Q 11-07-197JEE MainJEE Main 2017 (8 Apr)Easy
If the Earth has no rotational motion, the weight of a person on the equator is $W$. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh $\frac34W$. The radius of the Earth is $6400$ km and $g = 10\ \text{m s}^{-2}$.
Answer: (A) $0.63\times10^{-3}\ \text{rad s}^{-1}$
At the equator the apparent weight is $mg - m\omega^2R$:
$$mg - m\omega^2R = \frac34mg \;\Rightarrow\; \omega^2 = \frac{g}{4R}$$
$$\omega = \sqrt{\frac{10}{4\times6.4\times10^6}} = \sqrt{3.9\times10^{-7}} \approx 0.63\times10^{-3}\ \text{rad s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics