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Current Electricity

JEE Physics · 159 questions · Page 5 of 16 · Click an option or "Show Solution" to reveal answer

Q41
Two batteries with e.m.f 12 V and 13 V are connected in parallel across a load resistor of 10 Ω\Omega . The internal resistances of the two batteries are 1 Ω\Omega and 2 Ω\Omega respectively. The voltage across the load lies between :
A 11.7 V and 11.8 V
B 11.6 V and 11.7 V
C 11.5 V and 11.6 V
D 11.4 V and 11.5 V
Correct Answer
Option C
Solution

Let potential at S, T, R = V and potential at P, Q, U, = 0 Using kirchhoff's law at P : Current at P is ,

V−121+V−132+V−010=0{{V - 12} \over 1} + {{V - 13} \over 2} + {{V - 0} \over {10}} = 0

⇒\Rightarrow

   \,\,\,
V1{V \over 1}

+

V2{V \over 2}

+

V10{V \over 10}

= 12 +

132{{13} \over 2}

⇒\Rightarrow

   \,\,\,
10V+5V+V10{{10V + 5V + V} \over {10}}

=

372{{37} \over 2}

⇒\Rightarrow

   \,\,\,
16V10{{16V} \over {10}}

=

372{{37} \over 2}

⇒\Rightarrow

   \,\,\,

V = 11.56 Volt.

Q42
A heating element has a resistance of 100 Ω\Omega at room temperature. When it is connected to a supply of 220 V, a steady current of 2 A passes in it and temperature is 500oC more than room temperature. what is the temperature coefficient of resistance of the heating element ?
A 0.5 × \times 10−-4 oC−-1
B 5 × \times 10−-4 oC−-1
C 1 × \times 10−-4 oC−-1
D 2 × \times 10−-4 oC−-1
Correct Answer
Option D
Solution

When temperature increased by 500oC then, nrw registance Rt =

2202{{220} \over 2}

= 110

Ω\Omega

We know, Rt = R0 (1 + ∝\propto

Δ\Delta

t) ⇒\Rightarrow 110 = 100 (1 + ∝\propto ×\times 500) ⇒\Rightarrow ∝\propto =

10100×500=2×10−4{{10} \over {100 \times 500}} = 2 \times {10^{ - 4}}

oC−-1

Q43
In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Ω\Omega, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.
A 2.5 Ω\Omega
B 1 Ω\Omega
C 1.5 Ω\Omega
D 2 Ω\Omega
Correct Answer
Option C
Solution

Internal resistance of potentiometer, r =

(EV−1)×R\left( {{E \over V} - 1} \right) \times R

Initially when no current passes through the galvanometer then emf, E = K (52) here K = potential gradient After cell is shunted by a resistance 5

Ω\Omega

, then, Terminal voltage, V = K(40)

∴   \therefore\,\,\,

r =

(52K40K−1)\left( {{{52K} \over {40K}} - 1} \right)

×\times 5 =

(2620−1)\left( {{{26} \over {20}} - 1} \right)

×\times 5 = 1.5

Ω\Omega
Q44
A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 × \times 10–4 A passes through it, its needle ( pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of :
A 200 ohm
B 250 ohm
C 6200 ohm
D 6250 ohm
Correct Answer
Option A
Solution

Ig = 4 ×\times 10−-4 ×\times 25 = 10−-2 A 2.5 = (50 + R) 10−-2 ∴\therefore R = 200

Ω\Omega
Q45
A metal wire of resistance 3 Ω\Omega is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60° at the centre, the equivalent resistance between these two points will be :-
A 5 /2 Ω\Omega
B 12/5 Ω\Omega
C 7/2 Ω\Omega
D 5 / 3 Ω\Omega
Correct Answer
Option D
Solution
R=ρlA=ρl(V/l)=ρl2V(V→Volume of wire)R = {{\rho l} \over A} = {{\rho l} \over {\left( {V/l} \right)}} = {{\rho {l^2}} \over V}\left( {V{\rm{ }} \to {\rm{ }}Volume{\rm{ }}\,of{\rm{ }}\,wire} \right)

⇒\Rightarrow Final resistance = 3 × (B)2 = 12

Ω\Omega
Req=2Ω∥10Ω=53Ω{R_{eq}} = 2\Omega \parallel 10\Omega = {5 \over 3}\Omega
Q46
The resistance of a galvanometer is 50 ohm and the maximum current which can be passed through it is 0.002 A. What resistance must be connected to it in order to convert it into an ammeter of range 0 – 0.5 A ?
A 0.02 ohm
B 0.2 ohm
C 0.002 ohm
D 0.5 ohm
Correct Answer
Option B
Solution

We have Ig Rg = (0.5 – Ig) S ⇒\Rightarrow (0.002) (50) = (0.5 – Ig) S

⇒S≃0.002×500.5=0.2Ω\Rightarrow S \simeq {{0.002 \times 50} \over {0.5}} = 0.2\Omega
Q47
In a conductor, if the number of conduction electrons per unit volume is 8.5 × 1028 m–3 and mean free time is 25ƒs (femto second), it's approximate resistivity is :- (me = 9.1 × 10–31 kg)
A 10–8 Ω\Omega m
B 10–7 Ω\Omega m
C 10–5 Ω\Omega m
D 10–6 Ω\Omega m
Correct Answer
Option A
Solution
ρ=2mne2τ\rho = {{2m} \over {n{e^2}\tau }}

= 3.34 × 10–8

Ω\Omega

m

Q48
A current of 5 A passes through a copper conductor (resistivity = 1.7 × 10–8 Ω\Omega m) of radius of cross-section 5 mm. Find the mobility of the charges if their drift velocity is 1.1 × 10–3 m/s.
A 1.3 m2/Vs
B 1.0 m2/Vs
C 1.8 m2/Vs
D 1.5 m2/Vs
Correct Answer
Option B
Solution
μ=VdE      E=ρJ\mu = {{{V_d}} \over E}\,\,\,\,\,\,E = \rho J
=1.1×10−31.7×10−8×5π×25×0−6= {{1.1 \times {{10}^{ - 3}}} \over {1.7 \times {{10}^{ - 8}} \times {5 \over {\pi \times 25 \times {0^{ - 6}}}}}}
=1.1×10−3×π×25×0−61.7×10−8×5≈1.01 m2/Vs= {{1.1 \times {{10}^{ - 3}} \times \pi \times 25 \times {0^{ - 6}}} \over {1.7 \times {{10}^{ - 8}} \times 5}} \approx 1.01\,{m^2}/Vs
Q49
A moving coil galvanometer allows a full scale current of 10–4 A. A series resistance of 2 MΩ\Omega is required to convert the above galvanometer into a voltmeter of range 0-5 V. Therefore the value of shunt resistance required to convert the above galvanometer into an ammeter of range 0.10 mA is :
A 200 Ω\Omega
B 500 Ω\Omega
C 100 Ω\Omega
D None of the options are correct
Correct Answer
Option D
Solution

Given data,

I=10−4 A,RS=2MΩ=2×106Ω,Vmax⁡=5 V\begin{aligned} I =10^{-4} \mathrm{~A}, \\\\ R_S =2 \mathrm{M} \Omega=2 \times 10^6 \Omega, \\\\ V_{\max } =5 \mathrm{~V} \end{aligned}

Let internal resistance of galvanometer is RGR_G. Then,

I×RS+I×RG=Vmax⁡⇒2×106×10−4+10−4×RG=5⇒10−4RG=5−200=−195 or RG=−195×104Ω\begin{array}{lc} I \times R_S+I \times R_G=V_{\max } \\\\ \Rightarrow 2 \times 10^6 \times 10^{-4}+10^{-4} \times R_G=5 \\\\ \Rightarrow 10^{-4} R_G=5-200=-195 \\\\ \text{ or } R_G=-195 \times 10^4 \Omega \end{array}

Resistance cannot be negative. ∴\therefore No option is correct.

Q50
Space between two concentric conducting spheres of radii a and b (b > a) is filled with a medium of resistivity ρ\rho . The resistance between the two spheres will be :
A ρ2π(1a+1b){\rho \over {2\pi }}\left( {{1 \over a} + {1 \over b}} \right)
B ρ4π(1a+1b){\rho \over {4\pi }}\left( {{1 \over a} + {1 \over b}} \right)
C ρ2π(1a−1b){\rho \over {2\pi }}\left( {{1 \over a} - {1 \over b}} \right)
D ρ4π(1a−1b){\rho \over {4\pi }}\left( {{1 \over a} - {1 \over b}} \right)
Correct Answer
Option D
Solution
R=∫abρ dx4πx2R = \int\limits_a^b {{{\rho \,dx} \over {4\pi {x^2}}}}
=ρ4π(1a−1b)= {\rho \over {4\pi }}\left( {{1 \over a} - {1 \over b}} \right)
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