A bob of mass m, suspended by a string of length $l_{1},$ is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspended by a string of length $l_{2},$ which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio $l_{1}/l_{2}$ is

Step 1: Find the speed of the first bob at its highest point.

For a bob suspended by a string of length $l_1$ to complete a vertical circle, its minimum speed at the highest point occurs when the tension in the string is zero.

$$v_{\text{top}} = \sqrt{g l_1}$$

Step 2: Determine the velocity of the second bob after the collision.

Since the two bobs have equal mass $m$ and undergo a head-on elastic collision, they exchange their velocities.

$$v_2 = v_{\text{top}}$$ $$v_2 = \sqrt{g l_1}$$

Step 3: Determine the minimum velocity required for the second bob.

The second bob is initially at rest, meaning the collision occurs at the lowest point of its circular path of radius $l_2$.

$$v_{\text{bottom, min}} = \sqrt{5 g l_2}$$

Step 4: Equate the velocities to find the ratio of the lengths.

Set the velocity acquired by the second bob equal to the minimum velocity required for it to complete a full circle.

$$v_2 = v_{\text{bottom, min}}$$ $$\sqrt{g l_1} = \sqrt{5 g l_2}$$ $$g l_1 = 5 g l_2$$ $$\frac{l_1}{l_2} = 5$$

Answer: The ratio $\dfrac{l_1}{l_2}$ is $5$.