Question

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

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

Answer 1

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

We know that the work required to take a unit mass from $\text{P}$ to infinity is $-V_P$. i.e., $$W=-V_P$$ ​ Where, $V_P$ is the gravitational potential at $\text{P}$ due to the disc.

To calculate $V_P$ let us divide the disc into small elements, each of thickness $dr$ as shown in the figure. Consider one such element at a distance $r$ from the center of the disc.


Mass of the element,

$dm=\frac{M\left(2\pi rdr\right)}{\pi (4R{)}^2-\pi (3R{)}^2}=\frac{2Mrdr}{7R^2}$

So, the potential at point $\text{P}$ is,

$V_P=-{\int }_{3R}^{4R}\frac{Gdm}{\sqrt{(4R{)}^2+r^2}}$

Substituting value of $dm$,

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

Put $r^2+16R^2=x^2$, we get $2rdr=2xdx$ $\Rightarrow rdr=xdx$

When $r=3R;x^2=9R^2+16R^2;\Rightarrow x=5R$

When $r=4R;x^2=16R^2+16R^2;\Rightarrow x=4\sqrt{2}R$

$\Rightarrow V_P=-\frac{2GM}{7R^2}{\int }_{5R}^{4\sqrt{2}R}\frac{xdx}{\sqrt{x^2}}$

$\Rightarrow V_P=-\frac{2GM}{7R^2}{\int }_{5R}^{4\sqrt{2}R}dx$

$\Rightarrow V_P=-\frac{2GM}{7R}(4\sqrt{2}-5)$

From the equation $\left(1\right)$,

$W=\frac{2GM}{7R}(4\sqrt{2}-5)$

Hence, option (a) is the correct answer.

Why this question: To make the students understand the application of integration to calculete the potential of continuous mass distribution.