Question

A thin uniform annular disc(see the figure) of mass M has an outer radius $4$R and an inner radius $3$R. 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

The correct option is A $\frac{2GM}{7R}(4\sqrt{2}-5)$

Potential at a distance $x$ on the axis of a ring of radius $R$ and mass $M=\frac{-GM}{\sqrt{x^2+R^2}}$

Let us assume a thin ring of width $dr$ at a distance $r$ from the centre.

$\left(dV\right)$ due to the ring at $P=\frac{-GM}{\sqrt{r^2+{\left(4R\right)}^2}}$

$dM=\frac{M}{\pi [{\left(4R\right)}^2-{\left(3R\right)}^2]}\times 2\pi rdr$

$=\frac{2Mrdr}{{7R}^2}$

$V$ due to annular disc at $P={\int }_{3R}^{4R}\frac{-G\left(\frac{2Mr}{{7R}^2}\right)dr}{\sqrt{r^2+{16R}^2}}$

$=\frac{-GM}{7R^2}{\int }_{3R}^{4R}\frac{2r}{\sqrt{r^2+{16R}^2}}dr$

$=\frac{-2GM}{{7R}^2}{\left[\sqrt{r^2+{16R}^2}\right]}_{3R}^{4R}$

$=\frac{-2GM}{{7R}^2}[\sqrt{32}R-\sqrt{25}R]$

$=\frac{-2GM}{7R}(\sqrt{32}-5)$

Work done in taking a unit mass from $P$ to infinity $=0-$ [Potential $P$]

$=0-\left[\frac{-2GM}{7R}(4\sqrt{2}-5)\right]$

$=\frac{2GM}{7R}[4\sqrt{2}-5]$