A circular cylinder of 20 cm diameter and 120 cm along is open at

SOLUTION

Concept:

Height of parabola (y)

\(y = \frac{{{\omega ^2}{R^2}}}{{2g}}\)

Given:

\(R = \frac{{20}}{2} = 10\;cm = 0.1\;cm\)

Let ‘ω’ be the maximum angular speed at which no water spills out.

Now,

Rise of liquid level at ends = 120 – 90

Rise of liquid level at ends = 30 cm

Thus, Fall of liquid level at centre = 30 cm

Now,

Y = 30 + 30 = 60 cm = 0.6 m

\(y = 0.6 = \frac{{{\omega ^2}\left( {{{0.1}^2}} \right)}}{{2\left( {9.81} \right)}}\)

ω = 34.3 rad/s

\(\omega = \frac{{2\pi N}}{{60}}\)

∴ N = 327.63 rpm