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) (4sqrt2 -5)`
B. `-(2GM)/(7R)(4sqrt-5)`
C. `(GM)/(4R)`
D. `(2GM)/(5R)(sqrt2-1)`

Answer

A

Explanation

(a) Let us consider a circular elemental area of radius x and
thickness dx. The area of the sheded portion`= 2pixdx.`
Let dm be the mass of the shaded portion
`: (Mass)/(area) = (M)/(pi(4R^2) - pi(3R)^2`
`=(dm)/(2pixdx)j`
`:. Dm = (2M)/(7R^2) xdx`
The gravitational potential of the mass dm at P is
`dV = (-G dm)/(sqrt(4R)^2 +x^2 = -(G)/(sqrt16R^2 +x^2) xx(2M)/(7R^2) xdx`
`=(-2GM)/(7R^2)` (xdx)/(sqrt16R^2 + x^2)` (1)`
Suppose `16R^2 + x^2 = t^2`
`rArr 2xdx = 2tdt rArr xdx = tdt`
Also for ` x = 3R, t =5R`
and for `x = 4R, t = 4sqrt2R`
On intefrating equation (1), taking the above limits,
we get
`V = - int_(5R)^(4sqrt2R) (2GM)/(7R^2) dt = (-2GM)/(7R^2) [t]_(5R)^(4sqrt2R)`
`=(-2GM)/(7R^2) [4sqrt2R -5R] rArr V = (-2GM)/(7R) (4sqrt2 -5)`
`Now (W_Poo)/(1) = V_oo - V_P = -V_P [:. V_oo = 0]`
`:. W_(Poo) = (2GM)/(7R) (4sqrt2-5)`