1. Given data

• Mass of body $= m$
• Radius of vertical circle $= R$
• Speed at the top: $v_T = n\sqrt{gR}, \quad n \ge 1$

We need the ratio: $\frac{K_B}{K_T}$ where $K_B$ and $K_T$ are kinetic energies at bottom and top respectively.

2. Kinetic energy at the top

At the top, $v_T^2 = n^2 gR$ So, $K_T = \frac12 m v_T^2 = \frac12 m n^2 gR$

3. Use conservation of mechanical energy

From top to bottom, the body descends by a height of $2R$.

Hence loss in gravitational potential energy is: $\Delta U = mg(2R) = 2mgR$

So kinetic energy increases by $2mgR$: $K_B = K_T + 2mgR$

Substitute $K_T$: $K_B = \frac12 m n^2 gR + 2mgR$

Write $2mgR$ as $\frac12 m(4gR)$: $K_B = \frac12 m gR (n^2 + 4)$

4. Find the required ratio

$\frac{K_B}{K_T} = \frac{\frac12 m gR (n^2 + 4)}{\frac12 m gR n^2}$

Cancelling common terms, $\frac{K_B}{K_T} = \frac{n^2 + 4}{n^2}$

5. Check options

• A: $\frac{n^2 + 4}{n^2}$ ✅
• B: $\frac{n + 4}{n}$
• C: $\frac{n^2}{n^2 + 4}$
• D: $\frac{n}{n + 4}$

So the correct option is A.