Are these NEET questions free to attempt?

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.

Do I need an account to practise NEET 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.

NEET NEET Questions — Free MCQ Practice

● ● ● ● ●

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.

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

● ● ● ● ●

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

NEET Practice Questions (MCQs) with Answers & Solutions

Frequently Asked Questions