Q 11-09-131JEE MainJEE Main 2020 (2 Sep, Shift 1)Medium
A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is $5\ \text{cm}$ and the angular speed of rotation is $\omega\ \text{rad s}^{-1}$. The difference in the height, $h$ (in cm) of liquid at the centre of vessel and at the sides of the vessel will be:
Answer: (C) $\dfrac{25\omega^{2}}{2g}$
In a liquid rotating with angular speed $\omega$, the free surface is a paraboloid. Applying Bernoulli-type pressure balance in the rotating frame, the rise of the surface at radius $r$ above the centre is
$$h = \frac{\omega^{2}r^{2}}{2g}$$
With $r = 5$ cm: $h = \dfrac{25\,\omega^{2}}{2g}$ (with $g$ in cm s$^{-2}$).
Solution by Sreeraj P, M.Sc Physics