Question

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on ${60}^{\circ }$ latitude becomes zero is (Radius of earth = 6400 At the poles g = 10$m{s}^{-2}$)

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Effective acceleration due to gravity due to rotation of earth $g^{\prime }=g-{\omega }^{2}Rco{s}^2\lambda$

$\Rightarrow 0=g-w^{2}Rco{s}^2{60}^{\circ }\Rightarrow w=\sqrt{\frac{4g}{R}}=2\sqrt{\frac{g}{R}}=\frac{2}{800}\frac{rad}{sec}[As{g}^{\prime }=0and\lambda ={60}^{\circ }]\Rightarrow \omega =\frac{1}{400}=2.5\times {10}^{-3}\frac{rad}{sec}$