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### IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)

A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is A
$${{2GM} \over {7R}}(4\sqrt 2 - 5)$$
B
$$- {{2GM} \over {7R}}(4\sqrt 2 - 5)$$
C
$${{GM} \over {4R}}$$
D
$${{2GM} \over {5R}}(\sqrt 2 - 1)$$

## Explanation

We need to find gravitational potential energy of a unit mass placed at the point P. The surface mass density of the annular disc is

$$\sigma = M/(16\pi {R^2} - 9\pi {R^2}) = M/(7\pi {R^2})$$.

Consider a small ring of radius r and thickness dr. The mass of the ring is dm = 2$$\pi$$r$$\sigma$$dr. As distance of any point of the ring from P is same, the potential at P due to the ring is

$${V_P} = - {{G(2\pi r\sigma dr)} \over {\sqrt {16{R^2} + {r^2}} }}$$

Integrate from r = 3R to r = 4R to get the potential energy of the unit mass placed at P

$${V_P} = \int\limits_{3R}^{4R} { - {{GdM} \over {\sqrt {{{(4R)}^2} + {{(r)}^2}} }} = - {{GM2\pi } \over {7\pi {R^2}}}\int\limits_{3R}^{4R} {{{rdr} \over {\sqrt {16{R^2} + {r^2}} }}} }$$

Solving, we get

$${V_P} = - {{GM2\pi } \over {7\pi {R^2}}}\left[ {\sqrt {16{R^2} + {r^2}} } \right]_{3R}^{4R} = - {{2GM} \over {7R}}\left( {4\sqrt 2 - 5} \right)$$

Work done in moving a unit mass from P to $$\infty$$ = V$$\infty$$ $$-$$VP

$$= 0 - \left( {{{ - 2GM} \over {7R}}\left( {4\sqrt 2 - 5} \right)} \right) = {{2GM} \over {7R}}\left( {4\sqrt 2 - 5} \right)$$

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