Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert

A particle is kept on the surface of a uniform sphere of mass 1000 kg and radius 1 m. The work done per unit mass against the gravitational force between them is

G=6.67×10-11 N m2 Kg-2
PhysicsGravitationAP EAMCETAP EAMCET 2021 (19 Aug Shift 1)
Options:
  • A 3.35×1010J kg1
  • B 3.35×1010J kg1
  • C 6.67×108J kg1
  • D 3.35×108J kg1
Solution:
2141 Upvotes Verified Answer
The correct answer is: 6.67×108J kg1

Work done by a conservative force is the change in potential energy, therefore, W=Uf-Ui

Now, the gravitational potential energy per unit mass or gravitational potential of the particle is, V=-GMr

So, the initial potential energy, Vi=-6.67×10-1110001=6.67×10-8 J kg-1

The final potential energy to take the particle far away from the sphere is Vf=-GM=0

Work done per unit mass of the particle to take it far away from the spare against the gravitational force is, 

W=Vf-Vi=0--6.67×10-8 J kg-1W=6.67×10-8 J kg-1

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.