Prasanna

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 5 Two Dimensional Analytical Geometry – II Ex 5.2

12th Maths Exercise 5.2 Samacheer Kalvi Question 1.
Find the equation of the parabola in each of the cases given below:
(i) focus (4, 0) and directrix x = – 4.
(ii) passes through (2, -3) and symmetric about y-axis.
(iii) vertex (1, – 2) and focus (4, – 2).
(iv) end points of latus rectum(4, – 8) and (4, 8) .
Solution:
(i) Focus = F = (4, 0)
⇒ a = 4
Equation of directrix x = – 4
⇒ The curve open to the right. So the equation will be of the form y² = 4 ax
Here a = 4
⇒ y² = 4 (4) x (i.e.,)y² = 16x

(ii) The parabola is symmetric about y axis. So the equation will be of the form
x² = 4 ay
It passes through (2, -3)
⇒ 2² = 4a(-3)
4 = -12a ⇒ a = \(-\frac{1}{3}\) ⇒ 4a = \(-\frac{4}{3}\)
∴ Equation of parabola is x² = \(-\frac{4}{3}\) y
3x²= – 4y.

(iii) The distance between vertex and focus = 3
(ie.,) a = 3
Parabola is open to the right.
So equation will be of the form y² = 4ax
Here a = 3 ⇒ y² = 12x
but the vertex is (1, -2)
So equation of the parabola is
(y + 2)² = 12(x – 1)

(iv) Focus = (4, 0)
Equation of the parabola will be of the form .
y² = 4ax
Here a = 4
⇒ y² = 16x

12th Maths Exercise 5.2 Question 2.
ind the equation of the ellipse in each of the cases given below:
(i) foci (± 3, 0), e = \(\frac{1}{2}\)
(ii) foci (0, ± 4) and end points of major axis are(0, ± 5).
(iii) length of latus rectum 8, eccentricity = \(\frac{3}{5}\) and major axis on x -axis.
(iv) length of latus rectum 4 , distance between foci \(4 \sqrt{2}\) and major axis as y -axis.
Solution:
(i) Given ae = 3 and e = \(\frac{1}{2}\)
⇒ a(\(\frac{1}{2}\)) = 3 ⇒ a = 6
So a² = 36
b² = a²(1 – e²) = 36 (1 – \(\frac{1}{4}\)) = 36 × \(\frac{3}{4}\) = 27
Since Foci = (± 3, 0), major axis is along x-axis
So equation of ellipse is \(\frac{x^{2}}{36}+\frac{y^{2}}{27}\) = 1

(ii) From the diagram we see that major axis is along y-axis.

Also a = 5 and ae = 4
⇒ 5e = 4 ⇒ e = \(\frac{4}{5}\)
Now a = 5 ⇒ a² = 25
ae = 4 ⇒ ae² = 16
We know b² = a² (1 – e²) = a² – a²e² = 25 – 16 = 9
Equation of ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{25}\) = 1

(iv) Given \(\frac{2 b^{2}}{a}\) = 4 and 2ae = \(4 \sqrt{2}\)
Now \(\frac{2 b^{2}}{a}\) = 4 2b² = 4a
⇒ b² = 2a
2ae = \(4 \sqrt{2}\) ae = \(2 \sqrt{2}\)
So a²e² = 4(2) = 8
We know b² = a²(1 – e²) = a² – a²e²
⇒ 2a = a² – 8 ⇒ a² – 2a -8 = 0
⇒ (a – 4) (a +2) = 0 ⇒a = 4 or -2
As a cannot be negative
a = 4 So a² = 16 and b² = 2(4) = 8
Also major axis is along j-axis
So equation of ellipse is \(\frac{x^{2}}{8}+\frac{y^{2}}{16}\) = 1

12th Maths Exercise 5.2 8th Sum  Question 3.
Find the equation of the hyperbola in each of the cases given below:
(i) foci (± 2, 0), eccentricity = \(\frac{3}{2}\)
(ii) Centre (2,1), one of the foci (8,1) and corresponding directrix x = 4 .
(iii) passing through(5, -2) and length of the transverse axis along JC axis and of length 8 units.
Solution:
(i) Given
ae = 2 and e = \(\frac{3}{2}\)
a( \(\frac{3}{2}\)) = 2 ⇒ a = \(\frac{4}{3}\) So a² = \(\frac{16}{9}\)
b² = a²(e² – 1) = a² e² – a² = 4 – \(\frac{16}{9}\) = \(\frac{20}{9}\)
Since the foci are (± 2, 0), transverse axis is along x-axis
So equation of hyperbola is
\(\frac{x^{2}}{16 / 9}-\frac{y^{2}}{20 / 9}\) = 1 ⇒ \(\frac{9 x^{2}}{16}-\frac{9 y^{2}}{20}\) = 1

(ii) Given Centre = (2, 1)
ae = 6 (distance between (2, 1) and (8, 1)) ……………. (1)
Also \(\frac{a}{e}\) = 2 ⇒ a = 2e
Equation of directrix is x = 4 [(i.e.,) (x – 2 = 2) Since centre is (2, 1)]
⇒ \(\frac{a}{e}\) = 2
Given ae = 6 ⇒ a² e² = 36
(i.e.) (2e)² (e)² = 36
⇒ 4e⁴ = 36 ⇒ e⁴ = 9
⇒ e = \(\sqrt{3}\)
Now e = \(\sqrt{3}\) a = \(2\sqrt{3}\)
∴ a² = 4 × 3 = 12
b² = a² (e² – 1) = a² e² – a² = 36 – 12 = 24
So here Centre = (2, 1)
So equation of hyperbola is
\(\frac{(x-2)^{2}}{12}-\frac{(y-1)^{2}}{24}\) = 1

(iii) Length of the transverse axis = 8
2a = 8 ⇒ a = 4
Transverse axis is along x-axis
So of equation of hyperbola is will be

12th Maths 5th Chapter Samacheer Kalvi Question 4.
Find the vertex, focus, equation of directrix and length of the latus rectum of the following:
(i) y² = 16x
(ii) x² = 24y
(iii) y² = -8x
(iv) x² – 2x + 8y + 17 = 0
(v) y² – 4y – 8x + 12 = 0
Solution:
(i) y² = 16x
It is of the form y² = 4ax (type I)
Here 4a = 16 ⇒ a = 4
Vertex = (0, 0)
Focus = (a, 0) = (4, 0)
Equation of directrix x + 4 = 0 (or) x = – 4
Length of latus rectum = 4a = 16 .

(ii) x² = 24y
This is of the form x² = 4ay (type III)
4a = 24 ⇒ a = 6
Vertex = (0, 0)
Focus = (0, a) = (0, 6)
Equation of directrix is y + a= 0 (i.e.,) y + 6 = 0 (or) y = -6
Length of latus rectum = 4a = 24.

(iii) y² = -8x
This is of the form y² = – 4ax (type II)
Here 4a = 8 ⇒ a = 2
Vertex = (0, 0)
Focus = (- a, 0) = (-2, 0)
Equation of directrix is x – 2 = 0 (or) x = 2
Length of latus rectum = 4a = 8.

(iv) x² – 2x + 8y + 17 = 0
x² – 2x = -8y – 17
x² – 2x + 1 – 1 = – 8y — 17
(x – 1)² = – 8y – 17 + 1 = – 8y + 16
(x – 1)² = – 8 (y – 2)
Taking x – 1 = X and y – 2 = Y.
We get X² = – 8Y.
This is of the form x² = – 4ay (type IV)
Where 4a = 8 ⇒ a = 2

(v) y² – 4y = 8x – 12 = 0
y² – 4y + 4 = 8x – 12 + 4
(y – 2)² = 8x – 8 = 8 (x – 1)
Taking x – 1 = X and y – 2 = Y.
We get Y² = 8X.
This is of the form y² = 4ax (type IV)
Where 4a = 8 ⇒ a = 2

12th Maths 5.2 Exercise Question 5.
Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:

Solution:
(i) \(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1
It is of the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, which is an ellipse
Here a² = 25, b² = 9
a = 5, b = 3
e² = \(\frac{a^{2}-b^{2}}{a^{2}}=\frac{25-9}{25}=\frac{16}{25}\) ⇒ e = \(\frac{4}{5}\)
Now e = \(\frac{4}{5}\) and a = 5 ⇒ ae = 4 and \(\frac{a}{e}=\frac{5}{4 / 5}=\frac{25}{4}\)
Here the major axis is along x axis
∴ Centre = (0, 0)
Foci = (± ae, 0) = (± 4, 0)
Vertices = (± a, 0) = (±5, 0)
Equation of directrix x = ± \(\frac{a}{e}\) (ie.,) x = ± \(\frac{25}{4}\)

(ii) \(\frac{x^{2}}{3}+\frac{y^{2}}{10}\) = 1
It is an ellipse and here (always a >b)

Here the major axis is alongy-axis
Centre = (0, 0)
Foci = (0, ± ae) = (0, ± \(\sqrt{7}\))
Vertices = (0, ± a) = (0, ± \(\sqrt{10}\) )
Equation of directrices y = ± \(\frac{10}{\sqrt{7}}\)

(iii) \(\frac{x^{2}}{25}-\frac{y^{2}}{144}\) = 1

Now, Here transverse axis is along x-axis
Centre = (0, 0)
Vertices = (± a, 0) = (± 5, 0)
Foci = (± ae, 0) = (± 13, 0)
Equation of directrices x = ± \(\frac{a}{e}\) (ie.,) x = ± \(\frac{25}{13}\)

(iv) \(\frac{y^{2}}{16}-\frac{x^{2}}{9}\) = 1
It is a hyperbola. Here transverse axis is along y-axis

Now Centre = (0, 0)
Vertices = (0, ± a) = (0, ± 4)
Foci = (0, ± ae) = (0, ± 5)
Equation of directrices y = ± \(\frac{16}{5}\)

Class 12 Maths Chapter 5 Exercise 5.2 Question 6.
Prove that the length of the latus rectum of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 is \(\frac{2 b^{2}}{a}\)
Solution:
The latus rectum LL’ of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 passes through the focus(ae, 0)

12th Maths 5.2 Question 7.
Show that the absolute value of the difference of the focal distances of any point P on the hyperbola is the length of its transverse axis.
Solution:

∴ S’P – SP = (a + ex) – (ex – a)
a + ex – ex + a = 2a (transverse axis)

Question 8.
Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:

Solution:

(v) 18x² + 12y² – 144x + 48y + 120 = 0
(18x² – 144x) + (12y² + 48y) = -120
18(x² – 8x) + 12 (y² + 4y) = -120
18(x² – 8x + 16 – 16) + 12(y² + 4y + 4 – 4) = -120
18(x – 4)² – 288 + 12(y + 2)² – 48 = – 120
18(x – 4)² + 12(y + 2)² = -120 + 288 + 48 = 216

Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.2 Additional Problems

Question 1.
Find the equation of the ellipse if centre is (3, – 4), one of the foci is (3 + \(\sqrt{3}\), – 4) and e = \(\frac{\sqrt{3}}{2}\)
Solution:

Question 2.
Find the equation of the hyperbola if centre (1, -2); length of the transverse axis is 8; e = \(\frac{5}{4}\) and the transverse axis is parallel to X-axis.
Solution:

Here, centre = (1, -2) and transverse axis is parallel to X-axis.

12th Maths Chapter 5 Exercise 5.2 Question 3.
Find axis, vertex, focus and equation of directrix for y² + 8x – 6y + 1 = 0
Solution:
y² – 6y = – 8x – 1
y² – 6y + 9 = – 8x – 1 + 9
(y – 3)² = – 8x + 8 = – 8(x – 1)

Comparing this equation with Y² = – 4aX we get
4a = 8 or a = 2
Vertex is (0, 0)
x – 1 = 0 ⇒ x = 1, y – 3 = 0 ⇒ y = 3
Axis y – 3 = 0, Vertex = (1, 3)
Focus is (- a, 0) = (-2, 0) + (1, 3) = (-1, 3)
Equation of directrix is x – a = 0. i.e., X – 2 = 0
⇒ x – 1 – 2 = 0 ⇒ x – 3 = 0
Latus rectum x + a = 0 i.e., x – 1 + 2 = 0
x + 1 = 0
Length of latus rectum = 4a = 8

Question 4.
Find axis, Vertex focus and equation of directrix for x² – 6x – 12y – 3 = 0.
Solution:
x² – 6x = 12y + 3
x² – 6x + 9 = 12y + 3 + 9 = 12y + 12
(x – 3)² = 12(y + 1)

This equation is of the form X² = 4aY
4a = 12
⇒ a = 3
Vertex is x – 3 = 0 ; y + 1 = 0
⇒ x = 3 ; y = -1

Question 5.
Find the eccentricity, centre, foci and vertices of the following hyperbolas: x² – 4y² – 8x – 6y – 18 = 0
Solution:

Question 6.
Find the eccentricity, centre, foci, vertices of 9x² + 4y² = 36
Solution:

Ex 5.2 Class 12 Question 7.
Find the eccentricity, centre, foci and vertices of the following hyperbolas:
x² – 4y² + 6x + 16y – 11 = 0
Solution:

10th Maths Exercise 5.2 Samacheer Kalvi Question 8.
Find the eccentricity, centre, foci and vertices of the following hyperbolas:
x² – 3y² + 6x + 6y + 18 = 0
Solution:

Guys who are planning to learn and understand the topics of 10th Social Science History can grab this Tamilnadu State board solutions for Chapter 5 Social and Religious Reform Movements in the 19th Century Questions and Answers from this page for free of cost. Make sure you use them as reference material at the time of preparation & score good grades in the final exams.

Students who feel tough to learn concepts can take help from this Samacheer Kalvi 10th Social Science Book Solutions Guide Pdf, all the Questions and Answers can easily refer in the exams. Go to the below sections and get 10th Social Science History Chapter 5 Social and Religious Reform Movements in the 19th Century Tamilnadu State Board Solutions PDF.

Tamilnadu Samacheer Kalvi 10th Social Science History Solutions Chapter 5 Social and Religious Reform Movements in the 19th Century

Do you feel scoring more marks in the 10th Social Science History Grammar sections and passage sections are so difficult? Then, you have the simplest way to understand the question from each concept & answer it in the examination. This can be only possible by reading the passages and topics involved in the 10th Social Science History Board solutions for Chapter 5 Social and Religious Reform Movements in the 19th Century Questions and Answers. All the Solutions are covered as per the latest syllabus guidelines. Check out the links available here and download 10th Social Science History Chapter 5 textbook solutions for Tamilnadu State Board.

Social and Religious Reform Movements in the 19th Century Textual Exercise

I. Choose the correct answer.

Social And Religious Reform Movements 10th Class Question 1.
In which year was Sati abolished?
(a) 1827
(b) 1829
(c) 1826
(d) 1927
Answer:
(b) 1829

Social And Religious Reform Movements In The 19th Century Book Back Answers Question 2.
What was the name of the Samaj founded by Dayanand Saraswati?
(a) Arya Samaj
(b) Brahmo Samaj
(c) Prarthana Samaj
(d) Adi Brahmo Samaj
Answer:
(a) Arya Samaj

10th Social History 5th Lesson Question 3.
Whose campaign and work led to the enactment of Widow Remarriage Reform Act of 1856?
(a) Iswarchandra Vidyasagar
(b) Raja Rammohan Roy
(c) Annie Besant
(d) Jyotiba Phule
Answer:
(a) Iswarchandra Vidyasagar

Social And Religious Reform Movements 10th Class Question Answer Question 4.
Whose voice was Rast Goftar?
(a) Parsi Movement
(b) Aligarh Movement
(c) Ramakrishna Mission
(d) Dravida Mahajana Sabha
Answer:
(a) Parsi Movement

Social And Religious Reform Movements 10th Class Notes Question 5.
Who was the founder of Namdhari Movement?
(a) Baba Dayal Das
(b) Baba Ramsingh
(c) Gurunanak
(d) Jyotiba Phule
Answer:
(b) Baba Ramsingh

Question 6.
Who was Swami 5hradhananda?
(a) a disciple of Swami Vivekananda
(b) one who caused a split in the Brahmo Samaj of India
(c) one who caused a split in the Arya Samaj
(d) founder of Samathuva Samajam.
Answer:
(c) one who caused a split in the Arya Samaj

Question 7.
Who was the founder of Widow Remarriage Association?
(a) M.G. Ranade
(b) Devendranath Tagore
(c) Jyotiba Phule
(d) Ayyankali
Answer:
(a) M.G. Ranade

Question 8.
Who was the author of the book Satyarthaprakash?
(a) Dayananda Saraswathi
(b) Vaikunda Swamy
(c) Annie Besant
(d) Swami Shradanatha
Answer:
(a) Dayananda Saraswathi

II. Fill in the blanks.

1. ……… founded the Samarasa Vedha Sanmarga Sangam.
2. The founder of Poona Sarvajanik Sabha was ………
3. Satyashodak Samaj was launched by ………
4. Gulumgir was written by ………
5. Satyarthaprakash enumerates the positive principles of ………
6. Ramakrishna Mission was established by ………
7. ……… was the forerunner of Akali Movement.
8. ……… brought tremendous changes in the caste structure in Kerala.
9. Oru paisa Tamilan was started by ………
Answers:
1. Ramalinga Swamigal
2. M.G. Ranade
3. Jyotiba Phule
4. Jyotiba Phule
5. conduct
6. Swami Vivekanand
7. Singh Sabha
8. Narayan Guru
9. Iyothee Thassar

III. Choose the correct statement.

Question 1.
(i) Raja Rammohan Roy preached monotheism
(ii) He encouraged idolatry
(iii) He published tracts condemning social evils
(iv) Raja Rammohan Roy was supported by Governor General William Bentinck
(a) (i) is correct
(b) (i) and (ii) are correct
(c) (i) (ii) and (iii) are correct
(d) (i) (iii) and (iv) are correct
Answer:
(d) (i) (iii) and (iv) are correct

Question 2.
(i) Prarthana Samaj was founded by Dr. Atma Ram Pandurang.
(ii) Prarthana Samaj encouraged interdining and inter-caste marriage.
(iii) Jyotiba Phule worked for the upliftment of men.
(iv) Prarthana Samaj had it’s origin in the Punjab.
(a) (i) is correct
(b) (ii) is correct
(c) (i) and (ii) are correct
(d) (iii) and (iv) are correct
Answer:
(c) (i) and (ii) are correct

Question 3.
(i) Ramakrishna Mission was actively involved in social causes such as education, health care, relief in time of calamities.
(ii) Ramakrishna emphasised the spiritual union with god through ecstatic practices.
(iii) Ramakrishna established the Ramakrishna Mission
(iv) Ramakrishna opposed the Partition of Bengal
(a) (i) is correct
(b) (i) and (ii) are correct
(c) (iii) is correct (d)
(iv) alone correct
Answer:
(b) (i) and (ii) are correct

Question 4.
Assertion (A): Jyotiba Phule opened orphanages and homes for widows
Reason (R): Jyotiba Phule opposed child marriage and supported widow remarriage
(a) Assertion is correct but reason is not apt to the assertion
(b) Assertion is correct and the reason is apt to the assertion
(c) Both are wrong
(d) Reason is correct but assertion is irrelevant
Answer:
(b) Assertion is correct and the reason is apt to the assertion

IV. Match the following.

Answers:
1. (d)
2. (e)
3. (b)
4. (a)
5. (c)

V. Answer briefly.

Question 1.
Mention the four articles of faith laid down by Maharishi Debendranath Tagore?
Answer:
He laid down four articles of faith.

1. In the beginning there was nothing. The one Supreme Being alone existed who created the universe.
2. He alone is the God of Truth, Infinite wisdom, Goodness and power, eternal, omnipresent.
3. Our Salvation depends on belief in Him and in His worship in this world.
4. Belief consists in loving Him and doing His will.

Question 2.
Discuss Mahadev Govind Ranade’s contribution to social reforms.
Answer:
Mahadev Govind Ranade was a great social reformer. He advocated for inter-caste dining, inter-caste marriage, widow remarriage and improvement of women and depressed classes. He founded the Widow Marriage Association, the Poona Sarvajanik Sabha and Decean Education society.

Question 3.
Assess the role of Ayyankali in fighting for the cause of “untouchables.”
Answer:
Ayyankali, as a child faced the problems that made him turn into a leader of an anti-caste movement. He fought for the basic rights including access to public spaces and entry to schools. He wore clothes associated with upper castes and rode an ox-cart which was a ban for untouchables on public roads.

Question 4.
Write a note on reforms of Ramalinga Adigal.
Answer:
Ramalinga swamigal emphasised the bonds of responsibility and compassion between living beings. He empressed the view that those who lack compassion for suffering beings are hard hearted, their wisdom clonded. He showed his compassion and mercy on all living beings including plants. He established the Samarasa Vedha Sanmarga Sangam in 1865 and it was renamed in 1872 as ‘Samarasa Suddha Sanmarga Sathya Sangam’ which means ‘Society for pure truth in universal self-hood’. Ramalinga also estabilished a free feeding house for everyone.

Question 5.
What was the impact of Swami Vivekananda’s activist ideology?
Answer:
Vivekananda’s activist ideology breed a sense of confidence among Indians. It kindled the desire for political change among many western-education young Bengal. He advocated that service to humanity is service to God. His ideology allowed the lower caste Hindus to be allowed in the Hindu rituals.

Question 6.
What are the differences between Reformist Movements and Revival Movements?
Answer:
While the reformist movements strived to change the fundamental system and structures of the society, through gradual changes within the existing institutions; revivalist movements tended to revive formers customs or practices and thus take the society back to the glorious past.

Question 7.
List the social evils eradicated by Brahmo Samaj.
Answer:

1. Abolition of Sati in 1829, Child marriage and Polygamy.
2. Advocated Education for women, western science in schools and colleges.

Question 8.
Highlight the work done by Jyotiba Phule for the welfare of the poor and the marginalized.
Answer:
Jyotiba Phule worked for the upliftment of depressed classes and the causes of women. He opened orphanges and homes for widows. He opened the first school for ‘untouchable’ in 1852 in Poona. He launched the Satyashodhak Samaj (Truth-Seekers Society) in 1870 to stir the non-Brahman masses to self-respect and ambition. He supported widow remarriage, which was prohibited particularly among high-caste Hindus. His work, Gulamgiri (slavery) is an important work that condemned the inequlities of caste.

Question 9.
What was the impact of lyotheeThassar’s visit to SriLanka.
Answer:
He converted himself to Buddhism. He founded the Sakya Buddhist society at Madras. He believed that the revival of Buddhism could liberate the people from the evil of caste that afflicted the Hindu society. He called the ‘Untouchables, Sathi Petham Atra Dravidar and urged them to register as castelss Dravidians.

VI. Answer all the questions given under each caption:

Question 1.
Aligarh Movement

(i) What is the main aim of this Movement?
Answer:
The main aim of the Aligarh Movement was to persuade the Muslims to acquire modem knowledge and English language.

(ii) Who is considered the soul of this Movement?
Answer:
Sir Sayed Ahmed Khan is considered the soul of the Aligarh Movement.

(iii)Why were English books translated into Urudu?
Answer:
Many English Books were translated into Urdu in order to enable the Muslims to accept the western science and take up government services.

(iv) Name the college which was later raised to the status of a University?
Answer:
Aligarh Muhammadan Anglo-Oriental College.

Question 2.
Ramalinga Adigal

(i) What is Jeevakarunya?
Answer:
Ramalinga Adigal showed his compassion and mercy on all living beings including plants . This he called as Jeevakarunya.

(ii) What are the Songs of Grace?
Answer:
His voluminous songs were compiled and published under the title Thiruvarutpa which are called Songs of Grace.

(iii) Point out the major contribution of Samarasa Vedha Sanmarga Sathya Sangam?
Answer:
The Major contribution of Samarasa Vedha Sanmarga Sathya Sangam were to establish free feeding house for everyone irrespective of caste at Vadalur in 1867.

(iv) Where did he establish his free feeding house?
Answer:
The free feeding house is at Vadalur, established in 1867.

Question 3.
Deoband Movement

(a) Who were the organizers of this Movement?
Answer:
The Deoband Movement was organised be the orthodox Muslim Ulemas.

(b) What were the two main objectives of the Movement?
Answer:
The two main objectives of this Movement were:

• to propagate the pure teachings of the Quran and the Hadith
• to encourage the spirit of Jihad against the foreign and un-Ismalic elements

(c) Who founded the school at Deoband?
Answer:
The Ulemas under the leadership of Muhammad Qusim Wanoutavi and Rashed Ahmad Gangohi founded the school at Deoband in the Saharanpur district of Uttar Pradesh in 1866.

(d) Against whom the fatwa was issued by Deoband Ulema?
Answer:
In 1888 Deoband Ulemas issued the fatwa (a religious decree) against Syed Ahmed Khan’s Orgnisation called ‘The United Patriotic Association’ and ‘The Muhammadan Anglo-Oriental Association.

VII. Answer in detail.

Question 1.
Compare and contrast the contributions of Revivalist Movements with that of Reform Movements.
Answer:

1. Reform Movements were the Brahmo Samaj, the Prarthana Samaj and the Aligarh movement.
2. Revival movements were the Arya Samaj, the Ramakrishna mission, Deoband Movement.
3. To compare, both the reform and revival movements were in some way or other attack the social evils that prevailed in the society like sati, female infanticide, child marriages and various superstitious beliefs and complex rituals.
4. The reform movements strived to change the fundamental system and structure of the society, while the reformist movements strived to revive the earlier customs and practices, protecting the cultural heritage of the country.
5. Revival movements proved that the rich cultural heritage of India was found superior to the Western culture. That’s why, Arya Samaj advocated ‘Go back to the Vedas’.
6. The Reform movement responded to the changes in the time period and scientific thinking of the modem era.
7. For example, Brahmo Samaj, considers other religions such as Christianity, Islam as equal to Hinduism.
8. Arya Samaj believes in the superiority of Hinduism.
9. The Reform movements were influenced by the Western culture, whereas the Revival movements were influenced by Indian culture only.

Question 2.
Discuss the circumstances that led to the Reform movements of 19th century.
Answer:

1. The nineteenth century India was plagued with a number of social evils such as sati, child marriage, female infanticide, polygamy and so on.
2. Women were subjugated by men and were not allowed to get education. They were restricted to home and hearth. They were considered inferior to men.
3. The condition of the depressed classes was miserable. They were subject to untouchable. Their entry to schools, temples and other public places, meant for upper castes, was banned. Hence, there was no education among the people belonging to lower castes.
4. There were other evil practices prevalent in the Indian society such as excessive superstitious religious beliefs, animal sacrifice, which needed attention of the social reformers for the benefit of the common people of the society.
5. There was total absence of reason in the society. The system of child marriage was prevalent which resulted to child widows. These widows were never accepted in the family and were destined to lead a pathetic life. These were the circumstances that led to the Reform movements in the nineteenth century.

Question 3.
Evaluate the contributions of Ramakrishna Paramahamsa and Swami
Answer:
Vivekananda to regenerate Indian society.

1. Ramakrishna Paramahamsa was a simple priest of Kolkata gained popularity by emphasising the spiritual union with God by singing bhajans.
2. He declared that the manifestations of the divine mother were infinite and said ‘Jiva is Siva’ which means all living beings are God.
3. He insisted on not mercy, rather service. Service for Man as regarded as Service to God.
4. After his death in 1886, his disciples, especially Vivekananda established the Ramakrishna Mission, which undertook not only religious activities but also social causes.
5. Education, Health Care, Relief in times of calamities were the main social works of the mission.
6. Vivekananda emphasised a cultural nationalism and made a call to Indian youth to regenerate Hindu society.
7. His ideas gave a sense of self-confidence among Indians who felt inferior in relation to the materialistic achievements of the west.
8. He suggested the orthodox Hindus that the lower castes should be allowed to engaged in the Hindu rituals from which they were excluded.
9. Vivekananda’s activist ideology rekindled the desire for political change among many Western education young Bengalis.
10. His ideas were inspiring to many of the militant nationalist before Indian independence and even for today’s people.

Question 4.
Write an essay on the role played by the 19th century reformers towards the cause of Women.
Answer:
Several social reformers emerged during the 19th century India who played a big role towards the cause of women:

(i) Raja Rammohun Roy was deeply concerned with the prevanting customs of sati, child marriage and polygamy. He advocated the rights of widows to remarry. He wanted – polygamy to end. He raised voice against sati system and forced the Governor-General William Bentinck to abolish this social evil in 1829. He condemned the subjugation of women and opposed the prevailing ideas that women were inferior to men. He strongly advocated education for women.

(ii) Ishwar Chandra Vidyasagar was lead against the burning of widows. He supported the idea of widow remarriage. He played a leading role in promoting education of girls and helped them in setting up a number of schools. He dedicated his entire life for the betterment of the child widows of the Hindu society. He led a movement that resulted in the widows’ Remarriage Reform Act of 1856.

(iii) Jyotiba Phule worked for the cause of women. He opposed child marriage and supported widow remarriage, which was prohibited particularly their lives for the uplift of the depressed classes and women. Jyotiba opened orphanages and homes for widows.

(iv) Swami Dayanand Saraswati said that the prohibition of widow remarriage had no scriptural sanction.

(v) Reformers like R.C. Bhandarkar and Justice Mahadev Govind Ranade devoted themselves to activities such as widow remarriage and improvement of women and depressed classes. M.G. Ranade founded the widow Remarriage Association in 1861.

Social and Religious Reform Movements in the 19th Century Additional Questions

I. Choose the correct answer:

Question 1.
The pioneer of the reform movements was
(a) Raja Rammohan Roy
(b) Keshab ChandraSen
(c) Devendranath Tagore
Answer:
(a) Raja Rammohan Roy

Question 2.
……………… dedicated his whole life for the betterment of the child widows of the Hindu Society.
(a) M.G. Ranade
(b) Iswar Chandra Vidyasagar
(c) Atma Ram Pandurang
(d) Ramakrishna
Answer:
(b) Iswar Chandra Vidyasagar

Question 3.
Swami Dayananda Saraswathi started the
(a) Brahmo Samaj
(b) Arya Samaj
(c) Prarthana Samaj
Answer:
(b) Arya Samaj

Question 4.
The first age of consent Act was passed in the year ………………
(a) 1830
(b) 1840
(c) 1850
(d) 1860
Answer:
(d) 1860

Question 5.
Vallalar’s devotional songs are compiled in a volume called
(a) Devaram
(b) Ettuthogai
(c) Thiru Arutpa
Answer:
(c) Thiru Arutpa

Question 6.
……………… wrote the book Satyarthaprakash.
(a) Shraddhananda
(b) Dayananda Saraswathi
(c) Jyotiba Phule
(d) Narayana Guru
Answer:
(b) Dayananda Saraswathi

Question 7.
Sir Syed Ahamed Khan started a school at
(a) Alipore
(b) Allepey
(c) Ghazipur
Answer:
(c) Ghazipur

Question 8.
Inspired by Sree Narayan Guru, ……………… founded the Sadhu Jana Paripalana Sangam.
(a) Deoband
(b) Ayyankali
(c) Sayyid Ahmed Khan
(d) Ramalinga Adigal
Answer:
(b) Ayyankali

Question 9.
“Go back to Vedas” was said by
(a) Swami Dayananda Saraswathi
(b) Annie Besant
(c) Swami Vivekananda
Answer:
(a) Swami Dayananda Saraswathi

Question 10.
Ramalinga Adigal was popularly known as ………………
(a) Nirankar
(b) Vallalar
(c) Vaikunda Swami
(d) Iyothee Thassar
Answer:
(b) Vallalar

II. Fill in the blanks:

1. ……… was the pioneer of reform movements.
2. Raja Rammohan Roy was called the ……… India.
3. Prarthana Samaj was founded in Bombay in 1867 by ………
4. Swami Dayananda Saraswathi founded the ……… in 1875.
5. ……… was founded by Madame Blavatsky and Henry S Olcott.
6. Annie Besant started the ……… movement.
7. ……… was founded by Swami Vivekananda.
8. Samarasa Sudha Sanmarka Sangam was started by ………
9. ……… was started by Sir Syed Ahamed Khan.
10. Jyotiba Phule formed the ………
11. The Hindus who had been converted to other religions were reconverted by ………
12. Theosophy means ………
13. Home Rule movement was started by ………
14. The childhood name of Swami Vivekananda was ………
15. Sathya Gnana Sabai was established by ………
16. “The Hindus and Muslims are the two eyes of the beautiful bird that was India” was said by ………
17. “Satya Shodhak Samaj” was formed by ………
Answers:
1. Raja Rammohan Roy
2. Herald of New Age
3. Dr. Atmaram Pandurang
4. Arya Saniaj
5. The Theosophical Society
6. Home Rule
7. The Ramakrishna Mission
8. Ramalinga Adigal
9. The Aligarh movement
10. Satya Shodhak Samaj
11. Suddhi movement
12. Wisdom of God
13. Annie Besant
14. Narendranath Dutta
15. ValIalar
16. Syed Ahamed Khan
17. Jyotiba Phule

III. Match the following:

Answers:
1. (e)
2. (a)
3. (c)
4. (b)
5. (d)

Answer:
1. (d)
2. (c)
3. (a)
4. (e)
5. (b)

Answers:
1. (d)
2. (c)
3. (a)
4. (e)
5. (b)

IV. Answer the following briefly:

Question 1.
Write a note on the Age of Consent Act.
Answer:
The Age of Consent Act was enacted by the efforts of the reformer Vidyasagar in 1860. Initially it was fixed for ten years. Later it was raised to twelve years in 1891 and thirteen years in 1925.

Question 2.
Name the services rendered by the Arya Samaj.
Answer:

1. Arya Samaj opposed Child marriage, Polygamy, Purdha System, Casteism and Sati.
2. It advocated women education, inter-caste marriage and inter-dining.
3. It insisted on education of the women and upliftment of the depressed classes.
4. The Samaj started a number of schools all over the country to impart English and Vedic education. These schools are called Dayananda Anglo-Vedic schools and colleges.

Question 3.
Who opposed Arya Samaj for its doctrinal purity?
Answer:
Swami Shraddhananda opposed Arya Samaj for its doctrinal purity. He was the opinion that the group running the DAV schools are too westernised and thereby ignore founder’s ideology. Therefore he started own network of schools, Gurukulas, emphasising the study of vedas.

Question 4.
What are the teachings of Vallalar?
Answer:

1. Vallalar condemned the inequalities based on birth.
2. He advocated that feeding the poor is the highest form of worship.
3. He insisted compassion and mercy on all living things.
4. According to Vallalar “Service to mankind is the path of Moksha”
5. God is the personification of mercy and knowledge.
6. The path of compassion and mercy are the only way to reach God.

Question 5.
What are the Islamic Reform movements originated in India and what do they mainly focus on?
Answer:
Scientific society founded of Sir Sayyid Ahmed Khan was his first initiative. Later he founded the Aligarh Movement emphasising education and started Aligarh University. Deoband movement organised by The orthodox Muslim Ulema to spread the spirit of Muslims against un-islamic elements.

Question 6.
Name the followers of Swami Dayananda Saraswathi.
Answer:

1. Lala Lajpat Rai
2. Lala Hansraj
3. Pandit Guru Dutt
4. Bala Gangathara Tilak
5. Gopala Krishna Gokhale

Question 7.
Who started the Sadharan Samaj? Why?
Answer:
In 1866, Keshab Chandra Sen left the Brahmo Samaj and founded a new organisation called as Debendranath’s organisation. Later it was called as Adi Brahmo Samaj. When Keshab’s 14 yrs old daughter married to an Indian prince, the opponents of child marriage left Brahmo Samaj and started the SadharanSamaj.

Question 8.
Write a note on Keshab Chandra Sen.
Answer:

1. He was one of the followers of Raja Rammohan Roy.
2. After the death of Raja Rammohan Roy, the works of the Brahmo Samaj were carried on by Keshab Chandra Sen.
3. Because of his efforts, an act was passed in 1872. It abolished polygamy and child marriage.

Question 9.
What are the Social reforms of Jyotiba Phule?
Answer:

1. He revolted against the domination of the Brahmins.
2. He worked for the rights of the peasants and other low caste people.
3. He formed the “Satya Shadhak Samaj” in 1873, to liberate the people of lower caste from the suppression by Brahmins.
4. He started Orphanage for the unfortunate children.
5. He decided to construct a common bathing tank outside his house for the people of lower caste people.

Question 10.
Write a short note about ‘Sree Narayana Guru’.
Answer:

1. He was a great social reformer from Kerala.
2. He started ‘Sree Narayana Guru Dharma Paripalana Yogam’ in 1903.
3. It worked for the social economic and educational development of the Ezhava community and other backward people.
4. He condemned animal sacrifices, casteism and other social evils.

Question 11.
What is called ‘Jeeva Karunya’?
Answer:

1. St. Ramalinga had the view that love is the “Master key to Spirituality”.
2. He showed compassion and mercy not only on human beings, but also on plants, insects, birds and animals. This is called ‘Jeeva Karunya’.

V. Answer all the questions given under each Captions:

Question 1.
Brahmo Samaj

(a) When was it founded? Who was its founder?
Answer:
Brahmo Samaj was founded in 1828 by Raja Ram Mohan Roy.

(b) What was the important Legislation passed due to the efforts of. its founder?
Answer:
Abolition of Sati in 1829 was the important Legislation passed due to the efforts of its founder.

(c) What does the Samaj forbade?
Answer:
The Samaj forbade idol-worship and condemned meaning less religious rituals an ceremonies.

(d) Whow took over the Samaj leadership after the death of its founder?
Answer:
Maharishi Debendranath Tagore, carried on the work of the founder after his death in 1833.

Question 2.
Arya Samaj

(a) What was the original name of Swamy Dayananda Saraswathi?
Answer:
Mul Shankar

(b) Who was his guru”?
Answer:
Swamy Viijanand was his guru.

(c) What was his motto?
Answer:
His motto was “Go back to Vedas”.

(d) What did the Samaj advocate?
Answer:
The samaj advocated women education, inter-caste marriage and inter-dining.

Question 3.
Social Reform Movements in Kerala and Tamil nadu

(a) Name few Social Reformers from Kerala and Tamil Nadu.
Answer:
Sri Narayana Guru and Ayyarkali from Kerala and Ramalinga Adigal, Vaikunta Swamigal from Tamil Nadu.

(b) How was Vaikundar called by his followers. What was his cult called as?
Answer:
His followers called him respectfully as Ayya (father) and his cult was known as Ayya vazhi (The path of Ayya).

(c) What happened to the voluminous songs composed by Ramalinga Adigal?
Answer:
His voluminous songs were compiled and published under the title Thiruvarutpa (songs of Grace).

(d) Name the Association founded by Ayyankali inspired by Sree Narayana Guru?
Answer:
Ayyankali founded the Sadhu Jana Paripalana Sangam (Association for the protection of the poor) in 1907, inspired by Sree Narayana Guru.

Question 4.
Aligarh Movement

(a) Name the first religious movement of the Muslims.
Answer:
Aligarh Movement was the first religious movement of the Muslims.

(b) What did Sir Syed Ahmed Khan strongly believe?
Answer:
He strongly believed in the Hindu-Muslim unity.

(c) What was his greatest achievement?
Answer:
His greatest achievement was the establishment of the Mohammaden Anglo Oriental College at Aligarh in 1875.

(d) Name the newspaper published by him.
Answer:
Tahzil-ud-Akhlaq

VI. Answer the following questions in detail:

Question 1.
Write about Brahmo Samaj.

(a) The Foundation of the Brahmo Samaj:
Answer:

1. Raja Rammohan Roy founded “Atmiya Saba” in 1815.
2. Later it developed into “Brahmo Samaj” in 1828.

(b) Aim of the Brahmo Samaj:
Answer:
To cure Hindu society and religion from all its evils and set it on right footing.

(c) Principle of the Brahmo Samaj:
Answer:
The Brahmo Samaj believed in a ‘Universal religion’ based on the principles of one Supreme God.

(d) Services of the Brahmo Samaj:
Answer:

1. The samaj condemned idol worship, costly rites and rituals, cartle distinctions, untouchability and the practice of Sati.
2. Because of Raja Rammohan Roy’s hard work, William Bentinck passed Sati prohibition Act in 1829.
3. The samaj also fought against polygamy and child marriage and supported inter-caste marriage and widow remarriage.
4. It tried to obtain a respectable position for women in the Indian Society.
5. It encouraged the study of English language and the western science in India. By western studies superstitions and blind faiths were removed from India. After the death of Raja Rammohan Roy, the works of the samaj was carried on by great men like Keshab Chandra Sen and Devendranath Tagore.

Question 2.
Write a note on Social Reformers of Tamil Nadu in the 19th century.
Answer:
Ramalinga Swamigal: Ramalinga Adigal was popularly known as Vallalar. He showed his compassion and mercy on all living beings including plants. This he called as Jeevakarunya. He established Samarasa Vedha Sanmarga Sangam in 1865. He also established a free feeding house for everyone irrespective of caste at Vadalur in 1867. He wrote many songs which were compiled in the name of Thiruvarutpa.

Vaikunda Swamigal: His original name was Mudichudum Perumal which changed as Muthukutty because of the objection by upper class. He calls himself Vaikuntar. He preached the idea of equality and condemned the ’ worship of idols. He was against animal sacrifice in the name of religion. He was against caste differences and wanted social integration of the society.

Iyothee Thassaar: Iyothee Thassar was scholar , writer ,siddha medicine practitioner and also journalist and political activist. He campaigned for the social justice and worked for building up a casteless society and removal of untouchability. He established several schools for untouchables in Tamil Nadu. He published a weekly ‘Oru Paisa Tamilan’ in 1907 and it lasted till 1914. He worked also for the revival of Buddhism.

Question 3.
“Ramalinga Adigal played a prominent role in the social and Religious Reform Movements” – Explain it.
Answer:

1. St. Ramalinga tried his best to find solution to the social evils by inculcating spiritual awareness.
2. St. Ramalinga led to unity and solidarity of the Tamils.
3. He favoured the creation of a casteless society to be guided by God.
4. He believed God to be in the form of “Arul Perum Jothi”.
5. Ramalinga Adigal believed that hunger and poverty are the evils of the society.
6. He advocated that feeding the poor is the highest form of worship.
7. Ramalinga Adigal condemned the inequalities based on birth and promoted universal love and brother-hood.
8. As a result, he founded Sathya Dharma Sala at Vadalur for feeding the poor. It provides food to everyone irrespective of caste and creed throughout the year.
9. He showed compassion and mercy not only on human beings, but also on plants, insects, birds and animals. This is called “Jeeva Karunya”.
10. He opposed superstitious beliefs and rituals.
11. He emphasized on being vegetarian.
12. He forbade the killing of animals for the sake of food.
13. In 1870, he established Sathya Gnana Sabai.
14. His devotional songs are compiled in a volume called “Thiru Arutpa”.

Important years and events:

We think the data given here clarify all your queries of Chapter 5 and make you feel confident to attempt all questions in the examination. So, practice more & more from Tamilnadu State Board solutions for 10th Social Science History Chapter 5 Social and Religious Reform Movements in the 19th Century Questions and Answers & score well. Need any information regarding this then ask us through comments & we’ll give the best possible answers very soon.

Students who are preparing for the Science exam can download this Tamilnadu State Board Solutions for Class 10th Science Chapter 10 from here for free of cost. These Tamilnadu State Board Textbook Solutions PDF cover all 10th Science Types of Chemical Reactions Book Back Questions and Answers.

All these concepts of Chapter 10 Types of Chemical Reactions are explained very conceptually by the subject teachers in Tamilnadu State Board Solutions PDF as per the prescribed Syllabus & guidelines. You can download Samacheer Kalvi 10th Science Book Solutions Chapter 10 Types of Chemical Reactions State Board Pdf for free from the available links. Go ahead and get Tamilnadu State Board Class 10th Science Solutions of Chapter 10 Types of Chemical Reactions.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions

Kickstart your preparation by using this Tamilnadu State Board Solutions for Class 10th Science Chapter 10 Types of Chemical Reactions Questions and Answers and get the max score in the exams. You can cover all the topics of Chapter 10 easily after studying the Tamilnadu State Board Class 10th Science Textbook solutions pdf. Download the Tamilnadu State Board Science Chapter 10 Types of Chemical Reactions solutions of Class 10th by accessing the links provided here and ace up your preparation.

Samacheer Kalvi 10th Science Types of Chemical Reactions Textual Problems Solved

Chemical Reactions Questions And Answers Pdf Question 1.
Calculate the pH of 0.01 M HNO₃?
Solution:

Samacheer Kalvi Guru 10th Science Question 2.
The hydroxyl ion concentration of a solution is 1 × 10^-9 M. What is the pOH of the solution?
Solution:

Samacheer Kalvi Question 3.
A solution has a pOH of 11.76. What is the pH of this solution?
Solution:
pH = 14 – pOH
pH = 14 – 11.76 = 2.24.

10th Science Samacheer Kalvi Question 4.
Calculate the pH of 0.001 molar solution of HCl.
Solution:
HCl is a strong acid and is completely dissociated in its solutions according to the process:
\(\mathrm{HCl}_{(\mathrm{aq})} \rightarrow \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)
From this process it is clear that one mole of HCl would give one mole of H⁺ ions.
Therefore, the concentration of H⁺ ions would be equal to that of HCl, i.e., 0.001 molar or 1.0 × 10^-3 mol litre^-1.

Samacheer Kalvi 10th Science Question 5.
What would be the pH of an aqueous solution of sulphuric acid which is 5 × 10^-5 mol litre^-1 in concentration?
Solution:
Sulphuric acid dissociates in water as:
\(\mathrm{H}_{2} \mathrm{SO}_{4(\mathrm{aq})} \rightarrow 2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\)
Each mole of sulphuric acid gives two moles of H⁺ ions in the solution. One litre of H₂SO₄ solution contains 5 × 10^-5 moles of H₂SO₄ which would give 2 × 5 × 10^-5 = 10 × 10^-5 or 1.0 × 10^-4 moles of H⁺ ion in one litre of the solution.
Therefore,

10th Samacheer Kalvi Science Question 6.
Calculate the pH of 1 × 10^-4 molar solution of NaOH.
Solution:
NaOH is a strong base and dissociates in its solution as:
\(\mathrm{NaOH}_{(\mathrm{aq})} \rightarrow \mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
One mole of NaOH would give one mole of OH^– ions.
Therefore,

10th Science Solution Samacheer Kalvi Question 7.
Calculate the pH of a solution in which the concentration of the hydrogen ions is 1.0 × 10^-8 mol litre^-1.
Solution:
Here, although the solution is extremely dilute, the concentration given is not of an acid or a base but that of H⁺ ions. Hence, the pH can be calculated from the relation:

Samacheer Kalvi 10th Science Solutions Question 8.
If the pH of a solution is 4.5, what is its pOH?
Solution:
pH + pOH = 14
⇒ pOH = 14 – 4.5 = 9.5
⇒ pOH = 9.5.

Samacheer Kalvi 10th Science Types of Chemical Reactions Textual Evaluation Solved

I. Choose the correct answer.

Samacheer Kalvi 10th Science Solution Question 1.
\(\mathrm{H}_{2(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{g})} \rightarrow 2 \mathrm{HCl}_{(\mathrm{g})}\) is a ______.
(a) Decomposition Reaction
(b) Combination Reaction
(c) Single Displacement Reaction
(d) Double Displacement Reaction.
Answer:
(b) Combination Reaction
Hint: It is a combination reaction between H₂ and Cl₂ to form HCl as product.

Science Solution Class 10 Samacheer Kalvi Question 2.
Photolysis is a decomposition reaction caused by:
(a) heat
(b) electricity
(c) light
(d) mechanical energy
Answer:
(c) light

Samacheer Kalvi 10 Science Question 3.
A reaction between carbon and oxygen is represented by \(\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{CO}_{2(\mathrm{g})}\) + Heat. In which of the type(s), the above reaction can be classified?
(i) Combination Reaction
(ii) Combustion Reaction
(iii) Decomposition Reaction
(iv) Irreversible Reaction
(a) i and ii
(b) i and iv
(c) i, ii and iii
(d) i, ii and iv.
Answer:
(d) i, ii and iv.
Hint:

• It is a combination reaction with one product.
• It involves the combustion of carbon to form CO₂.
• The reaction is irreversible and we cannot reverse the reaction.

Samacheer Kalvi Chemistry Question 4.
The chemical equation
Na₂SO_4(aq) + BaCI_2(aq) → BaSO_4(s)↓ + 2NaCl_(aq) represents which of the following types of reaction?
(a) Neutralisation
(b) Combustion
(c) Precipitation
(d) Single displacement
Answer:
(c) Precipitation

Samacheer Kalvi 10th Science Solution Book Question 5.
Which of the following statements are correct about a chemical equilibrium?
(i) It is dynamic in nature
(ii) The rate of the forward and backward reactions are equal at equilibrium
(iii) Irreversible reactions do not attain chemical equilibrium
(iv) The concentration of reactants and products may be different
(a) i, ii and iii
(b) i, ii and iv
(c) ii, iii and iv
(d) i, iii and iv.
Answer:
(a) i, ii and iii
Hint: Chemical equilibrium is dynamic in nature for a reversible reaction.
At equilibrium, Rate of forwarding reaction = Rate of backward reaction.

Samacheer Kalvi Guru 10th Science Book Back Answers Question 6.
A single displacement reaction is represented by
X_(s) + 2HCl_(aq) → XCl_2(aq) + H_2(g). Which of the following(s) could be X?
(i) Zn
(ii) Ag
(iii) Cu
(iv) Mg. Choose the best pair.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Answer:
(d) (i) and (iv)

Samacheer Kalvi 10th Standard Science Question 7.
Which of the following is not an “element + element → compound” type reaction?
(a) \(\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{CO}_{2(\mathrm{g})}\)
(b) \(2 \mathrm{K}_{(\mathrm{s})}+\mathrm{Br}_{2(l)} \rightarrow 2 \mathrm{KBr}_{(\mathrm{s})}\)
(c) \(2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{g})}\)
(d) \(4 \mathrm{Fe}_{(\mathrm{s})}+3 \mathrm{O}_{2(\mathrm{g})} \rightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3(\mathrm{s})}\).
Answer:
(c) \(2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{g})}\)
Hint: It involves one reactant compound i.e., carbon monoxide (CO).
\(2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow 2 \mathrm{CO}_{2}\)

10th Standard Science Samacheer Kalvi Question 8.
Which of the following represents a precipitation reaction?
(a) \(\mathrm{A}_{(\mathrm{s})}+\mathrm{B}_{(\mathrm{s})} \rightarrow \mathrm{C}_{(\mathrm{s})}+\mathrm{D}_{(\mathrm{s})}\)
(b) \(\mathrm{A}_{(\mathrm{s})}+\mathrm{B}_{(\mathrm{aq})} \rightarrow \mathrm{C}_{(\mathrm{aq})}+\mathrm{D}_{(\mathrm{l})}\)
(c) \(\mathrm{A}_{(\mathrm{aq})}+\mathrm{B}_{(\mathrm{aq})} \rightarrow \mathrm{C}_{(\mathrm{s})}+\mathrm{D}_{(\mathrm{aq})}\)
(d) \(\mathrm{A}_{(\mathrm{aq})}+\mathrm{B}_{(\mathrm{s})} \rightarrow \mathrm{C}_{(\mathrm{aq})}+\mathrm{D}_{(\mathrm{l})}\).
Answer:
(c) \(\mathrm{A}_{(\mathrm{aq})}+\mathrm{B}_{(\mathrm{aq})} \rightarrow \mathrm{C}_{(\mathrm{s})}+\mathrm{D}_{(\mathrm{aq})}\)
Hint: It involves the formation of solid precipitation by mixing of two aqueous solutions.

10th Standard Samacheer Kalvi Science Question 9.
The pH of a solution is 3. Its [OH^–] concentration is:
(a) 1 × 10^-3 M
(b) 3 M
(c) 1 × 10^-11 M
(d) 11 M
Answer:
(c) 1 × 10^-11 M

Samacheer Kalvi 10th Science Book Question 10.
Powdered CaCO₃ reacts more rapidly than flaky CaCO₃ because of ______.
(a) large surface area
(b) high pressure
(c) high concentration
(d) high temperature.
Answer:
(a) large surface area
Hint: We know that greater the surface area, faster will be a chemical reaction. Hence powdered CaCO₃ reacts more rapidly.

II. Fill in the blanks.

Samacheer Kalvi Science 10th Question 1.
A reaction between an acid and a base is called ______.
Answer:
Neutralization.

Question 2.
When lithium metal is placed in hydrochloric acid, _______ gas is evolved.
Answer:
Hydrogen.

Question 3.
The equilibrium attained during the melting of ice is known as ______.
Answer:
Physical equilibrium.

Question 4.
The pH of a fruit juice is 5.6. If you add slaked lime to this juice, its pH ______ (increase/decrease)
Answer:
Increases.

Question 5.
The value of the ionic product of water at 25°C is ______.
Answer:
1 × 10^-14

Question 6.
The normal pH of human blood is ______.
Answer:
7.4.

Question 7.
Electrolysis is type of _______ reaction.
Answer:
Decomposition reaction.

Question 8.
The number of products formed in a synthesis reaction is ______.
Answer:
One.

Question 9.
Chemical volcano is an example for ______ type of reaction.
Answer:
Decomposition reaction.

Question 10.
The ion formed by dissolution of H⁺ in water is called ______.
Answer:
Hydronium ion.

III. Match the following

Question 1.
Identify the types of reaction.

Answer:
i – c, ii-a, iii – d, iv- b.

IV. True or False: (If false give the correct statement)

Question 1.
Silver metal can displace hydrogen gas from nitric acid.
Answer:
False.
Correct statement: In the activity series, any metals that are below hydrogen will not react with HNO₃.

Question 2.
The pH of rainwater containing dissolved gases like SO₃, CO₂, NO₂ will be less than 7.
Answer:
True.

Question 3.
At the equilibrium of a reversible reaction, the concentration of the reactants and the products will be equal.
Answer:
False.
Correct statement: At equilibrium rate of the forward reaction is equal to the rate of backward reaction.

Question 4.
Periodical removal of one of the products of a reversible reaction increases the yield.
Answer:
True.

Question 5.
On dipping a pH paper in a solution, it turns into yellow. Then the solution is basic.
Answer:
True.

V. Short Answer Questions

Question 1.
When an aqueous solution of potassium chloride is added to an aqueous solution of silver nitrate, a white precipitate is formed. Give the chemical equation of this reaction.
Answer:
\(\mathrm{KCl}_{(\mathrm{aq})}+\mathrm{AgNO}_{3(\mathrm{aq})} \rightarrow \mathrm{AgCl}_{(\mathrm{s})}+\mathrm{KNO}_{3(\mathrm{aq})}\)
Formation of white precipitate by the above reaction is due to formation of silver chloride (AgCl).

Question 2.
Why does the reaction rate of a reaction increase on raising the temperature?
Answer:
The rate of a reaction increases at higher temperature, because the heat to the reactants provide energy to breakup more bonds and speeds up the reaction.

Question 3.
Define a combination reaction. Give one example of an exothermic combination reaction.
Answer:
A chemical reaction in which 2 or more reactants combine to form a single product, the reaction is known as combination reaction. Most of the combination reaction are exothermic because they involve formation of new bonds. For example,
\(\mathrm{H}_{2(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{g})} \rightarrow 2 \mathrm{HCl}_{(\mathrm{g})}\)
\(\mathrm{SiO}_{2(\mathrm{s})}+\mathrm{CaO}_{(\mathrm{s})} \rightarrow \mathrm{CaSiO}_{3(\mathrm{s})}\).

Question 4.
Differentiate reversible and irreversible reactions.
Answer:

Reversible reaction	Irreversible reaction
1. Reactions can be reversed	1. The reaction cannot be reversed
2. It proceeds in both directions	2. It is unidirectional
3. It attains equilibrium	3. Equilibrium is not attained
4. It is relatively slow	4. It is fast

VI. Answer in detail.

Question 1.
What is called thermolysis reactions?
Answer:
A chemical reaction is a process in which old bond breaks up and new chemical bond get formed. Thermolysis chemical reactions is a special type of chemical reaction in which the reactant get decomposed by heat. For example,
\(\mathrm{CaCO}_{3(\mathrm{s})} \stackrel{\text { Heat }}{\rightleftharpoons} \mathrm{CaO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{g})}\)
\(2 \mathrm{HgO}_{(\mathrm{s})} \stackrel{\mathrm{Heat}}{\longrightarrow} 2 \mathrm{Hg}_{(\mathrm{l})}+\mathrm{O}_{2(\mathrm{g})}\)
In these reactions heat is supplied to break the bonds, so generally they are endothermic in nature.

Question 2.
Explain the types of double displacement reactions with examples.
Answer:
When two compounds react, if their ions are interchanged, then the reaction is called double displacement reactions. There are two types of double displacement reactions. They are
(i) Precipitation reactions : When aqueous solutions of two compounds are mixed, if they react to form an insoluble compound and a soluble compound, then it is called precipitation reaction. Because the insoluble compound, formed as one of the products, is a precipitate and hence the reaction is so called.

(ii) When aqueous solutions of potassium iodide and lead (II) nitrate are mixed, a double displacement reaction takes place between them.
Pb(NO₃)_2(aq) + 2KI_(aq) → PbI_2(s)↓ + 2KNO_3(aq).

(iii) Potassium and lead displace or replace one other and form a yellow precipitate of lead (II) iodide.

(iv) Neutralization reactions: When an acid reacts with the base to form a salt and water. It is called ‘neutralization reaction’ as both acid and base neutralize each other.

(v) Reaction of sodium hydroxide with hydrochloric acid is a typical neutralization reaction. Here, sodium replaces hydrogen from hydrochloric acid forming sodium chloride, a neutral soluble salt.
NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)

Question 3.
Explain the factors influencing the rate of a reaction.
Answer:
The factors influencing the rate of a reaction are,
(i) Nature of the reactants: The reaction of sodium with hydrochloric acid is faster than that with acetic acid. Do you know why? Hydrochloric acid is a stronger acid than acetic acid and thus more reactive. So, the nature of the reactants influences the reaction rate.
\(\begin{aligned} 2 \mathrm{Na}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} & \rightarrow 2 \mathrm{NaCl}_{(\mathrm{aq})}+\mathrm{H}_{2(\mathrm{g})}(\mathrm{fast}) \\ 2 \mathrm{Na}_{(\mathrm{s})}+2 \mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} & \rightarrow 2 \mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{aq})}+\mathrm{H}_{2(\mathrm{g})}(\mathrm{slow}) \end{aligned}\).

(ii) The concentration of the reactants: Changing the number of reactants also increases the reaction rate. The amount of the substance present in a certain volume of the solution is called ‘concentration’. More the concentration, more particles per volume exist in it and hence faster the reaction. Granulated zinc reacts faster with 2M hydrochloric acid than 1M hydrochloric acid.

(iii) Temperature: Most of the reactions go faster at a higher temperature. Because adding heat to the reactants provides energy to break more bonds and thus speed up the reaction. Calcium carbonate reacts slowly with hydrochloric acid at room temperature. When the reaction mixture is heated the reaction rate increases.

(iv) Pressure: If the reactants are gases, increasing their pressure increases the reaction rate. This is because on increasing the pressure the reacting particles come closer and collide frequently.

(v) Catalyst: A catalyst is a substance which increases the reaction rate without being consumed in the reaction. In certain reactions, adding a substance as catalyst speeds up the reaction. For example, on heating potassium chlorate, it decomposes into potassium chloride and oxygen gas, but at a slower rate. If manganese dioxide is added, it increases the reaction rate.

(vi) The surface area of the reactants: When solid reactants are involved in a reaction, their powdered form reacts more readily. For example, powdered calcium carbonate reacts more readily with hydrochloric acid than marble chips. Because powdering of the reactants increases the surface area and more energy is available on the collision of the reactant particles. Thus, the reaction rate is increased. You will study more about reaction rate in your higher classes.

Question 4.
How does pH play an important role in everyday life?
Answer:
Our body works within the pH range of 7.0 to 7.8. Living organisms can survive only in a narrow range of pH change. Different body fluids have different pH values.
Eg: pH of blood is ranging from 7.35 to 7.45. Any increase or decrease in this value leads to diseases. The ideal pH for blood is 7.4.

pH in our digestive system : It is very interesting to note that our stomach produces hydrochloric acid. It helps in the digestion of food without harming the stomach. During indigestion the stomach produces too much acid and this causes pain and irritation. pH of the stomach fluid is approximately 2.0.

pH changes as the cause of tooth decay : pH of the saliva normally ranges between 6.5 to 7.5. White enamel coating of our teeth is calcium phosphate, the hardest substance in our body. When the pH of the mouth saliva falls below 5.5, the enamel gets weathered. Toothpastes, which are generally basic ate used for cleaning the teeth that can neutralise the excess acid and prevent tooth decay.

pH of soil : In agriculture, the pH of the soil is very important. Citrus fruits require slightly alkaline soil, while rice requires acidic soil and sugarcane requires neutral soil.

pH of rain water : The pH of rain water is approximately 7, which means that it is neutral and also represents its high purity. If the atmospheric air is polluted with oxide gases of sulphur and nitrogen, they get dissolved in the rain water and make its pH less than 7. Thus, if the pH of rain water is less than 7, then it is called acid rain. When acid rain flows into the rivers it lowers the pH of the river water also. The survival of aquatic life in such rivers becomes difficult.

Question 5.
What is chemical equilibrium? What are its characteristics?
Answer:
Chemical equilibrium is the state for a reversible chemical reaction where the rate of forwarding direction is equally balanced by the rate of backward direction and the process seems like to be stopped.
Its characteristics are,

• The rate of the forward and backward reaction are equal in chemical equilibrium.
• The observable properties such as pressure, concentration, colour, density, viscosity, etc. of the system unchanged with time.
• In physical equilibrium, the volume of all phases remains constant.

VII. HOT Questions

Question 1.
A solid compound ‘A’ decomposes on heating into ‘B’ and a gas ‘C’ On passing the gas ‘C’ through water. It becomes acidic. Identify A, B and C.
\(\mathrm{CaCO}_{3(\mathrm{s})} \longrightarrow \mathrm{CaO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{g})}\)
Calcium carbonate decomposes into solid Calcium oxide and a gas CO₂. This CO₂ dissolves in water and forms carbonic acid.
\(\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}\) (Carbonic acid).

Question 2.
Can a nickel spatula be used to stir copper sulphate solution? Justify your answer.
Answer:
No, a nickel spatula cannot be used to stir CuSO₄ solution, because Nickel will displace copper from CuSO₃ solution and Cu gets deposited on the Ni spatula.

VIII. Solve the following problems.

Question 1.
Lemon juice has a pH = 2, what is the concentration of H⁺ ions?
Solution:

Question 2.
Calculate the pH of 1.0 × 10^-4 molar solution of HNO₃?
Solution:

Question 3.
What is the pH of 1.0 × 10^-5 molar solution of KOH?
Solution:

Question 4.
The hydroxide ion concentration of a solution is 1 × 10^-11. What is the pH of the solution?
Solution:

Samacheer Kalvi 10th Science Types of Chemical Reactions Additional Question Solved

I. Choose the best answer.

Question 1.
Methane + Oxygen → A + Water. Identify A ______.
(a) Carbon monoxide
(b) Carbon dioxide
(c) ethane
(d) LPG.
Answer:
(i) Carbon dioxide

Question 2.
The chemical reaction in which electricity is used to bring about the change is:
(a) Thermal decomposition
(b) Photo decomposition
(c) Single displacement reaction
(d) Electrolytic decomposition
Answer:
(d) Electrolytic decomposition

Question 3.
Decomposition of the molecule occurs on passing an electric current through its aqueous solution. This process is termed as ______.
(a) Thermolysis
(b) Photolysis
(c) Electrolysis
(d) None of these.
Answer:
(c) Electrolysis

Question 4.
The most reactive element in the activity series is:
(a) platinum
(b) potassium
(c) sodium
(d) gold
Answer:
(b) potassium

Question 5.
Which one of the following is a more reactive element?
(a) Cu
(b) Li
(c) Zn
(d) Pb.
Answer:
(b) Li

Question 6.
Which one of the following is the least reactive element?
(a) Au
(b) Fe
(c) Ca
(d) Na.
Answer:
(a) Au

Question 7.
Double displacement reaction is also called as ______.
(a) Metastasis
(b) Metathesis
(c) Methanolysis
(d) Metalysis.
Answer:
(b) Metathesis

Question 8.
The acidic solution among the following is:
(a) seawater
(b) coffee
(c) lime water
(d) antacid
Answer:
(b) coffee

Question 9.
The reaction of sodium hydroxide with hydrochloric acid is an example for ______.
(a) Combination reaction
(b) Thermolysis reaction
(c) Neutralization reaction
(d) Precipitation reaction.
Answer:
(c) Neutralization reaction

Question 10.
In agriculture, the nature of the soil for rice is:
(a) alkaline
(b) neutral
(c) acidic
(d) none of the above
Answer:
(c) acidic

Question 11.
The role of manganese dioxide in the decomposition of potassium chlorate is ______.
(a) act as a reactant
(b) act as a catalyst
(c) act as a reagent
(d) act as a reaction medium.
Answer:
(b) act as a catalyst

Question 12.
The ionic product of water K_w is ______.

Answer:
(c) \(\mathrm{K}_{\mathrm{W}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]\)

Question 13.
Unit of ionic product of water is ______.
(a) mol^-2 dm^-6
(b) mol² dm^-6
(c) mol² dm⁶
(d) mol dm^-3.
Answer:
(b) mol² dm^-6

Question 14.
What is the value of ionic product of water?
(a) 1 × 10^-4
(b) 1 × 10^-1
(c) 1 × 10¹⁴
(d) 1 × 10^-14.
Answer:
(d) 1 × 10^-14.

Question 15.
pH of milk of magnesia is ______.
(a) 10
(b) 15
(c) 14
(d) 10.5.
Answer:
(a) 10

Question 16.
What is the relationship between pH and pOH?
(a) pH – pOH = 14
(b) pH + pOH = 14
(c) pH/pOH = 14
(d) pH + pOH = 1.4.
Answer:
(b) pH + pOH = 14

Question 17.
\(\mathrm{Zn}+2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2} \uparrow\)
The above reaction is an example of ______.
(а) Combination reaction
(b) Double displacement reaction
(c) Displacement reaction
(d) Decomposition reaction.
Answer:
(c) Displacement reaction

Question 18.
A student tests the pH of pure water using a pH paper. It shows a green colour. If a pH paper is after adding lemon juice to water, what colour will he observe?
(a) Green
(b) Red
(c) Yellow
(d) Pink.
Answer:
(b) Red

Question 19.
When aqueous solution of silver nitrate and sodium chloride are mixed, ______ precipitate is immediately formed.
(a) white
(b) yellow
(c) red
(d) black.
Answer:
(a) white

Question 20.
pH = – log₁₀ [H⁺]. The pH of a solution containing hydrogen ion concentration of 0.001 M solution is ______.
(a) 3
(b) 11
(c) 14
(d) 15.
Answer:
(a) 3

Question 21.
Silver anklet has got tarnished due to the formation of ______.
(a) Ag₂O
(b) AgNO₃
(c) Ag₂S
(d) AgBr.
Answer:
(c) Ag₂S

Question 22.
The colour of the precipitate formed by the reaction of lead nitrate with potassium iodide is ______.
(a) white
(b) yellow
(c) black
(d) red.
Answer:
(b) yellow

Question 23.
The reaction of calcium oxide with water is a ______ reaction.
(a) exothermic
(b) endothermic
(c) isothermic
(d) adiabatic.
Answer:
(a) exothermic

Question 24.
The gas released when calcium carbonate reacts with dilute hydrochloric acid is ______.
(a) NO₂
(b) CO
(c) CO₂
(d) H₂.
Answer:
(c) CO₂

Question 25.
The chemical used in whitewashing on the walls is ______.
(a) CaO
(b) CaCl₂
(c) CaOCl₂
(d) Ca(OH)₂.
Answer:
(d) Ca(OH)₂.

Question 26.
The chemical formula of marble is ______.
(a) CaCO₃
(b) CaOCl₂
(c) CaSO₄. \(\frac { 1 }{ 2 }\) H₂O
(d) Ca(OH)₂.
Answer:
(a) CaCO₃

Question 27.
Combustion of coal is an example of _____ reaction.
(a) displacement
(b) reduction
(c) decomposition
(d) combination.
Answer:
(d) combination.

Question 28.
The colour change takes place when copper carbonate is strongly heated is ______.
(a) green to black
(b) green to blue
(c) blue to green
(d) blue to black.
Answer:
(a) green to black

Question 29.
The reaction of heat on copper carbonate into copper (II) oxide is _________ reaction.
(a) combination
(b) decomposition
(c) displacement
(d) redox.
Answer:
(b) decomposition

Question 30.
The dissolution of glucose in water is a _____ reaction.
(a) exothermic
(b) endothermic
(c) neutralisation
(d) combination.
Answer:
(b) endothermic

Question 31.
All combustion reactions are _______ reactions.
(a) combination
(b) exothermic
(c) endothermic
(d) neutralisation.
Answer:
(b) exothermic

Question 32.
The factor that affects the rate of the chemical reaction is ______.
(a) temperature
(b) concentration
(c) catalyst
(d) all the above.
Answer:
(d) all the above.

Question 33.
Magnesium ribbon reacts at a very faster rate with ______ acid.
(a) hydrochloric
(b) acetic
(c) formic
(d) oxalic.
Answer:
(a) hydrochloric

Question 34.
Our body metabolism is carried out by means of ______ secreted in our stomach.
(a) sulphuric acid
(b) hydrochloric acid
(c) nitric acid
(d) acetic acid.
Answer:
(b) hydrochloric acid

Question 35.
Which metals do not liberate gas on reaction with acids?
(a) Zn, Mg
(b) Ag, Cu
(c) Na, K
(d) Cr, Al.
Answer:
(b) Ag, Cu

Question 36.
When CO₂ is passed through lime water, it turns ______.
(a) milky
(b) black
(c) red
(d) blue.
Answer:
(a) milky

Question 37.
The physical form of calcium carbonate is ______.
(a) limestone
(b) chalk
(c) marble
(d) all the above.
Answer:
(d) all the above.

Question 38.
The colour change takes place when copper (II) oxide reacts with dilute hydrochloric acid is ______.
(a) blue to green
(b) black to green
(c) green to black
(d) green to blue.
Answer:
(b) black to green

Question 39.
Which of the following is a strong base?
(a) NH₄OH
(b) Ca(OH)₂
(c) Al(OH)₃
(d) NaOH.
Answer:
(d) NaOH.

Question 40.
Zinc reacts with sodium hydroxide to form ______.
(a) Zinc hydroxide + H₂O
(b) Sodium zincate + H₂
(c) Zinc oxide + H₂
(d) Zinc oxide + H₂O.
Answer:
(b) Sodium zincate + H₂

Question 41.
_____ reacts with sodium hydroxide.
(a) Cu
(b) Ag
(c) Cr
(d) Zn.
Answer:
(d) Zn.

Question 42.
________ is used as a medicine for stomach disorder.
(a) Sodium hydroxide
(b) Ammonium hydroxide
(c) Magnesium hydroxide
(d) Calcium hydroxide.
Answer:
(c) Magnesium hydroxide

Question 43.
The pH of stomach fluid is ______.
(a) 12
(b) 14
(c) 2
(d) 1.
Answer:
(c) 2

Question 42.
The hardest substance in the human body is ______.
(a) bone
(b) the enamel coating of teeth
(c) brain
(d) liver.
Answer:
(b) the enamel coating of teeth

Question 43.
Sugarcane requires _______ soil.
(a) acidic
(b) alkaline
(c) neutral
(d) amphoteric.
Answer:
(c) neutral

Question 44.
Rice requires _____ soil.
(a) acidic
(b) basic
(c) alkaline
(d) neutral.
Answer:
(a) acidic

Question 45.
______ is a double salt.
(a) Sodium chloride
(b) Washing soda
(c) Potash alum
(d) Bleaching powder.
Answer:
(c) Potash alum

II. Fill in the blanks.

Question 1.
During chemical changes _____ are formed and these changes are more _____ than physical changes.
Answer:
New products, permanent.

Question 2.
Calcium oxide reacts with water to produce ______ and the reaction is ______.
Answer:
Slaked lime, exothermic.

Question 3.
During whitewashing, ______ reacts slowly with carbon dioxide in the air to form a thin layer of _______ on the walls.
Answer:
Calcium hydroxide, calcium carbonate.

Question 4.
When copper carbonate is heated, the products formed are _____ , ______ and change of colour from ______ to ____ is observed.
Answer:
CuO, CO₂, green, black

Question 5.
When lead nitrate is heated, the gas liberated is ______ and its colour is ______.
Answer:
NO₂, Reddish – brown

Question 6.
Copper sulphate solution changes its blue colour into _____ colour when an iron nail is added to it and it acquires _____ colour.
Answer:
Green, brownish.

Question 7.
When Barium chloride reacts with sodium sulphate, the product formed is _____ and it is a ______ precipitate.
Answer:
Barium sulphate, white.

Question 8.
Powdered ____ reacts more quickly with hydrochloric acid than marble chips.
Answer:
CaCO₃

Question 9.
Calcium carbonate present in the marble reacts with hydrochloric acid at a faster rate at _______ temperature.
Answer:
Higher.

Question 10.
______ is a substance which furnishes H⁺ ions when dissolved in water and a ______ is a substance which furnishes OH^– ions in water.
Answer:
Acid, Base.

Question 11.
Acids present in plants and animals are ______ and the acids in rocks and minerals are ______.
Answer:
Organic acid, inorganic acid.

Question 12.
Metal displaces _______ gas from dilute acid and the flame goes off with a ______ sound.
Answer:
Hydrogen, popping.

Question 13.
_____ is used in the manufacturing of soap and ______ is used in whitewashing buildings.
Answer:
Sodium hydroxide, calcium hydroxide.

Question 14.
pH scale was introduced by ______ and the pH of the solution is 7, it is a ______ solution.
Answer:
S.P.L. Sorenson, neutral.

Question 15.
The pH of a normal healthy human skin is _____ and the pH of the stomach fluid is ______.
Answer:
7.4, 5.5.

Question 16.
The ideal pH for blood is _______ and pH of the mouth falls below ______.
Answer:
7.35 – 7.45, Sulphuric acid.

Question 17.
All photo decomposition reaction are _______ reactions.
Answer:
Endothermic.

Question 18.
Precipitation reactions give a _____ as the product.
Answer:
Insoluble salt.

Question 19.
Plants cannot grow in a ______ soil.
Answer:
Acidic.

Question 20.
Equilibrium is possible in a _____ system.
Answer:
Closed.

Question 21.
Pure water is a _______ electrolyte.
Answer:
Weak.

Question 22.
Most of the combination reactions are _______ in nature.
Answer:
Exothermic.

Question 23.
Silicon dioxide reacts with calcium oxide to form ______.
Answer:
Calcium silicate.

Question 24.
Our mobile phones get energy from its _____ battery by chemical reaction.
Answer:
Lithium.

Question 25.
A ______ is a substance which increases the reaction rate.
Answer:
Catalyst.

Question 26.
pH range of human saliva is ______.
Answer:
6 – 8.

Question 27.
pH range of fresh milk is ______.
Answer:
5.

Question 28.
The pH of a solution can be determined by using a _______ indicator.
Answer:
Universal.

Question 29.
pH is ______.
Answer:
-log₁₀ [H⁺].

Question 30.
The white enamel coating of our teeth is ______.
Answer:
Calcium phosphate.

III. Match the following.

Question 1.

(i) \(\mathrm{S}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{SO}_{2(\mathrm{g})}\)	(a) Electrolytic decomposition reaction
(ii) \(2 \mathrm{NaCl}_{(\mathrm{aq})} \rightarrow 2 \mathrm{Na}_{(\mathrm{s})}+\mathrm{Cl}_{2(\mathrm{g})}\)	(b) Combination reaction
(iii) \(2 \mathrm{AgBr}_{(\mathrm{s})} \rightarrow 2 \mathrm{Ag}_{(\mathrm{s})}+\mathrm{Br}_{2(\mathrm{g})}\)	(c) Single displacement reaction
(iv) \(\mathrm{Zn}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \rightarrow \mathrm{ZnCl}_{2(\mathrm{aq})}+\mathrm{H}_{2(\mathrm{g})}\)	(d) Photo decomposition reaction

Answer:
i – b, ii – a, iii – d, iv – c.

Question 2.

Sample	pH
(i) Oranges	(a) 8
(ii) Grapes	(b) 7
(iii) Pure water	(c) 3.5
(iv) Egg white	(d) 4

Answer:
i – c, ii – d, iii – b, iv – a.

Question 3.

Sample	pH
(i) Sour milk	(a) 4.2
(ii) Tomato juice	(b) 4.5
(iii) Seawater	(c) 12
(iv) Lime water	(d) 8

Answer:
i – b, ii – a, iii – d, iv – c.

Question 4.

(i) Citrus fruits	(a) Neutral soil
(ii) Rice	(b) pH = 7
(iii) Sugarcane	(c) alkaline soil
(iv) Rainwater	(d) acidic soil

Answer:
i – c, ii – d, iii – a, iv – b.

Question 5.

(i) Sodium hydroxide	(a) medicine
(ii) Calcium hydroxide	(b) stain remover
(iii) Ammonium hydroxide	(c) soap
(iv) Magnesium hydroxide	(d) whitewashing

Answer:
i – c, ii – d, iii – b, iv – a.

Question 6.

(i) Gain of electrons	(a) Oxidation reaction
(ii) Loss of electrons	(b) Combination reaction
(iii) Combustion of coal	(c) Redox reaction
(iv) Rusting of iron	(d) Reduction reaction

Answer:
i – d, ii-a, iii – b, iv – e.

IV. State true or false. If false, give the correct statement.

Question 1.
Chemical changes are reversible changes.
Answer:
False.
Correct statement: Chemical changes are irreversible changes.

Question 2.
The silver anklet has got tarnished when exposed to air due to the formation of silver oxide. (Ag₂O).
Answer:
False.
Correct statement: The silver anklet has got tarnished when exposed to air due to the formation of silver sulphide (Ag₂S).

Question 3.
Brisk effervescence takes place with the evolution of CO₂ when calcium carbonate reacts with dilute hydrochloric acid.
Answer:
True.

Question 4.
The chemical formula for marble is CaCl₂.
Answer:
False.
Correct statement: The chemical formula for marble is CaCO₃

Question 5.
A reaction in which a single product is formed from two or more reactants is known as displacement reaction.
Answer:
False.
Correct statement: A reaction in which a single product is formed from two or more reactants is known as a combination reaction.

Question 6.
Lead Nitrate on heating decomposes to give lead oxide with the evolution of reddish-brown gas NO₂ and O₂ gas.
Answer:
True.

Question 7.
Any reaction that produces a precipitate is called a redox reaction.
Answer:
False.
Correct statement: Any reaction that produces a precipitate is called a precipitation reaction.

Question 8.
The reaction in which a more reactive element displaces a less reactive element from its compound is called displacement reaction.
Answer:
True.

Question 9.
The chemical reactions which take place with the evolution of heat energy are called endothermic reaction.
Answer:
False.
Correct statement: The chemical reactions which take place with the evolution of heat energy are called exothermic reaction.

Question 10.
All combustion reactions are endothermic reactions.
Answer:
False.
Correct statement: All combustion reactions are exothermic reactions.

Question 11.
Our body metabolism is carried out by means of sulphuric acid secreted in our stomach.
Answer:
False.
Correct statement: Our body metabolism is carried out by means of hydrochloric acid secreted in our stomach.

Question 12.
Limestone, chalk and marble are different physical forms of calcium oxide.
Answer:
False.
Correct statement: Limestone, chalk and marble are different physical forms of calcium carbonate.

Question 13.
The atmosphere of earth is made up of thick white and yellowish clouds of sulphuric acid.
Answer:
False.
Correct statement: The atmosphere of Venus is made up of thick white and yellowish clouds of sulphuric acid.

Question 14.
Bond breaking releaser energy whereas bond formation absorbs energy.
Answer:
False.
Correct statement: Bond breaking absorbs energy whereas bond formation releases energy.

Question 15.
More active elements readily displace less active elements from their aqueous solution.
Answer:
True.

Question 16.
The irreversible reaction is relatively slow.
Answer:
False.
Correct statement: Irreversible reaction is fast.

V. Assertion and Reason

Question 1.
Assertion (A): The lustrous white colour of the silver anklet slowly changes into slightly back colour.
Reason (R): silver anklet reacts with H₂S in the air to form Ag₂S (Silver Sulphide) which is black in colour.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) Both (A) and (R) are wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 2.
Assertion (A): The reaction of calcium oxide with water is an exothermic reaction.
Reason (R): The reaction is accompanied by a hissing sound and formation of bubbles leading to the absorption of a considerable amount of heat.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(c) (A) is correct but (R) is wrong

Question 3.
Assertion (A): During the reaction of calcium carbonate with dilute hydrochloric acid, brisk effervescence takes place.
Reason (R): Brisk effervescence is due to the evolution of carbon dioxide gas.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct

Question 4.
Assertion: When copper carbonate is heated strongly, the green colour is changed into a black colour.
Reason (R): The colour change is due to the decomposition of copper carbonate into copper oxide.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 5.
Assertion (A): when lead nitrate is heated, the gas released has red – orange colour and it is lead oxide.
Reason (R): Lead nitrate on heating undergoes combination reaction.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are wrong

Question 6.
Assertion (A): iron js more reactive than copper.
Reason (R): iron js displaced from iron sulphate by copper.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are wrong.
Answer:
(b) (A) is correct but (R) is wrong

Question 7.
Assertion (A): Copper displaces zinc (or) lead from the salt solution.
Reason (R): Copper is more reactive than zinc and lead.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are wrong.
Answer:
(d) Both (A) and (R) are wrong.

Question 8.
Assertion (A): All combustion reactions are exothermic reactions.
Reason (R): During combustion reaction, heat energy is liberated.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are wrong.
Answer:
(a) Both (A) and (R) are correct

Question 9.
Assertion (A): when glucose is kept on our tongue, a cooling effect is felt.
Reason (R): It is an endothermic reaction in which heat is absorbed.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 10.
Assertion (A): Powdered calcium carbonate reacts more quickly with hydrochloric acid than marble chips.
Reason (R): Powdered calcium carbonate offers a large surface area than marble chips. Because the greater the surface area, the greater is the rate of the reaction.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct

Question 11.
Assertion (A): Combination reaction is exothermic in nature.
Reason (R): During the formation of new bonds, releases a huge amount of energy in the form of heat.
(a) (A) is correct and (R) explains (A)
(b) (A) is correct but (R) does not explain (A)
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) (A) is correct and (R) explains (A)

Question 12.
Assertion (A): When glucose is kept on our tongue, a cooling effect is felt.
Reason (R): It is an endothermic reaction in which heat is absorbed.
(a) (A) is correct and (R) explains (A)
(b) (A) is correct but (R) does not explain (A)
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) (A) is correct and (R) explains (A)

VI. Short Answer Questions.

Question 1.
What type of chemical reaction takes place when
(i) limestone is heated
(ii) a magnesium ribbon is burnt in air?
Answer:
(i) When limestone is heated, the reaction takes place is a decomposition reaction.

(ii) When a magnesium ribbon is burnt in air, it is an oxidation reaction.

Question 2.
What is a combustion reaction?
Answer:
A combustion reaction is one in which the reactant rapidly combines with oxygen to form one or more oxides and energy (heat). So in combustion reactions, one of the reactants must be oxygen.

Question 3.
Which of the following is a combustion?

1. Digestion of food
2. Rusting of iron.

Answer:

1. Digestion of food : Not a combustion reaction, because it is a endothermic process and where energy is utilized.
2. Rusting of iron : Combustion reaction.

Question 4.
When the lead powder is added to copper chloride solution, a displacement reaction occurs and solid copper is formed.
(i) Write the equation for the reaction.
(ii) Why does the displacement reaction occur?
Answer:

(ii) Copper is less reactive than lead. So lead has displaced copper from copper chloride. It is a displacement reaction.

Question 5.
What happens when lead nitrate reacts with potassium iodide solution?
Answer:
Lead nitrate solution when reacts with a potassium iodide solution, a deep yellow precipitate of Pbl₂ is formed.

Question 6.
Differentiate Exothermic reaction and Endothermic reaction.
Answer:

Exothermic reaction	Endothermic reaction
1. The chemical reactions which take place with the evolution of heat energy are called exothermic reactions.	1. The chemical reactions which take place with the absorption of heat energy a called endothermic reactions.
2. e.g. All combustion reactions are exothermic.	2. e.g. Dissolution of glucose is the endothermic reaction.
3. N2 + 3H2 → 2NH3 + Heat	3. 2NH3 + Heat → N2 + 3H2

Question 7.
Why the study of reaction rate is important?
Answer:
Faster the reaction, more will be the amount of the product in a specified time. So, the rate of a reaction is important for a chemist for designing a process to get a good yield of a product. Rate of reaction is also important for a food processor who hopes to slow down the reactions that cause food to spoil.

Question 8.
What is meant by decomposition reaction? Give an example.
Answer:
A single compound breaks down to produce two or more substances. Such type of reaction is called decomposition reaction.

Question 9.
Give reason.
(a) Granulated Zinc reacts faster with 2M HCl than 1M HCl.
Answer:
Reason: As the concentration of the reactant increases the rate of the reaction increases.

(b) Food kept at room temperature spoils faster than that kept in the refrigerator.
Answer:
Reason: In the refrigerator the temperature is lower than room temperature, so the reaction rate is less.

Question 10.
What is meant by double displacement reaction? Give an example.
Answer:
A double decomposition reaction is a reaction in which the exchange of ions between two reactants occur leading to the formation of two different products.

Question 11.
Why the toothpastes are generally basic in nature?
Answer:
The pH of the saliva is usually between 6.5 to 7.5, when the pH of the mouth saliva falls below 5.5 the enamel coating of our teeth calcium phosphate gets weathered. That is why the tooth pastes are generally basic in nature.

Question 12.
What happens to food containing fat and oil kept open for a long time?
Answer:
When food containing fat and oil is left as such for a long time, it becomes stale. The stale food develops bad taste and foul smell. This is very common in curd and cheese, particularly in summer. Oils and fats are slowly oxidised to certain foul – smelling compounds.

Question 13.
Define the rate of a chemical reaction.
Answer:
The rate of a chemical reaction is defined as the change in concentration of any one of the reactants or product per unit time.
\(\text { Rate }=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\).

Question 14.
Define catalyst.
Answer:
A substance which alters the rate of a reaction without undergoing any change in mass and composition is known as a catalyst.

Question 15.
What happens during a chemical reaction?
Answer:

• In a chemical reaction, the atoms of the reacting molecules or elements are rearranged to form new molecules.
• Old chemical bonds between atoms are broken and new chemical bonds are formed.
• Bond breaking absorbs energy whereas bond formation releases energy.

Question 16.
What; is a balanced chemical equation?
Answer:
A balanced chemical equation is the simplified representation of a chemical reaction which describes the chemical composition, the physical state of the reactants and the products, and the reaction conditions.

Question 17.
What are the main classes of decomposition reactions?
Answer:
There are three main classes of decomposition reactions. They are,

• Thermal decomposition reactions
• Electrolytic decomposition reactions
• Photo decomposition reactions

Question 18.
What are thermal decomposition reactions?
Answer:
In a thermal decomposition reaction, the reactant is decomposed by applying heat.
For example, on heating mercury (II) oxide is decomposed into mercury metal and oxygen gas.
\(2 \mathrm{HgO}_{(\mathrm{s})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{Hg}_{(\mathrm{l})}+\mathrm{O}_{2(\mathrm{g})}\).

Question 19.
What are Electrolytic decomposition reactions?
Answer:
In this type of reaction, the reactant is decomposed by applying electricity.
For example, decomposition of sodium chloride occurs on passing electric current through its aqueous solution.
\(2 \mathrm{NaCl}_{(\mathrm{aq})} \stackrel{\text { Electricity }}{\longrightarrow} 2 \mathrm{Na}_{(\mathrm{s})}+\mathrm{Cl}_{2(\mathrm{g})}\).

Question 20.
What is the photodecomposition reaction?
Answer:
In this type of reaction, the reactant is decomposed by applying light.
For example, when silver bromide is exposed to light, it breaks down into silver metal and bromine gas.
\(2 \mathrm{AgBr}_{(\mathrm{s})} \stackrel{\mathrm{Light}}{\longrightarrow} 2 \mathrm{Ag}_{(\mathrm{s})}+\mathrm{Br}_{2(\mathrm{g})}\).

Question 21.
What is Metathesis reaction?
Answer:
The ion of one compound is replaced by the ion of another compound. Ions of identical charges are only interchanged, i.e. a cation can be replaced by another cation. This reaction is called metathesis reaction.

Question 22.
What is a precipitation reaction?
Answer:
When aqueous solution of two compounds are mixed, if they react to form an insoluble compound and a soluble compound, then it is called precipitation reaction.
e.g„ \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2(\mathrm{aq})}+2 \mathrm{KI}_{(\mathrm{aq})} \longrightarrow \mathrm{PbI}_{2(\mathrm{s})} \downarrow+2 \mathrm{KNO}_{3(\mathrm{aq})}\).

Question 23.
How will you distinguish between combination and decomposition reactions?
Answer:

Combination Reactions	Decomposition Reactions
One or more reactants combine to form a single product	A single reactant is decomposed to form one or more products
Energy is released	Energy is absorbed
Elements or compounds may be the reactants	The single compound is the reactant

Question 24.
What is a neutralization reaction?
Answer:
It is another type of displacement reaction in which the acid reacts with the base to form a salt and water. It is called neutralization reaction as both acid and base neutralizes each other.

Question 25.
What is Combustion reaction?
Answer:
A combustion reaction is one in which the reactant rapidly combines with oxygen to form one or more oxides and energy (heat).

Question 26.
What are reversible and irreversible reactions?
Answer:
Reversible reaction : A reversible reaction is a reaction that can be reversed, i.e., the products can be converted back to the reactants.
e.g., \(\mathrm{PCl}_{5(\mathrm{g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{g})}\)
Irreversible reaction: The reaction that cannot be reversed is called irreversible reaction. The irreversible reactions are unidirectional.

Question 27.
Why is the reaction rate important?
Answer:
Faster the reaction, more will be the amount of the product in a specified time. So, the rate of a reaction is important for a chemist for designing a process to get a good yield of a product. Rate of reaction is also important for a food processor who hopes to slow down the reactions that cause food to spoil.

VII. Long Answer Questions.

Question 1.
Suggest a reason for each observation given below.
Answer:
(i) In fireworks, powdered Mg is used rather than Mg ribbon.
Powdered Mg will have larger surface area than Mg ribbon and the rate of the reaction increases.

(ii) Zn and dil. H₂SO₄ react much more quickly when a few drops of CuSO₄ solution are added.
When few drops of CuSO₄ are added to the solution containing Zn and dil. H₂SO₄, the rate of the reaction increases, because CuSO₄ acts as catalyst.

(iii) The reaction between MgCO₃ and dil. HCl speeds up when some con. HCl is added. As the concentration of the reactant increases, the rate of the reaction increases.

Question 2.
Observe the given chemical change and answer the following:

1. Identify ‘A’ and ‘B’.
2. Write the commercial name of calcium hydroxide.
3. Identify products ‘C’ and ‘D’ when HCl is allowed to oxide react with calcium oxide.
4. Say whether calcium oxide is acidic or basic.

Answer:

1. A is calcium carbonate. B is carbon-di-oxide.
2. Slaked lime is the commercial name of calcium hydroxide.
3. The products C and D are calcium chloride (CaCl₂) and water (H₂O).
4. Calcium oxide is basic in nature.

Question 3.
Take copper nitrate in a test tube and heat it over the flame.

1. What is the colour of cupric nitrate?
2. What do you observe?
3. Name the type of reaction that takes place.
4. Write the balanced equation.

Answer:

1. The colour of cupric nitrate is Blue.
2. When cupric nitrate is heated in a test tube, we can observe the evolution of reddish-brown gas (NO₂) Nitrogen dioxide.
3. The reaction takes place is a decomposition reaction.
4. 

Question 4.
Redox reactions are reactions during which electron transfer takes place. Here magnesium atom transfers two electrons one each to the two chlorine atoms.
(i) What are the products of this reaction?
(ii) Write the balanced equation for the complete reaction.
(iii) Which element is being oxidized?
(iv) Which element is being reduced?
(v) Write the reduction part of the reaction.
Answer:
(i) Magnesium atom is converted to Mg²⁺ ion. Two chlorine atoms are converted to 2Cl^– ions. So the products are Mg²⁺ ion and 2Cl^– ions.
(ii)

(iii) Mg atom is being oxidised by donating 2 electrons.
(iv) Cl₂ molecule is being reduced by accepting 2 electrons.
(v) Cl₂ + 2e^– → 2Cl^–. It is a reduction reaction in which gain of e^– take place.

Question 5.
Take Cu(NO₃)₂ in a test tube and heat it over the flame.
(i) What is the colour of Cu(NO₃)₂?
Answer:
Blue

(ii) What do you observe?
Answer:
Evolution of reddish-brown gas.

(iii) Name the type of reaction that takes place.
Answer:
Decomposition.

(iv) Write the balanced equation.
Answer:
Cu(NO₃)₂ → 2CuO + 4NO₂ + O₂↑

Question 6.
Sodium hydroxide and hydrochloric acid react as shown in this equation.
\(\mathbf{N a O H}_{(\mathrm{aq})}+\mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{NaCl}_{(\mathrm{aq})}+\mathbf{H}_{2} \mathbf{O}_{(\mathrm{l})}\)
(i) Which type of chemical reaction is this?
(ii) The reaction is exothermic. Explain what that means.
(iii) Differentiate exothermic reaction and endothermic reaction.
(iv) What happens to the temperature of the solution as the chemicals react?
Answer:
(i) It is a neutralisation reaction.
(ii) An exothermic reaction is a chemical reaction in which the evolution of heat energy takes place.
(iii)

Exothermic reaction	Endothermic reaction
1. The reaction In which heat energy is evolved is known as an exothermic process.	1. The reaction in which heat energy is absorbed is known as an endothermic process.
2. N2 + 3H2 → 2NH3 + Heat	2. 2NH3 + Heat → N2 + 3H2

(iv) When NaOH reacts with HCl to give NaCl and water, heat is evolved. So the solution’s temperature increases.

VIII. HOT Questions.

Question 1.
Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid is added to test tube A, while acetic acid is added to test tube B. The amount and concentration taken for both the acids are same. In which test tube does the reaction occur more vigorously and why?
Answer:
In the test tube, the reaction occurs more vigorously.
Comparing hydrochloric acid and acetic acid, HCl (hydrochloric acid) is a strong acid and more reactive whereas acetic acid is a weak organic acid and less reactive.
\(\mathrm{Mg}+2 \mathrm{HCl} \stackrel{\text { Fast reaction }}{\longrightarrow} \mathrm{MgCl}_{2}+\mathrm{H}_{2} \uparrow\).

Question 2.
Classify the following reactions based on the rate of the reactions as very fast or instantaneous slow and moderate reactions.
(a) AgNO_3(aq) + NaCl_(aq) → AgCl ↓ + NaNO_3(aq)
Answer:
Very fast reaction (or) Instantaneous reaction.

(b) 2 Na_(s) + 2H₂O_(l) → 2NaOH_(aq) + H_2(g)
Answer:
Very fast reaction.

(c) Rusting of iron.
Answer:
Very slow reaction.

(d) Inversion of cane sugar into Glucose and fructose.
Answer:
Moderately slow reaction.

(e) Fermentation of sugar into alcohol.
Answer:
Very slow reaction.

Question 3.
What is the chemical reaction taken place in the tarnishing of silver anklet?
Answer:
The lustrous white colour of the silver anklet slowly changes into a slightly black colour. It is due to the formation of silver sulphide (Ag₂S) as a result of the reaction between silver and hydrogen sulphide in the air.

Question 4.
Why toothpaste are basic in nature?
Answer:
The white enamel coating of our teeth is calcium phosphate, the hardest substance in our body. It does not dissolve in water. If the pH of mouth falls below 5.5, the enamel gets corroded. Toothpaste is generally basic and used for cleaning the teeth can neutralize the excess acid and prevent tooth decay.

Question 5.
Why the solution of slaked lime is used for whitewashing?
Answer:
A solution of slaked lime is used for whitewashing walls. Calcium hydroxide reacts slowly with the carbon dioxide in the air to form a thin layer of calcium carbonate on the walls. Calcium carbonate is formed after two to three days of whitewashing and gives a shiny finish to the walls. It is interesting to note that the chemical formula for marble is also CaCO₃.

Question 6.
Complete the following reactions.

Answer:

Question 7.
Let us consider the following two reactions.

Which reaction will not occur. Why?
Answer:
The first reaction involves the displacement of chlorine from NaCl, by fluorine. In the second reaction, chlorine displaces fluorine from NaF. Out of these two, the second reaction will not occur. Because fluorine is more active than chlorine and occupies the upper position in the periodic table. So, in displacement reactions, the activity of the elements and their relative position in the periodic table are the key factors to determine the feasibility of the reactions. More active elements readily displace less active elements from their aqueous solutions.

Question 8.
Which of the metals displaces hydrogen gas from hydrochloric acid? Silver or zinc. Give the chemical equation of the reaction and justify your answer.
Answer:
Zinc displaces hydrogen gas from hydrochloric, acid.
Zinc is more reactive than silver.
Zn + 2HCl → ZnCl₂ + H₂ ↑.

Question 9.
Foods kept at room temperature spoils faster than that kept in the refrigerator. Why?
Answer:
Food kept at room temperature spoils faster than that kept in the refrigerator. In the refrigerator, the temperature is lower than the room temperature and hence the reaction rate is less.

Question 10.
How will you enhance the rate of decomposition of potassium chlorate?
Answer:
On heating potassium chlorate, it decomposes into potassium chloride and oxygen gas, but at a slower rate. If manganese dioxide is added, it increases the reaction rate.

Here MnO₂ is the catalyst. Therefore the addition of MnO₂ enhances the rate of decomposition of potassium chlorate.

Question 11.
Powdered calcium carbonate reacts more readily with hydrochloric acid than marble chips. Why?
Answer:
Powdering of the reactants increases the surface area and more energy is available on the collision of the reactant particles. Thus, the reaction rate is increased. Hence powdered calcium carbonate reacts more readily with hydrochloric acid than marble chips.

Samacheer Kalvi 10th Science Types of Chemical Reactions Additional Problems Solved

Question 1.
The [OH^–] ion concentration of a solution is 1.0 × 10^-8 M. What is the pH of the solution?
Solution:
The concentration of hydroxide ion = [OH^–] = 1.0 × 10^-8 M.

Question 2.
The hydrogen ion concentration of a solution is 1.0 × 10^-9 m. What is the pH of the solution? Find out whether the given solution is acidic, basic or neutral.
Solution:

Question 3.
The hydroxide ion concentration of a solution is 0.001 m. What is the pH of the solution?
Solution:

pH = 14 – pOH
pH = 14 – 3 = 11
pH = 11.

Question 4.
The hydroxide ion concentration of a solution is 1.0 × 10^-9 m. What is the pH of the solution?
Solution:

Question 5.
The hydrogen ion concentration of a solution is 1 × 10^-4 m. Calculate the pH and pOH of that solution.
Solution:

Question 6.
Calculate the pH of sodium hydroxide solution having the concentration of OH^– 0.01 mL^-1.
Solution:

Believing that the Tamilnadu State Board Solutions for Class 10th Science Chapter 10 Types of Chemical Reactions Questions and Answers learning resource will definitely guide you at the time of preparation. For more details about Tamilnadu State Board Class 10th Science Chapter 10 textbook solutions, ask us in the below comments and we’ll revert back to you asap.

Students who are preparing for the Science exam can download this Tamilnadu State Board Solutions for Class 10th Science Chapter 9 from here for free of cost. These Tamilnadu State Board Textbook Solutions PDF cover all 10th Science Nuclear Physics Book Back Questions and Answers.

All these concepts of Chapter 9 Solutions are explained very conceptually by the subject teachers in Tamilnadu State Board Solutions PDF as per the prescribed Syllabus & guidelines. You can download Samacheer Kalvi 10th Science Book Solutions Chapter 9 Solutions State Board Pdf for free from the available links. Go ahead and get Tamilnadu State Board Class 10th Science Solutions of Chapter 9 Solutions.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions

Kickstart your preparation by using this Tamilnadu State Board Solutions for Class 10th Science Chapter 9 Solutions Questions and Answers and get the max score in the exams. You can cover all the topics of Chapter 9 easily after studying the Tamilnadu State Board Class 10th Science Textbook solutions pdf. Download the Tamilnadu State Board Science Chapter 9 Solutions solutions of Class 10th by accessing the links provided here and ace up your preparation.

Samacheer Kalvi 10th Science Solutions Textual Problems Solved

I. Problems based on Solubility.

10th Science Samacheer Kalvi Question 1.
1.5 g of solute is dissolved in 15 g of water to form a saturated solution at 298K. Find out the solubility of the solute at the temperature.
Solution:
Mass of the solute = 1.5 g
Mass of the solvent = 15 g
Solubility of the solute = \(\frac{\text { Mass of the solute }}{\text { Mass of the solvent }} \times 100\)
= \(\frac{1.5}{15} \times 100=10 \mathrm{g}\).

Samacheer Kalvi 10th Science Question 2.
Find the mass of potassium chloride would be needed to form a saturated solution in 60 g of water at 303 K? Given that solubility of the KCl is 37 / 100 g at this temperature.
Solution:
Mass of potassium chloride in 100 g of water in saturated solution = 37 g
Mass of potassium chloride in 60 g of water in saturated solution = \(\frac{37}{100} \times 60\) = 22.2 g.

You can Download Samacheer Kalvi 10th Science Guide PDF help you to revise the complete Syllabus and score more marks in your examinations.

Samacheer Kalvi Class 10 Science Solutions Question 3.
What is the mass of sodium chloride that would be needed to form a saturated solution in 50 g of water at 30°C. Solubility of sodium chloride is 36 g at 30°C?
Solution:
At 30°C, 36 g of sodium chloride is dissolved in 100 g of water.
Mass of sodium chloride that would be need for 100 g of water = 36 g
Mass of sodium chloride dissolved in 50 g of water = \(\frac{36 \times 50}{100}\) = 18 g.

10th Samacheer Kalvi Science Question 4.
The Solubility of sodium nitrate at 50°C and 30°C is 114 g and 96 g respectively. Find the amount of salt that will be thrown out when a saturated solution of sodium nitrate-containing 50 g of water is cooled from 50°C to 30°C?
Answer:
Amount of sodium nitrate dissolved in 100 g of water at 50°C is 114 g
Amount of sodium nitrate dissolving in 50 g of water at 50°C is = \(\frac{114 \times 50}{100}\) = 57 g
Similarly amount of sodium nitrate dissolving in 50 g of water at 30°C is = \(\frac{96 \times 50}{100}\) = 48 g
Amount of sodium nitrate thrown when 50 g of water is cooled from 50°C to 30°C is 57 – 48 = 9 g.

II. Problems based on Mass percentage.

Samacheer Kalvi 10th Science Solutions Question 1.
A solution was prepared by dissolving 25 g of sugar in 100 g of water. Calculate the mass percentage of solute.
Solution:
Mass of the solute = 25 g
Mass of the solvent = 100 g

Class 10 Science Samacheer Kalvi Question 2.
16 grams of NaOH is dissolved in 100 grams of water at 25°C to form a saturated solution. Find the mass percentage of solute and solvent.
Solution:
Mass of the solute (NaOH) = 16 g
Mass of the solvent H₂O = 100 g
(i) Mass percentage of the solute
= \(\frac{\text { Mass of the solute }}{\text { Mass of the solute }+\text { Mass of the solvent }} \times 100\)
= \(\frac{16 \times 100}{16+100}=\frac{1600}{116}\)
Mass percentage of the solute = 13.79 %.

(ii) Mass percentage of solvent = 100 – (Mass percentage of the solute) = 100 – 13.79 = 86.21 %.

Science Solution Class 10 Samacheer Kalvi Question 3.
Find the amount of urea which is to be dissolved in water to get 500 g of 10 % w/w aqueous solution?
Solution:
Mass percentage = \(\frac{\text { Mass of the solute }}{\text { Mass of the solution }} \times 100\)
\(10=\frac{\text { Mass of the urea }}{500} \times 100\)
Mass of urea = 50 g.

III. Problem based on Volume – Volume percentage.

Samacheer Kalvi 10th Science Book Solutions Question 1.
The solution is made from 35 ml of methanol and 65 ml of water. Calculate the volume percentage.
Solution:
The volume of the ethanol = 35 ml
The volume of the water = 65 ml

10th Science Solution Samacheer Kalvi Question 2.
Calculate the volume of ethanol in 200 ml solution of 20 % v/v aqueous solution of ethanol.
Solution:
Volume of aqueous solution = 200 ml
Volume percentage = 20 %

Activity

10th Std Science Solutions Samacheer Kalvi Activity 1.
Look at the following pictures. Label them as a dilute and concentrated solution and justify your answer.

Solution:
In the above picture, the first teacup is more concentrated and the first beaker is more concentrated CuSO₄ solution. The reason is the colour intensity is more in a concentrated solution.

Samacheer Kalvi 10th Science Solutions Textual Evaluation Solved

I. Choose the best answer.

Class 10 Science Solutions Samacheer Kalvi Question 1.
A solution is a ____ mixture.
(a) homogeneous
(b) heterogeneous
(c) homogeneous and heterogeneous
(d) non – homogeneous.
Answer:
(a) homogeneous

Science 10th Samacheer Kalvi Question 2.
The number of components in a binary solution is:
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Samacheer Kalvi 10th Science Answers Question 3.
Which of the following is the universal solvent?
(a) Acetone
(b) Benzene
(c) Water
(d) Alcohol.
Answer:
(c) Water

Science Class 10 Samacheer Kalvi Question 4.
A solution in which no more solute can be dissolved in a definite amount of solvent at a given temperature is called:
(a) Saturated solution
(b) Unsaturated solution
(c) Supersaturated solution
(d) Dilute solution
Answer:
(a) Saturated solution

Samacheer Class 10 Science Solutions Question 5.
Identify the non – aqueous solution ______.
(a) sodium chloride in water
(b) glucose in water
(c) copper sulphate in water
(d) sulphur in carbon-di-sulphide.
Answer:
(d) sulphur in carbon-di-sulphide.

Samacheer Kalvi Guru 10th Science Question 6.
When pressure is increased at a constant temperature the solubility of gases in liquid:
(a) No change
(b) increases
(c) decreases
(d) no reaction
Answer:
(b) increases

Science Samacheer Kalvi Question 7.
The solubility of NaCl in 100 ml water is 36 g. If 25 g of salt is dissolved in 100 ml of water how much more salt is required for saturation ______.
(a) 12 g
(b) 11 g
(c) 16 g
(d) 20 g.
Answer:
(b) 11 g

Samacheer Kalvi 10 Science Question 8.
A 25% alcohol solution means:
(a) 25 ml alcohol in 100 ml of water
(b) 25 ml alcohol in 25 ml of water
(c) 25 ml alcohol in 75 ml of water
(d) 75 ml alcohol in 25 ml of water
Answer:
(c) 25 ml alcohol in 75 ml of water

Samacheer Kalvi Guru 10th Science Book Pdf Download Question 9.
Deliquescence is due to ______.
(a) Strong affinity to water
(b) Less affinity to water
(c) Strong hatred of water
(d) Inertness to water.
Answer:
(a) Strong affinity to water

Samacheer Kalvi.Guru 10th Science Question 10.
Which of the following is hygroscopic in nature?
(a) ferric chloride
(b) copper sulphate penta hydrate
(c) silica gel
(d) none of the above
Answer:
(c) silica gel

II. Fill in the blanks.

10th Standard Science Samacheer Kalvi Question 1.
The component present in a lesser amount, in a solution, is called ______.
Answer:
Solute.

Samacheer Kalvi 10th Standard Science Question 2.
Example for liquid in solid type solution is ______.
Answer:
Sodium chloride dissolved in water.

Question 3.
Solubility is the amount of solute dissolved in _______ g of solvent.
Answer:
100.

Question 4.
Polar compounds are soluble in ______ solvents.
Answer:
Polar.

Question 5.
Volume percentage decreases with increases in temperature because of ______.
Answer:
Expansion of liquid.

III. Match the following.

Question 1.

Blue vitriol	(a) CaSO4.2H2O
Gypsum	(b) CaO
Deliquescence	(c) CuSO4.5H2O
Hygroscopic	(d) NaOH

Answer:
1 – (c), 2 – (a), 3 – (d), 4 – (b).

IV. True or False: (If false give the correct statement)

Question 1.
Solutions which contain three components are called binary solution.
Answer:
False.
Correct Statement: Solutions which contain two components are called binary solution.

Question 2.
In a solution, the component which is present in a lesser amount is called solvent.
Answer:
False.
Correct Statement:

• In a solution, the component which is present in a larger amount is called a solvent.
• In a solution, the component which is present in a lesser amount is called solute.

Question 3.
Sodium chloride dissolved in water forms a non-aqueous solution.
Answer:
False.
Correct Statement: Sodium chloride dissolved in water forms an aqueous solution.

Question 4.
The molecular formula of green vitriol is MgSO₄.7H₂O
Answer:
False.
Correct Statement: The molecular formula of green vitriol is FeSO₄ .7H₂O
Question 5.
When Silica gel is kept open, it absorbs moisture from the air, because it is hygroscopic in nature.
Answer:
True.

V. Short Answer Questions.

Question 1.
Define the term Solution.
Answer:
A solution is a homogeneous mixture of two or more substances.

Question 2.
What is mean by binary solution?
Answer:
Solutions which are made of one solute and one solvent, then it is called binary solution.

Question 3.
Give an example each

1. gas in liquid
2. solid in liquid
3. solid in solid
4. gas in gas.

Answer:

1. Carbon – di – oxide dissolved in water (Soda water).
2. Sodium chloride dissolved in water.
3. Copper dissolved in gold (Alloy).
4. A mixture of Helium – Oxygen gases.

Question 4.
What is aqueous and non-aqueous solution? Give an example.
Answer:
(i) Aqueous solution : The solution in which water acts as a solvent.
(ii) Non-aqueous solution : The solution in which any liquid other than water acts as a solvent.
Eg: Alcohol, benzene, CS₂ acetone.

Question 5.
Define Volume percentage.
Answer:
Volume percentage is defined as the percentage by volume of solute (in ml) present in the given volume of the solution.
Volume percentage = \(\frac{\text { Volume of the solute }}{\text { Volume of the solution }} \times 100\).

Question 6.
The aquatic animals live more in cold region Why?
Answer:
Aquatic animals live more in cold regions because the solubility of O₂ in water is more at low temperature and therefore the amount of dissolved O₂ is more in the water of cold regions.

Question 7.
Define Hydrated salt.
Answer:
The number of water molecules found in the crystalline substance or salts is called water of crystallization. Such salts are called hydrated salts.

Question 8.
A hot saturated solution of copper sulphate forms crystals as it cools. Why?
Answer:
A hot saturated solution of CuSO₄ forms crystal as it cools. Because on cooling the water molecules move closer together and there is less space for the solution to hold on to as much of the dissolved solid and so it forms crystals.

Question 9.
Classify the following substances into deliquescent, hygroscopic. Conc. Sulphuric acid, Copper sulphate penta hydrate, Silica gel, Calcium chloride, and Gypsum salt.
Answer:

1. Deliquescent substances: Calcium chloride
2. Hygroscopic substances: Conc Sulphuric acid, Copper sulphate penta hydrate, Silica gel and Gypsum salt.

VI. Long Answer Questions.

Question 1.
Write notes on
(i) saturated solution
(ii) unsaturated solution
Answer:
(i) Saturated solution : A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called saturated solution.
Eg: 36 g of sodium chloride in 100 g of water at 25°C forms saturated solution.

(ii) Unsaturated solution : Unsaturated solution is one that contains less solute than that of the saturated solution at a given temperature.
Eg: 10 g or 20 g or 30 g of Sodium chloride in 100 g of water at 25°C forms an unsaturated solution.

Question 2.
Write notes on various factors affecting solubility.
Answer:
Factors affecting solubility: There are three main factors which govern the solubility of the solute. They are
(i) Nature of the solute and solvent
(ii) Temperature
(iii) Pressure

(i) Nature of the solute and solvent: The nature of the solute and solvent plays an important role insolubility. Although water dissolves an enormous variety of substances, both ionic and covalent, it does not dissolve everything. The phrase that scientists often use when predicting solubility is “like dissolves like.” This expression means that dissolving occurs when similarities exist between the solvent and the solute. For example, Common salt is a polar compound and dissolves readily in polar solvent like water.

Non – polar compounds are soluble in non-polar solvents. For example, Fat dissolved in ether. But non-polar compounds, do not dissolve in polar solvents; polar compounds do not dissolve in non-polar solvents.

(ii) Effect of Temperature
The solubility of Solids in Liquid: Generally, the solubility of a solid solute in a liquid solvent increases with increase in temperature. For example, a greater amount of sugar will dissolve in warm water than in cold water.
In the endothermic process, solubility increases with increase in temperature. In the exothermic process, solubility decreases with increase in temperature.

The solubility of Gases in liquid: Solubility of gases in liquids decreases with increase in temperature. Generally, water contains dissolved oxygen. When water is boiled, the solubility of oxygen in water decreases, so oxygen escapes in the form of bubbles. Aquatic animals live more in cold regions because more amount of dissolved oxygen is present in the water of cold regions. This shows that the solubility of oxygen in water is more at low temperatures.

(iii) Effect of Pressure: Effect of pressure is observed only in the case of solubility of a gas in a liquid. When the pressure is increased, the solubility of a gas in liquid increases.
The common examples for solubility of gases in liquids are carbonated beverages, i.e. soft drinks, household cleaners containing an aqueous solution of ammonia, formalin aqueous solution of formaldehyde, etc.

Question 3.
(a) What happens when MgSO₄.7H₂O is heated? Write the appropriate equation.
(b) Define solubility.
Answer:
(a) MgSO₄.7H₂O has a water of crystallization is 7. When magnesium sulphate heptahydrate crystals are gently heated, it loses seven water molecules and becomes anhydrous magnesium sulphate.

(b) Solubility: It is defined as the number of grams of a solute that can be dissolved in 100 g of a solvent to form its saturated solution at a given temperature and pressure.

Question 4.
In what way, hygroscopic substances differ from deliquescent substances.
Answer:
Difference between hygroscopic and deliquescent substances is in the extent to which each material can absorb moisture. This is because both of these terms are very much related to each other and they refer to the property of observing and the retention of moisture from the air. However, they differ in the extent of absorption of moisture where hygroscopic materials absorb moisture but not to the extent the original substance dissolves in it, which is the case with deliquescence. Therefore deliquescence can be regarded as an extreme condition of hygroscopic activity.

Difference between hygroscopic substances and deliquescence

Hygroscopic Substances	Deliquescence Substances
When exposed to the atmosphere at ordinary temperature, they absorb moisture and do not dissolve.	When exposed to the atmospheric air at ordinary temperature, they absorb moisture and dissolve.
Hygroscopic substances do not change their physical state on exposure to air.	Deliquescent substances change its physical state on exposure to air.
Hygroscopic substances may be amorphous solids or liquids.	Deliquescent substances are crystalline solids.

Question 5.
A solution is prepared by dissolving 45 g of sugar in 180 g of water. Calculate the mass percentage of solute.
Answer:
Mass of sugar (solute) = 45 g
Mass of water (solvent) = 180 g.

Question 6.
3.5 litres of ethanol is present in 15 litres of an aqueous solution of ethanol. Calculate volume per cent of the ethanol solution.
Answer:
Volume of ethanol (solute) = 3.5 litre
Volume of aqueous solution = 15 litre
Volume of the solution = \(\frac{\text { Volume of the solute }}{\text { Volume of the solution }} \times 100\)
= \(\frac{3.5}{15} \times 100\)
Volume percentage = 23.33 %.

VII. HOT Questions.

Question 1.
Vinu dissolves 50 g of sugar in 250 ml of hot water, Sarath dissolves 50 g of same sugar in 250 ml of cold water. Who will get faster dissolution of sugar? and Why?
Answer:
Vinu will get faster dissolution of sugar. Because Vinu dissolves 50 g of sugar in 250 ml of hot water, whereas Sarath dissolves 50 g of sugar in 250 ml of cold water. Solubility of a solid in liquid increases with increase in temperature.

Question 2.
‘A’ is a blue coloured crystalline salt. On heating, it loses a blue colour and to give ‘B’. When water is added, ‘B’ gives back to ‘A’. Identify A and B, write the equation.
Answer:

• Blue coloured crystalline salt is copper sulphate pentahydrate (A)
• On heating Copper sulphate pentahydrate it loses blue colour and to give anhydrous copper sulphate (B).
• When water is added to the anhydrous copper sulphate (B) gives back to copper sulphate pentahydrate (A).
• 

Question 3.
Will the cool drinks give more fizz at top of the hills or at the foot? Explain.
Answer:
Carbonated cool drinks give more fizz at the foot hill because solubility of gases in liquid decrease with increase in temperature. At higher altitudes, the temperature is low and the dissolved CO₂ will not escape as fizz, whereas at the foot hill it does.

Samacheer Kalvi 10th Science Solutions Additional Questions Solved

I. Choose the best answer.

Question 1.
Find out the homogeneous mixture.
(a) Salt + Water
(b) Sand + Water
(c) Clay + Water
(d) Gold + Water.
Answer:
(a) Salt + Water

Question 2.
The aqueous solution is:
(a) S in CS₂
(b) I₂ in CCl₄
(c) Salt in H₂O
(d) None of the above
Answer:
(c) Salt in H₂O

Question 3.
In a solution, the component which is present in a larger amount is called ______.
(a) solvent
(b) dissolution
(c) solute
(d) mole.
Answer:
(а) solvent

Question 4.
The solubility of ammonia gas at 25°C is:
(a) 36 g
(b) 48 g
(c) 80 g
(d) 184 g
Answer:
(b) 48 g

Question 5.
Example for liquid in liquid binary solution is ______.
(a) Copper dissolved in gold
(b) Water vapour in the air
(c) Ethyl alcohol dissolved in water
(d) NaCl dissolved in water.
Answer:
(c) Ethyl alcohol dissolved in water

Question 6.
Which one of the following is an aqueous solvent?
(a) Benzene
(b) Acetone
(c) Alcohol
(d) Water.
Answer:
(d) Water.

Question 7.
Mass percentage of a solution is independent of:
(a) temperature
(b) amount of solute
(c) amount of solvent
(d) chemical nature of the solute
Answer:
(a) temperature

Question 8.
Solubility is equal to ______.

Answer:
(c) \(\frac{\text { Mass of the solute }}{\text { Mass of the solvent }} \times 100\)

Question 9.
In which case solubility increases with increase in temperature?
(a) Endothermic process
(b) Exothermic process
(c) Both (a) and (b)
(b) None of these.
Answer:
(a) Endothermic process

Question 10.
The factor/factors which affect the solubility of a solute:
(a) Nature of the solute and solvent
(b) Temperature
(c) Pressure
(d) All of these
Answer:
(d) All of these

Question 11.
The effect of pressure on the solubility of a gas in a liquid is given by ______.
(a) Hess’s law
(b) Ohm’s law
(c) Henry’s law
(d) Gases law.
Answer:
(c) Henry’s law

Question 12.
5% sugar solution means ______.
(a) 5 g of sugar in 95 g of water
(b) 50 g of sugar in 50 g of water
(c) 20 g of sugar in 80 g of water
(d) 95 g of sugar in 5 g of water.
Answer:
(a) 5 g of sugar in 95 g of water

Question 13.
The water of crystallisation present in white vitriol is:
(a) 2
(b) 7
(c) 5
(d) 24
Answer:
(b) 7

Question 14.
Mass percentage is independent of ______.
(a) density
(b) volume
(c) pressure
(d) temperature.
Answer:
(d) temperature.

Question 15.
Volume percentage is expressed as ______.
(a) v/v
(b) w/w
(c) v/w
(d) w/v.
Answer:
(a) v/v

Question 16.
When the temperature increases, volume percentage ______.
(a) Increases
(b) Decreases
(c) no change
(d) increases then decrease.
Answer:
(b) Decreases

Question 17.
The mass of NaCl needed to form saturated solution in 50 g of water at 30°C. When the solubility of NaCl is 36 g at 30°C is:
(a) 9 g
(b) 18 g
(c) 57 g
(d) 19 g
Answer:
(b) 18 g

Question 18.
Zinc sulphate heptahydrate is also called as ______.
(a) Blue Vitriol
(b) Epsom salt
(c) Green Vitriol
(d) White Vitriol.
Answer:
(d) white Vitriol.
Question 19.
Water of crystallization of blue vitriol is ______.
(a) 1
(b) 5
(c) 4
(d) 2.
Answer:
(b) 5

Question 20.
On heating copper sulphate pentahydrate crystals ______.
(a) blue colour changes into green
(b) green colour changes into blue
(c) blue colour changes into colourless
(d) no colour changes.
Answer:
(c) blue colour changes into colourless

Question 21.
Water of crystallization of Epsom salt is ______.
(a) 1
(b) 5
(c) 4
(d) 1.
Answer:
(d) 7.

Question 22.
Hygroscopic substances are used as ______.
(a) drying agents
(b) hydrating agents
(c) freezing agents
(d) reducing agents.
Answer:
(a) drying agents

Question 23.
If two liquids are mutually soluble, they are called _____ liquids.
(a) miscible
(b) immiscible
(c) insoluble
(d) viscous.
Answer:
(a) miscible

Question 24.
Aquatic species are more comfortable in cold water because of ______.
(a) as the temperature decreases, the solubility of dissolved oxygen increases
(b) as the temperature increases, the solubility of dissolved oxygen increases
(c) as the temperature increases, the solubility of dissolved oxygen decreases
(d) None of these.
Answer:
(a) as the temperature decreases, the solubility of dissolved oxygen increases

Question 25.
The process of food assimilation by man is in the form of ______.
(a) solution
(b) solid
(c) solute
(d) solvent.
Answer:
(a) solution

Question 26.
Common salt in water is an example of ______.
(a) Colloidal solution
(b) Binary solution
(c) Suspension
(d) all the above.
Answer:
(b) Binary solution

Question 27.
36 g of NaCl in 100 g of water is a ____ solution.
(a) Non – aqueous
(b) Unsaturated
(c) Supersaturated
(d) Saturated.
Answer:
(d) Saturated.
Question 28.
Nitrogen in soil is an example of _____ solution in nature.
(a) saturated
(b) unsaturated
(c) supersaturated
(d) aqueous.
Answer:
(a) saturated

Question 29.
The solubility of CuSO₄ in H₂O is ______ at 20°C.
(a) 36 g
(b) 20.7 g
(c) 10 g
(d) 95 g.
Answer:
(b) 20.7 g

Question 30.
The solubility of NaNO₃ in 100 g water at 25°C is ______.
(a) 36 g
(b) 95 g
(c) 184 g
(d) 92 g.
Answer:
(d) 92 g.

Question 31.
Which of the following affect solubility?
(a) Temperature
(b) Nature of solute and solvent
(c) Pressure
(d) all the above.
Answer:
(d) all the above.

Question 32.
In _____ process, solubility increases with increase in temperature.
(a) exothermic
(b) endothermic
(c) isothermic
(d) adiabatic.
Answer:
(b) endothermic

Question 33.
In ______ process, solubility decreases with increase in temperature.
(a) exothermic
(b) adiabatic
(c) endothermic
(d) isothermic.
Answer:
(a) exothermic

Question 34.
Increase in ______ increases the solubility of gases.
(a) temperature
(b) pressure
(c) no. of moles
(d) concentration.
Answer:
(b) pressure

Question 35.
The solubility of NaBr in H₂O is ______.
(a) 36 g
(b) 95 g
(c) 184 g
(d) 92 g.
Answer:
(b) 95 g

Question 36.
The gas – filled in soft drinks in ______.
(a) O₂
(b) N₂
(c) CO₂
(d) H₂
Answer:
(c) CO₂

Question 37.
The solubility of KNO₃ in water is ______ process.
(a) Exothermic
(b) endothermic
(c) Isothermic
(d) Adiabatic.
Answer:
(b) endothermic

Question 38.
The solubility of CaO in water is ______ process.
(a) Exothermic
(b) endothermic
(c) Adiabatic
(d) Isothermic.
Answer:
(a) Exothermic

Question 39.
The solubility of NaI in water is ______.
(a) 184 g
(b) 92 g
(c) 95 g
(d) 36 g.
Answer:
(a) 184 g

II. Fill in the blanks.

Question 1.
A solution is a _____ mixture of two or more substances.
Answer:
Homogeneous.

Question 2.
Common salt dissolved in water is an example for _____ solution.
Answer:
Binary.

Question 3.
In a solution, the component present in a lesser amount by weight is called ____ and in a large amount by weight is called ______.
Answer:
Solute, solvent.

Question 4.
The solution in which water acts as a solvent is called ______ and the solution in which Benzene acts as a solvent is called ______.
Answer:

1. The aqueous solution,
2. Non – aqueous solution.

Question 5.
20 g of NaCl in 100 g of water is _____ solution.
Answer:
Unsaturated.

Question 6.
36 g of NaCl in 100 g of water at room temperature forms a ____ solution.
Answer:
Saturated.

Question 7.
Nitrogen in soil is an example for _____ in nature.
Answer:
Saturated solution.

Question 8.
A solution which has more of solute than the saturated solution at a given temperature is called ______ solution.
Answer:
Supersaturated.

Question 9.
In _____ process, solubility increases with ______ in temperature.
Answer:
Endothermic, increase.

Question 10.
In an ______ process, solubility decreases with ______ in temperature.
Answer:
Exothermic, increase.

Question 11.
The solubility of KNO₃ increases with the _____ in temperature.
Answer:
Increase.

Question 12.
The solubility of CaO decreases with the ______ in temperature.
Answer:
Increase.

Question 13.
The solubility of oxygen is _____ in cold water.
Answer:
More.

Question 14.
A polar compound is insoluble in _____ solvent.
Answer:
Non – polar.

Question 15.
An increase in ______ increases the solubility of a gas in a liquid.
Answer:
Pressure.

Question 16.
______ gas is filled in soft drinks using the effect of pressure.
Answer:
CO₂

Question 17.
Hygroscopic substances are used as ______.
Answer:
Drying agents.

Question 18.
Anhydrous calcium chloride is an example of ______.
Answer:
hygroscopic substance.

Question 19.
Hygroscopic substances absorb moisture without changing their ______.
Answer:
Physical state.

Question 20.
Deliquescent substances lose their ______.
Answer:
Crystalline shape.

Question 21.
Ferric chloride is an example of ______.
Answer:
Deliquescent substances.

Question 22.
On heating, hydrated crystalline salts lose their ______.
Answer:
Water of crystallization.

Question 23.
The number of water molecules in blue vitriol is ______.
Answer:
Five.

Question 24.
The number of water molecules in magnesium sulphate heptahydrate is ______.
Answer:
Seven.

Question 25.
Mass percentage is _______ to temperature.
Answer:
Independent.

Question 26.
To qualify the solute in a solution, we can use the term ______.
Answer:
Concentration.

Question 27.
Based on the amount of solute, the solution is classified into ________ types.
Answer:
Three.

Question 28.
Zinc dissolved in copper is an example of ______ solution.
Answer:
Solid in solid.

Question 29.
______ is an example of gas in the liquid.
Answer:
Carbon dioxide dissolved in water.

Question 30.
______ and _______ are in the form of solution to decide the physiological activity of human beings.
Answer:
Blood, Lymph.

III. Match the following.

Question 1.

i. Sugar in water	(a) Exothermic process
ii. Sulphur in carbon sulphide	(b) Aqueous solution
iii. The solubility of CaO	(c) Endothermic process
iv. Solubility of KNO3	(d) Non-aqueous solution

Answer:
i – b, ii – d, iii – a, iv – c.

Question 2.

i. Saturated solution	(a) Sulphur in acetone
ii. Unsaturated solution	(b) 56 g NaCl in water
iii. Supersaturated solution	(c) 10 g of NaCl in water
iv. Non-aqueous solution	(d) 36 g of NaCl in water

Answer:
i – d, ii – c, iii – b, iv – a.

Question 3.

i. Solubility of KNO3	(a) decreases with increase in temperature
ii. The solubility of CaO	(b) increases with increase in pressure
iii. The solubility of oxygen	(c) increases with increase in temperature
iv. Solubility of CO2	(d) more in cold water

Answer:
i – c, ii – a, iii – d, iv – b.

Question 4.

i. Mercury with sodium	(a) Liquid in gas
ii. NaCl in water	(b) Liquid in solid
iii. CO2 in water	(c) Solid in liquid
iv. Water vapour in air	(d) Gas in liquid

Answer:
i – b, ii – c, iii – d, iv – a.

Question 5.

Solute	Solubility
i. CaCO3	(a) 80
ii. NH3	(b) 91
iii. NaOH	(c) 0.0013
iv. C6H12O6	(d) 48

Answer:
i – c, ii – d, iii – a, iv – b.

Question 6.

i. Epsom salt	(a) ZnSO4.7H2O
ii. Green Vitriol	(b) MgSO4.7H2O
iii. White Vitriol	(c) SiO2
iv. Silica gel	(d) FeSO4.7H2O

Answer:
i – b, ii – d, iii – a, iv – c.

IV. State whether true or false. If false, give the correct statement.

Question 1.
A solution is a heterogeneous mixture of two or more substances.
Answer:
False.
Correct statement: A solution is a homogeneous mixture of two or more substances.

Question 2.
A solution contains four components, it is called as a binary solution.
Answer:
False.
Correct statement: A solution contains two components, it is called as a binary solution.

Question 3.
The solution in which water acts as a solvent, it is called a non-aqueous solution.
Answer:
False.
Correct statement: The solution in which water acts as a solvent, it is called an aqueous solution.

Question 4.
The solution of sulphur in CS₂ is a suitable example of a non-aqueous solution.
Answer:
True.

Question 5.
Nitrogen in soil is an example of a supersaturated solution.
Answer:
False.
Correct statement: Nitrogen in the soil is an example of a saturated solution.

Question 6.
The solubility of CuSO₄ in H₂O is 36 g at 25°C.
Answer:
False.
Correct statement: Solubility of CuSO₄ in H₂O is 20.7 g at 20°C.

Question 7.
100 ml of water can dissolve 36 g of NaCl at 25°C to attain saturation.
Answer:
True.

Question 8.
In an endothermic process, solubility decreases with increase in temperature.
Answer:
False.
Correct statement: In an endothermic process, solubility increases with increase in temperature.

Question 9.
In an exothermic process, solubility decreases with increase in temperature.
Answer:
True.

Question 10.
The solubility of CaO increases with the increase in temperature.
Answer:
False.
Correct statement: The solubility of CaO decreases with the increase in temperature.

Question 11.
The solubility of oxygen is more in cold water.
Answer:
True.

Question 12.
SO₂ gas is filled in soft drinks using the effect of pressure.
Answer:
False.
Correct statement: CO₂ gas is filled in soft drinks using the effect of pressure.

Question 13.
Ether act as a universal solvent.
False.
Correct statement: Water acts universal solvent.

Question 14.
Covalent compounds are soluble in water.
Answer:
False.
Correct statement: Covalent compounds are insoluble in water.

Question 15.
36 g of sodium chloride in 100 g of water at 25°C forms a saturated solution.
Answer:
True.

Question 16.
40 g of sodium chloride in 100 g of water at 25°C forms an unsaturated solution.
Answer:
False.
Correct statement: 40 g of sodium chloride in 100 g of water at 25°C forms a supersaturated solution.

Question 17.
Aquatic animals live more in hot regions.
Answer:
False.
Correct statement: Aquatic animals live more in cold regions.

Question 18.
Mass percentage is dependent of temperature.
Answer:
False.
Correct statement: Mass percentage is independent of temperature.

Question 19.
Hygroscopic substances do not change their physical state on exposure to air.
Answer:
True.

Question 20.
Deliquescent substances do not change their physical state on exposure to air.
Answer:
False.
Correct statement: Deliquescent substances change its physical state on exposure to air.

V. Assertion and Reason

Question 1.
Assertion (A): Salt solution-common salt dissolved in water is an example of a binary solution.
Reason (R): A solution with two components is called a binary solution.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 2.
Assertion (A): Sugar in water is a true solution.
Reason (R): True solution is a homogeneous mixture that contains small solute particles that are dissolved throughout the solvent.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is wrong but (R) is correct
(d) (A) is correct but (R) is wrong.
Answer:
(b) Both (A) and (R) are correct

Question 3.
Assertion (A): Solution of sulphur in CS₂ is an example of a non-aqueous solution.
Reason (R): The solution in which any liquid other than water acts as a solvent is called the non – aqueous solution.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 4.
Assertion (A): Solubility of KNO₃ increases with increase in temperature.
Reason (R): Solubility of KNO₃ is an endothermic process.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct

Question 5.
Assertion (A): Aquatic species are more comfortable in cold water.
Reason (R): Solubility of oxygen is more in cold water.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 6.
Assertion (A): CO₂ gas is filled in soft drinks using the effect of pressure.
Reason (R): A decrease in pressure increases the solubility of a gas in a liquid.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(c) (A) is correct but (R) is wrong

Question 7.
Assertion (A): Aqueous copper sulphate is a binary solution.
Reason (R): Copper sulphate solution contains two components i.e., one solute-copper sulphate and one solvent-water.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 8.
Assertion (A): Water act as a universal solvent.
Reason (R): Most of the substances are soluble in water.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 9.
Assertion (A): When Blue vitriols are gently heated it turns colourless.
Reason (R): It loses five water molecules and becomes anhydrous compounds.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 10.
Assertion (A): Anhydrous calcium chloride is an example of a hygroscopic substance.
Reason (R): Anhydrous calcium chloride changes its physical state on exposure to air.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is wrong but (R) is correct
(d) (A) is correct but (R) is wrong
Answer:
(d) (A) is correct but (R) is wrong.

VI. Short Answer Questions.

Question 1.
State Henry’s Law.
Answer:
The solubility of a gas in liquid is directly proportional to the pressure of the gas over the solution at a definite temperature.

Question 2.
Define solute and solvent.
Answer:
In a solution, the component present in a lesser amount by weight is called solute and the component present in a large amount by weight is called solvent.

Question 3.
What is a
(i) saturated and
(ii) supersaturated solution?
Answer:
(i) Saturated : A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature.
(ii) Super saturated solution : A solution which contains more solute than the saturated solution at a given temperature.

Question 4.
Why water is known as a universal solvent?
Answer:
Most of the solutes (substances) are soluble in water and so it is called universal solvent.

Question 5.
Common salt dissolves in water easily. Give reason.
Answer:
Common salt is an electrolyte. It is easily dissociated into its ions like Na⁺ and Cl^– in polar solvent water. So NaCl (common salt) an inorganic compound readily dissolves in water.

Question 6.
What is meant by the ternary solution?
Answer:
Solutions which contain three components are called ternary solutions.
E.g. If salt and sugar are added to water, both dissolves in water forming a solution. Here two solutes are dissolved in one solvent. Therefore it is a ternary solution.

Question 7.
Name the type of solution formed in the following cases;

1. 20 g of NaCl in 100 g of water at 25°C
2. 36 g of NaCl in 100 g of water at 25°C
3. Nitrogen in soil
4. Sulphur in CS₂.

Answer:

1. Unsaturated
2. Saturated
3. Saturated
4. Non-aqueous solution

Question 8.
What are concentrated and dilute solutions?
Answer:
Two solutions having the same solute and solvent, the one which contains a higher amount of solute per the given amount of solvent is said to be concentrated solution and the other is said to be dilute solution.

Question 9.
Define Henry’s law.
Answer:
To quantify the solute in a solution, we can use the term concentration. The concentration of a solution may be defined as the amount of solute present in a given amount of solution or solvent.
(or)
The solubility of a gas in a liquid is directly proportional to the pressure of the gas over the solution at a definite temperature.

Question 10.
Define Mass percentage.
Answer:
Mass percentage of a solution is defined as the percentage by mass of the solute present in the solution.
Mass percentage,
\(=\frac{\text { Mass of the solute }}{\text { Mass of the solution }} \times 100\).

Question 11.
What is water crystallization?
Answer:
The number of water molecules found in the crystalline substance is called water of crystallization.

Question 12.
What are hygroscopic substances?
Answer:
Certain substances, when exposed to the atmospheric air at ordinary temperature, absorb moisture without changing their physical state. Such substances are called hygroscopic substances and this property is called hygroscopy.

Question 13.
List out the examples for hygroscopic substances.
Answer:

• Conc.Sulphuric acid (H₂SO₄).
• Phosphorus Pentoxide (P₂O₅).
• Quick lime (CaO).
• Silica gel (SiO₂).
• Anhydrous calcium chloride (CaCl₂).

Question 14.
What are the Deliquescent substances?
Answer:
Certain substances which are so hygroscopic, when exposed to the atmospheric air at ordinary temperatures, absorb enough water and get completely dissolved. Such substances are called deliquescent substances and this property is called deliquescence.

Question 15.
When is deliquescence is maximum?
Answer:

• The temperature is low
• The atmosphere is humid.

Question 16.
Give examples for deliquescent substances.
Answer:

• Calcium chloride (CaCl₂)
• Caustic soda (NaOH)
• Caustic potash (KOH)
• Ferric chloride (FeCl₃).

VII. Long Answer Questions.

Question 1.
From the table given below, furnish your points of inference.

Substance	Solubility at 25°C
NaCl	36 g
NaBr	95 g
NaI	184 g

Answer:
Inferences:

• At 25°C, 36g NaCl is dissolved in 100 g water to give a saturated solution.
• At 25°C, 95 g NaBr is dissolved in 100 g water to get a saturated solution.
• At 25°C, 184g Nal is dissolved in 100 g water to get a saturated solution.
• The solubility of a solute at a given solvent at a particular temperature is defined as the number of grams of solute necessary to saturate 100 g of the solvent at that temperature.
• In the above tabular column, we infer that the solubility of NaI is the highest and the solubility of NaCl is the lowest.
• All the NaCl, NaBr, Nal solution in water are called aqueous solutions.
• Depending upon the amount of the solute, the solutions are classified as a saturated and unsaturated solution.
• The above solutions are saturated solutions.

Question 2.
Distinguish between the saturated and unsaturated solution at a temperature of 25°C using the data given below (Note: Solubility of NaCl is 36 g).

1. 16 g NaCl in 100 g water
2. 36 g NaCl in 100 g water.

Answer:
The solubility of NaCl is 36g.

1. 16 g NaCl in 100 g water is an unsaturated solution.
2. 36 g NaCl in 100 g water is a saturated solution.

Saturated Solution	Unsaturated Solution
1. A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called a saturated solution.	1. A solution in which the solute is lesser in the amount in the solvent is called unsaturated solution.
2. In this solution, no more solute can be dissolved. If more of solute is added, it will not dissolve to give a supersaturated solution.	2. More of solute can be dissolved to get a saturated solution.

Question 3.
Write a note on the type of solution based on the amount of solute present in a solution.
Answer:
Based on the amount of solute, in the given amount of solvent, solutions are classified into the following types.
(i) Saturated solution : A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called saturated solution. Eg: 36 g of sodium chloride in 100 g of water at 25°C forms saturated solution.Further addition of sodium chloride, leave it undissolved.

(ii) Unsaturated solution : Unsaturated solution is one that contains less solute than that of the saturated solution at a given temperature. Eg: 10 g or 20 g or 30 g of Sodium chloride in 100 g of water at 25°C forms an unsaturated solution.

(iii) Super saturated solution : Supersaturated solution is one that contains more solute than the saturated solution at a given temperature. Eg: 40 g of sodium chloride in 100 g of water at 25°C forms super saturated solution. This state can be achieved by altering any other conditions liken temperature, pressure. Super saturated solutions are unstable, and the solute is reappearing as crystals when the solution is disturbed.

Question 4.
Find the concentration of a solution in terms of weight per cent if 20 g of common salt is dissolved in 50 g of water.
Answer:

VIII. HOT Questions.

Question 1.
Observe the diagram.

1. Which is a concentrated solution and why?
2. Which is dilute solution and why?
   

Answer:

1. Flask (b) is a concentrated solution. Because (b) flask contains a large number of solute particles than (a) flask.
2. Flask (a) is a dilute solution. Because (a) flask contains a lesser number of solute particles than (b) flask.

Question 2.
Where we use the phrase – “Like dissolves Like” and explain the meaning of the phrase?
Answer:
The phrase “like dissolves like” is often used for predicting solubility. This expression means that dissolving occurs when similarities exist between the solvent and the solute. For example, Common salt is a polar compound and dissolves readily in polar solvent like water.

Question 3.
Why bubbling occurs when water is boiled?
Answer:
The solubility of a gas in liquid decrease with increase in temperature. Generally, water contains dissolved oxygen. When water is boiled, the solubility of oxygen in water decreases, so oxygen escapes in the form of bubbles.

Question 4.
What happens when blue vitriol is heated? Explain.
Answer:
When blue vitriol (Copper sulphate pentahydrate) crystals are gently heated, it loses it five water molecules and becomes colourless anhydrous copper sulphate.

Samacheer Kalvi 10th Science Solutions Additional Problems Solved

Question 1.
Take 10 g of common salt and dissolve it in 40 g of water. Find the concentration of a solution in terms of weight per cent.
Solution:
Weight percent = \(\frac{\text { Weight of the solute }}{\text { Weight of solute }+\text { Weight of solvent }} \times 100\)
\(=\frac{10}{10+40} \times 100=20 \%\).

Question 2.
2 g of potassium sulphate was dissolved in 12.5 ml of water. On cooling, the first crystals appeared at 60°C. What is the solubility of potassium sulphate in water at 60°C? Solution. 12.5 ml of water weighs 12.5 g.
Solution:
In 12.5 g of water, the amount of potassium sulphate dissolved is 2 g.
In 1 g of water, the amount of potassium sulphate dissolved is 2 / 12.5 g.
Hence in 100 g of water, the amount of potassium sulphate dissolved is (2 × 100) / 12.5 = 16 g.
The solubility of potassium sulphate in the water at 60°C is 16 g.

Question 3.
50 g of a saturated solution of NaCl at 30°C is evaporated to dryness and 13.2 g of dry NaCl was obtained. Find the solubility of NaCl at 30°C in water.
Solution:
Mass of water in solution = 50 – 13.2 = 36.8 g
Solubility of NaCl
\(\begin{aligned} &=\frac{\text { Mass of } \mathrm{NaCl}}{\text { Mass of water }} \times 100 \\=& \frac{13.2}{36.8} \times 100=36 \mathrm{g} \end{aligned}\)
Solubility of NaCl = 36 g (approx).

Question 4.
An empty evaporating dish weighs 20.0 g. After adding a saturated solution of NaNO₃, the dish weighs 66.0 g. When evaporated to dryness, the dish with crystals weighs 42.5 g. Find the solubility of NaNO₃ at 20°C.
Solution:
Weight of saturated solution of NaNO₃ = (66.0 – 20.0) g = 46.0 g
Weight of crystals of NaNO₃ = (41.5 – 20.0) g = 21.5 g
Weight of water in saturated solution = (46.0 – 21.5) g = 24.5 g
Solubility of NaNO3 = \(\frac{\text { Weight of NaNO }_{3} \text { crystals }}{\text { Weight of water }} \times 100\)
\(=\frac{21.5}{24.5} \times 100=87.7 \mathrm{g}\)
Solubility of NaNO₃ at 20°C is = 87.7 g in 100 g H₂O.

Question 5.
30 g sugar is dissolved in 170g of water. Calculate the concentration of solution in terms of weight percent.
Solution:
Weight percent = \(\frac{\text { Weight of the solute }}{\text { Weight of solute }+\text { Weight of solvent }} \times 100\)
\(=\frac{30}{30+170} \times 10=15 \%\).

Question 6.
50 g common salt is dissolved in 150 g of water; Find oat the concentration of solution in terms of weight percent.
Solution:
Weight percent = \(\frac{\text { Weight of the solute }}{\text { Weight of solute }+\text { Weight of solvent }} \times 100\)
\(=\frac{50}{50+150} \times 100\)
= 25 %.

Believing that the Tamilnadu State Board Solutions for Class 10th Science Chapter 9 Solutions Questions and Answers learning resource will definitely guide you at the time of preparation. For more details about Tamilnadu State Board Class 10th Science Chapter 9 textbook solutions, ask us in the below comments and we’ll revert back to you asap.

Students those who are preparing for the 12th Physics exam can download this Samacheer Kalvi 12th Physics Book Solutions Questions and Answers for Chapter 3 Magnetism and Magnetic Effects of Electric Current from here free of cost. These Tamilnadu State Board Solutions for Class 12th Physics PDF cover all Chapter 3 Magnetism and Magnetic Effects of Electric Current Questions and Answers. All these concepts of Samacheer Kalvi 12th Physics Book Solutions Questions and Answers are explained very conceptually as per the board prescribed Syllabus & guidelines.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Kickstart your preparation by using this Tamilnadu State Board Solutions for 12th Physics Chapter 3 Magnetism and Magnetic Effects of Electric Current Questions and Answers and get the max score in the exams. You can cover all the topics of Chapter 3 Magnetism and Magnetic Effects of Electric Current Questions and Answers pdf. Download the Tamilnadu State Board Solutions by accessing the links provided here and ace up your preparation.

Samacheer Kalvi 12th Physics Magnetism and Magnetic Effects of Electric Current Textual Evaluation Solved

Samacheer Kalvi 12th Physics Magnetism and Magnetic Effects of Electric Current Multiple Choice Questions   

12th Physics Chapter 3 Book Back Answers Question 1.
The magnetic field at the center O of the following current loop is-


Answer:

12th Physics 3rd Lesson Book Back Answers Question 2.
An electron moves straight inside a charged parallel plate capacitor of uniform charge density σ The time taken by the electron to cross the parallel plate capacitor when the plates of the capacitor are kept under constant magnetic field of induction \(\vec { B } \) is-


Answer:

12th Physics Lesson 3 Book Back Answers Question 3.
The force experienced by a particle having mass m and charge q accelerated through a potential difference V when it is kept under perpendicular magnetic field \(\vec { B } \) is-

Answer:

Samacheer Kalvi Guru 12th Physics Question 4.
A circular coil of radius 5 cm and 50 turns carries a current of 3 ampere. The magnetic dipole moment of the coil is-
(a) 1.0 amp – m²
(b) 1.2 amp – m²
(c) 0.5 amp – m²
(d) 0.8 amp – m²
Answer:
(b) 1.2 amp – m²

Magnetic Effect Of Electric Current Class 12 Numericals Pdf Question 5.
A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current 1 = 8 m A (milli ampere). The radii of inside and outside turns are a = 50 mm and b = 100 mm respectively. The magnetic induction at the center of the spiral is
(a) 5 μT
(b) 7 μT
(c) 8 μT
(d) 10 μT
Answer:
(b) 7 μT

Samacheerkalvi.Guru 12th Physics Question 6.
Three wires of equal lengths are bent in the form of loops. One of the loops is circle, another is a semi-circle and the third one is a square. They are placed in a uniform magnetic field and same electric current is passed through them. Which of the following loop configuration will experience greater torque ?
(a) circle
(b) semi-circle
(c) square
(d) all of them
Answer:
(a) circle

Samacheer Kalvi.Guru 12th Physics Question 7.
Two identical coils, each with N turns and radius R are placed coaxially at a distance R as shown in the figure. If I is the current passing through the loops in the same direction, then the magnetic field at a point P which is at exactly at \(\frac { R }{ 2 }\) distance between two coils is-


Answer:

Samacheer Kalvi Guru Physics Question 8.
A wire of length l carries a current I along the Y direction and magnetic field is given by \(\vec { B } \) = \(\frac { β }{ √3 }\) \((\hat{i}+\hat{j}+\hat{k})\). The magnitude of Lorentz force acting on the wire is-

Answer:

Physics Solution Class 12 Samacheer Kalvi Question 9.
A bar magnet of length l and magnetic moment M is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be- (NEET 2014)

(a) M
(b) \(\frac { π }{ 3 }\) M
(c) \(\frac { 2 }{ π }\) M
(d) \(\frac { 1 }{ 2 }\) M
Answer:
(b) \(\frac { π }{ 3 }\) M

Samacheer Kalvi 12th Physics Question 10.
A non-conducting charged ring of charge q, mass m and radius r is rotated with constant angular speed ω. Find the ratio of its magnetic moment with angular momentum is
(a) \(\frac { q }{ m }\) M
(b) \(\frac { 2q }{ 3 }\) M
(c) \(\frac { q }{ 2m }\) M
(d) \(\frac { q }{ 4m }\) M
Answer:
(c) \(\frac { q }{ 2m }\) M

Question 11.
The B_H curve for a ferromagnetic material is shown in the figure. The material is placed inside a long solenoid which contains 1000 turns/ cm. The current that should be passed in the solenoid to demagnetize the ferromagnet completely is-

(a) 1.00 m A(milli ampere)
(b) 1.25 mA
(c) 1.50 mA
(d) 1.75 mA
Answer:
(b) 1.25 mA

Question 12.
Two short bar magnets have magnetic moments 1.20 Am² and 1.00 Am²respectively. They are kept on a horizontal table parallel to each other with their north poles pointing towards the south. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centers is (Horizontal components of Earth’s magnetic induction is 3.6 x 10^-5 Wb m^-2) (NSEP 2000-2001)
(a) 3.60 x 10^-5 Wb m^-1
(b) 3.5 x 10^-5 Wb m^-1
(c) 2.56 x 10^-4 Wb m^-1
(d) 2.2 x 10^-4 Wb m^-1
Answer:
(c) 2.56 x 10^-4 Wb m^-1

Question 13.
The vertical component of Earth’s magnetic field at a place is equal to the horizontal component. What is the value of angle of dip at this place?
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(b) 45°

Question 14.
A flat dielectric disc of radius R carries an excess charge on its surface. The surface charge density is σ. The disc rotates about an a × is perpendicular to its plane passing through the center with angular velocity ω. Find the magnitude of the torque on the disc if it is placed in a uniform magnetic field whose strength is B which is directed perpendicular to the a × is of rotation
(a) \(\frac { 1 }{ 4 }\) σωπ BR
(b) \(\frac { 1 }{ 4 }\) σωπ BR²
(c) \(\frac { 1 }{ 4 }\) σωπ BR³
(d) \(\frac { 1 }{ 4 }\) σωπ BR⁴
Answer:
(d) \(\frac { 1 }{ 4 }\) σωπ BR⁴

Question 15.
A simple pendulum with charged bob is oscillating with time period T and let θ be the angular displacement. If the uniform magnetic field is switched ON in a direction perpendicular to the plane of oscillation then-
(a) time period will decrease but θ will remain constant
(b) time period remain constant but θ will decrease
(c) both T and θ will remain the same
(d) both T and θ will decrease
Answer:
(c) both T and θ will remain the same

Magnetism and Magnetic Effects of Electric Current Short Answer Questions

Question 1.
What is meant by magnetic induction?
Answer:
The magnetic induction (total magnetic field) inside the specimen \(\vec { B } \) is equal to the sum of the magnetic field \(\vec { B } \)₀ produced in vacuum due to the magnetising field and the magnetic field \(\vec { B } \)_m due to the induced magnetisation of the substance.
\(\vec { B } \) = \(\vec { B } \)₀ + \(\vec { B } \)_m = µ₀\(\vec { H } \) + µ₀\(\vec { I } \) = µ₀ (\(\vec { H } \) + \(\vec { I } \)) = (\(\vec { H } \) + \(\vec { I } \))

Question 2.
Define magnetic flux.
Answer:
The number of magnetic field lines crossing per unit area is called magnetic flux B.
Φ_B = \(\vec { B } \) .\(\vec { A } \) = BA cos θ = B ⊥ A

Question 3.
Define magnetic dipole moment.
Answer:
The magnetic dipole moment is defined as the product of its pole strength and magnetic length.
\(\vec { P } \) = q_m \(\vec { d } \)

Question 4.
State Coulomb’s inverse law.
Answer:
The force of attraction or repulsion between two magnetic poles is directly proportional to the product of their pole strengths and inversely proportional to the square of the distance between them.
\(\overrightarrow{\mathrm{F}} \propto \frac{q_{m_{\mathrm{A}}} q_{m_{b}}}{r^{2}} \hat{r}\)

Question 5.
What is magnetic susceptibility?
Answer:
It is defined as the ratio of the intensity of magnetisation (\(\vec { M } \)) induced in the material due to the magnetising field (\(\vec { H } \))
\(\chi_{m}=\left|\frac{\overrightarrow{\mathrm{M}}}{\overrightarrow{\mathrm{H}}}\right|\)

Question 6.
State Biot-Savart’s law.
Answer:
The magnitude of magnetic field d\(\vec { B } \) at a point P at a distance r from the small elemental length taken on a conductor carrying current varies-
(i) directly as the strength of the current I
(ii) directly as the magnitude of the length element \(\vec { dl } \)
(iii) directly as the sine of the angle (say,θ) between \(\vec { dl } \) and \(\hat{r}\) .
(iv) inversely as the square of the distance between the point P and length element \(\vec { dl } \) This is expressed as
\(d \mathrm{B} \propto \frac{\mathrm{I} d l}{r^{2}} \sin \theta\)

Question 7.
What is magnetic permeability?
Answer:
Magnetic permeability:
The magnetic permeability can be defined as the measure of ability of the material to allow the passage of magnetic field lines through it or measure of the capacity of the substance to take magnetisation or the degree of penetration of magnetic field through the substance.

Question 8.
State Ampere’s circuital law.
Answer:
The line integral of magnetic field over a closed loop is µ₀ times net current enclosed by the loop.

Question 9.
Compare dia, para and ferromagnetism.
Answer:

Question 10.
What is meant by hysteresis?
Answer:
The phenomenon of lagging of magnetic induction behind the magnetising field is called hysteresis. Hysteresis means Tagging behind’.

Magnetism and Magnetic Effects of Electric Current Long Answer Questions

Question 1.
Discuss Earth’s magnetic field in detail.
Answer:
Gover suggested that the Earth’s magnetic field is due to hot rays coming out from the Sun. These rays will heat up the air near equatorial region. Once air becomes hotter, it rises above and will move towards northern and southern hemispheres and get electrified. This may be responsible to magnetize the ferromagnetic materials near the Earth’s surface.

The north pole of magnetic compass needle is attracted towards the magnetic south pole of the Earth which is near the geographic north pole. Similarly, the south pole of magnetic compass needle is attracted towards the geographic north pole of the Earth which is near magnetic north pole. There are three quantities required to specify the magnetic field of the Earth on its surface, which are often called as the elements of the Earth’s magnetic field. They are:

(a) Magnetic declination (D):
The angle between magnetic meridian at a point and geographical meridian is called the declination or magnetic declination (D).

(b) Magnetic dip or inclination (I):
The angle subtended by the Earth’s total magnetic field \(\vec { B } \) with the horizontal direction in the magnetic meridian is called dip or magnetic inclination (I) at that point.

(c) The horizontal component of the Earth’s magnetic field (B_H):
The component of Earth’s magnetic field along the horizontal direction in the magnetic meridian is called horizontal component of Earth’s magnetic field, denoted by B_H.
Let B_E be the net Earth’s magnetic field at a point P on the surface of the Earth. B_E can be resolved into two perpendicular components.
Horizontal component B_H = B_E cos I …… (1)
Vertical component B_V = B_E sin I …… (2)
Dividing the equation, we get tan I = \(\frac {{ B }_{V}}{{ B }_{H}}\) …….(3)

(i) At magnetic equator The Earth’s magnetic field is parallel to the surface of the Earth (i.e., horizontal) which implies that the needle of magnetic compass rests horizontally at an angle of dip, I = 0°.
B B_E = B_E
Bv = 0
This implies that the horizontal component is maximum at equator and vertical component is zero at equator.

(ii) At magnetic poles. The Earth’s magnetic field is perpendicular to the surface of the Earth (i.e., vertical) which implies that the needle of magnetic compass rests vertically at an angle of dip, I = 90°
Hence, B_H = 0
Bv = B_E
This implies that the vertical component is maximum at poles and horizontal component is zero at poles.

Question 2.
Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current.
Answer:
Magnetic field due to long straight conductor carrying current:
Consider a long straight wire NM with current I flowing from N to M. Let P be the point at a distance a from point O. Consider an element of length dl of the wire at a distance l from point O and \(\vec { r } \) be the vector joining the element dl with the point P. Let θ be the angle between \(\vec { dl } \) and \(\vec { r } \). Then, the magnetic field at P due to the element is d\(\vec { B } \) = \(\frac { { \mu }_{ 0 }I\vec { dl } }{ 4{ \pi r }^{ 2 } } \) sinθ (unit vector perpendicular to \(\vec { dl } \) and \(\vec { r } \)) …… (1)
The direction of the field is perpendicular to the plane of the paper and going into it. This can be determined by
taking the cross product between two vectors \(\vec { dl } \) and \(\vec { r } \) (let it be \(\hat{n}\) ). The net magnetic field can be determined by integrating equation with proper limits.

\(\vec { B } \) = \(\int { d\vec { B } } \)
From the figure, in a right angle triangle PAO,
tan (π – θ) = \(\frac { a }{ l }\)
l = \(\frac { a }{ tan θ }\) (since tan (π – θ) = -tan θ) ⇒ \(\frac { 1 }{ tan θ }\)
l = a cot θ and r = a cosec θ
differentiating,
dl = a cosec² θdθ

This is the magnetic field at a point P due to the current in small elemental length. Note that we have expressed the magnetic field OP in terms of angular coordinate i.e. θ. Therefore, the net magnetic field at the point P which can be obtained by integrating d\(\vec { B } \) by varying the angle from θ = (φ₁ to θ = φ₁ is

For a an infinitely long straight wire, 1 = 0 and 2 = , the magnetic field is
\(\vec { B } \) = \(\frac {{ µ }_{0}}{ 2πa }\) \(\hat{n}\) …… (3)
Note that here \(\hat{n}\) represents the unit vector from the point O to P.

Question 3.
Obtain a relation for the magnetic induction at a point along the axis of a circular coil carrying current.
Answer:
Magnetic field produced along the axis of the current carrying circular coil:
Consider a current carrying circular loop of radius R and let I be the current flowing through the wire in the direction. The magnetic field at a point P on the axis of the circular coil at a distance z from its center of the coil O. It is computed by taking two diametrically opposite line elements of the coil each of length \(\vec { dl } \) at C and D. Let \(\vec { r } \) be the vector joining the current element (1 \(\vec { dl } \)) at C to the point P.
PC = PD = T = \(\sqrt { { R }^{ 2 }+{ Z }^{ 2 } } \) and
angle ∠CPO = ∠DPO = θ
According to Biot-Savart’s law, the magnetic field at P due to the current element I \(\vec { dl } \) is
d\(\vec {B} \) = \(\frac {{ µ }_{0}}{ 4π }\) \(\frac{\mathrm{I} d \vec{l} \times \hat{r}}{r^{2}}\) ….. (1)
The magnitude of magnetic field due to current element l dl at C and D are equal because of equal distance from the coil. The magnetic field dB due to each current element I \(\vec { dl } \) is resolved into two components; dB sin θ along y-direction and dB cos θ along z-direction. Horizontal components of each current element cancels out while the vertical components (dB cos θ \(\hat{k}\)) alone contribute to total magnetic field at the point P.

If we integrate \(\vec { dl } \) around the loop, d\(\vec { B } \) sweeps out a cone, then the net magnetic field \(\vec { B } \) at point P is

Using Pythagorous theorem r² = R² + Z² and integrating line element from 0 to 2πR, we get

Note that the magnetic field \(\vec { B } \) points along the direction from the point O to P. Suppose if the current flows in clockwise direction, then magnetic field points in the direction from the point P to O.

Question 4.
Compute the torque experienced by a magnetic needle in a uniform magnetic field.
Answer:
Torque Acting on a Bar Magnet in Uniform Magnetic Field:
Consider a magnet of length 21 of pole strength q_m kept in a uniform magnetic field \(\vec { B } \) Each pole experiences a force of magnitude q_mB but acts in opposite direction.

Therefore, the net force exerted on the magnet is zero, so that there is no translatory motion. These two forces constitute a couple (about midpoint of bar magnet) which will rotate and try to align in the direction of the magnetic field \(\vec { B } \).

The force experienced by north pole, \(\vec { F } \)_N = q_m \(\vec { B } \) ……. (1)
The force experienced by south pole, \(\vec { F } \)_S = q_m \(\vec { B } \) …….. (2)
Adding equations (1) and (2), we get the net force acting on the dipole as
\(\vec { F } \) = \(\vec { F } \)_N + \(\vec { F } \)_S = \(\vec { O } \)

This implies, that the net force acting on the dipole is zero, but forms a couple which tends to rotate the bar magnet clockwise (here) in order to align it along \(\vec { B } \).
The moment of force or torque experienced by north and south pole about point O is

By using right hand cork screw rule, we conclude that the total torque is pointing into the paper. Since the magnitudes

The magnitude of total torque about point O
τ = l × q_m B sin θ +l × q_m B sin θ
τ = 2l xq_m B sin θ
τ = P_m B sin θ
(∴ q_m × 2l = P_m )
In vector notation, τ = p_m × \(\vec { B } \)

Question 5.
Calculate the magnetic induction at a point on the a×ial line of a bar magnet.
Answer:
Magnetic field at a point along the axial line of the magnetic dipole (bar magnet):
Consider a bar magnet NS. Let N be the North Pole and S be the south pole of the bar magnet, each of pole strength q_m and separated by a distance of 2l. The magnetic field at a point C (lies along the axis of the magnet) at a distance from the geometrical center O of the bar magnet can be computed by keeping unit north pole (q_MC = 1 A m) at C. The force experienced by the unit north pole at C due to pole strength can be computed using Coulomb’s law of magnetism as follows:
The force of repulsion between north pole of the bar magnet and unit north pole at point C (in free space) is

where r – l is the distance between north pole of the bar magnet and unit north pole at C. The force of attraction between South Pole of the bar magnet and unit North Pole at point C (in free space) is

where r + 1 is the distance between south pole of the bar magnet and unit north pole at C.

From equation (1) and (2), the net force at point C is \(\vec { F } \) = \(\vec { F } \)_N + \(\vec { F } \)_S.
From definition, this net force is the magnetic field due to magnetic dipole at a point C(\(\vec { F } \) = \(\vec { B } \))

Since, magnitude of magnetic dipole moment is \(\left| { \vec { P } }_{ m } \right| \) p_m = q_m. 2l the magnetic field point C equation (3) can be written as

If the distance between two poles in a bar magnet are small (looks like short magnet) compared to the distance between geometrical centre O of bar magnet and the location of point C i.e.,
r >>1 then, (r² – l²)² ≈ r⁴ ….. (5)
Therefore, using equation (5) in equation (4), we get

Question 6.
Obtain the magnetic induction at a point on the equatorial line of a bar magnet. Magnetic field at a point along the equatorial line due to a magnetic dipole (bar magnet)
Answer:
Consider a bar magnet NS. Let N be the north pole and S be the south pole of the bar magnet, each with pole strength q_m and separated by a distance of 2l. The magnetic field at a point C (lies along the equatorial line) at a distance r from the geometrical center O of the bar magnet can be computed by keeping unit north pole (q_mC = 1 A m) at C. The force experienced by the unit north pole at C due to pole strength N-S can be computed using Coulomb’s law of magnetism as follow’s:
The force of repulsion between North Pole of the bar magnet and unit north pole at point C (in free space) is

\(\vec { F } \)_N = -F_N cos θ \(\hat{i}\) + F_N sin θ \(\hat{j}\) …… (1)
Where F_N = \(\frac { { µ }_{ 0 } }{ 4\pi } \) \(\frac { { q }_{ m } }{ { r’ }^{ 2 } } \)
The force of attraction (in free space) between south pole of the bar magnet and unit north pole at point C is
\(\vec { F } \)_S = -F_S cos θ \(\hat{i}\) + F_S sin θ j …… (2)
Where \(\vec { F } \)_S = \(\frac { { µ }_{ 0 } }{ 4\pi } \) \(\frac { { q }_{ m } }{ { r’ }^{ 2 } } \)
From equation (1) and equation (2), the net force at point C is \(\vec { F } \) = F_N + F_S This net force is equal to the magnetic field at the point C.
\(\vec { B } \) = -(F_N + F_S) cos θ \(\hat{i}\)
Since, F_N = F_S


In a right angle triangle NOC as shown in the Figure 1

Substituting equation 4 in equation 3 We get

Since, magnitude of magnetic dipole moment is \(\left| { \vec { P } }_{ m } \right| \) = P_m = q_m. 2l and substituting in equation (5), the magnetic field at a point C is

If the distance between two poles in a bar magnet are small (looks like short magnet) when compared to the distance between geometrical center O of bar magnet and the location of point C i.e., r>> l, then,
(r² + l²)^{\(\frac { 3 }{ 2 }\)}
Therefore, using equation (7) in equation (6), we get ≈ r³ ……… (7)

Since P_m \(\hat{i}\) = \(\left| { \vec { P } }_{ m } \right| \)_m, in general, the magnetic field at equatorial point is given by

Note that magnitude of B_axial is twice that of magnitude of B_equatorial and the direction of are opposite.

Question 7.
Find the magnetic induction due to a long straight conductor using Ampere’s circuital law. Magnetic field due to the current carrying wire of infinite
Answer:
length using Ampere’s law:
Consider a straight conductor of infinite length carrying current I and the direction of magnetic field lines. Since the wire is geometrically cylindrical in shape C and symmetrical about its axis, we construct an Amperian loop in the form of a circular shape at a distance r from the centre of the conductor. From the Ampere’s law, we get


Where dl is the line element along the amperian loop (tangent to the circular loop). Hence, the angle between magnetic field vector and line element is zero. Therefore,

where I is the current enclosed by the Amperian loop. Due to the symmetry, the magnitude of the magnetic field is uniform over the Amperian loop, we can take B out of the integration.

For a circular loop, the circumference is 2πr, which implies,

In vector form, the magnetic field is
\(\vec { B } \) = \(\frac { { μ }_{ 0 }I }{ 2\pi r } \)\(\hat{n}\)
Where \(\hat{n}\) is the unit vector along the tangent to the Amperian loop. This perfectly agrees with the result obtained from Biot-Savarf s law as given in equation
\(\vec { B } \) = \(\frac { { μ }_{ 0 }I }{ 2\pi a } \)\(\hat{n}\)

Question 8.
Discuss the working of cyclotron in detail.
Answer:
Cyclotron:
Cyclotron is a device used to accelerate the charged particles to gain large kinetic energy. It is also called as high energy accelerator. It was invented by Lawrence and Livingston in 1934.

Principle:
When a charged particle moves normal to the magnetic field, it experiences magnetic Lorentz force.

Construction:
The particles are allowed to move in between two semicircular metal containers called Dees (hollow D – shaped objects). Dees are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the Dees. The two Dees are kept separated with a gap and the source S (which ejects the particle to be accelerated) is placed at the center in the gap between the Dees. Dees are connected to high frequency alternating potential difference.

Working:
Let us assume that the ion ejected from source S is positively charged. As soon as ion is ejected, it is accelerated towards a Dee (say, Dee – 1) which has negative potential at that time. Since the magnetic field is normal to the plane of the Dees, the ion undergoes circular path. After one semi-circular path in Dee-1, the ion reaches the gap between Dees. At this time, the polarities of the Dees are reversed so that the ion is now accelerated towards Dee-2 with a greater velocity. For this circular motion, the centripetal force of the charged particle q is provided by Lorentz force.
\(\frac {{ mv }^{2}}{ r }\) qvB ⇒ r = \(\frac { m }{ qb }\) v ⇒ r ∝ v
From the equation, the increase in velocity increases the radius of circular path. This process continues and hence the particle undergoes spiral path of increasing radius. Once it reaches near the edge, it is taken out with the help of deflector plate and allowed to hit the target T. Very important condition in cyclotron operation is the resonance condition. It happens when the frequency ƒ at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator f_osc From equation
f_osc = \(\frac { qB }{2πm}\) T = \(\frac { 1 }{{ f }_{osc}}\)
The time period of oscillation is
T = \(\frac { 2πm }{qB}\)
The kinetic energy of the charged paricle is
K E = \(\frac { 1 }{ 2 }\) mv² = \(\frac {{ q }^{2}}{{ B }^{2}}{{ r }^{2}}{ 2m }\)

Limitations of cyclotron:
(a) the speed of the ion is limited
(b) electron cannot be accelerated
(c) uncharged particles cannot be accelerated

Question 9.
What is tangent law? Discuss in detail.
Answer:
Tangent law:
Statement:
When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields.

Explanation:
Let B be the magnetic field produced by passing current through the coil of the tangent Galvanometer and B_H be the horizontal component of earth’s magnetic field. Under the action of two magnetic fields, the needle comes to rest making angle θ with B_H, such that

tan θ = \(\frac { B }{{ B }_{H}}\)
B = B_H tan θ ……. (1)
When no current is passed through the coil, the small magnetic needle lies along horizontal component of Earth’s magnetic field. When the circuit is switched ON, the electric current will pass through the circular coil and produce magnetic field. Now there are two fields which are acting mutually perpendicular to each other. They are:
(1) the magnetic field (B) due to the electric current in the coil acting normal to the plane of the coil.
(2) the horizontal component of Earth’s magnetic field (B_H)
Because of these crossed fields, the pivoted magnetic needle deflects through an angle θ. From tangent law, B = B_H tan θ When an electric current is passed through a circular coil of radius R having N turns, the magnitude of magnetic field at the center is
B = µ₀ \(\frac { NI }{ 2R }\) …….. (2)
From equation (1) and equation (2), we get
µ₀ \(\frac { NI }{ 2R }\) = B_H tan θ
The horizontal component of Earth’s magnetic field can be determined as
B = µ₀ \(\frac { NI }{ 2R tan θ }\) in tesla ……. (3)

Question 10.
Explain the principle and working of a moving coil galvanometer.
Answer:
Moving coil galvanometer:
Moving coil galvanometer is a device which is used to indicate the flow of current in an electrical circuit.

Principle:
When a current carrying loop is placed in a uniform magnetic field it experiences a torque.

Construction:
A moving coil galvanometer consists of a rectangular coil PQRS of insulated thin copper wire. The coil contains a large number of turns wound over a light metallic frame. A cylindrical soft-iron core is placed symmetrically inside the coil. The rectangular coil is suspended freely between two pole pieces of a horse-shoe magnet.

The upper end of the rectangular coil is attached to one end of fine strip of phosphor bronze and the lower end of the coil is connected to a hair spring which is also made up of phosphor bronze. deflection of the coil with the help of lamp and scale arrangement. The other end of the mirror is connected to a torsion head T. In order to pass electric current through the galvanometer, the suspension strip W and the spring S are connected to terminals.

Working:
Consider a single turn of the rectangular coil PQRS whose length be l and breadth b. PQ = RS = l and QR = SP = b. Let I be the electric current flowing through the rectangular coil PQRS. The horse-shoe magnet has hemi – spherical magnetic poles which produces a radial magnetic field.

Due to this radial field, the sides QR and SP are always parallel to to the B-field (magnetic field) and experience no force. The sides PQ and RS are always parallel to the B-field and experience force and due to this, torque is produced.
For single turn, the deflection couple.
τ = bF = bBIl= (lb) BI = ABI since, area of the coil
A = lb For coil with N turns, we get r = NABI …… (1)
Due to this deflecting torque, the coil gets twisted and restoring torque (also known as restoring couple) is developed. Hence the magnitude of restoring couple is proportional to the amount of twist θ. Thus τ = K θ ……. (2)
where K is the restoring couple per unit twist or torsional constant of the spring. At equilibrium, the deflection couple is equal to the restoring couple. Therefore by comparing equation (1) and (2), we get

NABI = Kθ ⇒ I = \(\frac { K }{ NAB }\)θ …….. (3)
(or) I = G θ
Where, G = \(\frac { K }{ NAB }\) is called is called galvanometer constant or current reduction factor of the galvanometer. Since, suspended moving coil galvanometer is very sensitive, we have to handle with high care while doing experiments. Most of the galvanometer we use arc pointer type moving coil galvanometer.

Question 11.
Discuss the conversion of galvanometer into an ammeter and also a voltmeter.
Answer:
Conversion of galvanometer into ammeter and voltmeter:
A galvanometer is very sensitive instrument to detect the current. It can be easily converted into ammeter and voltmeter.
(i) Galvanometer to an Ammeter:
Ammeter is an instrument used to measure current flowing in the electrical circuit. The ammeter must offer low resistance such that it will not change the current passing through it. So ammeter is connected in series to measure the circuit current.

A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer. This low resistance is called shunt resistance S. The scale is now calibrated in ampere and the range of ammeter depends on the values of the shunt resistance.

Let I be the current passing through the circuit. When current I reaches the junction A, it divides into two components. Let I_g be the current passing through the galvanometer of resistance R_g through a path AGE and the remaining current (I – I_g) passes along the path ACDE through shunt resistance S.

The value of shunt resistance is so adjusted that current I produces full scale deflection in the galvanometer. The potential difference across galvanometer is same as the potential difference across shunt resistance.

Since, the deflection in the galvanometer is proportional to the current passing through it.
θ = \(\frac { 1 }{ g }\) I_g ⇒ θ Ig ⇒ θ ∝ I
So, the deflection in the galvanometer measures the current I passing through the circuit (ammeter). Shunt resistance is connected in parallel to galvanometer. Therefore, resistance of ammeter can be determined by computing the effective resistance, which is

Since, the shunt resistance is a very low resistance and the ratio \(\frac { S }{{R}_{ g }}\) is also small. This means, R_g is also small, i.e., the resistance offered by the ammeter is small. So, when we connect ammeter in series, the ammeter will not change the resistance appreciably and also the current in the circuit. For an ideal ammeter, the resistance must be equal to zero. Hence, the reading in ammeter is always lesser than the actual current in the circuit. Let I_ideal be current measured from ideal ammeter and I_actual be the actual current measured in the circuit by the ammeter. Then, the percentage error in measuring a current through an ammeter is

(ii) Galvanometer to a voltmeter: A voltmeter is an instrument used to measure potential difference across any two points in the electrical circuits. It should not draw any current from the circuit otherwise the value of potential difference to be measured will change. Voltmeter must have high resistance and when it is connected in parallel, it will not draw appreciable current so that it will indicate the true potential difference.

A galvanometer is converted into a voltmeter by connecting high resistance R_h in series with galvanometer. The scale is now calibrated in volt and the range of voltmeter depends on the values of the resistance connected in series i.e. the value of resistance is so adjusted that only current I_g produces full scale deflection i nthe galvanometer.

Let Rg be the resistance of galvanometer and ‘g be the current with which the galvanometer produces full scale deflection. Since the galvanometer is connected in series with high resistance, the current in the electrical circuit is same as the current passing through the galvanometer.
Potential difference
I = I_g

Since the galvanometer and high resistance are connected in series, the total resistance or effective resistance gives the resistance of voltmeter. The voltmeter resistance is R_v R_g + R_h

Note that I_g ∝ V
The deflection in the galvanometer is proportional to current I . But current is proportional to the potential difference. Hence the deflection in the ga’anometer is proportional to potential difference. Since the resistance of voltmeter is very large, a voltmeter connected in an electrical circuit will draw least current in the circuit. An ideal voltmeter is one which has infinite resistance.

Question 12.
Calculate the magnetic field inside and outside of the long solenoid using Ampere’s circuital law.
Answer:
Magnetic field due to a long current carrying solenoid:
Consider a solenoid of length L having N turns. The diametre of the solenoid is assumed to be much smaller when compared to its length and the coil is wound very closely.

In order to calculate the magnetic field at any point inside the solenoid, we use Ampere’s circuital law. Consider a rectangular loop abed. Then from Ampere’s circuital law,


= µ₀ x (total current enclosed by Amperian loop)
The left hand side of the equation is

Since the elemental lengths along bc and da are perpendicular to the magnetic field which is along the axis of the solenoid, the integrals

Since the magnetic field outside the solenoid is zero, the integral

For the path along ab, the integral is

where the length of the loop ab is h. But the choice of length of the loop ab is arbitrary. We ean take very large loop such that it is equal to the length of the solenoid L. Therefore the integral is

let N I be the current passing through the solenoid of N turns, then

The number of turns per unit length is given by \(\frac { NI }{ L }\) = n, then
B = µ₀ \(\frac { nLI }{ L }\) = µ₀nI
Since n is a constant for a given solenoid and p0 is also constant. For a fixed current I, the magnetic field inside the solenoid is also a constant.

Magnetism and Magnetic Effects of Electric Current Numerical problems

Question 1.
A bar magnet having a magnetic moment \(\overrightarrow { M } \) is cut into four pieces i.e., first cut in two pieces along the axis of the magnet and each piece is further cut into two pieces. Compute the magnetic moment of each piece.
Solution:
Consider a bar magnet of magnetic moment \(\overrightarrow { M } \) . When a bar magnet first cut in two pieces

Their magnetic moment of each pieces \(\frac { \overrightarrow { M } }{ 4 } \)
Their magnetic moment of each pieces \(\overrightarrow { M } \)_new \(\frac { 1 }{4} \) \(\overrightarrow { M } \)

Question 2.
A conductor of linear mass density 0.2 g m^-1 suspended by two flexible wire as shown in figure. Suppose the tension in the supporting wires ¡s zero when it is kept inside the magnetic field of I T whose direction is into the page. Compute the current inside the conductor and also the direction of the current. Assume g = 10 m s^-2.

Solution:
Downward force, F = mg
Linear mass density, \(\frac { m }{ l }\) = 0.2 gm^-1
\(\frac { m }{ l }\) = 0.2 x 10^-3 kg m^-1
m = (0.2 x 10^-3 x l) kg m^-1
F = (0.2 x 10^-3 x l x 10) N
Upward magnetic force acting on the wire
F = BIl
0.2 x 10^-3 x l x 10 = 1 x I x l
I = 2 x 10^-3
I = 2 m A

Question 3.
A circular coil with cross-sectional area 0.1 cm² is kept in a uniform magnetic field of strength 0.2 T. If the current passing in the coil is 3 A and plane of the loop is perpendicular to the direction of magnetic field. Calculate
(a) total torque on the coil
(b) total force on the coil
(c) average force on each electron in the coil due to the magnetic field of the free electron density for the material of the wire is 10²⁸ m^-3
Solution:
Cross sectional area of coil, A = 0.1 cm²
A = 0.1 x 10^-4m²
Uniform magnetic field of strength, B = 0.2T
Current passing in the coil, I = 3A
Angle between the magnetic field and normal to the coil, θ = 0°
(a) Total torque on the coil,
τ = ABI sin θ = 0.1 x 10^-4 x 0.2 x 3 sin0° sin0° = 0
τ = 0

(b) Total force on the coil
F = BIl sin θ = 0.2 x 3 x l x sin 0°
F = 0

(c) Average force:
F = qV_dB
drift velocity, V_d = \(\frac { 1 }{ ne A }\)
[∵ q = e]
F = e \(\left( \frac { 1 }{ neA } \right) \)B
[∵n = 10²⁸ m^-3
\(\frac { IB }{ n A }\) = \(\frac{3 \times 0.2}{10^{28} \times 0.1 \times 10^{-4}}\) = 6 x 10^-24
F_av = 0.6 x 10^-23 N

Question 4.
A bar magnet is placed in a uniform magnetic field whose strength is 0.8 T. Suppose the bar magnet orient at an angle 30° with the external field experiences a torque of 0.2 N m. Calculate:
(i) the magnetic moment of the magnet
(ii) the work done by an applied force in moving it from most stable configuration to the most unstable configuration and also compute the work done by the applied magnetic field in this case.
Solution:
Uniform magnetic field strength B = 0.8T
Bar magnet orient an angle with magnetic field θ = 30°
Torque τ = 0.2 Nm
(i) Magnetic moment of the magnet,
Torque τ = P_m B Sin θ
∴ Magnetic moment, P_m = \(\frac { τ }{ B sin θ }\) = \(\frac { 0.2 }{ 0.8 × Sin 30° }\) = \(\frac { 0.2 }{ 0.4 }\)
P_m = 0.5 Am²

(ii) Work done by external torque is stored in the magnet as potential energy.
W = U = – P_m B Sin θ
Here, applied force acting on magnet its moving from most stable θ’ to most unstable θ.
θ’= 0° and θ = 180°
So, workdone W = U = – P_m B (Cosθ – Cosθ’)
= – P_m B (Cos 180° – Cos 0°) = – 0.5 x 0.8 ((-1) – 1) = – 0.4 (-2)
W = U = 0.8
W= 0.8 J

Question 5.
A non – conducting sphere has a mass of 100 g and radius 20 cm. A flat compact coil of wire with turns 5 is wrapped tightly around it with each turns concentric with the sphere. This sphere is placed on an inclined plane such that plane of coil is parallel to the inclined plane. A uniform magnetic field of 0.5 T exists in the region in vertically upward direction. Compute the current 1 required to rest the sphere in equilibrium.

Solution:
At equilibrium
f_s R-p_m B sin θ = 0
mgR = NBAI
I = \(\frac { mg R }{ N B A }\) = \(\frac { mg R }{{ N B π R }^{2}}\)
I = \(\frac { mg }{ π R N B }\)
Mass of the sphere, m= 100 g = 100 x 10^-3 kg
Radius of the sphere R = 20 cm = 20 x 10^-2 m
Number of turns n = 5
Uniform magnetic field B = 0.5 T
I = \(\frac{100 \times 10^{-3} \times 10}{\pi \times 20 \times 10^{-2} \times 5 \times 0.5}\) = \(\frac{1000 \times 10^{-3}}{\pi \times 50 \times 10^{-2}}\) = \(\frac{20 \times 10^{-1}}{\pi}\)
I = \(\frac { 2 }{ π }\) A.

Question 6.
Calculate the magnetic field at the center of a square loop which carries a current of 1.5 A, length of each loop is 50 cm.
Solution:

Current through the square loop, I = 1.5A
Length of each loop, l = 50cm 50 x 10^-2 m
According to Biot-Savart Law.
Magnetic field due to a current carrying straight wire

B =0.084866 x 10^-5 T
Magnetic field at a point p? of centre of current carrying square ioop
B’ =4 sides x B
= 4 x 0.08487 x 10^-5 = 0.33948 x 10^-5
B’ =3.4 x 10^-6 T

Question 7.
Show that the magnetic field at any point on the axis of the solenoid having n turns per unit length is B = \(\frac { 1 }{ 2 }\) µ₀ nI (cos θ₁ – cosθ₂)
Solution:
A solenoid is a cylindrical coil having number of circular turns. Consider a solenoid having radius R consists of n number’of turns per unit length.
Let ‘P’ be the point at a distance ‘x’ from the origin of the solenoid. The current carrying element dx at a distance x from origin and the distance r from point ‘P’.
r = \(\sqrt{\mathrm{R}^{2}+\left(x_{0}-x\right)^{2}}\)
The magnetic field due to current carrying circular coil at any axis is
dB = \(\frac{\mu_{0} \mathrm{IR}^{2}}{2 r^{3}}\) × N
Where N = ndx, then
dB = \(\frac {{ µ }_{0}}{ 2 }\) \(\frac{n \mathrm{IR}^{2} d x}{r^{3}}\) ……. (1)
sin θ = \(\frac { R }{ r }\)
r = R × \(\frac { 1 }{ sin θ }\) = R cosec θ ……. (2)
tan θ = \(\frac { R }{{ x }_{0}-x}\)
x₀ – x = R × \(\frac { 1 }{ tan θ }\) = R cot θ
\(\frac { dx }{ dθ }\) = R cosec² θ
⇒ dx = R cosec² θ dθ ……. (3)
From above three equations, wer get
dB = \(\frac {{ µ }_{0}}{ 2 }\) \(\frac{n \mathrm{IR}^{2}\left(\mathrm{R} \csc ^{2} \theta d \theta\right)}{\mathrm{R}^{3} \csc ^{3} \theta}\)
dB = \(\frac {{ µ }_{0}}{ 2 }\) n I sin θ dθ
Now total magnetic field can be obtained by integrating from θ₁ to θ₂, we get

Question 8.
Let I₁ and I₂ be the steady currents passing through a long horizontal wire XY and PQ respectively. The wire PQ is fixed in horizontal plane and the wire XY be is allowed to move freely in a vertical plane. Let the wire XY is in equilibrium at a height d over the parallel wire PQ as shown in figure.

Solution:
If the wire XY is slightly displaced and released, it executes simple uarmonic motion due to force of repulsion produced between the current carrying wire.
Acceleration of the wire, a = -ω²y
Time period of oscillation of the wire,
T = 2π \(\sqrt { \frac { d }{ g } } \)

Samacheer Kalvi 12th Physics Magnetism and Magnetic Effects of Electric Current Additional Questions Solved

I. Choose the Correct Answer

Question 1.
Source (s) of a magnetic field is (are)-
(a) an isolated magnetic pole
(b) a static electric charge
(c) a moving electric charge
(d) all of these
Answer:
(c) a moving electric charge

Question 2.
Magnetic field lines-
(a) cannot intersect at all
(b) intersect at infinity
(c) intersect within the magnet
(d) intersect at the neutral points
Answer:
(a) cannot intersect at all

Question 3.
A magnetic needle is kept in a non-uniform magnetic field. It experiences-
(a) a force and a torque
(b) a force but not a torque
(c) a torque but not a force
(d) neither a force nor a torque
Answer:
(a) a force and a torque

Question 4.
The SI unit of pole strength is-
(a) Am²
(b) Am^-1
(c) Am^-2
(d) Am
Answer:
(d) Am

Question 5.
Earth’s magnetic field always has a horizontal component except at-
(a) equator
(b) magnetic pole
(c) a latitude of 60°
(d) a latitude of 50°
Answer:
(b) magnetic pole

Question 6.
The angle of dip at the magnetic equator is-
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Answer:
(a) 0°

Question 7.
At a certain place the horizontal component of earth’s magnetic field is √3 times vertical component. The angle of dip at that place is-
(a) 75°
(b) 60°
(c) 45°
(d) 30°
Answer:
(d) 30°
Hint:
tan δ = \(\frac {{ B }_{V}}{{ B }_{H}}\) = \(\frac { 1 }{ √3 }\) ⇒ δ = tan^-1 \(\left( \frac { 1 }{ \sqrt { 3 } } \right) \) = 30°

Question 8.
At magnetic poles the angle of dip is-
(a) 45°
(b) 30°
(c) 0°
(d) 90°
Answer:
(d) 90°

Question 9.
The horizontal component of earth’s magnetic field at a place is 3.6 x 10^-5T. If the angle of dip at this place is 60°. the vertical components of earth’s field at this place is-
(a) 1.2 x 10^-5T
(b) 2.4 x 10^-5T
(c) 4 x 10^-5T
(d) 6.2 x 10^-5T
Answer:
(d) 6.2 x 10^-5T
Hint:
B_v = B_H tan δ = 3.6 x 10^-5 x tan 60°
B_v = 6.2 x 10^-5T

Question 10.
A bar magnet of magnetic moment M is cut into two parts of equal length. The magnetic moment of either part is-
(a) M
(b) 2M
(c) \(\frac { M }{ 2 }\)
(d) Zero
Answer:
(c) \(\frac { M }{ 2 }\)

Question 11.
A magnetic needle suspended by a silk thread is vibrating in the earths magnetic field, If the temperature of the needle is increased by 500°C, then-
(a) time period decreases
(b) time period remains unchanged
(c) time period increases
(d) the needle stops vibrating
Answer:
(c) time period increases
Hint:
Magnet moment decreases with temperature. Therefore the time period will increase.

Question 12.
Demagnetisation of a magnet can be done by-
(a) rough handling
(b) magnetising in the opposite direction
(c) heating
(d) all the above
Answer:
(d) all the above

Question 13.
All the magnetic materials lose their magnetic properties when-
(a) dipped in water
(b) dipped in oil
(c) brought near a piece of iron
(d) strongly heated
Answer:
(d) strongly heated

Question 14.
The relative permeability of a paramagnetic material is-
(a) greater than unity
(b) less than unity
(c) equal to unity
(d) negative
Answer:
(a) greater than unity

Question 15.
The relative permeability of a diamagnetic material is-
(a) greater than unity
(b) less than unity
(c) equal to unity
(d) negative
Answer:
(b) less than unity

Question 16.
Which of the following is most suitable for the core of an electromagnet?
(a) air
(b) soft iron
(c) steel
(d) Co-Ni alloy
Answer:
(b) soft iron

Question 17.
Soft Iron is used in many parts of electrical machines for-
(a) low hysteresis loss and low permeability
(b) low hysteresis less and high permeability
(c) high hysteresis loss and low permeability
(d) high hysteresis loss and high permeability
Answer:
(b) low hysteresis loss and high permeability

Question 18.
When a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is-
(a) attracted by the poles
(b)repel led by the poles
(c) attracted by the north pole and repelled by the south pole
(d) attracted by the south pole and repelled by the north pole
Answer:
(b) repelled by the poles

Question 19.
At Curie point, a ferromagnetic material becomes-
(a) non magnetic
(b) diamagnetic
(c) paramagnetic
(d) antiferromagnetic
Answer:
(c) paramagnetic

Question 20.
Magnetic permeability is maximum for-
(a) diamagnetic substances
(b) paramagnetic substances
(c) ferromagnetic substances
(d) all of these
Answer:
(c) ferromagnetic substances

Question 21.
The material of a permanent magnet has-
(a) high retentivity, low coercivity
(b) low retentivity, low coercivity
(c) low retentivity, low coercivity
(d) high retentivity, high eoercivity
Answer:
(d) high retentivity, high cuercivity

Question 22.
Which one of the following is not made of soft iron?
(a) electromagnet
(b) core of transformer
(c) core of dynamo
(d) magnet of loudspeaker
Answer:
(d) magnet of loudspeaker

Question 23.
A dip circle k at right angles to the magnetic meridian. The apparent dip is-
(a) 0°
(b) 30°
(c) 600
(d) 90°
Answer:
(d) 90°

Question 24.
A magnetic needle is placed in a uniform magnetic field. It experience-
(a) a force and a torque
(b) a force but not a torque
(c) a torque but not a force
(d) neither a force nor a torque
Answer:
(c) a torque but not a force

Question 25.
Relative permeability of iron is 5500. Its magnetic susceptibility is-
(a) 5501
(b) 5500 x 10^-7
(c) 5500 x 10⁷
(d) 5499
Answer:
(d) 5499
Hint:
µ_r = 1 + x
⇒ x = µ_r -1 = 5500 – 1 = 5499

Question 26.
The inherent property of all matter is / are-
(a) paramagnetism
(b) diamagnetism
(c) ferromagnetism
(d) all the above
Answer:
(b) diamagnetism

Question 27.
A magnetic needle suspended freely-
(a) orients itself in a definite direction
(b) remains in any direction
(c) become vertical with N-pole up
(d) become vertical with N-pole down
Answer:
(a) orients itself in a definite direction

Question 28.
The earth’s magnetic field at a given point is 0.5 x10^-5 Wb/m². This field is to be annulled by magnetic induction at the center of circular conducting loop of radius 5 cm. The current required to be flown in the loop is nearly-
(a) 0.2 A
(b) 0.4 A
(c) 4 A
(d) 40 A
Answer:
(b) 0.4 A
Hint:
B = \(\frac {{ µ }_{0}I}{ 2r }\) ⇒ I = \(\frac {2r B}{{ µ }_{0}}\) = \(\frac{2 \times 0.05 \times 0.5 \times 10^{-5}}{4 \pi \times 10^{-7}}\) = 04 A

Question 29.
A frog can be leviated in a magnetic field produced by a current in a vertical solenoid placed below the frog. This is possible because the body of the frog is-
(a) paramagnetic
(b) diamagnetic
(c) ferromagnetic
(d) antiferromagnetic
Answer:
(a) paramagnetic

Question 30.
The magnetic moment of a current carrying circular coil of ladius r varies as-
(a) \(\frac { 1 }{{ r }^{2}}\)
(b) \(\frac { 1 }{ r }\)
(c) r
(d) r²
Answer:
(d) r²
Hint:
Magnetic moment, M = 1A = I(πr²) = M αr²

Question 31.
For a paramagnetic material, the dependence of the magnetic susceptibility % on the absolute temperature T is given as-
(a) x ∝ r
(b) x ∝ \(\frac { 1 }{{T}^{ 2 }}\)
(c) x ∝ \(\frac { 1 }{ T }\)
(d) x ∝ T²
Answer:
(c) x ∝ \(\frac { 1 }{ T }\)

Question 32.
A charged particle (charge q) is moving in a circle of radius R with uniform speed V. The associated magnetic moment is given by-
(a) qVR²
(b) \(\frac {{qVR}^{ 2 }}{2}\)
(c) qVR
(d) \(\frac { qVR }{ 2 }\)
Answer:
(d) \(\frac { qVR }{ 2 }\)
Hint:
Magnetic moment, M = IA = \(\frac { q }{ T }\) (π R²)
M = \(\frac { qv }{ 2πR }\) (π R²) = \(\frac { qvR }{ 2 }\)

Question 33.
Nickel shows ferromagnetic property at room temperature. If the temperature is increased . beyond Curie temperature, then it will show-
(a) antiferromagnetism
(b) no magnetic property
(c) diamagnetism
(d) para magnetism
Answer:
(d) paramagnetism

Question 34.
Aproton enters a magnetic field of flux density 1.5 Wb/m² with a speed of 2x 10⁷ m/s at angle of 30° with the field. The force on the proton will be-
(a) 0.24 x 10^-12 N
(b) 2.4 x 10^-12 N
(c) 24 x 10^-12 N
(d) 0.024 x 10^-12 N
Answer:
(b) 2.4 x 10^-12 N
Hint:
F = Bqv sin θ =1.5 x 1.6 x 10^-19 x 2 x 10⁷ x sin 30° = 2.4 x 10^-12 N

Question 35.
A moving charge produces-
(a) an electric field only
(b)a magnetic field only
(c) both electric and magnetic fields
(d) neither an electric nor a magnetic field
Answer:
(c) both electric and magnetic fields

Question 36.
A straight conductor carrying a current I, is split into a circular loop of radius r as shown in the figure. The magnetic field at the centre O of the circle, in tesla is-

(a) \(\frac {{ µ }_{0}I}{ 2r }\)
(b) \(\frac {{ µ }_{0}I}{ 2πr }\)
(c) \(\frac {{ µ }_{0}I}{ πr }\)
(d) zero
Answer:
(d) zero
Hint:
Field due to the upper and lower semicircles will cancel out.

Question 37.
In a moving coil galvanometer the current ‘i’ is related to the deflection θ as-
(a) i α θ
(b) i α tan θ
(c) i α θ²
(d) i α √θ
Answer:
(a) i α θ

Question 38.
The magnetic field due to a current carrying circular coil on the axis, at a large distance r from the centre of the coil, varies approximately as-
(a) \(\frac { 1 }{ r }\)
(b) \(\frac { 1 }{ { r }^{ \frac { 3 }{ 2 } } } \)
(c) \(\frac { 1 }{{ r }^{3}}\)
(d) \(\frac { 1 }{{ r }^{2}}\)
Answer:
(c) \(\frac { 1 }{{ r }^{3}}\)

Question 39.
A moving charge is subjected to an external magnetic field. The change in the kinetic energy of the particle-
(а) increases with the increase in the field strength
(b) decreases with the increase in the field strength
(c) is always zero
(d) depends upon whether the field is uniform or non-uniform
Answer:
(c) is always zero

Question 40.
Lorentz force is given by-
(a) q (E + V × B)
(b)q (E -V × B)
(c) q (E + V.B)
(d) q (E × B × V)
Answer:
(a) q (E + V × B)

Question 41.
A circular loop has radius R and a current I flows through it. Another circular loop has radius 2R and a current 21 flows through it. Ratio of the magnetic fields at their centres is-
(a) \(\frac { 1 }{ 4 }\)
(b) 1
(c) 2
(d) 4
Answer:
(b) 1
Hint:
B₁ = \(\frac {{ µ }_{0}I}{ 2r }\) and B₂ = \(\frac { µ(2I) }{ 2(2R) }\) ; \(\frac {{ B }_{1}}{ { B }_{2}}\) = 1

Question 42.
The magnetic field inside a solenoid is-
(a) directly proportional to current
(b) inversely proportional to current
(c) directly proportional to its length
(d) inversely proportional to the total number of turns
Answer:
(a) directly proportional to current

Question 43.
A circular loop of area 0.01 m² and carrying a current of 10A is placed parallel to a magnetic field of intensity 0.1T. The torque acting on the loop, in Nm is-
(a) 1.1
(b) 0.8
(c) 0.001
(d) 0.01
Answer:
(d) 0.01
Hint:
τ = BIA = 0.1 x 10 x 0.01 = 0.01 Nm

Question 44.
In a current carrying long solenoid the field produced does not depend upon-
(a) number of turns per unit length
(b) current flowing
(c) radius of the solenoid
(d) all of the above
Answer:
(c) radius of the solenoid

Question 45.
When a charged particle enters a uniform magnetic field its kinetic energy-
(a) remains constant
(b) increases
(c) decreases
(d) becomes zero
Answer:
(a) remains constant

II Fill in the blanks

Question 1.
At Curie point, a ferromagnetic material becomes ………………
Answer:
paramagnetic

Question 2.
Electromagnets are made of soft iron because soft iron has ………………
Answer:
high susceptibility and low retentivity

Question 3.
The word magnetism is derived from iron ore ………………
Answer:
haematetite

Question 4.
The chemical formula of magneties is ………………
Answer:
Fe₃ O₄

Question 5.
The iron ore magnetite was found in the island of ………………
Answer:
Magnesia

Question 6.
……………… suggested that earth behaves as a giant bar magnet
Answer:
Gilbert

Question 7.
The field at the surface of the earth is approximately equal to
Answer:
10^-4T

Question 8.
The natural magnets have ………………
Answer:
irregular shape and they are weak

Question 9.
Pieces of iron or steel that acquires magnetic properties when it is rubbed with a magnet are called ………………
Answer:
artificial magnet

Question 10.
Artificial magnet in the form of a rectangular or cylinderical bar is called ………………
Answer:
bar magnet

Question 11.
In magnet, the magnetic attraction is maximum at the ……………… of the magnet.
Answer:
poles

Question 12.
The unit of pole strength is ………………
Answer:
ampere meter

Question 13.
The unit of magnetic flux density is ………………
Answer:
Weber metre² (or) tesla

Question 14.
The value of po is equal to ………………
Answer:
4π x 10^-7 Hm^-1

Question 15.
A unit pole experiences a force of ………………
Answer:
10^-7N

Question 16.
At neutral points the resultant magnetic field due to the magnet and earth is ………………
Answer:
magnetic field of the earth

Question 17.
The circular scale of deflection magneto meter is divided into ……………… quadrants?
Answer:
4

Question 18.
Each quadrant of the circular scale of deflection magnetometer is graduated from ………………
Answer:
0 – 90°

Question 19.
The sensitivity of the deflection magneto meter is more at ………………
Answer:
45°

Question 20.
The magnetic field used to magnetise a material is called the ………………
Answer:
magnetic intensity

Question 21.
The unit of magnetising field or magnetic intensity is ………………
Answer:
ampere metre 1

Question 22.
The pole strength per unit area of the cross-section of the material is termed as
Answer:
intensity of magnetisation ………………

Question 23.
The magnetic moment per unit volume of the material is termed as ………………
Answer:
intensity of magnetisation

Question 24.
In diamagnetic materials the net magnetic moment of atoms is ………………
Answer:
zero

Question 25.
The susceptibility of diamagnetic materials has a ……………… value.
Answer:
low negative

Question 26.
The susceptibility (x_m) of bismuth is ………………
Answer:
0.00017

Question 27.
The relative permeability of diamagnetic material is ………………
Answer:
less than one

Question 28.
Ferromagnetic substances have ………………magnetic moment.
Answer:
spontaneous net

Question 29.
As the temperature increase, the value of susceptibility of the ferromagnetic substance ………………
Answer:
decrease

Question 30.
The phenomenon of lagging of magnetic induction behind the magnetising field is called ………………
Answer:
hysteresis

Question 31.
The direction of the magnetic field in a current carrying conductor is given by ………………
Answer:
Maxwell’s right cork screw rule

Question 32.
The relative permeability (µ_r) for air is ………………
Answer:
one

Question 33.
The instrument used for measuring current is ………………
Answer:
Tangent galvanometer

Question 34.
Tangent galvanometer works on the principle of ………………
Answer:
Tangent Law

Question 35.
The tangent law is applied in ………………
Answer:
Tangent

III Match the following

Question 1.

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

Question 2.

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

Question 3.

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

Question 4.

(i) → (d)
(ii) → (c)
(in) →(a)
(iv) → (b)

Question 5.

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

IV Assertion and Reason Questions

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

Question 1.
Assertion: The poles of magnet cannot be separated by breaking into two pieces.
Reason: The magnetic moment will be reduced to half when a magnet is broken into two equal pieces.
Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
Explanation: As we know every atom of a magnet acts as a dipole, so poles cannot be separated. When magnet is broken into two equal pieces, magnetic moment of each part will be half of the original magnet.

Question 2.
Assertion: When radius of circular loop carrying current is doubled, its magnetic moment becomes four times.
Reason: Magnetic moment depends on area of the loop.
Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
Explanation: Magnetic moment M = IA = I (πr²)
New magnetic moment Ml = I π (2r)² = 4πIr² = 4M

Question 3.
Assertion: The ferromagnetic substance do not obey Curie’s Law.
Reason: At Curie point a ferromagnetic substance start behaving as a paramagnetic substance.
Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
Explanation: The susceptibility of ferromagnetic substance decreases with the rise of temperature in a complicated manner. After Curie point the susceptibility ferromagnetic substance varies inversely with its absolute temperature.

Question 4.
Assertion: Soft iron is used as transformer core.
Reason: Soft iron has narrow hysteresis loop.
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
Explanation: For high efficiency of transformer, the energy loss will be lesser if the hysteresis loop is lesser area, i.e narrow.

Question 5.
Assertion: Cyclotron does not accelerate electron.
Reason: Mass of the electron is very small
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
Explanation: Cyclotron is suitable for accelerating heavy particles like protons, x-particles. etc. and not for electrons because of low mass.

Question 6.
Assertion: Cyclotron is a device which is used to accelerate the positive ion.
Reason: Cyclotron frequency depends upon the velocity.
Answer:
(c) If assertion is true but reason is false.
Explanation: Cyclotron is utilised to accelerate the positive ion. And cyclotron frequency is given by v =\(\frac { Be }{ 2πm }\). It means cyclotron frequency doesn’t depends upon velocity.

Magnetism and Magnetic Effects of Electric Current 2-mark Questions

Question 1.
Define magnetic declination (D).
Answer:
The angle between magnetic meridian at a point and geographical meridian is called the declination or magnetic declination (D).

Question 2.
Define magnetic inclination (I).
Answer:
The angle subtended by the Earth’s total magnetic field B with the horizontal direction in the magnetic meridian is called dip or magnetic inclination (I) at the point.

Question 3.
Define magnetic field.
Answer:
The magnetic field \(\vec { B } \) at a point is defined as a force experienced by the bar magnet of unit pole strength.
B = \(\frac { 1 }{{ q }_{m}}\) \(\vec { F } \) Its unit is N A^-1 m^-1

Question 4.
Define magnetic flux density.
Answer:
The magnetic flux density is defined as the number of magnetic field lines crossing unit area kept normal to the direction of line of force. Its unit is Wb m^-2 or tesla.

Question 5.
State ‘Tangent Law’.
Answer:
When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields.

Question 6.
Define in terms of magnetising field.
Answer:
The magnetic field which is used to magnetize a sample or specimen is called the magnetising field. Magnetising field is a vector quantity and it denoted by \(\vec { H } \) and its unit is A m^-1 .

Question 7.
State Right hand thumb rule.
Answer:
Answer:
If we curl the fingers of right hand in the direction of current in the loop, then the stretched thumb gives the direction of the mag netic moment associated with the loop.

Question 8.
State ‘One ampere’.
Answer:
One ampere is defined as that current when it is passed through each of the two infinitely long parallel straight conductors kept at a distance of one meter apart in vacuum causes each conductor to experience a force of 2 x 10^-7 newton per meter length of conductor.

Magnetism and Magnetic Effects of Electric Current 3-Mark Questions

Question 1.
Explain Curie’s Law of magnetism.
Answer:
When temperature is increased, thermal vibration will upset the alignment of magnetic dipole moments. Therefore, the magnetic susceptibility decreases with increase in temperature. In many cases, the susceptibility of the materials is This relation is called Curie’s law.
x_m ∝\(\frac { 1 }{ T }\) or x_m = \(\frac { C }{ T }\)
This relation is called Curie’s law.

Question 2.
Write down the concept of Maxwell’s right hand cork screw rule.
Answer:
This rule is used to determine the direction of the magnetic field. If we rotate a right-handed screw using a screw driver, then the direction of current is same as the direction in which screw advances and the direction of rotation of the screw gives the direction of the magnetic field.

Question 3.
Define in terms of Voltage Sensitivity of a galvanometer.
Answer:
It is defined as the deflection produced per unit voltage applied across it.

Question 4.
Define in terms of Current Sensitivity of a galvanometer.
Answer:
It is defined as the deflection produced per unit current flowing through it.
I_s = \(\frac { θ }{ I }\) = \(\frac { N A B }{ K }\) ⇒ I_s = \(\frac { 1 }{ G }\)

Magnetism and Magnetic Effects of Electric Current 5-Marks Questions

Question 1.
Derive an expression for potential energy of a bar magnet in a uniform magnetic field.
Answer:
When a bar magnet (magnetic dipole) of dipole moment \(\vec { B } \)_m is held at an angle 0 with the direction of a uniform magnetic field B , the magnitude of the torque acting on the dipole is

\(\left| { \vec { \tau } }_{ B } \right| \) = \(\left| { \vec { P } }_{ B } \right| \) \(\left| \vec { B } \right| \) sin θ ….. (1)
If the dipole is rotated through a very small angular displacement dθ against the torque τ_B at constant angular velocity, then
the work done by external torque (\({ \vec { \tau } }_{ ext }\)) for this small angular displacement is given by
dW = (\({ \vec { \tau } }_{ ext }\)) dθ …… (2)
The bar magnet has to be moved at constant angular velocity, which implies that
\(\left| { \vec { \tau } }_{ B } \right| \) = \(\left| { \vec { \tau } }_{ ext } \right| \)
dW = P_m B sin θ d θ
Total work done in rotating the dipole from θ’ to θ is e e

W = -P_m B (cosθ – cosθ’) …… (3)
This work done is stored as potential energy in bar magnet at an angle θ when it is rotated from θ’ to θ and it can be written as
U = -P_m B (cosθ – cosθ’) …(4)
In fact, the equation (4) gives the difference in potential energy between the angular positions θ’ and θ. We can choose the reference point θ’ = 90°, so that second term in the equation becomes zero and the equation 4 can be written as
U = -P_m B(cosθ) …(5)
U = -P_m • \(\vec { B } \)
Case 1:
(i) If θ = 0°, then
U= —P_m B (cos 0° ) = P_m B
(ii) If θ = 180°, then
U=-p_m B(cos 180°) = p_m B
The potential energy stored in a bar magnet in a uniform magnetic field is given by

Question 2.
Write down the application of Hysteresis loop.
Answer:
Applications of hysteresis loop:
The significance of hysteresis loop is that it provides information such as retentivity, coercivity, permeability, susceptibility and energy loss during one cycle of magnetisation for each ferromagnetic material. Therefore, the study of hysteresis loop will help us in selecting proper and suitable material for a given purpose. Some examples:

(i) Permanent magnets:
The materials with high retentivity, high coercivity and high permeability are suitable for making permanent magnets.
Examples:
Steel and Alnico

(ii) Electromagnets:
The materials with high initial permeability, low retentivity, low coercivity and thin hysteresis loop with smaller area are preferred to make electromagnets.
Examples:
Soft iron and Mumetal (Nickel Iron alloy).

(iii) Core of the transformer:
The materials with high initial permeability, large magnetic induction and thin hysteresis loop with smaller area are needed to design transformer cores.
Examples:
Soft iron

Question 3.
Write down the difference between soft and hard ferromagnetic materials.
Answer:

Question 4.
State and explain the Biot – Savart Law.
Answer:
Biot and Savart experimentally observed that the magnitude of magnetic field d\(\vec { B } \) at a point P at a distance r from the small elemental length taken on a conductor carrying current varies
(i) directly as the strength of the current I
(ii) directly as the magnitude of the length element \(\vec { dl } \)
(iii) directly as the sine of the angle (say,θ) between \(\vec { dl } \) and \(\hat{r}\).
(iv) inversely as the square of the distance between the point P and length element \(\vec { dl } \). This is expressed as

dB ∝ \(\frac { I dl }{{r}^{ 2 }}\) sin θ
dB = k \(\frac { I dl }{{r}^{ 2 }}\) sin θ
Where K = \(\frac {{μ }_{0}}{ 4π }\) in SI units and k = 1 in CGS units.
In vector notation,
d\(\vec { B } \) = \(\frac {{μ }_{0}}{ 4π }\) \(\frac{\mathrm{I} d \vec{l} \times \hat{r}}{r^{2}}\) ….. (1)
Here vector d\(\vec { B } \) is perpendicular to both I \(\vec { dl } \) (pointing current carrying conductor the direction of current flow) and the unit vector and \(\hat{r}\) directed from \(\vec { dl } \) toward point P The equation 1 is used to compute the magnetic field only due to a small elemental length \(\vec { dl } \) of the conductor. The net magnetic field at P due to the conductor is obtained from principle of superposition by considering the contribution from all current elements I \(\vec { dl } \). Hence integrating equation (1), we get
\(\vec { B } \) = ∫ d\(\vec { B } \) = \(\frac {{ µ }_{0}I}{ 4π }\) ∫ \(\frac { \vec { dl } \times \quad \hat { r } }{ { r }^{ 2 } } \) ……. (2)
where the integral is taken over the entire current distribution.
Case:
1. If the pont P lies on the ckonductor, then θ = 0°. Therefore, d\(\vec { B } \) is zero.
2. If the point lies perpendicular to the conductor, then θ = 90°. Therefore, d\(\vec { B } \) is maximum and is given by d\(\vec { B } \) = \(\frac { I dl }{{ r }{2}}\) \(\hat{n}\).
where \(\hat{n}\) is the unit vector perpendicular to both I \(\vec { dl } \) and \(\hat{r \).

Question 5.
Write down the difference between Coulomb’s Law and Biot-Savart Law
Answer:

Question 6.
Obtain an expression for the magnetic dipole moment of a revolving electron.
Answer:
Magnetic dipole moment of revolving electron:
Suppose an electron undergoes circular motion around the nucleus. The circulating electron in a loop is like current in a circular loop (since flow of charge is current). The magnetic dipole moment due to current carrying circular loop is

\(\vec { µ } \)_L = \(\overrightarrow { A } \) ….. (1)
In magnitude, µ = I A
If T is the time period of an electron, the current due to circular motion of th electron is
I = \(\frac { -e }{ T }\) ……. (2)
where -e is the charge of an electron. If R is the radius of the circular orbit and v is the velocity of the electron in the circular orbit, then
T = \(\frac { 2πR }{ v }\) …… (3)
Using equation (2) and equation (3) in equation (1), we get
µ_L = –\(\frac{\frac{e}{2 \pi \mathrm{R}}}{v}\) πR² = \(\frac { evR }{ 2 }\) ……. (4)
Where A = πR² is the area of the circular loop. By definetion, angular moment of the electron about O is
\(\overrightarrow { L } \) = \(\overrightarrow { R } \) x \(\overrightarrow { P } \)
In magnitude,
L = PR = mvR …….. (5)
Using equation (4) and equation (5), we get
\(\frac { { \mu }_{ L } }{ L } \) = –\(\frac { evR }{ \frac { 2 }{ mvR } } \) = \(\frac { e }{ 2m }\) \(\overrightarrow { L } \) ……. (6)
The negative sign indicates that the magnetic moment and angular momentum are in opposite direction.
In magnitude,
\(\frac { { \mu }_{ L } }{ L } \) = \(\frac { e }{ 2m }\) = \(\frac{1.60 \times 10^{-19}}{2 \times 9.11 \times 10^{-31}}\) = 0. 0878 x 10¹²
\(\frac { { \mu }_{ L } }{ L } \) = 8.78 x 10¹⁰ C kg^-1 = content
The ration \(\frac { { \mu }_{ L } }{ L } \) is a constant and also known as gyro-magnetic ration \(\frac { { \mu }_{ L } }{ L } \). It must be noted that the gyro-magnetic ratio is a constant of proportionality which connects angular momentum of the electron and the magnetic moment of the electron. According to Neil’s Bohr quantization rule, the angular momentum of an electron moving in a stationary orbit is quantized, which means,
l = nh = n \(\frac { h }{ 2π }\)
where, h is the planck’s constant (h = 6.63 x 10^-34 J s) and number n takes natural numbers (i.e., n = 1,2,3,…..). Hence,

µ_L = n x 9.27 x 10^-24 A m²
The minimum magnetic moment can be obtained by subsituting by sustituting n = 1,
µ_L = n x 9.27 x 10^-24 A m² = 9.27 x 10^-24 J T^-1 = (µ_L)_min = µ_B
Where, µ_B = \(\frac { eh }{ 4πm }\) = 9.27 x 10^-24 A m²
is called Bohr magneton. This is a convenient unit with which one can measure atomic magnetic moments.

Question 7.
Apply Ampere’s Circuital Law to find the magnetic field both inside and outside of a toroidal solenoid.

Answer:
A solenoid is bent in such a way its ends are joined together to form a closed ring shape, is called a toroid. The magnetic field has constant magnitude inside the toroid whereas in the interior region (say, at point P) and exterior region (say, at point Q), the magnetic field is zero.

(a) Open space interior to the toroid:
Let us calculate the magnetic field B_p at point P. We construct an Amperian loop 1 of radius r₁ around the point P. For simplicity, we take circular loop so that the length of the loop is its circumference.

L₁ = 2π₁
Ampere’s circuital law for the loop 1 is

Scince, the loop 1 encloses no current, I_enclosed = 0

This is possible only if the magnetic field at point P vanishes i.e.
\(\vec { B } \)_p = 0

(b) Open space exterior to the toroid:
Let us calculate the magnetic field B_Q at point Q. We construct an Amperian loop 3 of radius r₃ around the point Q. The length of the loop is L₃= 2πr₃ Ampere’s circuital law for the loop 3 is

Since, in each turn of the toroid loop, current coming out of the plane of paper is cancelled by the current going into the plane of paper. Thus, I_enclosed = 0

This is possible only if the magnetic field at point Q vanishes i.e. \(\vec { B } \)_Q = 0

(c) Inside the toroid:
Let us calculate the magnetic field B_S at point S by constructing an Amperian loop 2 of radius r₂ around the point S. The length of the loop is L₂= 2πr₂ Ampere’s circuital law for the loop 2 is

Let I be the current passing through the toroid and N be the number of turns of the toroid, then I_enclosed = NI

The number of turns per unit length is n = \(\frac { n }{{ 2πr }_{2}}\) then the magnetic field at point S is B_S = μ₀nI

Question 8.
Obtain an expression for force on a moving charge in a magnetic field.
Answer:
When an electric charge q is moving with velocity \(\vec { v } \) in the magnetic field \(\vec { B } \), it experiences a force, called magnetic force \(\vec { F } \)_m. After careful experiments, Lorentz deduced the force experienced by a moving charge in the magnetic field \(\vec { F } \)_m

\(\vec { F } \)_m = q(\(\vec { v } \) x \(\vec { B } \)) ……. (1)
In magnitude, F_m=qvB sin θ ……. (2)
The equations (1) and (2) imply

1. \(\vec { F } \)_m is directly proportional to the magnetic field \(\vec { B } \)
2. \(\vec { F } \)_m is directly proportional to the velocity \(\vec { v } \)
3. \(\vec { F } \)_m is directly proportional to sine of the angle between the velocity and magnetic field
4. \(\vec { F } \)_m is directly proportional to the magnitude of the charge q
5. The direction of \(\vec { F } \)_m is always perpendicular to \(\vec { v } \) and g as \(\vec { F } \)_m is ti’e cross product of \(\vec { v } \) and \(\vec { B } \)
6. The direction of jprn is on negative charge is opposite to the direction of F charge provided other factors are identical.
7. If velocity v of the charge q is along magnetic field \(\vec { B } \) then, \(\vec { F } \)_m is zero.

Magnetism and Magnetic Effects of Electric Current Numerical Problems

Question 1.
The radius of the first orbit of hydrogen atom is 0.5 Å. The electron moves in an orbit with a uniform speed of 2.2 x 10⁶ ms^-1. What is the magnetic field produced at the centre of the nucleus due to the motion of this electron? Use µ₀ = 4π x 10^-7 Hm^-1 and electric charge = 1.6 x 10^-19C
Solution:
Here, r = 0.5 Å = 0.5 x 10^-10m
v = 2.2 x 10⁶ ms^-1
Period of revolution of electron

= 1.12 x 10^-3 A
Magnetic field produced at the centre of the nucleus,

B = 14. 07 T

Question 2.
A positive charge of 1.5 µC is moving with a speed of 2 x 10^-6 ms^-1 along the positive X-axis. A magnetic field, \(\overrightarrow { B } \) = (o.2 \(\hat{j}\) + 0.4 \(\hat{k}\)) tesla acts in space. Find the magnetic force acting on the charge.
Solution:
Here q = 1.5 µC = 1.5 x 10^-6C
\(\vec { v } \) = 2 x 10^-6 \(\hat{i}\) ms^-1 ; \(\overrightarrow { B } \) = (o.2 \(\hat{j}\) + 0.4 \(\hat{k}\)) T
Magnetic force on the positive charge is
\(\vec { F } \) = (\(\vec { v } \) x \(\overrightarrow { B } \))
= 1.5 x 10^-6 [2 x 10⁶\(\hat{i}\) x (o.2 \(\hat{j}\) + 0.4 \(\hat{k}\))]
3.0 [o.2 \(\hat{j}\) x \(\hat{j}\) + 0.4 \(\hat{i}\) x \(\hat{k}\)]
= (0.6 \(\hat{k}\) – 1.2 \(\hat{j}\)) N
[∴ i x \(\hat{j}\) = \(\hat{k}\), \(\hat{i}\) x \(\hat{k}\) = –\(\hat{j}\)]

Question 3.
Copper has 8.0 x 10²⁸ electrons per cubic metre carrying a current and lying at right angle to a magnetic field of strength 5 x 10^-3 T. experiences a force of 8.0 x 10^-2 N. Calculate the drift velocity of free electrons in the wire.
Solution:
n = 8 x 10²⁸ m^-3 ; l = lm
A= 8 x 10²⁸ m^-2 ; e = 1.6 x 10^-9C
Total charge contained in the wire
q = volume of wire x ne = Alne = 8 x 10^-6 x 1 x 8 x10²⁸ x 1.6 x 10^-19C
= 102.4 x 10³ C
If v_d is the drift speed of electrons, then
F = qv_d B sin 90° = qv_d B
∴ v_d = \(\frac { F }{ qB }\) = \(\frac{8.0 \times 10^{-2}}{102.4 \times 10^{3} \times 5 \times 10^{-3}}\) ms^-1
v_d = 1.56 x 10^-4 ms^-1

Question 4.
An electron is moving at 10⁶ ms^-1 in a direction parallel to a current of 5A flowing through an infinitely long straight wire, separated by a perpendicular distance of 10cm in air. Calculate the magnitude of the force experienced by the electron.
Solution:
Magnetic field of the straight wire carrying a current of 2 A, at a distance of 10cm or 0. lm from it is
B = \(\frac {{ µ }_{0} I}{ 2πr }\) = \(\frac{4 \pi \times 10^{-7} \times 5}{2 \pi \times 0.1}\) = 10^-5T
This field acts perpendicular to the direction of the electron. So magnetic force on the electron is
F = qv B sin 90° = 1.6 x 10^-19 x 10⁶ 10^-5 x 1
F = 1.6 x 10^-18 N

Common Errors and Its Rectifications

Common Errors:

1. Students may not know about arrow mark in the figures, (diagram)
2. Students wrongly mention the unit of magnetic induction. Eg. 0.5 x 10^-5 Tesla. Most of the children makes the mistakes in this area.
3. Student may confuse the direction of current versus produced magnetic induction.

Rectifications:

1. Arrow mark is importance to show where the electric and magnetic fields are directed.
2. The correct unit of magnetic induction of T (or) tesla.
3. If the current moves in upward the magnetic induction is anti-clock. If the current in downward, the magnetic induction is clockwise.

We believe that the shared knowledge about Tamilnadu State Board Solutions for 12th Physics Chapter 3 Magnetism and Magnetic Effects of Electric Current Questions and Answers learning resource will definitely guide you at the time of preparation. For more details visit Tamilnadu State Board Solutions for 12th Physics Solutions and for more information ask us in the below comments and we’ll revert back to you asap.

Students can Download Tamil Chapter 2.1 இன்பத்தமிழ்க் கல்வி Questions and Answers, Summary, Notes Pdf, Samacheer Kalvi 7th Tamil 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 7th Tamil Solutions Term 2 Chapter 2.1 இன்பத்தமிழ்க் கல்வி

மதிப்பீடு

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
பெண்க ளுக்கு நிகராகப் பாரதிதாசன் கூறுவது ……
அ) மயில்
ஆ) குயில்
இ) கிளி
ஈ) அன்னம்
Answer:
அ) மயில்

Question 2.
பின்வருவனவற்றுள் மலை’ யைக் குறிக்கும் சொல்
அ) வெற்பு
ஆ) காடு
இ) கழனி
ஈ) புவி
Answer:
அ) வெற்பு

Question 3.
‘ஏடெடுத்தேன்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ….
அ) ஏடெடு + தேன்
ஆ) ஏட்டு + எடுத்தேன்
இ) ஏடு + எடுத்தேன்
ஈ) ஏ + டெடுத்தேன்
Answer:
இ) ஏடு + எடுத்தேன்

Question 4.
‘துயின்றிருந்தார்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ….
அ) துயின்று + இருந்தார்
ஆ) துயில் + இருந்தார்
இ) துயின்றி + இருந்தார்
ஈ) துயின் + இருந்தார்
Answer:
அ) துயின்று + இருந்தார்

Question 5.
என்று + உரைக்கும் என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் …..
அ) என்று உரைக்கும்
ஆ) என்றிரைக்கும்
இ) என்றரைக்கும்
ஈ) என்றுரைக்கும்
Answer:
ஈ) என்றுரைக்கும்

பொருத்துக

1. கழனி – கதிரவன்
2. நிகர் – மேகம்
3. பரிதி – சமம்
4. முகில் – வயல்
Answer:

1. கழனி – வயல்
2. நிகர் – சமம்
3, பரிதி – கதிரவன்
4. முகில் – மேகம்

குறுவினா

Question 1.
பாரதிதாசனின் மனத்தைக் கவர முயன்ற இயற்கைப் பொருள்கள் யாவை?
Answer:
வானம், நீரோடை, தாமரை மலர்கள், காடு, வயல், மேகம் முதலியன பாரதிதாசனின் மனத்தைக் கவர முயன்ற இயற்கைப் பொருள்களாகும்.

Question 2.
தமிழ்மொழிக் கல்வி பயில்வதால் உண்டாகும் நன்மைகள் எவையெனப் பாரதிதாசன் குறிப்பிடுகிறார்?
Answer:
தமிழ்மொழிக் கல்வி பயில்வதால் உண்டாகும் நன்மைகள் :
வாழ்வில் துன்பங்கள் நீங்கும்.
நெஞ்சில் தூய்மை உண்டாகும்.
வீரம் வரும்.

சிறுவினா

‘இன்பத்தமிழ்க் கல்வி’ – பாடலின் மையக்கருத்தை உங்கள் சொந்த நடையில் எழுதுக.
Answer:

• வானமும், நீரோடையும், தாமரை மலர்களும் எங்களைக் கவிதையாக இயற்றுக என்றன.
• காடும், வயலும், மேகமும் கண்ணில் புகுந்து கவிதையாய் மாற முயற்சித்தன.
• ஆடும் மயிலுக்கு நிகரான பெண்கள் அன்பினைக் களிச்சித்திரம் ஆக்கக் கூறினர்.
• குளிர் தென்றலும், ஆடும் மயிலும், அன்னமும், சுடர்விடும் கதிரோனும், வேல் ஏந்திய வீரர்களும் தங்கள் அழகினை எழுதுங்கள் என்றனர்.
• இதற்கிடையில் அறியாமையால் துன்பப்படும் தமிழ்நாட்டு மக்களின் காட்சி என்னுள் இரக்கத்தை உண்டாக்கி என் உயிரில் கலந்தது.
• இத்துன்பம் நீங்கிட அனைவரும் இன்பத் தமிழ்க் கற்றவர் என்ற நிலை வர வேண்டும். அந்நிலையிலேயே துன்பம் நீங்கி, நெஞ்சில் தூய்மை உண்டாகி வீரம் வந்து சேரும் என்று இன்பத் தமிழ்க்கல்வி வழியே பாரதிதாசன் கூறுகிறார்.

சிந்தனை வினா

தமிழ்மொழிக் கல்வி பயில்வதால் ஏற்படும் நன்மைகளாக நீங்கள் கருதுவனவற்றைத் தொகுத்து எழுதுக.
Answer:
தமிழ்மொழிக் கல்வி பயில்வதால் ஏற்படும் நன்மைகள் :

• தமிழ்மொழியில் கல்வி பயிலும் குழந்தைகள் மிகவும் ஆழமாகக் கல்வி கற்பார்கள்.
• மாணவர்களால் தாங்கள் கற்பதைப் பேசுவதுடன் ஒப்பிட்டுப் பார்க்க முடியும்.
• சமூகத்துடன் ஒப்பிட்டுப் பார்ப்பதனால் எளிமையாகக் கற்க இயலும்.
• கற்கும் திறனைப் பெறுவார்கள்.
• சுய சிந்தனை வளரும்.
• எளிமையாக உரையாட முடியும்.
• கலந்துரையாடல், விவாதித்தல், பேச்சுப் போட்டி, கட்டுரைப்போட்டி இவற்றை எளிதாக எதிர்கொள்வார்கள்.
• தன்னம்பிக்கை வளர்கிறது.

கற்பவை கற்றபின்

Question 1.
இயற்கைக்காட்சி குறித்து நான்கு வரிகளில் கவிதை எழுதுக,
Answer:
இயற்கை :
பூக்களின் வலி மலர்வதால் மறைகிறது
மூங்கிலின் வலி இன்னிசையால் மறைகிறது
மேகத்தின் வலி மழைத்துளியால் மறைகிறது
மனிதனே உன்வலி இயற்கையை நீ போற்றுவதால் மறைந்து விடும் மறவாதே

Question 2.
‘தாய்மொழி வழிக் கல்வியே சிறந்தது’ என்பது குறித்து வகுப்பில் கலந்துரையாடுக.
Answer:
மாணவன் – 1 ; தாய்மொழி வழிக்கல்வியே சிறந்தது’ என்பது என் கருத்து.
முதலில் நாம் ஒன்றைத் தெளிவுபடுத்திக் கொள்ள வேண்டும். என்னவெனில் தாய்மொழியைக் கற்பது, தாய்மொழி வழி கற்பது ஆகிய இவ்விரண்டிற்குமிடையேயுள்ள வேறுபாட்டை அறியாமலேயே  வாதிட்டுக் கொண்டுள்ளோம்.
மாணவன் -2 : என்ன வேறுபாடு? தெளிவாகக் கூறு.
மாணவன் -1 : தாய்மொழியில் உள்ள இலக்கண இலக்கியங்களைக் கற்றுப் புலமை
பெறுவது தாய்மொழியைக் கற்றலாகும். தாய்மொழியின் வழியாய் அறிவியல், வரலாறு, புவியியல், கணிதம் ஆகிய பாடங்களைக் கற்றுத்  தெளிவது தாய்மொழி வழிக் கற்றலாகும்.

மாணவன் – 2 : பிறமொழியில் கற்பதனால் என்ன இழப்பு எனக் கூறு?
மாணவன் -1 : தாய்மொழியின் வாயிலாகக் கற்பவன்தான் சுயமாகச் சிந்திக்கிறான். சிந்தனை ஆற்றலைப் பெறமுடியும். எண்ணியதை எண்ணியவாறு வெளிப்படுத்த முடியும். :
மாணவன் -2 : பிறமொழியைக் கற்றவனால் அம்மொழியில் எழுதப்படிக்கத் தெரிந்தபோது, சிந்திக்க முடியாதா?
மாணவன் -1 : கண்டிப்பாக முடியாது. எடுத்துக்காட்டாக “உனக்குப் பிடித்த ஒன்றைப்
பற்றிக் கூறு என்றால் அவன் முதலில் தன் தாய்மொழியில் சிந்தித்து விட்டுப் பிறகுதான் அவன் அறிந்த வேறுமொழியில் பதிலளிப்பான் – தாய்மொழி என்றால் உடனே பதிலளிப்பான். கல்விச் சிந்தனையாளர்களும் இதனையே வலியுறுத்துகின்றனர்.

மாணவன் -2 : அப்படியென்றால்.
மாணவன் -1 : என்ன அப்படியென்றால் என்று இழுக்கிறாய்? உண்மையான கல்வியறிவைப் பெற வேண்டுமெனில் தாய்மொழி வழிக் கல்விதான் அவசியம். அவர்களால் மட்டுமே சிறந்த சிந்தனையாளர்களாகவும், படைப்பாளர்களாகவும் கண்டுபிடிப்பாளர்களாகவும் உருவாக முடியும்.

மாணவன் -2 : நீ சொல்வதும் சரிதான். மேனாள் குடியரசுத் தலைவர் அப்துல்கலாம்,
சர்.சி.வி. இராமன் , கணிதமேதை இராமானுஜம் போன்றோர்களும் அவர்கள் தாய்மொழி வழியில் கற்றதனால்தான் பல சாதனைகளை நிகழ்த்த முடிந்தது. இரவீந்தரநாத தாகூர் அவருடைய தாய்மொழியான வங்க மொழியில்தான் கீதாஞ்சலியை எழுதினார். மகாத்மாகாந்தியடிகள் அவருடைய சுயசரிதையைத் தம் தாய்மொழியில்தான் எழுதினார்.
மாணவன் -1 : சரியாகக் கூறினாய். நான் கூற வந்த செய்தியும் இதுதான். நீயே
கூறி விட்டாய். நாமும் நம் தாய்மொழி வழியிலே கல்வி கற்போம். சிறப்புறுவோம்! நன்றி!

கூடுதல் வினாக்கள்

சொல்லும் பொருளும் :

1. எத்தனிக்கும் – முயலும்
2. வெற்பு – மலை
3. கழனி – வயல்
4. நிகர் – சமம்
5. பரிதி – கதிரவன்
6. அன்னதோர் – அப்படி ஒரு
7. கார்முகில் – மழைமேகம்
8. துயின்றிருந்தார் – உறங்கியிருந்தார்

நிரப்புக.

Question 1.
பரிதி என்பதன் பொருள்
Answer:
கதிரவன்

Question 2.
வெற்பு என்பதன் பொருள் ……….
Answer:
மலை

Question 3.
பாரதிதாசன் எழுதிய …………… என்னும் நாடக நூலுக்குச் சாகித்திய அகாடமி விருது அளிக்கப்பட்டது.
Answer:
பிசிராந்தையார்

விடையளி :

Question 1.
பாரதிதாசன் எழுதிய நூல்கள் யாவை?
Answer:

• பாண்டியன் பரிசு
• அழகின் சிரிப்பு
• இசையமுது
• குடும்ப விளக்கு
• கண்ண கி புரட்சிக் காப்பியம்
• இருண்ட வீடு

பாடலின் பொருள்

கவிதை எழுத ஏடு ஒன்று எடுத்தேன். என்னைக் கவிதையாக எழுதுக என்று வானம் கூறியது. நீரோடையும் தாமரை மலர்களும் எங்களைக் கவி ஓவியமாகத் தீட்டுக என்றன. காடும் வயல்களும் கருநிற மேகங்களும் என் கண்களைக் கவர்ந்து, கவிதையில் இடம் பெற முயன்றன. ஆடும் மயில் போன்ற பெண்கள் அன்பினைக் கவிதையாக எழுதுக என்றனர்.

சோலையின் குளிர்ந்த தென்றல் வந்தது. பசுமையான தோகையையுடைய மயில் வந்தது. அன்னம் வந்தது. மாணிக்கம் போல் ஒளி வீசி மாலையில் மேற்குத் திசையில் மறைகின்ற கதிரவனும் வந்தான். வேல் ஏந்திய வீரர்கள், மலை போன்ற எங்களது தோள்களின் அழகினை எழுதுங்கள் என்றனர். இவ்வாறு அழகிய காட்சிகள் எல்லாம் பெருந்திரளாக வந்து தங்களைக் கவிதையாக எழுதுமாறு கூறின.

ஆனால் துன்பத்தில் கிடக்கும் என் தமிழ்நாட்டு மக்கள் அறியாமையில் தூங்கிக் கிடக்கிறார்கள். அந்தக் காட்சி என் மனத்தில் இரக்கத்தை உண்டாக்கி என் உயிரில் வந்து கலந்து விட்டது. இத்துன்பம் நீங்க அனைவரும் இன்பத்தமிழ்க் கல்வியைக் கற்றவர்கள் என்னும் நிலை ஏற்பட வேண்டும். அந்நிலை ஏற்பட்டால் வாழ்வில் துன்பங்கள் நீங்கிடும். நெஞ்சில் தூய்மை உண்டாகிவிடும். வீரம் வரும்.

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 3 Algebra Ex 3.7

10th Maths Exercise 3.7 Samacheer Kalvi Question 1.
Find the square root of the following rational expressions.

Solution:

Ex 3.7 Class 10 Samacheer Question 2.
Find the square root of the following
(i) 4x² + 20x + 25
(ii) 9x² – 24xy + 30xz – 40yz + 25z² + 16y²
(iii) 1 + \(\frac{1}{x^{6}}+\frac{2}{x^{3}}\)
(iv) (4x² – 9x + 2) (7x² – 13x – 2) (28x² – 3x – 1)
(v) ( 2x² + \(\frac{17}{6} x\) + 1) (\(\frac{3}{2} x^{2}\) + 4x + 2)(\(\frac{4}{3} x^{2}+\frac{11}{3} x\) + 2)
Solution:

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 2 Numbers and Sequences Ex 2.9

10th Maths Exercise 2.9 Samacheer Kalvi Question 1.
Find the sum of the following series
(i) 1 + 2 + 3 + … + 60
(ii) 3 + 6 + 9 + … + 96
(iii) 51 + 52 + 53 + … + 92
(iv) 1 + 4 + 9 + 16 + … + 225
(v) 6² + 7² + 8² + … + 21²
(vi) 10³ + 11³ + 12³ + … + 20³
(vii) 1 + 3 + 5 + … + 71
Solution:
(i) 1 + 2 + 3 + ……… + 60

= 4278 – 1275 = 3003
(iv) 1 + 4 + 9 + 16 + … + 225
= 1² + 2² + 3² + 4² + ……… + 15²
\(\sum_{1}^{n} n^{2}=\frac{n(n+1)(2 n+1)}{6}\)

Ex 2.9 Class 10 Samacheer Question 2.
If 1 + 2 + 3 + … + k = 325, then find 1³ + 2³ + 3³ + …………. + K³.
Solution:
1 + 2 + 3 + … + K = 325

If 1 + 2 + 3 … + k = 325
1³ + 2³ + 3³ + … + K³ = (325)² = 105625

10th Maths Exercise 2.9 Answers Question 3.
If 1³ + 2³ + 3³ + … + K³ = 44100 then find 1 + 2 + 3 + … + k.
Solution:
If 1³ + 2³ + 3³ + … + K³ = 44100
1 + 2 + 3 + … + K = \(\sqrt { 44100 }\)
= 210

Exercise 2.9 Class 10 Maths Question 4.
How many terms of the series 1³ + 2³ + 3³ + … should be taken to get the sum 14400?
Solution:
1³ + 2³ + 3³ + ……… + n³ = 14400
\(\left(\frac{n(n+1)}{2}\right)^{2}\) = 14400 = (120)²
\(\frac{n(n+1)}{2}\) = \(\sqrt { 14400 }\) = 120
n(n + 1) = 240
Method 1:
n² + n – 240 = 0
n² + 16n – 15n – 240 = 0
n(n + 16) – 15(n + 16) = 0
(n + 16)(n – 15) = 0
n = -16, 15
∴ 15 terms to be taken to get the sum 14400.
Method 2:
n² + n – 240 = 0

Exercise 2.9 Class 10 Maths Samacheer Question 5.
The sum of the squares of the first n natural numbers is 285, while the sum of their cubes is 2025. Find the value of n.
Solution:
1² + 2² + 3² + …… + n² = 285
1³ + 2³ + 3³ + …… + n³ = 2025

10th Maths Exercise 2.9 Question 6.
Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm,…, 24 cm. How much area can be decorated with these colour papers?
Solution:
10² + 11² + 12² + … + 24²
= (1² + 2² + … + 24²) – (1² + 2² + … + 9²)

∴ Rekha has 4615 cm² colour papers. She can decorate 4615 cm² area with these colour papers.

Ex 2.9 Class 10 Question 7.
Find the sum of the series (2³ – 1) + (4³ – 3³) + (6³ – 15³) +… to (i) n terms (ii) 8 terms.
Solution:
(2³ – 1) + (4³ – 3³) + (6³ – 15³) + ……… n

= 4n³ + 3n² = sum of ‘n’ terms.
When n = 8
Sum = 4 × 8³ + 3 × 8²
= 2048 + 192 = 2240

You can Download Samacheer Kalvi 9th 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 9th Maths Solutions Chapter 1 Set Language Ex 1.5

Exercise 1.5 Class 9 Maths Samacheer Question 1.
Using the adjacent venn diagram, find the following sets:
(i) A – B
(ii) B – C
(iii) A’ ∪ B’
(vi) A’ ∩ B’
(v) (B ∪ C)’
(vi) A – (B ∪ C)
(vii) A – (B ∩ C)

Solution:
(i) A – B = {3, 4, 6}
(ii) B – C = {-1, 5, 7}
(iii) A’ ∪ B’
A’ = {1, 2, 0, -3, 5, 7, 8}
B’ = {-3, 0, 1, 2, 3, 4, 6)
A’ ∪ B’ = {-3, 0, 1, 2, 3, 4, 5, 6, 7, 8)

(iv) A’ ∩ B’
A’ ∩ B’ = {-3, 0, 1, 2}

(v) B ∪ C = {-3, -2, -1, 0, 3, 5, 7, 8}
(B ∪ C)’ = U – (B ∪ C)
= {-3, -2, -1, 0, 1,2, 3, 4, 5, 6, 7, 8} – {-3, -2, -1, 0, 3, 5, 7, 8}
(B ∪ C)’ = {1, 2, 4, 6}

(vi) A – (B ∪ C) = {-2, -1, 3, 4,6} – {-3, -2, -1, 0, 3, 5, 7, 8} = {4, 6}
A – (B ∩ C)
B ∩ C = {-2, 8}
A- (B ∩ C) = {-2, -1, 3, 4, 6} – {-2, 8} = {-1, 3, 4, 6}

9th Maths Exercise 1.5 Samacheer Kalvi Question 2.
If K = {a, b, d, e,f}, L = {b, c, d, g} and M {a, b, c, d, h} then find the following:
(i) K ∪ (L ∩ M)
(ii) K ∩ (L ∪ M)
(iii) (K ∪ L) ∩ (K ∪ M)
(iv) (K ∩ L) ∪ (K ∩ M) and verify distributive laws.
Solution:
K = {a, b, d, e, f}, L = {b, c, d, g} and M {a, b, c, d, h}
(i) K ∪ (L ∩ M)
L ∩ M = {b, c, d, g} ∩ {a, b, c, d, h} = {b, c, d}
K ∪ (L ∩ M) = {a, b, d, e, f } ∪ {b, c, d) = {a, b, c, d, e, f}

(ii) K ∩(L ∪ M)
L ∪ M = {a, b, c, d, g, h}
K ∩ (L ∪ M) = {a, b, d, e, f} ∩ {a, b, c, d, g, h} = {a, b, d}

(iii) (K ∪ L) ∩ (K ∪ M)
K ∪ L = {a, b, c, d, e, f, g}
K ∪ M = {a, b, c, d, e, f, h}
(K ∪ L) ∩ (K ∪ M) = {a, b, c, d, e,f}

(iv) (K ∩ L) ∪ (K ∩ M)
(K ∩ L) = {b, d)
(K ∩ M) = {a,b,d}
(K ∩ L) ∪ (K ∩ M) = {b, d} ∪ {a, b, d} = {a, b, d}

Distributive laws
K ∪ (L ∩ M) = (K ∪ L) ∩ (K ∪ M)
{a, b, c, d, e, f) = {a, b, c, d, e, f, g} ∩ {a, b, c, d, e, f, h}
= {a, b, c, d, e, f}
Thus Verified.
K ∩ (L ∪ M) = (K ∩ L) ∪ (K ∩ M)
{a, b, d} = {a, b, c, d, e, f, g} ∪ {a, b, c, d, e, f, h}
= {a, b, d}
Thus Verified.

9th Maths Set Language Exercise 1.5 Question 3.
If A = {x : x ∈ Z, -2 < x ≤ 4}, B = {x : x ∈ W, x ≤ 5}, C ={-4, -1, 0, 2, 3, 4}, then verify A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
Solution:
A = {x : x ∈ Z, -2 < x ≤ 4} = {-1, 0, 1, 2, 3, 4}
B = {x : x ∈ W, x ≤ 5} = {0, 1, 2, 3, 4, 5}
C = {-4, -1, 0, 2, 3, 4}
A ∪ (B ∩ C)
B ∩ C = {0, 1, 2, 3, 4, 5} ∩ {-4, -1, 0, 2, 3, 4} = {0, 2, 3, 4}
A ∪ (B ∩ C) = {-1, 0, 1, 2, 3, 4} ∪ (0, 2, 3, 4} ={-1, 0, 1, 2, 3, 4} …………. (1)
(A ∪ B) ∩ (A ∪ C)
A ∩ B = {0, 1, 2, 3, 4}
A ∩ C = {-1, 0, 2, 3, 4}
(A ∩ B) ∪ (A ∩ C) = {0, 1, 2, 3, 4} ∪ {-1, 0, 2, 3, 4}= {-1, 0, 1, 2, 3, 4} …………. (2)
From (1) and (2), it is verified that
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

9th Maths Exercise 1.5 Question 4.
Verify A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) using Venn diagrams.
Solution:
L.H.S A ∪ (B ∩ C)

From (2) and (5), it is verified that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Class 9 Maths Chapter 1 Samacheer Kalvi Question 5.
If A = {b, c, e, g, h}, B = {a, c, d, g, i} and C = {a, d, e, g, h}, then show that A – (B ∩ C) = (A – B) ∪ (A – C).
Solution:
A = {b, c, e, g, h}
B = {a, c, d, g, i}
C = {a, d, e, g, h}
B ∩ C = {a, d, g}
A – (B ∩ C) = {b, c, e, g, h} – {a, d, g} = {b, c, e, h} ……..… (1)
A- B = {b, c, e, g, h} – {a, c, d, g, i} = {b, e, h}
A – C = {b, c, e, g, h} – {a, d, e, g, h} = {b, c}
(A – B) ∪ (A – C) = {b, c, e, h} ………..… (2)
From (1) and (2) it is verified that
A – (B ∩ C) = (A – B) ∪ (A – C)

Samacheerkalvi.Guru 9th Maths Question 6.
If A = {x : x = 6 n ∈ W and n < 6}, B = {x : x = 2n, n ∈ N and 2 < n ≤ 9} and C = {x : x = 3n, n ∈ N and 4 ≤ n < 10}, then show that A – (B ∩ C) = (A – B) ∪ (A – C)
Solution:
A = {x : x = 6n, n ∈ W, n < 6}
x = 6n
n = {0, 1, 2, 3, 4, 5}
⇒ x = 6 × 0 = 0
x = 6 × 1= 6
x = 6 × 2 = 12
x = 6 × 3 = 18
x = 6 × 4 = 24
x = 6 × 5 = 30
∴ A = {0, 6, 12, 18, 24, 30}

B = { x : x = 2n, n ∈ N, 2 < n ≤ 9}
n = {3, 4, 5, 6, 7, 8, 9}
x = 2 n
⇒ x = 2 × 3 = 6
2 × 4 = 8
2 × 5 = 10
2 × 6 = 12
2 × 7 = 14
2 × 8 = 16
2 × 9 = 18
∴ B {6, 8, 10, 12, 14, 16, 18}

C = { x : x = 3n, n ∈ N, 4 ≤ n < 10}
N = { 4, 5, 6, 7, 8, 9}
x = 3 × 4 = 12
⇒ x = 3 × 5 = 15
x = 3 × 6 = 18
x = 3 × 7 = 21
x = 3 × 8 = 24
x = 3 × 9 = 27
x = 2 × 9 = 18
∴ C = {12, 15, 18, 21, 24, 27}

A – (B ∩ C) = (A – B) ∪ (A – C)
L.H.S R.H.S
B ∩ C = {12,18}
A – (B ∩ C) = {0, 6, 12, 18, 24, 30} – {12, 18} = {0, 6, 24, 30} ……….…. (1)
(A – B) = {0, 24, 30}
(A – C) = {0, 6, 30}
(A – B) ∪ (A – C) = {0, 6, 24, 30} …………. (2)
From (1) and (2), it is verified that
A – (B ∩ C) = (A – B) ∪ (A – C).

Samacheer Kalvi Guru 9th Maths Question 7.
If A = {-2, 0, 1, 3, 5}, B = {-1, 0, 2, 5, 6} and C = {-1, 2, 5, 6, 7}, then show that A – (B ∪ C) = (A – B) ∩ (A – C).
Solution:
A = {-2, 0, 1, 3, 5},
B = {-1, 0, 2, 5, 6}
C ={-1, 2, 5, 6, 7}
B ∪ C = {-1, 0, 2, 5, 6, 7}
A – (B ∪ C) = {-2, 1, 3} …………. (1)
(A – B) = {-2, 1, 3}
(A – C) = {-2, 0, 1, 3}
(A – B) ∩ (A – C) = {-2, 1, 3} ………..… (2)
From (1) and (2), it is verified that . A – (B ∪ C) = (A – B) ∩ (A – C)

9th Standard Maths Exercise 1.5 Question 8.
if A={y: y = \(\frac{a+1}{2}\), a ∈ W and a ≤ 5},B = {y: y=\(\frac{2 n-1}{2}\),n ∈ W and n < 5} and C={−1,\(-\frac{1}{2}\), 1, \(\frac{3}{2}\), 2} then show that A – (B ∪ C) = (A – B) ∩ (A – C).
Solution:


(A – B) ∩ (A – C) = {3} …………. (2)
From (1) and (2), it is verified that A – (B ∪ C) = (A – B) ∩ (A – C).

9th Maths Exercise 1.5 In Tamil Question 9.
Verify A – (B ∩ C) = (A – B) ∪ (A – C) using Venn diagrams.
Solution:

∴ A – (B ∩ C) = (A – B) ∪ (A – C)
Hence it is proved.

Kalvi Guru 9th Maths Question 10.
If U = {4, 7, 8, 10, 11, 12, 15, 16}, A = {7, 8, 11, 12} and B = {4, 8, 12, 15}, then verify De Morgan’s Laws for complementation.
U = {4, 7, 8, 10, 11, 12, 15, 16}
A = {7, 8, 11, 12}, B = {4, 8, 12, 15}
De Morgan’s Laws for complementation.
(A ∪ B)’ = A’ ∩ B’
A ∪ B = {4, 7, 8, 11, 12, 15}
(A ∪ B)’ = {4, 7, 8, 10, 11, 12, 15, 16} – {4, 7, 8, 11, 12, 15}
= {10, 16} ……………. (1)
A’ = {4, 10, 15, 16}
B’ = {7, 10, 11, 16}
A’ ∩ B’ = {10, 16} ………………(2)
From (1) and (2) it is verified that (A ∪ B)’ = A’ ∩ B’.

9th Class Math Exercise 1.5 Solution Question 11.
Verify (A ∩ B)’ = A’ ∪ B’ using Venn diagrams.
Solution:
(A ∩ B)’ = A’ ∪ B’

(2) = (5)
∴ (A ∩ B)’ = A’ ∪ B’

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: