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Current Electricity

JEE Physics · 159 questions · Page 15 of 16 · Click an option or "Show Solution" to reveal answer

Q141
A 220220 volt, 10001000 watt bulb is connected across a 110110 voltvolt mains supply. The power consumed will be
A 750750 watt
B 500500 watt
C 250250 watt
D 10001000 watt
Correct Answer
Option C
Solution

We know that

R=Vrated2Prated=(220)21000R = {{V_{rated}^2} \over {{P_{rated}}}} = {{{{\left( {220} \right)}^2}} \over {1000}}

When this bulb is connected to

110110

volt mains supply we get

P=V2R=(110)2×1000(220)2=10004=250WP = {{{V^2}} \over R} = {{{{\left( {110} \right)}^2} \times 1000} \over {{{\left( {220} \right)}^2}}} = {{1000} \over 4} = 250W
Q142
A moving coil galvanometer has 150150 equal divisions. Its current sensitivity is 1010- divisions per milliampere and voltage sensitivity is 22 divisions per millivolt. In order that each division reads 11 volt, the resistance in ohmsohms needed to be connected in series with the coil will be -
A 105{10^5}
B 103{10^3}
C 99959995
D 9999599995
Correct Answer
Option C
Solution

KEY CONCEPT : Resistance of Galvanometer,

G=Current   sensitivityVoltage   sensityvity⇒G=102=5ΩG = {{Current\,\,\,sensitivity} \over {Voltage\,\,\,sensityvity}} \Rightarrow G = {{10} \over 2} = 5\Omega

Here

ig={i_g} =

Full scale deflection current

=15010=15  mA= {{150} \over {10}} = 15\,\,mA
V=V=

voltage to be measured

=150=150

volts (such that each division reads

11

volt)

⇒R=15015×10−3−5=9995Ω\Rightarrow R = {{150} \over {15 \times {{10}^{ - 3}}}} - 5 = 9995\Omega
Q143
A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is :
A 0.50 W
B 0.072 W
C 0.10 W
D 0.125 W
Correct Answer
Option C
Solution

PR = 0.5 W ⇒\Rightarrow i2R = 0.5 W iR = 2.5 ⇒\Rightarrow i = 0.2 A & R = 12.5

Ω\Omega

Also, V = E – ir ⇒\Rightarrow 2.5 = 3 – (0.2)r ⇒\Rightarrow r = 2.5

Ω\Omega

Power dissipated in internal resistance = i2r = (0.2)2(2.5) = 0.1 W

Q144
When 5V5V potential difference is applied across a wire of length 0.10.1 m,m, the drift speed of electrons is 2.5×10−4  ms−1.2.5 \times {10^{ - 4}}\,\,m{s^{ - 1}}. If the electron density in the wire is 8×1028  m−3,8 \times {10^{28}}\,\,{m^{ - 3}}, the resistivity of the material is close to :
A 1.6×10−6Ωm1.6 \times {10^{ - 6}}\Omega m
B 1.6×10−5Ωm1.6 \times {10^{ - 5}}\Omega m
C 1.6×10−8Ωm1.6 \times {10^{ - 8}}\Omega m
D 1.6×10−7Ωm1.6 \times {10^{ - 7}}\Omega m
Correct Answer
Option B
Solution
V=IR=(neAvd)ρℓAV = IR = \left( {neA{v_d}} \right)\rho {\ell \over A}

∴\therefore

ρ=VVdln⁡e\rho = {V \over {{V_d}\ln e}}

Here

V=V=

potential difference

l=l =

length of wire

n=n=

no. of electrons per unit volume of conductor.

e=e=

no. of electrons Placing the value of above parameters we get resistivity

ρ=58×1028×1.6×10−19×2.5×10−4×0.1\rho = {5 \over {8 \times {{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times 2.5 \times {{10}^{ - 4}} \times 0.1}}
=1.6×10−5Ωm= 1.6 \times {10^{ - 5}}\Omega m
Q145
A copper rod of cross-sectional area A carries a uniform current I through it. At temperature T, if the volume charge density of the rod is ρ\rho , how long will the changes take to travel a distance d ?
A 2ρ d AI{{2\rho \,d\,A} \over {\rm I}}
B 2ρ d AI T{{2\rho \,d\,A} \over {{\rm I}\,T}}
C ρ d AI {{\rho \,d\,A} \over {{\rm I}\,}}
D ρ d AI T{{\rho \,d\,A} \over {{\rm I}\,T}}
Correct Answer
Option C
Solution

Given : Volume charge density of rod = ρ\rho. We know that current

I=neAvdI = neA{v_d}

; where n is number of electrons, e is electronic charge, A is area, vd is drift velocity Since volume charge density is ρ\rho = ne.

Therefore,

I=ρAvd⇒vd=IρAI = \rho A{v_d} \Rightarrow {v_d} = {I \over {{\rho _A}}}

Now, time =

distan⁡cespeed{{dis\tan ce} \over {speed}}

⇒\Rightarrow time

=dvd=dI/ρA=ρAdI= {d \over {{v_d}}} = {d \over {I/\rho A}} = {{\rho Ad} \over I}

Thus, time required to travel distance

d=ρdAId = {{\rho dA} \over I}
Q146
A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot and the other is kept cold, then, an electric current will
A flow from Antimony to Bismuth at the hot junction
B flow from Bismuth to Antimony at the cold junction
C now flow through the thermocouple
D flow from Antimony to Bismuth at the cold junction
Correct Answer
Option D
Solution

At cold junction, current flows from Antimony to Bismuth (because current flows from metal occurring later in the series to metal occurring earlier in the thermoelectric series).

Q147
In a potentiometer experiment the balancing with a cell is at length 240240 cm.cm. On shunting the cell with a resistance of 2Ω,2\Omega , the balancing length becomes 120120 cmcm. The internal resistance of the cell is
A 0.5Ω0.5\Omega
B 1Ω1\Omega
C 2Ω2\Omega
D 4Ω4\Omega
Correct Answer
Option C
Solution

The internal resistance of the cell,

r=(ℓ1−ℓ2ℓ2)×Rr = \left( {{{{\ell _1} - {\ell _2}} \over {{\ell _2}}}} \right) \times R
=240−120120×2=2Ω= {{240 - 120} \over {120}} \times 2 = 2\Omega
Q148
The resistance of the series combination of two resistances is S.S. When they are jointed in parallel the total resistance is P.P. If S=nPS = nP then the Minimum possible value of nn is
A 22
B 33
C 44
D 11
Correct Answer
Option C
Solution
S=R1+R2S = {R_1} + {R_2}

and

P=R1R2R1+R2P = {{{R_1}{R_2}} \over {{R_1} + {R_2}}}
S=nP⇒R1+R2=n(R1R2)(R1+R2)S = nP \Rightarrow {R_1} + {R_2} = {{n\left( {{R_1}{R_2}} \right)} \over {\left( {{R_1} + {R_2}} \right)}}
⇒(R1+R2)2=nR1R2\Rightarrow {\left( {{R_1} + {R_2}} \right)^2} = n{R_1}{R_2}
⇒n=R11+R22+R1R2R1R2\Rightarrow n = {{R_1^1 + R_2^2 + {R_1}{R_2}} \over {{R_1}{R_2}}}
n=R1R2+R2R1+2n = {{{R_1}} \over {{R_2}}} + {{{R_2}} \over {{R_1}}} + 2

Arithmetic mean

>>

Geometric mean Minimum value of

nn

is

44
Q149
Resistance of the wire is measured as 2 Ω\Omega and 3 Ω\Omega at 10∘^\circC and 30∘^\circC respectively. Temperature co-efficient of resistance of the material of the wire is :
A 0.033 ∘^\circC−-1
B −-0.033 ∘^\circC−-1
C 0.011 ∘^\circC−-1
D 0.055 ∘^\circC−-1
Correct Answer
Option A
Solution

R10 = 2 = R0(1 + α\alpha ×\times 10) R30 = 3 = R0(1 + α\alpha ×\times 30) On solving α\alpha = 0.033/

∘^\circ

C

Q150
On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 kΩ\Omega . How much was the resistance on the left slot before interchanging the resistances?
A 910 Ω\Omega
B 990 Ω\Omega
C 505 Ω\Omega
D 550 Ω\Omega
Correct Answer
Option D
Solution
Xl{X \over l}

=

1000−X100−l{{1000 - X} \over {100 - l}}

. . . . . (1) When X and Y are interchanged.

1000−Xl−10{{1000 - X} \over {l - 10}}

=

X100−l+10{X \over {100 - l + 10}}

⇒\Rightarrow

   \,\,\,
1000−Xl−10{{1000 - X} \over {l - 10}}

=

X110−l{X \over {110 - l}}

. . . . . (2) From (1) and (2) we get,

l100−l{l \over {100 - l}}

=

110−ll−10{{110 - l} \over {l - 10}}

⇒\Rightarrow

   \,\,\,
ll

2 −- 10 = 11000 −- 100

ll

−- 110

ll

+

ll

2 ⇒\Rightarrow

   \,\,\,

200

ll

= 11000

ll

= 55 cm Putting this value of

ll

, in equation (1)

X55{X \over {55}}

=

1000−X100−55{{1000 - X} \over {100 - 55}}

⇒\Rightarrow

   \,\,\,

45X = 55000 −- 55X ⇒\Rightarrow

   \,\,\,

100X = 55000 ⇒\Rightarrow

   \,\,\,

X = 550

Ω\Omega
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