Question

If a man at the equator would weigh ${\left(\frac{3}{5}\right)}^{th}$ of his weight, the angular speed of the earth is.

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

Answer 1

The correct option is B $\sqrt{\frac{2g}{5R}}$

Weight of the body at equator $=\frac{3}{5}$ of initial weight
$\therefore g^{{}^{\prime }}=\frac{3}{5}g\left(becausemassremainsconstant\right)$
$g^{{}^{\prime }}=g-{\omega }^{2}Rco{s}^2\lambda \Rightarrow \frac{3}{5}g=g-{\omega }^{2}Rco{s}^2\left(0^{\circ }\right)$
$\Rightarrow {\omega }^2=\frac{2g}{5R}\Rightarrow \omega =\sqrt{\frac{2g}{5R}}$