Variation of G Due to the Rotation of Earth

Q. In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be $(g=10m{s}^{-2}andradiusofearthis6400km)$

1. $0radse{c}^{-1}$
2. $\frac{1}{800}radse{c}^{-1}$
3. $\frac{1}{80}radse{c}^{-1}$
4. $\frac{1}{8}radse{c}^{-1}$

Q. Let $\omega$ be the angular velocity of the earth’s rotation about its axis. Assume that the acceleration due to gravity on the earth’s surface has the same value at the equator and at the poles. An object weighed at the equator gives the same reading as a reading taken at a depth $d$ below earth’s surface at a pole $(d<<R)$.The value of $d$ is

1. $\frac{{\omega }^2{R}^2}{g}$
2. $\frac{{\omega }^2{R}^2}{2g}$
3. $\frac{2{\omega }^2{R}^2}{g}$
4. $\frac{{\omega }^2{R}^2}{6g}$

Q.

Consider a particle of mass m suspended vertically by a string at the equator. Let R and M denote the radius and the mass of the earth respectively. If $\omega$ is the angular velocity of earth's rotation about its own axis, the tension in the string is equal to

1. $\frac{GmM}{2R^2}$

2. $\frac{GmM}{R^2}$

3. $\frac{GmM}{R^2}-m{\omega }^{2}R$

4. $\frac{GmM}{R^2}+m{\omega }^{2}R$