Question

A liquid kept in a cylindrical vessel is rotated about vertical axis through the centre of circular base. The difference in the heights of the liquid at the centre of vessel and its edge is \( R \) (radius of vessel), \( \omega \) (angular velocity of rotation), and \( g \) (acceleration due to gravity)

• A. \( \frac{R^2 \omega^2}{g} \)
• B. \( \frac{R \omega}{g} \)
• C. \( \frac{R \omega}{2g} \)
• D. \( \frac{R^2 \omega^2}{2g} \)

Hint:

For rotating liquids in a cylindrical vessel, the height difference due to centrifugal force is directly proportional to the square of the radius and angular velocity, and inversely proportional to gravity.

Answer 1

The Correct Option is D

Step 1: Understanding the problem.
The difference in heights of the liquid at the centre and at the edge of a rotating liquid surface can be derived using the concept of centrifugal force acting on the liquid. The pressure difference in the liquid due to this force results in a height difference.
Step 2: Using the equation for height difference.
The height difference \( \Delta h \) can be related to the centrifugal force by the formula: \[ \Delta h = \frac{R^2 \omega^2}{2g} \] where \( R \) is the radius of the vessel, \( \omega \) is the angular velocity, and \( g \) is the acceleration due to gravity.
Step 3: Conclusion.
Thus, the correct answer is (D) \( \frac{R^2 \omega^2}{2g} \).