JEE Physics 2025 Question with Solution

Given: A body moves in a vertical circle of radius $R$. At the top, $v_{\text{top}} = \sqrt{n g R}$, so

$$v_{\text{top}}^2 = n g R$$

Find: The ratio $\dfrac{K_{\text{bottom}}}{K_{\text{top}}}$.

Using conservation of mechanical energy between the top and bottom positions, the loss in gravitational potential energy from top to bottom is

$$mg(2R) - mg(-0) = 2mgR$$

so the kinetic energy increases by $2mgR$.

Hence,

$$\frac{1}{2} m v_{\text{bottom}}^2 = \frac{1}{2} m v_{\text{top}}^2 + 2 m g R$$

Substituting $v_{\text{top}}^2 = n g R$,

$$v_{\text{bottom}}^2 = n g R + 4 g R = (n + 4) g R$$

Now,

$$K_{\text{top}} = \frac{1}{2} m n g R$$

and

$$K_{\text{bottom}} = \frac{1}{2} m (n + 4) g R$$

Therefore,

$$\frac{K_{\text{bottom}}}{K_{\text{top}}} = \frac{\frac{1}{2} m (n + 4) g R}{\frac{1}{2} m n g R} = \frac{n + 4}{n}$$

So the correct option is C. The solution contains inconsistent statements claiming option A, but the working shown gives $\dfrac{n+4}{n}$.

Given: $v_{\text{top}} = \sqrt{n g R}$.

At the top,

$$K_{\text{top}} = \frac{1}{2} m v_{\text{top}}^2 = \frac{1}{2} m n g R$$

From top to bottom, the body descends through a height of $2R$. Therefore the decrease in potential energy is

$$\Delta U = 2 m g R$$

This appears as an increase in kinetic energy:

$$K_{\text{bottom}} = K_{\text{top}} + 2 m g R$$

So,

$$K_{\text{bottom}} = \frac{1}{2} m n g R + 2 m g R$$ $$K_{\text{bottom}} = \frac{1}{2} m g R (n + 4)$$

Now divide by $K_{\text{top}}$:

$$\frac{K_{\text{bottom}}}{K_{\text{top}}} = \frac{\frac{1}{2} m g R (n + 4)}{\frac{1}{2} m g R n} = \frac{n + 4}{n}$$

Therefore, the ratio of kinetic energies at bottom and top is $\frac{n + 4}{n}$.