Physics MCQs for NEET — Practice Questions with Answers

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NEET 2025

Two gases A and B are filled at the same pressure in separate cylinders with movable pistons of radius $r_A$ and $r_B$, respectively. On supplying an equal amount of heat to both the systems reversibly under constant pressure, the pistons of gas A and B are displaced by 16 cm and 9 cm, respectively. If the change in their internal energy is the same, then the ratio $r_A/r_B$ is equal to:

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Explanation

Same $Q$ and same $\Delta U\Rightarrow$ same $W = P\Delta V$. With equal $P$: $\pi r_A^2(16) = \pi r_B^2(9)\Rightarrow\dfrac{r_A}{r_B} = \sqrt{\dfrac{9}{16}} = \dfrac{3}{4}$.

NEET 2025

A physical quantity $P$ is related to four observations $a$, $b$, $c$ and $d$ as follows: $P = \dfrac{a^3 b^2}{c\sqrt{d}}$. The percentage errors of measurement in $a$, $b$, $c$ and $d$ are 1%, 3%, 2%, and 4% respectively. The percentage error in the quantity $P$ is:

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Explanation

$\dfrac{\Delta P}{P} = 3(1\%) + 2(3\%) + 1(2\%) + \tfrac12(4\%) = 3+6+2+2 = 13\%$.

NEET 2025

The intensity of transmitted light when a polaroid sheet, placed between two crossed polaroids at $22.5^\circ$ from the polarization axis of one of the polaroid, is ($I_0$ is the intensity of polarised light after passing through the first polaroid):

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Explanation

$I = I_0\cos^2 22.5^\circ\cos^2 67.5^\circ = I_0\cos^2 22.5^\circ\sin^2 22.5^\circ = I_0\left(\tfrac12\sin45^\circ\right)^2 = \dfrac{I_0}{8}$.

NEET 2025

Two identical point masses P and Q, suspended from two separate massless springs of spring constants $k_1$ and $k_2$, respectively, oscillate vertically. If their maximum speeds are the same, the ratio $(A_Q/A_P)$ of the amplitude $A_Q$ of mass Q to the amplitude $A_P$ of mass P is:

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Explanation

$v_{max} = A\omega = A\sqrt{k/m}$. Equal masses and equal $v_{max}$: $A_P\sqrt{k_1} = A_Q\sqrt{k_2}\Rightarrow\dfrac{A_Q}{A_P} = \sqrt{\dfrac{k_1}{k_2}}$.

NEET 2025

A pipe open at both ends has a fundamental frequency $f$ in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to:

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Explanation

Open pipe: $f = \dfrac{v}{2L}$. Half-submerged becomes a closed pipe of length $L/2$: $f' = \dfrac{v}{4(L/2)} = \dfrac{v}{2L} = f$.

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