NEET Practice Questions (MCQs) with Answers & Solutions

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NEET 2025

To an ac power supply of 220 V at 50 Hz, a resistor of $20\ \Omega$, a capacitor of reactance $25\ \Omega$ and an inductor of reactance $45\ \Omega$ are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively:

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Explanation

$Z = \sqrt{R^2+(X_L-X_C)^2} = \sqrt{20^2+20^2} = 20\sqrt2 \approx 28.3\ \Omega$. $I = \dfrac{220}{28.3} \approx 7.8$ A; $\tan\phi = \dfrac{20}{20} = 1\Rightarrow\phi = 45^\circ$.

NEET 2025

The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.

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Explanation

Angular momentum conserved: $I\omega = \text{const}$, $I = \tfrac25 MR^2\propto R^2$, so $T\propto R^2$. $T_2 = 27\times2^2 = 108$ days.

NEET 2025

A model for quantized motion of an electron in a uniform magnetic field $B$ states that the flux passing through the orbit of the electron is $n(h/e)$ where $n$ is an integer, $h$ is Planck's constant and $e$ is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be ($m$ is the mass of the electron):

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Explanation

Flux $B\pi r^2 = n\tfrac{h}{e}$; for $n=1$, $Br^2 = \dfrac{h}{\pi e}$. With $r = \dfrac{mv}{eB}$, $\mu = \tfrac12 evr = \dfrac{e^2Br^2}{2m} = \dfrac{e^2}{2m}\cdot\dfrac{h}{\pi e} = \dfrac{he}{2\pi m}$.

NEET 2025

Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity $2K$ while that in the middle has thermal conductivity $K$. The left end of the combination is maintained at temperature $3T$ and the right end at $T$. The rods are thermally insulated from outside. In steady state, temperature at the left junction is $T_1$ and that at the right junction is $T_2$. The ratio $T_1/T_2$ is:

3T 2K K 2K T₁ T₂ T
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Explanation

Equal heat current: $2(3T-T_1) = (T_1-T_2) = 2(T_2-T)$. Solving gives $T_1 = \tfrac{5T}{2}$, $T_2 = \tfrac{3T}{2}$, so $\dfrac{T_1}{T_2} = \dfrac{5}{3}$.

NEET 2025

The plates of a parallel plate capacitor are separated by $d$. Two slabs of different dielectric constant $K_1$ and $K_2$ with thickness $\frac{3}{8}d$ and $\frac{d}{2}$, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If $K_1 = 1.25\,K_2$, the value of $K_1$ is:

K₁ 3d/8 K₂ d/2 d/8
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Explanation

Series of three regions (air gap $= d/8$): $\dfrac{d}{2} = \dfrac{3d}{8K_1}+\dfrac{d}{2K_2}+\dfrac{d}{8}$. With $K_2 = 0.8K_1$: $\dfrac{3}{8K_1}+\dfrac{0.625}{K_1} = \dfrac{3}{8}\Rightarrow K_1 = \dfrac{8}{3} \approx 2.66$.

NEET 2025

Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every $T$ min. A girl is driving scooty with a speed of 60 km/h in the direction X to Y notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period $T$ of the bus service and the speed (assumed constant) of the buses.

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Explanation

$\dfrac{vT}{v-60} = 30$, $\dfrac{vT}{v+60} = 10$. Dividing: $\dfrac{v+60}{v-60} = 3\Rightarrow v = 120$ km/h; then $T = 15$ min.

NEET 2025

A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of $60^\circ$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take $g = 10\ \text{m/s}^2$):

60° 5 m, 20 kg
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Explanation

Rod makes $30^\circ$ with floor. Torque about the floor end: $N_w\,l\sin30^\circ = mg\,\tfrac{l}{2}\cos30^\circ\Rightarrow N_w = \dfrac{mg}{2}\cot30^\circ$... giving wall normal $= 100\sqrt3$ N. Friction $f = N_w = 100\sqrt{3}$ N.

NEET 2025

In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency $\omega(t)$ and average amplitude $A(t)$ of the system change with time $t$. Which one of the following options schematically depicts these changes correctly?

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Explanation

As sand leaks, mass $m$ decreases, so $\omega = \sqrt{k/m}$ increases. The adiabatic invariant $E/\omega$ is constant and $E = \tfrac12 kA^2$, giving $A\propto\sqrt{\omega}$, so $A$ also increases.

NEET 2025

A balloon is made of a material of surface tension $S$ and its inflation outlet (from where gas is filled in it) has small area $A$. It is filled with a gas of density $\rho$ and takes a spherical shape of radius $R$. When the gas is allowed to flow freely out of it, its radius $r$ changes from $R$ to 0 (zero) in time $T$. If the speed $v(r)$ of gas coming out of the balloon depends on $r$ as $r^a$ and $T\propto S^\alpha A^\beta \rho^\gamma R^\delta$ then:

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Explanation

Excess pressure $\Delta P = \tfrac{4S}{r}$; Bernoulli gives $v = \sqrt{8S/(\rho r)}\propto r^{-1/2}$ so $a=-\tfrac12$. From $Av = -4\pi r^2\tfrac{dr}{dt}$, integrating $r$ from $R$ to 0 gives $T\propto R^{7/2}A^{-1}S^{-1/2}\rho^{1/2}$, i.e. $\alpha=-\tfrac12,\beta=-1,\gamma=\tfrac12,\delta=\tfrac72$.

NEET 2025

Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero of V.S. is at $x = 0.1$ cm when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is M = 5 cm and the number of coinciding vernier division is 8, the measured diameter after zero error correction, is:

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Explanation

LC $= 1\,\text{MSD} - 1\,\text{VSD} = 0.1 - 0.09 = 0.01$ cm. Observed $= 5 + 8(0.01) = 5.08$ cm. Positive zero error $= +0.10$ cm. Corrected $= 5.08 - 0.10 = 4.98$ cm.

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