A tube of length 1 m is filled completely with an ideal liquid of mass 2 M , and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is F then angular velocity of the tube is \sqrt\fracF\alpha M in SI unit. The value of \alpha is \underline\hspace2cm .

Circular Motion PYQ — JEE Main 2025 · 22 Jan · Shift 2 · Q75 with Solution

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Correct answer: 1

We need to find $\omega$ in the form

\omega = \sqrt{\frac{F}{\alpha M}}

and then determine $\alpha$ .

Since the liquid is uniformly distributed in the tube,

\lambda = \frac{m}{L} = \frac{2M}{1} = 2M

So mass of a small element of length $dr$ at distance $r$ from the axis is

dm = \lambda \, dr = 2M\,dr

A small liquid element at distance $r$ rotates with angular speed $\omega$ , so required centripetal force is

dF = dm\,\omega^2 r

Substituting $dm = 2M\,dr$ ,

dF = 2M\omega^2 r\,dr

The force exerted by the liquid at the outer end must provide the centripetal force for the entire liquid column. Thus,

F = \int_0^L 2M\omega^2 r\,dr

With $L=1$ ,

F = 2M\omega^2 \int_0^1 r\,dr

F = 2M\omega^2 \left[\frac{r^2}{2}\right]_0^1

F = 2M\omega^2 \cdot \frac{1}{2} = M\omega^2

So,

\omega^2 = \frac{F}{M}

which gives

\omega = \sqrt{\frac{F}{M}}

Comparing with

\omega = \sqrt{\frac{F}{\alpha M}}

we get

\alpha = 1

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