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Physics · Gravitation

JEE Main 2025 — 3 April, Morning Shift — Question 46

A particle is released from height SS above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.

  1. Option A:

    S2,3gS2\frac{S}{2}, \sqrt{\frac{3 g S}{2}}

  2. Option B:

    S4,3gS2\frac{S}{4}, \sqrt{\frac{3 g S}{2}}

    Correct
  3. Option C:

    S4,3gS2\frac{S}{4}, \frac{3 g S}{2}

  4. Option D:

    S2,3gS2\frac{S}{2}, \frac{3 g S}{2}

Answer: B

Step-by-step solution

PE=mgh\mathrm{PE}=m g h

 KE =mg(S−h) given mg(S−h)=3mghS=4hh=S4U=2g(3S)4=3gS2\begin{aligned} & \text { KE }=m g(S-h)\\ & \text { given } m g(S-h)=3 m g h \\ & S=4 h \\ & h=\frac{S}{4} \\ & U=\sqrt{\frac{2 g(3 S)}{4}}=\sqrt{\frac{3 g S}{2}} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
3 April, Morning Shift
Subject
Physics
Chapter
Gravitation
Topic
Gravitational Potential Energy and Potential

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