HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current

# HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current | Physics Class 11 - NEET PDF Download

``` 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
< >
```

## Physics Class 11

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## FAQs on HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current - Physics Class 11 - NEET

 1. What is the thermal effect of current?
Ans. The thermal effect of current refers to the phenomenon where the flow of electric current through a conductor produces heat. This is due to the resistance offered by the conductor, which causes the conversion of electrical energy into thermal energy.
 2. How is the thermal effect of current used in everyday life?
Ans. The thermal effect of current is used in various everyday applications such as electric heaters, toasters, electric stoves, and incandescent bulbs. These devices utilize the heat generated by the flow of current to produce warmth or light.
 3. What is the chemical effect of current?
Ans. The chemical effect of current occurs when electric current passing through an electrolyte causes a chemical reaction. This reaction leads to the decomposition of the electrolyte or the deposition of a metal on an electrode.
 4. How is the chemical effect of current used in electroplating?
Ans. The chemical effect of current is utilized in electroplating, where a thin layer of metal is deposited onto an object. This process involves the use of an electrolyte solution and passing an electric current through it, causing the metal ions to be attracted to the object and form a coating.
 5. What is the difference between the thermal and chemical effects of current?
Ans. The thermal effect of current refers to the production of heat when current flows through a conductor, whereas the chemical effect of current involves the chemical reactions that occur when current passes through an electrolyte. The thermal effect is primarily associated with the resistance of the conductor, while the chemical effect is due to the interaction between the current and the electrolyte.

## Physics Class 11

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