A particle is released from a height of \(S\) above the surface of the earth. At a certain height, its kinetic energy is three times its potential energy. The distance from the earth's surface and the speed of the particle at that instant are respectively:
1. \(\frac{S}{2},\frac{\sqrt{3gS}}{2}\) 2. \(\frac{S}{4}, \sqrt{\frac{3gS}{2}}\)
3. \(\frac{S}{4},\frac{3gS}{2}\) 4. \(\frac{S}{4},\frac{\sqrt{3gS}}{3}\)
Subtopic:  Gravitational Potential Energy |
 70%
Level 2: 60%+
NEET - 2021
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The work done to raise a mass \(m\) from the surface of the earth to a height \(h\), which is equal to the radius of the earth, is:
1. \(\dfrac{3}{2}mgR\) 2. \(mgR\)
3. \(2mgR\) 4. \(\dfrac{1}{2}mgR\)
Subtopic:  Gravitational Potential Energy |
 67%
Level 2: 60%+
NEET - 2019
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Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final - initial) of an object of mass \(m\) when taken to a height \(h\) from the surface of the earth (of radius \(R\) and mass \(M\)), is given by:

1. \(-\frac{GMm}{R+h}\) 2. \(\frac{GMmh}{R(R+h)}\)
3. \(mgh\) 4. \(\frac{GMm}{R+h}\)
Subtopic:  Gravitational Potential Energy |
 64%
Level 2: 60%+
NEET - 2019
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