Question

A thin uniform annular disc (see figure) of mass $M$ has outer radius $4R$ and inner radius $3R$. If the work required to take a unit mass from point P on its axis to infinity is $\frac{2GM}{nR}(4\sqrt{2}-5)$, then $n$ = ($n$ is an integer)

Answer 1

Surface mass density $\sigma =\frac{M}{\pi (4R{)}^2-\pi (3R{)}^2}=\frac{M}{7\pi R^2}$

Consider an elemental ring of radius $dr$ at a distance $r$ from the centre of the disc.


Then, mass of the ring

$dm=\sigma 2\pi rdr$

Gravitational potential due to the ring

$U_p=-{\int }_{3R}^{4R}\frac{Gdm\left(1\right)}{r^{\prime }}$
where $r^{\prime }$ is the distance from the ring to the point $P$

$=-{\int }_{3R}^{4R}\frac{Gdm\left(1\right)}{{(16R^2+r^2)}^{1/2}}$
$=-{\int }_{3R}^{4R}\frac{2\pi G\sigma rdr}{{(16R^2+r^2)}^{1/2}}$

After solving $U_p=-\frac{2GM}{7R^2}(\sqrt{r^2+16R^2}$

After applying limits,
$U_p=-\frac{2GM}{7R}(4\sqrt{2}-5)$

Work done by external agent $=U_{\infty }-U_P$
$=\frac{2GM(4\sqrt{2}-5)}{7R}$