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# Chapter 33 : Thermal and Chemical Effects of Current - HC Verma Solution, Physics Class 11 Notes | EduRev

## Physics Class 11

Created by: Ambition Institute

## JEE : Chapter 33 : Thermal and Chemical Effects of Current - HC Verma Solution, Physics Class 11 Notes | EduRev

``` Page 1

33.1
CHAPTER – 33
THERMAL AND CHEMICAL EFFECTS OF ELECTRIC CURRENT
1. i = 2 A, r = 25 ?,
t = 1 min = 60 sec
Heat developed = i
2
RT = 2 × 2 × 25 × 60 = 6000 J
2. R = 100 ?, E = 6 v
Heat capacity of the coil = 4 J/k ?T = 15°c
Heat liberate ?
Rt
E
2
= 4 J/K × 15
? t
100
6 6
?
?
= 60 ? t = 166.67 sec = 2.8 min ?
3. (a) The power consumed by a coil of resistance R when connected across a supply v is P =
R
v
2
The resistance of the heater coil is, therefore R =
P
v
2
=
500
) 250 (
2
= 125 ??
(b) If P = 1000 w then R =
P
v
2
=
1000
) 250 (
2
= 62.5 ?
4. ƒ = 1 × 10
–6
?m P = 500 W E = 250 v
(a) R =
P
V
2
=
500
250 250 ?
= 125 ?
(b) A = 0.5 mm
2
= 0.5 × 10
–6
m
2
= 5 × 10
–7
m
2
R =
A
l ƒ
= l =
ƒ
RA
=
6
7
10 1
10 5 125
?
?
?
? ?
= 625 × 10
–1
= 62.5 m
(c) 62.5  = 2 ?r × n, 62.5 = 3 × 3.14 × 4 × 10
–3
× n
? n =
3
10 4 14 . 3 2
5 . 62
? ? ?
? n =
14 . 3 8
10 5 . 62
3
?
?
?
˜ 2500 turns ?
5. V = 250 V P = 100 w
R =
P
v
2
=
100
) 250 (
2
= 625 ??
Resistance of wire R =
A
l ƒ
= 1.7 × 10
–8
×
6
10 5
10
?
?
= 0.034 ?
? The effect in resistance = 625.034 ?
? The current in the conductor =
R
V
= ?
?
?
?
?
?
034 . 625
220
A
? The power supplied by one side of connecting wire = 034 . 0
034 . 625
220
2
? ?
?
?
?
?
?
? The total power supplied = 2 034 . 0
034 . 625
220
2
? ? ?
?
?
?
?
?
= 0.0084 w = 8.4 mw
6. E = 220 v P = 60 w
R =
P
V
2
=
60
220 220 ?
=
3
11 220 ?
?
(a) E = 180 v P =
R
V
2
=
11 220
3 180 180
?
? ?
= 40.16 ˜ 40 w
10 cm
Page 2

33.1
CHAPTER – 33
THERMAL AND CHEMICAL EFFECTS OF ELECTRIC CURRENT
1. i = 2 A, r = 25 ?,
t = 1 min = 60 sec
Heat developed = i
2
RT = 2 × 2 × 25 × 60 = 6000 J
2. R = 100 ?, E = 6 v
Heat capacity of the coil = 4 J/k ?T = 15°c
Heat liberate ?
Rt
E
2
= 4 J/K × 15
? t
100
6 6
?
?
= 60 ? t = 166.67 sec = 2.8 min ?
3. (a) The power consumed by a coil of resistance R when connected across a supply v is P =
R
v
2
The resistance of the heater coil is, therefore R =
P
v
2
=
500
) 250 (
2
= 125 ??
(b) If P = 1000 w then R =
P
v
2
=
1000
) 250 (
2
= 62.5 ?
4. ƒ = 1 × 10
–6
?m P = 500 W E = 250 v
(a) R =
P
V
2
=
500
250 250 ?
= 125 ?
(b) A = 0.5 mm
2
= 0.5 × 10
–6
m
2
= 5 × 10
–7
m
2
R =
A
l ƒ
= l =
ƒ
RA
=
6
7
10 1
10 5 125
?
?
?
? ?
= 625 × 10
–1
= 62.5 m
(c) 62.5  = 2 ?r × n, 62.5 = 3 × 3.14 × 4 × 10
–3
× n
? n =
3
10 4 14 . 3 2
5 . 62
? ? ?
? n =
14 . 3 8
10 5 . 62
3
?
?
?
˜ 2500 turns ?
5. V = 250 V P = 100 w
R =
P
v
2
=
100
) 250 (
2
= 625 ??
Resistance of wire R =
A
l ƒ
= 1.7 × 10
–8
×
6
10 5
10
?
?
= 0.034 ?
? The effect in resistance = 625.034 ?
? The current in the conductor =
R
V
= ?
?
?
?
?
?
034 . 625
220
A
? The power supplied by one side of connecting wire = 034 . 0
034 . 625
220
2
? ?
?
?
?
?
?
? The total power supplied = 2 034 . 0
034 . 625
220
2
? ? ?
?
?
?
?
?
= 0.0084 w = 8.4 mw
6. E = 220 v P = 60 w
R =
P
V
2
=
60
220 220 ?
=
3
11 220 ?
?
(a) E = 180 v P =
R
V
2
=
11 220
3 180 180
?
? ?
= 40.16 ˜ 40 w
10 cm
Thermal & Chemical Effects of Electric Current
33.2
(b) E = 240 v P =
R
V
2
=
11 220
3 240 240
?
? ?
= 71.4 ˜ 71 w
7. Output voltage = 220 ± 1% 1% of 220 V = 2.2 v
The resistance of bulb R =
P
V
2
=
100
) 220 (
2
= 484 ?
(a) For minimum power consumed V
1
= 220 – 1% = 220 – 2.2 = 217.8
? i =
R
V
1
=
484
8 . 217
= 0.45 A
Power consumed = i × V
1
= 0.45 × 217.8 = 98.01 W
(b) for maximum power consumed V
2
= 220 + 1% = 220 + 2.2 = 222.2
? i =
R
V
2
=
484
2 . 222
= 0.459
Power consumed = i × V
2
= 0.459 × 222.2 = 102 W
8. V = 220 v P = 100 w
R =
P
V
2
=
100
220 220 ?
= 484 ?
P = 150 w V = PR = 22 22 150 ? ? = 150 22 = 269.4 ˜ 270 v ?
9. P = 1000 V = 220 v R =
P
V
2
=
1000
48400
= 48.4 ?
Mass of water = 1000
100
1
? = 10 kg
Heat required to raise the temp. of given amount of water = ms ?t = 10 × 4200 × 25 = 1050000
Now heat liberated is only 60%. So % 60 T
R
V
2
? ? = 1050000
? T
100
60
4 . 48
) 220 (
2
? ? = 1050000 ? T =
60
1
6
10500
? nub = 29.16 min. ?
10. Volume of water boiled = 4 × 200 cc = 800 cc
T
1
= 25°C T
2
= 100°C ? T
2
– T
1
= 75°C
Mass of water boiled = 800 × 1 = 800 gm = 0.8 kg
Q(heat req.) = MS ?? = 0.8 × 4200 × 75 = 252000 J.
1000 watt – hour = 1000 × 3600 watt-sec = 1000× 3600 J
No. of units =
3600 1000
252000
?
= 0.07 = 7 paise
(b) Q = mS ?T = 0.8 × 4200 × 95 J
No. of units =
3600 1000
95 4200 8 . 0
?
? ?
= 0.0886 ˜ 0.09
Money consumed = 0.09 Rs = 9 paise.
11. P = 100 w V = 220 v
Case I : Excess power = 100 – 40 = 60 w
Power converted to light =
100
60 60 ?
= 36 w
Case II : Power =
484
) 220 (
2
= 82.64 w
Excess power = 82.64 – 40 = 42.64 w
Power converted to light =
100
60
64 . 42 ? = 25.584 w
Page 3

33.1
CHAPTER – 33
THERMAL AND CHEMICAL EFFECTS OF ELECTRIC CURRENT
1. i = 2 A, r = 25 ?,
t = 1 min = 60 sec
Heat developed = i
2
RT = 2 × 2 × 25 × 60 = 6000 J
2. R = 100 ?, E = 6 v
Heat capacity of the coil = 4 J/k ?T = 15°c
Heat liberate ?
Rt
E
2
= 4 J/K × 15
? t
100
6 6
?
?
= 60 ? t = 166.67 sec = 2.8 min ?
3. (a) The power consumed by a coil of resistance R when connected across a supply v is P =
R
v
2
The resistance of the heater coil is, therefore R =
P
v
2
=
500
) 250 (
2
= 125 ??
(b) If P = 1000 w then R =
P
v
2
=
1000
) 250 (
2
= 62.5 ?
4. ƒ = 1 × 10
–6
?m P = 500 W E = 250 v
(a) R =
P
V
2
=
500
250 250 ?
= 125 ?
(b) A = 0.5 mm
2
= 0.5 × 10
–6
m
2
= 5 × 10
–7
m
2
R =
A
l ƒ
= l =
ƒ
RA
=
6
7
10 1
10 5 125
?
?
?
? ?
= 625 × 10
–1
= 62.5 m
(c) 62.5  = 2 ?r × n, 62.5 = 3 × 3.14 × 4 × 10
–3
× n
? n =
3
10 4 14 . 3 2
5 . 62
? ? ?
? n =
14 . 3 8
10 5 . 62
3
?
?
?
˜ 2500 turns ?
5. V = 250 V P = 100 w
R =
P
v
2
=
100
) 250 (
2
= 625 ??
Resistance of wire R =
A
l ƒ
= 1.7 × 10
–8
×
6
10 5
10
?
?
= 0.034 ?
? The effect in resistance = 625.034 ?
? The current in the conductor =
R
V
= ?
?
?
?
?
?
034 . 625
220
A
? The power supplied by one side of connecting wire = 034 . 0
034 . 625
220
2
? ?
?
?
?
?
?
? The total power supplied = 2 034 . 0
034 . 625
220
2
? ? ?
?
?
?
?
?
= 0.0084 w = 8.4 mw
6. E = 220 v P = 60 w
R =
P
V
2
=
60
220 220 ?
=
3
11 220 ?
?
(a) E = 180 v P =
R
V
2
=
11 220
3 180 180
?
? ?
= 40.16 ˜ 40 w
10 cm
Thermal & Chemical Effects of Electric Current
33.2
(b) E = 240 v P =
R
V
2
=
11 220
3 240 240
?
? ?
= 71.4 ˜ 71 w
7. Output voltage = 220 ± 1% 1% of 220 V = 2.2 v
The resistance of bulb R =
P
V
2
=
100
) 220 (
2
= 484 ?
(a) For minimum power consumed V
1
= 220 – 1% = 220 – 2.2 = 217.8
? i =
R
V
1
=
484
8 . 217
= 0.45 A
Power consumed = i × V
1
= 0.45 × 217.8 = 98.01 W
(b) for maximum power consumed V
2
= 220 + 1% = 220 + 2.2 = 222.2
? i =
R
V
2
=
484
2 . 222
= 0.459
Power consumed = i × V
2
= 0.459 × 222.2 = 102 W
8. V = 220 v P = 100 w
R =
P
V
2
=
100
220 220 ?
= 484 ?
P = 150 w V = PR = 22 22 150 ? ? = 150 22 = 269.4 ˜ 270 v ?
9. P = 1000 V = 220 v R =
P
V
2
=
1000
48400
= 48.4 ?
Mass of water = 1000
100
1
? = 10 kg
Heat required to raise the temp. of given amount of water = ms ?t = 10 × 4200 × 25 = 1050000
Now heat liberated is only 60%. So % 60 T
R
V
2
? ? = 1050000
? T
100
60
4 . 48
) 220 (
2
? ? = 1050000 ? T =
60
1
6
10500
? nub = 29.16 min. ?
10. Volume of water boiled = 4 × 200 cc = 800 cc
T
1
= 25°C T
2
= 100°C ? T
2
– T
1
= 75°C
Mass of water boiled = 800 × 1 = 800 gm = 0.8 kg
Q(heat req.) = MS ?? = 0.8 × 4200 × 75 = 252000 J.
1000 watt – hour = 1000 × 3600 watt-sec = 1000× 3600 J
No. of units =
3600 1000
252000
?
= 0.07 = 7 paise
(b) Q = mS ?T = 0.8 × 4200 × 95 J
No. of units =
3600 1000
95 4200 8 . 0
?
? ?
= 0.0886 ˜ 0.09
Money consumed = 0.09 Rs = 9 paise.
11. P = 100 w V = 220 v
Case I : Excess power = 100 – 40 = 60 w
Power converted to light =
100
60 60 ?
= 36 w
Case II : Power =
484
) 220 (
2
= 82.64 w
Excess power = 82.64 – 40 = 42.64 w
Power converted to light =
100
60
64 . 42 ? = 25.584 w
Thermal & Chemical Effects of Electric Current
33.3
?P = 36 – 25.584 = 10.416
Required % = 100
36
416 . 10
? = 28.93 ˜ 29% ?
12. R
eff
= 1
8
12
? =
2
5
i =
? ? 2 / 5
6
=
5
12
Amp.
i ? 6 = (i – i ?)2 ? i ? 6 = i 2 2
5
12
? ?
8i ? =
5
24
? i ? =
8 5
24
?
=
5
3
Amp
i – i ? =
5
3
5
12
? =
5
9
Amp
(a) Heat = i
2
RT = 60 15 2
5
9
5
9
? ? ? ? = 5832
2000 J of heat raises the temp. by 1K
5832 J of heat raises the temp. by 2.916K.
(b) When 6 ? resistor get burnt R
eff
= 1 + 2 = 3 ??
i =
3
6
= 2 Amp.
Heat = 2 × 2 × 2 ×15 × 60 = 7200 J
2000 J raises the temp. by 1K
7200 J raises the temp by 3.6k
13. ? = 0.001°C a = – 46 × 10
–6
v/deg, b = –0. 48 × 10
–6
v/deg
2
Emf = a
BlAg
? +(1/2) b
BlAg
?
2
= – 46 × 10
–6
× 0.001 – (1/2) × 0.48 × 10
–6
(0.001)
2
= – 46 × 10
–9
– 0.24 × 10
–12
= – 46.00024 × 10
–9
= – 4.6 × 10
–8
V ?
14. E = a
AB
? + b
AB
?
2
a
CuAg
= a
CuPb
– b
AgPb
= 2.76 – 2.5 = 0.26 ?v/°C
b
CuAg
= b
CuPb
– b
AgPb
= 0.012 – 0.012 ?vc = 0
E = a
AB
? = (0.26 × 40) ?V = 1.04 × 10
–5
V
15. ? = 0°C
a
Cu,Fe
= a
Cu,Pb
– a
Fe,Pb
= 2.76 – 16.6 = – 13.8 ?v/°C
B
Cu,Fe
= b
Cu,Pb
– b
Fe,Pb
= 0.012 + 0.030 = 0.042 ?v/°C
2
Neutral temp. on –
b
a
=
042 . 0
8 . 13
°C = 328.57°C
16. (a) 1eq. mass of the substance requires 96500 coulombs
Since the element is monoatomic, thus eq. mass = mol. Mass
6.023 × 10
23
atoms require 96500 C
1 atoms require
23
10 023 . 6
96500
?
C = 1.6 × 10
–19
C
(b) Since the element is diatomic eq.mass = (1/2) mol.mass
? (1/2) × 6.023 × 10
23
atoms 2eq. 96500 C
? 1 atom require =
23
10 023 . 6
2 96500
?
?
= 3.2 × 10
–19
C
17. At Wt. At = 107.9 g/mole
I = 0.500 A
E
Ag
= 107.9 g [As Ag is monoatomic]
Z
Ag
=
f
E
=
96500
9 . 107
= 0.001118
M = Zit = 0.001118 × 0.5 × 3600 = 2.01
1 ? ?
i
i-i ?
6
6 ? ?
2 ? ?
Page 4

33.1
CHAPTER – 33
THERMAL AND CHEMICAL EFFECTS OF ELECTRIC CURRENT
1. i = 2 A, r = 25 ?,
t = 1 min = 60 sec
Heat developed = i
2
RT = 2 × 2 × 25 × 60 = 6000 J
2. R = 100 ?, E = 6 v
Heat capacity of the coil = 4 J/k ?T = 15°c
Heat liberate ?
Rt
E
2
= 4 J/K × 15
? t
100
6 6
?
?
= 60 ? t = 166.67 sec = 2.8 min ?
3. (a) The power consumed by a coil of resistance R when connected across a supply v is P =
R
v
2
The resistance of the heater coil is, therefore R =
P
v
2
=
500
) 250 (
2
= 125 ??
(b) If P = 1000 w then R =
P
v
2
=
1000
) 250 (
2
= 62.5 ?
4. ƒ = 1 × 10
–6
?m P = 500 W E = 250 v
(a) R =
P
V
2
=
500
250 250 ?
= 125 ?
(b) A = 0.5 mm
2
= 0.5 × 10
–6
m
2
= 5 × 10
–7
m
2
R =
A
l ƒ
= l =
ƒ
RA
=
6
7
10 1
10 5 125
?
?
?
? ?
= 625 × 10
–1
= 62.5 m
(c) 62.5  = 2 ?r × n, 62.5 = 3 × 3.14 × 4 × 10
–3
× n
? n =
3
10 4 14 . 3 2
5 . 62
? ? ?
? n =
14 . 3 8
10 5 . 62
3
?
?
?
˜ 2500 turns ?
5. V = 250 V P = 100 w
R =
P
v
2
=
100
) 250 (
2
= 625 ??
Resistance of wire R =
A
l ƒ
= 1.7 × 10
–8
×
6
10 5
10
?
?
= 0.034 ?
? The effect in resistance = 625.034 ?
? The current in the conductor =
R
V
= ?
?
?
?
?
?
034 . 625
220
A
? The power supplied by one side of connecting wire = 034 . 0
034 . 625
220
2
? ?
?
?
?
?
?
? The total power supplied = 2 034 . 0
034 . 625
220
2
? ? ?
?
?
?
?
?
= 0.0084 w = 8.4 mw
6. E = 220 v P = 60 w
R =
P
V
2
=
60
220 220 ?
=
3
11 220 ?
?
(a) E = 180 v P =
R
V
2
=
11 220
3 180 180
?
? ?
= 40.16 ˜ 40 w
10 cm
Thermal & Chemical Effects of Electric Current
33.2
(b) E = 240 v P =
R
V
2
=
11 220
3 240 240
?
? ?
= 71.4 ˜ 71 w
7. Output voltage = 220 ± 1% 1% of 220 V = 2.2 v
The resistance of bulb R =
P
V
2
=
100
) 220 (
2
= 484 ?
(a) For minimum power consumed V
1
= 220 – 1% = 220 – 2.2 = 217.8
? i =
R
V
1
=
484
8 . 217
= 0.45 A
Power consumed = i × V
1
= 0.45 × 217.8 = 98.01 W
(b) for maximum power consumed V
2
= 220 + 1% = 220 + 2.2 = 222.2
? i =
R
V
2
=
484
2 . 222
= 0.459
Power consumed = i × V
2
= 0.459 × 222.2 = 102 W
8. V = 220 v P = 100 w
R =
P
V
2
=
100
220 220 ?
= 484 ?
P = 150 w V = PR = 22 22 150 ? ? = 150 22 = 269.4 ˜ 270 v ?
9. P = 1000 V = 220 v R =
P
V
2
=
1000
48400
= 48.4 ?
Mass of water = 1000
100
1
? = 10 kg
Heat required to raise the temp. of given amount of water = ms ?t = 10 × 4200 × 25 = 1050000
Now heat liberated is only 60%. So % 60 T
R
V
2
? ? = 1050000
? T
100
60
4 . 48
) 220 (
2
? ? = 1050000 ? T =
60
1
6
10500
? nub = 29.16 min. ?
10. Volume of water boiled = 4 × 200 cc = 800 cc
T
1
= 25°C T
2
= 100°C ? T
2
– T
1
= 75°C
Mass of water boiled = 800 × 1 = 800 gm = 0.8 kg
Q(heat req.) = MS ?? = 0.8 × 4200 × 75 = 252000 J.
1000 watt – hour = 1000 × 3600 watt-sec = 1000× 3600 J
No. of units =
3600 1000
252000
?
= 0.07 = 7 paise
(b) Q = mS ?T = 0.8 × 4200 × 95 J
No. of units =
3600 1000
95 4200 8 . 0
?
? ?
= 0.0886 ˜ 0.09
Money consumed = 0.09 Rs = 9 paise.
11. P = 100 w V = 220 v
Case I : Excess power = 100 – 40 = 60 w
Power converted to light =
100
60 60 ?
= 36 w
Case II : Power =
484
) 220 (
2
= 82.64 w
Excess power = 82.64 – 40 = 42.64 w
Power converted to light =
100
60
64 . 42 ? = 25.584 w
Thermal & Chemical Effects of Electric Current
33.3
?P = 36 – 25.584 = 10.416
Required % = 100
36
416 . 10
? = 28.93 ˜ 29% ?
12. R
eff
= 1
8
12
? =
2
5
i =
? ? 2 / 5
6
=
5
12
Amp.
i ? 6 = (i – i ?)2 ? i ? 6 = i 2 2
5
12
? ?
8i ? =
5
24
? i ? =
8 5
24
?
=
5
3
Amp
i – i ? =
5
3
5
12
? =
5
9
Amp
(a) Heat = i
2
RT = 60 15 2
5
9
5
9
? ? ? ? = 5832
2000 J of heat raises the temp. by 1K
5832 J of heat raises the temp. by 2.916K.
(b) When 6 ? resistor get burnt R
eff
= 1 + 2 = 3 ??
i =
3
6
= 2 Amp.
Heat = 2 × 2 × 2 ×15 × 60 = 7200 J
2000 J raises the temp. by 1K
7200 J raises the temp by 3.6k
13. ? = 0.001°C a = – 46 × 10
–6
v/deg, b = –0. 48 × 10
–6
v/deg
2
Emf = a
BlAg
? +(1/2) b
BlAg
?
2
= – 46 × 10
–6
× 0.001 – (1/2) × 0.48 × 10
–6
(0.001)
2
= – 46 × 10
–9
– 0.24 × 10
–12
= – 46.00024 × 10
–9
= – 4.6 × 10
–8
V ?
14. E = a
AB
? + b
AB
?
2
a
CuAg
= a
CuPb
– b
AgPb
= 2.76 – 2.5 = 0.26 ?v/°C
b
CuAg
= b
CuPb
– b
AgPb
= 0.012 – 0.012 ?vc = 0
E = a
AB
? = (0.26 × 40) ?V = 1.04 × 10
–5
V
15. ? = 0°C
a
Cu,Fe
= a
Cu,Pb
– a
Fe,Pb
= 2.76 – 16.6 = – 13.8 ?v/°C
B
Cu,Fe
= b
Cu,Pb
– b
Fe,Pb
= 0.012 + 0.030 = 0.042 ?v/°C
2
Neutral temp. on –
b
a
=
042 . 0
8 . 13
°C = 328.57°C
16. (a) 1eq. mass of the substance requires 96500 coulombs
Since the element is monoatomic, thus eq. mass = mol. Mass
6.023 × 10
23
atoms require 96500 C
1 atoms require
23
10 023 . 6
96500
?
C = 1.6 × 10
–19
C
(b) Since the element is diatomic eq.mass = (1/2) mol.mass
? (1/2) × 6.023 × 10
23
atoms 2eq. 96500 C
? 1 atom require =
23
10 023 . 6
2 96500
?
?
= 3.2 × 10
–19
C
17. At Wt. At = 107.9 g/mole
I = 0.500 A
E
Ag
= 107.9 g [As Ag is monoatomic]
Z
Ag
=
f
E
=
96500
9 . 107
= 0.001118
M = Zit = 0.001118 × 0.5 × 3600 = 2.01
1 ? ?
i
i-i ?
6
6 ? ?
2 ? ?
Thermal & Chemical Effects of Electric Current
33.4
18. t = 3 min = 180 sec w = 2 g
E.C.E = 1.12 × 10
–6
kg/c
? 3 × 10
–3
= 1.12 × 10
–6
× i × 180
? i =
180 10 12 . 1
10 3
6
3
? ?
?
?
?
=
2
10
72 . 6
1
? ˜ 15 Amp.
19.
L 4 . 22
H
2
? 2g 1L ?
4 . 22
2
m = Zit
4 . 22
2
= T 5
96500
1
? ? ? T =
5
96500
4 . 22
2
? = 1732.21 sec ˜ 28.7 min ˜ 29 min.
20. w
1
= Zit ? 1 = 3600 5 . 1 2
96500 3
mm
? ? ?
?
? mm =
3600 5 . 1 2
96500 3
? ?
?
= 26.8 g/mole
2
1
E
E
=
2
1
w
w
?
?
?
?
?
?
?
3
mm
9 . 107
=
1
w
1
? w
1
=
8 . 26
3 9 . 107 ?
= 12.1 gm
21. I = 15 A Surface area = 200 cm
2
, Thickness = 0.1 mm
Volume of Ag deposited = 200 × 0.01 = 2 cm
3
for one side
For both sides, Mass of Ag = 4 × 10.5 = 42 g
Z
Ag
=
F
E
=
96500
9 . 107
m = ZIT
? 42 = T 15
96500
9 . 107
? ? ? T =
15 9 . 107
96500 42
?
?
= 2504.17 sec = 41.73 min ˜ 42 min
22. w = Zit
2.68 = 60 10 i
96500
9 . 107
? ? ?
? I =
6 9 . 107
965 68 . 2
?
?
= 3.99 ˜ 4 Amp
Heat developed in the 20 ? resister = (4)
2
× 20 × 10 × 60 = 192000 J = 192 KJ ?
23. For potential drop, t = 30 min = 180 sec
V
i
= V
f
+ iR ? 12 = 10 + 2i ? i = 1 Amp
m = Zit = 60 30 1
96500
9 . 107
? ? ? = 2.01 g ˜ 2 g
24. A= 10 cm
2
× 10
–4
cm
2
t = 10m = 10 × 10
–6
Volume = A(2t) = 10 × 10
–4
× 2 × 10 × 10
–6
= 2 × 10
2
× 10
–10
= 2 × 10
–8
m
3
Mass = 2 × 10
–8
× 9000 = 18 × 10
–5
kg
W = Z × C ? 18 × 10
–5
= 3 × 10
–7
× C
? q =
7
5
10 3
10 18
?
?
?
?
= 6 × 10
2
V =
q
W
= ? W = Vq = 12 × 6 × 10
2
= 76 × 10
2
= 7.6 KJ
? ? ? ? ?
20
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```
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## Physics Class 11

129 videos|243 docs|151 tests

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