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Physics MCQs for NEET — Practice Questions with Answers

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NEET 2025

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Consider a water tank shown in the figure. It has one wall at $x = L$ and can be taken to be very wide in the $z$ direction. When filled with a liquid of surface tension $S$ and density $\rho$, the liquid surface makes angle $\theta_0\ (\theta_0 \ll 1)$ with the $x$-axis at $x = L$. If $y(x)$ is the height of the surface, then the equation for $y(x)$ is: (take $\theta(x) = \sin\theta(x) = \tan\theta(x) = \frac{dy}{dx}$, $g$ is the acceleration due to gravity)

A. $\dfrac{d^2y}{dx^2} = \dfrac{\rho g}{S}\, x$ B. $\dfrac{d^2y}{dx^2} = \dfrac{\rho g}{S}\, y$ C. $\dfrac{d^2y}{dx^2} = \sqrt{\dfrac{\rho g}{S}}$ D. $\dfrac{dy}{dx} = \sqrt{\dfrac{\rho g}{S}}\, x$

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Explanation

Balancing the Laplace pressure of the curved surface, $S\,\dfrac{d^2y}{dx^2} = \rho g\,y$ for small slopes, i.e. $\dfrac{d^2y}{dx^2} = \dfrac{\rho g}{S}\,y$.

NEET 2025

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A microscope has an objective of focal length 2 cm, eyepiece of focal length 4 cm and the tube length of 40 cm. If the distance of distinct vision of eye is 25 cm, the magnification in the microscope is:

A. 100 B. 125 C. 150 D. 250

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Explanation

$m = \dfrac{L}{f_o}\cdot\dfrac{D}{f_e} = \dfrac{40}{2}\times\dfrac{25}{4} = 20\times 6.25 = 125$.

NEET 2025

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An electron (mass $9\times10^{-31}$ kg and charge $1.6\times10^{-19}$ C) moving with speed $c/100$ ($c$ = speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9\times10^{-4}$ T perpendicular to its direction of motion. We wish to apply a uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then ($c = 3\times10^8\ \text{ms}^{-1}$):

A. $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27\times10^4\ \text{V m}^{-1}$ B. $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27\times10^2\ \text{V m}^{-1}$ C. $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27\times10^2\ \text{V m}^{-1}$ D. $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27\times10^4\ \text{V m}^{-1}$

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Explanation

For zero deflection (velocity selector) the electric force must balance the magnetic force: $qE = qvB$ with $\vec{E}\perp\vec{B}$. $E = vB = \dfrac{c}{100}\times B = 3\times10^6\times 9\times10^{-4} = 2.7\times10^3 = 27\times10^2\ \text{V m}^{-1}$.

NEET 2025

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There are two inclined surfaces of equal length ($L$) and same angle of inclination $45^\circ$ with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on the rough surface than on the smooth surface. The coefficient of kinetic friction $(\mu_k)$ between the object and the rough surface is close to:

A. 0.25 B. 0.40 C. 0.5 D. 0.75

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Explanation

For equal $L$: $a_{smooth}t_1^2 = a_{rough}t_2^2$ with $t_2 = 2t_1\Rightarrow a_{smooth} = 4a_{rough}$. $g\sin45^\circ = 4g(\sin45^\circ-\mu\cos45^\circ)\Rightarrow \mu = \tfrac{3}{4}\tan45^\circ = 0.75$.

NEET 2025

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The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m. If $F_A$ and $F_B$ are the forces applied by the breaks on cars A and B, respectively, then the ratio $F_A/F_B$ is:

A. $\dfrac{2}{3}$ B. $\dfrac{3}{2}$ C. $\dfrac{1}{3}$ D. $\dfrac{1}{2}$

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Explanation

Work–energy: $F = \dfrac{KE}{d}$. $F_A = \dfrac{100}{1000} = 0.1$, $F_B = \dfrac{225}{1500} = 0.15$; $\dfrac{F_A}{F_B} = \dfrac{0.1}{0.15} = \dfrac{2}{3}$.

NEET 2025

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The current passing through the battery in the given circuit, is:

A. 2.0 A B. 0.5 A C. 2.5 A D. 1.5 A

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Explanation

The 6 Ω arm bridges the balanced Wheatstone network, so it carries no current. The remaining resistors reduce to an effective resistance giving $I = \dfrac{5\ \text{V}}{2.5\ \Omega} = 2.0\ \text{A}$ through the battery.

NEET 2025

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A bob of heavy mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$ as shown in figure. If the string gets slack at some point $P$ making an angle $\theta$ from the horizontal, the ratio of the speed $v$ of the bob at point $P$ to its initial speed $v_0$ is:

A. $(\sin\theta)^{1/2}$ B. $\left(\dfrac{1}{2+3\sin\theta}\right)^{1/2}$ C. $\left(\dfrac{\cos\theta}{2+3\sin\theta}\right)^{1/2}$ D. $\left(\dfrac{\sin\theta}{2+3\sin\theta}\right)^{1/2}$

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Explanation

String slack at $P$: $mg\sin\theta = \dfrac{mv^2}{l}\Rightarrow v^2 = gl\sin\theta$. Energy from bottom to $P$ (height $l(1+\sin\theta)$): $v^2 = v_0^2 - 2gl(1+\sin\theta)$. Hence $v_0^2 = gl(2+3\sin\theta)$ and $\dfrac{v}{v_0} = \left(\dfrac{\sin\theta}{2+3\sin\theta}\right)^{1/2}$.

NEET 2025

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The output (Y) of the given logic implementation is similar to the output of an/a ________ gate.

A. AND B. NAND C. OR D. NOR

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Explanation

Each NOR gate gives $(A+B)'$. Feeding both identical outputs to a NAND: $Y = \overline{(A+B)'\cdot(A+B)'} = \overline{(A+B)'} = A+B$, i.e. an OR gate.

NEET 2025

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The electric field in a plane electromagnetic wave is given by $E_z = 60\cos(5x + 1.5\times10^9\,t)\ \text{V/m}$. Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field):

A. $B_y = 2\times10^{-7}\cos(5x + 1.5\times10^9\,t)\ \text{T}$ B. $B_x = 2\times10^{-7}\cos(5x + 1.5\times10^9\,t)\ \text{T}$ C. $B_z = 60\cos(5x + 1.5\times10^9\,t)\ \text{T}$ D. $B_y = 60\sin(5x + 1.5\times10^9\,t)\ \text{T}$

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Explanation

$B_0 = \dfrac{E_0}{c} = \dfrac{60}{3\times10^8} = 2\times10^{-7}\ \text{T}$. With $\vec E$ along $z$ and propagation along $-x$, $\vec B$ is along $y$: $B_y = 2\times10^{-7}\cos(5x+1.5\times10^9 t)$.

NEET 2025

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A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take $g = 9.8\ \text{m/s}^2$):

A. 21 NS B. 7 NS C. 0 D. 84 NS

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Explanation

$v_{down} = \sqrt{2g(40)} = 28\ \text{m/s}$, $v_{up} = \sqrt{2g(10)} = 14\ \text{m/s}$. Impulse $= m(v_{up}+v_{down}) = 0.5(14+28) = 21\ \text{N·s}$.

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