1. S/4, 3gS/2 2. S/4, √(3gS)/2 3. S/2, √(3gS)/2 4. S/4, √(3gS/2)

Answer: 4. S/4, √(3gS/2) Key idea: total mechanical energy mgS is conserved (taking the surface as zero of potential energy). At height h: U=mgh , K=mg(S-h) . Given K=3U : S-h=3h⇒ h=S/4 . The particle has fallen 3S/4 , so v²=2g·3S/4=3gS/2 , v=√(3gS/2) . Option 4.

Topic: Potential energy of spring, conservation of mechanical energy, conservative and non-conservative forces· Work, Energy and Power

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

1. \(\frac{S}{4}, \frac{3gS}{2}\)

2. \(\frac{S}{4}, \frac{\sqrt{3gS}}{2}\)

3. \(\frac{S}{2}, \frac{\sqrt{3gS}}{2}\)

4. \(\frac{S}{4}, \sqrt{\frac{3gS}{2}}\)

Doubtly Asked 30 Oct 2026 NEET UG 2021 Practice test neet Save WhatsApp Report

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Answer: 4. \(\frac{S}{4}, \sqrt{\frac{3gS}{2}}\)

Key idea: total mechanical energy \(mgS\) is conserved (taking the surface as zero of potential energy).

At height h: \(U=mgh\), \(K=mg(S-h)\). Given \(K=3U\): \(S-h=3h\Rightarrow h=S/4\).
The particle has fallen \(3S/4\), so \(v^2=2g\cdot\dfrac{3S}{4}=\dfrac{3gS}{2}\), \(v=\sqrt{\dfrac{3gS}{2}}\).

Option 4.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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

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