All Courses
Get the App Signup / Login
Sign in Open App
NEET Exam  >  NEET Notes  >  Physics Class 11  >  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

Download, print and study this document offline
Please wait while the PDF view is loading
 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
< >
Read More
102 videos|411 docs|121 tests
Join Course for Free

Top Courses for NEET

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.
Related Exams
About this Document
1.1K Views
4.82/5 Rating
Nov 15, 2024 Last updated
Document Description: HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current for NEET 2024 is part of Physics Class 11 preparation. The notes and questions for HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current have been prepared according to the NEET exam syllabus. Information about HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current covers topics like and HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current.
Introduction of HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current in English is available as part of our Physics Class 11 for NEET & HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current in Hindi for Physics Class 11 course. Download more important topics related with notes, lectures and mock test series for NEET Exam by signing up for free. NEET: HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current | Physics Class 11 - NEET
Description
Full syllabus notes, lecture & questions for HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current | Physics Class 11 - NEET - NEET | Plus excerises question with solution to help you revise complete syllabus for Physics Class 11 | Best notes, free PDF download
Information about HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current
In this doc you can find the meaning of HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current defined & explained in the simplest way possible. Besides explaining types of HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current theory, EduRev gives you an ample number of questions to practice HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current tests, examples and also practice NEET tests
102 videos|411 docs|121 tests
Join Course for Free
Download as PDF
Explore Courses for NEET exam

Top Courses for NEET

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

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

,

Exam

,

Previous Year Questions with Solutions

,

pdf

,

Objective type Questions

,

Viva Questions

,

study material

,

shortcuts and tricks

,

Extra Questions

,

Sample Paper

,

practice quizzes

,

Semester Notes

,

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

,

video lectures

,

Important questions

,

ppt

,

past year papers

,

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

,

MCQs

,

Free

,

Summary

,

mock tests for examination

;
Additional Information about HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current for NEET Preparation

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

The HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current is an invaluable resource that delves deep into the core of the NEET exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current now and kickstart your journey towards success in the NEET exam.

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

The importance of HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current cannot be overstated, especially for NEET aspirants. This document holds the key to success in the NEET exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

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

HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current Notes on EduRev are your ultimate resource for success.

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

The "HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current NEET Questions" guide is a valuable resource for all aspiring students preparing for the NEET exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current on the App

Students of NEET can study HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of HC Verma Solutions: Chapter 33 - Thermal & Chemical Effects of Current is prepared as per the latest NEET syllabus.
© EduRev
Education Revolution
Follow Us
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup