JEE Main 2018 — Gravitation Question with Solution
From: JEE Main 2018 (Online) 15th April Morning Slot
Question
Take the mean distance of the moon and the sun from the earth to be km and km respectively. Their masses are kg and kg respectively. The radius of the earth is km. Let be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to is :
Choose an option
Show full solutionCorrect option: A
Correct answer
A
Step-by-step explanation
As gravitational force of attraction,
F =
Force of attraction berween earth and moon
F1 =
Force of attraction between earth and sun,
F2 =
F1 = r1
F2 = r2
=
=
= diameter of the earth = 2 Rearth
Given
m = 8 1022 kg
Ms = 2 1030 kg
r1 = 0.4 106 km
r2 = 150 106 km
= 2
F =
Force of attraction berween earth and moon
F1 =
Force of attraction between earth and sun,
F2 =
F1 = r1
F2 = r2
=
=
= diameter of the earth = 2 Rearth
Given
m = 8 1022 kg
Ms = 2 1030 kg
r1 = 0.4 106 km
r2 = 150 106 km
= 2
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This is a previous-year question from JEE Main 2018, covering the Gravitation chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.