JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Laws of Motion

Let the tube have length $\ell =1$ m and total liquid mass $2M$, so its linear mass density is $\lambda =\frac{2M}{\ell }$.

Consider a liquid element at distance r from the rotation axis (the fixed end), of length dr. Rotating at angular velocity $\omega$, it requires a centripetal force
$dF=(\lambda dr){\omega }^{2}r$

Integrating over the length of the tube, the total outward force exerted on the far end is
$F={\int }_0^{\ell }\lambda {\omega }^{2}rdr=\lambda {\omega }^2\frac{{\ell }^2}{2}=\frac{2M}{\ell }\cdot {\omega }^2\cdot \frac{{\ell }^2}{2}=M{\omega }^2\ell$

Solving for $\omega$:
$\omega =\sqrt{\frac{F}{M\ell }}=\sqrt{\frac{F}{M}}$ since $\ell =1$ m.

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