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

• Correct answer
  A

  $\frac{2 G M}{7 R}(4 \sqrt{2}-5)$
• B

  $-\frac{2 G M}{7 R}(4 \sqrt{2}-5)$
• C

  $\frac{G M}{4 R}$
• D

  $\frac{2 G M}{5 R}(\sqrt{2}-1)$

Solution

$$ \text { } \begin{aligned} W=\Delta U & =U_f-U_i=U_{\infty}-U_p \\ & =-U_P \\ & =-m V_P \\ & =-V_P \quad(\text { as } m=1) \end{aligned} $$ Potential at point $P$ will be obtained by integration as given below
Let $d M$ be the mass of small ring as shown $$ \begin{aligned} d M & =\frac{M}{\pi(4 R)^2-\pi(3 R)^2}(2 \pi r) d r \\ & =\frac{2 M r d r}{7 R^2} \\ d V_P & =-\frac{G \cdot d M}{\sqrt{16 R^2+r^2}} \\ & =-\frac{2 G M}{7 R^2} \int_{3 R}^{4 R} \frac{r}{\sqrt{16 R^2+r^2}} \cdot d r \\ & =-\frac{2 G M}{7 R}(4 \sqrt{2}-5) \\ \therefore \quad W & =+\frac{2 G M}{7 R}(4 \sqrt{2}-5) \end{aligned} $$ $\therefore$ correct option is (a).

Hence, the correct answer is option: 'A'.