Question

The acceleration due to gravity on the earth's surface at the poles is $g$ and angular velocity of the earth about the axis passing through the pole is $\omega$. An object is weighed at the equator and at a height $h$ above the poles by using a spring balance. If the weights are found to be same, then $h$ is $:( h < < R ,$ where $R$ is the radius of the earth)

• A. $\frac{ R ^{2} \omega^{2}}{8 g }$
• B. $\frac{ R ^{2} \omega^{2}}{4 g }$
• C. $\frac{ R ^{2} \omega^{2}}{ g }$
• D. $\frac{ R ^{2} \omega^{2}}{2 g }$

Answer 1

Answer: (D)

$g_{e}=g-R \omega^{2}$
$g_{2}=g\left(1-\frac{2 h}{R}\right) g_{1}=g e$
$g_{2}=g-\frac{2 g h}{R}$
Now $R \omega^{2}=\frac{2 g h}{R}$
$h=\frac{R^{2} \omega^{2}}{2 g}$