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

Answer 1

Answer: (C)

$m g'=m g-m R \omega^{2} \cos ^{2} \phi$
Now, $\frac{3}{5} m g=m g-m R \omega^{2}$
or $m R \omega^{2}=m g-\frac{3}{5} m g=\frac{2}{5} m g$
or $\omega=\sqrt{\frac{2 g}{5 R}}$