Question

The angular velocity of the earth's rotation about its axis is $\omega$. An object weighed by a spring balance gives the same reading at the equator as at height $h$ above the poles, the value of $h$ will be:

• A. $\frac{{\omega }^2{R}^2}{g}$
• B. $\frac{{\omega }^2{R}^2}{2g}$
• C. $\frac{{2\omega }^2{R}^2}{g}$
• D. $\frac{{2\omega }^2{R}^2}{3g}$

Answer 1

Answer: (B)

At equational it experience a pseudo force $m{w}^{2}R$ outwards due to the the rotation, which is absent in poles.

$\therefore w_{pole}=w_{equator}$

$m{g}_h=mg-m{w}^{2}R$

$mg(1-\frac{2h}{R})=mg-m{w}^{2}R$

On rearranging,

$\frac{2hg}{R}=w^{2}R$

$\therefore h=\frac{w^2{R}^2}{2g}$