Q. No.Starting from the centre of the earth, having radius \(R,\) the variation of \(g\) (acceleration due to gravity) is shown by:
| 1. |  |
2. |  |
| 3. |  |
4. |  |
| 1. | \(\dfrac R {n^2}\) | 2. | \(\dfrac {R~(n-1)} n\) |
| 3. | \(\dfrac {Rn} { (n-1)}\) | 4. | \(\dfrac R n\) |
The density of a newly discovered planet is twice that of Earth. If the acceleration due to gravity on its surface is the same as that on Earth, and the radius of Earth is \(R,\) what will be the radius of the new planet?
| 1. | \(4R\) | 2. | \(\dfrac{1}{4}R\) |
| 3. | \(\dfrac{1}{2}R\) | 4. | \(2R\) |
Radii and densities of two planets are \(R_1, R_2\) and \(\rho_1, \rho_2\) respectively. The ratio of accelerations due to gravity on their surfaces is:
1. \(\frac{\rho_1}{R_1}:\frac{\rho_2}{R_2}\)
2. \(\frac{\rho_1}{R^2_1}: \frac{\rho_2}{R^2_2}\)
3. \(\rho_1 R_1 : \rho_2R_2\)
4. \(\frac{1}{\rho_1R_1}:\frac{1}{\rho_2R_2}\)
| 1. | \(g' = 3g\) | 2. | \(g' = 9g\) |
| 3. | \(g' = \frac{g}{9}\) | 4. | \(g' = 27g\) |
The height of a point vertically above the earth’s surface, at which the acceleration due to gravity becomes \(1\%\) of its value at the surface is: (Radius of the earth = \(R\))
1. \(8R\)
2. \(9R\)
3. \(10R\)
4. \(20R\)
\(1\) kg of sugar has maximum weight:
1. at the pole.
2. at the equator.
3. at a latitude of \(45^{\circ}.\)
4. in India.
| 1. | \(32\) N | 2. | \(56\) N |
| 3. | \(72\) N | 4. | zero |
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 |
Acceleration due to gravity is:
| 1. | independent of the mass of the earth. |
| 2. | independent of the mass of the body. |
| 3. | independent of both the mass of the earth and the body. |
| 4. | dependent on both the mass of the earth and the body. |
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