1. | \(3.5\) | 2. | \(5.9\) |
3. | \(16.3\) | 4. | \(110.8\) |
The escape velocity for a rocket from the earth is \(11.2\) km/s. Its value on a planet where the acceleration due to gravity is double that on the earth and the diameter of the planet is twice that of the earth (in km/s) will be:
1. | \(11.2\) | 2. | \(5.6\) |
3. | \(22.4\) | 4. | \(53.6\) |
For the moon to cease as the earth's satellite, its orbital velocity has to be increased by a factor of:
1. | \(2\) | 2. | \(\sqrt{2}\) |
3. | \(1/\sqrt{2}\) | 4. | \(4\) |
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. \(20R\)
If the density of the earth is increased \(4\) times and its radius becomes half of what it is, our weight will be:
1. four times the present value
2. doubled
3. the same
4. Halved
The escape velocity for the Earth is taken \(v_d\). Then, the escape velocity for a planet whose radius is four times and the density is nine times that of the earth, is:
1. | \(36v_d\) | 2. | \(12v_d\) |
3. | \(6v_d\) | 4. | \(20v_d\) |
The value of \(g\) at a particular point is \(9.8~\text{m/s}^2\). Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass then value of \(g\) at the same point will now become: (assuming that the distance of the point from the centre of the earth does not shrink)
1. | \(4.9~\text{m/s}^2\) | 2. | \(3.1~\text{m/s}^2\) |
3. | \(9.8~\text{m/s}^2\) | 4. | \(19.6~\text{m/s}^2\) |
1. | decrease by \(1\%\) | 2. | increase by \(1\%\) |
3. | increase by \(2\%\) | 4. | remain unchanged |
The change in the potential energy, when a body of mass \(m\) is raised to a height \(nR\) from the Earth's surface is: (\(R\) = Radius of the Earth)
1. \(mgR\left(\frac{n}{n-1}\right)\)
2. \(nmgR\)
3. \(mgR\left(\frac{n^2}{n^2+1}\right)\)
4. \(mgR\left(\frac{n}{n+1}\right)\)
A satellite whose mass is \(m\), is revolving in a circular orbit of radius \(r\), around the earth of mass \(M\). Time of revolution of the satellite is:
1. \(T \propto \frac{r^5}{GM}\)
2. \(T \propto \sqrt{\frac{r^3}{GM}}\)
3. \(T \propto \sqrt{\frac{r}{\frac{GM^2}{3}}}\)
4. \(T \propto \sqrt{\frac{r^3}{\frac{GM^2}{4}}}\)