Practice Start Test

Current Electricity

JEE Physics · 159 questions · Page 4 of 16 · Click an option or "Show Solution" to reveal answer

Q31
The supply voltage to room is 120V.120V. The resistance of the lead wires is 6Ω6\Omega . A 6060 WW bulb is already switched on. What is the decrease of voltage across the bulb, when a 240240 WW heater is switched on in parallel to the bulb?
A zero
B 2.92.9 Volt
C 13.313.3 Volt
D 10.0410.04 Volt
Correct Answer
Option D
Solution

Power of bulb

=60W=60W
(given)\left( {given} \right)

Resistance of bulb

=120×12060=240Ω= {{120 \times 120} \over {60}} = 240\Omega
[  \left[ {\,\,} \right.
 P=V2R ]\left. {\,P = {{{V^2}} \over R}\,} \right]

Power of heater

=240W=240W

(given) Resistance of heater

=120×120240=60Ω= {{120 \times 120} \over {240}} = 60\Omega

Voltage across bulb before heater is switched on,

V1=240246×120=117.73  {V_1} = {{240} \over {246}} \times 120 = 117.73\,\,

volt Voltage across bulb after heater is switched on,

V2=4854×120=106.66{V_2} = {{48} \over {54}} \times 120 = 106.66

volt Hence decrease in voltage

V1−V2=117.073−106.66=10.04{V_1} - {V_2} = 117.073 - 106.66 = 10.04

Volt (approximately)

Q32
This questions has Statement - I{\rm I} and Statement - I{\rm I}I{\rm I}. Of the four choices given after the Statements, choose the one that best describes into two Statements. Statement - I{\rm I} : Higher the range, greater is the resistance of ammeter. Statement - I{\rm I}I{\rm I} : To increase the range of ammeter, additional shunt needs to be used across it.
A Statement - I{\rm I} is true, Statement - II{\rm II} is true, Statement - II{\rm II} is the correct explanation of statement - I{\rm I}.
B Statement - I{\rm I} is true, Statement - II{\rm II} is true, Statement - II{\rm II} is not the correct explanation of statement - I{\rm I}.
C Statement - I{\rm I} is true, Statement - II{\rm II} is false
D Statement - I{\rm I} is false, Statement - II{\rm II} is true
Correct Answer
Option D
Solution

Statements

I{\rm I}

is false and Statement

I{\rm I}
I{\rm I}

is true For ammeter, shunt resistance,

S=IgGI−IgS = {{{{\rm I}_g}G} \over {{\rm I} - {{\rm I}_g}}}

Therefore for

I{\rm I}

to increase,

SS

should decrease, So additional

SS

can be connected across it.

Q33
The resistance of a wire is 55 ohm at 50∘C{50^ \circ }C and 66 ohm at 100∘C.{100^ \circ }C. The resistance of the wire at 0∘C{0^ \circ }C will be
A 33 ohm
B 22 ohm
C 11 ohm
D 44 ohm
Correct Answer
Option D
Solution

KEY CONCEPT : We know that

Rt=R0(1+αt),{R_t} = R{}_0\left( {1 + \alpha t} \right),

where

Rt{R_t}

is the resistance of the wire at

t∘C,{t^ \circ }C,
R0{R_0}

is the resistance of the wire at

0∘C{0^ \circ }C

and α\alpha is the temperature coefficient of resistance

⇒R50=R0(1+50α)            ...(i)\Rightarrow {R_{50}} = {R_0}\left( {1 + 50\alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)
R100=R0(1+100α)             ...(ii){R_{100}} = {R_0}\left( {1 + 100\alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

From

(i), R50−R0=50αR0                 ...(iii)\left( i \right),\,{R_{50}} - {R_0} = 50\alpha {R_0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)

From

(ii),R100−R0=100αR0            ...(iv)\left( {ii} \right),{R_{100}} - {R_0} = 100\alpha {R_0}\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iv} \right)

Dividing

(iii)(iii)

by

(iv),(iv),

we get

R50−R0R100−R0=12{{{R_{50}} - {R_0}} \over {{R_{100}} - R{}_0}} = {1 \over 2}

Here,

R50{{R_{50}}}
=5Ω= 5\Omega

and

R100=6Ω{{R_{100}} = 6\Omega }

∴\therefore

5−R06−R0=12{{5 - {R_0}} \over {6 - {R_0}}} = {1 \over 2}

or,

6−R0=10−2R06 - {R_0} = 10 - 2{R_0}

or,

R0=4Ω.{R_0} = 4\Omega .
Q34
A constant voltages is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be :
A Doubled
B Halved
C Unchanged
D Increased 8 times
Correct Answer
Option D
Solution

Since rate of heat

=V2R⇒= {{{V^2}} \over R} \Rightarrow

rate of heat

∝1R\propto {1 \over R}

, where

R=ρLA=ρLπr2R = {{\rho L} \over A} = {{\rho L} \over {\pi {r^2}}}

; here r is radius of wire and L is length of wire. Therefore,

R∝Lr2R \propto {L \over {{r^2}}}

. Thus, rate of heat

∝r2L\propto {{{r^2}} \over L}

. When length in halved and radius is doubled then, rate of heat

∝(2r)2L/2=8r2L\propto {{{{(2r)}^2}} \over {L/2}} = {{8{r^2}} \over L}

Therefore, rate of heat developed in wire will be increased 8 times.

Q35
A moving coil galvanometer has resistance 50Ω\Omega and it indicates full deflection at 4mA current. A voltmeter is made using this galvanometer and a 5 kΩ\Omega resistance. The maximum voltage, that can be measured using this voltmeter, will be close to :
A 15 V
B 10 V
C 40 V
D 20 V
Correct Answer
Option D
Solution

G = 50

Ω\Omega

S = 5000

Ω\Omega

Ig = 4 × 10–3 V = ig (G + S) V = 4 × 10–3 (50 + 5000) = 4 × 10–3 (5050) = 20.2 volt

Q36
The resistance of an electrical toaster has a temperature dependence given by R(T) = R0 [1 + α\alpha (T − T0)] in its range of operation. At T0 = 300 K, R = 100 Ω\Omega and at T = 500 K, R = 120 Ω\Omega . The toaster is connected to a voltage source at 200 V and its temperature is raised at a constant rate from 300 to 500 K in 30 s. The total work done in raising the temperature is :
A 400 ln⁡ 1.51.3 J\ln \,{{1.5} \over {1.3}}\,J
B 200 ln⁡ 23 J\ln \,{{2} \over {3}}\,J
C 60000 ln⁡ 65 J\ln \,{{6} \over {5}}\,J
D 300 J
Correct Answer
Option C
Solution

Given, R = R6 [1 + α\alpha (T −- t0)] 120 = 100 [1 + α\alpha (500 −- 300)] ⇒\Rightarrow 200 α\alpha =

15{1 \over 5}

⇒\Rightarrow α\alpha = 10−-3 oC−-1 Temperature of the toaster raised from 300 K to 500 K in 30 s.

∴\therefore Increment in the temperature in time t,

Δ\Delta

T =

500−30030t{{500 - 300} \over {30}}t

=

2003t{{200} \over 3}t

=

203t{{20} \over 3}t

Total work done in raising the temperature =

∫0tV2R(t) dt\int\limits_0^t {{{{V^2}} \over {R\left( t \right)}}\,dt}

=

∫0tV2R0(1+αΔt) dt\int\limits_0^t {{{{V^2}} \over {{R_0}\left( {1 + \alpha \Delta t} \right)}}} \,dt

=

∫030(200)2100(1+10−3×203t) dt\int\limits_0^{30} {{{{{\left( {200} \right)}^2}} \over {100\left( {1 + {{10}^{ - 3}} \times {{20} \over 3}t} \right)}}} \,dt

=

40000100∫030dt(1+t150){{40000} \over {100}}\int\limits_0^{30} {{{dt} \over {\left( {1 + {t \over {150}}} \right)}}}
400×150[ln⁡(1+t150)]030400 \times 150\left[ {\ln \left( {1 + {t \over {150}}} \right)} \right]_0^{30}
=60000[ln⁡(1+30150)−ln⁡1]= 60000\left[ {\ln \left( {1 + {{30} \over {150}}} \right) - \ln 1} \right]
=60000ln⁡(65) J= 60000\ln \left( {{6 \over 5}} \right)\,J
Q37
Which of the following statements is false?
A Kirchhoff’s second law represents energy conservation.
B Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude.
C In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed.
D A rheostat can be used as a potential divider.
Correct Answer
Option C
Solution

There is no change in null point, if the cell and the galvanometer are exchanged in a balanced wheatstone bridge.

Q38
When a current of 5 mA is passed through a galvanometer having a coil of resistance 15Ω\Omega , it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 – 10V is:
A 4.005 × 103 Ω\Omega
B 1.985 × 103 Ω\Omega
C 2.535 × 103 Ω\Omega
D 2.045 × 103 Ω\Omega
Correct Answer
Option B
Solution

Given : Current through the galvanometer, ig = 5 × 10–3 A Galvanometer resistance, G = 15

Ω\Omega

Let resistance R to be put in series with the galvanometer to convert it into a voltmeter.

V = ig (R + G) 10 = 5 × 10–3 (R + 15) ∴\therefore R = 2000 – 15 = 1985 = 1.985 × 103

Ω\Omega
Q39
A galvanometer with its coil resistance 25 Ω\Omega requires a current of 1 mA for its full deflection. In order to construct an ammeter to read upto a current of 2 A, the approximate value of the shunt resistance should be :
A 2.5×10−3 Ω2.5 \times {10^{ - 3}}\,\Omega
B 1.25×10−2Ω1.25 \times {10^{ - 2}}\Omega
C 1.25×10−3Ω1.25 \times {10^{ - 3}}\Omega
D 2.5×10−2Ω2.5 \times {10^{ - 2}}\Omega
Correct Answer
Option B
Solution

Given, Ig = 1 ma I −- Ig = 2 A Rg = 25

Ω\Omega
∴   \therefore\,\,\,

Ig Rg = (I −- Ig) S ⇒\Rightarrow

   \,\,\,

S =

10−3×252{{{{10}^{ - 3}} \times 25} \over 2}

⇒\Rightarrow

   \,\,\,

S = 1.25 ×\times 10−-2

Ω\Omega
Q40
A galvanometer having a coil resistance of 100 Ω100\,\Omega gives a full scale deflection, when a currect of 11 mAmA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 1010 A,A, is :
A 0.1 Ω0.1\,\Omega
B 3 Ω3\,\Omega
C 0.01 Ω0.01\,\Omega
D 2 Ω2\,\Omega
Correct Answer
Option C
Solution
IgG=(I−Ig)s{\rm I}gG = \left( {{\rm I} - {\rm I}g} \right)s

∴\therefore

10−3×100=(10−10−3)×S{10^{ - 3}} \times 100 = \left( {10 - {{10}^{ - 3}}} \right) \times S

∴\therefore

S≈0.01  ΩS \approx 0.01\,\,\Omega
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