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 cm and the angular speed of rotation is ω rad s –1 . The difference in the height, h (in cm) of liquid at the centre of vessel and at the side will be :

Solution

Physics sir · 4 steps

The free surface is a parabola: balance the centripetal pressure rise against the extra height of liquid.

1. Pressure along a radius. A thin horizontal strip of liquid at radius r moves in a circle, so the pressure must rise outward to supply ρω² r per unit volume. Integrating dP/dr = ρω² r from the axis to the wall gives an extra pressure of 1/2ρω² R² at the wall.
2. Height supplies it. The surface is open to the atmosphere everywhere. So the extra pressure at the wall must come from a taller liquid column, ρ g h.

   ρ g h = 1/2ρω² R² ⇒ h = (ω² R²)/2g

3. Put in numbers. The diameter is 10 cm, so R = 5 cm and R² = 25 cm². Then h = 25ω²/2g, in cm when g is taken in cm s⁻².
4. So. The height difference is h = 25ω²/2g, which is option (c).

The idea to keep. For liquid rotating in a vessel, the rise at the wall over the centre is h = (ω² R²)/2g. Take R as the radius, not the diameter, and remember it goes as R².

Answer from the NTA final answer key. Solution by PrepWiser.