JEE Main 2025 — Circular Motion Question with Solution

As the string becomes slack when the bob reaches at the point D so tension is zero at this point. And the bob just completes the circle.

We know, to complete a vertical loop of radius R, the minimum velocity required at the bottom of loop is $\sqrt{5gR}$.

So, $$V_0=\sqrt{5gl}=V_A$$ (as R = $l$)

So, using energy conservation,

$$\frac{1}{2}m{v}_A^2=\frac{1}{2}m{v}_B^2+mgh$$

$$\Rightarrow \frac{1}{2}m(5gl)=\frac{1}{2}m{v}_B^2+mg\left(\frac{l}{2}\right)$$

$$\Rightarrow K{E}_B=\frac{mgl}{2}(5-1)=2mgl$$

For point C,

$$\frac{1}{2}m{v}_A^2=\frac{1}{2}m{v}_c^2+mg\left(l+\frac{l}{2}\right)$$

$$\Rightarrow \frac{5}{2}mgl=K{E}_C+\frac{3mgl}{2}$$

$$K{E}_C=\frac{mgl}{2}(5-3)=mgl$$

So, $$\frac{K{E}_B}{K{E}_C}=\frac{2mgl}{mgl}=2$$