Blog Log in Get Started Free

NEET Practice Questions (MCQs) with Answers & Solutions

Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

● ● ● ● ●

A voltmeter has a resistance of G ohm and range V volts. The value of resistance used in series to convert it into a voltmeter of range nV volts is :

(n-1)G

G/n

G/(n-1)

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

We know that;

$R = \frac{V}{I_{g}} - G$

The voltmeter gives the full-scale deflection for potential difference V. Its resistance is G.

Hence,

$I_{g} = \frac{V}{G}$

Given that V = nV. Therefore,

$R = \frac{n V}{(V / G)} - G = (n - 1) G$

● ● ● ● ●

An ammeter is obtained by shunting a 30 $Ω$ galvanometer with a 30 $Ω$ resistance. What additional shunt should be connected across it to double the range?

15 $Ω$

10 $Ω$

5 $Ω$

none of these

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

The total range is doubled, i.e., $4 I_{g}$ .

Now shunt required is :

$S = \frac{I_{g}}{4 I_{g} - I_{g}} \times G = 10 Ω This is the resultant shunt . So, \frac{1}{30} + \frac{1}{x} = \frac{1}{10} o r x = 15 Ω$

● ● ● ● ●

Two similar headlight lamps are connected in parallel to each other. Together, they consume 48 W from a 6 V battery. What is the resistance of each filament?

6 $Ω$

4 $Ω$

3.0 $Ω$

1.5 $Ω$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

The power consumed by each lamp is 24 W.

Hence, using R = $(V^{2} / P)$ , we find R = (36/24) = 1.5 $Ω$ .

● ● ● ● ●

Two electric bulbs, rated for the same voltage, have powers of 200 W and 100 W, respectively. If their resistances are $r_{1} a n d r_{2}$ , respectively then :

$r_{1} = 2 r_{2}$

$r_{2} = 2 r_{1}$

$r_{2} = 4 r_{1}$

$r_{1} = 4 r_{2}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$P = \frac{V^{2}}{R}, \frac{P_{1}}{P_{2}} = \frac{R_{2}}{R_{1}} o r r_{2} = 2 r_{1}$

● ● ● ● ●

If the current in an electric bulb decreases by 0.5%, the power in the bulb decreases by approximately :

0.5%

0.25%

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$P = i^{2} R, \frac{d P}{P} = 2 \frac{d I}{I} = 2 \times 0.5 % = 1 %$

● ● ● ● ●

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 $Ω$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

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 :

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

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

R is reduced by a factor of 4.

So, P is increased by a factor of 4.

● ● ● ● ●

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 $Ω$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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 $Ω$ .

● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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.

● ● ● ● ●

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 :

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

The current required by each bulb is :

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

If n bulbs are joined in parallel, then,

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

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.