Get Answers to all your Questions

header-bg qa

A particle of mass 10 g is kept on the surface of  a uniform sphere of mass 100 Kg  and radius  10 cm  Find the work to be done against the gravitational force between them to take the particle far away from the sphere.

(\left ( You\: might\: take\: G= 6.67\times 10^{-11}Nm^{2}/kg^{2} \right )

  • Option 1)

    6.67\times 10^{-9}J

  • Option 2)

    6.67\times 10^{-10}J

  • Option 3)

    13.34\times 10^{-10}J

  • Option 4)

    3.33\times 10^{-10}J


Answers (1)


As we learnt in 

Change of potential energy -

\Delta U=GMm\left [ \frac{1}{r_{1}}-\frac{1}{r_{2}} \right ]


\Delta U\rightarrow change of energy

r_{1},r_{2}\rightarrow distances

- wherein

if body is moved from  r_{1} to r_{2}  use this equation


 dw=\int dr=\frac{Gm_{1}m_{2}}{r^{2}} dr

\int dw=Gm_{1}m_{2}\int_{r}^{\infty }\frac{dr}{r^{2}}=Gm_{1}m_{2}\left [ \frac{1}{r} \right ]^\infty _r


W=\frac{\left ( 6.67\times10 ^{-11} \right )\left ( 100\times \right )\left ( 10\times 10^{-3} \right )}{10\times 10^{-2}}

=6.67\times 10^{-10}J

Option 1)

6.67\times 10^{-9}J

Incorrect Option

Option 2)

6.67\times 10^{-10}J

Correct option

Option 3)

13.34\times 10^{-10}J

Incorrect Option

Option 4)

3.33\times 10^{-10}J

Incorrect Option

Posted by


View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE