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Question: Let to be the angular velocity of the earth's rotation about its axis. Assume that the acceleration ...

Let to 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 the poles. An object weighed by a spring balance gives the same reading at the equator as at a height h above the poles (h << R). The value of h is

A

B

ω2R22g\frac { \omega ^ { 2 } R ^ { 2 } } { 2 g }

C

2ω2R2g\frac { 2 \omega ^ { 2 } R ^ { 2 } } { g }

D

Answer

ω2R22g\frac { \omega ^ { 2 } R ^ { 2 } } { 2 g }

Explanation

Solution

Apparent weights at the equator = mg - mω2R

Weight at a height h above the pole = mg.

Putting mg - mω2R = mg .

or ω2R =