Q 11-07-163JEE MainJEE Main 2020 (5 Sep, Shift 2)Medium
The acceleration due to gravity on the earth's surface at the poles is $g$ and the angular velocity of the earth about the axis passing through the poles is $\omega$. An object is weighed at the equator and at a height $h$ above the poles using a spring balance. If the weights are found to be the same, then $h$ is: ($h \ll R$, where $R$ is the radius of the earth)
Answer: (A) $\dfrac{R^2\omega^2}{2g}$
At the equator, rotation reduces the effective gravity: $g_e = g - \omega^2R$.
At height $h$ above the pole (no rotational effect): $g_h = g\left(1 - \dfrac{2h}{R}\right)$.
Setting $g_e = g_h$:
$$\omega^2 R = \frac{2gh}{R} \Rightarrow h = \frac{R^2\omega^2}{2g}$$
Solution by Sreeraj P, M.Sc Physics