A body of mass 'm' connected to a massless and unstretchable string goes in verticle circle of radius 'R' under gravity g. The other end of the string is fixed at the center of circle. If velocity at top of circular path is $$n\sqrt{gR}$$, where, $$n /geq$$, then ratio of kinetic energy of the body at bottom to that at top of the circle is

The body moves in a vertical circle of radius $$R$$ under gravity $$g$$. The velocity at the top is given as $$v_{\text{top}} = n \sqrt{gR}$$, where $$n \geq 1$$ to ensure the body completes the circle.

The kinetic energy at the top is:

$$KE_{\text{top}} = \frac{1}{2} m v_{\text{top}}^2 = \frac{1}{2} m (n \sqrt{gR})^2 = \frac{1}{2} m n^2 gR$$

To find the kinetic energy at the bottom, use conservation of mechanical energy. Set the reference point for potential energy at the bottom of the circle.

At the bottom, height = 0, so potential energy = 0.

At the top, height = $$2R$$, so potential energy = $$mg \cdot 2R = 2mgR$$.

By conservation of energy:

Total energy at bottom = Total energy at top

$$\frac{1}{2} m v_{\text{bottom}}^2 + 0 = \frac{1}{2} m v_{\text{top}}^2 + 2mgR$$

Simplify:

$$\frac{1}{2} v_{\text{bottom}}^2 = \frac{1}{2} v_{\text{top}}^2 + 2gR$$

Multiply both sides by 2:

$$v_{\text{bottom}}^2 = v_{\text{top}}^2 + 4gR$$

Substitute $$v_{\text{top}} = n \sqrt{gR}$$:

$$v_{\text{bottom}}^2 = (n \sqrt{gR})^2 + 4gR = n^2 gR + 4gR = gR(n^2 + 4)$$

Kinetic energy at the bottom is:

$$KE_{\text{bottom}} = \frac{1}{2} m v_{\text{bottom}}^2 = \frac{1}{2} m gR (n^2 + 4)$$

The ratio of kinetic energy at the bottom to that at the top is:

$$\frac{KE_{\text{bottom}}}{KE_{\text{top}}} = \frac{\frac{1}{2} m gR (n^2 + 4)}{\frac{1}{2} m gR n^2} = \frac{n^2 + 4}{n^2}$$

Comparing with the options:

A. $$\frac{n^{2}}{n^{2}+4}$$

B. $$\frac{n^{2}+4}{n^{2}}$$

C. $$\frac{n+4}{n}$$

D. $$\frac{n}{n+4}$$

Option B matches the result.

Thus, the ratio is $$\frac{n^{2}+4}{n^{2}}$$.