A small mirror of mass m is… - JEE Main 2025 Physics PYQ

JEE Main 2025 — Physics Modern Physics

2025mcqhard

A small mirror of mass $m$ is suspended by a massless thread of length $l$ . Then the small angle through which the thread will be deflected when a short pulse of laser of energy $E$ falls normal on the mirror ( $\mathrm{c}=$ speed of light in vacuum and $\mathrm{g}=$ acceleration due to gravity)

Official previous-year question

Held on 4 Apr 2025 · Verified 6 Jul 2026.

Options

A
$\theta=\frac{3 \mathrm{E}}{4 \mathrm{mc} \sqrt{\mathrm{g} l}}$
B
$\theta=\frac{\mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{g} l}}$
C
$\theta=\frac{\mathrm{E}}{2 \mathrm{mc} \sqrt{\mathrm{gl}}}$
D
$\theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{g} l}}$

Did you get this right?

Solution

JEE Main 2025 Physics, Dual Nature of Matter — solution figure 1

Force due to beam assuming complete reflection $\mathrm{F}=\frac{2 \mathrm{P}}{\mathrm{C}}=\frac{2}{\mathrm{C}} \frac{\mathrm{dE}}{\mathrm{dt}} ; \mathrm{P}$ is power

So change in momentum of mirror.

$\mathrm{m}(\mathrm{~V}-0)=\int \mathrm{Fdt}=\frac{2}{\mathrm{C}} \int \mathrm{dE}=\frac{2 \mathrm{E}}{\mathrm{C}}$

Now using work energy theorem ...(1) JEE Main 2025 Physics, Dual Nature of Matter — solution figure 2

$\begin{aligned}
& \mathrm{W}_{\mathrm{g}}=\Delta \mathrm{k} \
& -\mathrm{mg} \ell(1-\cos \theta)=0-\frac{1}{2} \mathrm{mv}^2 \
& \mathrm{~g} \ell\left(2 \sin ^2 \frac{\theta}{2}\right)=\frac{\mathrm{v}^2}{2}
\end{aligned}$

as $\theta$ is small

$\begin{aligned} & \mathrm{g} \ell 2\left(\frac{\theta}{2}\right)^2=\frac{1}{2} \frac{4 \mathrm{E}^2}{\mathrm{~m}^2 \mathrm{c}^2} \\ & \mathrm{~g} \ell \theta^2=\frac{4 \mathrm{E}^2}{\mathrm{~m}^2 \mathrm{c}^2} \\ & \theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{g} \ell}}\end{aligned}$

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Options

$\theta=\frac{3 \mathrm{E}}{4 \mathrm{mc} \sqrt{\mathrm{g} l}}$

$\theta=\frac{\mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{g} l}}$

$\theta=\frac{\mathrm{E}}{2 \mathrm{mc} \sqrt{\mathrm{gl}}}$

$\theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{g} l}}$

Solution

JEE Main 2025 Physics, Dual Nature of Matter — solution figure 1

Force due to beam assuming complete reflection $\mathrm{F}=\frac{2 \mathrm{P}}{\mathrm{C}}=\frac{2}{\mathrm{C}} \frac{\mathrm{dE}}{\mathrm{dt}} ; \mathrm{P}$ is power

So change in momentum of mirror.

$\mathrm{m}(\mathrm{~V}-0)=\int \mathrm{Fdt}=\frac{2}{\mathrm{C}} \int \mathrm{dE}=\frac{2 \mathrm{E}}{\mathrm{C}}$

Now using work energy theorem ...(1) JEE Main 2025 Physics, Dual Nature of Matter — solution figure 2

as $\theta$ is small

Options

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics