# If the acceleration due to gravity at the surface of the earth is g ,the work done in slowly lifting a body of mass m from the earths surface to a height R equal to the radius of the earth is.a)1/2mgrb)2mgrc)mgrd)none of these

Hello,

Here force applied is variable because acceleration due to gravity change with height.

{h = height from earth surface which is variable}

So force applied

Small work done by applied force to lift height dh

So option (a) is correct.

• 41

=work done=force*displacement

=gravitational acceleration at height equal to the radius of the earth

=g(r)^2/(r+h)^2

=g(r)^2/(2r)^2

g(r)^2/4r^2

g/4

therfore gravitational acceleration there=

g/4

therefore =gravitational accleration will be equal to g/4 therefore potential energy stored at point=work done

mgr/4=potential energy

which equals work done

therefore work done =mgr/4

therfore option none of these is correct

option mgr cannot be correct because gravitational acceleration at a height equal to the radius of the earth is g/4 and since P.E=Work done=mgr/4 so option d is correct

• -15

wrong. the correct ans is mgr/2

• -13

thank you vijay sir!

• 0

Your question is really very nice.

You should ask such type of question.

• -8

Thank you sir but it would be really nice if you would view my other ques as some of their answers are not commong according to what you solved.

• 1
What are you looking for?