If earth is supposed to be a sphere of radius R, if $g_{30}$ is value of acceleration due to gravity at latitude of $ 30^\circ$ and g at the equator, the value of $g - {g_{{{30}^o}}}$ is
$\frac{1}{4}{\omega ^2}R$
$\frac{3}{4}{\omega ^2}R$
${\omega ^2}R$
$\frac{1}{2}{\omega ^2}R$
Given below are two statements :
Statement$-I:$ Acceleration due to gravity is different at different places on the surface of earth.
Statement$-II:$ Acceleration due to gravity increases as we go down below the earth's surface.
In the light of the above statements, choose the correct answer from the options given below
At a height of $10 \,km$ above the surface of earth, the value of acceleration due to gravity is the same as that of a particular depth below the surface of earth. Assuming uniform mass density for the earth, the depth is ............. $km$
If a planet consists of a satellite whose mass and radius were both half that of the earth, the acceleration due to gravity at its surface would be ......... $m/{\sec ^2}$ ($ g$ on earth $= 9.8\, m/sec^2$ )
If the Earth losses its gravity, then for a body
A simple pendulum doing small oscillations at a place $\mathrm{R}$ height above earth surface has time period of $T_1=4 \mathrm{~s}$. $T_2$ would be it's time period if it is brought to a point which is at a height $2 R$ from earth surface. Choose the correct relation $[R=$ radius of Earth]: