NEET Practice Questions (MCQs) with Answers & Solutions

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NEET 2025

A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:

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Explanation

The displacement current produces a magnetic field that is non-zero everywhere and is maximum on the cylindrical surface joining the plate edges (at $r = R$).

NEET 2025

An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster's angle. Then—

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Explanation

$\tan\theta_B = 1.73 = \sqrt3\Rightarrow\theta_B = 60^\circ$. At Brewster's angle the reflected light is completely (plane) polarized and the angle of reflection equals $\theta_B = 60^\circ$.

NEET 2025

Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is $q$ and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (radii of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:

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Explanation

After touching A: $q_A = q/2$. Touching B (now $q/2 + q = 3q/2$ shared): $q_B = 3q/4$. New $F\propto\left(\tfrac{q}{2}\right)\left(\tfrac{3q}{4}\right) = \tfrac{3}{8}q^2$, i.e. $\dfrac{3F}{8}$.

NEET 2025

A container has two chambers of volumes $V_1 = 2$ litres and $V_2 = 3$ litres separated by a partition made of a thermal insulator. The chambers contain $n_1 = 5$ and $n_2 = 4$ moles of ideal gas at pressures $p_1 = 1$ atm and $p_2 = 2$ atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of:

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Explanation

Internal energy conserved $\Rightarrow P_f = \dfrac{p_1V_1 + p_2V_2}{V_1+V_2} = \dfrac{1(2)+2(3)}{5} = \dfrac{8}{5} = 1.6$ atm.

NEET 2025

A particle of mass $m$ is moving around the origin with a constant force $F$ pulling it towards the origin. If Bohr model is used to describe its motion, the radius $r$ of the $n^{th}$ orbit and the particle's speed $v$ in the orbit depend on $n$ as:

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Explanation

$F = \dfrac{mv^2}{r}$ (constant) and $mvr = \dfrac{nh}{2\pi}$. Eliminating: $v^3\propto n\Rightarrow v\propto n^{1/3}$ and $r\propto \dfrac{n}{v}\propto n^{2/3}$.

NEET 2025

The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?

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Explanation

Kepler: $T\propto r^{3/2}$. $T_{Mer} = 687\times\left(\tfrac14\right)^{3/2} = \dfrac{687}{8} \approx 86 \approx 88$ days.

NEET 2025

A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:

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Explanation

$W' = W\left(\dfrac{R}{R+R/3}\right)^2 = 48\left(\dfrac{3}{4}\right)^2 = 48\times\dfrac{9}{16} = 27$ N.

NEET 2025

A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:

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Explanation

Each piece $= R/8$. Four in parallel $= \dfrac{R/8}{4} = \dfrac{R}{32}$. Two such sets in series $= \dfrac{R}{32}+\dfrac{R}{32} = \dfrac{R}{16}$.

NEET 2025

De-Broglie wavelength of an electron orbiting in the $n = 2$ state of hydrogen atom is close to (Given Bohr radius $= 0.052$ nm):

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Explanation

$r_2 = n^2a_0 = 4(0.052) = 0.208$ nm. $2\pi r = n\lambda\Rightarrow\lambda = \dfrac{2\pi r_2}{2} = \pi(0.208) \approx 0.67$ nm.

NEET 2025

An electric dipole with dipole moment $5\times10^{-6}$ Cm is aligned with the direction of a uniform electric field of magnitude $4\times10^5$ N/C. The dipole is then rotated through an angle of $60^\circ$ with respect to the electric field. The change in the potential energy of the dipole is:

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Explanation

$\Delta U = pE(1-\cos60^\circ) = (5\times10^{-6})(4\times10^5)(1-0.5) = 2\times0.5 = 1.0$ J.

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