1
JEE Main 2026 (Online) 5th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is $\_\_\_\_$ - ( $S$ is the surface tension of mercury).

A

$8 \pi r^2 \mathrm{~S}$

B

$16 \pi r^2 S$

C

$64 \pi r^2 S$

D

$4 \pi r^2 \mathrm{~S}$

2
JEE Main 2026 (Online) 5th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

An ideal gas at pressure $P$ and temperature $T$ is expanding such that $P T^3=$ constant. The coefficient of volume expansion of the gas is $\_\_\_\_$ .

A

$\frac{2}{T}$

B

$\frac{1}{T}$

C

$\frac{4}{T}$

D

$\frac{3}{T}$

3
JEE Main 2026 (Online) 5th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

$$ \text { Match List - I with List - II. } $$

$$
\text { List - I }
$$
$$
\text { List - II }
$$
A. $$
\sin ^2 \omega t
$$
I. Periodic with time period $T=\frac{\pi}{\omega}$ but not simple harmonic motion (SHM)
B. $$
\sin ^3(2 \omega t)
$$
II. Periodic with time period $T=\frac{2 \pi}{\omega}$ but Not SHM
C. $$
\sin (\omega t)+\cos (\pi \omega t)
$$
III. Periodic with time period $T=\frac{\pi}{\omega}$ and SHM
D. $$
\cos \omega t+\cos 2 \omega t
$$
IV. Non-periodic

Choose the correct answer from the options given below :

A

A-III, B-I, C-IV, D-II

B

A-II, B-I, C-III, D-IV

C

A-III, B-II, C-IV, D-I

D

A-II, B-I, C-IV, D-III

4
JEE Main 2026 (Online) 5th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A metal rod of length $L$ rotates about one end at origin with a uniform angular velocity $\omega$. The magnetic field radially falls off as $B(\mathrm{r})=B_{\mathrm{o}} \mathrm{e}^{-\lambda r} ; \lambda$ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :

A

$$ B_o \omega\left[\frac{1}{\lambda^2}-e^{-\lambda L}\left(\frac{1}{\lambda^2}+\frac{L}{\lambda}\right)\right] $$

B

$$ B_o \omega\left[\frac{1}{\lambda^2}+e^{-\lambda L}\left(\frac{1}{\lambda^2}+\frac{L}{\lambda}\right)\right] $$

C

$$ B_o \omega\left[\frac{4}{\lambda^2}-e^{-2 \lambda L}\left(\frac{1}{\lambda^2}+\frac{2 L}{\lambda}\right)\right] $$

D

$$ B_0 \omega\left[\frac{3}{\lambda^2}-e^{-3 \lambda L}\left(\frac{3}{\lambda^2}+\frac{L}{\lambda}\right)\right] $$

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