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Physics MCQs for NEET — Practice Questions with Answers

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An electric bulb rated for 500 W at 100 V is used in a circuit having a 200 V supply. The resistance R that must be put in series with the bulb, so that the bulb draws 500 W, is :

18 $Ω$

20 $Ω$

40 $Ω$

700 $Ω$

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A wire, when connected to a 220 V mains supply, has power dissipation $P_{1}$ . Now, the wire is cut into two equal pieces, which are connected in parallel to the same supply. Power dissipation in this case is $P_{2}$ . Then $P_{2} : P_{1}$ is :

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Explanation

$P = V^{2} / R$ .

R is reduced by a factor of 4.

So, P is increased by a factor of 4.

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The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of 100 W and 200 V lamp when not in use?

14,000 $Ω$

400 $Ω$

40 $Ω$

4 $Ω$

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Explanation

When the filament is hot, $R = \frac{V^{2}}{P} = \frac{200 \times 200}{100} = 400 Ω$

Hence, when cold, the resistance is 40 $Ω$ .

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Two electric bulbs A and B are rated 60 W and 100 W, respectively. If they are connected in parallel to the same source, then,

both the bulbs draw the same current

bulb A draws more current than bulb B

bulb B draws more current than bulb A

currents drawn in the bulbs are in the ratio of their resistances

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Explanation

Given $P_{A} = 60 W, P_{B} = 100 W$ . We know that current flowing through the bulb is I = P/V. We also know that as both the bulbs are connected in parallel, potential difference (V) across both the bulbs is the same. Thus, $I α P$ . Since the power of bulb B is greater than that of bulb A, bulb B draws more current than bulb A.

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A factory is served by a 220 V supply line. In a circuit protected by a fuse marked 10 A, the maximum number of 100 W lamps in parallel that can be turned on is :

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Explanation

The current required by each bulb is :

$I = \frac{P}{V} = \frac{100}{220} A$

If n bulbs are joined in parallel, then,

$n I = I_{f u s e} o r n \times \frac{100}{220} = 10 o r n = 22$ .

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A 100 W bulb designed to operate on 100 V is to be connected across a 500 V source. Find the resistance to be put in series so that bulb consumes 100 W only.

100 $Ω$

500 $Ω$

400 $Ω$

300 $Ω$

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A current wire is hidden in a wall. Its position can be located with the help of a

Moving coil galvanometer

Voltmeter

Hotwire ammeter

Magnetic needle

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Explanation

Magnetic needle deflect in the presence of a magnetic field produced by the wire.

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The resistance of an ideal voltmeter is

Zero

Very low

Very large

Infinite

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Explanation

The resistance of an ideal voltmeter is considered as infinite.

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The net resistance of a voltmeter should be large to ensure that :

It does not get overheated

It does not draw excessive current

It can measure large potential difference

It does not appreciably change the potential difference to be measured

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Explanation

The resistance of voltmeter is too high, so that it draws negligible current from the circuit, hence potential drop in the external circuit is also negligible.

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A voltmeter of resistance 1000 Ω gives full-scale deflection when a current of 100 mA flow through it. The shunt resistance required across it to enable it to be used as an ammeter reading 1 A at full-scale deflection is :

10000 Ω

9000 Ω

222 Ω

111 Ω

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Explanation

By using $\frac{i}{i_{g}} = 1 + \frac{G}{S}$

⇒ $\frac{i}{100 \times 10^{- 3}} = 1 + \frac{1000}{S}$

⇒ $S = \frac{1000}{9} = 111 Ω$

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