Current Electricity- 3 Practice Questions - DPP for NEET

 Page 1


1. (a) Potential gradient 
( )
.
h
eR
x
R R rL
=
++
( )
3
2
0.2102
4.9.
490 01
10
R
R
R
-
-
´
Þ = ´ Þ =W
++
2. (c) Let S be larger and R be smaller resistance connected
in two gaps of meter bridge.
100 100 20
4
20
l
S R RR
l
-- æö
\= ==
ç÷
èø
           ......(i)
When 
15W
resistance is added to resistance R, then
( ) ( )
100 40 6
15 15
404
S RR
- æö
= + =+
ç÷
èø
           .......(ii)
From equations (i) and (ii) 
9 R=W
3. (b)
1
2
240
12 12
120
l
rR
l
æö
æö
= -= -= W
ç÷ ç÷
èø
èø
4. (a) Potential difference per unit length
2
0.5/
4
V
Vm
L
= ==
5. (d)
( )
55
. 0.4
( ) 5 45 0 10
8
h
eR
E ll
R R rL
lm
= ´ Þ = ´´
+ + ++
Þ=
6. ( d) Current through 2 W = 1.4 
( )
( )( )
255
10 2 25 5
ìü +
ïï
íý
+ ++
ïï
îþ
 = 1A
7. (c) Post office box is based on the principle of
Wheatstone’s bridge.
8. (b) Using
1
2
150
1 2 11
100
l
rR
l
æö
æö
= -= -=W
ç÷ ç÷
èø
èø
9. (a) Since the given bridge is balanced, hence there will be
no current through 9 W resistance. This resistance has
no effect and must be ignored in the calculations.
5 W
10 W 8 W
4 W
1.4 A
14 W
9 W
R
AB
 = 
9 18
27
´
= 6W
10. (c) Potential gradient ()
7
2
6
0.1 10
10/
10
-
-
-
r´
===
i
x Vm
A
11. (d)
100
.
l
SR
l
- æö
=
ç÷
èø
Initially, 
100
30 10 25
l
l cm
l
- æö
= ´ Þ=
ç÷
èø
Finally, 
100
10 30 75
l
l cm
l
- æö
= ´ Þ=
ç÷
èø
So, shift =50cm.
12. (d)
1 12
2 12
(6 2)2
(6 2)1
E ll
E ll
+ +
= ==
--
13. (c)
1 12
2 12
58 293
58 291
E ll
E ll
+ +
= ==
--
14. (d)
2 10
0.4 0.16
( ) (10 40 0) 1
= ´= ´ ´=
+ + ++
h
eR
El
R R rL
V
15. (c)
Resistance of the part 
AC
0.1 40 4
AC
R = ´ =W and 0.1 606
CB
R = ´ =W
In balanced condition 
4
4
66
= Þ =W
X
X
Equivalent resistance 
5
eq
R =W
so current drawn from battery 
5
1
5
iA ==
.
16. (a)
12
2
55 50
101
50
æö - - æö
= ´ ¢Þ = ´ =W
ç÷
ç÷
èø
èø
ll
r Rr
l
Page 2


1. (a) Potential gradient 
( )
.
h
eR
x
R R rL
=
++
( )
3
2
0.2102
4.9.
490 01
10
R
R
R
-
-
´
Þ = ´ Þ =W
++
2. (c) Let S be larger and R be smaller resistance connected
in two gaps of meter bridge.
100 100 20
4
20
l
S R RR
l
-- æö
\= ==
ç÷
èø
           ......(i)
When 
15W
resistance is added to resistance R, then
( ) ( )
100 40 6
15 15
404
S RR
- æö
= + =+
ç÷
èø
           .......(ii)
From equations (i) and (ii) 
9 R=W
3. (b)
1
2
240
12 12
120
l
rR
l
æö
æö
= -= -= W
ç÷ ç÷
èø
èø
4. (a) Potential difference per unit length
2
0.5/
4
V
Vm
L
= ==
5. (d)
( )
55
. 0.4
( ) 5 45 0 10
8
h
eR
E ll
R R rL
lm
= ´ Þ = ´´
+ + ++
Þ=
6. ( d) Current through 2 W = 1.4 
( )
( )( )
255
10 2 25 5
ìü +
ïï
íý
+ ++
ïï
îþ
 = 1A
7. (c) Post office box is based on the principle of
Wheatstone’s bridge.
8. (b) Using
1
2
150
1 2 11
100
l
rR
l
æö
æö
= -= -=W
ç÷ ç÷
èø
èø
9. (a) Since the given bridge is balanced, hence there will be
no current through 9 W resistance. This resistance has
no effect and must be ignored in the calculations.
5 W
10 W 8 W
4 W
1.4 A
14 W
9 W
R
AB
 = 
9 18
27
´
= 6W
10. (c) Potential gradient ()
7
2
6
0.1 10
10/
10
-
-
-
r´
===
i
x Vm
A
11. (d)
100
.
l
SR
l
- æö
=
ç÷
èø
Initially, 
100
30 10 25
l
l cm
l
- æö
= ´ Þ=
ç÷
èø
Finally, 
100
10 30 75
l
l cm
l
- æö
= ´ Þ=
ç÷
èø
So, shift =50cm.
12. (d)
1 12
2 12
(6 2)2
(6 2)1
E ll
E ll
+ +
= ==
--
13. (c)
1 12
2 12
58 293
58 291
E ll
E ll
+ +
= ==
--
14. (d)
2 10
0.4 0.16
( ) (10 40 0) 1
= ´= ´ ´=
+ + ++
h
eR
El
R R rL
V
15. (c)
Resistance of the part 
AC
0.1 40 4
AC
R = ´ =W and 0.1 606
CB
R = ´ =W
In balanced condition 
4
4
66
= Þ =W
X
X
Equivalent resistance 
5
eq
R =W
so current drawn from battery 
5
1
5
iA ==
.
16. (a)
12
2
55 50
101
50
æö - - æö
= ´ ¢Þ = ´ =W
ç÷
ç÷
èø
èø
ll
r Rr
l
DPP/ P 38
109
17. (c) Potential gradient
()
h
V eR
x
L R R rL
==
++
3
2.2
2.2 10 1 ' 990
(10)
h
R
R
-
Þ ´ = ´Þ =W
+
18. (a)
3
4
2.4 10
4 10
1.25
-
-
´
= = r Þ = = =´
r´
E
ExliliA
l
19. ( b) Give circuit is a balanced Wheaststone bridge circuit,
hence it can be redrawn as follows
R
AB
 = 
( )
126
126
´
+
 = 4 W.
20. (a) Balancing length is independent of the cross sectional
area of the wire.
21. (a) In meter bridge experiment, it is assumed that the
resistance of the L shaped plate is negligible, but
actually it is not so. The error created due to this is
called, end error. To remove this the resistance box and
the unknown resisance must be interchanged and then
the mean reading must be taken.
22. (a) Ammeter is always connected in series with circuit.
23. (a) In balanced Wheastone bridge, the arms of
galvanometer and cell can be interchanged without
affecting the balance of the bridge.
24. (d)
W
W
W
W
If P is slightly icnreased, potential of C will decrease.
Hence current will from A to C.
If Q is slightly increased, potential of C will increase.
Hence current will flow from C to A.
25-27
We have
sc
fs
V 10.0V
R R 20.0 9980
I 0.00100A
= - = - W=W
At full-scale deflection, V
ab
 = 10.0V, voltage across the
meter is 0.0200 V , voltage across R
s
 is 9.98 V , and current
through the voltmeter is 0.00100 A. In this case most of the
voltage appears across the series resistor.
The equivalent meter resistance is R
eq
 = 20.0 W + 9980 W
= 10,000 W . Such a meter is described as a "1,000 ohms-
per-volt meter" referring to the ratio of resistance to full-
scale deflection. In normal operation the current through
the circuit element being measured is much greater than
0.00100 A, and the resistance between points a and b in
the circuit is much less than 10,000 W. So the voltmeter
draws off only a small fraction of the current and disturbs,
only slightly the circuit being measured.
25. (d), 26. (c),  27. (b)
28. (d) The resistance of the galvanometer is fixed. In meter
bridge experiments, to protect the galvanometer from
a high current, high resistance is connected to the
galvanometer in order to protect it from damage.
29. (a) Sensitivity 
1
Potential gradiant
µµ (Length of wire)
30. (a) If either the e.m.f. of the driver cell or potential
difference across the whole potentiometer wire is lesser
than the e.m.f. of the experimental cell, then balance
point will not obtained.