JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Rotational Mechanics

Take $R=10\text{cm}$, each disc mass $m=600\text{g}$, rod mass $600\text{g}$ and rod length $30\text{cm}$. From the figure the axis $AB$ is perpendicular to the rod, lying $10\text{cm}$ from the centre of the left disc and $20\text{cm}$ from the centre of the right disc. Each disc lies in a plane perpendicular to the rod, so the vertical axis through its own centre is a diameter, about which its moment of inertia is $\frac{1}{4}m{R}^2$; the parallel-axis theorem then shifts this to $AB$.

Left disc: $\frac{1}{4}(600)(10^2)+600(10^2)=75000{\text{g⋅cm}}^2$.

Right disc: $\frac{1}{4}(600)(10^2)+600(20^2)=255000{\text{g⋅cm}}^2$.

The centre of the rod lies $5\text{cm}$ from the axis, so $I_{\text{rod}}=\frac{1}{12}(600)(30^2)+600(5^2)=60000{\text{g⋅cm}}^2$.

The total moment of inertia about $AB$ is


With the applied torque $\tau =43\times 10^5\text{dyne⋅cm}$,