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Current Electricity

NEET Physics · 105 questions · Page 6 of 11 · Click an option or "Show Solution" to reveal answer

Q51
A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of (i) infinity, (ii) 9.5 Ω\Omega the balancing lengths on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is
A 0.25 Ω\Omega
B 0.95 Ω\Omega
C 0.5 Ω\Omega
D 0.75 Ω\Omega
Correct Answer
Option C
Solution

Internal resistance of the cell,

r=(E−VV)R=(ℓ1−ℓ2ℓ2)Rr = \left( {{{E - V} \over V}} \right)R = \left( {{{{\ell _1} - {\ell _2}} \over {{\ell _2}}}} \right)R
=(3−2.852.85)×(9.5)Ω=0.5Ω= \left( {{{3 - 2.85} \over {2.85}}} \right) \times \left( {9.5} \right)\Omega = 0.5\Omega
Q52
Ten identical cells connected in series are needed to heat a wire of length one meter and radius 'r' by 10 o C in time 't'. How many cells will be required to heat the wire of length two meter of the same radius by the same temperature in time 't' ?
A 20
B 30
C 40
D 10
Correct Answer
Option A
Solution

Let ρ\rho be resistivity of the material of the wire and r be radius of the wire. Therefore, resistance of 1 m wire is

R=ρ(1)πr2=ρπr2R = {{\rho \left( 1 \right)} \over {\pi {r^2}}} = {\rho \over {\pi {r^2}}}
(∵R=ρlA)\left(\because {R = {{\rho l} \over A}} \right)

Let

ε\varepsilon

be emf of each cell. In first case, 10 cells each of emf

ε\varepsilon

are connected in series to heat the wire of length 1 m by

Δ\Delta

T(= 10°C) in time t.

∴(10ε)Rr=msΔT\therefore {{\left( {10\varepsilon } \right)} \over R}r = ms\Delta T

...(i) In second case, Resistance of same wire of length 2 m is

R′=ρ(2)πr2=2ρπr2=2RR' = {{\rho \left( 2 \right)} \over {\pi {r^2}}} = {{2\rho } \over {\pi {r^2}}} = 2R

Let n cells each of emf

ε\varepsilon

are connected in series to heat the same wire of length 2 m, by the same temperature

Δ\Delta

T (= 10°C) in the same time t.

∴(nε)2t2R=(2m)sΔT\therefore {{{{\left( {n\varepsilon } \right)}^2}t} \over {2R}} = \left( {2m} \right)s\Delta T

...(ii) Divide (ii) by (i), we get

n2200=2⇒n2=400{{{n^2}} \over {200}} = 2 \Rightarrow {n^2} = 400

∴\therefore n = 20

Q53
Two rods are joined end to end, as shown. Both have a cross-sectional area of 0.01 cm 2 . Each is 1 meter long. One rod is of copper with a resistivity of 1.7 × \times 10 −-6 ohm-centimeter, the other is of iron with a resistivity of 10 −-5 ohm-centimeter. How much voltage is required to produce a current of 1 ampere in the rods?
A 0.00145 V
B 0.0145 V
C 1.7 × \times 10 −-6 V
D 0.117 V
Correct Answer
Option D
Solution

Copper rod and iron rod are joined in series.

∴R=RCu+RFe=(ρ1+ρ2)ℓA\therefore R = {R_{Cu}} + {R_{Fe}} = \left( {{\rho _1} + {\rho _2}} \right){\ell \over A}
(∵R=ρℓA)\left(\because {R = \rho {\ell \over A}} \right)

From ohm’s law V = RI = (1.7 × 10 –6 × 10 –2 + 10 –5 × 10 –2 )

÷\div

0.01 × 10 –4 volt = 0.117 volt (∵\because I = 1A)

Q54
A 12 cm wire is given a shape of a right angled triangle ABC having sides 3 cm, 4 cm and 5 cm as shown in the figure. The resistance between two ends (AB, BC, CA) of the respective sides are measuread one by one ratio
A 9 : 16 : 25
B 27 : 32 : 35
C 21 : 24 : 25
D 3 : 4 : 5
Correct Answer
Option B
Solution

Resistance is directly proportional to length

1RAB=13+14+5=(4+5)+3(3)(4+5){1 \over {{R_{AB}}}} = {1 \over 3} + {1 \over {4 + 5}} = {{\left( {4 + 5} \right) + 3} \over {\left( 3 \right)\left( {4 + 5} \right)}}
RAB=3×(4+5)3+(4+5)=2712{R_{AB}} = {{3 \times \left( {4 + 5} \right)} \over {3 + \left( {4 + 5} \right)}} = {{27} \over {12}}

Similarly,

RBC=4×(3+5)4+(3+5)=3212{R_{BC}} = {{4 \times \left( {3 + 5} \right)} \over {4 + \left( {3 + 5} \right)}} = {{32} \over {12}}
RAC=5×(3+4)5+(3+4)=3512{R_{AC}} = {{5 \times \left( {3 + 4} \right)} \over {5 + \left( {3 + 4} \right)}} = {{35} \over {12}}

∴\therefore R AB : R BC : R AC = 27 : 32 : 35

Q55
The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the celll will be
A 0.1 A
B 2.0 A
C 1.0 A
D 0.2 A
Correct Answer
Option D
Solution

From the question, total resistance of Wheatstone bridge is: (40) × (120)/(40 + 120) = 30

Ω\Omega

Now the current through the cell is: = 7V/(5+30)

Ω\Omega

= (1/5)A = 0.2A

Q56
A wire of resistance 4 Ω\Omega is stretched to twice its original length. The resistance of stretched wire would be
A 8 Ω\Omega
B 16 Ω\Omega
C 2 Ω\Omega
D 4 Ω\Omega
Correct Answer
Option B
Solution

Resistance of a wire,

R=ρlA=4ΩR = \rho {l \over A} = 4\Omega

...(i) When wire is stretched twice, its new length be

l′l'

. Then

l′=2ll' = 2l

On stretching volume of the wire remains constant.

∴lA=l′A′\therefore lA = l'A'

where A' is the new cross-sectional area

⇒A′=ll′A=l2lA=A2\Rightarrow A' = {l \over {l'}}A = {l \over {2l}}A = {A \over 2}

∴\therefore Resistance of the stretched wire is

R′=ρl′A′=ρ2l(A/2)=4ρlAR' = \rho {{l'} \over {A'}} = \rho {{2l} \over {\left( {A/2} \right)}} = 4\rho {l \over A}
=4(4Ω)=16Ω= 4\left( {4\Omega } \right) = 16\Omega
Q57
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω\Omega is
A 0.8 Ω\Omega
B 1.0 Ω\Omega
C 0.2 Ω\Omega
D 0.5 Ω\Omega
Correct Answer
Option D
Solution
I=εR+rI = {\varepsilon \over {R + r}}
⇒IR+Ir=ε\Rightarrow IR + Ir = \varepsilon

Here,

R=10Ω,r=?,ε=2.1V,I=0.2AR = 10\Omega ,r = ?,\varepsilon = 2.1V,I = 0.2A

∴\therefore 0.2 × 10 + 0.2 × r = 2.1 2 + 0.2r = 2.1 0.2r = 0.1 ⇒\Rightarrow r =

12{1 \over 2}

= 0.5

Ω\Omega
Q58
The power dissipated in the circuit shown in the figure is 30 watts. The value of R is
A 20 Ω\Omega
B 15 Ω\Omega
C 10 Ω\Omega
D 30 Ω\Omega
Correct Answer
Option C
Solution

The power dissipated in the circuit

P=V2ReqP = {{{V^2}} \over {{R_{eq}}}}

...(i) v = 10 volt

1Req=1R+15=5+R5R{1 \over {{R_{eq}}}} = {1 \over R} + {1 \over 5} = {{5 + R} \over {5R}}
Req=(5R5+R){R_{eq}} = \left( {{{5R} \over {5 + R}}} \right)

P = 30 W Substituting the values in equation (i)

30=(10)2(5R5+R)30 = {{{{\left( {10} \right)}^2}} \over {\left( {{{5R} \over {5 + R}}} \right)}}
15R5+R=10{{15R} \over {5 + R}} = 10

15R = 50 + 10R 5R = 50 R = 10

Ω\Omega
Q59
In the circuit shown the cells A and B have negligible resistances. For V A = 12 V, R 1 = 500 Ω\Omega and R = 100 Ω\Omega the galvanometer (G) shows no deflection. The value of V s is
A 4 V
B 2 V
C 12 V
D 6 V
Correct Answer
Option B
Solution

Since the galvanometer shows no deflection so current will flow as shown in the figure. Current,

I=VAR1+R=12V(500+100)Ω=12600AI = {{{V_A}} \over {{R_1} + R}} = {{12V} \over {\left( {500 + 100} \right)\Omega }} = {{12} \over {600}}A
VB=IR=(12600A)(100Ω)=2V{V_B} = IR = \left( {{{12} \over {600}}A} \right)\left( {100\Omega } \right) = 2V
Q60
If voltage across a bulb rated 220 volt-100 watt drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is
A 20%
B 2.5%
C 5%
D 10%
Correct Answer
Option C
Solution

Resistance of bulb is constant

P=V2R⇒Δpp=2ΔVV+ΔRRP = {{{V^2}} \over R} \Rightarrow {{\Delta p} \over p} = {{2\Delta V} \over V} + {{\Delta R} \over R}
∴Δpp=2×2.5+0=5%\therefore {{\Delta p} \over p} = 2 \times 2.5 + 0 = 5\%
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