1
JEE Main 2025 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1

The fractional compression $\left( \frac{\Delta V}{V} \right)$ of water at the depth of 2.5 km below the sea level is __________ %. Given, the Bulk modulus of water = $2 \times 10^9$ N m$^{-2}$, density of water = $10^3$ kg m$^{-3}$, acceleration due to gravity $g = 10$ m s$^{-2}$.

A

1.0

B

1.25

C

1.75

D

1.5

2
JEE Main 2025 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain.

Reason (R) : Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa.

In the light of the above statements, choose the most appropriate answer from the options given below :

A

Both (A) and (R) are true but (R) is not the correct explanation of (A).

B

(A) is true but (R) is false.

C

Both (A) and (R) are true and (R) is the correct explanation of (A).

D

(A) is false but (R) is true.

3
JEE Main 2025 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Two projectiles are fired with same initial speed from same point on ground at angles of $(45^\circ - \alpha)$ and $(45^\circ + \alpha)$, respectively, with the horizontal direction. The ratio of their maximum heights attained is :
A

$ \frac{1+\sin\alpha}{1-\sin\alpha} $

B

$ \frac{1+\sin2\alpha}{1-\sin2\alpha} $

C

$ \frac{1-\tan\alpha}{1+\tan\alpha} $

D

$ \frac{1-\sin2\alpha}{1+\sin2\alpha} $

4
JEE Main 2025 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1

An electric dipole of mass $m$, charge $q$, and length $l$ is placed in a uniform electric field $\vec{E} = E_0\hat{i}$. When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be :

A

$ \frac{1}{2\pi} \sqrt{\frac{ml}{2qE_0}} $

B

$ 2\pi \sqrt{\frac{ml}{2qE_0}} $

C

$ 2\pi \sqrt{\frac{ml}{qE_0}} $

D

$\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{~m} l}{\mathrm{q} \mathrm{E}_0}}$

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