JEE Main 2018PhysicsGravitationHardMCQ

JEE Main 2018Gravitation Question with Solution

JEE Main 2018 (15 Apr)

Question

Take the mean distance of the moon and the sun from the earth to be 0.4×106 km and 150×106 km, respectively. Their masses are  8×1022 kg and 2×1030 kg, respectively. The radius of the earth is 6400 km. Let ΔF1 be the difference in the forces exerted by the moon at the nearest and farthest point on the earth, and ΔF2 be the difference in the forces exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to ΔF1ΔF2 is,

Choose an option

Show full solutionCorrect option: C
Correct answer
C2

Step-by-step explanation

F1=GMemr13,F2=GMeMsr22 

ΔF1= (-)2GMemr13 Δr1 , ΔF2=(-)2GMeMsr22 Δr2

ΔF1ΔF2= mΔr1r13 r23MsΔr2=mMsr23r13Δr1Δr2

Using Δr1=Δr2=2Rearth

m=8×1022 kg

Ms= 2×1030 kg

r1=0.4×106 km

r2=150×106 km

We get ΔF1ΔF2=2.

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Gravitation chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

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.