1. | \(9.8 ~\text{ms}^{-2}\) | 2. | \(4.9 ~\text{ms}^{-2}\) |
3. | \(3.92 ~\text{ms}^{-2}\) | 4. | \(19.6~\text{ms}^{-2}\) |
1. | \(\dfrac{\pi RG}{12g}\) | 2. | \(\dfrac{3\pi R}{4gG}\) |
3. | \(\dfrac{3g}{4\pi RG}\) | 4. | \(\dfrac{4\pi G}{3gR}\) |
Assuming the earth to be a sphere of uniform density, its acceleration due to gravity acting on a body:
1. | increases with increasing altitude. |
2. | increases with increasing depth. |
3. | is independent of the mass of the earth. |
4. | is independent of the mass of the body. |
A body weighs \(72~\text{N}\) on the surface of the earth. What is the gravitational force on it at a height equal to half the radius of the earth?
1. \(32~\text{N}\)
2. \(30~\text{N}\)
3. \(24~\text{N}\)
4. \(48~\text{N}\)
1. | \(\dfrac R {n^2}\) | 2. | \(\dfrac {R~(n-1)} n\) |
3. | \(\dfrac {Rn} { (n-1)}\) | 4. | \(\dfrac R n\) |
A body weighs \(200\) N on the surface of the earth. How much will it weigh halfway down the centre of the earth?
1. | \(100\) N | 2. | \(150\) N |
3. | \(200\) N | 4. | \(250\) N |
A mass falls from a height \(h\) and its time of fall \(t\) is recorded in terms of time period \(T\) of a simple pendulum. On the surface of the earth, it is found that \(t=2T\). The entire setup is taken on the surface of another planet whose mass is half of that of the Earth and whose radius is the same. The same experiment is repeated and corresponding times are noted as \(t'\) and \(T'\). Then we can say:
1. \(t' = \sqrt{2}T\)
2. \(t'>2T'\)
3. \(t'<2T'\)
4. \(t' = 2T'\)