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)

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

$m{g}^{\prime }=mg-mR{\omega }^{2}co{s}^2\varphi$
Now,
$\frac{3}{5}mg=mg-mR{\omega }^2$
or $mR{\omega }^2=mg-\frac{3}{5}mg$
or $mR{\omega }^2=\frac{2}{5}mg$
or $\omega =\sqrt{\frac{2g}{5R}}$