JEE Physics 2025 Question with Solution

Given: Tube length is $L = 1 \, \text{m}$, total mass of liquid is $2M$, and angular velocity is $\omega$.

Find: The value of $\alpha$ if the force at the other end is written in the form proportional to $M\omega^2$.

Take a small element of liquid of length $dx$ at a distance $x$ from the axis of rotation.

Linear mass density is

$$\lambda = \frac{2M}{L}$$

Since $L = 1 \, \text{m}$,

$$\lambda = 2M$$

Mass of the small element is

$$dm = \lambda \, dx = 2M \, dx$$

The required centripetal force for this element is

$$dF = \omega^2 x \, dm$$

So,

$$dF = \omega^2 x (2M \, dx) = 2M\omega^2 x \, dx$$

Integrating from $x = 0$ to $x = 1$,

$$F = \int_0^1 2M\omega^2 x \, dx$$ $$F = 2M\omega^2 \left[\frac{x^2}{2}\right]_0^1$$ $$F = M\omega^2$$

Hence, in the relation $F = \alpha M\omega^2$, we get

$$\alpha = 1$$

Therefore, the value of $\alpha$ is $1$.

Given: A liquid of total mass $2M$ fills a tube of length $1 \, \text{m}$ rotating with angular velocity $\omega$ about one end.

Find: The constant $\alpha$ associated with the end force.

Each liquid element at distance $x$ needs centripetal acceleration $\omega^2 x$. Therefore the net inward force needed on an element $dm$ is

$$dF = dm \, \omega^2 x$$

Using uniform linear mass density,

$$dm = \frac{2M}{1} \, dx = 2M \, dx$$

Hence,

$$dF = 2M\omega^2 x \, dx$$

The total force transmitted to the closed end is the sum over the whole liquid column:

$$F = \int_0^1 2M\omega^2 x \, dx = M\omega^2$$

So the coefficient is

$$\alpha = 1$$

The second provided approach concludes $\alpha = M$, but that is dimensionally inconsistent for a numerical-value answer and contradicts the direct integration above. Therefore the defensible extracted answer is $1$.