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

● ● ● ● ●

Two parallel metal plates having charges +Q and -Q faces each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will

A.

become zero

B.

increase

C.

decrease

D.

remain same

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

Explanation

When the parallel metal plates are dipped in kerosene oil, the electric field between the plates decreases. This is because the kerosene oil has a higher permittivity than vacuum or air, which reduces the effective electric field according to the relation E = Eâ‚€/Îº, where Îº is the relative permittivity.

● ● ● ● ●

The mean free path of electrons in a metal is $4 \times 10^{- 8} m .$ The electric field which can give on an average 2 eV energy to an electron in the metal will be in a unit of $V m^{- 1}$ :

A.

$8 \times 10^{7}$

B.

$5 \times 10^{- 11}$

C.

$8 \times 10^{- 11}$

D.

$5 \times 10^{7}$

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

Explanation

Energy=2eV

$e V_{0} = 2 e V$

$\Rightarrow V_{0} = 2$

Now, electric field E= $\frac{2}{4 \times 10^{- 8}}$

= $0.5 \times 10^{8}$

= $5 \times 10^{7} V m^{- 1}$

● ● ● ● ●

Four particles each having charge q are placed at the vertices of a square of side a. The value of the electric potential at the midpoint of one of the side will be

A.

0

B.

$\frac{1}{4 π ϵ_{0}} \frac{2 q}{a} (2 + \frac{2}{\sqrt{5}})$

C.

$\frac{1}{4 π ϵ_{0}} \frac{2 q}{a} (2 - \frac{2}{\sqrt{5}})$

D.

$\frac{1}{4 π ϵ_{0}} \frac{2 q}{a} (1 + \frac{1}{\sqrt{5}})$

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

● ● ● ● ●

If W be the amount of heat produced in the process of charging an uncharged capacitor then the amount of energy stored in it is

A.

2W

B.

$\frac{W}{2}$

C.

W

D.

zero

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

Explanation

Energy stored in capacitor = Heat produced in process of charging

● ● ● ● ●

A metallic sphere of capacitance $C_{1}$ , charged to electric potential $V_{1}$ is connected by a metal wire to another metallic sphere of capacitance $C_{2}$ charged to electric potential $V_{2}$ . The amount of heat produced in the connecting wire during the process is

A.

$\frac{C_{1} C_{2}}{2 (C_{1} + C_{2})} {(V_{1} + V_{2})}^{2}$

B.

$\frac{C_{1} C_{2}}{2 (C_{1} + C_{2})} {(V_{1} - V_{2})}^{2}$

C.

$\frac{C_{1} C_{2}}{C_{1} + C_{2}} {(V_{1} - V_{2})}^{2}$

D.

zero

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

Explanation

Initial charge on the spheres:

$Q_{1} = C_{1} V_{1} & Q_{2} = C_{2} V_{2} Initial potential energy of the spheres - U_{1} = \frac{1}{2} C_{1} V_{1}^{2} a n d, U_{2} = \frac{1}{2} C_{2} V_{2}^{2} Charges will flow between the spheres until they have same electric potential . Let q charge flows from sphere 1 to 2 . Then, V_{1}' = V_{2}' \frac{Q_{1} - q}{C_{1}} = \frac{Q_{2} + q}{C_{2}} \Rightarrow q = \frac{Q_{1} C_{2} - Q_{2} C_{1}}{C_{1} + C_{2}} = \frac{C_{1} V_{1} C_{2} - C_{2} V_{2} C_{1}}{C_{1} + C_{2}} \Rightarrow q = \frac{C_{1} C_{2}}{(C_{1} + C_{2})} (V_{1} - V_{2}) \Rightarrow V_{1}' = V_{2}' = \frac{1}{C_{1}} [Q_{1} - \frac{C_{1} C_{2}}{(C_{1} + C_{2})} (V_{1} - V_{2})] = \frac{1}{C_{1}} [\frac{C_{1} V_{1} (C_{1} + C_{2}) - C_{1} C_{2} (V_{1} - V_{2})}{(C_{1} + C_{2})}] = \frac{1}{C_{1}} [\frac{C_{1}^{2} V_{1} - C_{1} C_{2} V_{2}}{C_{1} + C_{2}}] = \frac{(C_{1} V_{1} - C_{2} V_{2})}{(C_{1} + C_{2})} Final Potential enrergy of the spheres - U_{1}' = \frac{1}{2} C_{1} {V_{1}^{'}}^{2} = \frac{1}{2} C_{1} {[\frac{C_{1} V_{1} - C_{2} V_{2}}{C_{1} + C_{2}}]}^{2} U_{2}' = \frac{1}{2} C_{2} {V_{2}^{'}}^{2} = \frac{1}{2} C_{2} {[\frac{C_{1} V_{1} - C_{2} V_{2}}{C_{1} + C_{2}}]}^{2}$

$Heat produced = Loss in potential energy = Initial potential enrgy - Final potential enrgy = (U_{f} - U_{i}) = [(U_{1} + U_{2}) - (U_{1}' + U_{2}')] = [\frac{1}{2} (C_{1} V_{1}^{2} + C_{2} V_{2}^{2}) - \frac{1}{2} (C_{1} + C_{2}) {[\frac{C_{1} V_{1} - C_{2} V_{2}}{C_{1} + C_{2}}]}^{2}] = \frac{1}{2} [\frac{(C_{1} + C_{2}) (C_{1} V_{1}^{2} + C_{2} V_{2}^{2})}{(C_{1} + C_{2})} - {(C_{1} V_{1} - C_{2} V_{2})}^{2}] = \frac{- 1}{2} [\frac{C_{1}^{2} V_{1}^{2} + C_{2}^{2} V_{2}^{2} - 2 C_{1} V_{1} C_{2} V_{2} - C_{1}^{2} V_{1}^{2} - C_{1} C_{2} V_{2}^{2} - C_{1} C_{2} V_{1}^{2} - C_{2}^{2} V_{2}^{2}}{(C_{1} + C_{2})}] = \frac{1}{2 (C_{1} + C_{2})} [C_{1} C_{2} V_{1}^{2} + C_{1} C_{2} V_{2}^{2} + 2 C_{1} C_{2} V_{1} V_{2}] = \frac{C_{1} C_{2}}{2 (C_{1} + C_{2})} {(V_{1} + V_{2})}^{2}$

● ● ● ● ●

The electric potential at the surface of a charged solid sphere of insulator is 20V. The value of electric potential at its centre will be

A.

30V

B.

20V

C.

40V

D.

Zero

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

Explanation

For a solid charged non-conducting sphere, for an inside point at distance r from center

$V = \frac{k Q}{2 R^{3}} (3 R^{2} - r^{2}) A t c e n t e r, r = 0 V_{c e n t e r} = \frac{3 k Q}{2 R}$

$A t s u r f a c e r = R, V_{s} = \frac{k Q}{R} ∴ V_{c e n t e r} = \frac{3 V_{s}}{2} = \frac{3}{2} \times 20 = 30 V$

● ● ● ● ●

The capacitance of a parallel plate capacitor is C. If a dielectric slab of thickness equal to one-fourth of the plate separation and dielectric constant K is inserted between the plates, then new capacitance become

A.

$\frac{K C}{2 (K + 1)}$

B.

$\frac{2 K C}{K + 1}$

C.

$\frac{5 K C}{4 K + 1}$

D.

$\frac{4 K C}{3 K + 1}$

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

Explanation

$C_{1} = \frac{k ε_{0} A}{d} \times 4 C_{2} = \frac{k ε_{0} A}{3 d} \times 4 C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} C_{e q} = \frac{4 K C}{3 K + 1}$

● ● ● ● ●

The electric potential at a point in space due to charge Q is Q $\times 10^{12}$ V. The value of an electric field at that point will be

A.

$\frac{Q}{9} \times 10^{12} \frac{N}{C}$

B.

$\frac{Q}{9} \times 10^{15} \frac{N}{C}$

C.

$\frac{Q}{9} \times 10^{9} \frac{N}{C}$

D.

$Q \times 10^{15} \frac{N}{C}$

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

Explanation

$\begin{aligned} (2) \\ V = \frac{KQ}{r} \\ Q \times 10^{12} = \frac{9 \times 10^{9} Q}{r} \\ r = 9 \times 10^{- 3} \\ E = \frac{KQ}{r^{2}} = \frac{9 \times 10^{9} Q}{9 \times 9 \times 10^{- 6}} \\ E = \frac{Q \times 10^{15}}{9} N / C \end{aligned}$

● ● ● ● ●

The electric potential at a point at distance 'r' from a short dipole is proportional to

A.

$r^{2}$

B.

$r^{- 1}$

C.

$r^{- 2}$

D.

$r^{1}$

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

Explanation

$V = \frac{K P}{r^{2}}$

● ● ● ● ●

A hollow charged metal spherical shell has radius R. If the potential difference between its surface and a point at a distance 3R from the center is V, then the value of electric field intensity at a point at distance 4R from the center is

A.

$\frac{3 V}{19 R}$

B.

$\frac{V}{6 R}$

C.

$\frac{3 V}{32 R}$

D.

$\frac{3 V}{16 R}$

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

Explanation

3.

$V = \frac{Kq}{R} - \frac{Kq}{3 R} = \frac{2 Kq}{3 R} E = \frac{Kq}{16 R^{2}} = \frac{3 RV}{2 \times 16 R^{2}} E = \frac{3 V}{32 R}$

Physics MCQs for NEET — Practice Questions with Answers

Frequently Asked Questions