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MCQMediumJEE Main 2025 · 4 April, Shift 1Kinetic Energy & Work-Energy Theorem

Physics Question from JEE Main 2025 · 4 April, Shift 1

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=c= speed of light in vacuum and g=g= acceleration due to gravity).

  • A

    θ=3E4mcgl\theta=\frac{3 \mathrm{E}}{4 \mathrm{mc} \sqrt{g l}}

  • B

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

  • C

    θ=E2mcgl\theta=\frac{\mathrm{E}}{2 \mathrm{mc} \sqrt{\mathrm{gl}}}

  • D

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

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