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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
Options:
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Verified Answer
The correct answer is:
$\frac{2 G M}{7 R}(4 \sqrt{2}-5)$
$\frac{2 G M}{7 R}(4 \sqrt{2}-5)$
$$
\text { } \begin{aligned}
W=\Delta U & =U_f-U_i=U_{\infty}-U_p \\
& =-U_P \\
& =-m V_P \\
& =-V_P \quad(\text { as } m=1)
\end{aligned}
$$
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).
\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).
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