Question

 If a body of mass m is taken out from a point below the surface of earth equal to half the radius of earth, R, to a height R above the earths surface, then work done on it will be

(5/6) mgR

(6/7) mgR

(7/8) mgR

(8/9) mgR


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Answer 1

Answer:

c

Explanation:

it is smalllerso it the weighi is smalallest

Answer 2

Answer:

The work done on the body will be $\frac{7mgR}{8}$

Explanation:

Given that,

A body of mass m is taken out from a point below the surface of the earth equal to half the radius of the earth, R, to a height R above the earth's surface and we are required to find the work done in this context.

We know that the potential of a body at a distance r from the center is

$V=\frac{GM}{2R}(\frac{r^2}{R^2} -3)$

The acceleration due to gravity is $g=\frac{GM}{R^2}$ hence the relation can be written as

$V=\frac{gR}{2}(\frac{r^2}{R^2}-3)$

Now potential at a height R from the surface of the earth is

$V_{r=2R}=\frac{gR}{2} (\frac{(2R)^2}{R^2} -3)=-\frac{gR}{2}$

Similarly, The potential at depth $\frac{R}{2}$ below the surface of the earth is

$V_{r=\frac{r}{2}}=\frac{gR}{2} (\frac{(\frac{R}{2})^2}{R^2} -3)=-\frac{11gR}{8}$

The work done is the product of mass and potential difference.

Therefore,

$W=m(V_{2R}-V_{\frac{R}{2}})\\=m(-\frac{gR}{2} -(-\frac{11gR}{8}))\\=\frac{7mgR}{8}$

Therefore,

The work done on the body will be $\frac{7mgR}{8}$

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