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

● ● ● ● ●

When a 12 Ω resistor is connected in parallel with a moving coil galvanometer then its deflection reduces from 50 divisions to 10 divisions. The resistance of the galvanometer is :

24 Ω

36 Ω

48 Ω

60 Ω

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

Explanation

$i_{g} = \frac{i S}{S + G}$ ⇒ $10 = \frac{50 \times 12}{12 + G}$ ⇒ $12 + G = 60$ ⇒ $G = 48 Ω$

● ● ● ● ●

A galvanometer can be used as a voltmeter by connecting a :

High resistance in series

Low resistance in series

High resistance in parallel

Low resistance in parallel

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

Explanation

To convert a galvanometer into a voltmeter, a high value resistance is to be connected in series with it.

● ● ● ● ●

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

$(n - 1) G$

$\frac{G}{n}$

$\frac{G}{(n - 1)}$

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

Explanation

To convert a voltmeter of range V to a higher range nV, a resistance (n-1)G should be connected in series with the existing resistance G. This increases the total resistance, reducing the current and allowing the voltmeter to measure n times higher voltage without damaging the meter.

● ● ● ● ●

Which of the following statement is wrong:

Voltmeter should have high resistance

Ammeter should have low resistance

Ammeter is placed in parallel across the conductor in a circuit

Voltmeter is placed in parallel across the conductor in a circuit

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

Explanation

Ammeter is always connected in series with circuit.

● ● ● ● ●

A moving coil galvanometer has a resistance of 50 Ωand gives full scale deflection for 10 mA. How could it be converted into an ammeter with a full scale deflection for 1A :

50/99 Ω in series

50/99 Ω in parallel

0.01 Ω in series

0.01 Ω in parallel

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

Explanation

$S = \frac{i_{g} \times G}{i - i_{g}} = \frac{10 \times 10^{- 3} \times 50}{1 - 10^{- 3} \times 10} = \frac{50}{99} Ω$ in parallel.

● ● ● ● ●

The resistance of a galvanometer is 50 ohms and the current required to give full-scale deflection is $100 μ A$ . In order to convert it into an ammeter, reading up to 10A, it is necessary to put a resistance of :

$5 \times 10^{- 3} Ω$ in parallel

$5 \times 10^{- 4} Ω$ in parallel

$10^{5} Ω$ in series

99,950 Ω in series

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

Explanation

Resistance in parallel $S = \frac{G i_{g}}{i - i_{g}} = \frac{50 \times 100 \times 10^{- 6}}{(10 - 100 \times 10^{- 6})}$

⇒ $S = 5 \times 10^{- 4} Ω$

● ● ● ● ●

If only 2% of the main current is to be passed through a galvanometer of resistance G, then the resistance of shunt will be

$\frac{G}{50}$

$\frac{G}{49}$

$50 G$

$49 G$

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

Explanation

$i_{g} = 2 %$ of $i = \frac{i}{50}$ ⇒ $S = \frac{G}{(n - 1)} = \frac{G}{(50 - 1)} = \frac{G}{49}$

● ● ● ● ●

The resistance of an ideal voltmeter is

Zero

Very low

Very large

Infinite

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

Explanation

The resistance of an ideal voltmeter is considered as infinite.

● ● ● ● ●

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

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

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.

● ● ● ● ●

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

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

Explanation

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

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

⇒ $S = \frac{1000}{9} = 111 Ω$

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.