Question

The rotation of the earth of radius $R$ about its axis speeds up to a value such that a man at latitude angle ${60}^{\circ }$ feels weightlessness. The duration of the day in such case will be

• A. $\pi \sqrt{\frac{R}{g}}$
• B. $\frac{\pi }{2}\sqrt{\frac{R}{g}}$
• C. $\frac{\pi }{3}\sqrt{\frac{R}{g}}$
• D. $\pi \sqrt{\frac{g}{R}}$

Answer 1

The correct option is A $\pi \sqrt{\frac{R}{g}}$

The acceleration due to gravity at latitude $\lambda$ is given by
$$g^{\prime }=g-{\omega }^{2}R{\cos }^2\lambda$$ Since, the man feels weightless, so $g^{\prime }=0$

$\Rightarrow 0=g-{\omega }^{2}R{\cos }^2{60}^{\circ }$

$\Rightarrow 0=g-\frac{{\omega }^{2}R}{4}$

$\Rightarrow {\omega }^2=\frac{4g}{R}$

$\Rightarrow \omega =2\sqrt{\frac{g}{R}}$

Time period of rotation of earth, $T=\frac{2\pi }{\omega }$

$\Rightarrow T=\frac{2\pi }{2\sqrt{\frac{g}{R}}}$

$\therefore T=\pi \sqrt{\frac{R}{g}}$

Hence, option (a) is the correct answer.

The acceleration due to gravity at altitude $\lambda$ is given by $$g^{\prime }=g-{\omega }^{2}R{\cos }^2\lambda$$ Why this question? To familiarize with the acceleration due to gravity at any point and the effect of earth's rotation on it.