JEE Physics 2025 Question with Solution

Given: A bob of mass $m$ moves in a vertical circle of radius $l$. The initial speed at point $A$ satisfies $v_0^2 = 5g l$, and the string becomes slack at point $D$, so $v_D^2 = g l$.

Find: The ratio $\frac{KE_B}{KE_C}$.

Using conservation of mechanical energy between points $A$ and $B$,

$$\frac{1}{2} m v_0^2 = \frac{1}{2} m v_B^2 + mg\frac{l}{2}$$

Substituting $v_0^2 = 5 g l$,

$$\frac{1}{2} m(5 g l) = \frac{1}{2} m v_B^2 + mg\frac{l}{2}$$

Hence,

$$KE_B = \frac{1}{2} m v_B^2 = 2mgl$$

At point $C$, using energy relation with point $D$,

$$\frac{1}{2} m v_C^2 = \frac{1}{2} m v_D^2 + mg\frac{l}{2}$$

Using $v_D^2 = g l$,

$$\frac{1}{2} m v_C^2 = \frac{1}{2} m g l + mg\frac{l}{2}$$

Therefore,

$$KE_C = mgl$$

Now,

$$\frac{KE_B}{KE_C} = \frac{2mgl}{mgl} = 2$$

Therefore, the ratio of kinetic energies at $B$ and $C$ is $2:1$, so the correct option is A.

Given: The bob executes vertical circular motion with points $A, B, C, D$ as shown. The initial condition gives $v_0^2 = 5g l$. At the top point $D$, the string just becomes slack, so $v_D^2 = g l$.

Find: The ratio of kinetic energies at $B$ and $C$.

Step 1: Apply energy conservation from $A$ to $B$:

$$\frac{1}{2} m v_0^2 = \frac{1}{2} m v_B^2 + mgh_B$$

With $h_B = \frac{l}{2}$,

$$\frac{1}{2} m (5 g l) = \frac{1}{2} m v_B^2 + m g \frac{l}{2}$$

So,

$$\frac{1}{2} m v_B^2 = 2mgl$$

Thus,

$$KE_B = 2mgl$$

Step 2: Apply energy conservation from $C$ to $D$:

$$\frac{1}{2} m v_C^2 = \frac{1}{2} m v_D^2 + mgh_C$$

With $v_D^2 = g l$ and $h_C = \frac{l}{2}$,

$$\frac{1}{2} m v_C^2 = \frac{1}{2} m g l + m g \frac{l}{2}$$

Hence,

$$KE_C = mgl$$

Step 3: Take the ratio:

$$\frac{KE_B}{KE_C} = \frac{2mgl}{mgl} = 2$$

Therefore, the required ratio is $2$.