin{array}{rrrr}
1 & -3 & 4 & 7 \\
9 & 1 & 2 & 0
\end{array}\right)\)
Order of A is 2 × 4. So ρ(A) ≤ 2
Consider the second order minor
\(\left|\begin{array}{cc}
1 & -3 \\
9 & 1
\end{array}\right|\) = 1 + 27 = 28 ≠ 0
There is a minor of order 2 which is not zero.
So ρ(A) = 2

Question 2.
Find the rank of the matrix A = \(\left(\begin{array}{rrrr}
-2 & 1 & 3 & 4 \\
0 & 1 & 1 & 2 \\
1 & 3 & 4 & 7
\end{array}\right)\)
Solution:
Given A = \(\left(\begin{array}{rrrr}
-2 & 1 & 3 & 4 \\
0 & 1 & 1 & 2 \\
1 & 3 & 4 & 7
\end{array}\right)\)
Order of A is 3 × 4. So ρ(A) ≤ 3
Consider the third order minor
\(\left|\begin{array}{ccc}
-2 & 1 & 3 \\
0 & 1 & 1 \\
1 & 3 & 4
\end{array}\right|\)
= -2(4 – 3) – 1(0 – 1) + 3(0 – 1)
= -2 + 1 – 3 =
= -4 ≠ 0
There exists a minor of order 3 which is not zero. So ρ(A) = 3

Question 3.
Find the rank of the matrix A = \(\left(\begin{array}{llll}
4 & 5 & 2 & 2 \\
3 & 2 & 1 & 6 \\
4 & 4 & 8 & 0
\end{array}\right)\)
Solution:
Given A = \(\left(\begin{array}{llll}
4 & 5 & 2 & 2 \\
3 & 2 & 1 & 6 \\
4 & 4 & 8 & 0
\end{array}\right)\)
Order of A is 3 × 4. So ρ(A) ≤ 3
Consider the third order minor,
\(\left|\begin{array}{lll}
4 & 5 & 2 \\
3 & 2 & 1 \\
4 & 4 & 8
\end{array}\right|\)
= 4(16 – 4) – 5(24 – 4) + 2 (12 – 8)
= 48 – 100 + 8
= -44 ≠ 0
There is a minor of order 3 which is not zero.
ρ(A) = 3

Question 4.
Examine the consistency of the system of equations:
x + y + z = 7, x + 2y + 3z = 18, y + 2z = 6
Solution:
The given system is x + y + z = 7 , x + 2y + 3z = 18, y + 2z = 6.
The matrix equation corresponding to the given system


The last equivalent matrix is in the echelon form. [A, B] has 3 non-zero rows and [A] has 2 non- zero rows.
ρ([A, B]) = 3, ρ(A) = 2
ρ(A) ≠ ρ([A, B])
Hence the given system is inconsistent and has no solution.

Question 5.
Find k if the equations 2x + 3y – z = 5, 3x – y + 4z = 2, x + 7y – 6z = k are consistent.
Solution:
The given system is 2x + 3y – z = 5, 3x – y + 4z = 2, x + 7y – 6z = k
It is also given that the system is consistent.
The matrix equation corresponding to the given system is

ρ(A) = 2; ρ([A, B]) = 2 or 3
For the equations to be consistent, ρ([A, B]) = ρ(A) = 2
k – 8 = 0 ⇒ k = 8

Question 6.
Find k if the equations x + y + z = 1, 3x – y – z = 4, x + 5y + 5z = k are inconsistent.
Solution:
The given system is x + y + z = 1, 3x – y – z = 4, x + 5y + 5z = k and it is inconsistent.
The matrix equation corresponding to the given system is

ρ(A) = 2, since the equivalent matrix has 2 non zero rows.
For the equations to be inconsistent
ρ([A, B]) ≠ ρ(A)
ρ([A, B]) ≠ 2 ⇒ k ≠ 0
So k can assume any real value other than 0.

Question 7.
Solve the equations x + 2y + z = 7, 2x – y + 2z = 4, x + y – 2z = -1 by using Cramer’s rule.
Solution:
The given system is x + 2y + z = 7, 2x – y + 2z = 4, x + y – 2z = -1
Here ∆ = \(\left|\begin{array}{ccc}
1 & 2 & 1 \\
2 & -1 & 2 \\
1 & 1 & -2
\end{array}\right|\)
= 1(2 – 2) -2(4 – 2) + 1(2 + 1)
= 12 + 3
= 15 ≠ 0
We can apply Cramer’s ruleand the system is consistent and it has unique solution.

Question 8.
The cost of 2kg. of wheat and 1kg. of sugar is ₹ 100. The cost of 1kg. of wheat and 1kg. of rice is ₹ 80. The cost of 3kg. of wheat, 2kg. of sugar and 1kg of rice is ₹ 220. Find the cost of each per kg., using Cramer’s rule.
Solution:
Let the cost of wheat per kg be ₹ x, cost of sugar per kg be ₹ y and cost of rice per kg be ₹ z, respectively.
It is given that 2x + y = 100, x + z = 80, 3x + 2y + z = 220
Here ∆ = \(\left|\begin{array}{lll}
2 & 1 & 0 \\
1 & 0 & 1 \\
3 & 2 & 1
\end{array}\right|\)
= 2 (-2) – 1(-2 )
= -2 ≠ 0
we can apply Cramer’s rule and the system is consistent and its has unique solution.

Hence the cost of wheat is ₹ 30/kg, cost of sugar is ₹ 40/kg and the cost of rice is ₹ 50/kg.

Question 9.
A salesman has the following record of sales for three months for three items A,B and C, which have different rates of commission.

Find out the rate of commission on the items A,B and C by using Cramer’s rule.
Solution:
Let the rate of commission on items A, B and C be ₹ x, ₹ y and ₹ z per unit respectively.
According to the problem we have,
January, 90x + 100y + 20z = 800
February, 130x + 50y + 40z = 900
March, 60x + 100y + 30z = 850

Question 10.
The subscription department of a magazine sends out a letter to a large mailing list inviting subscriptions for the magazine. Some of the people receiving this letter already subscribe to the magazine while others do not. From this mailing list, 60% of those who already subscribe will subscribe again while 25% of those who do not now subscribe will subscribe. On the last letter it was found that 40% of those receiving it ordered a subscription. What percent of those receiving the current letter can be expected to order a subscription?
Solution:
Let X represent people who subscribe for the magazine and Y represent people who do not subscribe for the magazine. Then
(X X) ⇒ those who already subscribed will do it again.
(X Y) ⇒ those who already subscribed will not do it again.
(Y X) ⇒ those who have not subscribed will do it now.
(Y Y) ⇒ those who have not subscribed already will not do it now also.
From the question,


Thus 39% of those receiving the current letter can be expected to order a subscription.