JEE Physics 2026 Question with Solution

Given: Two discs each of mass $0.6 \, \text{kg}$ and radius $0.1 \, \text{m}$ are connected by a rod of length $0.3 \, \text{m}$ and mass $0.6 \, \text{kg}$. The applied torque is $43\times10^{-7} \, \text{dyne\cdot cm}$.

Find: Angular acceleration about axis $AB$.

Concept:

$$\alpha = \frac{\tau}{I}$$

where $\tau$ is the applied torque and $I$ is the moment of inertia of the system about the given axis.

Step 1: Convert given quantities into SI units

$$\text{Mass of each disc} = 600 \, \text{g} = 0.6 \, \text{kg}$$ $$\text{Mass of rod} = 600 \, \text{g} = 0.6 \, \text{kg}$$ $$r = 10 \, \text{cm} = 0.1 \, \text{m}$$

Given torque:

$$43\times10^{-7} \, \text{dyne\cdot cm}$$

Since

$$1 \, \text{dyne\cdot cm} = 10^{-7} \, \text{N\cdot m}$$ $$\tau = 43\times10^{-14} \, \text{N\cdot m}$$

Step 2: Moment of inertia of each disc about axis $AB$ The axis $AB$ passes through the midpoint of the rod. Each disc centre is at a distance of $0.2 \, \text{m}$ from the axis.

Moment of inertia of a disc about its own central axis:

$$I_{\text{disc,cm}} = \frac{1}{2}mr^2 = \frac{1}{2}(0.6)(0.1)^2 = 0.003 \, \text{kg m}^2$$

Using parallel axis theorem:

$$I_{\text{disc}} = I_{\text{disc,cm}} + md^2 = 0.003 + 0.6(0.2)^2 = 0.003 + 0.024 = 0.027$$

For two discs:

$$I_{\text{discs}} = 2\times0.027 = 0.054$$

Step 3: Moment of inertia of the rod The rod rotates about an axis through its centre and perpendicular to its length:

$$I_{\text{rod}} = \frac{1}{12}ML^2 = \frac{1}{12}(0.6)(0.3)^2 = 0.0045$$

Step 4: Total moment of inertia

$$I_{\text{total}} = 0.054 + 0.0045 = 0.0585 \, \text{kg m}^2$$

Step 5: Calculate angular acceleration

$$\alpha = \frac{43\times10^{-14}}{0.0585} \approx 11 \, \text{rad s}^{-2}$$

Therefore, the angular acceleration is $11 \, \text{rad s}^{-2}$ and the correct option is D.