JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Work, Power & Energy

For a bob to complete a vertical circle, the minimum velocity at the bottom is:

$u=\sqrt{5g{l}_1}$

At the highest point, the velocity is:

$v_{top}=\sqrt{u^2-2g(2l_1)}=\sqrt{5g{l}_1-4g{l}_1}=\sqrt{g{l}_1}$

In an elastic collision between two equal masses where one is at rest, the velocities are exchanged.

After collision, the second bob has velocity $\sqrt{g{l}_1}$ at the highest point of the first bob's circle.

For the second bob to complete its own circle, this velocity must be the minimum velocity at the highest point of its circle:

$\sqrt{g{l}_1}=\sqrt{g{l}_2}$ (minimum velocity at top for completing circle)

But the minimum velocity at the top for completing a circle is $\sqrt{g{l}_2}$, and at the bottom it would be $\sqrt{5g{l}_2}$.

Wait, the second bob starts at the highest point of the first bob's circle with velocity $\sqrt{g{l}_1}$. For it to complete its own circle of radius $l_2$, this must be the minimum velocity at the bottom of its circle:

$\sqrt{g{l}_1}=\sqrt{5g{l}_2}$

$g{l}_1=5g{l}_2$

$\frac{l_1}{l_2}=5$

The answer is 5.