Question

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

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

Answer 1

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

The acceleration due to gravity at latitude $\lambda$ is given by

$g^{\prime }=g-{\omega }^{2}R{\cos }^2\lambda .....\left(1\right)$

Since the man feels weightless at $\lambda ={60}^{\circ }$,
So the weight of man at that point is zero which means $g^{\prime }=0$

So, putting the value in equation $\left(1\right)$, we get

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

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

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

And we know that time period $T$ is given by

$T=\frac{2\pi }{\omega }=2\pi \times \frac{1}{2}\sqrt{\frac{R}{g}}$

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

Hence, option $\left(a\right)$ is the correct answer.