Acceleration due to gravity on the surface of earth is ' g '. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is \underline\hspace2cm g.

Gravitation PYQ — JEE Main 2025 · 24 Jan · Shift 2 · Q74 with Solution

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Correct answer: 9

On the surface of a planet,

g = \frac{GM}{R^2}

where:

The diameter is reduced to one third of the original value.

If original diameter is $D$ , then new diameter is

D' = \frac{D}{3}

Since radius is half of diameter, the new radius also becomes one third:

R' = \frac{R}{3}

So the new acceleration due to gravity is

g' = \frac{GM}{(R')^2} = \frac{GM}{\left(\frac{R}{3}\right)^2}

g' = \frac{GM}{\frac{R^2}{9}} = 9\frac{GM}{R^2}

But

\frac{GM}{R^2} = g

So,

g' = 9g

The blank is filled by:

\boxed{9}