Student Login

Home›Chapters›Gravitation PYQs›Acceleration Due to Gravity›Question 26

JEE Main 2024 (27 Jan, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from JEE Main 2024 (27 Jan, Shift 1)

Q26JEE Main 2024 · 27 JAN · SHIFT 1Easy

Acceleration Due to Gravity

The acceleration due to gravity on the surface of earth is $g$ . If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be:

Show answer and solution

Answer: (d)

$g = \frac{GM}{R^2}$ ; with $M$ fixed, $g \propto \frac{1}{R^2}$ .
Diameter half $\Rightarrow R' = \frac{R}{2} \Rightarrow g' = 4g$
So the answer is (d) $4g$ .

Report a mistake

More questions on Acceleration Due to Gravity

If ' R ' is the radius of earth & ' g ' is acceleration due to gravity on earth's…MHT CET 2024 (4 May, Shift 1)
A boy weighs 72 N on the surface of earth. The gravitational force on a body due to earth…MHT CET 2024 (16 May, Shift 2)
A simple pendulum has a periodic time ' T₁ ' when it is on the surface of earth of radius…MHT CET 2024 (16 May, Shift 1)
The height ' h ' above the earth's surface at which the value of acceleration due to…MHT CET 2024 (4 May, Shift 1)
The height at which the weight of the body becomes (1/16)th of its weight on the surface…MHT CET 2024 (3 May, Shift 2)
The height above the earth's surface at which the acceleration due to gravity becomes…MHT CET 2024 (2 May, Shift 1)
A 90 kg body placed at 2R distance from surface of earth experiences gravitational pull…JEE Main 2024 (4 Apr, Shift 2)
The depth 'd' at which the value of acceleration due to gravity becomes 1/n-1 times the…MHT CET 2024 (11 May, Shift 1)