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A body of mass m is placed on earth surface which is taken from earth surface to a height of h=3 R, then change in gravitational potential energy is

Answer»

`(mgR)/(4)`
`(2)/(3)mgR`
`(3)/(4)mgR`
`(3)/(4)mgR`

Solution :Gravitational potential energy on earth.s surface`=-(GM m)/(R )` where M and R are the MASS the RADIUS of the earth respectively, m is the mass of the body and G is the universal gravitational constant.
Gravitational potential energy at a height h=3R
`=-(GM m)/(R+h)=-(GM m)/(R+3R)=-(GM m)/($r)`
`:.` Change in potential energy
`=-(GM m)/(4R)-(-(GM m)/(R ))=-(GM m)/(4R )+(GM m)/(R )=(3)/(4)(GMm)/(r )`
Again, we have, `(GM m)/(R^(2))=MG`
(where g is acceleration due to GRAVITY on earth.s surface).
`:. (GM m)/(R^(2))=mgR`.
`:.` Change in potential energy `=(3)/(4)mgR`.


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