Physics · Gravitation

JEE Main 2025 — 24 January, Evening Shift — Question 68

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 $\qquad$ g.

Answer: 9

Numerical answer — enter this value.

Step-by-step solution

Now since diameter is reduced to $1 / 3^{\text {rd }}$ , radius also reduces to

$1 / 3^{\text {rd }}$ , keeping mass constant

New value of acceleration due to gravity on Earth's

surface is

$g^{\prime}=\frac{G M}{\left(\frac{R_{e}}{3}\right)^{2}}=9 \frac{G M e}{R_{e}^{2}}=9 g$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Gravitation
Topic: Gravitational Field and Gravity