Blog Log in Get Started Free

Physics MCQs for NEET — Practice Questions with Answers

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

All Physics Chemistry Botany Zoology

Chapter Language Difficulty Show only questions I answered incorrectly Register free to filter questions

● ● ● ● ●

A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Wb/m². The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30^o with the direction of the field, the torque required to keep the coil in stable equilibrium will be:

0.15 Nm

0.20 Nm

0.24 Nm

0.12 Nm

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

● ● ● ● ●

A bar magnet of length l and magnetic dipole moment M is bent to form an arc which subtends an angle of $120 °$ at centre. The new magnetic dipole moment will be

$\frac{3 M}{2 π}$

$\frac{3 \sqrt{3} M}{2 π}$

$\frac{3 M}{π}$

$\frac{2 M}{π}$

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

Explanation

When a bar magnet is bent into an arc, its magnetic dipole moment changes. The new magnetic dipole moment is directly proportional to the original dipole moment and the angle subtended by the arc at the center. For an angle of 120Â°, the new dipole moment is (3âˆš3/2Ï€)M.

● ● ● ● ●

The unit of pole strength is

$A m^{2}$

$\frac{A^{2}}{m}$

$\frac{A^{2}}{m^{2}}$

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

Explanation

If we define magnetic moment like electric dipole moment

Magnetic moment ( Ampere X $m^{2}$ ) = Pole strength X Distance (m)

Unit of pole strength = Ampere X meter

● ● ● ● ●

Magnetic field lines

Cannot intersect

Are always closed curves

Can pass through vacuum

All of these

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

Explanation

Magnetic field lines cannot intersect, are always closed curves, and can pass through a vacuum. Therefore, option o4 which states 'All of these' is the correct answer. Magnetic field lines represent the direction of the magnetic force at a given point and must follow certain rules to accurately depict the magnetic field.

● ● ● ● ●

$If a magnet is suspended at angle 30 ° to the magnetic meridian, the dip needle makes an angle of 45 ° with the horizontal . The value of re a l dip angle at the place is$

$\tan^{- 1} \frac{\sqrt{3}}{2}$

$\tan^{- 1} 1$

$\tan^{- 1} \frac{1}{2}$

$\tan^{- 1} (\frac{2}{\sqrt{3}})$

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

● ● ● ● ●

Two small bar magnets are placed in air at a distance r apart. The magnetic force between them is proportional to

$r^{2}$

$r^{- 2}$

$r^{- 3}$

$r^{- 4}$

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

Explanation

$F = \frac{μ_{0}}{4 π} \frac{6 M_{1} M_{2}}{r^{4}} F α \frac{1}{r^{4}}$

● ● ● ● ●

The vertical component of the earths magnetic field is zero at

Magnetic poles

Magnetic equator

Geographic poles

Geographic equator

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

Explanation

$V = B \sin ϕ A t e q u a t o r ϕ = 0 s o V = 0$

● ● ● ● ●

A short magnetic dipole is placed at the origin with its dipole movement directed along the x-axis. If magnetic field induction at a point P (r, 0) is B $\hat{i}$ , the magnetic field induction at point Q(0,2r) will be

$\frac{B}{16} \hat{i}$

$- \frac{B}{8} \hat{j}$

$\frac{B}{16} \hat{j}$

$- \frac{B}{16} \hat{j}$

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

Explanation

$At axial point : B = \frac{μ_{0}}{4 π} \frac{2 M}{r^{3}} At equitorial point (at same distance) : B' = \frac{μ_{0}}{4 π} \frac{M}{r^{3}} = \frac{1}{2} \frac{μ_{0}}{4 π} \frac{2 M}{r^{3}} = \frac{1}{2} B At distance 2 r : B' = \frac{μ_{0}}{4 π} \frac{M}{{(2 r)}^{3}} = = \frac{1}{8} \frac{μ_{0}}{4 π} \frac{M}{r^{3}} = \frac{1}{8} \frac{B}{2} = \frac{B}{16}$

● ● ● ● ●

The magnetic field at a point x on the axis of a small bar magnet is equal to the field at a point y on the equator of the same magnet. The ratio of the distances of x and y from the centre of the magnet is

$2^{- 3}$

$2^{\frac{- 1}{3}}$

$2^{3}$

$2^{\frac{1}{3}}$

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

Explanation

The magnetic field due to a bar magnet follows an inverse cube law, meaning the field strength is inversely proportional to the cube of the distance from the center. If the field is equal at points x and y, their distances from the center must follow the ratio (x/y)^3 = 2. Therefore, the ratio x/y = 2^(1/3).

● ● ● ● ●

A magnet of magnetic moment 20 C.G.S. units are freely suspended in a uniform magnetic field of intensity 0.3 C.G.S. units. The amount of work done in deflecting it by an angle of $30^{o}$ in C.G.S. units is

$3 \sqrt{3}$

$3 (2 - \sqrt{3})$

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

Explanation

$W = MB (1 - cosθ) = 20 \times 0.3 \times (1 - \cos 30^{0}) = 6 (1 - \frac{\sqrt{3}}{2}) = 3 (2 - \sqrt{3})$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Physics 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.