Are these Physics questions free to attempt?

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.

Do I need an account to practise Physics MCQs?

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

Are these previous year NEET questions?

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

Physics NEET Questions — Free MCQ Practice

● ● ● ● ●

A metallic sphere is given some charge. Electric energy is stored

A.

Only inside the sphere

B.

Only outside the sphere

C.

Both inside and outside

D.

Neither inside nor outside

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

Explanation

Charge always resides on the outer surface.

● ● ● ● ●

Capacitors $C_{1} = 10 μ F a n d C_{2} = 30 μ F$ are connected in series across a source of emf 20KV. The potential difference across $C_{1}$ will be

A.

5 KV

B.

15 KV

C.

10 KV

D.

20 KV

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

Explanation

In a series combination of capacitors, the potential difference across each capacitor is inversely proportional to its capacitance. Since C2 = 30Î¼F is thrice C1 = 10Î¼F, the potential drop across C1 will be thrice that across C2. With a total potential of 20KV, the drop across C1 is 15KV.

● ● ● ● ●

Two metallic spheres of radii 2cm and 3cm are given charges 6mC and 4mC respectively. The final charge on the smaller sphere will be if they are connected by a conducting wire

A.

4mC

B.

6mC

C.

5mC

D.

10mC

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

Explanation

Finally, both spheres need to have equal potential. Let final charges be $q_{1} a n d q_{2}$

$\frac{k q_{1}}{2} = \frac{k q_{2}}{3} \frac{q_{1}}{q_{2}} = \frac{2}{3} A l s o, q_{1} + q_{2} = 10 S o l v i n g, q_{1} = 4 m C$

● ● ● ● ●

When a proton at rest is accelerated by a potential difference V, its speed is found to be v. The speed of an $α$ particle when accelerated by the same potential difference from rest will be

A.

v

B.

$\frac{v}{\sqrt{2}}$

C.

$v \sqrt{2}$

D.

2v

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

Explanation

$q V = \frac{1}{2} m v^{2}$

assuming charge on Proton as q and mass as m, above is the equation based on conservation of energy (decrease in potential energy is equal to the increase in kinetic energy here)

therefore, $v_{p r o t o n}$ = $\sqrt{\frac{2 q V}{m}}$

As α particle has 2 protons and 2 neutrons, so charge = 2q and mass = 4m and the equation becomes

^{$2 q V =$ $\frac{1}{2} \times 4 m \times {v_{α}}^{2}_{^{}}$}

So, $v_{α}$ = √(qV/m)

Therefore, $v_{α}$ = $\frac{v_{p r o t o n}}{\sqrt{2}}$

● ● ● ● ●

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

A.

0

B.

$\frac{\sqrt{2} Q^{2}}{4 π ε_{0} a}$

C.

$\frac{\sqrt{2} Q^{2}}{π ε_{0} a}$

D.

$\frac{Q^{2}}{2 π ε_{0} a}$

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

A.

Zero

B.

180°

C.

90°

D.

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.

a/b

B.

b/a

C.

a²/b²

D.

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

A.

$V \sqrt{e / m}$

B.

$\sqrt{e V / m}$

C.

$\sqrt{2 e V / m}$

D.

$2 e V / m$

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)

A.

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

B.

$V = \frac{p \cos θ}{4 π ε_{0} r}$

C.

$V = \frac{p \sin θ}{4 π ε_{0} r}$

D.

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

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

A.

0

B.

$\frac{q}{8 π ε_{0} x_{0} \ln 2}$

C.

∞

D.

$\frac{q \ln 2}{4 π ε_{0} x_{0}}$

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

Physics MCQs for NEET — Practice Questions with Answers

Frequently Asked Questions