1. Interpret the motion

   The bob moves in a vertical circle of radius $l$.

   Let us take the usual positions:

   • $A$ = lowest point
   • $B$ = point at angle $90^\circ$ from $A$ (side point)
   • $C$ = top point
   • $D$ = point where string just becomes slack

   Since the string becomes slack at point $D$, the tension there is zero.

2. Condition for slackness

   At any point making angle $\theta$ from the lowest point, radial equation is $T - mg\cos\theta = \frac{mv^2}{l}$ (with proper sign convention toward center).

   At the top point $C$, for the string to just become slack, $T_C = 0$ so $\frac{mv_C^2}{l} = mg$ $v_C^2 = gl$

   Hence kinetic energy at $C$ is $K_C = \frac{1}{2}m v_C^2 = \frac{1}{2}mgl$

3. Find speed at point $B$ using energy conservation

   From $B$ to $C$, the bob rises by height $l$.

   Therefore, $K_B = K_C + mgl$

   Substitute $K_C = \frac{1}{2}mgl$: $K_B = \frac{1}{2}mgl + mgl = \frac{3}{2}mgl$

4. Required ratio

   $\frac{K_B}{K_C} = \frac{\frac{3}{2}mgl}{\frac{1}{2}mgl} = 3$

5. Check options

   • A: $4$ ✗
   • B: $1$ ✗
   • C: $2$ ✗
   • D: $3$ ✓

So, the correct ratio is $3$.