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. \(T \propto \sqrt{\frac{r^3}{\frac{GM^2}{4}}}\)
Suppose the gravitational force varies inversely as the \(n^{th}\)
1. \(R^{\left(\frac{n+1}{2}\right)}\)
2. \(R^{\left(\frac{n-1}{2}\right)}\)
3. \(R^n\)
4. \(R^{\left(\frac{n-2}{2}\right)}\)
Time period of a satellite revolving above Earth’s surface at a height equal to \(R\) (the radius of Earth) will be:
(\(g\) is the acceleration due to gravity at Earth’s surface)
1. \(2 \pi \sqrt{\frac{2 R}{g}}\)
2. \(4 \sqrt{2} \pi \sqrt{\frac{R}{g}}\)
3. \(2 \pi \sqrt{\frac{R}{g}}\)
4. \(8 \pi \sqrt{\frac{R}{g}}\)
A rocket of mass \(M\) is launched vertically from the surface of the earth with an initial speed \(v\). Assuming the radius of the earth to be \(R\) and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is:
1. \(\frac{R}{\left(\frac{gR}{2v^2}-1\right)}\)
2. \(R\left({\frac{gR}{2v^2}-1}\right)\)
3. \(\frac{R}{\left(\frac{2gR}{v^2}-1\right)}\)
4. \(R{\left(\frac{2gR}{v^2}-1\right)}\)