Question:
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 \(\omega\) rad s\(^{-1}\). The difference in the height, \(h\) (in cm) of liquid at the centre of vessel and at the side will be :
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 \(\omega\) rad s\(^{-1}\). The difference in the height, \(h\) (in cm) of liquid at the centre of vessel and at the side will be :
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In the rotating liquid the surface is a paraboloid with \(h = \frac{\omega^2 r^2}{2g}\); put \(r = 5\) cm.
Updated On: Nov 5, 2026
- \(\frac{2\omega^2}{25g}\)
- \(\frac{5\omega^2}{2g}\)
- \(\frac{25\omega^2}{2g}\)
- \(\frac{2\omega^2}{5g}\)
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The Correct Option is C
Solution and Explanation
For a liquid rotating with angular speed \(\omega\), the surface height at distance \(r\) from the axis rises as \(y = \frac{\omega^2 r^2}{2g}\) above the centre level.
The vessel has diameter 10 cm, so the side is at \(r = 5\) cm.
\(h = \frac{\omega^2 (5)^2}{2g} = \frac{25\omega^2}{2g}\) (with \(h\) in cm and \(g\) in cm s\(^{-2}\), as the options use \(r\) in cm).
Answer: (3) \(\frac{25\omega^2}{2g}\)
The vessel has diameter 10 cm, so the side is at \(r = 5\) cm.
\(h = \frac{\omega^2 (5)^2}{2g} = \frac{25\omega^2}{2g}\) (with \(h\) in cm and \(g\) in cm s\(^{-2}\), as the options use \(r\) in cm).
Answer: (3) \(\frac{25\omega^2}{2g}\)
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