The height '\(h\)' at which the weight of a body will be the same as that at the same depth '\(h\)' from the surface of the earth is (Radius of the earth is \(R\) and effect of the rotation of the earth is neglected) :
1. \( \frac{\sqrt{5} R-R}{2} \)
2. \( \frac{\sqrt{5}}{2} R-R \)
3. \( \frac{R}{2} \)
4. \( \frac{\sqrt{3} R-R}{2}\)

The value of the acceleration due to gravity is \(g_1\) at a height \(h=\frac{R}{2}\) (\(R\) = radius of the earth) from the surface of the earth. It is again equal to \(g_1\) at a depth \(d\) below the surface of the earth. The ratio \(\frac{d}{R}\) equals:
1. \( \frac{7}{9} \)
2. \( \frac{4}{9} \)
3. \( \frac{1}{3} \)
4. \( \frac{5}{9} \)

A body weighs \(49\) N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator? (use \(g=\frac{GM}{R^2}=9.8~\mathrm{ms^{-2}}\) radius of earth, \(R=6400\) km.]
1. \(49~\mathrm{N}\)
2. \(48.83~\mathrm{N}\)
3. \(49.83~\mathrm{N}\)
4. \(49.17~\mathrm{N}\)

In the reported figure of earth, the value of acceleration due to gravity is same at point \(A\) and \(C\) but it is smaller than that of its value at point B (surface of the earth). The value of \(OA:OB\) will be:

  
1. \(4:5 \)
2. \(5:4\)
3. \(2:3\)
4. \(3:2\)