Question

What should be the angular velocity of rotation of the earth about its own axis so that the weight of a body at the equator reduces to $3/5$ or its present value?(Take R as the radius of the earth).

• A. $\sqrt{\frac{g}{3R}}$
• B. $\sqrt{\frac{2g}{3R}}$
• C. $\sqrt{\frac{2g}{5R}}$
• D. $\sqrt{\frac{2g}{7R}}$

Answer 1

Answer: (C)

$\sqrt{\frac{2g}{5R}}$

Weight at equator $W=mg$ where m is mass and g is acceleration due to gravity.

Given: $W^1=\frac{3}{5}mg$

We know, $m{g}^1=mg-mR{\omega }^2\cos \lambda$

where $\lambda =0$ equator

$\Rightarrow mR{\omega }^2=\frac{2}{5}mg\Rightarrow \omega =\sqrt{\frac{2g}{5R}}$