A thin uniform disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass for point P on its axis to infinity is

Options

A. `(2GM)/(7R) (4 sqrt(2) - 5)`
B. `-(2GM)/(7R) (4sqrt(2) - 5)`
C. `(GM)/(4R)`
D. `(2GM)/(5R) (sqrt(2) -1)`

Answer

A

Explanation

Consider an element of the disc of radius `x` and thichness `dx`. Mass of the element is

`dM = (M xx 2pix dx)/([pi(4R)^(2) -pi(3R)^(2)]) = (2M)/(7R^(2) xdx`
Gravitational potential at `P` due to element of the
disc `dV = - G((2M)/(7R^(2))x dx) xx (1)/(sqrt(x^(2) + (4R)^(2)))`
Workdone in taking a unit mass from unit `P` to infinity is
`W = - V = +int_(3R)^(4R) G(2M)/(7R^(2)) x dx (1)/(sqrt(x^(2) + (4R)^(2))`
`= (2GM)/(7R^(2))int_(3R)^(4R) (x dx)/([x^(2) + (4R)^(2)]^(1//2))`
Let `sqrt(x^(2) + (4R)^(2))= z`
Differenting it, we have
`(1)/(2)[x^(2) + (4R)^(2)]^(-1//2) xx 2 x dx = dz`
or `(x dx)/([x^(2) + (4R)^(2)]^(2)) = dz`
When, `x = 3R`
`z = sqrt((3R)^(2) + (4R)^(2) = 5R)`
When, `x = 4R`
`z = sqrt((4R)^(2) + (4R)^(2)) = 4sqrt(2)R`
`:. W = (2GM)/(7R^(2))int_(5R)^(4sqrt(2)R) dz = (2GM)/(7R^(2)) [z]_(5R)^(4sqrt(2)R)`
`= (2GM)/(7R^(2)) [4sqrt(2)R - 5R] = (2GM)/(7R) [4sqrt(2) - 5]`