JEE Physics 2026 Question with Solution

Given: Masses of the two blocks are $M$ and $m$. The pulley consists of a thin rim of mass $M$ and two rods, each of mass $M$ and length equal to the diameter $2R$.

Find: The magnitude of acceleration of the blocks.

For an Atwood machine with a massive pulley,

$$a = \frac{(M-m)g}{M + m + \frac{I}{R^2}}$$

So we first calculate the moment of inertia of the pulley.

For the thin rim,

$$I_{\text{rim}} = MR^2$$

For each rod of length $2R$ rotating about its midpoint,

$$I_{\text{rod}} = \frac{1}{12}M(2R)^2 = \frac{1}{3}MR^2$$

Since there are two rods,

$$I = I_{\text{rim}} + 2I_{\text{rod}} = MR^2 + 2\left(\frac{1}{3}MR^2\right) = \frac{5}{3}MR^2$$

Now,

$$\frac{I}{R^2} = \frac{5}{3}M$$

Substituting in the acceleration formula,

$$$$ $$a = \frac{(M-m)g}{\frac{8}{3}M + m}$$

Therefore, the correct option is B.

Given: The pulley has rotational inertia, so it contributes an effective translational mass of $\frac{I}{R^2}$.

Find: The acceleration using the effective-mass approach.

The total effective mass resisting motion is

$$M + m + \frac{I}{R^2}$$

The pulley has

• rim contribution: $MR^2$
• two rod contributions: $2 \times \frac{1}{3}MR^2$

So,

$$I = MR^2 + \frac{2}{3}MR^2 = \frac{5}{3}MR^2$$

Hence,

$$\frac{I}{R^2} = \frac{5}{3}M$$

Thus,

$$a = \frac{(M-m)g}{M + m + \frac{5}{3}M} = \frac{(M-m)g}{\frac{8}{3}M + m}$$

This works because the pulley's rotational inertia appears as an additional inertia term in the denominator. Therefore, the correct option is B.