Question

A thin uniform annular disc of mass $M$ , has an outer radius of $4R$ and an inner radius of $3R$ as shown in the figure. The work required to take a unit mass from point $P$ on its axis to infinity is

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

Answer 1

Answer: (A)

$\text{W}=\Delta \text{U}={\text{U}}_{\text{f}}-{\text{U}}_{\text{i}}={\text{U}}_{\infty }-{\text{U}}_{\text{P}}$
$=-U_P=-m{V}_P=-V_P($ as $m=1)$
Upon integration, potential at point P will be obtained as given below
Let dM be the mass of small ring as shown
$\text{dM}=\frac{\text{M}}{\pi {\left(4\text{R}\right)}^2-\pi {\left(3\text{R}\right)}^2}\left(2\pi \text{r}\right)\text{dr}=\frac{2\text{Mr}\text{dr}}{7{\left(\text{R}\right)}^2}$
${\text{dV}}_{\text{P}}=-\frac{\text{G}\text{d}\text{M}}{\sqrt{16{\text{R}}^2+{\text{r}}^2}}$

$=-\frac{2\text{GM}}{7{\text{R}}^2}{\int }_{3\text{R}}^{4\text{R}}\frac{\text{r}}{\sqrt{16{\text{R}}^2+{\text{r}}^2}}\text{dr}$
$=-\frac{2GM}{7R}(4\sqrt{2}-5)$
$\therefore \text{W}=+\frac{2\text{GM}}{7\text{R}}\left(4\sqrt{2}-5\right)$