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Physics Question on Escape Speed

The angular velocity of earth's rotation about its axis is ′ω′.'\omega'. An object weighed by a spring balance gives the same reading at the eiquator as at a height 'h' above the poles. The value o f 'h' will be

A

ω2 R2g\frac{\omega^2\,R^2}{g}

B

ω2 R22g\frac{\omega^2\,R^2}{2g}

C

2ω2 R2g\frac{2\omega^2\,R^2}{g}

D

2ω2 R23g\frac{2\omega^2\,R^2}{3g}

Answer

ω2 R22g\frac{\omega^2\,R^2}{2g}

Explanation

Solution

mg−mω2R=mg(1−2hR)mg - m \omega^2R = mg \left( 1 - \frac{2h}{R} \right) ⇒ ω2R=2ghR ⇒ h=ω2R22g\Rightarrow \, \omega^2 R =\frac{2gh}{R} \, \Rightarrow \, h =\frac{\omega^2R^2}{2g}