Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.2

Question 1.
Solve the following . equations by using Cramer’s rule
(i) 2x + 3y = 7; 3x + 5y = 9
(ii) 5x + 3v = 17; 3x + 7y = 31
(iii) 2x + y – z = 3; x + y + z – 1; x – 2y – 3z = 4
(iv) x + y + z = 6; 2x + 3y – z = 56; x – 2y – 3z = -7
Solution:
The equations are
2x + 3y = 7
3x + 5y = 9
Here Δ = \(\left|\begin{array}{ll}
2 & 3 \\
3 & 5
\end{array}\right|\) = 10 – 9 = 1
≠ 0
∴ We can apply Cramer’s Rule

∴ x = 8; y = -3

(ii) The equations are
5x + 3y = 17
3x + 7y = 31
Here Δ = \(\left|\begin{array}{ll}
5 & 3 \\
3 & 7
\end{array}\right|\) = 35 – 9 = 26
≠ 0
We can apply Cramer’s Rule

∴ x = 1; y = 4

(iii) The equations are
2x + y – z = 3
x + y + z = 1
x – 2y – 3z = 4
Here Δ = \(\left|\begin{array}{ccc}
2 & 1 & -1 \\
1 & 1 & 1 \\
1 & -2 & -3
\end{array}\right|\)
= 2 (-3 + 2) – 1 (-3 – 1) – 1 (-2 – 1)
= 2 (-1) -1 (-4) -1 (-3)
= -2 + 4 + 3
Δ = 5 ≠ 0
∴ We can apply Cramer’s Rule and the system is consistant and it has unique solution.
Δ_x = \(\left|\begin{array}{ccc}
3 & 1 & -1 \\
1 & 1 & 1 \\
4 & -2 & -3
\end{array}\right|\)
= 3 (-3 + 2) – 1 (-3 – 4) – 1 (-2 – 4)
= 3 (-1) -1 (-7) -1 ( 6)
= -3 + 7 + 6 = 10
Δ_y = \(\left|\begin{array}{ccc}
2 & 3 & -1 \\
1 & 1 & 1 \\
1 & 4 & -3
\end{array}\right|\)
= 2 (- 3 – 4) -3 (-3 – 1) -1 (4 – 1)
= 2 (-7) – 3 (-4) – 1 (3)
= -14 + 12 – 3
= -5
Δ_z = \(\left|\begin{array}{ccc}
2 & 1 & 3 \\
1 & 1 & 1 \\
1 & -2 & 4
\end{array}\right|\)
= 2 (4 + 2) – 1 (4 – 1) + 3 (- 2 – 1)
= 2 (6) – 1(3) + 3 (-3)
= 12 – 3 – 9 = 0
∴ By Cramer’s rule

∴ (x, y, z) = (2, -1, 0)

(iv) The equations are
x + y + z = 6
2x + 3y – z = 5
6x – 2y – 3z = -7
Here Δ = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 3 & -1 \\
6 & -2 & -3
\end{array}\right|\)
= 1 (-9 – 2) -1 (-6 + 6) + 1 (-4 – 18)
= 1 (-11) – 1 (0) +1 (-22)
= -11 – 22
= -33 ≠ 0
∴ We can apply Cramer’s Rule and the system is consistent and it has unique solution.
Now,
Δ_x = \(\left|\begin{array}{ccc}
6 & 1 & 1 \\
5 & 3 & -1 \\
-7 & -2 & -3
\end{array}\right|\)
= 6 (- 9 – 2) – 1 (- 15 – 7) + 1 (- 10 + 21)
= 6 (-11) – 1 (-22)+ 1 (11)
= -66 + 22 + 11
= -33
Δ_y = \(\left|\begin{array}{ccc}
1 & 6 & 1 \\
2 & 5 & -1 \\
6 & -7 & -3
\end{array}\right|\)
= 1 (-15 – 7) – 6 (- 6 + 6) + 1 (- 14 – 30)
= 1 (-22) – 6 (0) + 1 (-44)
= -66
Δ_z = \(\left|\begin{array}{ccc}
1 & 1 & 6 \\
2 & 3 & 5 \\
6 & -2 & -7
\end{array}\right|\)
= 1 (- 21 + 10) – 1 (- 14 – 30) + 6 (- 4 – 18)
= 1 (-11) – 1 (-44) + 6 (-22)
= -11 + 44 – 132
= -99
∴ By Cramer’s rule

(x, y, z) = (1, 2, 3)

(v) The equations are
x + 4y + 3z = 2
2x – 6y + 6z = -3
5x – 2y + 3z = -5
Here Δ = \(\left|\begin{array}{ccc}
1 & 4 & 3 \\
2 & -6 & 6 \\
5 & -2 & 3
\end{array}\right|\)
= 1 (-18 + 12) -4 (6 -30) + 3 (-4 + 30)
= 1 (-6) – 4 (-24) + 3 (26)
= -6 + 96 + 78
= 168 ≠ 0
We can apply Cramer’s Rule and the system is consistent and it has unique solution.
Δ_x = \(\left|\begin{array}{ccc}
2 & 4 & 3 \\
-3 & -6 & 6 \\
-5 & -2 & 3
\end{array}\right|\)
= 2 (-18 + 12) – 4 (-9 + 30) + 3 (6 – 30)
= 2 (-6) -4 (21) + 3 (-24)
= -12 – 84 – 72
= -168
Δ_y = \(\left|\begin{array}{ccc}
1 & 2 & 3 \\
2 & -3 & 6 \\
5 & -5 & 3
\end{array}\right|\)
= 1 (-9 + 30) – 2 (6 – 30) + 3 (-10 + 15)
= 1(21) – 2 (-24) + 3 (5)
= 21 + 48 + 15
= 84
Δ_z = \(\left|\begin{array}{ccc}
1 & 4 & 2 \\
2 & -6 & -3 \\
5 & -2 & -5
\end{array}\right|\)
= 1 (30 – 6) – 4 (-10 +15) + 2 (-4 + 30)
= 1 (24) – 4 (5) + 2 (26)
= 24 – 20 + 52
= 56
∴ By Cramer’s rule

(x, y, z) = (-1, \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 3 }\))

Question 2.
A commodity was produced by using 3 units of labour and 2 units of capital, the total cost is Rs 62. If the commodity had been produced by using 4 units of labour and one unit of capital, the cost is Rs 56. What is the cost per unit of labour and capital? (Use determinant method).
Solution:
Let ‘x’ be the cost per unit of labour
Let ‘y’ be the capital
∴ 3x + 2y = 62
4x + y = 56

∴ The cost per unit of labour is Rs 10 and Cost per unit of capital is Rs 16

Question 3.
A total of Rs 8,600 was invested in two accounts. One account earned 4\(\frac { 3 }{ 4 }\)% annual interest and the other earned 6\(\frac { 1 }{ 2 }\)% annual interest. If the total interest for one year was Rs 431.25, how much was invested in each accout? (Use determinant method)
Solution:
Let the two investments be x and y
Given that
x + y = 8600 ……. (1)

Multiply by 4 (both sides)
19x + 26y = 172500 ……….. (2)
Here Δ = \(\left|\begin{array}{ll}
1 & 1 \\
19 & 26
\end{array}\right|\) = 26 – 19 = 7 ≠ 0
∴ We can apply Cramer’s Rule
Δ_x = \(\left|\begin{array}{ll}
8600 & 1 \\
172500 & 26
\end{array}\right|\)
= 223600 – 172500
= 51100
Δ_y = \(\left|\begin{array}{ll}
1 & 8600 \\
19 & 172500
\end{array}\right|\)
= 172500 – 163400
= 9100
By Cramer’s Rule
x = \(\frac { Δ_x }{ Δ }\) = \(\frac { 51100 }{ 7 }\) = 7300
y = \(\frac { Δ_y }{ Δ }\) = \(\frac { 9100 }{ 7 }\) = 1300
∴ Amount invested at 4\(\frac { 3 }{ 4 }\)% is Rs 7,300 and
Amount invested at 6\(\frac { 1 }{2 }\)% is Rs 1,300

Question 4.
At marina two types of games viz., Horse riding and Quad Bikes riding are available on hourly rent. Keren and Bentia spent Rs 780 and Rs 560 during the month of May

Find the hourly charges for the two games (rides). (Use determinant method).
Solution:
let the hourly charges for horse riding be x and let the hourly charges for Quad Bikes riding be y.
For Keren
3x + 4y = 780 ……… (1)
For Benita
2x + 3y = 560 ……… (2)
Here Δ = \(\left|\begin{array}{ll}
3 & 4 \\
2 & 3
\end{array}\right|\) = 9 – 8 = 1 ≠ 0
∴ We can apply Cramer’s Rule

∴ The hourly charges for horse riding is Rs 100 and Quad Bikes riding is Rs 120

Question 5.
In a market survey three commodities A, B and C were considered. In finding out the index number some fixed weights were assigned to the three varieties in each of the commodities. The table below provides the information regarding the consumption of three commodities according to the three varieties and also the total weight received by the commodity.

Find the weights assigned to the three varieties by using Cramer’s Rule.
Solution:
Let x, y and z be are consumption of three commodities A, B and C respectively.
Given that
x + 2y + 3z = 11 ……… (1)
2x + 4y + 5z = 21 ……… (2)
3x + 5y + 6z = 27 ……… (2)
Here Δ = \(\left|\begin{array}{ccc}
1 & 2 & 3 \\
2 & 4 & 5 \\
3 & 5 & 6
\end{array}\right|\)
= 1 (24 – 25) – 2 (12 – 15) + 3 (10- 12)
= 1 ( 1) 2 ( 3) + 3 (- 2)
= -1 + 6 – 6 = -1 ≠ 0
∴ We can apply Cramer’s Rule
Now Δ_x = \(\left|\begin{array}{ccc}
11 & 2 & 3 \\
21 & 4 & 5 \\
27 & 5 & 6
\end{array}\right|\)
= 11 (24 – 25) -2 (126 – 135) + 3 (105 – 108)
= 11(-1) -2 (-9)+ 3 (-3)
= -11 + 18 – 9
= -2
Δ_y = \(\left|\begin{array}{ccc}
1 & 11 & 3 \\
2 & 21 & 5 \\
3 & 27 & 6
\end{array}\right|\)
= 1 (126 – 135) -11 (12 – 15) + 3 (54 – 63)
= 1 (-9) -11 (-3) + 3 (-9)
= -9 + 33 – 27
= -3
Δ_z = \(\left|\begin{array}{ccc}
1 & 2 & 11 \\
2 & 4 & 21 \\
3 & 5 & 27
\end{array}\right|\)
= 1 (108 – 105) – 2 (54 – 63) + 11 (10 – 12)
= 1(3) – 2 (-9) + 11 (-2)
= 3 + 18 – 22 = -1
By Cramer’s Rule

∴ (x, y, z) = (2, 3, 1)

Question 6.
A total of Rs 8,500 was invested in three interest-earning accounts. The interest rates were 2%, 3%, and 6%, if the total simple interest for one year was Rs 380 and the amount invested at 6% was equal to the sum of the amounts in the other two accounts, then how much was invested in each account? (use Cramer’s rule)
Solution:
Let the amount invested at 2%, 3%, and 6% are x, y, and z respectively.
x + y + z = 8500 ………. (1)

2x + 3y + 6z = 38000 ………. (2)
x + y = z
x + y – z = 0 ……… (3)
Here Δ = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 3 & 6 \\
1 & 1 & -1
\end{array}\right|\)
= 1 (-3 – 6) -1 (-2 – 6) + 1 (2 – 3)
= 1 (-9) -1 (-8) + 1 (-1)
= -9 + 8 – 1 = -2
≠ 0
∴ We can apply Cramer’s Rule
Δ_x = \(\left|\begin{array}{ccc}
8500 & 1 & 1 \\
38000 & 3 & 6 \\
0 & 1 & -1
\end{array}\right|\)
= 8500 (-3 – 6) -1 (-38000 – 0) + 1 (38000 – 0)
= 8500 (-9) + 38000 + 38000
= -76500 + 76000
= -500
Δ_y = \(\left|\begin{array}{ccc}
1 & 8500 & 1 \\
2 & 38000 & 6 \\
1 & 0 & -1
\end{array}\right|\)
= 1(-38000 – 0) – 8500 (-2 – 6) + 1 (0 – 38000)
= -38000 + 68000-38000
= 68000 – 76000
= -8000
Δ_z = \(\left|\begin{array}{ccc}
1 & 1 & 8500 \\
2 & 3 & 38000 \\
1 & 1 & 0
\end{array}\right|\)
= (0 – 38000) -1 (0 – 38000) + 8500 (2 – 3)
= -38000 + 38000 – 8500
= -8500
By Cramer’s Rule

∴ Amount invested at 2% is Rs 250; Amount invested at 3% is Rs 4,000 and Amount invested at 6% is Rs 4,250

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.1

Question 1.
Find the rank of each of the following matrices
(i) \(\left(\begin{array}{ll}
5 & 6 \\
7 & 8
\end{array}\right)\)
Solution:
Let A = \(\left|\begin{array}{ll}
5 & 6 \\
7 & 8
\end{array}\right|\)
Order of A is 2 × 2 [∴ P(A) ≤ 2]
Consider the second order minor = \(\left|\begin{array}{ll}
5 & 6 \\
7 & 8
\end{array}\right|\)
= 40 – 42
|A| = -2 ≠ 0
There is a minor of order 2, which is not zero
ρ(A) = 2

(ii) \(\left(\begin{array}{ll}
1 & -1 \\
3 & -6
\end{array}\right)\)
Solution:
Let A = \(\left(\begin{array}{ll}
1 & -1 \\
3 & -6
\end{array}\right)\)
Order of A is 2 × 2 [∴ ρ(A) ≤ 2]
Consider the second order minor = \(\left|\begin{array}{ll}
1 & -1 \\
3 & -6
\end{array}\right|\)
= -6 – (-3)
= -6 + 3 = -3
|A| ≠ 0
There is a minor of order 2, which is not zero
∴ ρ(A) = 2

(iii) \(\left(\begin{array}{ll}
1 & 4 \\
2 & 8
\end{array}\right)\)
Solution:
Let A = \(\left(\begin{array}{ll}
1 & 4 \\
2 & 8
\end{array}\right)\)
Order of A is 2 × 2 [∴ ρ(A) ≤ 2]
Consider the second order minor = \(\left|\begin{array}{ll}
1 & 4 \\
2 & 8
\end{array}\right|\)
= 8 – 8 = 0
Since the second are minor vanishes, ρ(A) ≠ 2
Consider a first order minor |1| ≠ 0
There is a minor of order 1, which is not zero
∴ ρ(A) = 1

(iv) \(\left[\begin{array}{ccc}
2 & -1 & 1 \\
3 & 1 & -5 \\
1 & 1 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
2 & -1 & 1 \\
3 & 1 & -5 \\
1 & 1 & 1
\end{array}\right]\)
Order of A is 3 × 3
∴ ρ(A) ≤ 2
Consider the third order minor \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
3 & 1 & -5 \\
1 & 1 & 1
\end{array}\right|\)
= 2(1 + 5) – (-1) (3 + 5) + 1 (3 – 1)
= 2 (6) + 1(8) + 1(2)
= 12 + 8 + 2
= 22 ≠ 0
There is a minor of order 3, which is not zero
∴ ρ(A) = 3

(v) \(\left[\begin{array}{ccc}
-1 & 2 & -2 \\
4 & -3 & 4 \\
-2 & 4 & -4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
-1 & 2 & -2 \\
4 & -3 & 4 \\
-2 & 4 & -4
\end{array}\right]\)
Order of A is 3 × 3
∴ ρ(A) ≤ 3
Consider the third order minor \(\left|\begin{array}{ccc}
-1 & 2 & -2 \\
4 & -3 & 4 \\
-2 & 4 & -4
\end{array}\right|\)
= -1(12 – 16) – 2 (-16 + 8) – 2 (16 – 6)
= -1(-4) – 2 (-8) – 2 (10)
= 4 + 16 – 20
= 0
Since the third order minor vanishes, therefore
∴ ρ(A) ≠ 3
Consider a second order minor \(\left|\begin{array}{ll}
-1 & 2 \\
4 & -3
\end{array}\right|\)
= 3 – 8
= -5 ≠ 0
There is a minor of order 2, which is not zero.
∴ ρ(A) = 2

The order of A is 3 × 4
∴ ρ(A) ≤ 3

The number of non zero rows is 2
∴ ρ(A) = 2

The order of A is 3 × 4
∴ ρ(A) ≤ 3
Consider the third order minor \(\left|\begin{array}{ccc}
3 & 1 & -5 \\
1 & -2 & 1 \\
1 & 5 & -7
\end{array}\right|\)
= 3 (14 – 5) – 1 (- 7 – 1) – 5 (5 + 2)
= 3 (9) – 1 (-8) – 5 (7)
= 27 + 8 – 35
= 0
\(\left|\begin{array}{ccc}
1 & -5 & 1 \\
-2 & 1 & -5 \\
5 & -7 & 2
\end{array}\right|\)
= 1 (2 – 35) + 5 (-4 + 25) – 1 (14 – 5)
= 1 (-33) + 5(21) – 1 (9)
= -33 + 105 – 9
= 63 ≠ 0
There is a minor of order 3, which is not zero.
∴ ρ(A) = 3

Order of A is 3 × 4
∴ ρ(A) ≤ 3
Consider the third order minor \(\left|\begin{array}{ccc}
1 & -2 & 3 \\
-2 & 4 & -1 \\
-1 & 2 & 7
\end{array}\right|\)
= 1 (28 + 2)+ 2 (-14 – 1) + 3 (-4 + 4)
= 1 (30) + 2 (- 15) + 3 (0)
= 0
\(\left|\begin{array}{ccc}
-2 & 3 & 4 \\
4 & -1 & -3 \\
2 & 7 & 6
\end{array}\right|\)
= -2 (-6 + 21) – 3 (24 + 6) + 4 (28 + 2)
= -2(15) – 3 (30)+ 4 (30)
= 0
\(\left|\begin{array}{ccc}
1 & 3 & 4 \\
-2 & -1 & -3 \\
-1 & 7 & 6
\end{array}\right|\)
= 1 (-6 + 21) – 3 (-12 – 3) + 4 (-14 – 1)
= 1 (15) -3 (-15) + 4 (-15)
= 15 + 45 – 60
= 0
\(\left|\begin{array}{ccc}
1 & -2 & 4 \\
-2 & 4 & -3 \\
-1 & 2 & 6
\end{array}\right|\)
= 1 (24 + 6) + 2 (-12 – 3) + 4 (-4 + 4)
= 1 (30) + 2 (-15) + 4(0)
= 30 – 30
= 0
Since all third order minors vanishes, ρ(A) ≠ 3
Now, let us consider the second order minors.
Consider one of the second order minors
\(\left|\begin{array}{ll}
-2 & 3 \\
4 & -1
\end{array}\right|\) = 2 – 12 = -10 ≠ 0
There is a minor of order 2 which is not zero
∴ ρ(A) = 2

Question 2.
If A = \(\left(\begin{array}{ccc}
1 & 1 & -1 \\
2 & -3 & 4 \\
3 & -2 & 3
\end{array}\right)\) and B = \(\left(\begin{array}{ccc}
1 & -2 & 3 \\
-2 & 4 & -6 \\
5 & 1 & -1
\end{array}\right)\)
find the rank of A B and the rank of B A.
Solution:

To find the rank of AB
Order of AB is 3 × 3
∴ ρ(AB) ≤ 3
Consider the third order minor |AB| = \(\left|\begin{array}{ccc}
-6 & 3 & -2 \\
28 & -12 & 20 \\
22 & -11 & 18
\end{array}\right|\)
= -6(-216 + 220) -3(504 – 440) – 2(-308 + 264)
= – 6(4) – 3(64) – 2(-44)
= -24 – 192 + 88
= 128 ≠ 0
There is a minor of order 3, which is not zero.
∴ ρ(AB) = 3
To find the rank of BA
Order of BA is 3 × 3
∴ ρ(BA) ≤ 3
Consider the third order minor |BA| = \(\left|\begin{array}{ccc}
6 & 1 & 0 \\
-12 & -2 & 0 \\
4 & 4 & -4
\end{array}\right|\)
= 6(8 – 0) – 1(48 – 0) + 0(-48 + 8)
= 6(8) – 1(48) + 0(-40)
= 48 – 48 + 0
= 0
Since the third order minor vanishes, therefore P(BA) ≠ 3
Consider a second order minor \(\left|\begin{array}{ll}
-12 & -2 \\
4 & 4
\end{array}\right|\) = -48 + 8 = -40 ≠ 0
There is a minor of order 2 which is not zero
∴ ρ(BA) = 2

Question 3.
Solve the following system of equations by rank method
x + y + z = 9, 2x + 5y + 7z = 52, 2x – y – z = 0
Solution:
x + y + z = 9
2x + 5y + 7z = 52
2x – y – z = 0
The matrix equation corresponding to the given system is

Obviously the last equivalent matrix is in the echelon form. It has three non-zero rows.
P(A) = P (A, B) = 3 = Number of unknowns
The given system is consistent and has unique solution.
To find the solution, let us rewrite the above ech-elon form into the matrix form.
\(\left[\begin{array}{lll}
1 & 1 & 1 \\
0 & 3 & 5 \\
0 & 0 & 2
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] =\left[\begin{array}{c}
9 \\
34 \\
16
\end{array}\right]\)
x + y + z = 9 ……..(1)
3y + 5z = 34 ……… (2)
2z = 16 ……… (3)
z = \(\frac { 16 }{ 2 }\) = 8
z = 8
Substitute z = 8 in eqn (2)
3y + 5 (8) = 34
3y + 40 = 34
3y = 34 – 40
3y = -6
y = -2
Substitute y = -2 and z = 8 in eqn (1)
x = (-2) + 8 = 9
x + 6 = 9
x = 9 – 6
x = 3
∴ x = 3, y = -2, z = 8

Question 4.
Show that the equations 5x + 3y + 7z = 4, 3x + 26y + 2z = 9, 7x + 2y + 10z = 5 are consistent and solve them by rank method.
Solution:
5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
The matrix equation corresponding to the given systematic


Obviously, the last equivalent matrix is in the echelon form. It has two non-zero rows.
P([A, B]) = 2 ; P(A) = 2
P(A) = P ([A, B]) = 2 < Number of unknowns
The given system is consistent and has infinitely many solutions.
The given system is equivalent to the matrix equation
\(\left[\begin{array}{lll}
1 & 5 & 1 \\
0 & -11 & 1 \\
0 & 0 & 0
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] =\left[\begin{array}{c}
2 \\
-3 \\
0
\end{array}\right]\)
x + 5y + z = 2 ……. (1)
-11y + z = -3 ……… (2)
let z = k

Where K ε R

Question 5.
Show that the following system of equations have unique solutions:
x + y + z = 3, x + 2y + 3z = 4, x + 4y + 9z = 6 by rank method.
Solution:
x + y + z = 3
x + 2y + 3z = 4
x + 4y + 9z = 6
The matrix equation corresponding to the given system is

The last equivalent matrix is in the echelon form [A, B] has 3 non-zero rows and [A] has 3 non-zero rows.
p([A,B]) = 3; ρ(A) = 3
ρ([A, B]) = ρ(A) = No. of unknowns
∴ The system of equations have unique solution.

Question 6.
For what values of the parameter λ, will the following equations fail to have unique solution:
3x – y + λz = 1, 2x + y + z = 2, x + 2y – λz = -1
Solution:
3x – y + λz = 1
2x + y + z = 2
x + 2y – λz = -1
The matrix equation corresponding to the given system is

If the equations fail to have unique solution.
ρ(A) ≠ ρ(A, B)
ρ(A, B) = 3
ρ(A) ≠ 3
Therefore 7 + 2λ = 0
2λ = -7 and λ = -7/2

Question 7.
The price of three commodities. X, Y, and Z are x,y, and z respectively Mr. Anand purchases 6 units of Z and sells 2 units of Y. Mr. Amar purchases a unit of Y and sells 3 units of X and 2 units of Z. Mr. Amit purchases a unit of X and sells 3 units of Y and a unit of Z. In the process they earn Rs 5,000/-, Rs 2,000/- and Rs 5,500/- respectively. Find the prices per unit of three commodities by the rank method.
Solution:
Let the equations for Mr. Anand, Mr. Amar, and Mr. Amit are
2x + 3y – 6z = 5000
3x – y + 2z = 2000
-x + 3y + z = 5500 respectively
The matrix equation corresponding to the given system is

∴ The given system is equivalent to the matrix equation
\(\left[\begin{array}{lll}
-1 & 4 & -8 \\
0 & 1 & -2 \\
0 & 0 & 7
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] =\left[\begin{array}{c}
3000 \\
1000 \\
3500
\end{array}\right]\)
-x + 4y – 8z = 3000 …….. (1)
y – 2z = 1000 ………. (2)
7z = 3500 ……….. (3)
Eqn (3) ⇒ z = \(\frac { 3500 }{ 7 }\)
∴ z = 500
Eqn (2) ⇒ y – 2(500) = 1000
y = 1000 + 1000
∴ y = 2000
Eqn (1) ⇒ -x + 4 (2000) – 8 (500) = 3000
-x + 8000 – 4000 = 3000
-x + 4000 = 3000
-x = 3000 – 4000
-x = – 1000
∴ x = 1000
The price of three commodities, x, y and z are Rs. 1000; Rs. 2000 and Rs. 500 respectively.

Question 8.
An amount of Rs 5,000/- is to be deposited in three different bonds bearing 6%, 7%, and 8% per year respectively. Total annual income is Rs 358/-, If the income from the first two investments is Rs 70/- more than the income from the third, then find the amount of investment in each bond by the rank method.
Solution:
Let the three different bonds be x, y, and z
x + y + z = 5000 …….. (1)

6x + 7y = 8z + 7000
6x + 7y – 8z = 7000 ……… (3)
The matrix equation corresponding to the given system is

∴ The given system is equivalent to the matrix equation
\(\left[\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 2 \\
0 & 0 & -16
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] =\left[\begin{array}{c}
5000 \\
5800 \\
-28800
\end{array}\right]\)
x + y + z = 5000 ……… (1)
y + 2z = 5800 …….. (2)
-16z = -28800 ……… (3)
eqn (3) ⇒ z = \(\frac { -28800 }{ -16 }\)
∴ z = 1800
eqn (2) ⇒ y + 2(1800) = 5800
y + 3600 = 5800
y = 5800 – 3600
∴ y = 2200
eqn (1) ⇒ x + 2200 + 1800 = 5000
x + 4000 = 5000
x = 5000 – 4000
∴ x = 1000
The amount of investment in each bond is Rs 1000, Rs 2200 and Rs 1800.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 13 Organic Nitrogen Compounds Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds

12th Chemistry Guide Organic Nitrogen Compounds Text Book Back Questions and Answers

Part – I Text Book Evaluation

I. Choose the Correct Answer

Question 1.
Which of the following reagent can be used to convert nitrobenzene to aniline
a) Sn/HCl
b) ZnHg/NaOH
c) LiAlH₄
d) All of these
Answer:
a) Sn/HCl

Question 2.
The method by which aniline cannot be prepared is
a) degradation of benzamide with Br₂/NaOH
b) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution.
c) Hydrolysis of phenylcyanide with acidic solution.
d) reduction of nitrobenzene by Sn/HCl
Answer:
b) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution.

Question 3.
Which one of the following will not undergo Hofmann bromamide reaction
a) CH₃CONHCH₃
b) CH₃CH₂CONH₂
c) CH₃CONH₂
d) C₆H₅CONH₂
Answer:
a) CH₃CONHCH₃
Solution: Only primary amides undergo Hofmann bromamide reaction

Question 4.
Assertion: Acetamide on reaction with KOH and bromine gives acetic acid
Reason: Bromine catalyses hydrolysis of acetamide.
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) assertion is true but reason is false
d) both assertion and reason are false
Answer:
d) both assertion and reason are false.

Question 5.

a) bromomethane
b) α -Bromo sodium acetate
c) methanamine
d) acetamide
Answer:
c) methanamine
Solution:

Question 6.
Which one of the following nitro compounds does not react with nitrous acid?
a) CH₃-CH₂-CH₂-N0₂
b) (CH₃)₂ CH-CH₂ NO₂
c) (CH₃)₃CNO₂

Answer:
c) (CH₃)₃CNO₂
Solution:
3⁰ Nitroalkane

Question 7.

a) Friedel – Crafts reaction
b) HVZ reaction
c) Schotten – Baumann reaction
d) None of these
Answer:
c) Schotten – Baurnann reaction

Question 8.
The product formed by the reaction of an aldehyde with a primary amine (NEET)
a) carboxylic acid
b) aromatic acid
c) Schiff’s base
d) ketone
Answer:
c) Schiff’s base

Question 9.
Which of the following reaction is not correct

d) None of these
Answer:

Solution:
p – nitrosation takes places, the product is im7

Question 10.
When aniline reacts with acetic anhydride the product formed is
a) o – amirìoacetophenone
b) m – aminoacetophenone
c) p – aminoacetophenone
d) acetanilide
Answer:
d) acetanilide
Solution:

Question 11.
The order of basic strength for methyl substituted amines in aqueous solution is
a) N(CH₃)₃> N(CH₃)₂ H > N(CH₃)H₂> NH₃
b) N(CH₃)H,> N(CH₃)₂H > N(CH₃)₃ > NH₃
c) NH₃> N(CH₃)H₂> N(CH₃)₂H > N(CH₃)₃
d) N(CH₃)₂ H > N(CH₃)H₂> N(CH₃)₃ > NH₃
Answer:
d) N(CH₃)₂ H > N(CH₃)H₂> N(CH₃)₃ > NH₃

Question 12.

a) H₃PO₂ and H₂O
b) H⁺/H₂O
c) HgSO₄ / H₂SO₄
d) Cu₂Cl₂
Answer:
a) H₃PO₂ and H₂0

Question 13.

a) C₆H₅-OH
b) C₆H₅-CH₂OH
c) C₆H₅-CHO
d) C₆H₅NH₂
Answer:
a) C₆H₅-OH
Solution:

Question 14.
Nitrobenzene on reaction with Con HNO₃ / H₂SO₄ at 80- 100^oC forms which one of the following products?
a) 1, 4 – dinitrobenzene
b) 2, 4, 6 – tirnitrobenzene
c) 1, 2 – dinitrobenzene
d) 1, 3 – dinitrobenzene
Answer:
d) 1, 3 -dinitrobenzene
Solution:

Question 15.
C₅H₁₃N reacts with HNO₂ to give an optically active compound – The compound is
a) pentan-1 -amine
b) pentan-2-amine
c) N, N – dimethyipropan -2-amine
d) N-methylbutan-2-amine
Answer:
d) N-methylbutan-2-amine

Question 16.
Secondary nitro alkanes react with nitrous acid to form
a) red solution
b) blue solution
c) green solution
d) yellow solution
Answer:
b) blue solution

Question 17.
Which of the following amines does not undergo acetylation?
a) t-butylamine
b) ethylamine
c) diethylamine
d) triethylamine
Answer:
d) triethylamine (3^oamine)

Question 18.
Which one of the following is most basic?
a) 2, 4 – dichloroaniline
b) 2, 4 – dimethylaniline
c) 2, 4 – dinitroaniline
d) 2, 4 – dibromoaniline
Answer:
b) 2,4-dimethylaniline
Solution: CH₃ is a+I group, all other – I group +T group increase the electron density on NH₂ and hence increase the basis nature.

Question 19.

a) Ethanol, hydroxylamine hydrochloride
b) Ethanol, ammonium hydroxide
c) Ethanol, NH₂OH
d) C₃H₅NH₂, H₂O
Answer:
a) Ethanol, hydroxylamine hydrochloride

Question 20.
UPAC name for the amine

a) 3 – Bimethy lamino – 3 – methyl pentane
b) 3(N,N – Triethyl) – 3 – amino pentane
c) 3-N,N – trimethyl pentanamine
d) 3 – (N,N – Dimethyl amino) – 3- methyl pentane
Answer:
d) 3 – (N,N – Dimethyl amino) -3- methyl pentane

Question 21.


Answer:
b)

Question 22.
Ammonium salt of benzoic acid is heated strongly with P₂O₅ and the product so formed is reduced and then treated with NaNO₂/ HCl at low temperature. The final compound formed is
a) Benzene diazonium chloride
b) BenzYl alcohol
c) Phenol
d) Nitrosobenzene
Answer:
b) Benzyl alcohol
Solution:

Question 23.
Identify X in the sequence given below.


Answer:

Question 24.
Among the following, the reaction that proceeds through an electrophilic substitution, is:

Answer:

Explanation:
a) Nucleophilic substitution
b) Electrophilic substitution
c) Addition Reaction
d) Nucleophilic substitution

Question 25.
The major product of the following reaction



Answer:

Solution:

II. Short Answer Questions

Question 1.
Write down the possible isomers of the C₄H₉NO₂ and give their IUPAC names
Answer:

Question 2.
There are two isomers with the formula CH₃ NO₂. How will you distinguish between them?
Answer:

• Primary and secondary nitroalkanes with a-H atom exhibit tautomerism.
• Tertiary amines do not exhibit tautomerism due to CH =N the absence of a-H atom.
• Nitromethane exists in two tautomeric forms namely \itro form Aci form nitroform and aciform.

Question 3.
What happens when
Answer:
i) 2 – Nitropropane boiled with HCl
ii). Nitrobenzene electrolytic reduction in strongly acidic medium.
iii). Oxidation of tert – butylamine with KMnO₄
iv). Oxidation of acetone oxime with trifluoromethoxy acetic acid.

Question 4.
How will you convert nitrobenzene into
Answer:

• 1, 3, 5 – trinitrobenzene
• o and p – nitrophenol
• m – nitro aniline
• azoxvbenzcne
• hvdrazohenzene
• N – phenvl hyd roxylamine
• aniline

Question 5.
Identify compounds A, B, and C in the following sequence of reactions.
Answer:

Question 6.
Write short notes on the following
Answer:

• Hofmann’s bromide reaction
• Ammonolysis
• Gabriel phthalimide synthesis
• Schotten – Baurnann reaction
• Carhvlamine reaction
• Mustard oil reaction
• Coupling reaction
• Diazotisation
• Gorenberg reaction

I Hofmann’s bromamide reaction:

In Hofnianns degradation acid amide is converted into an amine with one carbon less by Br₂/ KOH.

ii. Ammonolysis:

In Hoffmann’s ammonolysis alkyl halides are heated with alcoholic ammonia in a sealed tube, mixtures of 1^o, 2^o and 3^o amines and quarternary ammonium salts are obtained.

iii. Gabriel phthalimide synthesis

Phthalimide on treatment with alcoholic KOH forms potassium phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives primary amine. Aniline can not be prepared by this method.

iv. Schotten – Baumann reaction:

Benzoylation of amines to give N-alkyl benzamide in presence of NaOH is known as Schotten Baumann reaction.

v. Carbylamine reaction

Primary’ amines react with chloroform and alcoholic KOH to form isocyanides called carbylamines with unpleasant smell. This is used to identify primary amines.

vi. Mustard oil reaction
When primary amines are treated with carbon disulphide, N-alkyl dithiocarbamic acid is formed which on treatment with HgCl₂ gives an alkyl isothiocyanate.

vii. Coupling reaction:

Benzenediazonium chloride reacts with electron-rich aromatic compounds like phenol, undergoing
electrophilic substitution at para position.

viii. Diazotisation:

Aniline reacts with nitrous acid at 273-278K to form benzene diazonium chloride.

ix. Gomberg reaction:

Benienediazonium chloride reacts with benzene in presence of NaOH to give biphenyl.

Question 7.
How will you distinguish between Primary, Secondary, and tertiary aliphatic amines
Answer:

Question 8.
Account for the following
Answer:

• Aniline does not undergo Friedel – Crafts reaction
• Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
• pK_b of aniline is more than that of methylamine
• Gabriel phthalimide synthesis is preferred for synthesising_primary amines
• Ethvlamine is soluble in water whereas aniline is not
• Amines are more basic than amides
• Although amino group is o – and p – directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m – nitroaniline.

i) Aniline does not undergo Friedel – crafts reaction, because Aniline is basic in nature
Aniline donates its lone pair to the lewis acid Alcl₃ to form an adduct which inhibits further electrophilic substitution reaction

ii) The stability of diazonium salts of aromatic amines is due to the dispersal of the positive charge over the benzene ring.

iii) In aniline the NH
group is directly attached to the benzene ring.
The lone pair of electron on nitrogen atom in aniline gets delocalised over the benzene ring.
Hence it is less available for protonation.
So aniline is less basic than methylamine and pK_b of aniline is more than that of methylamine.

iv) Gabriel phthalimide synthesis results in the formation of primary amines only. Secondary and tertiary amines are not formed in this synthesis. Thus a pure primary amine can be obtained.
Therefore, Gabriel phthalimide synthesis is preferred for synthesizing primary amines.

v) Ethvlamine forms intermolecular hydrogen bonds with water. Hence it is soluble in water.
But aniline does not form hydrogen bond with water to a very large extent due to the presence of a large hydrophobic C₆H₅ – group. Hence aniline is insoluble in water

vi) In amides the carbonyl group (C = 0) is highly electronegative and draws the electrons towards it.
This makes the lone pair of amide nitrogen less available to accept a proton.
Hence amides are less basic than amines.
But in amines the alkyl groups being electron releasing groups. the electron pair on amine nitrogen is readily available to accept a proton.

vii) Hence amines are more basic than amides.
In strong acid medium aniline is protonated to form anilinium ion which is rn-directing.
Hence a substantial amount of rn-nitroaniline is formed during nitration of aniline.

Question 9.
Arrange the following
Answer:
i) In increasing order of solubility in water C₆H₅NH₂,(C₂H₅)₂NH,C₂H₅NH₂
ii) In increasing order of basic strength
a) aniline, p – toludine and p – nitroaniline
b) C₆H₅NH₂ C₆H₅NHCH₃, C₆H₅NH₂, p-Cl-C₆H₄-NH₂
iii) In decreasing order of basic strength in gas phase
(C₂H₅)NH₂, (C₂H₅)NH, (C₂H₅)₃N and NH₃
iv) In increasing order of boiling point C₆H₅OH, (CH₃)₂NH,C₂H₅NH₂
v) In decreasing order of the pK_b values C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and CH₃NH₂
vi) Increasing order of basic strength C₆H₅NH₂, C₆H₅N(CH₃) H₅3)₂, (C₂H₅)₂ NH and CH₃ NH₂
vii) In decreasing order of basic strength

i) Increasing order of solubility in water C₆H₅NH₂ < (C₂H₅) NH < C₂H₅ NH₂
Hint
Aromatic amine
aliphatic amine
Solubility α \(\frac{1}{\text { Molarmass }}\)
ii) Increasing order of basic strength
a) p-nitro aniline
aniline
p-toludine
b) p – Cl- C₆H₄– NH₂ < C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂
Hint > p-nitrogroup (-I) electron with drawing, hence decreases the basic strength.
Hint > p-methyl group (+I) electron releasing, hence increases the basic strength.
iii) Decreasing order of basic strength in gas phase
(C₂H₅)₃N > (C₂H₅)₂ NH > C₂H₅ NH₂> NH₃
Hint > In the gas-phase there is no solvation effect. As a result the basic strength mainly depends on the + I effect. Higher the number of alkyl groups higher will be the +1 effect, stronger is the base.
iv) Increasing order of boiling point.
(CH₃)₂NH <C₂H₅NH₂ <C₆H₅OH
Hint > Amines generally have lower boiling point than alcohols of comparable molar mass. Since amines have weaker H-bonds.
Hint
Secondary amines
primary amines
v) Decreasing order of pK values
C₆H₅NHCH₃> C₂H₅NH₂> CH₃ NH₂> (C₂H₅)₂ NH
Hint > Basic character increases. pKb value decreases. Higher the basic nature, lower will be the pkb’ value.
vi) Increasing order of basic strength
C₆H₅NH₂ < C₆H₅-N(CH₃)₂ < CH₃ NH₂ < (C₂H₅)₂NH
vii) Decreasing order of basic strength

Question 10.
How will you prepare propan – 1 – amine from
Answer:
i) butane nitrile
ii) propanamide
iii) 1-nitropropane

Question 11.

Answer:

Question 12.
How will you convert diethylamine into
Answer:
i) N,N – dimethylacetamide
ii) N – nitrosódiethylamine

Question 13.
Identify A, B and C
Answer:

Question 14.
Identify A,B,C and D
Answer:

Question 15.
Complete the following reaction

Answer:

Question 16.
Predict A,B, C and D for the following reaction

Answer:

Question 17.
A dibromo derivative (A) on treatment with KCN followed by acid hydrolysis and heating gives a monobasic acid (B) along with liberation of CO₂ . (B) on heating with liquid ammonia followed by treating with Br₂ / KOH gives (c) which on treating with NaNO₂ and HC1 at low temperature followed by oxidation gives a monobasic acid (D) having molecular mass 74. Identify A to D.
Answer:
D is a monobasic and with molecular mass 74.
M.F of D is CnH_2n+1COOH
MolarMassis 12 × n + 2n × 1 + 1 × 1 + 12 + 2 × 16 + 1 × 1 = 74
14n + 46 = 74
14n=28
n=2

Question 18.
Identify A to E in the following frequency of reactions

Answer:

III. Evaluate Yourself

Question 1.
Write all possible isomers for the following compounds.
Answer:
i) C ₂H ₅ – NO ₂
ii) C ₃H ₇ – NO ₂

Question 2.
Find out the product of the following reactions.
Answer:

Question 3.
Predict the major product that would be obtained on nitration of the following compounds?
Answer:

Question 4.
Draw the structure of the following compounds
i. Neopentylaniine
ii. Tert – butylamine
iii. α- amino propionaldehyde
iv. Tribenzylamine
v. N – ethyl – N – methyihexan – 3- amine
Answer:

Question 5.
Give the correct IUPAC names for the following amines
Answer:

12th Chemistry Guide Organic Nitrogen Compounds Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

Question 1.
Nitro ethane and ethyl nitrite are
a) chain isomers
b) position isomers
c) functional isomers
d) tautomers
Answer:
c) functional isomers

Question 2.
1-nitrobutane and 2-nitrobutane are
a) chain isomers
b) Position isomers
c) functional isomers
d) tautomers
Answer:
b) Position isomers

Question 3.
1-nitrobutane and 2- methyl-1-nitro propane are
a) chain isomers
b) Position isomers
c) funtional isomers
d) tautomers
Answer:
a) chain isomers

Question 4.

a) chain isomers
b) position isomers
c) functional isomers
d) tautorners
Answer:
d) tautomers

Question 5.
Which among the following is a tertiai nitroalkane?
a) 2-nitro propane
b) 2-methyl-.1—nitro propane
C) 2-methyl – 2-nitropropane
d) nitro ethane
Answer:
c) 2-methyl – 2-nitropropan

Question 6.
Which among the following does not sho tautomerism?
a) 2-nitro propane
b) 2-methyl-1-nitropropane
c) 2-methyl-2-nitropropane
d) nitro ethane
Answer:
c) 2-methyl-2-nitropropane
Hint : Tertiary Nitroalkanes do not exhibit tautomerism due to the absence of Oc-H atom.

Question 7.
Which nitro alkane does not dissolve in NaOH?
a) 2-nitro propane
b) 2-methyl – 1 – nitro propane
c) 2 – methvl-2-nitropropane
d) nitroethane
Answer:
c) 2 – methyl- 2 – nitropropane
Hint : Tertiary mitroalkanes do not react with NaOH due to the absence of ∝-H atom.

Question 8.
Which among the following does not react with Nitrous acid.
a) 2 – nitro propane
h) 2-methyl-1-nitropropane
c) 2 – methyl – 2 – nitro propane
d) nitro ethane
Answer:
c) 2 – methyl – 2 – nitro propane
Hint : Tertiary mitroalkanes do not react with nitrous acid due to the absence of ∝- H atom.

Question 9.
The correct order of acidic nature of nitro alkanes is
a) nitro propane> nitroethane > nitro methane
b) nitro propane c) nitro methane < nitro ethane < nitro propane d) nitro ethane> nitro methane> nitro propane
Answer:
b) nitro propane < nitroethane < nitromethane
Hint: Length of alkyl group attached to the ∝- carbon increases, acidic nature decreases

Question 10.

The above reaction follows the mechanism
a) E₁
b) E₂
c) S_N¹
d) S_N²
Answer:
d) S_N²

Question 11.
Acetaldoxime is converted into nitroethane by reaction with
a) HNO₂
b) CF₃COOOH
c) KMnO₄
d) HNO₃/H₂SO₄
Answer:
b) CF₃COOOH

Question 12.
Oil of mirbane is
a) nitro ethane
b) nitro propane
c) nitro benzene
d) nitro aniline
Answer:
c) nitrobenzene

Question 13.
Direct nitration of nitro benzene gives
a) o – dinitro benzene
b) m – dinitro benzene
c) p – dinitro benzene
d) All the above
Answer:
b) m – dinitro benzerte

Question 14.
P – diamino benzene can be converted to p – dinitro benzene by
a) Caros acid
b) persuiphuric acid
c) peroxy trifluoro acetic acid
d) all the above
Answer:
d) all the above

Question 15.
Nef carbonyl synthesis given by
a) C₆H₅CHO

c) C₆H₅CH₂NO₂
d) All of these
Answer:
a) C₆H₅CHO

Question 16.
In electrophilic substititution reaction of nitro benzene, the – NO₂ group acts as
a) activating group and m – directing
b) deactivating group and rn-directing
c) activating group and o,p – directing
d) deactivating group and o,p -directing
Answer:
b) deactivating group and m-directing

Question 17.
Identify X in the following sequence of reaction

a) (CH₃)₂CHCOCl
b) CH₃CH₂CH(OH)CH₂Cl
c) CH₃CH₂CH₂COCl
d) ClCH₂CH₂CH₂CHO
Answer:
c) CH₃CH₂CH₂COCl

Question 18.
The reducing agent used to reduce alkylcyanides to primary amines is
a)H₂/Ni
b)LiAlH₄
c) Na/C₂H₅OH
d) all the above
Answer:
d) all the above

Question 19.
The reducing agent used in Mendius reaction is
a) H₂/Ni
b) LiAlH₄
c) Na/C₂H₅OH
d) NaBH₄
Answer:
c) Na/C₂H₅OH

Question 20.
Reduction of alkyl iso cyanides give
a) primary amines
b) secondary amines
c) tertiary amines
d) none of the above
Answer:
b)secondary amines

Question 21.

a) CH₃NH₂, CH₃CH₂NH₂
b) CH₃OH, CH₂CH₂OH
c) CH₃CH₂NH₂, CH₃NH₂
d) CH₃CN, CH₃CH₂CN
Answer:
c) CH₃CH₂NH₂, CH₃NH₂

Question 22.
Gabriel phthalimide synthesis is used for the preparation of
a) aliphatic primary amines
b) aromatic primary amines
c) both (a) & (b)
d) none of the above
Answer:
a)aliphatic primary amines

Question 23.
Which among the following can not be prepared by Gabriel phthalimide synthesis?
a) Methanamine
b) ethanamine
c) benzenamine
d) propananiine
Answer:
c)benzenamine

Question 24.
Two molecules of propannitrile in the presence of Na/Ether to form 3-imino- 2-methylpentanenitrile. This reaction is known as
a) Baltz – schiemann reaction
b) Thorpe nitrile condensation
c) Gomberg reaction
d) hotten – Baurnann reaction
Answer:
b) Thorpe nitrile condensation

Question 25.
In Hoffmann’s ammonolysis, if excess alkyl halide is taken, the final product is
a) primary amine
b) secondary amine
c) tertiary amine
d) quaternary ammonium salt
Answer:
d) quaternary ammonium salt

Question 26.
Hoffmann’s ammonolysis reaction is an example of
a) nucleophilic addition
b) nucleophilic substitution
c) electrophilic addition
d) electrophilic substitution
Answer:
b) nucleophilic substitution

Question 27.
In Hoffmann’s ammonolysis the order of reactivity of alkyihalides with amines is
a) RI > RBr> RCl
b) RI < RBr < RCl c) RI > RBr < RCl
d) RI < RBr> RCl
Answer:
a)RI>RBr>RCl

Question 28.

A and B are respectively
a) CH₃Na, CH₃CN
b) CH₃NH₂,CH₃N₃
c) CH₃N₃ CH₃NH₂
d) CH₃NH₂,CH₃CN
Answer:
c) CH₃N₃ CH₃NH₂

Question 29.
The correct ascending order of basic strength of alkylamines in aqueous solution is.
a) R₂NH > RNH₂ > R₃N > NH₃
b) NH₃ > R₃N > RNH₂ > R₂NH
c) NH₃ < R₃N < RNH₂ < R₂NH
d) R₂NH < RNH₂ < R₃N < NH₃
Answer:
c) NH₃ < R₃N < RNH₂ < R₂NH

Question 30.
Among the three types of amines, secondary amine is more basic due to
a) +1 effect
b) steric effect
c) hydration effect
d) all the above
Answer:
d) all the above

Question 31.
In case of sustituted aniline, the groups which increase the basic strength are
(i) -CH₃
(ii) -NO₂
(iii) -OCH₃
(iv) -Cl
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
b)(i) & (iii)

Question 32.
The correct order of basic strength of amines is
a) C₆ H₅ – NH₂ > C₆ H₅ CH₂ NH₂> NH₃ > C₆ H₅ CH₂ NHCH₃ > C₂ H₅ NH₂

b) C₆ H₅ CH₂ NH₂ > C₆ H₅ CH₂ NHCH₃ > NH₃ > C₆ H₅ NH₂ > C₂H₅NH₂

c) C₂ H₅ NH₂ > C₆H₅ CH₂ NH₂ > NH₃ > C₆ H₅ CH₂ NHCH₃ > C₆ H₅ NH₂

d) C₆ H₅CH₂ NH CH₃ > C₆ H₅ CH₂ NH₂ > NH₃ > C₂ H₅ NH₂ > C₆ H₅ NH₂
Answer:
c) C₂ H₅ NH₂ > C₆H₅ CH₂ NH₂ > NH₃ > C₆ H₅ CH₂ NHCH₃ > C₆ H₅ NH₂
Hint : Alkylamine > Aralkvlamine > Ammonia > N – Aralkylamine > Arylamine

Question 33.
Schotten Baumann reaction is an example of
a) nucleophilic substitution
b) nucleophilic addition
c) electrophilic substitution
d) electropholic addition
Answer:
a)nucleophilic substitution

Question 34.
The product ‘D’ of the reaction

a) CH₃CH₂NH₂
b) CH₃CN
c) HCONH₂
d) CH₃CONH₂
Answer:
d) CH₃CONH₂

Question 35.
The test used to identify a primary amine is
a) iodoform test
b) silver mirror test
c) libermann’s nitroso test
d) carbylamine test
Answer:
d) carbylamine test

Question 36.
The test used to identify a secondary amine is
a) iodo form test
b) silver mirror test
c) Libermann’s nitroso test
d) carbylamine test
Answer:
c) Libermann’s nitroso test

Question 37.
When ethylamine is treated with chloroform and alcoholic KOH an unpleasant smell is evolved due to the formation of
a) ethylcyanide
b) ethylisocyanide
c) ethylisocyanate
d) ethylcyanate
Answer:
b) ethylisocyanide

Question 38.
C₆H₅N + 2Cl^– + H₃PO₂ + H₂O → C₆H₆ + H₃PO₃ + HCl + N₂ This reaction proceeds through
a) SN¹ mechanism
b) SN² mechanism
c) free radical mechanism
d) elimination reaction
Answer:
c) free radical mechanism

Question 39.
Benzene diazonium chloride is converted into chloro benzene by
(i) Sandmeyer reaction
(ii) Gatter mann reaction
(iii) Gomberg reaction
a) (i) &(ii)
b) (i.) & (iii)
c) (ii) & (iii)
d) (i) (ii) & (iii)
Answer:
a)(i) &(ii)

Question 40.


Answer:

Question 41.
Hydrogen cyanide and hydrogen isocyanide are
a) chain isomers
b) position isomers
c) tautomers
d) Geometrical isomers
Answer:
c) tautomers

Question 42.
The IUPAC name of C₆H₅CN is
a) phenyl cyanide
b) benzene carbonitrile
c) benzonitrile
d) benzene cyanide
Answer:
b) benzene carbonitrile

Question 43.

a) CH₃CN
b) CH₃NH₂
c) CH₃CONH₂
d) CH₃CH₂CONH₂
Answer:
d) CH₃CH₂CONH₂

Question 44.
Methyl magnesium chloride is converted into ethanenitrile by treating with
a) cyanogen
b) cyanogen chloride
c) methyl cyanide
d) hydrogen cyanide
Answer:
b) cyanogen chloride

Question 45.
Which one of the following compound is a strong base?

Answer:

Question 46.
Which among the following will not undergo diazotisation?
a) m- toluidine
b) aniline
c) p – amino phenol
d) benzylamine
Answer:
d) benzylamine
Hint: Aromatic amines in which – NH₂ group is directly attached to the benzene ring undergo diazotisation

Question 47.
An organic compound (A) C₃H₉N when treated with nitrous acid, gave an alcohol (B) and N₂ gas. (A) on warming with CHCl₃ and Caustic potash gave (C) which on reduction gave isopropyl methylamine. Organic compound (A) is
a) CH₃CH₂NH CH₃
b) CH₃ CH₂ CH₂ NH₂

Answer:

Question 48.

a) Ethanamide, Ethanamine, Ethanol
b) Ethanarnine, Ethanol, Ethanoic acid
c) Ethanol, Ethanal, Ethanoic acid
d) Ethanamine, Ethanal, Ethanoic acid
Answer:
b) Ethanamine, Ethanol, Ethanoic acid
Hints:

Question 49.
Which of the following nitro compounds when reacted with nitrous acid followed by treatment with alkali produces blue colour?
a) 2-methyl-2-nitropropane
b) 2-methyl-1-nitropropane
c) 2-nitropropane
d) nitrobezene
Answer:
c) 2-nitropropane
Hunt: Nitro group attached to a secondary carbon atom will give pseudonitrol with HNO₂ which on further reaction with alkali gives blue colour.

II. Assertion and reason

Question 1.
Assertion (A) : Aniline when exposed to air becomes coloured
Reason (R) : Aniline when exposed to air undergoes oxidation
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct
Answer:
a) Both A and R are correct, R explians A

Question 2.
Assertion (A) : Aniline is more basic than ammonia
Reason (R) : The lone pair of electron on nitrogen atom in aniline gets delocalised over the benzene ring.
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct
Answer:
d) A is wrong but R is correct.
Correct Assertion A: Aniline is less basic than ammonia.

Question 3.
Assertion (A) : Nitroalkanes are more acidic than aldehvdes, ketones and cyanides.
Reason (R) : a-H atom of 10 and 20 nitroalkanes, show acidic character because of the electron releasing effect of NO₂ group.
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
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.
Correct Reason (R) : a- H atom of 1^o and 2^o nitroalkanes show acidic character because of the electron withdrawing effect of NO₂ group.

Question 4.
Assertion (A) : Nitro benzene does not undergo Friedel crafts reactions:
Reason (R) : NO₂ group present in the benzene ring is strongly deactivating in nature.
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct
Answer:
a) Both(A) and (R) are correct, R explains A.

III. Pick out the correct statement

Question 1.
(i) In amines the nitrogen atom is Sp₃ hybridised.
(ii) In amines, the fourth Sp₃ hybridised orbital contains a lone pair of electron.
(iii) Due to the presence of lone pair of electron C- N-C bond angle is more than the normal tetrehedral bond angle.
(iv) The C-N-C bond angle of trimethylamine is 1090 which is greater than tetrahedral angle,
(a) (i) & (ii)
(b) (ii)&(iii)
(c) (iii)&(iv)
(d) (I&(iv)
Answer:
(a)(i) & (ii)
Correct statement:
(iii) Due to the presence of lonepair of electron C-N-C bond angle is less than the normal tetrahedral bond angle.
(iv) The C-N-C bond angle of trimethylamine is 1080 which is lower than tetrahedral angle.

Question 2.
(i) Tertiary amines form inter molecular hydrogen bond.
(ii) Amines have higher boiling points than alcohols.
(iii) Nitrogen being less electronegative than oxygen, the N-H bond in amines is less polar than O-H bond in alcohols
(iv) Lower aliphatic amines are soluble in water, since they form hydrogen bond with water,
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer:
(c) (iii) & (iv)
Correct Statement:
(i) Tertiary amines do not form intermolecular hydrogen bond.
(ii) Amines have lower boiling point than alcohols.

Question 3.
(i) In Hofmann’s degradation reaction primary amines with one carbon less than the parent amides are formed.
(ii) Gabriel phthalimide synthesis is useful for the preparation of three types of amines.
(iii) Alkyl halides can be converted to primary amines by treating with sodium azide followed by reduction using LiAlH₄
(iv) When phenol reacts with ammonia in presence of anhydrous ZnCl₂ gives p -amino phenol.
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (i) & (iv)
Answer:
c)(i) & (iii)
Correct statement:
(ii) Gabriel phthalimide synthesis is useful for the preparation of aliphatic primary amines
(iv) When phenol reacts with ammonia in presence of anhydrous ZnCl₂ gives aniline.

Question 4.
(i) Alkyl cyanides on reduction with LiAlH₄ give secondary amines.
(ii) Alkyl cyanides on hydrolysis with alkali or dilute mineral acid give carboxylic acids.
(iii) Nitriles containing a- H atom undergo condensation with esters in presence of sodamide in ether to form ketonitriles.
(iv) Alkycyanides have lower boiling points than analogous acetylenes
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d (i) & (iv)
Answer:
b)(ii) & (iii)
Correct statement:
(i) Alkylcyanides on reduction with LiAlH₄ give primary amines.
(iv) Alkylcyanides have higher boiling points than analogous acetylenes.

IV. Pick out the incorrect statement

Question 1.
(i) Bromobenzene can be converted into nitrobenzene by reacting with KNO₂
(ii) Nitroethane is suspected to cause genetic damage and be harmful to the nervous system.
(iii) 2-methyl – 2-nitro propane reacts with HCl undergoing hydrolysis to form 2-methyl – 2 – propanol.
(iv) Ethyl nitrite on reduction with Sn / HCl gives ethanol.
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (iii) & (iv)
Answer:
c) (i) & (iii)
Correct statement:
(i) Bromo benzene can not be converted into nitrobenzene by reacting with KNO₂
(iii) 2-methyl -2-nitropropane does not undergo hydrolysis with HCl to form 2-methyl – 2 – propanol

Question 2.
(i) Nitroalkanes exhibit functional isomerism with alkyl nitrites.
(ii) Tertiary nitroalkanes exhibit tautomerism due to the absence of a-H atom.
(iii) The electrical conductivity of aciform of nitro alkanes is high.
(iv) When the number of alkyl group attached to the a- carbon of nitroalkane increases, its acidity increases.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
c) (ii) & (iv)
Correct statemet:
(ii) Tertiary nitroalkanes do not exhibit tautomerism due to the absence of a- H atom.
(iv) When the number of alkyl group attached to the a- carbon of nitro alkane increases its acidity decreases.

Question 3.
(i) Benzoylation of aniline is known as Schotten. Baumann reaction
(ii) The reaction of ethylamine with nitrous acid is known as diazotisation.
(iii) The reaction of primary amine with nitrous acid is known as Libermann’s nitroso test.
(iv) The reaction of primary amine with CS₂ and HgCl₂ is known as Mustard oil reaction.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statement:
(ii) The reaction of aniline with nitrous acid is known as diazotisation.
(iii) The reaction of secondary amine with nitrous acid is known as Libermann’s nitroso test.

Question 4.
(i) Diazo compounds obtained from the coupling reactions of diazonium salts are coloured and are used as dyes.
(ii) Aryl fluorides and iodides can not be prepared by direct halogenation but can be prepared by using benzene diozonium chloride.
(iii) In Sandmeyer reaction and Gattermann reaction the catalyst used are Cu/HCl and Cu₂Cl₂ / HCZ respectively .
(iv) Benzene diazonium chloride when added with boiling water gives benzene.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correct statement:
(iii) In Sandmeyer reaction and Gattermann reaction the catalyst used are Cu₂Cl₂/ HCl and Cu / HCl respectively
(iv) Benzene diazonium chloride when added with boiling water gives phenol.

V. Match the following

Question 1.

Answers:
1. – Phenylhydroxylamine
2. – Azobenzene
3. – Aniline
4. – Hydrazo benzene
5. – Nitrosobenzene

Question 2.


Answer:
1. – Prop – 2-en-1-amine
2. – N – methylpropan – 2 – amine
3. – Hexan -1, 6 – diamine
4. – Phenylmethanamine
5. – N, N – dimethylben zenamine

Question 3.

Answer:
1. – N-phenylbenzamide
2. – Benzene diazonium chloride
3. – Methyl isocyanide
4. – Methylisothiocyanate
5. – p-hydroxyazobenzene

Question 4.

Answer:
1. – insecticide
2. – diuretic
3. – TNT
4. – fuel additive
5. – Solvent in perfume industry

VI. Two Mark Questions

Question 1.
Write a note on acidic nature of nitroalkanes.
Answer:

• α-H atom of 1⁰and 2⁰nitroalkanes show acidic character.
• This is due to the electron withdrawing effect of – NO₂group.
• Nitroalkanes are more acidic than aldehydes, Ketones, esters and cyanides.
• Due to acidic nature nitroalkanes dissolve in NaOH to form a salt.
• Aci-nitro form is more acidic than nitro form.
• As the number of alkyl group attached to a- Carbon increases, acidity decreases due to + I effect of alkyl groups.

Question 2.
How is the oil of mirbane prepared?
Answer:
When benzene is heated at 330K with the nitrating mixture (Con HNO₃ + Con H₂S0₄), electrophilic substitution takes place to form nitrobenzene (oil of mirbane)

Question 3.
How is P – dinitrobenzene prepared?
Answer:

• Direct nitration is nitrobenzene gives m-dinitro benzene.
• The following indirect method is adopted to prepare p-dinitrobenzene.
  

Question 4.
Convert p-diamino benzene into p-dinitro benzene.
Answer:
Amino group can be directly converted into nitro group using Caro’s acid (H₂SO₅) or persulphuric acid (H₂S₂O₈) or peroxytrifluro acetic acid (CF₃COOOH) as oxidising agent

Question 5.
Write the various reduction stages of nitromethane.
Answer:

Question 6.
Write about the reduction of nitro methane in acid medium and neutral medium.
Answer:

Question 7.
What happpens when ethylnitrite is reduced by Sn / Hcl?
Answer:

• Ethyl nitrite on reduction with Sn / HCl gives ethanol.

Question 8.
What happens when ethyl nitrite undergoes hydrolysis with acid or base?
Answer:

• Ethyl nitrite undergoes hydrolysis with acid or base to give ethanol.

Question 9.
What is Nef Carbonyl Synthesis?
Answer:

Question 10.
Write about the electrolytic reduction of nitro benzene?
Answer:

Question 11.
Convert m- dinitrobenzene into m – nitroaniline
Answer:
Selective reduction of rn – dinitrobenzene with ammonium suiphide gives m – nitroaniline

Question 12.
What happens when
(i) methyl cyanide and
(ii) methyl isocyanide are reduced?
Answer:
(i) Methyl cyanide on reduction gives a primary amine ethanamine.

(ii) Methylisocyanide on reduction gives a secondary amine N-methylmcthanamine

Question 13.
How can you convert acetamide into
(i) ethanamine
(ii) methanamine.
Answer:
Acetamide on reduction with LiAlH₄ gives ethanamine.

Acetamide undergoes Hofmanns degradation with Br₂/KOH to form methanamine

Question 14.
Write about Sabatier – Mailhe method of preparing three types of Amines
Answer:

• Vapour of an alcohol and ammonia are passed over alumina, W₂O₅ or Silica as 400_o Call the types of amines are formed.
• This method is known as Sabatier-Mailhe method.

Question 15.
Write a note on basic strength of aniline.
Answer:

• Aniline is less basic than alkylamines and ammonia.
• In aniline, the NH2 group is directly attached to the benzene ring.
• The lone pair of electron in nitrogen atom in aniline gets delocalised over the benzene ring.
• Hence it is less available for protonation.
• In case of subtituted aniline, election releasing groups like – CH₃, -OCH₃, -NH₂ increase the basic strength.
• Electron with drawing groups like -NO₂, -X, -COOH decrease the basic strength.

Question 16.
How is sulphanilic acid prepared?
Answer:

• Aniline reacts with conc H₂SO₄to form anilinium hydrogen sulphate.
• This on heating with H₂SO₄ at 453-473K gives p-aminobenzene suiphonic acid known as
  sulphanilic acid

Question 17.
How will you convert aliphatic carboxylic acid into aromatic carboxylic acid?
Answer:

• When benzene diazonium fluoroborate is heated with acetic acid, benzoic acid is obtained.
• This reaction is used to convert aliphatic carboxylic acid into aromatic carboxylic acid.

Question 18.
Write about the isomerisation of alkylisocyanides.
Answer:
When alkyl isocyanides are heated at 250_oC, they change into the more stable isomeric cyanides.

Question 19.
Write the uses of nitroalkanes.
Answer:

• They are good solvents for a large number of organic compounds including vinyl polymers, cellulose esters, synthetic rubbers, oils, fats, waxes and dyes.
• Used in organic synthesis.
  Nitromethane reacts with halogen in presence of alkali to form trihalogen derivative, (e.g.,) with chlorine it forms chloropicrin, CCl₃NO₂ which is used as soil sterilizing agent.
• Nitroethane is also used as a fuel additive and precursor to Rocket propellants.

Question 20.
How is chloropicrin prepared ?
Answer:
Nitromethane on treatment with Cl2 in presence of NaOH gives chloropicrin.

Chloropicrin (Trichloronitro methane)

VII. Three mark questions

Question 1.
How are nitro compounds classified?
Answer:
Classification of nitrocompounds

Question 2.
Convert (i) acetaldelyde into nitro ethane
(ii) acetone into 2-nitropropane
Answer:

Question 3.
Write about the hydrosysis of nitro alkanes
Answer:

• Hydrolysis can be effected using conc: HCI or conc H2S04
• Primary nitro alkanes give carboxylic acids.
• Secondary nitro alkanes give ketones.
• Tertiary nitro alkanes have no reaction.

Question 4.
Write about the classification of amines
Answer:

Question 5.
Write notes on structure of amines
Answer:

• Like ammonia, nitrogen atom of amines is trivalent.
• It carries a lone pair of electron.
• Nitrogen atom is sp³ hybridised.
• Out of four sp³ hybridised orbitals, three sp3 orbitals overlap with hydrogen or carbon of alkyl groups.
• Fourth sp³ orbital contains a lone pair of electron.
• Amines posses pyramidal geometry.
• Due to the presence of lone pair of electron the C-N-H or C-N-C bond angle in less than the normal
  tetra hedral bond angle 109.5^O
• In trimethyl amine the C-N-C bond angle is 108^O which is lower than the tetrahedral angle and higher than the H-N-H bond angle of 107^O in ammonia.
• This increase is due to the repulson between the bulky methyl groups.
  

Question 6.
Prepare aniline from
(i) chloro benzene
(ii) Phenol
(iii) nitro benzene
Answer:

Question 7.
Derive an expression for basic strength of amines.
Answer:

• In aqueous solutions the following equilibrium exists.
• It lies far to the left, hence amines are weak bases compared to NaOH.

• Basicity constant K _bis a measure of the extent to which the amine accepts the hydrogen – ion H⁺ from water.
• Larger the value of K_b,, smaller is the value of p^K_b and stronger is the base.

Question 8.
An organic compound (A) on reduction gives compound (B). (B) on treatment with CHCl₃ and alcoholic KOH gives (C) . (C) on catalytic reduction gives N-methyl aniline. Identify A, B, C and write its equation.
Answer:

Question 9.
Convert aniline into
(i) p-bromo aniline
(ii) p-nitro aniline
Answer:
(i) To get mono bromo derivatives, -NH₂ is first acylated to reduce its activity

(ii) To get para product, the -NH₂ group is protected by acetylation with acetic anhydride. Then the nitrated product is hydrolysed to form the product

VIII. Five mark questions

Question 1.
Explain structural isomerism exhibited by nitro alkanes
Answer:

Question 2.
Explain the reduction reactions of nitro benzene in various medium.
Answer:

Question 3.
An Organic compound (A) – C₇H₇NO on treatment with Br₂ and KOH gives an amine (B), which gives carbylamine test. (B) upon diazotization to give (C). (C) on coupling_with p-cresol to give
compound (D). Identify A, B, C and D with necessary reaction.
Hofmann’s degradation reaction
Compound (A is Benzamide
Compound B is Aniline (primary amine)
Diazotization reaction:

Question 4.
Write a short note an Libermann’s nitroso test ?
Answer:
Alkyl and aryl secondary amines react with nitrous acid to give N-nitroso amine

Question 5.
An Organic compound (A) – CNCl react with methyl magnesium Bromide to give, compound B – (C₂H₃N). B – upon catalytic reduction to give compound C – (C₂H₇N). C gives carbylamine test, Identify compound A, B and C and write the reactions.
Answer:
An organic Compound A is cyanogon chloride react with methylmagnesium Bromide to give compound B – ethanenitrite – (C₂H₃N)

B – upon catalytic reduction to give compound C – Ethananxuie (C₂H₇N)

Question 6.
Explain the action of nitrous acid on three types of amine.
Answer:
Three classes of amines react differently with nitrous acid which is prepared in situ from a mixture of NaNO₂ and Hcl.
(a) Primary amines:
Ethylamine reacts with nitrous acid to form ethyldiazonium chloride which is unstable. It is converted into ethanol by liberating N₂.

• Aniline reacts with nitrous acid at low temperature (273 – 278K) to give benzene diazonium chloride.
• Benzene diazonium chloride is stable for short time and slowly decomposes even at low
  temperatures.

b) Secondary amines:
Alkyl and aryl secondary amines react with nitrous acid to give N-nitroso amine, which is insoluble in water.

c) Tertiary amines: :
Aliphatic tertiary amine reacts with nitrous acid to form trialkyl ammonium nitrite salt which is
insoluble in water.

Aromatic tertiary amine reacts with nitrous acid at 273K to give p-nitroso compound

Question 7.
Explian:
(i) Sand mayer reaction
(ii) Gaftermann reaction
(iii) Baltz – schiemann reaction
Answer:
(i) Sand mayer reaction:

(ii) Gattermann reaction:

(iii) Baltz – Schiemann Reaction:

Question 8.
Convert benzene diazonium chloride into:
(i) benzene
(ii) iodo benzene
(iii) phenol
(iv) nitrobenzene
(v) phenyl hydrazine

Question 9.
Explain coupling reactions of benzene diazonium chlorid
Answer:

• Benzene diazonium chloride reacts with electron rich aromatic compounds like phenol, aniline to form brightly coloured azo compounds.
• Coupling generally occurs at the para position.
• If para position is occupied then coupling occurs at the ortho position.
• Coupling tendency is enhanced if an electron donating group is present at the para-position to – N₂ Cl group.
• This is an electrophilic substitution.

Question 10.
Write notes on
(i) Thrope nitrile condensation
(ii) Levine and Hauser acetylation
Answer:
(i) Thrope nitrile condensation :
Self condensation of two molecules of alkylnitrile containing α-H atom in the presence of sodium to form iminonitrile.

Nitriles containing α-H atom undergo condensation with esters in the presence of sodamide in ether to form ketonitriles,
This is known as Levine – Hauser acetylation.
This reaction involves the replacement of ethoxy (OC₂H₅) group by methylnitrile (-CH₂ CN) group and is called as cyanomethylation reaction.

Question 11.
An aromatic nitro compound (A) on reduction with Sn/HCl gives compound (B) C₆H₇N which on treatment with Benzoyl chloride in the presence of pyridine to give compound (C). Compound (B) on treatment with CH₃Br to give compound (D) which further reacts with NaNO₂ /HCl to give compound (E) with yellow oil liquid. Identify (A) to (E) and write the
reactions.
Compound (A) is nitro benzene on reduction with Sn/HCl gives compound (B) is Aniline

Compound (B) is Aniline reacts with Benzoyl chloride in the presence of pyridine to give compound
(C) is N – phenyl benzamide
Schotten- Baumann reaction

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf Chapter 12 Employee Training Method Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 12 Employee Training Method

12th Commerce Guide Employee Employee Training Method Text Book Back Questions and Answers

I. Choose The Correct Answer.

Question1.
Off the job training is given
a) In the classroom
b) On off days
c) Outside the factory
d) In the playground
Answer:
c) Outside the factory

Question 2.
Vestibule training is provided
a) On the job
b) In the classroom
c) In a situation similar to the actual working environment
d) By the committee
Answer:
c) In a situation similar to the actual working environment

Question 3.
Improves Skill Levels of employees to ensure better job performance
a) Training
b) Selection
c) Recruitment
d) Performance appraisal
Answer:
a) Training

Question 4.
When trainees are trained by a supervisor or by a superior at the job is called
a) Vestibule training
b) Refresher training
c) Roleplay
d) Apprenticeship training
Answer:
d) Apprenticeship training

Question 5.
……… is useful to prevent skill obsolescence of employees
a) Training
b) Job analysis
c) Selection
d) Recruitment
Answer:
a) Training

Question 6.
Training methods can be classified into training …….. training.
a) Job rotation and Job enrichment
b) On the Job and Off the Job
c) Job analysis and Job design
d) Physical and mental
Answer:
b) On the job and Off the Job

Question 7.
Elaborate discussion on specific topic comes under ……………. method of training.
a) Understudy
b) Coaching
c) Conferences
d) Counselling
Answer:
c) Conferences

II. Very Short Answer Questions.
Question 1.
What is meant by training?
Answer:
Training is the act of increasing/enhancing the new skill of problem-solving activity and technical knowledge of an employee for doing the jobs themselves.

Question 2.
What is Mentoring training method?
Answer:

• Mentoring is the process of sharing knowledge and experience of an employee.
• The focus in this training is on the development of attitude of trainees.
• It is always done by senior person, it is also one-to-one interaction like coaching.
• The mentor is responsible for the providing necessary support to trainees, and feedback

Question 3.
What is Roleplay?
Answer:
Under this method, trainees have explained the situation and assigned roles. They have to act out the roles assigned to them without any rehearsal.

Question 4.
State e-learning method.
Answer:

• E-learning is the use to technological process of a traditional classroom or office.
• It refers to online learning or web-based training.
• Web-based training is any where and any time information can pass over the internet.

III. Short Answer Questions.

Question 1.
What is vestibule training?
Answer:
Vestibule training is the training of employees in an environment similar to the actual work environment. This type of training is given to avoid any damage or loss to machinery in the actual place by trainees.

Question 2.
What do you mean by on the job training?
Answer:

• On the job training refers to the training which is given to the employee at the workplace by his immediate supervisor.
• It is based on the principle of “Learning by Doing and Learning while Earning”.
• It is suitable for imparting skills that can be learnt in a relatively short period of time.

Question 3.
Write down various steps in a training programme.
Answer:

Question 4.
Write short note on trainer and trainee.
Answer:
A person who is learning and practicing the skills of a particular job is called a trainee. Trainees should be selected on the basis of self-interest and recommendation by the supervisor.

A trainer is a person who teaches skills to the employee and prepares them for job activity. Trainers may be supervisor, co-workers, HR staffs, specialists in the other parts of the company, outside consultants.

IV. Long Answer Questions

Question 1.
Discuss various types of training.
Answer:
According to Edwin B. Flippo” Training is the act of increasing the knowledge and skills of an employee for doing particular jobs”. Training may be mainly divided into:

• On the job training
• Off the job training

(a) On the Job Training: On the job training refers to the training which is given to the employee at the workplace. The following are the on the job training methods.

1. Coaching Method: In this method of training, the superior teaches or guides the new employee about the knowledge and skills.
2. Mentoring Method: Mentoring is the process of sharing the knowledge and experience of an employee.
3. Internship Training Method: A superior gives training to subordinates or understudy like an assistant to a manager.

(b) Off the Job Training: It is the training method wherein the workers learn the job role away from the actual workplace. The following are types of off the job training:

1. Lecture Method: Under this method, trainees are educated about concepts, theories, principles in any particular area.
2. E-learning Method: E-learning is the use of technological processes to access a traditional classroom or office.

Question 2.
What are the differences between on the job training and off the job training? [MALS] [CDM]
Answer:

Question 3.
Explain the benefits of training
Answer:
1. Benefits to the Organization:

• It improves the skill of employees and increases productivity.
• It reduces wastages of materials and idle time.
• It minimizes the time for supervision.
• It reduces the frequent accidents at the workplace and consequent payment of compensation.

2. Benefits to the Employees:

• It increases the knowledge, skill of the employees.
• It enables him to gain promotion in a shorter time.
• It improves the employee’s productivity.
• Employees get higher earnings through incentives and rewards.

3. Benefits of Customer:

• Customers get better quality of product and service.
• Customers get innovative products or value-added or feature-rich products.

12th Commerce Guide Employee Training Method Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
…………………………… training refers to the training which is given to the employee at the workplace by his immediate superior.
a) On the job
b) Off the job
c) PMKVY
d) NOTA
Answer:
a) On the job

Question 2.
…………………………… training is the training wherein the workers learn the job role away from the actual work floor.
a) Off the job
b) On the job
c) STEP
d) All of these
Answer:
a) Off the job

Question 3.
On the job training is given to the employee at the ____________
(a) college
(b) home
(c) workplace
(d) none of the above
Answer:
(c) workplace

Question 4.
Pick the odd one out:
a) Job Rotation
b) Job Instruction Techniques
c) Apprenticeship
d) Field Trip
Answer:
d) Field Trip

Question 5.
Which one of the following is not correctly matched?
a) Conference method – Seminar
b) Role play method – Assigned Roles
c) Coaching method – Subordinate Teaches Superior
d) Committee Assignment – Particular issue assigned
Answer:
c) Coaching method – Subordinate Teaches Superior

Question 6.
Which is the correct statement?
i) Training improves the employee’s productivity
ii) Employees get rewards and incentives.
iii) It decrease the morale of the employees
a) (i) is correct
b) (ii) is correct
c) (i) and (ii) are correct
d) (ii) and (iii) are correct

II. Match The Following.

Question 1.
Match List-I with List-II

a) (i) – 4 (ii) – 3 (iii) -1 (iv) – 2
b) (i) – 3 (ii) – 2 (iii) – 4 (iv) -1
c) (i) – 2 (ii) – 4 (iii)-3 (iv) – 1
d) (i) – 1 (ii) – 3 (iii) – 2 (iv) – 4
Answer :
a) (i) – 4 (ii) – 3 (iii) -1 (iv) – 2

Question 2.

Match List-I with List-II

a) (i) – 3 (ii) – 4 (iii) – 2 (iv) – 1
b) (i) – 2 (ii) – 3 (iii) – 1 (iv) – 4
c) (i) – 4 (ii) – 3 (iii) – 2 (iv) – 1
d) (i) – 1 (ii) – 2 (iii) – 3 (iv) – 4
Answer :
b) (i) – 2 (ii) – 3 (iii) -1 (iv) – 4

III. Assertion and Reason.

Question 1.
Assertion (A): Under this method, trainees have explained the situation and assigned roles
Reason (R): They have to act out the roles assigned to them without any rehearsal
a) (A) is True (R) is False
b) (A) is False (R) is True
c) Both (A) and(R) are False
d) Both (A) and (R) are True
Answer :
d) Both (A) and (R) are True

IV. Very Short Answer Questions.

Question 1.
What is the Job Rotation method?
Answer:

• Job Rotation is an important method of training for broadening the knowledge of executives.
• Under this method, a trainee is periodically shifted from one work (Department) to another work (Department).

Question 2.
Write a note on internship training.
Answer:
When a superior gives training to subordinates, it is called internship training. The subordinates learn through experience and observation by participating in handling day-to-day problems.

Question 3.
What is the Internal ship [under study] Training method?
Answer:

• A superior gives training to subordinates or understudy like an assistant to a manager or director.
• The subordinate learns through experience and observation by participating in handling day-to-day problems.

Question 4.
What is the case study method?
Answer:

• Trainees have described a situation which stimulate their interest to find solution.
• They have to use their theoretical knowledge and practical knowledge to find solution to the problems presented.

Question 5.
What is the Group Discussion method?
Answer:

• Under this method participants are divided into various groups.
• Each group has to prepare solutions (issue provided to them) after deep discussion with their group members.

Question 6.
Demonstration Training method – Explain.
Answer:

• It is a visual display of how something works or how to do something.
• Demonstration involves showing by reason or proof explaining or making clear by use of examples or experiments.

V. Long Answer Questions.

Question 1.
Explain any three purposes of training.
Answer:
The purposes of training are as follows:

1. Improved Quality of Work- Training helps to focus on a specific area and enables the employees to increase the quality of work carried out by them.
2. Prevention of obsolescence- Training helps to learn more knowledge regarding the latest techniques and trends.
3. Improved Safety Measures- Employees gain awareness about the risks involved in the job and safety measures to be adopted through a proper training program.

Question 2.
What are the steps in Designing a Training Programme?
Answer:
1) Whom to Train?

• Training Department has to determine the candidates few whom Training should be imparted.

2) Who is a Trainee?

• A person who is learning and practicing the skills of a particular job is called a “Trainee”.

3) Who are the Trainers?

• A trainer is a person who teaches skills to the employee and prepare them for a job.

4) What method will be used?

• The training segment should decide the appropriate method of training among the various methods of training available.

5) What should be the level of training?

• Training Department should decide the level of training to be imparted to the employees.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 9 Electro Chemistry Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 9 Electro Chemistry

12th Chemistry Guide Electro Chemistry Text Book Questions and Answers

Part – I Text Book Evaluation

I. Choose the correct answer

Question 1.
The number of electrons that have a total charge of 9650 coulombs is ………..
(a) 6.22 × 10²³
(b) 6.022 × 10²⁴
(c) 6.022 × 10²²
(d) 6.022 × 10^-34
Answer:
(c) 6.022 × 10²²
Hint: IF = 96500 C = 1 mole of e^– = 6.023 × 10²³ e^–
9650 C = \(\frac{6.22 \times 10^{23}}{96500} \times 9650\) = 6.022 × 10²²

Question 2.
Consider the following half cell reactions:
Mn²⁺ + 2e^– → Mn E° = – 1.18V
Mn²⁺ → Mn³⁺ + e^– E = – 1.51V
The E for the reaction 3Mn²⁺ → Mn+2Mn³⁺, and the possibility of the forward reaction are respectively.
(a) 2.69V and spontaneous
(b) – 2.69 and non spontaneous
(c) 0.33V and Spontaneous
(d) 4.18V and non spontaneous
Answer:
(b) – 2.69 and non spontaneous
Hint: Mn⁺ + 2e^– → Mn(E⁰_red) = 1.18V
2[Mn²⁺ → Mn³⁺ + e^–] (E⁰_ox) = – 1.51V
3Mn²⁺+ → Mn³⁺ + 2Mn³⁺ + (E⁰_cell) = ?
E⁰_red = (E⁰_ox) + (E⁰_cell)
= – 1.51 – 1.18 and non spontaneous
= – 2.69 V
Since E° is – ve ∆G is +ve and the given forward cell reaction is non – spontaneous.

Question 3.
The button cell used in watches function as follows
Zn_(s) + Ag₂O_(s) + H₂O₍₁₎ \(\rightleftharpoons\) 2Ag_(s) + Zn²⁺_(aq) + 2OH^–_(aq) the half cell potentials are
Ag₂O_(s) + H₂O₍₁₎ + 2e^– →2Ag_(S) + 2OH^–_(aq) E° = 034V. The cell potential will be
(a) 0.84V
(b) 1.34V
(c) 1.10V
(d) 0.42V
Answer:
(c) 1.10V
Hint: Anodic oxidation: (Reverse the given reaction)
(E⁰_ox ) = 0.76V cathodic reduction
E⁰_cell = (E⁰_ox) + (E⁰_red) = 0.76 + 0.34 = 1.1V

Question 4.
The molar conductivity of a 0.5 mol dm^-3 solution of AgNO₃ with electrolytic conductivity of 5.76 × 10^-3S cm^-1at 298 K is ………….
(a) 2.88 S cm² mo1^-1
(b) 11.52 S cm² mol^-1
(c) 0.086 S cm² mol^-1
(d) 28.8 S cm² mol^-1
Answer:
(b) 11.52 S cm² mol^-1
Solution:

Question 5.

Calculate A⁰_HOAC using appropriate molar conductances of the electrolytes listed above at infinite dilution in water at 25°C.
(a) 517.2
(b) 552.7
(c) 390.7
(d) 217.5
Answer:
(c) 390.7
Hint:

Question 6.
Faradays constant is defined as
(a) charge carried by I electron
(b) charge carried by one mole of electrons
(c) charge required to deposit one mole of substance
(d) charge carried by 6.22 × 10¹⁰ electrons
Answer:
(b) charge carried by one mole of electrons
Solution:
IF = 96500 C = 1 charge of mole of e^– = charge of 6.022 × 10²³ e^–

Question 7.
How many faradays of electricity are required for the following reaction to occur
MnO^–₄ → Mn²⁺
(a) 5F
(b) 3F
(C) IF
(d) 7F
Answer:
(a) 5F
Hint:
7MnO^–₄ + 5e^– → Mn²⁺ + 4H₂O
5 moles of electrons i.e., 5F charge is required.

Question 8.
A current strength of 3.86 A was passed through molten Calcium oxide for 41 minutes and 40 seconds. The mass of Calcium in grams deposited at the cathode is (atomic mass of Ca is 40g / mol and IF = 96500C).
(a) 4
(b) 2
(c) 8
(d) 6
Answer:
(b) 2
Solution:
m = ZIt
41mm 40sec = 2500 seconds
= \(\frac { 40 × 3.86 × 2500 }{ 2 x 96500 }\)
Z = \(\frac { m }{ n × 96500 }\) = \(\frac { 40 }{ 2 × 96500 }\)
= 2g

Question 9.
During electrolysis of molten sodium chloride, the time required to produce 0.1 mol of chlorine gas using a current of 3A is ………..
(a) 55 minutes
(b) 107.2 minutes
(c) 220 minutes
(d) 330 minutes
Answer:
(b) 107.2 minutes
Solution:
\(\frac { m }{ ZI }\) (mass of 1 mole of Cl₂ gas = 71)
t = \(\frac { m }{ ZI }\) mass of 0.1 mole of Cl₂ gas = 7.1 g mol^-1)

Question 10.
The number of electrons delivered at the cathode during electrolysis by a current of 1 A in 60 seconds is (charge of electron = 1.6 × 10^-19C)
(a) 6.22 × 10²³
(b) 6.022 × 10²⁰
(c) 3.75 × 10²⁰
(d) 7.48 × 10²³
Answer:
(c) 3.75 × 10²⁰
Solution:
Q = It
= 1A × 60S
96500 C charge 6.022 × 10²³ electrons
60 C charge = \(\frac{6.022 \times 10^{23}}{96500} \times 960\)
= 3.744 × 10²⁰ electrons

Question 11.
Which of the following electrolytic solution has the least specific conductance?
(a) 2N
(b) 0.002N
(c) 0.02N
(d) 0.2N
Answer:
(b) 0.002N
Solution:
In general, specific conductance of an electrolyte decreases with dilution. SO, 0.002N solution has least specific conductance.

Question 12.
While charging lead storage battery
(a) PbSO₄ on cathode is reduced to Pb
(b) PbSO₄ on anode is oxidised to PbO₄
(c) PbSO₄ on anode is reduced to Pb
(d) PbSO₄ on cathode is oxidised to Pb
Answer:
(c) PbSO₄ on anode is reduced to Pb.
Solution:
Charging: anode: PbSO₄(s) + 2e^– → Pb (s) + SO₄^-2 (aq)
Cathode: PbSO₄(s) + 2H₂O (1) → PbO₂ (s) + SO₄^-2 (aq) + 2e_–

Question 13.
Among the following cells
I. Leclanche cell
II. Nickel – Cadmium cell
III. Lead storage battery
IV. Mercury cell
Primary cells are …………
(a) I and IV
(b) I and III
(c) III and IV
(d) II and III
Answer:
(a) I and IV

Question 14.
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
(a) Zinc is lighter than iron
(b) Zinc has lower melting point than iron
(c) Zinc has lower negative electrode potential than iron
(d) Zinc has higher negative electrode potential than iron
Answer:
(d) Zinc has higher negative electrode potential than iron
Solution:
E⁰_Zn⁺|Zn = – 0.76V and E⁰_Fe²⁺|Fe = 0.44V. Zinc has higher negative electrode potential than iron, iron cannot be coated on zinc.

Question 15.
Assertion: pure iron when heated in dry air is converted with a layer Of rust.
Reason: Rust has the composition Fe₃O₄
(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) assertion is true but reason is false
(d) both assertion and reason are false.
Answer:
(d) both assertion and reason are false.
Solution:
Both are false

1. Dry air has no reaction with iron
2. Rust has the composition Fe₂O₃ x H₂O

Question 16.
In H₂ – O₂ fuel cell the reaction occur at cathode is ……….
(a) O₂(g) + 2H₂O (l) + 4e^– → 4OH^–(aq)
(b) H⁺(aq) + OH^–(aq) → H₂O (l)
(c) 2H₂(g) + O₂(g) → 2H₂O (g)
(d) H⁺ + e^– → 1/2 H₂
Answer:
(a) O₂(g) + 2H₂O (l) + 4e^– → 4OH^–(aq)
Solution:
(a) O₂(g) + 2H₂O (l) + 4e^– → 4OH^–(aq)

Question 17.
The equivalent conductance of M/36 solution of a weak monobasic acid is 6mho cm² and at infinite dilution is 400 mho cm². The dissociation constant of this acid is ………….
(a) 1.25 x 10^-16
(b) 6.25 x 10 ^-6
(c) 1.25 x 10^-4
(d) 6.25 x 10^-5
Answer:
(b) 6.25 x 10 ^-6
Hint: α = \(\frac { 6 }{ 400 }\)
K_a = α²C = \(\frac { 6 }{ 400 }\) x \(\frac { 6 }{ 400 }\) x \(\frac { 1 }{ 36 }\)
= 6.25 x 10^-6

Question 18.
A conductivity cell has been calibrated with a 0.01M, 1:1 electrolytic solution (specific conductance (K = 1.25 x 10^-3 S cm^-1 ) in the cell and the measured resistance was 800Ω at 25⁰ C . The cell constant is,
(a) 10^-1 cm^-1
(b) 10^-1 cm^-1
(c) 1 cm^-1
(d) 5.7 x 10^-12
Answer:
(c) 1 cm^-1
Hint: R = p.\(\frac { 1 }{ A }\)
Cell constant = \(\frac { R }{ ρ }\) = k.R \((\frac { 1 }{ ρ } =k)\) = 1.25 x 10^-3 f^-1cm^-1 x 800Ω = 1cm^-1

Question 19.
Conductivity of a saturated solution of a sparingly soluble salt AB (1:1 electrolyte) at 298K is 1.85 x 10^-5 S m^-1. Solubility product of the saltAB at 298K (Λ⁰_m)_AB = 14 x 10^-3 S m² mol^-1.
(a) 5.7 x 10^-2
(b) 1.32 x 10¹²
(c) 7.5 x 10^-12
(d) 1.74 x 10^-12
Answer:
(d) 1.74 x 10^-12
Solution:

Question 20.
In the electrochemical cell: Zn|ZnSO₄ (0.01M)||CuSO₄ (1.0M)|Cu , the emf of this Daniel cell is E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that CuSO₄ changed to 0.0 1M, the emf changes to E₂. From the followings, which one is the relationship between E₁ and E₂?
(a) E₁ < E₂
(b) E₁ > E₂
(c) E₂ = 0↑E₁
(d) E₁ = E₂
Answer:
(b) E₁ > E₂
Solution:

Question 21.
Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below:

Then the species undergoing disproportional is …………..
(a) Br₂
(b) BrO₄^–
(c) BrO₃^–
(d) HBrO
Answer:
(d) HBrO
Hint:

(E_cell)_A = – 1.82 + 1.5 = – 0.32V
(E_cell)_B = – 1.5 + 1.595 = + 0.095V
(E_cell)_C = 1.595 + 1.0652 = – 0.529V
The species undergoing disproportionation is HBrO

Question 22.
For the cell reaction
2Fe³⁺(aq) + ^2I(aq) → 2Fe²⁺ (aq) + I₂(aq)
EC⁰_cell = 0.24V at 298K. The standard Gibbs energy (∆ G⁰ ) of the cell reactions is …………
(a) – 46.32 KJ mol^-1
(b) – 23.16 KJ mol^-1
(c) 46.32 KJ mol^-1
(d) 23.16 KJ mor^-1
Answer:
(a) – 46.32 KJ mol^-1
Solution:
n = 2; E⁰_cell = 0.24V; ∆G = ?; F = 96500C
∆G⁰ = – nFE°
∆G⁰= – 2 × 96500 × 0.24
∆G⁰ = – 46320 J mol^-1
∆G⁰ = – 46.32 KJ mol^-1

Question 23.
A certain current liberated 0.504gm of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution?
(a) 31.75
(b) 15.8
(c) 7.5
(d) 63.5
Answer:
(b) 15.8
Solution:
m₁ = 0.504 g m₂ = ?
e₁ = 1.008 e₂ = 31.77
\(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\mathrm{e}_{1}}{\mathrm{e}_{2}}\)
∴ \(\mathrm{m}_{2}=\frac{\mathrm{m}_{1}}{\mathrm{e}_{1}} \cdot \mathrm{e}_{2}=\frac{0.504 \times 31.77}{1.008}=15.885 \mathrm{~g}\)

Question 24.
A gas X at 1 atm is bubble through a solution containing a mixture of 1MY^– and 1MZ^-1 at 25°C . If the reduction potential of Z > Y> X, then
(a) Y will oxidize X and not Z
(b) Y will oxidize Z and not X
(c) Y will oxidize both X and Z
(d) Y will reduce both X and Z
Answer:
(a) Y will oxidize X and not Z
Solution:

Zoxidises Yand X
Y oxidises X and reduces Z (does not oxidise Z)
X nduces Y and Z

Question 25.
Cell equation: A²⁺ + 2B^– → A²⁺ + 2B
A²⁺ + 2e^– → AE° = + 0.34V and log₁₀ K = 15.6 at 300K for cell reactions find E° for
B¹ + e^– → B
(a) 0.80
(b) 1.26
(c) – 0.54
(d) – 10.94
Answer:
(a) 0.80
Solution:
A²⁺ + 2B^– → A²⁺ + 2B E°_cell
Half reaction anode A → A²⁺ + 2e^–
E°_ox = -.034V [Given : A²⁺ + 2e^– → A E° = +0.34V]
Cathode 2B⁺ + 2e^– → 2B E°_red = ?]
log₁₀K = 156; T = 300K; n = 2;
F = 96500C;
R 8.314 JK^-1 mol^-1
∆G = – 2.303 RT log K
:. -nFE° = – 2.303 RT log K

∴ E°_red = E°_cell – E°_oxid
= 0.4643 – (-0.34)
= 0.4643 + 0.34
E° 0.8043 V = 0.80V

II. Short Answer Questions

Question 1.
Define anode and cathode
Answer:

1. Anode: The electrode at which the oxidation occur is called anode.
2. Cathode: The electrode at which the reduction occur is called cathode.

Question 2.
Why does conductivity of a solution decrease on dilution of the solution?
Answer:
Conductivity always decreases with decrease in concentration (on dilution of the solution) both for weak as much as for strong electrolytes. ¡t is because the number of ions per unit volume that carry the current is a solution decreases on dilution.

Question 3.
State Kohlrausch Law. How is it useful to determine the molar conductivity of weak electrolyte at infinite dilution.
Answer:
Kohlrausch’s law:
It is defined as, at infinite dilution the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions.

Determination of the molar conductivity of weak electrolyte at infinite dilution.
It is impossible to determine the molar conductance at infinite dilution for weak electrolytes experimentally. However, the same can be calculated using Kohlraushs Law. For example, the molar conductance of CH₃COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCI, NaCI and CH₃COONa.
Λ°_CH3COONa = λ°_Na⁺ + λ°_CH3COONa …….(1)
Λ°_HCl = λ°_H⁺ + λ°_Cl^– ………………(2)
Λ°_NaCl = λ°_Na⁺ + λ°_Cl^– …………….(3)
Equation (1) + Equation (2) – Equation (3) gives,
(Λ°_CH3COONa) + (Λ°_HCl) – (Λ°_NaCl) = λ°_H⁺ + λ°_CH3COONa = Λ°_CH3COONa

Question 4.
Describe the electrolysis of molten NaCI using inert electrodes.
Answer:
1. The electrolytic cell consists of two iron electrodes dipped in molten sodium chloride and they are connected to an external DC power supply via a key.

2. The electrode which is attached to the negative end of the power supply is called the cathode and the one is which attached to the positive end is called the anode.

3. Once the key is closed, the external DC power supply drives the electrons to the cathode and at the same time
pull the electrons from the anode.

Cell reactions:
Na⁺ ions are attracted towards cathode, where they combines with the electrons and reduced to liquid sodium.
Cathode (reduction)
Na⁺_(I) + e^–Na₍₁₎
E⁰ = – 2.7 1V
Similarly, Cl^– ions are attracted towards anode where they losses their electrons and oxidised to chlorine gas. Anode (oxidation)
2Cl^–₍₁₎ Cl_2(g) + 2e^–
E° = – 1 .36V
The overall reaction is,
2Na⁺_(l) + 2Cl^–_(l) → 2Na_(l) + Cl_2(g) (g)
E⁰ = 4.07 V

The negative E° value shows that the above reaction is a non spontaneous one. Hence, we have to supply a voltage greater than 4.07V to cause the electrolysis of molten NaCI. In electrolytic cell, oxidation occurs at the anode and reduction occur at the cathode as in a galvanic cell, but the sign of the electrodes is the reverse i.e., in the electrolytic cell cathode is -ve and anode is +ve.

Question 5.
State Faraday’s Laws of electrolysis.
Answer:
Faraday’s laws of electrolysis:
1. First law:
The mass of the substance (M) liberated at an electrode during electrolysis is directly proportional to the quantity of charge (Q) passed through the cell. M α Q

2. Second law:
When the same quantity of charge is passed through the solutions of different electrolytes, the amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents. M α Z

Question 6.
Describe the construction of Daniel cell. Write the cell reaction.
Answer:
The separation of half reaction is the basis for the construction of Daniel cell. It consists of two half cells.

Oxidation half cell:
The metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker.

Reduction half cell:
A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker.

Joining the half cells :
The zinc and copper strips are externally connected using a wire through a switch (K) and a load (ex : volt meter)

The electrolytic solution in the cathodic and anodic compartment are connected using an inverted U tube containing agar-agar gel mixed with an inert electrolyte like Kcl, Na₂SO₄ etc.,
This acts as the salt bridge.
When the switch (K) closes the circuit, the electrons flow from zinc strip to copper strip.
This is due to the following redox reactions.

Anodic oxidation:
Zinc strip acts as the anode.
Here oxidation occurs.
Zinc is oxidised to Zn²⁺ ions and electrons.
Zn²⁺ ions enter the solution and electrons enter the zinc metal strip; then flow through the external wire and enter the copper strip.
Electrons are liberated at zinc electrode and hence it is negative.

Cathodic reduction:
Copper strip acts as the cathode.
Here reduction occurs. .
Cu²⁺ ions are reduced to copper metal.
Cu²⁺ ions in the solution accepts electrons flowing through the circuit from zinc to copper strip and gets deposited on the electrode as copper metal.
Here electrons are consumed and hence it is positive.

Salt bridge:

The electrolytes present in two half cells are connected using salt bridge. To maintain electrical neutrality Cl- ions (from KCl) move from the salt bridge.
Due to cathodic reduction, cathodic compartment contains more number of SO₄^2- ions in the solution and becomes
negatively charged. To maintain electrical neutrality K⁺ ions (from KCl) move from the salt bridge.

Completion of circuit:

Electrons flow from negatively charged zinc anode into positively charged copper cathode through the external wire.
At the same time anions move towards anode and cations move towards cathode.
This completes the circuit.

Consumption of electrodes:

As the Daniel cell operates, the mass of zinc electrode gradually decreases, while the mass of copper electrode increases.
Hence the cell will function until the entire metallic zinc is converted into Zn²⁺ ions or the entire Cu²⁺ ions are converted into metallic copper.

Question 7.
Why is anode in galvanic cell considered to be negative and cathode positive electrode?
Answer:
A galvanic cell works basically in reverse to an electrolytic cell. The anode is the electrode where oxidation takes place, in a galvanic cell, it is the negative electrode, as when oxidation occurs, electrons are left behind on the electrode.

The anode is also the electrode where metal atoms give up their electrons to the metal and go into solution. The electron left behind on it render ¡t effectively negative and the electron flow goes from it through the wire to the cathode.

Positive aqueous ions in the solution are reduced by the incoming electrons on the cathode. This why the cathode is a positive electrode, because positive ions are reduced to metal atoms there.

Question 8.
The conductivity of a 0.01%M solution of a 1:1 weak electrolyte at 298K is 1.5 x 10^-4 S cm^-1.

1. molar conductivity of the solution
2. degree of dissociation and the dissociation constant of the weak electrolyte

Given that
λ⁰cation = 248.2 S cm² mol^-1
λ⁰anion = 51.8 S cm² mol^-1
Answer:
1. Molar conductivity
C = 0.01M
k = 1.5 × 10^-4 S cm^-1
(or)
K = 1.5 × 10^-2 S m^-1
\(\frac{\kappa \times 10^{-3}}{\mathrm{C}}\) S m^-1 mol^-1 m³ = \(\frac{1.5 \times 10^{-2} \times 10^{-3}}{0.01}\) S m² mol^-1
Λ_m = 1.5 × 10^-3 s m^-1

2. Degree of dissociation
α = \(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\infty}^{\circ}}\) (or) α = \(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\)
= (248.2 + 51.8)S cm² mol^-1
= 300 S cm² mol^-1
K_a = \(\frac{\alpha^{2} C}{1-\alpha}\)
= \(\frac{(0.05)^{2}(0.01)}{1-0.05}\)
K_a = 2.63 × 10^-5

Question 9.
Which of 0.1M HCl and 0.1 M KCl do you expect to have greater molar conductance and why?
Answer:
Compare to 0.1M HCl and 0.1 M KCl, 0.1M HCl has greater molar conductance.

1. Molar conductance of 0.1M HCl = 39.132 × 10^-3 S m² mol^-1.
2. Molar conductance of 0.1 M KCl = 12.896 × 10^-3 S m² mol^-1

Because, H⁺ ion in aqueous solution being smaller size than K⁺ ion and H⁺ ion have greater mobility than K ion. When mobility of the ion increases, conductivity of that ions also increases. Hence, 0. 1M HCI solution has greater molar conductance than 0.1 M KCI solution.

Question 10.
Arrange the following solutions in the decreasing order of specific conductance.

1. 0.01M KCI
2. 0.005M KCI
3. 0.1M KCI
4. 0.25 M KCI
5. 0.5 M KCI

Answer:
0.005M KCl > 0.01M KCI > 0.1M KCI > 0.25KCl > 0.5 KCI.
Specific conductance and concentration of the electrolyte. So if concentration decreases, specific conductance increases.

Question 11.
Why is AC current used instead of DC in measuring the electrolytic conductance?
Answer:
1. AC current to prevent electrolysis of the solution.

2. If we apply DC current to the cell the positive ions will be attracted to the negative plate and the negative ions to the positive plate. This will cause the composition of the electrolyte to change while measuring the equivalent conductance.

3. So DC current through the conductivity cell will lead to the electrolysis of the solution taken in the cell. To avoid such a electrolysis, we have to use AC current for measuring equivalent conductance.

Question 12.
0.1M NaCI solution is placed in two different cells having cell constant 0.5 and 0.25cm^-1 respectively. Which of the two will have greater value of specific conductance.
Answer:
The specific conductance values are same. Because the reaction (cation) of cell constant does not change.

Question 13.
A current of 1 .608A is passed through 250 mL of 0.5M solution of copper sulphate for 50 minutes. Calculate the strength of Cu²⁺ after electrolysis assuming volume to be constant and the current efficiency is 100%.
Answer:
Given, I = I .608A
t = 50 min (or) 50 × 60 = 3000 S
V = 250 mL
C = 0.5M
η = 100%
The number of Faraday’s of electricity passed through the CuSO₄ solution
Q = It
= Q = 1.608 × 3000
Q = 4824C
Number of Faraday’s of electricity = \(\frac { 4824C }{ 96500C }\) = 0.5F
Electrolysis of CuSO₄
Cu²⁺_(aq) + 2e^– → Cu_(s)
The above equation shows that 2F electricity will deposit 1 mole of Cu²⁺
0.5F electicity will deposit \(\frac { 1mol }{ 2F }\) × 0.5F = 0.025 mol
Initial number of molar of Cu²⁺ in 250 ml of solution = \(\frac { 1mol }{ 250mL }\) × 250mL = 0.125 mol
Number of molar of Cu²⁺ after electrolysis 0.125 – 0.025 = 0.1 mol
Concentration of Cu²⁺ = \(\frac { 0.1mol }{ 250mL }\) × 1000 mL = 0.4 M

Question 14.
Can Fe³⁺ oxidises Bromide to bromine under standard conditions?
Given: \(\mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^{\mathrm{o}}=0.771\)
\(\mathrm{E}_{\mathrm{Br}_{2} \mid \mathrm{Br}}^{\mathrm{O}}=1.09 \mathrm{~V}\)
Answer:
Required half cell reaction

E⁰_cell = (E⁰_ox) + (E⁰_red) = – 1.09 + 0.771 = – 0.319V
We know that ∆G° = – nFE⁰_cell
If E⁰_cell is -ve; ∆G is +ve and the cell reaction is non-spontaneous.
Hence, Fe³⁺ cannot oxidise Bromide to Bromine.

Question 15.
Is it possible to store copper sulphate in an iron vessel for a long time?
Given:

Answer:
E⁰_cell = (E⁰_ox) + (E⁰_red) = 0.44 V + 0.34V = 0.78V
These +ve E⁰_cell values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.

Question 16.
Two metals M₁ and M₂ have reduction potential values of – xV and + yV respectively. Which will liberate H₂ in H₂SO₄?
Answer:
Metals having negative reduction potential acts as powerful reducing agent. Since M₁ has – xV, therefore M1 easily liberate H₂ in H₂SO₄. Metals having higher oxidation potential will liberate H₂ from H₂SO₄. Hence, the metal M₁ having + xV, oxidation potential will liberate H₂ from H₂SO₄.

Question 17.
Reduction potential of two metals M₁ and M₂ are
\(E_{\mathrm{M}_{1}^{2+}}^{0}=-2.3 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{M}_{1}^{2+}}^{0}=0.2 \mathrm{~V}\)
Predict which one Is better for coating the surface of iron.
Given:
\(\mathbf{E}_{\mathrm{Fe}^{2+} \mid \mathbf{F e}}^{0}=-\mathbf{0 . 4 4} \mathbf{V}\)
Answer:
Oxidation potential of M₁ is more +ve than the oxidation potential of Fe which indicates that it will prevent iron from rusting.

Question 18.
Calculate the standard emf of the cell: Cd | Cd²⁺|| Cu²⁺ | Cu and determine the cell reaction. The standard reduction potentials of Cu²⁺ | Cu and Cd²⁺ | Cd are 0.34V and – 0.40 volts respectively. Predict the feasibility of the cell reaction.
Answer:

emf is +ve, so ∆G is (-)ve, the reaction is feasible.

Question 19.
In fuel cell H₂ and O₂ react to produce electricity. In the process, H₂ gas is oxidised at the anode and O₂ at cathode. If 44.8 litre of H₂ at 25°C and also pressure reacts in 10 minutes, what is average current produced? If the entire current is used for electro deposition of Cu from Cu²⁺, how many grams of Cu deposited?
Answer:
Oxidation at anode:
2H₂(g) + 4OH^– (aq) → 4H₂O (1) + 4e^–
1 mole of hydrogen gas produces 2 moles of electrons at 25°C and 1 atm pressure, 1 mole of hydrogen gas occupies = 22.4 litres
∴no. of moles of hydrogen gas produced = \(\frac{1 \mathrm{mole}}{22.4 \text { litres }}\) x 44.8 litres = 2 moles of hydrogen
∴2 of moles of hydrogen produces 4 moles of electro i.e., 4F charge. We know that Q = It
I = \(\frac { Q }{ t }\) = \(\frac{4 \mathrm{F}}{10 \mathrm{mins}}\) = \(\frac { 4×96500 }{ 10x60s }\)
I = 643.33 A
Electro deposition of copper
Cu²⁺_(aq) + 2e^– → Cu_(s)
2F charge is required to deposit
1 mole of copper i.e., 63.5 g
If the entire current produced in the fuel cell i.e., 4 F is utilised for electrolysis, then 2 x 63.5 i.e., 127.0 g copper will be deposited at cathode.

Question 20.
The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate and chromium nitrate respectively. If 2.935g of Ni was deposited in the first cell. The amount of Cr deposited in the another cell? Given: molar mass of Nickel and chromium are 58.74 and 52gm^-1 respectively.
Answer:

Question 21.
0.1M copper sulphate solution in which copper electrode is dipped at 25C. Calculate the electrode potential of copper.
Given : \(\left.\mathrm{E}_{\mathrm{Cu}^{2+}}^{\mathrm{o}}\right|_{\mathrm{Cu}}=0.34\)
Answer:
Given that
[Cu²⁺] = 0.1 M
E⁰_Cu²⁺|Cu = 0.34
E_cell = ?
Cell reaction is Cu²⁺(aq) + 2e^– → Cu (s)
E_cell = E⁰ – \(\frac { 0.0591 }{ n }\) log \(\frac{[\mathrm{Cu}]}{\left[\mathrm{Cu}^{2+}\right]}\) = 0.34 – \(\frac { 0.0591 }{ 2 }\) log \(\frac { 1 }{ 0.1 }\)
= 0.34 – 0.0296 = 0.31 V

Question 22.
For the cell Mg(s) | Mg²⁺(aq) || Ag⁺ (aq) | Ag (s), calculate the equilibrium constant at 25°C and maximum work that can be obtained during operation of cell. Given:
\(\mathrm{E}_{\mathrm{Mg}^{2+} \mid \mathrm{Mg}}^{\mathrm{o}}=-2.37 \mathrm{~V}\) and \(\mathrm{E}_{\mathrm{Ag}^{2+} \mid \mathrm{Ag}}^{\mathrm{o}}=0.80 \mathrm{~V}\)
Answer:
Oxidation at anode
Mg → Mg²⁺ + 2e^–……………(1) (E^o_oxi = 2.37 V
Reduction at cathode
Ag⁺ + e^– → Ag……………(2) (E^o_red = 0.80 V
E⁰_cell = (E⁰_ox) + (E⁰_red) = 2.37 + 0.80 = 3.17 V
Overall reaction
Mg + 2Ag⁺ → Mg²⁺ + 2Ag
∆G° = -nFE°
= – 2 × 96500 × 3.17
= – 6.118 × 10⁵ J
We know that W_max = ∆G°
W_max = + 6.118 × 10⁵ J
Relationship between ∆G° and K_eq is,
∆G = – 2.303 RT logK_eq
∆G = – 2.303 × 8.314 × 298 log K_eq [25°C = 298 K]
log K_eq = \(\frac{6.118 \times 10^{5}}{2.303 \times 8.314 \times 298}\) = \(\frac{6.118 \times 10^{5}}{5705.84}\)
log K_eq = 107.223
K_eq = Antilog (107.223)
K_c = 1.5849

Question 23.
8.2 × 10¹² litres of water is available in a lake. A power reactor using the electrolysis of water in the lake produces electricity at the rate of 2 × 10⁶ Cs^-1 at an appropriate voltage. How many years would it like to completely electrolyse the water in the lake. Assume that there is no loss of water except due to electrolysis.
Answer:
Hydrolysis of water:
At anode: 2H₂O → 4H⁺ + O₂ + 4e^– …………..(1)
At cathode: 2H₂O + 2e^– → H₂ + 2OH^–
Overall reaction: 6H₂O → 4H^– + 4OH^– +2H₂ + O₂
(or)
Equation (1) + (2) × 2
= 2H₂O → 2H₂ + O₂
According to Faraday’s Law of electrolysis, to electrolyse two mole of Water
(36g ≃ 36 mL. of H₂O), 4F charge is required alternatively, when 36 mL of water is electrolysed,
the charge generated = 4 × 96500 C.
When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to

Given that in 1 second, 2 × 10⁶ C is generated therefore, the time required to generate
96500 × 10¹⁵ C is = \(\frac{1 \mathrm{S}}{2 \times 10^{6} \mathrm{C}}\) × 96500 × 10¹⁵C = 48250 × 10⁹ S
Number of year = \(\frac{48250 \times 10^{9}}{365 \times 24 \times 60 \times 60}\) 1 year = 365 days
= 1.5299 × 10⁶
= 365 × 24 hours
= 365 × 24 × 60 min
= 365 × 24 × 60 × 60 sec

Question 24.
Derive an expression for Nernst equation.
Answer:
Nernst equation is the one which relates the cell potential and the concentration of the species involved in an electrochemical reaction.
Let us consider an electrochemical cell for which the overall redox reaction is,
Answer:
xA + yB = lC + mD
The reaction quotient Q is,
\(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\)
We know that,
∆G = ∆G⁰ + RT ln Q
∆G = – nFE_cell
∆G⁰ = -nFE⁰_cell
equation (1) becomes
– nFE_cell = -nFE⁰_cell + RT ln Q
Subsitute the Q value in equation (2)
– nFE_cell = – nFE⁰_cell + RT ln \(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\). ………..(3)
Divide the whole equation (3) by – nF
E_cell = E°_cell – \(\frac { RT }{ nF }\) ln \(\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)\)
E_cell = E°_cell – \(\frac { 2.303RT }{ nF }\) log \(\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)\) ……………(4)
This is called the Nernst equation.
At 25°C (298 K) equation (4) becomes,
E_cell = E°_cell – \(\frac { 2.303×8.314×298 }{ nx96500 }\) log \(\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)\)
E_cell = E°_cell – \(\frac { 0.0591 }{ n }\) log \(\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)\)

Question 25.
Write a note on sacrificial protection.
Answer:
In this method, the metallic structure to be protected is made cathode by connecting it with more active metal (anodic metal). So that all the corrosion will concentrate only on the active metal. The artificially made anode thus gradually gets corroded protecting the original metallic structure. Hence this process is otherwise known as sacrificial anodic protection. Al, Zn and Mg are used as sacrificial anodes.

Question 26.
Explain the function of H₂ – O₂ fuel cell.
Answer:
In this case, hydrogen act as a fuel and oxygen as an oxidant and the electrolyte is aqueous KOH maintained at 200°C and 20 – 40 atm. Porous graphite electrode containing Ni and NiO serves as the inert electrodes. Hydrogen and oxygen gases are bubbled through the anode and cathode, respectively.
Oxidation occurs at the anode:
2H_2(g)+ 4OH_-(aq) → 4H₂O₍₁₎ + 4e^–
Reduction occurs at the cathode O₂(g) + 2 H₂O(1) + 4e^– → 4 OH^– (aq)
The overall reaction is 2H₂(g) + O₂(g) → 2H₂O(1)
The above reaction is the same as the hydrogen combustion reaction, however, they do not react directly ie., the oxidation and reduction reactions take place separately at the anode and cathode respectively like H₂ – O₂ fuel cell. Other fuel cell like propane – O₂ and methane O₂ have also been developed.

Question 27.
Ionic conductance at infinite dilution of Al³⁺ and SO₄^2- are 189 and 160 mho cm² equiv^-1. Calculate the equivalent and molar conductance of the electrolyte Al₂(SO₄) at infinite dilution.

Solution:

III. Evaluate Yourself

Question 1.
Calculate the molar conductance of 0.01M aqueous KCI solution at 25°C . The specific conductance of KCl at 25°C is 14.114 x 10^-2 Sm^-1.
Answer:
Concentration of KCI solution = 0.01 M.
Specific conductance (K) = 14.114 × 10^-2 S m^-1
Molar conductance (Λ_m) = ?
Λ_m = \(\frac{\kappa \times 10^{-3}}{M}\) = \(\frac{14.114 \times 10^{-2} \times 10^{-3}}{0.01}\)
S m^-1 mol^-1 m³
Λ_m = 14.114 × 10^-5 × 10² = 14.114 × 10^-3 Sm² mol^-1

Question 2.
The emf of the following cell at 25°C is equal to 0.34v. Calculate the reduction potential of copper electrode.
Pt(s) | H₂(g,1atm) | H⁺ (aq,1M) || Cu²⁺(aq,1M) | Cu(s)
Answer:
SHE Value is zero
E°_cell = E°_R – E°_L
= 0.34 – 0 = 0.34V
The reduction potential of copper electrode = 0.34V

Question 3.
Using the calculated emf value of zinc and copper electrode, calculate the emf of the following cell at 25°C.
Zn (s) | Zn²⁺ (aq, 1M) || Cu²⁺(aq, 1M) | Cu (s)
Answer:
E°_cell = E°_R – E°_L
E_zn/zn²⁺ = 0.76V
E_Cu/Cu²⁺ = 0.76V
E°_cell = 0.76 – (- 0.34V)
E°_cell = 0.76 – (- 0.34)
E°_cell = + 1.1 V

Question 4.
Write the overall redox reaction which takes place in the galvanic cell,
Pt(s) | Fe²⁺(aq),Fe²⁺(aq) || MnO^–₄(aq), H⁺(aq), Mn²⁺(aq) || Pt(s)
Answer:
At Anode half cell – 5Fe²⁺_(aq) → 5Fe³⁺_(aq) + 5e^–
At cathode half cell – MnO^–_4(aq) + 8H⁺_(aq) + 5e^– → Mn²⁺_(aq) + 4H₂O₍₁₎
Overall redox reaction – 5Fe²⁺_(aq) + MnO^–₄(aq) + 8H⁺(aq) → 5Fe³⁺_(aq) + Mn²⁺_(aq) + 4H₂O₍₁₎

Question 5.
The electrochemical cell reaction of the Daniel cell is
Zn (s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu_(s)
What is the change in the cell voltage on increasing the ion concentration in the anode compartment by a factor 10?
Answer:
Zn (s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu_(s)
ln the case E°_cell = 1.1V
Reaction quotient Q for the above reaction is, Q = \(\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\)
E_cell = E°_cell – \(\frac { 0.0591 }{ n }\) log \(\frac{\left[Z n^{2+}\right]}{\left[C u^{2+}\right]}\)
If suppose concentration of Cu²⁺ is 1 .OM then the concentration of Zn²⁺ is 10M (why because,
ion concentration in the anode compartment increased by a 10 factor)
E_cell = 1.1 – \(\frac { 0.0591 }{ n }\) log \((\frac { 10 }{ 1 })\)
= 1.1 – 0.02955             ……………….(1)
= 1.070 V (cell voltage decreased)
Thus, the initial voltage is greater than E° because Q < 1. As the reaction proceeds, [Zn²⁺] in the anode compartment increases as the zinc electrode dissolves, while [Cu²⁺] in the cathode compartment decreases as metallic copper is deposited on the electrode.

During this process, the Q = [Zn²⁺] [Cu²⁺] steadily increases and the cell voltage therefore steadily decreases. [Zn²⁺] will continue to increase in the anode compartment and [Cu²⁺] will continue to decrease in the cathode compartment. Thus the value of Q will increase further leading to a further decrease in value.

Question 6.
A solution of a salt of metal was electrolysed for 150 minutes with a current of 0.15 amperes. The mass of the metal deposited at the cathode is 0.783g. calculate the equivalent mass of the metal.
Answer:
Given,
I = 0.15 amperes
t = 150 mins
= t = 150 × 6Osec
= t = 9000sec
Q = It
= Q = 0.15 × 9000 coulombs
= Q = 1350 coulombs
Hence, 135 coulombs of electricity deposit is equal to 0.783g of metal.
96500 coulombs of electricity, \(\frac { 0.783 × 96500 }{ 1350 }\) = 55.97 gm of metal
Hence equivalent mass of the metal is 559.7

12th Chemistry Guide Electro Chemistry Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

Question 1.
When electric current is passed through an electrolytic solution, charge is carried by
a) electrons
b) ions
c) atoms
d) molecules
Answer:
b) ions

Question 2.
Ohm’s law
a) I = VR
b) I = \(\frac{R}{V}\)
c) I = \(\frac{V}{R}\)
d) V = \(\frac{I}{R}\)
Answer:
c) I = \(\frac{V}{R}\)

Question 3.
The cell constant of a conductivity cell with platinum electrodes at a distance of 0.5 cm and area of cross section 5 cm² is
a) 10 cm^-1
b) 0.1 cm^-1
c) 2.5 cm^-1
d) 0.25 cm^-1
Answer:
b) 0.1 cm^-1
Cell constant = \(\frac{l}{A}=\frac{0.5}{5}\) = 0.1cm^-1

Question 4.
Specific conductance (or) conductivity =
a) 1/ρ
b) \(\frac{1}{R} \frac{A}{l}\)
c) \(\mathrm{R} \frac{\mathrm{A}}{l}\)
d) both (a) and (b)
Answer:
d) both (a) & (b)

Question 5.
For 1 : 1 electrolyte like NaCl equivalent conductance is
a) less than molar conductance
b) greater than molar conductance
c) equal to molar conductance
d) zero
Answer:
c) equal to molar conductance

Question 6.
The relationship between molar conductance and equivalent conductance of 1M H₂SO₄ is
\(a) \Lambda_{\mathrm{m}}=\frac{\Lambda}{2}
b) \Lambda_{\mathrm{m}}=2 \Lambda
c) \Lambda_{\mathrm{m}}=\Lambda
d) \frac{\Lambda_{\mathrm{m}}}{\Lambda}=0\)
Reason :

Question 7.
As concentration of the electrolyte decreases the specific conductance of the solution
a) decreases
b) increases
c) remains the same
d) becomes zero
Answer:
a) decreases

Question 8.
As concentration of the electorlyte decreases the molar conductance and equivalent conductance of the solution
a) decreases
b) increases
c) remains the same
d) becomes zero
Answer:
b) increases

Question 9.
In the measurement of conductivity of an electrolyte using wheatstone bridge arrangement which is correct?
a) PQ = RS
\(b) \frac{Q}{P}=\frac{R}{S}
c) \frac{P}{Q}=\frac{R}{S}
d) \frac{P}{Q}=\frac{S}{R}\)
Answer:
c) \(\frac{P}{Q}=\frac{R}{S}\)

Question 10.
Which among the following solution of NaCl has the maximum molar conductance value?
a) 10^-4 M
b) 10^-3 M
c) 10^-2 M
d) 10^-1 M
Answer:
a) 10^-4 M
Reason : As dilution increases, molar conductance increases.

Question 11.
Molar conductance of an electrolyte approaches a limiting value in
a) highly concentrated solution
b) concentrated solution
c) very dilute solution
d) dilute solution
Answer:
c) very dilute solution

Question 12.
For a weak electrolyte there is a sudden increase in molar conductance as the concentration approaches
a) infinity
b) maximum value
c) zero
d) negative value
Answer:
c) zero

Question 13.
Limiting molar conductivity values of strong electrolytes can be determined by
a) Kohlrausch’s law
b) Faradays’ law
c) Nernst equation
d) extrapolating the straight line
Answer:
d) extrapolating the straight line

Question 14.
Limiting molar conductivity values of weak electrolytes can be determined by
a) Kohlrausch’s law
b) Faradays’ law
c) Nernst equation
d) extrapolating the straight line
Answer:
a) Kohlrausch’s law

Question 15.
Debye and Huckel derived an equation for the conductivity of strong electrolytes by assuming
a) partial dissociation
b) incomplete dissociation
c) complete dissociation
d) negligible dissociation
Answer:
c) complete dissociation

Question 16.
The basis for Kohlrausch’s law is
a) molar conductance
b) specific conductance
c) specific resistance
d) limiting molar conductance
Answer:
d) limiting molar conductance

Question 17.
Degree of dissociation of weak electrolytes can be calculated using the expression
\(a) [latex]\alpha=\frac{\mathrm{K}_{\mathrm{m}}}{\mathrm{K}_{\mathrm{m}}^{\mathrm{o}}}
b) \quad \alpha=\frac{\mathrm{K}_{\mathrm{m}}^{\mathrm{O}}}{\mathrm{K}_{\mathrm{m}}}
c) \alpha=\frac{\Lambda_{\mathrm{m}}^{\mathrm{o}}}{\Lambda_{\mathrm{m}}}
d) \alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\mathrm{O}}}\)
Answer:
d) \(\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\mathrm{O}}}\)

Question 18.
The device which converts chemical energy into electrical energy is known as
a) battery
b) Galvanic cell
c) Voltaic cell
d) all the above
Answer:
d) all the above

Question 19.
The device which converts electrical energy into chemical energy is known as
a) Galvanic cell
b) Voltaic cell
c) electrolytic cell
d) all the above
Answer:
c) electrolytic cell

Question 20.
In Daniel cell zinc undergoes
a) oxidation
b) reduction
c) hydrolysis
d) galvanisation
Answer:
a) oxidation

Question 21.
In Daniel cell copper ions undergo
a) oxidation
b) reduction
c) hydrolysis
d) galvanisation
Answer:
b) reduction

Question 22.
In Daniel cell the anode and cathode half cells are respectively
a) Cu | Cu²⁺, Zn | Zn²⁺
b) ZZn | Zn²⁺, Cu | Cu²⁺
c) Zn | Zn²⁺, Cu²⁺ | Cu
d) Cu²⁺ | Cu, Zn²⁺ | Zn
Answer:
c) Zn | Zn²⁺, Cu²⁺ | Cu

Question 23.
In Daniel cell, the number of electrons transfered in the redox reaction
Zn + CuSO₄ → ZnSO₄+ Cu is
a) 1
b) 2
c) 3
d) 4
Answer:
b) 2

Question 24.
Which of the following statement is correct with respect to electrolytic conductance? ULffi
a) Conductivity increases with the decrease ’ in Viscosity
b) Conductivity increases with increase in temperature
c) Molar conductance of a solution decreases with increase in dilution
d) Conductance decrease with increase in temperature.
Answer:
c) Molar conductance of a solution decreases with increase in dilution

Question 25.
In Daniel cell the Electronic solutions in the two half cells are connected using a
a) wire
c) copper strip
b) zinc strip
d) salt bridge
Answer:
d) salt bridge

Question 26.
A salt bridge is containing an inverted U tube
a) agar – agar gel with Kcl
b) agar – agar gel with Na₂SO₄
c) both (a) & (b)
d) none of the above
Answer:
c) both (a) & (b)

Question 27.
The general representation of a fuel cell is
a) Fuel / Electrode / Electrolyte / Electrode / Oxidant
b) Oxidant /Electrode / Electrolyte / Fuel
c) Fuel / Electrode / Electrolyte / Electrode / Reductant
d) Oxidant /Electrode / Electrolyte /Reductant
Answer:
a) Fuel / Electrode / Electrolyte / Electrode / Oxidant

Question 28.
Which of the following is correct?
a) 1 C = 1J × 1V
b) 1V = 1C × 1J
c) 1J = 1C × 1V
d) 1V = \(\frac{1 C}{1 \mathrm{~J}}\)

Question 29.
Ecell is equal to
a) (E_ox)_anode + (E_red)_cathode
b) (E_red)_cathode – (E_red)_anode
c) (E_red)_anode + (E_red)_cathode
d) both (a) & (b)
Answer:
d) both (a) & (b)

Question 30.
The emf of standard hydrogen electrode is
assigned an arbitrary value of
a) zero
b) one
c) infinity
d) negative
Answer:
a) zero

Question 31.
Electrical charge carried by one electron is

Answer:
d) Both (a) & (b)

Question 32.
A cell reaction is spontaneous when AG° and ocell E are respectively
a) + ve and – ve
b) – ve and + ve
c) 1 and 0
d) 0 and 1
Answer:
b) – ve and + ve

Question 33.
Electrolysis is a
a) photochemical reaction
b) spontaneous reaction
c) non-spontaneous reaction
d) substitution reaction
Answer:
c) non-spontaneous reaction

Question 34.
If the atomic mass of an ion Mⁿ⁺ is A, its electrochemical equivalent is
a) \(\frac{\mathrm{nA}}{\mathrm{F}}\)
b) \(\frac{\mathrm{F}}{\mathrm{nA}}\)
c) \(\frac{\mathrm{nF}}{\mathrm{A}}\)
d) \(\frac{A}{n F}\)
Answer:
d) \(\frac{A}{n F}\)
Reason Z = \(\frac{\text { Atomic mass }}{\text { Valency } \times \text { Faraday constant }}\)

Question 35.
When Q coulomb electric charge is passed through a series of aqueous solution of AgNO₃, AlCl₃ and CUSO₄, the increasing order of mass of the metals liberated is
a) Ag > Cu > Al
b) AZ < Cu < Ag
c) Cu > Ag > Al
d) Ag < Cu < Al
Ans:
b) AZ < Cu < Ag
Reason: m a equivalent mass eAl < eCu < eAg \ mAl < mCu < mAg

Question 36.
Leclanche cell is a
a) primary battery
b) secondary battery
c) rechargeable battery
d) electrolytic cell
Answer:
a) primary battery

Question 37.
Lithium ion battery is a
a) primary battery
b) secondary battery
c) non – rechargeable battery
d) none of the above
Answer:
b) secondary battery

Question 38.
In cellular phones/ laptop Computers the battery used is
a) Lechlanche cell
b) Lead storage battery
c) Lithium – ion battery
d) H₂ – O₂ fuel cell
Answer:
c) Lithium – ion battery

Question 39.
Rusting of iron is
a) a hydrolysis reaction
b) an electrochemical redox reaction
c) an oxidation reaction
d) a reduction reaction
Answer:
b) an electrochemical redox reaction

Question 40.
The formula of rust is
a) PeO
b) Fe₃O₄
c) Fe (OH)₃
d) Fe₂O₃. × H₂O
Answer:
d) Fe₂O₃. × H₂O

Question 41.
The most convenient method to protect the bottom of ship made of iron is
a) coating it with red lead oxide
b) white tin plating
c) connecting it with Mg block
d) connecting it with Pb block
Answer:
c) connecting it with Mg block
Reason: As Mg is more reactive metal than iron, it gets corroded in preference to iron. This is known as cathodic protection.

Question 42.
Standard reduction potential of three metals A, B and C are – 1.5 V, + 1 V and – 2 V respectively. The decreasing order of reducing power of these metals is
a) B > C > A
b) A > B > C
c) C > A > B
d) C > B > A
Answer:
c) C > A > B
– 2 V – 1.5 V + 1 V
Reason: lower the E° value higher is the reducing power.

Question 43.
Limiting molar conductivity of NH4OH (A?n)NH4OH is equal to

Answer:
d

Question 44.
A hydrogen gas electrode in made by dipping platinum wire in a solution of pH = 10 and by passing hydrogen gas around the platinum wire at one atmosphere pressure. The oxidation potential of the electrode would be
a) 0.59 V
b) 0.118 V
c) 1.18 V
d) 0.059 V
Answer:
a) 0.59 V

Reason: H₂ —> 2H⁺ + 2e^–
Concentration 1 atm

Question 45.
A button cell used in watches functions as following:
Zn(s) + Ag₂O + H₂O_(l) ⇌ 2Ag_(s)+ Zn²⁺_(aq) + 2OH^–_(aq)
If half cell potentials are :
Zn²⁺_(aq) + 2e^–_(aq) → Zn_(s) ; E° = -0.76 V
Ag₂O + H₂O(l) + 2e^– → 2Ag_(s) + 2OH^–_(aq) E° = – 0.34 V
The cell potential will be
a) 0.42 V
b) 0.84 V
c) 1.34 V
d)1.10V
Answer:
d) 1.10 V
Reason : E_cell – (E_ox)_anode + (E_red)_cathode
= + 0.76 + 0.34 = + 1.10V

Question 46.
How many grams of cobalt metal will be deposited when a solution of Cobalt (II) chloride is electrolysed with a current of 10 amperes for 109 minutes? (1 Faraday = 96500 C; Atomic mass of (Co = 59 u)
a) 0.66
b) 4.0
c) 20.0
d) 40.0
Answer:
c) 20.0

Question 47.
Consider the half-cell reactions
Mn²⁺ + 2e^– → Mn; E° = – 1.18 V
Mn²⁺ → Mn³⁺ + e^– ; E° = – 1.51 V
The E° for the reaction 3Mn²⁺ → Mn + 2Mn³⁺ and the possibility of the forward reaction are respectively.
a) – 2.69 and not possible
b) – 4.18 V and possible
c) + 0.33 V and possible
d) + 2.69 V and not possible
Answer:
a) – 2.69 and not possible
Reason : E° = E°_oxid + E°_red
= – 1.51 +(-1.18) E°= – 2.69V
= 1.51 -1.18 ΔG°= -nFE°,
E° is negative, ΔG° is positive and hence the forward reaction is not possible.

Question 48.
The electrolyte used in Lechlanche cell is
a) Paste of KOH and ZnO
b) 38 % solution of H₂SO₄
c) Moist of paste NH₄Cl and ZnCl₂
d) Moist sodium hydroxide
Answer:
c) Moist of paste NH₄Cl and ZnCl₂

Question 49.
During the electrolysis of fused NaCl, which reaction occurs at anode.
a) chloride ions are oxidised
b) chloride ions are reduced
c) Sodium ions are oxidised
d) sodium ions are reduced
Answer:
a) chloride ions are oxidised

Question 50.
Amount of electricity that can deposit 108 gm of silver from AgNO₃ solution is
a) 1 ampere
b) 1 coulomb
c) 1 Faraday
d) 1 Volt
Answer:
c) 1 Faraday
Reason: 1 g eq.mass of Ag (108 g) will be liberated by 1 Faraday

II. Match the following:

Question 1.

Answer:
d) Sm^-1
e) S
b) m^-1
a) Sm² mol^-1
c) Sm²g equiv^-1

III. Pick out the correct statements

Question 1.
i) For strong electrolyte the plot ∧mVs√C is not a linear one.
ii) For a strong electrolyte, at high concentration the number of constituent ions in a given volume is high.
iii) At high concentration the ions experience a viscous drag due to greater solvation.
iv) At infinite dilution the ions are so close and the interaction between them becomes significant.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)

Correct Statement:
i) For strong electrolytes the plot ∧mVs√C is a straight line.
iv) At infinite dilution the ions are so apart and the interaction between them becomes insignificant. ‘

Question 2.
i) In Daniel cell electrons are liberated at : inc electrode and hence it is negative.
ii) Electrons flow from copper cathode to zinc anode.
iii) Through salt bridge ions can move into (or) out of the half cells.
iv) The zinc and copper strips are connected externally through salt bridge.

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

Correct Statement:
(ii) Electrons flow from, zinc anode to copper cathode.
(iv) The zinc and copper strips are connected externally through a wire.

Question 3.
i) Electrolysis is carried out in an electrolytic cell by connecting it to an AC power supply.
ii) The device which is used to carry out the electrolysis is called the electrolytic cell.
iii) In the electrolytic cell cathode is + ve and anode is – ve.
iv) The electrochemical process occuring in the electrolytic cell and galvanic cell are the reverse of each other.
a) (i) & (ii) b) (ii) & (iii)
c) (iii) & (iv) d) (ii) & (iv)
Answer:
d) (ii) & (iv)

Correct Statement :
(i) Electrolysis is carried out in an electrolytic cell by connecting it to a DC power supply.
(iii) In the electrolytic cell cathode is – ve and anode is + ve.

Question 4.
i) The amount of substance deposited when 1 ampere current passed for 1 second is its equivalent mass.
ii) When same quantity of electric charge is passed through different solutions the masses liberated are inversely proportional to their equivalent masses.
iii) One gram equivalent mass of any substance can be liberated by passing one Faraday electricity.
iv) One Faraday of electric charge is carried by Avogadro number of electrons.

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

Correct Statement :
(i) The amount of substance deposited when 1 ampere current passed for 1 second is its electrochemical equivalent.
(ii) When same quantity of electric charge is passed through different solutions the masses liberated are directly proportional to their equivalent masses.

IV. Pick out the incorrect statements

Question 1.
i) At constant temperature, the current flowing through the cell (I) is directly proportional to the resistance of the cell.
ii) Resistance is the opposition that a cell offers to the flow of electric current through it.
iii) The conductivity of the electrolyte is measured using a conductivity cell.
iv) Resistance of an electrolytic solution is directly proportional to the cross sectional area and
inversely proportional to the length.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)

Correct Statement :
(i) At constant temperature, the current flowing through the cell (I) is directly proportional to the voltage across the cell (V).
(iv) Resistance of an electrolytic solution is directly proportional to the length and inversely proportional to the cross sectional area.

Question 2.
i) The conductance of lm3 electrolytic solution is called the specific conductance.
ii) Conductivity increases with increase in viscosity.
iii) Specific conductivity increases with increase in dilution.
iv) Conductivity of a cell can be measured using wheatstone bridge arrangement.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)

Correct Statement:
(ii) Conductivity increases with decrease in viscosity.
(iii) Specific conductivity decreases with increase in dilution.

Question 3.
i) Molar conductivity is due to the independent migration of cations in one direction and anions in the opposite direction.
ii) At infinite dilution each constituent ion of the electrolyte makes a definite contribution towards the molar conductance.
iii) It is possible to determine the molar conductance at infinite dilution for weak electrolytes experimentally.
iv) The solubility product of a sparingly soluble salt can not be determined using conductivity measurements.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)

Correct Statement:
(iii) It is impossible to determine the molar conductance at infinite dilution for weak electrolytes experimentally.
(iv) The solubility product of a sparingly soluble salt can be determined using conductivity measurements.

Question 4.
i) Corrosion of iron is called rusting.
ii) Coating of iron over zinc is called galvanisation.
iii) The standard reduction potential (E°) is a measure of the oxidising tendency of the species.
iv) Greater the E° values in the spectrochemical series greater is the tendency of the species to donate electrons and undergo oxidation.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
c) (ii) & (iv)

Correct Statement:
(ii) Coating of zinc over iron is called galvanisation.
(iv) Greater the E° values in the spectrochemical series greater is the tendency of the species to accept electrons and undergo reduction.

V. Assertion and reason

Question 1.
Assertion (A): If the temperature of the electrolytic solution increases, conductance also increases.
Reason (R): As temperature increases, the kinetic energy of the ions decreases and the attractive force between the ions increases.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
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. Correct Reason R: As temperature increases, the kinetic energy of the ions increases and the attractive force between the ions decreases.

Question 2.
Assertion (A): In the wheatstone bridge arrangement AC power supply is used in the measurement of conductivity of an electrolyte in conductivity cell.
Reason (R): DC power supply will lead to the electrolysis of the solution taken in the cell.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
a) Both A& R are correct, R explains A

Question 3.
Assertion (A): In lead acid storage battery, when, a potential greater than 2V is applied across electrodes the cell reactions of discharge process are reverse.
Reason (R): The electrochemical reactions which take place in galvanic cell may be reversed by applying a potential slightly greater than the emf generated by the cell.

a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
a) Both A & R are correct, R explains A

Question 4.
Assertion (A): Magnesium is used as a sacrificial anode to protect iron from rusting. Reason (R): The reduction potential of magnesium is higher than iron hence magnesium has lesser tendency to undergo corrosion in preference to iron.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
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 Correct Reason R: The reduction potential of magnesium is less than iron and hence magnesium has higher tendency to undergo corrosion in preference to iron.

VI. Two Mark Question

Question 1.
State Ohm’s law.
Answer:
At constant temperature, the current flowing through the cell (1) is directly proportional to the voltage across the cell (V).
ie) I ∝ V (or) I = V/R (or) V = IR

Question 2.
Define resistivity.
Answer:
Resistivity is defined as the resistance of an electrolyte confined between two electrodes having unit cross sectional area (1m2) and are separated by unit distance (1m).

Question 3.
Define specific conductance (or) conductivity.
Answer:
Specific conductance is defined as the conductance of a cube of an electrolytic solution of unit dimension, (lm3).

Question 4.
Define molar conductivity (∧_m)
Answer:
Conductivity of a cell in which the electrodes are lm apart containing 1 mole of electrolyte in V m3 of electrolytic solution is called as molar conductivity.
Relation between specific conductance and molar conductance.
\(\Lambda_{\mathrm{m}}=\frac{\mathrm{K}\left(\mathrm{sm}^{-1}\right) \times 10^{3}}{\mathrm{M}} \mathrm{mot}^{-1} \mathrm{~m}^{3}\)

Question 5.
Define equivalent conductance (∧)
Answer:
Conductivity of a cell in which the electrodes are lm apart and containing 1 gram equivalent of electrolyte in Vm3 of electrolytic solution is called as equivalent conductance (∧)
\(\Lambda=\frac{\kappa \times 10^{-3}}{N} \mathrm{Sm}^{2} \mathrm{~g} \text { equiv }^{-1}\)
K = Specific conductance
N Normality

Question 6.
Define Electrochemical equivalent.
Answer:
Electrochemical equivalent is defined as the amount of substance deposited at the electrode by the passage of 1 coulomb charge.
Electrochemical equivalent
Z = \(\frac{\text { Equivalent mass }}{96500}\)

Question 7.
What is the role of salt bridge in Galvanic cell?
Answer:
To maintain the electrical neutrality in both the compartments, the non reactive anions Cl^– (from KCl taken in the salt bridge) move from the salt bridge and enter into the anodic compartment, at the same time some of the K⁺ ions move from the salt bridge into the cathodic compartment.

Question 8.
What are the applications of Kohlrausch’s law?
Answer:
It is used to calculate the molar conductance of a weak electrolyte at infinite dilution.
\((\mathrm{eg}) \Lambda_{\mathrm{CH}_{\mathrm{COOH}}}^{\mathrm{o}}=\Lambda_{\mathrm{CH}_{2} \mathrm{COONa}}^{\mathrm{o}}+\Lambda_{\mathrm{HCl}}^{\mathrm{o}}-\Lambda_{\mathrm{NaCl}}^{\circ}\)
It is used to calculate the degree of dissociaton of weak electrolytes.
\(\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\mathrm{o}}}\)
It is used to calculate the solubility product of sparingly soluble salts.
For 1:1 electrolyte
\(K_{S P}=\left(\frac{K \times 10^{-3}}{\Lambda_{m}^{o}}\right)^{2}\)

Question 9.
Define emf of a-cell.
Answer:

• The force that pushes the electrons away from the anode and pulls them towards the cathode is called the electromotive force (emf) or the cell potential.
• SI unit of emf is volt (V).

Question 10.
Define Joule.
Answer:
When there is one volt difference in electrical potential between the anode and cathode, the energy released for each coulomb of charge that moves between them is defined as one joule of energy.
1 J = 1 C × 1 V.

Question 11.
Write the factors affecting cell voltage.
Answer:

• Nature of the electrodes.
• Concentration of the electrolytes.
• Temperature at which the cell is operated.

Question 12.
What is the IUPAC definition of electrode potential (E)?
Answer:
Electrode potential is the electromotive force of a cell in which the electrode on the left is a standard hydrogen electrode and the electrode on the right is the electrode in question.

Question 13.
What is the IUPAC definition of standard electrode potential (E°)?
Answer:
Standard electrode potential (E°) is the value of the standard emf of a cell in which molecular hydrogen under standard pressure is oxidised to solvated protons at the left hand electrode.

Question 14.
Define electrolysis.
Answer:

• Electrolysis is a process in which the electrical energy is used to cause a non- spontaneous chemical reaction to occur.
• The energy is often used to decompose a compound into its elements.

Question 15.
What is galvanisation?
Answer:

• Coating of zinc over iron is called galvanisation.
• Galvanisation of iron prevents rusting.

Question 16.
How cathodic protection helps to protect the metal corrosion
Answer:

• Unlike galvanisation in this method the entire surface of iron is not covered with zinc.
• Metals like Mg or Zn which corrode more easily than iron can be used as a sacrificial anode.
• Iron material acts as a cathode.
• So iron is protected but Mg or Zn is corroded during rusting.
• This is known as cathodic protection.

Question 17.
How metals can be protected from corrosion?
Answer:

• Coating metal surface by paint.
• Galvanisation.
• Cathodic protection.
• Passivation.
• Alloy formation.

Question 18.
What is electrochemical series?
How is it useful to predict corrosion?
Answer:

• The series in which the standard aqueous electrode potential at 298 K for various metal – metal ion electrodes are arranged in the decreasing order of their standard reduction potential values is called electrochemical series.
• Standard reduction potential E° is a measure of the oxidising tendency of The species.
• Greater the E° value, greater is the tendency shown by the species to accept electrons and undergo reduction.
• Higher the E° value lower is the tendency to undergo corrosion..

VII.Three Mark Questions

Question 1.
The molar conductivity of a strong electrolyte and a weak electrolyte increases with dilution, why?
Answer:
Strong electrolytes:

• When dilution increases, the ions are far apart.
• Attractive force between the ions decreases.
• The interaction between the ions becomes insignificant.
• Hence molar conductivity increases and reaches a maximum value at infinite dilution.

Weak electrolytes:

• According to Ostwald’s dilution law, as dilution increases the dissociation of weak electrolyte increases.
• As dissociation increases the number of ions increases.
• Hence molar conductivity increases with dilution.

Question 2.
Write a note on Debye-Huckel and onsagar equation.
Answer:

• At infinite dilution, the interaction between the ions in the electrolyte solution is negligible.
• Except this condition, electrostatic interaction between the ions alters the properties of the solution from those expected from the free – ions – value.
• The influence of ion-ion interactions on the conductivity of strong electrolytes was studied by Debye and Huckel.
• They considered that each ion is surrounded by an ionic atmosphere of opposite sign.
• They derived an expression relating the molar conductance of strong electrolytes with the concentration by assuming complete dissociation.
• Later, this equation was further developed by onsagar.
• For a uni-univalent electrolyte the Debye, Huckel and onsagar equation is given as,
  
  A and B are constants which depend only on the nature of the solvent and temperature.
  D – dielectric constant of the medium, r) – viscosity of the medium.
  T – temperature in Kelvin.

Question 3.
From the following data, prove that each constituent ion of the electrolyte makes a definite contribution towards the molar conductance irrespective of the nature of other ion with which it is associated.
Answer:

Solution:

• This value 23.41 for the difference between the (λ°_m) values of K⁺ and Na⁺ is constant irrespective of the anion (Cl^–, Br^– or NO₃^–) with which they are associated.
• This proves that at infinite dilution each constituent ion of the electrolyte makes a definite contribution towards the molar conductance of the electrolyte irrespective of the nature of the other ion with which it is Associated
• Thus Kohlraush’s law can be proved.
• Similarly we can conclude that

Question 4.
What is known as intercalation?
Answer:

• In lithium-ion battery, during discharge, the Li+ ions produced at the anode move towards cathode through the non-aqueous electrolyte.
• When a potential greater than the emf produced by the cell, is applied across the electrode, the cell reaction is reversed.
• Now the Li+ ions move from cathode to anode where they become embedded on the porous electrode.
• This is known as intercalation.

Question 5.
Write a note on standard Hydrogen Electrode (SHE).
Answer:

• Standard Hydrogen Electrode is used as a reference electrode.
• Its emf is assigned an arbitrary value of zero volt.
• It consists of a platinum electrode in contact with 1 M HC1 and 1 atm hydrogen gas.
• The hydrogen gas is bubbled through the solution at 25°C.
• SHE can act as a cathode as well as an anode. If SHE is used as a cathode, the reduction reaction is
  
  2H⁺_(aq,1M) + 2e^– → H₂ (g, latm) E° = 0 volt
• If SHE is used as an anode, the oxidation reaction
  is H₂, (g, 1atm) →2H⁺(aq,1M) + 2e^– E°= 0 volt

Question 9.
Write a note on lithium – ion battery.
Answer:

• Lithium – ion battery is a secondary battery (rechargeable)
• Anode – Porous graphite
• Cathode – transition metal oxide such as Co02.
• Electrolyte – Lithium salt in an organic solvent.
• Oxidation at anode :
  Li_(s) → Li⁺_(aq) + e^– ………………(1)
• Reduction at cathode :
  Li⁺+ CoO_2(s) + e^– → LiCoO_2(s) ……………..(2)
• Equation (1) + (2) gives the overall redox reaction.
  Li_(s) + CoO_2(s) – LiCoO_2(s) ………………. (3)
• Both electrodes allow Li⁺ ions to move in and out of their structures.
• Discharge process – Li⁺ ions produced at the anode move towards cathode through the non-aqueous electrolyte.
• Recharge process – A potential greater than
  the emf produced by the cell is applied across the electrode, the cell reaction is reversed. .
• Li⁺ ions move from cathode to anode where they become embedded on the porous electrode. This is known as intercalation.
• Uses – cellular phones, laptop computers, digital cameras.

VIII. Five Mark Questions.

Question 1.
What are the factors affecting electrolytic conductance
Answer:

• As inter ionic attraction increases, conductance decreases.
• Solvent of high dielectric constant show high conductance in solution.
• As viscosity of the medium increases, conductance decreases.
• An temperature of the electrolytic solution increases, conductance increases because kinetic energy of ions increase and the attractive force between them decrease.
• As dilution increases specific conductance decreases as the number of ions for unit volume decreases.
• As dilution increases, molar conductance and equivalent conductance of a strong electrolyte increases, because inter ionic attractive force between the ions decreases with dilution.
• As dilution increases, molar conductance and equivalent conductance of a weak electrolyte increases, because degree of dissociation increases with dilution.

Question 2.
How is the conductivity of an electrolytic solution determined using a Wheatstone bridge?

Answer:

• Conductivity of an electrolytic solution is determined by using a Wheatstone bridge.
• Here one of the resistance is replaced by a conductivity cell containing the electrolytic solution of unknown conductivity.
• If DC power supply is used, it will lead to the electrolysis of the electrolytic solution.
• So, AC power supply is used to prevent electrolysis.
• A Wheatstone bridge is constituted using conductance and the curve is almost parallel to Am axis.
• This is because as the dilution increases dissociation of the weak electrolyte also increases.
• ∧°_m values for strong electrolytes can be obtained by extrapolating the straight line.
• But this is not applicable for weak electrolytes, as the plot is not a linear one.
• ∧°_m values of the weak electrolytes can be determined using Kohlrausch’s law.

Question 4.
Explain the thermodynamics of cell reactions.
Answer:
In a galvanic cell, the chemical energy is converted into electrical energy.
Electrical energy produced by the cell = Total charge of electrons x emf of the cell.
If ‘n’ is the number of moles of electrons exchanged between the oxidising and reducing agents in the overall cell reaction, the electrical energy produced by cell is Electrical energy = charge of ‘n’ mole of electrons x E_cell …………… (1)

Charge of 1 mole of electrons = 1 F
∴ Charge of n mole of electrons = n F
∴ Equation (1) becomes
Electrical energy = nFE_cellE0u……………(2)

This energy is ueto do work
∴ Maximum work obtaJed from a galvanic cell is
(W_max))cell nFE_cell ………………….(3)
The (-). sign indicates work is done by the system on the surroundings.

According to ‘second aw of thermodynamics
Maximum work done by the system =
change -in-Gibbs free energyof the.’system
(W_cell = ∆G ………………..(4)
Combining’ (3) and (4)
∆G = -nFE_cell ………………(5)

If Ecell is positive, ∆G is negative and the cefi reaction is spontaneous.

When all the cell components are m their
standard state, tbe above equation becomes
∆G° = -nFE°_cell ……………..(6)
Also ∆G° = – RT In K_eq ……………..(7)

Question 5.
Write about Lechlanche cell
Answer:

• Lechlanche cell is a primary battery (non – rechargeable)
• Anode : Zinc container
• Cathode : Graphite rod in contact with MnO₂.
• Electrolyte : NH₄Cl & ZnCl₂ in water.
• EMF: 1.5 V

Oxidation at anode :

Equation (1) + (2) +(3) gives the overall redox reaction

Ammonia produced at the cathode combines with Zn²⁺ to form a complex ion [Zn (NH₃)₄]²⁺.
As the reaction proceeds, the concentration of NH₄⁺ will decrease and the aqueous NH₃ will increase which lead to the decrease in the emf of the cell.

Question 6.
Write about mercury button cell.
Answer:

• Mercury button cell is a primary battery (non – rechargeable)
• Anode : Zinc amalgamated with mercury.
• Cathode : HgO mixed with graphite.
• Electrolyte : Paste of KOH and ZnO.
• EMF :1.35 V
• Oxidation at anode :
  
• It has higher capacity and longer life.
• Uses : In pacemakers, electronic watches, cameras, etc.,

Question 7.
Write about lead storage battery.
Answer:

• Electrochemical reactions which take place in a galvanic cell may be reversed by applying a potential slightly greater than the emf . generated by the cell.
• This principle is used in secondary batteries to regenerate the original reactants.
• Lead storage battery is a secondary battery (reachargeable)
• Anode: Spongy lead.
• Cathode : Lead plate bearing PbO₂.
• Electrolyte : 38 % by mass of H₂SO₄ with density 1.2 g / mL.
• EMF : emf of a single cell is 2 V. Usually six cells are combined in series to produce 12 volt.

Oxidation at anode:
\(\mathrm{Pb}_{(\mathrm{s})} \rightarrow \mathrm{Pb}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \)………..1
\(\mathrm{Pb}^{2+} ions combine with \mathrm{SO}_{4}^{2-} to form PbSO₄ precipitate.
\mathrm{Pb}_{(\mathrm{aq})}^{2+}+\mathrm{SO}_{4}^{2} \underset{(\mathrm{aq})} \rightarrow \mathrm{PbSO}_{4(\mathrm{~s})}\) ………………..(2)

Reduction at cathode;
\(\mathrm{PbO}_{2(\mathrm{~s})}+4 \mathrm{H}_{(\mathrm{aq})}+2 \mathrm{e} \rightarrow \mathrm{Pb}_{(\mathrm{aq})}^{2+}+2 \mathrm{H}_{2} \mathrm{O}_{(1)}\) …………(3)
Pb²⁺ ions combine with \(\mathrm{SO}_{4}^{2-}\) from H₂SO₄ to form PbSO₄ precipitate..
\(\mathrm{Pb}_{(\mathrm{aq})}^{2+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-} \rightarrow \mathrm{PbSO}_{4(\mathrm{~s})}\)………………..(4)

Equation (1) + (2) + (3) + (4) gives the overall redox reaction.
\(\begin{array}{r}
\mathrm{Pb}_{(s)}+\mathrm{PbO}_{2(\mathrm{~s})}+4 \mathrm{H}_{(\mathrm{aq})}^{+}+2 \mathrm{SO}_{4(\mathrm{aq})}^{2-} \rightarrow \\
2 \mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{H}_{2} \mathrm{O}_{(1)}
\end{array}\)
Emf of the cell depends on the concentration of H₂SO₄
As the cell reaction uses \(\mathrm{SO}_{4}{ }^{2}\) ions, the concentration of H₂SO₄ decreases.
When the cell potential decreases to about 1.8 V, the cell has to be recharged.

Recharge of the cell:
A potential greater than 2 V is applied across the electrodes, the cell reactions of the discharge process are reversed.
During recharge, the role of anode and cathode is reversed and H2S04 is regenerated.
Oxidation occurs at the cathode (now act as anode)

\(\begin{aligned}
\mathrm{PbSO}_{4(s)}+& 2 \mathrm{H}_{2} \mathrm{O}_{(1)} \rightarrow \\
& \mathrm{PbO}_{2(\mathrm{~s})}+4 \mathrm{H}_{(\mathrm{aq})}+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+2 \mathrm{e}
\end{aligned}\)
Reduction occurs at the anode (now act as cathode)
\(\mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}_{(s)}+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\)
Overall redox reaction is
\(\begin{array}{c}
2 \mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{H}_{2} \mathrm{O}_{(1)} \rightarrow \\
\mathrm{Pb}_{(\mathrm{s})}+\mathrm{PbO}_{2(\mathrm{~s})}+4 \mathrm{H}_{(\mathrm{aq})}+2 \mathrm{SO}_{4}^{2-}
\end{array}\)

Thus the overall cell reaction is exactly the reverse of the redox reaction which takes place while discharging. .
Uses : In automobiles, trains, inverters etc.,

Question 8.
Explain the electrochemical mechanism of corrosion.
Answer:

• The redox process which causes deterioration of metal by oxygen and moisture is called corrosion.
• Rusting or corrosion of iron is an electrochemical process.
• Rusting requires both oxygen and water.
• Since it is an electrochemical redox process, it requires an anode and cathode in different places on the iron.
• Iron surface and a droplet of water on the surface form a tiny galvanic cell as shown in the figure.
• The region enclosed by water is exposed to low amount of oxygen and it acts as anode.
• The remaining area has high amount of oxygen and it acts as cathode.
• Based on the oxygen content, an electrochemical cell is formed. 

Corrosion occurs at the anode ie, the region covered by water.
Oxidation at anode : Iron dissolves in the anode region.
\(2 \mathrm{Fe}_{(\mathrm{s})} \rightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+4 \mathrm{e}^{-} \quad \mathrm{E}^{\circ}=0.44 \mathrm{~V}\) ……………………. (1)
Electrons move through the iron metal from the anode to the cathode area.
At the cathode oxygen dissolved in water is reduced to water.

Reduction at cathode:
Atmospheric carbon dioxide and water react to give carbonic acid which furnishes the H+ions for reduction.
O_2(g) + 4HM⁺+_(aq) + 4e^– → 2H₂O_(l)
E° = 1.23V ……………..(2)
Electrical circuit is completed by the migration of ions through water droplet.
The overall redox reaction is, (1) + (2)
\(2 \mathrm{Fe}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{H}_{(\mathrm{aq})}^{+} \rightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+2 \mathrm{H}_{2} \mathrm{O}_{(l)}\)
E° = 0.44 + 1.23 = 1.67 V
The positive emf value shows that the reaction is spontaneous.
Fe²⁺ ions are further oxidised to Fe3+, which on further reaction with oxygen form rust.

VIII. Additional problems:

Problems based on conductivity :

Question 1.
A conductivity cell has two platinum electrodes separated by a distance 1.5cm and the cross sectional area of each electrode is 4.5sq cm. Using this cell, the resistance of 0.5 N electrolytic solution was measured as 15Ω. Find the specific conductance of the solution.
Solution:
l = 1.5 cm = 1.5 × 10^-2 m
A = 4.5 cm² = 4.5 × (10^-4)m²
R = 15Ω

\(K=\frac{1}{R}\left(\frac{\ell}{A}\right)\)
\(K=\frac{1}{15 \Omega} \times \frac{1.5 \times 10^{-2} \mathrm{~m}}{4.5 \times\left(10^{-4}\right) \mathrm{m}^{2}}\)
= 2.22 Sm^-1

Question 2.
0.04 N solution of a weak add has a specific conductance 4.23 × 10^-4 Scm^-1. The degree of dissociation of acid at this dilution is 0.0612. Calculate the equivalent conductance of weak acid at infinite dilution.
Solution:

Problems based on Kohlrausch’s Law

Question 1.
Equivalent conductances of NaCl, HCl and CH₃COONa at infinite dilution are 126.45,426.16 and 91 S cm2g equiv^-1. Calculate the equivalent conductance of CH₃COOH at infinite dilution.
Solution:

= 91 + 426.16 – 126.45
= 517.16 -126.45
∧°_CH3COOH = 390.71Sm²gequ^-1

Question 2.
The equivalent conductance of M/36 solution of a Weak monobasic acid is 6 mho cm² equivalent and infinite dilution is 400 mho cm² equivalent^-1. Calculate the dissociation constant of this acid.
Solution:

Problems based on Faraday’s law

Question 1.
If 50 milli ampere of current is passed through copper coulometer for 60 minutes, calculate the amount of copper deposited.
Solution:
I = 50 mA = 50 × 10^-3 A;
t = 60 min = 60 × 60 = 3600 seconds

Question 2.
0.1978 g of copper is deposited by a current of 0.2 ampere in 50 minuets. What is the electro chemical equivalent of copper?
Solution:
m = 0.1978 g; I = 0.2 A;
t = 50 mins = 50 × 60 = 3000 sec
Z = ?
m = ZIt
∴ \(Z=\frac{m}{I t}\)
\(=\frac{0.1978}{0.2 \times 3000}\)
Z = 3.297 × 10^-4gC^-1

Question 3.
What current strength in amperes will be required to liberate 10 g of iodine from potassium iodide solution in one hour?
Solution:
l = ?m = 10g;
t = 1 hr = 60 × 60 = 3600 sec

Question 4.
An electric current is passed through three cells in series containing respectively solutions of copper sulphate, silver nitrate and potassium iodide. What weight of silver and iodine will be liberated while 1.25 g copper is being deposited?
Solution:

Question 5.
On passing a current of 1 ampere for 16 min 5 sec through 1 litre solution of CuCl₂ all copper of the solution was deposited at cathode. Calculate the normality of CuCl₂ solution.
Solution:
1 = 1 ampere
t = 16 min 5 sec = 16 * 60 + 5 = 965 sec
N = ? m = ZIt

Problems based on emf

Question 1.
The reaction \(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{Co}^{2+} \rightleftharpoons \mathrm{Co}_{(\mathrm{s})}+\mathrm{Zn}^{2+}\) occurs in a cell Compute the standard emf of the cell.
Solution:
Given that
\(E^{0} z_{n} / Z n^{2+}\) = +0.76 V and \(E_{\mathrm{Co} / \mathrm{Co}}^{2+}\) = +0.28 V
Zn_(s) + Co²⁺ Co_(s) +Zn²⁺
Oxidation potential of Co ion is change into reduction potential
+0.28 V = -0.28 V
E° Cell = E° _{oxidation (2n/Zn²⁺)} + E° _{reduction(Co²⁺/Co)}
E° cell = + 0.76 + (-0.28)
E°cell = +0.48 V
This reaction is feasible.

Question 2.
What is the E°el, of the reaction
\(\mathrm{Cu}_{(\mathrm{aq})}^{2+}+\mathrm{Sn}_{(\mathrm{aq})}^{2+} \rightleftharpoons \mathrm{Cu}_{(s)}+\mathrm{Sn}_{(\mathrm{aq})}^{4+}\) at 25°C, If the equilibrium constant for the reaction is 1 × 10⁶.
Solution :
K_eq= 1 × 10⁶ T = 25° C + 273 = 298 K
R = 8.314 JK^-1 mol^-1
Cu²⁺ + 2e^– → Cu
Sn²⁺ → Sn⁴⁺ + 2e^–

Question 3.
Cu⁺_(aq) | is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction.
\(2 \mathrm{Cu}_{(\mathrm{aq})}^{+} \rightleftharpoons \mathrm{Cu}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(s)}\)
Calculate E° for the above reaction
\({ }^{\mathrm{E}^{\mathrm{O}}} \mathrm{Cu}^{2+} \mid \mathrm{Cu} = 0.34\) and \(\mathrm{E}_{\mathrm{Cu}^{2+} \mathrm{Cu}^{+}}^{\mathrm{o}}=0.15 \mathrm{~V}\)
Solution:

Equation (1) = Equation (2) + Equation (3)
ΔG°₁ = ΔG°₂ + ΔG°₃
– 0.68 F= 0.15F—1E°₃F
-0.68 = -0.15 – E°₃
E°₃ = 0.68 — 0.15 = 0.53 V
\(2 \mathrm{Cu}_{(\mathrm{aq})}^{+} \rightleftharpoons \mathrm{Cu}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(\mathrm{s})}\)
Half reactions are
Cu⁺ → Cu²⁺ + e^–
(Oxidation at anode)
E = 0.15V ……………….(4)
(Reduction at cathode)
Cu⁺ + e^–→ Cu
E°_red =E°₃= 0.53V ………………..(5)
Overall reaction is
2Cu⁺ → Cu²⁺ + Cu
:.E° = (E°_oxid)_Anode + (E°_red)_cathode
= – 0.15 + 0.53
E°_cell = 0.38V

Question 4.
For the redox reaction
\(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.1 \mathrm{M}) \rightleftharpoons \mathrm{Zn}^{2+}(1 \mathrm{M})+\mathrm{Cu}_{(\mathrm{s})}\)
if E°_cell = 1.1V calculate E_cell of the reaction
Solution:
E°ell = 1.1V; n = 2; Ecell=?
[Cu²⁺] = 0.1 M; [Zn²⁺] = 1M

Question 5.
Calculate the emf of the cell Zn | Zi²⁺ (0.001 M | Ag⁺ (0.1 M) | Ag. The standard potential of Ag⁺ | Ag half cell is + 0.80 V and Zn | Zn²⁺ is + 0.76 V.
Anode : Zn → Zn²⁺ + 2e^– E° = + 0.76 V
Cathode : 2Ag⁺ + 2e^– → 2Ag E° = + 0.80 V
Cell reaction : Zn + 2Ag⁺ ⇌ Zn²⁺ + 2Ag
E° = + 0.76 + 0.80 = + 1.56 V

E_cell = 1.56 – 0.02955 log 10^-1
E_cell = 1.56 + 0.02955 = 1.58955 V

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 3 Magnetism and Magnetic Effects of Electric Current Text Book Back Questions and Answers, Notes.

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

12th Physics Guide Magnetism and Magnetic Effects of Electric Current Text Book Back Questions and Answers

Part – 1

Text Book Evaluation:

I. Multiple choice questions:

Question 1.
The magnetic field at the center O of the following current loop is

a) \(\frac{\mu_{0} I}{4 r} \otimes\)
b) \(\frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \mathrm{r}} \odot\)
c) \(\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{r}} \otimes\)
d) \(\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{r}} \odot\)
Answer:
a) \(\frac{\mu_{0} I}{4 r} \otimes\)
Solution:

Question 2.
An electron moves straight line inside a charged parallel plate capacitor of uniform charge density σ. The time taken by the electron to cross the parallel plate capacitor undeflected when the plates of the capacitor are kept under constant magnetic field of induction B is

a) \(\varepsilon_{0} \frac{\mathrm{elB}}{\sigma}\)
b) \(\varepsilon_{\mathrm{O}} \frac{\mathrm{lB}}{\sigma l}\)
c) \(\varepsilon_{\mathrm{o}} \frac{\mathrm{lB}}{\mathrm{e} \sigma}\)
d) \(\varepsilon_{0} \frac{\mathrm{B}}{\sigma}\)
Answer:
\(\varepsilon_{0} \frac{\mathrm{B}}{\sigma}\)
Solution:
\(\mathrm{E}=\frac{\sigma}{\varepsilon_{\mathrm{o}}}\)
[∴Parallel plate capacitors]

Question 3.
A particle having mass m and charge q accelerated through a potential difference V. Find the force experienced when it is kept under perpendicular magnetic field B is
a) \(\sqrt{\frac{2 q^{3} B V}{m}}\)
b) \(\sqrt{\frac{q^{3} B^{2} V}{2 m}}\)
c) \(\sqrt{\frac{2 q^{3} B^{2} V}{m}}\)
d) \(\sqrt{\frac{2 q^{3} B V}{m^{3}}}\)
Answer:
\(\sqrt{\frac{2 q^{3} B^{2} V}{m}}\)
Solution:

Question 4.
A circular coil of radius 5 cm arid has 50 turns carries a current of 3 amperes. The magnetic dipole moment of the coil is
a) 1.0 Am²
b) 1.2 Am²
c) 0.5 Am²
d) 0.8 Am²
Answer:
b) 1.2 Am²
Solution:
r = 5 cm = 5 × 10^-2m
N = 50 turns
I = 3A
M = NIA = 50 × 3 × 3.14 × 25 × 104
M = 1.2amp – m²

Question 5.
A thin insulated wire forms a plane spiral of N=100 tight turns carrying a current I = 8mA (milliampere). The radii of inside and outside turns are a = 50mm 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
Solution:

Question 6.
Three wires of equal lengths are bent in the form of loops. One of the loops is a circle, another is a semi-circle and the third one is a square. They are placed in a uniform magnetic field and the 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
Solution:
Circle has greater torque because torque is directly proportional to area.

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 at a distance of R/2 from the center of each coil

a) \(\frac{8 \mathrm{~N} \mu_{\mathrm{o}} \mathrm{I}}{\sqrt{5} \mathrm{R}}\)
b) \(\frac{8 \mathrm{~N} \mu_{\mathrm{o}} \mathrm{I}}{5^{3 / 2} \mathrm{R}}\)
c) \(\frac{8 \mathrm{~N} \mu_{\mathrm{O}} \mathrm{I}}{5 \mathrm{R}}\)
d) \(\frac{4 \mathrm{~N} \mu_{\mathrm{o}} \mathrm{I}}{\sqrt{5} \mathrm{R}}\)
Answer:
b) \(\frac{8 \mathrm{~N} \mu_{\mathrm{o}} \mathrm{I}}{5^{3 / 2} \mathrm{R}}\)
Solution:

Question 8.
A wire of length l carrying a current I along the Y direction is kept in a and magnetic field is given by
\(\mathrm{B}=\frac{\beta}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k}) T\).
The magnitude of Lorentz force acting on the wire is
a) \(\sqrt{\frac{2}{\sqrt{3}}} \beta \mathrm{I} l\)
b) \(\sqrt{\frac{1}{\sqrt{3}}} \beta I l\)
c) √2βIl
d) \(\sqrt{\frac{1}{2}} \beta \mathrm{I} l\)
Answer:
a) \(\sqrt{\frac{2}{\sqrt{3}}} \beta \mathrm{I} l\)
Solution:

Question 9.
A bar magnet of length l and magnetic moment p_m is bent in the form of an arc as shown in Figure. The new magnetic dipole moment will be

a) p_m
b) \(\frac{3}{\pi} p_{m}\)
c) \(\frac{2}{\pi} \mathrm{p}_{\mathrm{m}}\)
d) \(\frac{1}{2} \mathrm{p}_{\mathrm{m}}\)
Answer:
b) \(\frac{3}{\pi} p_{m}\)
Solution:

Question 10.
A non-conducting charged ring of charge of q, mass m and radius r is rotated about its axis with constant angular speed to. Find the ratio of its magnetic moment with angular momentum is
a) \(\frac{q}{m}\)
b) \(\frac{2 q}{m}\)
c) \(\frac{\mathrm{q}}{2 \mathrm{~m}}\)
d) \(\frac{q}{4 m}\)
Answer:
c) \(\frac{\mathrm{q}}{2 \mathrm{~m}}\)
Solution:

Question 11.
The BH 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 mA
b) 1.25 mA
c) 1.50 mA
d) 1.75 mA
Answer:
c) 1.50 mA

Question 12.
Two short bar magnets have magnetic moments 1.20 Am² and 1.00 Am², respectively. 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 × 10^-5 Wbm^-2)
a) 3.60 × 10^-5 Wbm^-2
b) 3.5 × 10^-5 Wbm^-2
c) 2.56 × 10^-4 Wbm^-2
d) 2.2 × 10^-4 Wbm^-2
Answer:
c) 2.56 × 104^-5 Wbm²
Solution:
\(B=\frac{\mu_{0}}{4 \pi} \frac{M}{\left(d^{2}+L^{2}\right)^{3 / 2}}\)
M₁ = 1.2 Am²
M₂ = 1 Am²
2L = 0.01m, 2d = 0.2m
\(B_{1}=\frac{4 \pi x 10^{-7} \times 1.2}{4 \pi\left[(0.1)^{2}+(0.005)^{2}\right]^{3 / 2}}\)
B₁ = 1.1955 × 10^-4 Wbm^-2
\(\mathrm{B}_{2}=\frac{4 \pi \times 10^{-7} \times 1}{4 \pi\left[(0.10)^{2}+(0.005)^{2}\right]^{3 / 2}}\)
B₂ = 1.1963 × 10^-4 Wbm^-2
B = B₁ + B₂ + B_earth
= (1.1955 + 1.1963 + 0.36) × 10^-4
B = 2.56 × 10^-4 Wbm^-2

Question 13.
The vertical component of Earth’s magnetic field at a place is equal to the horizontal component. What is the value of the angle of dip at this place?
a) 30°
b) 45°
c) 60°
d) 90°
Answer:
b) 45°
Solution:
B_H = B_E cos I
B_V = B_E sin I
B_E cos I = B_E sin I
\(\frac{\sin I}{\cos I}=1\)
tan I = 1
I = 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 axis 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 axis of rotation
a) \(\frac{1}{4}\)σωπBR
b) \(\frac{1}{4}\)σωπBR²
c) \(\frac{1}{4}\)σωπBR³
d) \(\frac{1}{4}\)σωπBR⁴
Answer:
d) \(\frac{1}{4}\)σωπBR⁴
Solution:

Question 15.
The potential energy òf magnetic dipole whose dipole moment is \(\vec{p}_{m}=(-0.5 \hat{i}+0.4 \hat{j}) \mathrm{Am}^{2}\) Am² kept in uniform magnetic field \(\vec{B}=0.2 \hat{i} \mathrm{~T}\)
a -0.1 J
b) -0.8 J
c) 0.1 J
d) 0.8 J
Answer:
c) 0.1 J

II. 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:

1. The number of magnetic field lines crossing per unit area.
2. φ_B = \(\overrightarrow{\mathrm{B}}\) . \(\overrightarrow{\mathrm{A}}\) = BA cos θ
3. S.I unit: weber (Wb); Dimensional formula [ML²T^-2A^-1]

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 distance between them,
\(\overrightarrow{\mathrm{F}} \alpha \frac{\mathrm{q}_{\mathrm{m}_{\mathrm{A}}} \mathrm{q}_{\mathrm{m}_{\mathrm{B}}}}{\mathrm{r}^{2}} \hat{\mathrm{r}}\) ;
\(\mathrm{F}=\frac{\mathrm{K} \mathrm{q}_{\mathrm{m}} \mathrm{A} \quad \mathrm{q}_{\mathrm{m}} \mathrm{B}}{\mathrm{r}^{2}}\)
q_mA & q_mB – pole strengths of two poles,
r – distance between two magnetic poles.

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\(\overrightarrow{\mathrm{B}\) at a point P at a distance r from the small length on a current-carrying conductor varies,
i) directly as the strength of the current I
ii) directly as the magnitude of the length element dI and r
iii) directly as the sine of the angle between dI and r̂
iv) inversely as the square of the distance between the point P and length element dl

Question 7.
What is magnetic permeability?
Answer:
Magnetic permeability: Magnetic permeability can be defined as the measure of the ability of the material to allow the passage of magnetic field lines through it or a measure of the capacity of the substance to take magnetisation or the degree of penetration of the 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 loop.
\(\oint_{C} \vec{B} \cdot \vec{l}=\mu_{o} I_{\text {enclosed }}\)
I_enclosed → net current in the closed loop C.

Question 9.
Compare dia, para and ferro-magnetism.
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’.

Question 11.
Define magnetic declination and inclination?
Answer:
Magnetic declination:

1. The angle between magnetic meridian at a point and geographical meridian.
2. Higher altitude – Greater declination
3. Equator – Smaller declination.
4. Magnetic inclination: The angle subtended by the Earth’s total magnetic field \(\overrightarrow{\mathrm{B}}\) with the horizontal direction in the magnetic meridian.

Question 12.
What is the resonance condition in a cyclotron?
Answer:

1. In cyclotron operation when the frequency f at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator f_osc. This is called resonance condition.
   f_osc = \(\frac{q B}{2 \pi m}\)
2. The time period of oscillation is T = \(\frac{2 \pi \mathrm{m}}{\mathrm{qB}}\)
3. The kinetic energy of charged particle is \(\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{q}^{2} \mathrm{~B}^{2} \mathrm{r}^{2}}{2 \mathrm{~m}}\)

Question 13.
Define ampere.
Answer:
Current 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 × 10^-7 newton per meter length of the conductor.

Question 14.
State Fleming’s left-Hand rule.
Answer:

1. Stretch forefinger, the middle finger, and the thumb of the left hand such that they are in mutually perpendicular directions.
2. The forefinger points the direction of the magnetic field.
3. The middle finger points in the direction of the electric current.
4. Thumb will point the direction of the force experienced by the conductor.

Question 15.
Is an ammeter connected in series or parallel in a circuit? Why.
Answer:

1. An ammeter is a low resistance instrument and it is always connected in series to the circuit.
2. Such that it will change the current passing through it.

Question 16.
Explain the concept of the selector.
Answer:

1. For a given magnitude of E field and B field, the forces act only on the particle moving with particular speed p.
   \(v=\frac{E}{B}\)
2. This speed is independent of mass and charge.

Question 17.
Why is the path of a charged particle not a circle when its velocity is not perpendicular to the magnetic field?
Answer:

1. If a charged particle moves in a region of a uniform magnetic field such that its velocity is not perpendicular to the magnetic field, then the velocity of the particle is split up into two components.
2. One component is parallel to the field while the other component perpendicular to the field.
3. The component of velocity parallel to find remains unchanged and the component perpendicular to the field keeps changing due to Lorentz force.
4. Hence the path of the particle is not a circle; it is helical around the field lines.
   
   Helical path of the electron in a uniform.

Question 18.
Give the properties of dia/para/ferromagnetic materials.
Answer:

Question 19.
What happens to the domains in a ferromagnetic material in the presence of an external magnetic field?
Answer:

1. The domains having magnetic moments parallel to the field glow bigger in size.
2. The other domains (not parallel to the field) are rotated so that they are aligned with the field.

Question 20.
How is a galvanometer converted into (i) an ammeter and (ii) a voltmeter?
Answer:

1. A galvanometer can be converted into an ammeter of a given range by connecting of suitable low resistance S called shunt in parallel to the galvanometer.
2. A galvanometer can be converted into suitable high resistance. R_h in series with the given galvanometer.

III. Long Answer Questions:

Question 1.
Discuss Earth’s magnetic field in detail.
Answer:

1. William Gilbert proposed that Earth behaves like a gigantic powerful bar magnet.
2. The temperature inside the earth is very high, so it is not possible for a magnet to retain its magnetism.
3. Gover suggested that the Earth’s magnetic field is due to hot rays coming out from the sun.
4. The branch of physics which deals with the Earth’s magnetic field is called Geomagnetism or Terrestrial magnetism.
5. Three quantities required to specify the Earth’s magnetic field are
   • magnetic declination (D)
   • magnetic dip or inclination (I)
   • the horizontal component of the Earth’s magnetic field (B_H)
6. B_E → Earth’smagneticfield.
   B_H → Horizontal component
   B_V → Vertical component
   Horizontal component B_H = B_E cos I
   Vertical component B_V= B_E sin I;
   \(\tan I=\frac{B_{V}}{B_{H}}\)
   i) At magnetic equator B_H = B_E [I = 0°]
   B_V = 0
   ii) At magnetic poles B_H = 0 [I = 90°]
   B_V = B_E

Question 2.
Deduce the relation for the magnetic field at a point due to an infinitely long straight conductor carrying current.
Answer:

1. Let NM be a long straight wire carrying a current I.
2. Let dl be the elemental length at a distance l from O.
3. Let P be a point at a distance ‘a’ from ‘o’.
4. According to Biot savart law, Magnetic Induction at P is
   \(\overrightarrow{\mathrm{dB}}=\frac{\mu_{\mathrm{o}} \mathrm{I\overrightarrow{\mathrm{dl}}}}{4 \pi \mathrm{r}^{2}} \sin \theta\) ………….(1)

The direction of the field is perpendicular to the plane of the paper and going into it.

From the figure, In △PAO,

tan(π – θ) = \(\frac{a}{l}\)
∵ tan(π – θ) = -tan θ
-tan θ = \(\frac{a}{l}\)
l = \(\frac{-a}{\tan \theta}\)
l = -a cot θ ………………(2)

From △PAO
sin θ = \(\frac{a}{r}\)
\(r=\frac{a}{\sin \theta}\)
∴r = a cosec θ …………….(3)
Differentiating (2)
dl = a cosec²θ dθ …………..(4)

Question 3.
Obtain a relation for the magnetic field at a point along the axis of a circular coil carrying current.
Answer:

1. Let R be the radius of a current-carrying circular loop.
2. I be the current flowing through the wire.
3. Let P be a point on the axis of the circular coil at a distance z from its center ‘o’.
4. Take two diametrically opposite element d\(\overrightarrow{\mathrm{I}\) at C and D.
5. Let \(\overrightarrow{\mathrm{r}}\) be the vector joining the current I\(\mathrm{d} \overrightarrow{\mathrm{l}}\) at C.
   
   Magnetic field due to current carrying circular loop using Biot-Savart’s law.
   
6. PC = PD = r = \(\sqrt{R^{2}+Z^{2}}\)
   ∠CPO =∠DPO = θ
7. According to Biot – Savart’s law,
   \(\mathrm{d} \overrightarrow{\mathrm{B}}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{Id} \overrightarrow{\mathrm{l}} \times \hat{\mathrm{r}}}{\mathrm{r}^{2}}\)
8. \(\mathrm{d} \overrightarrow{\mathrm{B}}\) is resolved into two components
   dB sin θ – along y – direction
   dB cos θ – along z – direction
9. Horizontal components cancel each other vertical component contribute to total magnetic field.
10. Integrating d\(\overrightarrow{\mathrm{I}\) around the ioop, d\(\overrightarrow{\mathrm{B}\) sweeps out a cone.
    
    (Using pythagorous)
    ∵\(\cos \theta=\frac{R}{\sqrt{R^{2}+Z^{2}}}\)
    r² = R² + Z²
    \(B=\frac{\mu_{0} I}{4 \pi} \int \frac{d l}{R^{2}+Z^{2}} \frac{R}{\sqrt{R^{2}+Z^{2}}} \hat{k}\)
    

Question 4.
Compute the torque experienced by a magnetic needle in a uniform magnetic field.
Answer:

1. Consider a magnetic needle or bar magnet of length 2l.
2. Pole strength → q_m.
3. Each pole experience a force q_mB, but in opposite direction.
4. The net force is zero.
5. The magnetic needle will rotate and constitute a couple.
6. Force experienced by the north pole
7.  Force experienced by south pole

\(\overrightarrow{\mathrm{F}}_{\mathrm{N}}=\mathrm{q}_{\mathrm{m}} \overrightarrow{\mathrm{B}}\)
\(\begin{array}{l}
\overrightarrow{\mathrm{F}}_{\mathrm{S}}=-\mathrm{q}_{\mathrm{m}} \overrightarrow{\mathrm{B}} \\
\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_{\mathrm{N}}+\overrightarrow{\mathrm{F}}_{\mathrm{S}}
\end{array}\)
\(\overrightarrow{\mathrm{F}}\) = 0
\(\begin{aligned}
\text { Torque } \vec{\tau} &=\mathrm{O} \overrightarrow{\mathrm{N}} \times \overrightarrow{\mathrm{F}}_{\mathrm{N}}+\mathrm{O} \overrightarrow{\mathrm{S}} \times \overrightarrow{\mathrm{F}}_{\mathrm{S}} \\
&=\mathrm{O} \overrightarrow{\mathrm{N}} \mathrm{x} \mathrm{q}_{\mathrm{m}} \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{OS}} \times\left(-\mathrm{q}_{\mathrm{m}} \overrightarrow{\mathrm{B}}\right)
\end{aligned}\)
Using right hand cork screw rule, total torque points into the paper.

Magnetic dipole kept in a uniform magnetic field

\(\begin{array}{l}
|\mathrm{ON}|=|\mathrm{OS}|=l \\
\left|\mathrm{q}_{\mathrm{m}} \overrightarrow{\mathrm{B}}\right|=\mid-\mathrm{q}_{\mathrm{m}} \overrightarrow{\mathrm{B}}
\end{array}\)
∴\(\overrightarrow{\mathrm{τ}}\) = l × q_mBsin θ + l × q_mBsin θ

Question 5.
Calculate the magnetic field at a point on the axial line of a bar magnet.
Answer:

1. Consider a bar magnet NS.
2. N – Northpole, S – Southpole
3. Pole strength – q_m.
4. Distance – 2l
5. C be a point along the axis of the magnet.
6. By Coulomb’s law of magnetism,
7. The force of repulsion
   
   Magnetic field at a point along the axial line due to magnetic dipole
   \(\overrightarrow{\mathrm{F}}_{\mathrm{N}}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{q}_{\mathrm{m}}}{(\mathrm{r}-l)^{2}} \hat{\mathrm{i}}\) …………..(1)
8. r – l → is the distance between north pole of the bar magnet and unit north pole at C.
9. The force of attraction,
   \(\overrightarrow{\mathrm{F}}_{\mathrm{S}}=\frac{-\mu_{0}}{4 \pi} \frac{\mathrm{q}_{\mathrm{m}}}{(\mathrm{r}+l)^{2}} \hat{\mathrm{i}}\) ………………(2)
10. r + l → is the distance between south pole of the bar magnet and unit north pole at C.
11. Net force at point C.
    \(\overrightarrow{\mathrm{F}}\) = \(\overrightarrow{\mathrm{F_N}}\) + \(\overrightarrow{\mathrm{F_S}}\)
    \(\overrightarrow{\mathrm{F}}\) = \(\overrightarrow{\mathrm{B}}\)
    
    

Question 6.
Obtain the magnetic field at a point on the equatorial line of a bar magnet.
Answer:

1. Consider a bar magnet NS.
2. N – Northpole, S – Southpole
3. Pole strength – q.
4. Distance between two poles – 2l
5. C be a point along the equatorial line
6. By Coulomb’s law of magnetism,
7. The force of repulsion,

Magnetic field at a point along the equatorial line due to a magnetic dipole

\(\overrightarrow{\mathrm{F}}_{\mathrm{N}}=-\mathrm{F}_{\mathrm{N}} \cos \theta \hat{\mathrm{i}}+\mathrm{F}_{\mathrm{N}} \sin \hat{\theta} \hat{\mathrm{j}}\)
where,
\(\mathrm{F}_{\mathrm{N}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{q}_{\mathrm{m}}}{\mathrm{r’}^{2}}\) …………(1)

The force of attraction,
\(\overrightarrow{\mathrm{F}}_{\mathrm{S}}=-\mathrm{F}_{\mathrm{S}} \cos \theta \hat{\mathrm{i}}-\mathrm{F}_{\mathrm{S}} \sin \hat{\theta} \hat{\mathrm{j}}\)
where,
\(\mathrm{F}_{\mathrm{S}}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{q}_{\mathrm{m}}}{\mathrm{r}^{\prime 2}}\) ……………(2)

The net force at point C,
\(\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_{\mathrm{N}}+\overrightarrow{\mathrm{F}}_{\mathrm{S}}\)
\(\overrightarrow{\mathrm{B}}\) = – (F_N + F_S) cosθî
Here,
\(\overrightarrow{\mathrm{F}}_{\mathrm{N}}=\overrightarrow{\mathrm{F}}_{\mathrm{S}}\)
∴ \(\begin{array}{l}
\overrightarrow{\mathrm{B}}=-\frac{2 \mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{q}_{\mathrm{m}}}{\mathrm{r}^{\prime 2}} \cos \theta \hat{\mathrm{i}} \\
\overrightarrow{\mathrm{B}}=-\frac{2 \mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{q}_{\mathrm{m}}}{\left(\mathrm{r}^{2}+1^{2}\right)} \cos \theta \hat{\mathrm{i}}
\end{array}\)

Question 7.
Find the magnetic field due to a long straight conductor using Ampere’s circuital law.
Answer:

1. I be the current flowing in a straight conductor of infinite length.
2. The wire is geometrically cylindrical in shape.
3. Symmetrical about an axis.
4. Amperian loop is constructed in the form of a circular shape at a distance r from the centre of
5. From Ampere’s law, \(\oint\) \(\overrightarrow{\mathrm{B}}\) . \(\overrightarrow{\mathrm{dl}}\) = μ₀I

Amerian loop for current-carrying straight line

\(\overrightarrow{\mathrm{dl}}\) is the line element along the loop.
The angle between magnetic field and line element is zero.
∴\(\oint\) B.dl = μ₀I
B is uniform over the loop
∴ B\(\oint\)dl = μ₀I
The circumference of the loop = 2πr
∴ \(\mathrm{B} \int_{0}^{2 \pi \mathrm{r}} \mathrm{dl}\) = μ₀I
B.2πr = μ₀I
\(\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi \mathrm{r}}\)
In vector form,
\(\vec{B}=\frac{\mu_{0} I}{2 \pi r} \hat{n}\)
n̂ – unit vector along the tangent to the loop.

Question 8.
Discuss the working of a cyclotron in detail.
Answer:
Cyclotron:

1. A device used to accelerate the charged particles to gain large kinetic energy.
2. It is also called as high energy accelerator.

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

Construction:

1. The particles are allowed to move in between two semi-circular metal containers called Dees.
2. Uniform magnetic field is controlled by an electromagnet.
3. The direction of magnetic field is normal to the plane of the Dees.
4. Source ‘S’ is kept between the two Dees.
5. Dees are connected to high frequency alternating potential difference.

Working:

1. The ion ejected from the source is positively charged.
2. It is accelerated towards a Dee-1 which has negative potential at that time.
3. The ion undergoes a circular path.
4. After one semi-circular path, the ion reaches the gap between the Dees.
5. The polarities of the Dees are reversed, so that ion accelerated towards Dee-2 with greater velocity.
6. The centripetal force is provided by Lorentz force 0 ; Bqv
   ∴ \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{Bqv}\)
   \(r=\frac{m}{B q} v\)
   i.e. r α v
   
   
7. If the radius of the circular path increases, velocity also increases.
8. Particles undergo a spiral path with increasing radius.
9. Particles took out with the help of the deflector plate and hit the target T.

Limitations :

1. The speed of the ion is limited.
2. An electron cannot be accelerated.
3. Unchanged particles cannot be accelerated.

Question 9.
What is tangent law? Discuss in detail.
Answer:
Tangent law:-
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.
B = B_H tan θ
Construction:

1. A circular coil of wire wound over brass or wood.
2. The circular turn table is adjusted through levelling screws.
3. Compass box at the center
4. Pivoted magnet and aluminium pointer are found.
5. The circular scale graduated with four quadrants.

Theory:

1. Two magnetic fields are perpendicular to each other.
2. Magnetic induction due to the current in the coil acting normal to the plane of the coil.
   
   B_H – horizontal component of Earth’s magnetic field.
   Magnetic induction at the centre of a coil
   B = \(\mathrm{B}=\mu_{\mathrm{o}} \frac{\mathrm{NI}}{2 \mathrm{R}}\) …………(2)
   ∴ Substituting (2) in (1)
   \(\begin{aligned}
   \mu_{0} \frac{\mathrm{NI}}{2 \mathrm{R}} &=\mathrm{B}_{\mathrm{H}} \tan \theta \\
   \mathrm{B}_{\mathrm{H}} &=\mu_{\mathrm{o}} \frac{\mathrm{NI}}{2 \mathrm{R}} \frac{1}{\tan \theta}
   \end{aligned}\) in tesla.

Question 10.
Derive the expression for the torque on a current-carrying coil in a magnetic field.
Answer:
Consider a rectangular loop PQRS carrying current I is placed in a uniform magnetic field B. Let a and b be the length and breadth rectangular loop respectively. The unit vector n̂ normal to the plane of the loop makes an angle θ with the magnetic field.

Rectangular coil placed in a magnetic field

1. The magnitude of the magnetic force acting on the current-carrying arm PQ is F_PQ = IaBsin (π/2) = IaB.
2. The magnitude of the force on the arm QR is F_QR =IbBsin \(\left(\frac{\pi}{2}-\theta\right)\) = IbB cos θ and its direction.
3. The magnitude of the force on the arm RS is F_RS =IaBsin( π/2) = IaB and its direction is downwards.
4. The magnitude of the force acting on the arm SP is F_SP = lbB\(\left(\frac{\pi}{2}-\theta\right)\) = IbB cos θ and its direction.
5. Since the forces F_QR and F_SP are equal, opposite, and collinear, they cancel each other. But the forces F_PQ and F_RS.

Side view of the current loop

6. The magnitude of torque acting on the arm PQ about AB is τ_PQ = (1/2 sinθ) and points in the direction of AB. The magnitude of the torque acting on the arm. RS about AB is τ_RS = (b/2 sinθ) IaB and points also in the same direction AB.
7. The total torque acting on the entire loop about an axis AB is given by
\(\tau=\left(\frac{b}{2} \sin \theta\right) F_{P Q}+\left(\frac{b}{2} \sin \theta\right) F_{R S}\)
= Ia(bsinθ)B
τ = IABsinθ along the direction B.
In vector form,
\(\overrightarrow{\mathrm{τ}}\) = (\(\overrightarrow{\mathrm{IA}}\)) × \(\overrightarrow{\mathrm{B}}\).
8. The above equation can also be written in terms of magnetic dipole moment.
\(\overrightarrow{\mathrm{τ}}\) = \(\overrightarrow{\mathrm{p_m}}\) × \(\overrightarrow{\mathrm{B}}\) where \(\overrightarrow{\mathrm{p_m}}\) = \(\overrightarrow{\mathrm{IA}}\)
9. If there are N turns in the rectangular loop, the torque is given by
τ = NIAB sinθ
Special cases:
a) When θ =90° or the plane of the loop is parallel to the magnetic field, the torque on the current loop is maximum.
τ_max = IAB
b) When θ = 0°/180° or the plane of the loop is perpendicular to the magnetic field, the torque on the current loop is zero.

Question 11.
Discuss the conversion of a galvanometer into an ammeter and also a voltmeter.
Answer:
Galvanometer to an ammeter:
1. The ammeter must offer low resistance, no change in current.
2. Galvanometer is converted to an ammeter by connecting low resistance in parallel.
3. Low resistance – Shunt resistance.
4. I – current in the circuit.
5. I_g – Galvanometer current.
6. R_g – Galvanometer resistance.
7. Current through the shunt, I_s = I – I_g
8. V_galvanometer = V_shunt
I_g R_g = (I – I_g)S
\(\mathrm{S}=\frac{\mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{g}}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}\) ………..(1)
\(\mathrm{I}_{\mathrm{g}}=\frac{\mathrm{S}}{\mathrm{S}+\mathrm{R}_{\mathrm{g}}} \mathrm{I}\)
I_g α I

9. The deflection in the galvanometer is proportional to current θ α I_g
\(\begin{array}{l}
\frac{1}{\mathrm{R}_{\text {eff }}}=\frac{1}{\mathrm{R}_{\mathrm{g}}}+\frac{1}{\mathrm{~s}} \\
\frac{1}{\mathrm{R}_{\text {eff }}}=\frac{\mathrm{R}_{\mathrm{g}} \mathrm{S}}{\mathrm{R}_{\mathrm{g}}+\mathrm{S}}=\mathrm{R}_{\mathrm{a}}
\end{array}\)
10. R_a is very low.
11. An ideal ammeter has zero resistance.

Galvanometer to a voltmeter:

1. The voltmeter must have high resistance.
2. A galvanometer is converted to a voltmeter by connecting high resistance in series.
3. R_g – Galvanometer resistance.
4. I_g – Galvanometer current.
5. R_h – High resistance
6. Since it is connected in series, the current is the same.
   
   i.e. I = I_g
   I_g = \(\frac{\text { Potential difference }}{\text { total resistance }}\)
   R_v = R_g + R_h
   \(\begin{array}{l}
   \mathrm{Ig}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{g}}+\mathrm{R}_{\mathrm{h}}} \\
   \mathrm{R}_{\mathrm{h}}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{R}_{\mathrm{g}}
   \end{array}\)
   ∴I_g α V
7. An ideal voltmeter has infinite resistance.

Question 12.
Calculate the magnetic field inside and outside of the long solenoid using Ampere’s circuital law.
Answer:

1. Consider a solenoid of length L having N turns.
2. Solenoid’s diameter is much smaller compared to its length.
   The magnetic field inside the solenoid:
3. Consider a rectangular loop abcd.
4. From Ampere’s circuital law,
5. Elemental length bc and da are perpendicular to the magnetic field.
   ∴ \(\int_{b}^{c} \vec{B} \cdot d \vec{l}=\int_{b}^{c}|\vec{B}| d \vec{l} \mid \cos 90^{\circ}=0 \quad \int_{d}^{a} \vec{B} \cdot d \overrightarrow{1}=0\)
   
   Amperian loop for solenoid.
   \(\begin{array}{l}
   \oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{\mathrm{l}}=\mu_{\mathrm{O}} \mathrm{I}_{\text {enclosed }} \\
   \mathrm{C} \\
   \oint_{\mathrm{C}} \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{\mathrm{l}}=\int_{\mathrm{a}}^{b} \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{1}+\int_{\mathrm{b}}^{c} \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{\mathrm{l}}+\int_{\mathrm{c}}^{\mathrm{d}} \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{\mathrm{l}}+\int_{\mathrm{d}}^{\mathrm{a}} \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{1}
   \end{array}\)
   

Question 13.
Derive the expression for the force between two parallel, current-carrying conductors.
Answer:
1. Two long parallel current-carrying conductors separated by a distance V in air.
2. I₁, I₂ – current passing through the conductors A and B in the same direction.
3. Magnetic field due to current I2 of elemental length dl in A
\(\begin{array}{l}
\overrightarrow{\mathrm{B}}_{1}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \pi \mathrm{r}}(-\hat{\mathrm{i}}) \\
\overrightarrow{\mathrm{B}}_{1}=\frac{-\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \pi \mathrm{r}}(\hat{\mathrm{i}}
\end{array}\)
4. The direction of magnetic field is perpendicular to the plane of the paper and outwards.
5. A force on a small elemental length dl in B, experience magnetic field B₁.



6. Attractive force – direction of electric current is same.
7. Repulsive force – direction of electric current is opposite.

Question 14.
Give an account of magnetic Lorentz force.
Answer:

Direction of the Lorentz force on
(a) positive charge
(b) negative charge
1. An electric charge q is moving with velocity \(\overrightarrow{\mathrm{v}}\) in the magnetic field \(\overrightarrow{\mathrm{B}}\), experience a force called magnetic force \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) deduced by Lorentz.
\(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) = q (\(\overrightarrow{\mathrm{v}}\) × \(\overrightarrow{\mathrm{B}\)) …………(1)
\(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) = qvBsin θ ……………(2)
Equations (1) & (2) implies

2. \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) is directly proportional to the magnetic field \(\overrightarrow{\mathrm{B}}\).

3. \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) is directly proportional to the velocity \(\overrightarrow{\mathrm{V}}\).

4. \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) is directly proportional to sine of the angle between the velocity and magnetic field.

5. \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) is directly proportional to the magnitude of the charge q.

6. \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) is always perpendicular to \(\overrightarrow{\mathrm{V}}\) and \(\overrightarrow{\mathrm{B}}\).

7. The direction of \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) on negative charge is opposite to the direction of \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) on positive charge.

8. If velocity \(\overrightarrow{\mathrm{v}}\) of the charge is along magnetic field, \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}\) = 0.

Question 15.
Compare the properties of soft and hard ferromagnetic materials.
Answer:

Question 16.
Derive an expression for the force acting on a current-carrying conductor placed in a magnetic field.
Answer:

1. The force experienced is equal to the sum of Lorentz forces on the individual charge carriers in the wire.
2. Consider a small length of wire ‘dl’, area – A, Current – I.
3. The free electrons drift opposite to the direction of the current.
4. The relation between I & V_d is given by
   I = neAV_d
   
5. In magnetic field, the force experienced is given by
   \(\overrightarrow{\mathrm{F}}\) = -e(\(\overrightarrow{\mathrm{v}}_{\mathrm{d}}\) × \(\overrightarrow{\mathrm{B}}\))
   n → number of free electrons per unit volume.
   n = \(\frac{\mathrm{N}}{\mathrm{V}}\)
   V = A dl
6. Lorentz force = n A dl × – e(\(\overrightarrow{\mathrm{v}}_{\mathrm{d}}\) × \(\overrightarrow{\mathrm{B}}\))
   dF = -enA dl (\(\overrightarrow{\mathrm{v}}_{\mathrm{d}}\) × \(\overrightarrow{\mathrm{B}}\))
7. Current element, I\(\overrightarrow{\mathrm{dl}}\) = – enA\(\overrightarrow{\mathrm{v}}_{\mathrm{d}}\)
   ∴ d\(\overrightarrow{\mathrm{F}}\) = (I\(\overrightarrow{\mathrm{dl}}\) × \(\overrightarrow{\mathrm{B}}\))
   \(\overrightarrow{\mathrm{F}}\) = (I \(\overrightarrow{\mathrm{l}}\) × \(\overrightarrow{\mathrm{B}}\))
   F = BIl sin θ
   Case (i):
   If the conductor is along the magnetic field; θ = 0° ; F = 0.
   Case (ii):
   If the conductor is perpendicular to the magnetic field; θ = 90°; F = BIl.

IV. Numerical problems:

Question 1.
A bar magnet having a magnetic moment p_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.
Answer:
If cut along the axis of the magnet of length ‘l’ into 4 pieces,
New pole strength M’= \(\frac{\mathrm{m}}{4}\)
New length l’ = l

Question 2.
A conductor of linear mass density 0.2 g m1 suspended by two flexible wire as shown in figure. Suppose the tension in the supporting wires is 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 for the current.

Answer:
To have zero tension in the wires, the magnetic force per unit length must be upwards and equal to the weight per unit length.
∴\(\left|\frac{F_{\mathrm{m}}}{\mathrm{L}}\right|=\mathrm{BI}=\frac{\mathrm{mg}}{\mathrm{L}}\)
I = \(\frac{\left(\frac{\mathrm{m}}{\mathrm{L}}\right) \mathrm{g}}{\mathrm{B}}\)
\(\frac{\mathrm{m}}{\mathrm{L}}\) = 0.2 gm^-1
= 0.2 × 10^-3 kgm^-1
B = It
g = 10ms^-2
∴I = \(\frac{0.2 \times 10^{-3} \times 10}{1}\) = 2 × 10^-3 A
I = 2 mA

Question 3.
A circular coil with a cross-sectional area of 0.1 cm² is kept in a uniform magnetic field of strength 0.2 T. If the current passing in the coil is 3A and the 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. The free electron density for the material of the wire is 10²⁸ m^-3.
Answer:
N= 1
A = 0.1 × 10^-4 m²
B = 0.2 T
I = 3A
θ = 0° [Plane is perpendicular to the field]
n = 10²⁸ m^-3
a) Torque, τ = NIB A sinθ
= l × 3 × 0.2 × 0. 1 × 10 × sin 0°
Torque = 0
b) Total force on a current loop is always zero in a magnetic field.
c) for free electron, drift velocity,
v_d = q(\(\overrightarrow{\mathrm{v}}\) × \(\overrightarrow{\mathrm{B}}\))
= qv_dB sin 90°
F = Bqv_d

= \(\frac{0.6}{0.1}\) × 10²⁸ × 10⁴
= 6 × 10^-24
Average Force, F = 0.6 × 10^-23 N.

Question 4.
A bar magnet is placed in a uniform magnetic field whose strength is 0.8 T. If the bar magnet is oriented at an angle 30° with the external field experiences a torque of 0.2 Nm. Calculate
i) the magnetic moment of the magnet
ii) the work is done by the magnetic field in moving it from the most stable configuration to the most its configuration and also compute the work done by the applied magnetic field in this case.
Answer:
i) τ = 0.2 Nm
τ = P_mB sinθ

ii) W = U_f – U_i
U_f = μ_B cos 180°
U_i = -μ_B cos 0°
∴W = μ_B cos 180° – (μ_B cos 0°)
= μ_B + μ_B
W = 2μ_B;
W = 2 P_mB
W = 2 × 0.5 × 0.8
W = 0.85 J and
W_mag = -0.85 J

Question 5.
A non-conducting sphere has a mass of 100g and a radius of 20cm. A flat compact coil of wire with turns 5 is wrapped tightly around it with each turn concentric with the sphere. This sphere is placed on an inclined plane such that the plane of the coil is parallel to the inclined plane. A uniform magnetic field of 0.5 T exists in the region in a vertically upward direction. Compute the current I required to rest the sphere in equilibrium.
Answer:
The sphere is in translational equilibrium,
f_a – Mg sinθ = 0 ……………..(1)
The sphere is in rotational equilibrium.
torque = μB sinθ (Produces by magnetic field clockwise)
Frictional force (anticlockwise torque) = f_sR
R – radius of the sphere
f_sR – μB sinθ = 0 ……………..(2)
Substitute (1) in (2)
f_s = mg sinθ
∴ mg sinθ R – μB sinθ = 0
mg sinθ R = μB sinθ

μB = mgR ……………(3)
μ = NIA
μ = NIπr² …………….(4)
NIπr² = mgR
I = \(\frac{\mathrm{mgR}}{\mathrm{BN} \pi \mathrm{R}^{2}}\)
I = \(\frac{\mathrm{mg}}{\mathrm{BN} \pi \mathrm{R}}\)
m = 100g = 100 ×10^-3 kg
= 0.1 kg
R = 20 cm = 0.2m
B = 0.5 T
N = 5 turns
g = 10 m/s²

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

Similarly for four sides BC, CD, DA
B = \(\frac{4 \mu_{0} I}{\sqrt{2} \pi L}\)
Here,
I = 1.5A
L = 50 cm
L = 0.5 m
B = \(\frac{4 \mu_{0} I}{\sqrt{2} \pi L}\)
\(B=\frac{4 \times 4 \pi \times 10^{-7} \times 1.5}{\sqrt{2} \pi \times 0.5}\)
= \(\frac{24 \times 10^{-7}}{70.7 \times 10^{-2}}\)
= 0.3394 × 10^-7 × 10²
B = 3.39 × 10^-6 T
B = 3.4 × 10^-6 T

Part II:

12th Physics Guide Magnetism and Magnetic Effects of Electric Current Additional Questions and Answers

I. Choose the incorrect pair:

Question 1.
1. Aurora borealis – northern lights
2. Aurora australis – southern lights
3. Bluish-Orange – collision
4. Purplish – red – Nitrogen molecules
Answer:
Bluish orange – collision

Question 2.
1. Anticlockwise current – North pole
2. Clockwise current – South pole
3. Angular momentum – \(\frac{\mathrm{nh}}{2 \pi}\)
4. Bohr magneton – \(\frac{4 \pi \mathrm{m}}{\mathrm{eh}}\)
Answer:
Bohr magneton – \(\frac{4 \pi \mathrm{m}}{\mathrm{eh}}\)

Question 3.
1. Tangent law – B_H tan θ
2. Tangent galvanometer – Joule’s law
3. Magnetic polarity – End rule
4. Ampere’s – circuital law – Biot – Savart law
Answer:
Tangent galvanometer – Joule’s law

II. Choose the odd man out:

Question 1.
High permeability, high retentivity, low coercivity, thin hysteresis loop.
Answer:
High retentivity – property of permanent magnet. Others are properties of electromagnets.

Question 2.
William Gilbert, Gover, Hans Christian Oersted, George Simon ohm.
Answer:
George Simon Ohm – Electricity. Others explain the magnetic effect.

Question 3.
Aluminium, Platinum, Chromium, Bismuth
Answer:
Bismuth – Diamagnetic. Others are paramagnetic.

Question 4.
Phosphor bronze, suspension strip, torsion head, Dees
Answer:
Dees – Parts of the cyclotron. Others are parts of moving coil galvanometer.

III. Fill in the blanks:

Question 1.
When a charged particle enters in a uniform magnetic field, its kinetic energy ________________
Answer:
remain constant.

Question 2.
Work done by the magnetic Lorentz force is ________________
Answer:
zero.

Question 3.
The direction of the force on a current-carrying conductor placed in a magnetic field is given by ________________
Answer:
Fleming’s left-hand rule.

Question 4.
The tangent galvanometer is most sensitive at a deflection of ________________
Answer:
45°

IV. Match the following:

Question 1.

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

Question 2.

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

Question 3.

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

Question 4.

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

V. Choose the correct pair:

Question 1.
1) Right-hand thumb rule – a. Direction of the magnetic moment
2) Gauss’s law – b. Electrostatics
3) Biot – Savart law – c. Direction of current
4) Maxwell’s right-hand cork screw rule – d. The direction of the magnetic field.
Answer:
Biot – Savart law – Direction of current

Question 2.
1. Radius (r) = \(\frac{\mathrm{qB}}{\mathrm{mV}}\)
2. Frequency (f_osc) = \(\frac{2 \pi \mathrm{m}}{\mathrm{qB}}\)
3. Time period (T) = \(\frac{\mathrm{qB}}{2 \pi \mathrm{m}}\)
4. Kinetic energy (K.E) = \(\frac{q^{2} B^{2} r^{2}}{2 m}\)
Answer:
Kinetic energy (K.E) = \(\frac{q^{2} B^{2} r^{2}}{2 m}\)

Question 3.
1. Figure of merit – deflection of two-scale division
2. Current sensitivity – Deflection produced per unit time
3. Voltage sensitivity – Deflection produced per unit voltage
4. Phosphor – bronze wire – Large couple per unit twist
Answer:
Voltage sensitivity – Deflection produced per unit voltage

Question 4.
1. Ammeter – measures the potential difference
2. Voltmeter – measure current
3. Shunt resistance – Parallel connection with a galvanometer
4. Ideal Ammeter – infinite resistance
Answer:
Shunt resistance – Parallel connection with galvanometer