Student Login

Home›Chapters›Gravitation PYQs›Acceleration Due to Gravity›Question 33

JEE Main 2024 (4 Apr, Shift 2) · Acceleration Due to Gravity

Acceleration Due to Gravity question from JEE Main 2024 (4 Apr, Shift 2)

Q33JEE Main 2024 · 4 APR · SHIFT 2Easy

Acceleration Due to Gravity

A $90\text{ kg}$ body placed at $2R$ distance from surface of earth experiences gravitational pull of: ( $R =$ Radius of earth, $g = 10\text{ m s}^{-2}$ )

Show answer and solution

Answer: (c)

Distance from the centre $r = R + 2R = 3R$ , so $g' = g\left(\frac{R}{3R}\right)^2 = \frac{g}{9}$ .
$F = mg' = 90 \times \frac{10}{9}$
$F = 100\ \text{N}$
So the answer is (c) 100 N.

Report a mistake

More questions on Acceleration Due to Gravity

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)
Assuming the earth to be a sphere of uniform mass density, a body weighed 300 N on the…JEE Main 2024 (6 Apr, Shift 2)
The height ' h ' from the surface of the earth at which the value of ' g ' will be…MHT CET 2024 (10 May, Shift 2)
An object of mass 100 kg falls from point A to B as shown in figure. The change in its…NEET 2024
If R is the radius of the earth and the acceleration due to gravity on the surface of…JEE Main 2024 (1 Feb, Shift 1)
A simple pendulum is placed at a place where its distance from the earth's surface is…JEE Main 2024 (30 Jan, Shift 2)
A pendulum is oscillating with frequency ' n ' on the surface of earth. If it is taken to…MHT CET 2024 (4 May, Shift 2)
The weights of an object are measured in a coal mine of depth ' h₁ ', then at sea level…MHT CET 2024 (15 May, Shift 1)