Current Electricity

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

Q101

The resistance of each arm of the Wheatstone's bridge is 10 ohm. A resistance of 10 ohm is connected in series with a galvanometer then the equivalent resistance across the battery will be

A 10 ohm

B 15 ohm

C 20 ohm

D 40 ohm

Correct Answer

Option A

Solution

Here, P = Q = S = R = 10

\Omega

since

{P \over Q} = {S \over R} = 1

, so it is balanced Wheatstone bridge and no current pass through galvanometer.

$\therefore$ Equivalent resistance is given by R =

{{\left( {10 + 10} \right)\left( {10 + 10} \right)} \over {\left( {10 + 10} \right) + \left( {10 + 10} \right)}}

= {{20 \times 20} \over {40}} = 10\Omega

Q102

The potentiometer is best for measuring voltage, as

A it has a sensitive galvanometer and gives null deflection

B it has wire of high resistance

C it measures p.d. in closed circuit

D it measures p.d in open circuit.

Correct Answer

Option A

Solution

When we measure the emf of a cell by the potentiometer then no current draws in the circuit in zero-deflection condition i.e., cell is in open circuit.

Thus, in this condition the actual value of a cell is found.

In this way potentiometer is equivalent to an ideal voltmeter of infinite resistance.

Note: The emf by the potentiometer is measured from null method in which zero deflection position is found on the wire.

Q103

Two bulbs of (40 W, 200 V), and (100 W, 200 V). Then correct relation for their resistances

A R 40 < R 100

B R 40 > R 100

C R 40 = R 100

D no relation can be predicted.

Correct Answer

Option B

Solution

P = {{{V^2}} \over R}

\Rightarrow R \propto {1 \over P}

\therefore {R_{40}} > {R_{100}}

Q104

A car battery of emf 12 V and internal resistance

5 \times {10^{ - 2}}

\Omega

., receives a current of 60 amp, from external source, then terminal potential difference of battery is

A 12 V

B 9 V

C 15 V

D 20 V

Correct Answer

Option C

Solution

{{V - E} \over r} = I \Rightarrow {{V - 12} \over {5 \times {{10}^{ - 2}}}} = 60

\Rightarrow V = 15V

Q105

The net resistance of the circuit between A and B is

{8 \over 3}\Omega

{14 \over 3}\Omega

{{16} \over 3}\Omega

{{22} \over 3}\Omega

Correct Answer

Option B

Solution

This is a balanced Wheatstone’s bridge so no current flows through the 7

\Omega

resistor. $\therefore$

{1 \over {{R_{eq}}}} = {1 \over {4 + 3}} + {1 \over {6 + 8}}

\Rightarrow {R_{eq}} = {{14} \over 3}\Omega

Ready for a full NEET mock test? Timed · full syllabus · instant results

Take a Mock Test →