An electric fan and a heater are marked as 100 W, 220 V and 1000 W, 220 V respectively. The resistance of heater is

equal to that of fan
lesser than that of fan
greater than that of fan
zero

Solution

As R α V²/P or R α 1/P,so resistance of heater is less than that of fan.

In a metre bridge, the balancing length from the left end(standard resistance of one ohm is in the right gap) is found to be 20 cm. The value of the unknown resistance is

0.8 Ω
0.5 Ω
0.4 Ω
0.25 Ω

The infinity resistance plug in a post-office box has

an air gap only
a resistance coil of infinite resistance
largest resistance available in box
resistance of the coil 5000Ω

For measuring voltage of any circuit, potentiometer is preferred to voltmeter because

the potentiometer is cheap and easy to handle.
calibration in the voltmeter is sometimes wrong .
the potential draws no current during measurement.
range of the voltmeter is not as wide as that of the potentiometer.

Coils in the resistance boxes are made from doubled up insulated wires

to cancel the effect of self induction
to nullify the heating effect
to nullify the Peltier effect
to reduce effective length of the wire

Potentiometer measures potential more accurately because

it measures potential in open circuit
it uses sensitive galvanometer for null deflection
it uses high resistance potentiometer wire
it measures potential in closed circuit

A flow of 10⁷ electrons per second in a conducting wire constitutes a current of

1.6 × 10^-26 A
1.6 × 10²⁶ A
1.6 × 10^-12 A
1.6 × 10¹² A

Solution

I = q/t = 10⁷ × 1.6 × 10^-19/1 = 1.6 × 10^-12 A

When a current I is set up in a wire of radius r,the drift velocity is v_d. If the same current is set up through a wire of radius 2 r, the drift velocity will be

4 v_d
2 v_d
v_d/2
v_d/4

Solution

I= n A e v_d or v_d α 1/π r²

A conductor carries a current of 50 μ A. If the area of cross-section of the conductor is 50 mm², then value of the current density in Am^–2 is

0.5
1
10^–3
10^–6

Solution

Current density J = I/A

= 50 × 16 ^-6/ 50 × 10^-6 = 1 Am^-2

The emf developed by a thermocouple is measured with the help of a potentiometer and not by a moving coilmillivoltmeter because

the potentiometer is more accurate than the voltmeter
the potentiometer is more sensitive than voltmeter
the potentiometer makes measurement with outdrawing any current from the thermocouple
measurement using a potentiometer is simpler than with a voltmeter

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