Question

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

• A. $\frac{2GM}{7R}(4\sqrt{2}-5)$
• B. $-\frac{2GM}{7R}(4\sqrt{2}-5)$
• C. $\frac{GM}{4R}$
• D. $\frac{2GM}{5R}(\sqrt{2}-1)$

Answer 1

Answer: (A)

$W=\Delta U=U_f-U_i=U_{\infty }-U_P=-U_P=-m{V}_P$
$=-V_P(asm=1)$
Potential at point P will be obtained by integration as given below.
Let dM be the mass of small ring as shown
$dM=\frac{M}{\pi (4R{)}^2-\pi (3R{)}^2}(2\pi r)dr=\frac{2Mrdr}{7R^2}$
$d{V}_P=-\frac{GdM}{\sqrt{16R^2+r^2}}=-\frac{2GM}{7R^2}\underset{3R}{\overset{4R}{\int }}\frac{r}{\sqrt{16R^2+r^2}}dr$
$=-\frac{2GM}{7R}(4\sqrt{2}-5)$
$\therefore W=+\frac{2GM}{7R}(4\sqrt{2}-5)$
$\therefore$ Correct option is (a).