JEE Main 2025 Jan 22 Shift 2, Physics Q24: Centre of Mass

The tube rotates about one end with angular velocity $\omega$. Treat the liquid column as a uniform rod of mass $2M$ and length $\ell =1\text{m}$, with its centre of mass at $\ell /2$ from the rotation axis.

The force at the far (outer) end equals the centripetal force required to keep the entire liquid in circular motion. Equivalently, the inward pull on the liquid from the closed far end provides centripetal force on the whole column, whose centre of mass is at distance $\ell /2$:

$F=(2M){\omega }^2\frac{\ell }{2}=M{\omega }^2\ell$.

With $\ell =1\text{m}$: $F=M{\omega }^2\Rightarrow \omega =\sqrt{\frac{F}{M}}$.

Comparing with $\omega =\sqrt{\frac{F}{\alpha M}}$: $\alpha =1$.