NEET  >  HC Verma Solutions: Chapter 32 - Electric Current in Conductors

# HC Verma Solutions: Chapter 32 - Electric Current in Conductors - Notes | Study Physics Class 11 - NEET

``` Page 1

32.1
ELECTRIC CURRENT IN CONDUCTORS
CHAPTER - 32
1. Q(t) = At
2
+ Bt + c
a) At
2
= Q
? A =
1 1
2 2
Q A'T'
A T
t T
?
?
? ?
b) Bt = Q
? B =
Q A 'T'
A
t T
? ?
c) C = [Q]
? C = A ?T ?
d) Current t =
? ?
2
dQ d
At Bt C
dt dt
? ? ?
= 2At + B = 2 ? 5 ? 5 + 3 = 53 A.
2. No. of electrons per second = 2 ? 10
16
electrons / sec.
Charge passing per second = 2 ? 10
16
? 1.6 ? 10
–9
coulomb
sec
= 3.2 ? 10
–9
Coulomb/sec
Current = 3.2 ? 10
–3
A.
3. i ? = 2 ?A, t = 5 min = 5 ? 60 sec.
q = i t = 2 ? 10
–6
? 5 ? 60
= 10 ? 60 ? 10
–6
c = 6 ? 10
–4
c
4. i = i
0
+ ?t, t = 10 sec, i
0
= 10 A, ? = 4 A/sec.
q =
t t t t
0 0
0 0 0 0
idt (i t)dt i dt tdt ? ? ? ? ? ?
? ? ? ?
= i
0
t +
2
t 10 10
10 10 4
2 2
?
? ? ? ? ?
= 100 + 200 = 300 C.
5. i = 1 A, A = 1 mm
2
= 1 ? 10
–6
m
2
f ? cu = 9000 kg/m
3
Molecular mass has N
0
atoms
= m Kg has (N
0
/M ? m) atoms =
0
3
N AI9000
63.5 10
?
?
No.of atoms = No.of electrons
n =
0 0
N Af N f No.of electrons
Unit volume mAI M
? ?
=
23
3
6 10 9000
63.5 10
?
? ?
?
i = V
d
n A e.
? V
d
=
23
6 19
3
i 1
nAe 6 10 9000
10 1.6 10
63.5 10
? ?
?
?
? ?
? ? ?
?
=
3
23 6 19
63.5 10
6 10 9000 10 1.6 10
?
? ?
?
? ? ? ? ?
=
3
26 19 6
63.5 10
6 9 1.6 10 10 10
?
? ?
?
? ? ? ? ?
Page 2

32.1
ELECTRIC CURRENT IN CONDUCTORS
CHAPTER - 32
1. Q(t) = At
2
+ Bt + c
a) At
2
= Q
? A =
1 1
2 2
Q A'T'
A T
t T
?
?
? ?
b) Bt = Q
? B =
Q A 'T'
A
t T
? ?
c) C = [Q]
? C = A ?T ?
d) Current t =
? ?
2
dQ d
At Bt C
dt dt
? ? ?
= 2At + B = 2 ? 5 ? 5 + 3 = 53 A.
2. No. of electrons per second = 2 ? 10
16
electrons / sec.
Charge passing per second = 2 ? 10
16
? 1.6 ? 10
–9
coulomb
sec
= 3.2 ? 10
–9
Coulomb/sec
Current = 3.2 ? 10
–3
A.
3. i ? = 2 ?A, t = 5 min = 5 ? 60 sec.
q = i t = 2 ? 10
–6
? 5 ? 60
= 10 ? 60 ? 10
–6
c = 6 ? 10
–4
c
4. i = i
0
+ ?t, t = 10 sec, i
0
= 10 A, ? = 4 A/sec.
q =
t t t t
0 0
0 0 0 0
idt (i t)dt i dt tdt ? ? ? ? ? ?
? ? ? ?
= i
0
t +
2
t 10 10
10 10 4
2 2
?
? ? ? ? ?
= 100 + 200 = 300 C.
5. i = 1 A, A = 1 mm
2
= 1 ? 10
–6
m
2
f ? cu = 9000 kg/m
3
Molecular mass has N
0
atoms
= m Kg has (N
0
/M ? m) atoms =
0
3
N AI9000
63.5 10
?
?
No.of atoms = No.of electrons
n =
0 0
N Af N f No.of electrons
Unit volume mAI M
? ?
=
23
3
6 10 9000
63.5 10
?
? ?
?
i = V
d
n A e.
? V
d
=
23
6 19
3
i 1
nAe 6 10 9000
10 1.6 10
63.5 10
? ?
?
?
? ?
? ? ?
?
=
3
23 6 19
63.5 10
6 10 9000 10 1.6 10
?
? ?
?
? ? ? ? ?
=
3
26 19 6
63.5 10
6 9 1.6 10 10 10
?
? ?
?
? ? ? ? ?
Electric Current in Conductors
32.2
=
3 3
63.5 10 63.5 10
6 9 1.6 10 6 9 16
? ?
? ?
?
? ? ? ? ?
= 0.074 ? 10
–3
m/s = 0.074 mm/s.
6. ? = 1 m, r = 0.1 mm = 0.1 ? 10
–3
m
R = 100 ?, f = ?
? R = f ? / a
? f =
6
Ra 100 3.14 0.1 0.1 10
1
?
? ? ? ?
?
?
= 3.14 ? 10
–6
= ? ? 10
–6
?-m.
?
7. ? ? = 2 ?
volume of the wire remains constant.
A ? = A ? ? ?
? A ? = A ? ? 2 ?
? A ? = A/2
f = Specific resistance
R =
f
A
?
; R ? =
f '
A '
?
100 ? =
f2 4f
A / 2 A
?
? ?
= 4R
? 4 ? 100 ? = 400 ? ?
8. ? = 4 m, A = 1 mm
2
= 1 ? 10
–6
m
2
I = 2 A, n/V = 10
29
, t = ?
i = n A V
d
e
? e = 10
29
? 1 ? 10
–6
? V
d
? 1.6 ? 10
–19
?
d
29 6 19
2
V
10 10 1.6 10
? ?
?
? ? ?
=
4
1 1
8000 0.8 10
?
?
t =
d
4
4 8000
V 1/ 8000
? ? ?
?
= 32000 = 3.2 ? 10
4
sec.
9. f
cu
= 1.7 ? 10
–8
?-m
A = 0.01 mm
2
= 0.01 ? 10
–6
m
2
R = 1 K ? = 10
3
?
R =
f
a
?
? 10
3
=
8
6
1.7 10
10
?
?
? ? ?
?
3
10
1.7
? ? = 0.58 ? 10
3
m = 0.6 km. ?
10. dR, due to the small strip dx at a distanc x d = R =
2
fdx
y ?
…(1)
tan ? =
y a b a
x L
? ?
?
?
y a b a
x L
? ?
?
? L(y – a) = x(b – a)
x ?
b–a
dx
a ?
y ?
? ? Y–a
b ?
Page 3

32.1
ELECTRIC CURRENT IN CONDUCTORS
CHAPTER - 32
1. Q(t) = At
2
+ Bt + c
a) At
2
= Q
? A =
1 1
2 2
Q A'T'
A T
t T
?
?
? ?
b) Bt = Q
? B =
Q A 'T'
A
t T
? ?
c) C = [Q]
? C = A ?T ?
d) Current t =
? ?
2
dQ d
At Bt C
dt dt
? ? ?
= 2At + B = 2 ? 5 ? 5 + 3 = 53 A.
2. No. of electrons per second = 2 ? 10
16
electrons / sec.
Charge passing per second = 2 ? 10
16
? 1.6 ? 10
–9
coulomb
sec
= 3.2 ? 10
–9
Coulomb/sec
Current = 3.2 ? 10
–3
A.
3. i ? = 2 ?A, t = 5 min = 5 ? 60 sec.
q = i t = 2 ? 10
–6
? 5 ? 60
= 10 ? 60 ? 10
–6
c = 6 ? 10
–4
c
4. i = i
0
+ ?t, t = 10 sec, i
0
= 10 A, ? = 4 A/sec.
q =
t t t t
0 0
0 0 0 0
idt (i t)dt i dt tdt ? ? ? ? ? ?
? ? ? ?
= i
0
t +
2
t 10 10
10 10 4
2 2
?
? ? ? ? ?
= 100 + 200 = 300 C.
5. i = 1 A, A = 1 mm
2
= 1 ? 10
–6
m
2
f ? cu = 9000 kg/m
3
Molecular mass has N
0
atoms
= m Kg has (N
0
/M ? m) atoms =
0
3
N AI9000
63.5 10
?
?
No.of atoms = No.of electrons
n =
0 0
N Af N f No.of electrons
Unit volume mAI M
? ?
=
23
3
6 10 9000
63.5 10
?
? ?
?
i = V
d
n A e.
? V
d
=
23
6 19
3
i 1
nAe 6 10 9000
10 1.6 10
63.5 10
? ?
?
?
? ?
? ? ?
?
=
3
23 6 19
63.5 10
6 10 9000 10 1.6 10
?
? ?
?
? ? ? ? ?
=
3
26 19 6
63.5 10
6 9 1.6 10 10 10
?
? ?
?
? ? ? ? ?
Electric Current in Conductors
32.2
=
3 3
63.5 10 63.5 10
6 9 1.6 10 6 9 16
? ?
? ?
?
? ? ? ? ?
= 0.074 ? 10
–3
m/s = 0.074 mm/s.
6. ? = 1 m, r = 0.1 mm = 0.1 ? 10
–3
m
R = 100 ?, f = ?
? R = f ? / a
? f =
6
Ra 100 3.14 0.1 0.1 10
1
?
? ? ? ?
?
?
= 3.14 ? 10
–6
= ? ? 10
–6
?-m.
?
7. ? ? = 2 ?
volume of the wire remains constant.
A ? = A ? ? ?
? A ? = A ? ? 2 ?
? A ? = A/2
f = Specific resistance
R =
f
A
?
; R ? =
f '
A '
?
100 ? =
f2 4f
A / 2 A
?
? ?
= 4R
? 4 ? 100 ? = 400 ? ?
8. ? = 4 m, A = 1 mm
2
= 1 ? 10
–6
m
2
I = 2 A, n/V = 10
29
, t = ?
i = n A V
d
e
? e = 10
29
? 1 ? 10
–6
? V
d
? 1.6 ? 10
–19
?
d
29 6 19
2
V
10 10 1.6 10
? ?
?
? ? ?
=
4
1 1
8000 0.8 10
?
?
t =
d
4
4 8000
V 1/ 8000
? ? ?
?
= 32000 = 3.2 ? 10
4
sec.
9. f
cu
= 1.7 ? 10
–8
?-m
A = 0.01 mm
2
= 0.01 ? 10
–6
m
2
R = 1 K ? = 10
3
?
R =
f
a
?
? 10
3
=
8
6
1.7 10
10
?
?
? ? ?
?
3
10
1.7
? ? = 0.58 ? 10
3
m = 0.6 km. ?
10. dR, due to the small strip dx at a distanc x d = R =
2
fdx
y ?
…(1)
tan ? =
y a b a
x L
? ?
?
?
y a b a
x L
? ?
?
? L(y – a) = x(b – a)
x ?
b–a
dx
a ?
y ?
? ? Y–a
b ?
Electric Current in Conductors
32.3
? Ly – La = xb – xa
?
dy
L 0 b a
dx
? ? ? (diff. w.r.t. x)
?
dy
L b a
dx
? ?
? dx =
Ldy
b a ?
…(2)
Putting the value of dx in equation (1)
dR =
2
fLdy
y (b a) ? ?
? dR =
2
fI dy
(b a) y ? ?
?
R b
2
0 a
fI dy
dR
(b a) y
?
? ?
? ?
? R =
fI (b a) fl
(b a) ab ab
?
?
? ? ?
.
11. r = 0.1 mm = 10
–4
m
R = 1 K ? = 10
3
?, V = 20 V
a) No.of electrons transferred
i =
3
V 20
R 10
? = 20 ? 10
–3
= 2 ? 10
–2
A
q = i t = 2 ? 10
–2
? 1 = 2 ? 10
–2
C.
No. of electrons transferred =
2 17
19
2 10 2 10
1.6 1.6 10
? ?
?
? ?
?
?
= 1.25 ? 10
17
.
b) Current density of wire
=
2
6
8
i 2 10 2
10
A 3.14 10
?
?
?
?
? ? ?
? ?
= 0.6369 ? 10
+6
= 6.37 ? 10
5
A/m
2
.
12. A = 2 ? 10
–6
m
2
, I = 1 A
f = 1.7 ? 10
–8
?-m
E = ?
R =
8
6
f 1.7 10
A 2 10
?
?
? ?
?
?
? ?
V = IR =
8
6
1 1.7 10
2 10
?
?
? ? ?
?
?
E =
8
2
6
dV V 1.7 10 1.7
10 V /m
dL I 2 2 10
?
?
?
? ?
? ? ? ?
?
?
?
= 8.5 mV/m. ?
13. I = 2 m, R = 5 ?, i = 10 A, E = ?
V = iR = 10 ? 5 = 50 V
E =
V 50
I 2
? = 25 V/m.
14. R ?
Fe
= R
Fe
(1 + ?
Fe
??), R ?
Cu
= R
Cu
(1 + ?
Cu
???)
R ?
Fe
= R ?
Cu
? R
Fe
(1 + ?
Fe
??), = R
Cu
(1 + ?
Cu
???)
Page 4

32.1
ELECTRIC CURRENT IN CONDUCTORS
CHAPTER - 32
1. Q(t) = At
2
+ Bt + c
a) At
2
= Q
? A =
1 1
2 2
Q A'T'
A T
t T
?
?
? ?
b) Bt = Q
? B =
Q A 'T'
A
t T
? ?
c) C = [Q]
? C = A ?T ?
d) Current t =
? ?
2
dQ d
At Bt C
dt dt
? ? ?
= 2At + B = 2 ? 5 ? 5 + 3 = 53 A.
2. No. of electrons per second = 2 ? 10
16
electrons / sec.
Charge passing per second = 2 ? 10
16
? 1.6 ? 10
–9
coulomb
sec
= 3.2 ? 10
–9
Coulomb/sec
Current = 3.2 ? 10
–3
A.
3. i ? = 2 ?A, t = 5 min = 5 ? 60 sec.
q = i t = 2 ? 10
–6
? 5 ? 60
= 10 ? 60 ? 10
–6
c = 6 ? 10
–4
c
4. i = i
0
+ ?t, t = 10 sec, i
0
= 10 A, ? = 4 A/sec.
q =
t t t t
0 0
0 0 0 0
idt (i t)dt i dt tdt ? ? ? ? ? ?
? ? ? ?
= i
0
t +
2
t 10 10
10 10 4
2 2
?
? ? ? ? ?
= 100 + 200 = 300 C.
5. i = 1 A, A = 1 mm
2
= 1 ? 10
–6
m
2
f ? cu = 9000 kg/m
3
Molecular mass has N
0
atoms
= m Kg has (N
0
/M ? m) atoms =
0
3
N AI9000
63.5 10
?
?
No.of atoms = No.of electrons
n =
0 0
N Af N f No.of electrons
Unit volume mAI M
? ?
=
23
3
6 10 9000
63.5 10
?
? ?
?
i = V
d
n A e.
? V
d
=
23
6 19
3
i 1
nAe 6 10 9000
10 1.6 10
63.5 10
? ?
?
?
? ?
? ? ?
?
=
3
23 6 19
63.5 10
6 10 9000 10 1.6 10
?
? ?
?
? ? ? ? ?
=
3
26 19 6
63.5 10
6 9 1.6 10 10 10
?
? ?
?
? ? ? ? ?
Electric Current in Conductors
32.2
=
3 3
63.5 10 63.5 10
6 9 1.6 10 6 9 16
? ?
? ?
?
? ? ? ? ?
= 0.074 ? 10
–3
m/s = 0.074 mm/s.
6. ? = 1 m, r = 0.1 mm = 0.1 ? 10
–3
m
R = 100 ?, f = ?
? R = f ? / a
? f =
6
Ra 100 3.14 0.1 0.1 10
1
?
? ? ? ?
?
?
= 3.14 ? 10
–6
= ? ? 10
–6
?-m.
?
7. ? ? = 2 ?
volume of the wire remains constant.
A ? = A ? ? ?
? A ? = A ? ? 2 ?
? A ? = A/2
f = Specific resistance
R =
f
A
?
; R ? =
f '
A '
?
100 ? =
f2 4f
A / 2 A
?
? ?
= 4R
? 4 ? 100 ? = 400 ? ?
8. ? = 4 m, A = 1 mm
2
= 1 ? 10
–6
m
2
I = 2 A, n/V = 10
29
, t = ?
i = n A V
d
e
? e = 10
29
? 1 ? 10
–6
? V
d
? 1.6 ? 10
–19
?
d
29 6 19
2
V
10 10 1.6 10
? ?
?
? ? ?
=
4
1 1
8000 0.8 10
?
?
t =
d
4
4 8000
V 1/ 8000
? ? ?
?
= 32000 = 3.2 ? 10
4
sec.
9. f
cu
= 1.7 ? 10
–8
?-m
A = 0.01 mm
2
= 0.01 ? 10
–6
m
2
R = 1 K ? = 10
3
?
R =
f
a
?
? 10
3
=
8
6
1.7 10
10
?
?
? ? ?
?
3
10
1.7
? ? = 0.58 ? 10
3
m = 0.6 km. ?
10. dR, due to the small strip dx at a distanc x d = R =
2
fdx
y ?
…(1)
tan ? =
y a b a
x L
? ?
?
?
y a b a
x L
? ?
?
? L(y – a) = x(b – a)
x ?
b–a
dx
a ?
y ?
? ? Y–a
b ?
Electric Current in Conductors
32.3
? Ly – La = xb – xa
?
dy
L 0 b a
dx
? ? ? (diff. w.r.t. x)
?
dy
L b a
dx
? ?
? dx =
Ldy
b a ?
…(2)
Putting the value of dx in equation (1)
dR =
2
fLdy
y (b a) ? ?
? dR =
2
fI dy
(b a) y ? ?
?
R b
2
0 a
fI dy
dR
(b a) y
?
? ?
? ?
? R =
fI (b a) fl
(b a) ab ab
?
?
? ? ?
.
11. r = 0.1 mm = 10
–4
m
R = 1 K ? = 10
3
?, V = 20 V
a) No.of electrons transferred
i =
3
V 20
R 10
? = 20 ? 10
–3
= 2 ? 10
–2
A
q = i t = 2 ? 10
–2
? 1 = 2 ? 10
–2
C.
No. of electrons transferred =
2 17
19
2 10 2 10
1.6 1.6 10
? ?
?
? ?
?
?
= 1.25 ? 10
17
.
b) Current density of wire
=
2
6
8
i 2 10 2
10
A 3.14 10
?
?
?
?
? ? ?
? ?
= 0.6369 ? 10
+6
= 6.37 ? 10
5
A/m
2
.
12. A = 2 ? 10
–6
m
2
, I = 1 A
f = 1.7 ? 10
–8
?-m
E = ?
R =
8
6
f 1.7 10
A 2 10
?
?
? ?
?
?
? ?
V = IR =
8
6
1 1.7 10
2 10
?
?
? ? ?
?
?
E =
8
2
6
dV V 1.7 10 1.7
10 V /m
dL I 2 2 10
?
?
?
? ?
? ? ? ?
?
?
?
= 8.5 mV/m. ?
13. I = 2 m, R = 5 ?, i = 10 A, E = ?
V = iR = 10 ? 5 = 50 V
E =
V 50
I 2
? = 25 V/m.
14. R ?
Fe
= R
Fe
(1 + ?
Fe
??), R ?
Cu
= R
Cu
(1 + ?
Cu
???)
R ?
Fe
= R ?
Cu
? R
Fe
(1 + ?
Fe
??), = R
Cu
(1 + ?
Cu
???)
Electric Current in Conductors
32.4
? 3.9 [ 1 + 5 ? 10
–3
(20 – ?)] = 4.1 [1 + 4 x 10
–3
(20 – ?)]
? 3.9 + 3.9 ? 5 ? 10
–3
(20 – ?) = 4.1 + 4.1 ? 4 ? 10
–3
(20 – ?)
?? 4.1 ? 4 ? 10
–3
(20 – ?) – 3.9 ? 5 ? 10
–3
(20 – ?) = 3.9 – 4.1
? 16.4(20 – ?) – 19.5(20 – ?) = 0.2 ? 10
3
? (20 – ?) (–3.1) = 0.2 ? 10
3
? ? – 20 = 200 ?
? ? = 220°C. ?
15. Let the voltmeter reading when, the voltage is 0 be X.
1 1
2 2
I R V
I R V
?
?
1.75 14.4 V 0.35 14.4 V
2.75 22.4 V 0.55 22.4 V
? ?
? ? ?
? ?
?
0.07 14.4 V 7 14.4 V
0.11 22.4 V 11 22.4 V
? ?
? ? ?
? ?
? 7(22.4 – V) = 11(14.4 – V) ? 156.8 – 7V = 158.4 – 11V
? (7 – 11)V = 156.8 – 158.4 ? –4V = –1.6
? V = 0.4 V.
16. a) When switch is open, no current passes through the ammeter. In the upper part of
the circuit the Voltmenter has ? resistance. Thus current in it is 0.
? Voltmeter read the emf. (There is not Pot. Drop across the resistor).
b) When switch is closed current passes through the circuit and if its value of i.
? – ir = 1.45
? 1.52 – ir = 1.45
? ir = 0.07
? 1 r = 0.07 ? r = 0.07 ?. ?
17. E = 6 V, r = 1 ?, V = 5.8 V, R = ?
I =
E 6
R r R 1
?
? ?
, V = E – Ir
? 5.8 =
6
6 1
R 1
? ?
?
?
6
R 1 ?
= 0.2
? R + 1 = 30 ? R = 29 ?. ?
18. V = ? + ir
? 7.2 = 6 + 2 ? r
? 1.2 = 2r ? r = 0.6 ?. ?
19. a) net emf while charging
9 – 6 = 3V
Current = 3/10 = 0.3 A
b) When completely charged.
Internal resistance ‘r’ = 1 ?
Current = 3/1 = 3 A ?
20. a) 0.1i
1
+ 1 i
1
– 6 + 1i
1
– 6 = 0
? 0.1 i
1
+ 1i
1
+ 1i
1
= 12
? i
1
=
12
2.1
ABCDA
? 0.1i
2
+ 1i – 6 = 0
? 0.1i
2
+ 1i
?? ?
V
A
r ?
r ?
2 ?
6 ?
1 ?
0.1
6 ?
1 ?
i 1 ?
6 ?
Page 5

32.1
ELECTRIC CURRENT IN CONDUCTORS
CHAPTER - 32
1. Q(t) = At
2
+ Bt + c
a) At
2
= Q
? A =
1 1
2 2
Q A'T'
A T
t T
?
?
? ?
b) Bt = Q
? B =
Q A 'T'
A
t T
? ?
c) C = [Q]
? C = A ?T ?
d) Current t =
? ?
2
dQ d
At Bt C
dt dt
? ? ?
= 2At + B = 2 ? 5 ? 5 + 3 = 53 A.
2. No. of electrons per second = 2 ? 10
16
electrons / sec.
Charge passing per second = 2 ? 10
16
? 1.6 ? 10
–9
coulomb
sec
= 3.2 ? 10
–9
Coulomb/sec
Current = 3.2 ? 10
–3
A.
3. i ? = 2 ?A, t = 5 min = 5 ? 60 sec.
q = i t = 2 ? 10
–6
? 5 ? 60
= 10 ? 60 ? 10
–6
c = 6 ? 10
–4
c
4. i = i
0
+ ?t, t = 10 sec, i
0
= 10 A, ? = 4 A/sec.
q =
t t t t
0 0
0 0 0 0
idt (i t)dt i dt tdt ? ? ? ? ? ?
? ? ? ?
= i
0
t +
2
t 10 10
10 10 4
2 2
?
? ? ? ? ?
= 100 + 200 = 300 C.
5. i = 1 A, A = 1 mm
2
= 1 ? 10
–6
m
2
f ? cu = 9000 kg/m
3
Molecular mass has N
0
atoms
= m Kg has (N
0
/M ? m) atoms =
0
3
N AI9000
63.5 10
?
?
No.of atoms = No.of electrons
n =
0 0
N Af N f No.of electrons
Unit volume mAI M
? ?
=
23
3
6 10 9000
63.5 10
?
? ?
?
i = V
d
n A e.
? V
d
=
23
6 19
3
i 1
nAe 6 10 9000
10 1.6 10
63.5 10
? ?
?
?
? ?
? ? ?
?
=
3
23 6 19
63.5 10
6 10 9000 10 1.6 10
?
? ?
?
? ? ? ? ?
=
3
26 19 6
63.5 10
6 9 1.6 10 10 10
?
? ?
?
? ? ? ? ?
Electric Current in Conductors
32.2
=
3 3
63.5 10 63.5 10
6 9 1.6 10 6 9 16
? ?
? ?
?
? ? ? ? ?
= 0.074 ? 10
–3
m/s = 0.074 mm/s.
6. ? = 1 m, r = 0.1 mm = 0.1 ? 10
–3
m
R = 100 ?, f = ?
? R = f ? / a
? f =
6
Ra 100 3.14 0.1 0.1 10
1
?
? ? ? ?
?
?
= 3.14 ? 10
–6
= ? ? 10
–6
?-m.
?
7. ? ? = 2 ?
volume of the wire remains constant.
A ? = A ? ? ?
? A ? = A ? ? 2 ?
? A ? = A/2
f = Specific resistance
R =
f
A
?
; R ? =
f '
A '
?
100 ? =
f2 4f
A / 2 A
?
? ?
= 4R
? 4 ? 100 ? = 400 ? ?
8. ? = 4 m, A = 1 mm
2
= 1 ? 10
–6
m
2
I = 2 A, n/V = 10
29
, t = ?
i = n A V
d
e
? e = 10
29
? 1 ? 10
–6
? V
d
? 1.6 ? 10
–19
?
d
29 6 19
2
V
10 10 1.6 10
? ?
?
? ? ?
=
4
1 1
8000 0.8 10
?
?
t =
d
4
4 8000
V 1/ 8000
? ? ?
?
= 32000 = 3.2 ? 10
4
sec.
9. f
cu
= 1.7 ? 10
–8
?-m
A = 0.01 mm
2
= 0.01 ? 10
–6
m
2
R = 1 K ? = 10
3
?
R =
f
a
?
? 10
3
=
8
6
1.7 10
10
?
?
? ? ?
?
3
10
1.7
? ? = 0.58 ? 10
3
m = 0.6 km. ?
10. dR, due to the small strip dx at a distanc x d = R =
2
fdx
y ?
…(1)
tan ? =
y a b a
x L
? ?
?
?
y a b a
x L
? ?
?
? L(y – a) = x(b – a)
x ?
b–a
dx
a ?
y ?
? ? Y–a
b ?
Electric Current in Conductors
32.3
? Ly – La = xb – xa
?
dy
L 0 b a
dx
? ? ? (diff. w.r.t. x)
?
dy
L b a
dx
? ?
? dx =
Ldy
b a ?
…(2)
Putting the value of dx in equation (1)
dR =
2
fLdy
y (b a) ? ?
? dR =
2
fI dy
(b a) y ? ?
?
R b
2
0 a
fI dy
dR
(b a) y
?
? ?
? ?
? R =
fI (b a) fl
(b a) ab ab
?
?
? ? ?
.
11. r = 0.1 mm = 10
–4
m
R = 1 K ? = 10
3
?, V = 20 V
a) No.of electrons transferred
i =
3
V 20
R 10
? = 20 ? 10
–3
= 2 ? 10
–2
A
q = i t = 2 ? 10
–2
? 1 = 2 ? 10
–2
C.
No. of electrons transferred =
2 17
19
2 10 2 10
1.6 1.6 10
? ?
?
? ?
?
?
= 1.25 ? 10
17
.
b) Current density of wire
=
2
6
8
i 2 10 2
10
A 3.14 10
?
?
?
?
? ? ?
? ?
= 0.6369 ? 10
+6
= 6.37 ? 10
5
A/m
2
.
12. A = 2 ? 10
–6
m
2
, I = 1 A
f = 1.7 ? 10
–8
?-m
E = ?
R =
8
6
f 1.7 10
A 2 10
?
?
? ?
?
?
? ?
V = IR =
8
6
1 1.7 10
2 10
?
?
? ? ?
?
?
E =
8
2
6
dV V 1.7 10 1.7
10 V /m
dL I 2 2 10
?
?
?
? ?
? ? ? ?
?
?
?
= 8.5 mV/m. ?
13. I = 2 m, R = 5 ?, i = 10 A, E = ?
V = iR = 10 ? 5 = 50 V
E =
V 50
I 2
? = 25 V/m.
14. R ?
Fe
= R
Fe
(1 + ?
Fe
??), R ?
Cu
= R
Cu
(1 + ?
Cu
???)
R ?
Fe
= R ?
Cu
? R
Fe
(1 + ?
Fe
??), = R
Cu
(1 + ?
Cu
???)
Electric Current in Conductors
32.4
? 3.9 [ 1 + 5 ? 10
–3
(20 – ?)] = 4.1 [1 + 4 x 10
–3
(20 – ?)]
? 3.9 + 3.9 ? 5 ? 10
–3
(20 – ?) = 4.1 + 4.1 ? 4 ? 10
–3
(20 – ?)
?? 4.1 ? 4 ? 10
–3
(20 – ?) – 3.9 ? 5 ? 10
–3
(20 – ?) = 3.9 – 4.1
? 16.4(20 – ?) – 19.5(20 – ?) = 0.2 ? 10
3
? (20 – ?) (–3.1) = 0.2 ? 10
3
? ? – 20 = 200 ?
? ? = 220°C. ?
15. Let the voltmeter reading when, the voltage is 0 be X.
1 1
2 2
I R V
I R V
?
?
1.75 14.4 V 0.35 14.4 V
2.75 22.4 V 0.55 22.4 V
? ?
? ? ?
? ?
?
0.07 14.4 V 7 14.4 V
0.11 22.4 V 11 22.4 V
? ?
? ? ?
? ?
? 7(22.4 – V) = 11(14.4 – V) ? 156.8 – 7V = 158.4 – 11V
? (7 – 11)V = 156.8 – 158.4 ? –4V = –1.6
? V = 0.4 V.
16. a) When switch is open, no current passes through the ammeter. In the upper part of
the circuit the Voltmenter has ? resistance. Thus current in it is 0.
? Voltmeter read the emf. (There is not Pot. Drop across the resistor).
b) When switch is closed current passes through the circuit and if its value of i.
? – ir = 1.45
? 1.52 – ir = 1.45
? ir = 0.07
? 1 r = 0.07 ? r = 0.07 ?. ?
17. E = 6 V, r = 1 ?, V = 5.8 V, R = ?
I =
E 6
R r R 1
?
? ?
, V = E – Ir
? 5.8 =
6
6 1
R 1
? ?
?
?
6
R 1 ?
= 0.2
? R + 1 = 30 ? R = 29 ?. ?
18. V = ? + ir
? 7.2 = 6 + 2 ? r
? 1.2 = 2r ? r = 0.6 ?. ?
19. a) net emf while charging
9 – 6 = 3V
Current = 3/10 = 0.3 A
b) When completely charged.
Internal resistance ‘r’ = 1 ?
Current = 3/1 = 3 A ?
20. a) 0.1i
1
+ 1 i
1
– 6 + 1i
1
– 6 = 0
? 0.1 i
1
+ 1i
1
+ 1i
1
= 12
? i
1
=
12
2.1
ABCDA
? 0.1i
2
+ 1i – 6 = 0
? 0.1i
2
+ 1i
?? ?
V
A
r ?
r ?
2 ?
6 ?
1 ?
0.1
6 ?
1 ?
i 1 ?
6 ?
Electric Current in Conductors
32.5
? i – 6 + 6 – (i
2
– i)1 = 0
? i – i
2
+ i = 0
? 2i – i
2
= 0 ? –2i ± 0.2i = 0
? i
2
= 0.
b) 1i
1
+ 1 i
1
– 6 + 1i
1
= 0
? 3i
1
= 12 ? i
1
= 4
DCFED
? i
2
+ i – 6 = 0 ? i
2
+ i = 6
ABCDA,
i
2
+ (i
2
– i) – 6  = 0
? i
2
+ i
2
– i = 6 ? 2i
2
– i = 6
? –2i
2
± 2i = 6 ? i = –2
i
2
+ i = 6
? i
2
– 2 = 6  ? i
2
= 8
1
2
i 4 1
i 8 2
? ? .
c) 10i
1
+ 1i
1
– 6 + 1i
1
– 6 = 0
? 12i
1
= 12 ? i
1
= 1
10i
2
– i
1
– 6 = 0
? 10i
2
– i
1
= 6
? 10i
2
+ (i
2
– i)1 – 6 = 0
? 11i
2
= 6
? –i
2
= 0
21. a) Total emf = n
1
E
in 1 row
Total emf in all news = n
1
E
Total resistance in one row = n
1
r
Total resistance in all rows =
1
2
n r
n
Net resistance =
1
2
n r
n
+ R
Current =
1 1 2
1 2 1 2
n E n n E
n /n r R n r n R
?
? ?
b) I =
1 2
1 2
n n E
n r n R ?
for I = max,
n
1
r + n
2
R = min
?
? ?
2
1 2 1 2
n r n R 2 n rn R ? ? = min
it is min, when
1 2
n r n R ?
? n
1
r = n
2
R
I is max when n
1
r = n
2
R.
i ?
i 2 ?
A ?
0.1 ? ?
1 ? ?
i 2 –i ?
6 ?
B ? C ?
1 ? ?
6 ?
E ?
D
F ?
1 ?
1 ?
6 ?
1 ?
i 1 ?
6 ?
i ?
i 2 ?
E ?
1 ? ?
1 ? ?
i 2 –i ?
6 ?
D C ?
1 ? ?
6 ?
B ?
F ?
A ?
1 ?
1 ?
6 ?
1 ?
i 1 ?
6 ?
i ?
i 2 ?
E ?
10 ? ?
1 ? ?
i 2 –i ?
6 ?
D ? C ?
1 ? ?
6 ?
B ?
F ?
A ?
n 1
r ? r ? r ?
r ? r ? r ?
R
```

## Physics Class 11

127 videos|464 docs|210 tests

## Physics Class 11

127 videos|464 docs|210 tests

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