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

Answer

A

Explanation

First we need to calculate gravitational potential at a point which is at a distance x from the centre of annular disc on its axis. Let us select a ring segment of radius r and radial thickness dr on the surface of disc. Mass of this ring segment can be written as follows :

Potential at the desired point due to this ring can be written as follows :
`dV = -G(dm)/(sqrt(x^(2) + r^(2)))`
`dV = -G(2Mrdr)/(7R^(2)sqrt(x^(2) + r^(2)))`
`V = -(2MG)/(7R^(2)) underset(3R)overset(4R)int(r dr)/(sqrt(x^(2) + r^(2)))`
`V = -(2MG)/(7R^(2)) underset(3R)overset(4R)int (rdr)/(sqrt(x^(2) + r^(2)))`
On integrating and substituting x = 4R, we get the following potential at point P.
`V_(P) = -(2GM)/(7R)(4sqrt(2) - 5)`
Potential difference between two points can be defined as follows :
`V_(2) - V_(1) = (W_("ext"))/(m)`
Here initial point is P and final point is infinity and potential at infinity is zero.
Hence by substituting in above equation, we get the following :
`rArr V_(oo) - V_(P) = (W_("ext"))/(m)`
`rArr 0 - V_(P) = (W_("ext"))/(m)`
`rArr 0 - [-(2GM)/(7R)(4sqrt(2) - 5)] = (W_("ext"))/(m)`
`rArr (W_("ext"))/(m) = (2GM)/(7R) (4sqrt(2) - 5)`
Above result gives the value of work per unit mass , hence option (a) is correct.