Two identical thin rings each of radius R are coaxially placed at a distance R. If the rings have a uniform mass distribution and each has mass m1 and m2 respectively, then the work done in moving a mass m from centre of one ring to that of the other is

The correct option is Gm(m1-m2)(√2-1)/R√2.

To find the work done in moving a mass m from the center of the first ring to the center of the second ring, we use the relation between work done, potential energy, and gravitational potential.
The work done $W$ in moving a mass $m$ from point A (center of ring 1) to point B (center of ring 2) is given by:
$W=m(V_B-V_A)$
where $V_A$ is the gravitational potential at the center of the first ring (Ring 1) and $V_B$ is the gravitational potential at the center of the second ring (Ring 2).

Step 1: Calculate the potential at the center of Ring 1 ($V_A$)
Let Ring 1 have mass $m_1$ and Ring 2 have mass $m_2$. Both rings have radius $R$ and are coaxially separated by a distance $R$.
The gravitational potential $V_A$ at the center of Ring 1 is the sum of:
1. The potential due to Ring 1 itself at its own center:
$V_{11}=-\frac{G{m}_1}{R}$
2. The potential due to Ring 2 at the center of Ring 1. Since Ring 2 is at an axial distance $R$, every point on Ring 2 is at a distance of $\sqrt{R^2+R^2}=R\sqrt{2}$ from the center of Ring 1:
$V_{12}=-\frac{G{m}_2}{R\sqrt{2}}$
Thus, the total potential at the center of Ring 1 is:
$V_A=-\frac{G{m}_1}{R}-\frac{G{m}_2}{R\sqrt{2}}$

Step 2: Calculate the potential at the center of Ring 2 ($V_B$)
Similarly, the gravitational potential $V_B$ at the center of Ring 2 is due to Ring 2 itself and Ring 1:
$V_B=-\frac{G{m}_2}{R}-\frac{G{m}_1}{R\sqrt{2}}$

Step 3: Calculate the work done
Now we find the difference in potential $(V_B-V_A)$:
$V_B-V_A=\left(-\frac{G{m}_2}{R}-\frac{G{m}_1}{R\sqrt{2}}\right)-\left(-\frac{G{m}_1}{R}-\frac{G{m}_2}{R\sqrt{2}}\right)$
Rearranging the terms:
$V_B-V_A=\frac{G}{R}\left(m_1-m_2\right)+\frac{G}{R\sqrt{2}}\left(m_2-m_1\right)$
Factoring out $\frac{G(m_1-m_2)}{R}$:
$V_B-V_A=\frac{G(m_1-m_2)}{R}\left(1-\frac{1}{\sqrt{2}}\right)$
Simplifying the term inside the brackets:
$V_B-V_A=\frac{G(m_1-m_2)}{R}\left(\frac{\sqrt{2}-1)}{\sqrt{2}}\right)=\frac{G(m_1-m_2)(\sqrt{2}-1)}{R\sqrt{2}}$

Finally, multiplying by the mass $m$ gives the total work done:
$W=\frac{Gm(m_1-m_2)(\sqrt{2}-1)}{R\sqrt{2}}$