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JEE Main 20254 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Morning Shift
Subject
Physics
Chapter
Dual Nature of Matter

JEE Main 2025 · 4 April · Morning Shift · Physics · Dual Nature of Matter

A small mirror of mass m is suspended by a massless thread of...

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

Options

  1. Aθ=3E4mcgl\theta=\frac{3 \mathrm{E}}{4 \mathrm{mc} \sqrt{\mathrm{g} l}}
  2. Bθ=Emcgl\theta=\frac{\mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{g} l}}
  3. Cθ=E2mcgl\theta=\frac{\mathrm{E}}{2 \mathrm{mc} \sqrt{\mathrm{gl}}}
  4. Dθ=2Emcgl\theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{g} l}}

Correct answer

D. θ=2Emcgl\theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{g} l}}

Step-by-step solution

A small mirror of mass m is suspended by a massless thread of length l . Then the small angle through which the thread
Force due to beam assuming complete reflection F=2PC=2CdEdt;P\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.
m( V−0)=∫Fdt=2C∫dE=2EC\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)A small mirror of mass m is suspended by a massless thread of length l . Then the small angle through which the thread
<br/>Wg=Δk−mgℓ(1−cos⁡θ)=0−12mv2 gℓ(2sin⁡2θ2)=v22<br/>\begin{aligned}<br/>& \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}<br/>\end{aligned}
as θ\theta is small
gℓ2(θ2)2=124E2 m2c2 gℓθ2=4E2 m2c2θ=2Emcgℓ\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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