Question

Two rings having masses $M$ and $2M$, respectively having same radius are placed coaxially as shown in figure.


If the mass distribution on both the rings is non-uniform, then gravitational potential at point $\text{P}$ is

• A. $-\frac{GM}{R}[\frac{1}{\sqrt{2}}+\frac{1}{2}]$
• B. $-\frac{GM}{R}[\frac{1}{\sqrt{2}}+\frac{2}{\sqrt{5}}]$
• C. $-\frac{GM}{R}[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{5}}]$
• D. Zero

Answer 1

The correct option is B $-\frac{GM}{R}[\frac{1}{\sqrt{2}}+\frac{2}{\sqrt{5}}]$

As all the points on the periphery of either ring are at same distance from point $\text{P}$, the potential at point $\text{P}$ due to whole ring can be calculated as
$$V=-\frac{GM}{\sqrt{R^2+x^2}}$$ Where $x$ is axial distance from the centre of the ring.

This expression is independent of the fact whether the distribution of mass is uniform or non-uniform.

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

$V=-\frac{GM}{\sqrt{R^2+R^2}}-\frac{G\times 2M}{\sqrt{R^2+(3R-R{)}^2}}$

$\Rightarrow V=-\frac{GM}{R\sqrt{2}}-\frac{G\times 2M}{R\sqrt{5}}$

$\Rightarrow V=-\frac{GM}{R}[\frac{1}{\sqrt{2}}+\frac{2}{\sqrt{5}}]$

Hence, option (a) is the correct answer.

Key Concept: The gravitational potential due to a ring on its axis at distance $r$ is given by $$\frac{GM}{\sqrt{R^2+r^2}}$$ Why this question: Eventhough the ring has non uniform mass distribution, the potential expression remains the same. This is because potential is scalar quantity and all points on the ring are equidistant from the point on the axis.