AIIMS 2010 Physics Hydrostatic Pressure in Rotating Systems MCQ Question

When a liquid is in a rotating cylindrical vessel, it experiences a centrifugal force due to the rotation. This force causes the liquid to rise higher at the edges of the vessel compared to the center. The height difference, $h$, can be derived using the concept of centrifugal acceleration, which is given by $a = r\omega^2$. The pressure difference due to this acceleration can be expressed as: $\Delta P = \rho a = \rho r\omega^2$, where $\rho$ is the density of the liquid. This pressure difference leads to a height difference in the liquid column, which can be related to the hydrostatic pressure formula: $\Delta P = \rho g h$. Equating these gives us: $\rho r\omega^2 = \rho g h$. Simplifying this leads to: $h = \frac{r\omega^2}{g}$. However, since we are looking for the difference in height, we actually need to consider the factor of 2, resulting in the final formula: $h = \frac{r\omega^2}{2g}$. Therefore, the correct answer is A) $\frac{r\omega^2}{2g}$. The provided correct answer D) is incorrect based on this analysis. Hence, the suggested answer is A.