Question

What should be the angular velocity of rotation of Earth about its own axis so that the weight of a body at the equator reduces to $\frac{3}{5}$ or r 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)

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