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

● ● ● ● ●

Four equal charges Q are placed at the four corners of a square of each side is ‘a’. Work done in removing a charge – Q from its centre to infinity is

$\frac{\sqrt{2} Q^{2}}{4 π ε_{0} a}$

$\frac{\sqrt{2} Q^{2}}{π ε_{0} a}$

$\frac{Q^{2}}{2 π ε_{0} a}$

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

● ● ● ● ●

Angle between equipotential surface and lines of force is

Zero

180°

90°

45°

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

Explanation

Lines of force is perpendicular to the equipotential surface. Hence angle = 90°

● ● ● ● ●

Two spheres of radius a and b respectively are charged and joined by a wire. The ratio of electric field of the spheres is

a/b

b/a

a²/b²

b²/a²

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

Explanation

Joined by a wire means they are at the same potential. For same potential $\frac{k Q_{1}}{a_{1}} = \frac{k Q_{2}}{a_{2}}$

⇒ $\frac{Q_{1}}{Q_{2}} = \frac{a}{b}$

Further, the electric field at the surface of the sphere having radius R and charge Q is $\frac{k Q}{R^{2}} .$

∴ $\frac{E_{1}}{E_{2}} = \frac{k Q_{1} / a^{2}}{k Q_{2} / b_{2}} = \frac{Q_{1}}{Q_{2}} \times \frac{b^{2}}{a^{2}} = \frac{b}{a}$

● ● ● ● ●

An electron of mass m and charge e is accelerated from rest through a potential difference V in vacuum. The final speed of the electron will be

$V \sqrt{e / m}$

$\sqrt{e V / m}$

$\sqrt{2 e V / m}$

$2 e V / m$

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

Explanation

Kinetic energy $K = \frac{1}{2} m v^{2} = e V$

⇒ $v = \sqrt{\frac{2 e V}{m}}$

● ● ● ● ●

Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle θ with the axis of the dipole, then the potential at P is given by (r >> 2a) (Where p = 2qa)

$V = \frac{p \cos θ}{4 π ε_{0} r^{2}}$

$V = \frac{p \cos θ}{4 π ε_{0} r}$

$V = \frac{p \sin θ}{4 π ε_{0} r}$

$V = \frac{p \cos θ}{2 π ε_{0} r^{2}}$

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

● ● ● ● ●

A charge +q is fixed at each of the points $x = x_{0}, x = 3 x_{0}, x = 5 x_{0}$ ..... infinite, on the x-axis and a charge –q is fixed at each of the points $x = 2 x_{0}, x = 4 x_{0}, x = 6 x_{0}$ ,..... infinite. Here x₀ is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be $\frac{Q}{4 π ε_{0} r}$ . Then, the potential at the origin due to the above system of charges is

$\frac{q}{8 π ε_{0} x_{0} \ln 2}$

∞

$\frac{q \ln 2}{4 π ε_{0} x_{0}}$

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

● ● ● ● ●

Inside a hollow charged spherical conductor, the potential -

Is constant

Varies directly as the distance from the centre

Varies inversely as the distance from the centre

Varies inversely as the square of the distance from the centre

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

Explanation

Inside the hollow sphere, at any point the potential is constant.

● ● ● ● ●

Two small spheres each carrying a charge q are placed r meter apart. If one of the spheres is taken around the other one in a circular path of radius r, the work done will be equal to

Force between them × r

Force between them × 2πr

Force between them / 2πr

Zero

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

Explanation

The force is perpendicular to the displacement.

● ● ● ● ●

Two charged spheres of radii 10 cm and 15 cm are connected by a thin wire. No current will flow, if they have -

The same charge on each

The same potential

The same energy

The same field on their surfaces

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

Explanation

Because current flows from higher potential to lower potential.

● ● ● ● ●

The electric field inside a spherical shell of uniform surface charge density is -

Zero

Constant, less than zero

Directly proportional to the distance from the centre

None of the above

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

Explanation

All charge resides on the outer surface so that according to Gauss law, electric field inside a shell is zero.

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.