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Q. No.| 1. | \(\frac{S}{2},\frac{\sqrt{3gS}}{2}\) | 2. | \(\frac{S}{4}, \sqrt{\frac{3gS}{2}}\) |
| 3. | \(\frac{S}{4},\frac{3gS}{2}\) | 4. | \(\frac{S}{4},\frac{\sqrt{3gS}}{3}\) |
| 1. | \(\dfrac{3}{2}mgR\) | 2. | \(mgR\) |
| 3. | \(2mgR\) | 4. | \(\dfrac{1}{2}mgR\) |
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. | \(\frac{GMm}{R+h}\) |
A satellite of mass \(m\) is orbiting the earth (of radius \(R\)) at a height \(h\) from its surface. What is the total energy of the satellite in terms of \(g_0?\)
(\(g_0\) is the value of acceleration due to gravity at the earth's surface)
| 1. | \(\dfrac{mg_0R^2}{2(R+h)}\) | 2. | \(-\dfrac{mg_0R^2}{2(R+h)}\) |
| 3. | \(\dfrac{2mg_0R^2}{(R+h)}\) | 4. | \(-\dfrac{2mg_0R^2}{(R+h)}\) |
A body of mass \(m\) is taken from the Earth’s surface to the height equal to twice the radius \((R)\) of the Earth. The change in potential energy of the body will be:
| 1. | \(\frac{2}{3}mgR\) | 2. | \(3mgR\) |
| 3. | \(\frac{1}{3}mgR\) | 4. | \(2mgR\) |
If a body of mass \(m\) placed on the earth's surface is taken to a height of \(h = 3R,\) then the change in gravitational potential energy is:
1. \(\dfrac{mgR}{4}\)
2. \(\dfrac{2}{3} mgR\)
3. \(\dfrac{3}{4} mgR\)
4. \(\dfrac{mgR}{2}\)
With what velocity should a particle be projected so that its height becomes equal to the radius of the earth?
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