Question

Assertion: If the radius of the earth is decreased keeping its mass constant, the effective value of $g$ may increase or decrease at the pole.
Reason: Value of $g$ on the surface of earth is given by $g = \dfrac{{GM}}{{{R^2}}}$:
A) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C) Assertion is correct but Reason is incorrect
D) Both Assertion and Reason are incorrect

Answer 1

The rotational motion has an impact on different parts of the earth differently, as the earth is not a perfect sphere. The effective value of $g$ with constant mass but different radius will be affected by its motion as though the mass remains the same but the moment of inertia may change.

Formulae Used:
The object of mass $m$ at a distance $R$ from a body of mass $M$ feels a gravitational force
$F = \dfrac{{GMm}}{{{R^2}}}$
where, $G$ is the universal gravitational constant.
If the earth has a mass ${M_e}$, the radius of the earth is $R$ and the angular velocity at that part of the earth is $\omega$ hence the change of the gravitational acceleration changes as
$g' = g - R{\omega ^2}{\cos ^2}\theta$

Complete step by step solution:
Step 1:
The mass of the earth remains constant. Only the radius decreases. Hence you can understand that due to the change in radius, the angular velocity also changes.
So, in eq (2) both the $R$ and $\omega$ changes.
But, as you know at poles, $\theta = {90^ \circ }$
Hence, eq (2) becomes
$g' = g - 0 = g$
So, the assertion is false.

Step 2:
The force on any object of mass $m$on the surface of the earth can be related to the eq (1) as
$mg = \dfrac{{GMm}}{{{R^2}}} \\\ g = \dfrac{{GM}}{{{R^2}}} \\\$
But again this is not uniform throughout the surface as the earth is not a perfect sphere and there are effects of movement as well.
Hence the value of $g$ shall be greater in the pole region than the equators.

$\therefore$ Both the assertion and the reason is incorrect. So, option (D) is correct.

Note:
The $\theta$ should be taken as the latitude that is the angle with the equatorial plane. The change in radius will change the moment of inertia which will eventually change the angular velocity. Because the effect of the rotation in the pole is zero, the change is found to be zero.