QJEEMP_2025_12: JEE Main 2025 Physics - Class 11 JEEMP_2025_12 Solution: A tube of length 1 m is filled completely with an ideal liquid…

To find the angular velocity of the tube, we need to determine the force exerted by the liquid on the closed end during uniform horizontal rotation.1. Physical AnalysisWhen the tube is rotated, every element of the liquid undergoes circular motion. To maintain this motion, a centripetal force is required. This force is provided by the pressure gradient within the liquid.Consider a small element of liquid of length $dx$ at a distance $x$ from the axis of rotation (the fixed end).Total mass of liquid: $2M$Length of tube: $L = 1 \text{ m}$Linear mass density ($\lambda$): $\frac{2M}{L} = \frac{2M}{1} = 2M$The mass of the small element is $dm = \lambda \, dx = 2M \, dx$.2. Centripetal Force CalculationThe centripetal force $dF$ required for this small element to rotate with angular velocity $\omega$ is:The total force exerted by the liquid at the far end ($x = L$) is the cumulative effect of the centripetal forces required for all elements from $x = 0$ to $x = L$.3. Integrating to find Total ForceThe pressure/force at the outer end is the integral of these differential forces:4. Solving for Angular Velocity ($\omega$)Given that $L = 1 \text{ m}$, the equation simplifies to:5. Finding the value of $\alpha$The problem states that the angular velocity is in the form:Comparing our derived result $\omega = \sqrt{\frac{1 \cdot F}{M}}$ with the given expression:ConclusionThe value of $\alpha$ is 1.