A particle of mass m was transferred the centre of the base of a uniform hemisphere of mass M and radius r into infinity. If the work performed in the process by the gravitational force exerted on the particle by the hemisphere is $\frac{-x}{2}\frac{GmM}{R}thenxis\mathrm{..........}$

One can imagine that the uniform hemisphere is made up of thin hemisphere layers of radii ranging from 0 to R. Let us consider such a layer (see figure) Potential at point O due to this layer is 
$d\varphi =\frac{\gamma dm}{r}=\frac{3\gamma m}{r^3}rdr,wheredm=\frac{M}{(2/3)\pi R^3}\left(\frac{4\pi r^2}{2}\right)dr$ 

(This is because all points of each hemisphere shell are equidistant from O)
Hence  $\varphi =\int d\varphi =-\frac{3\gamma M}{R^3}\underset{0}{\overset{R}{\int }}rdr=-\frac{3\gamma M}{2R}$
Hence the work done by the gravitational field force on the particle of mass m, to move it to infinity is given by the formula. $A=m\varphi \sin ce\varphi =0$  at infinity
Hence the sought work, $A_{0\to \infty }=\frac{-3\gamma mM}{2R}$ (The work done by the external agent is = A)