A liquid is kept in a cylindrical vessel which is rotated about its axis. The liquid rises at the sides. If the radius of the vessel is 0.05 m and the speed of rotation is 2 rev s − 1 , the difference

Question

A liquid is kept in a cylindrical vessel which is rotated about its axis. The liquid rises at the sides. If the radius of the vessel is 0.05 m and the speed of rotation is 2 rev s − 1 , the difference in height of the liquid at the centre of vessel and its sides is-

TestHub · Accepted Answer

The pressure difference in a rotating fluid is given by $P_{2} - P_{1} = \frac{1}{2} ρ ω^{2} (r_{2}^{2} - r_{1}^{2})$ .

Also, $P_{2} - P_{1} = ρ g h$ .

Equating these, $h = \frac{ω ^{2} r ^{2}}{2 g}$ .

Given $r = 0.05$ m, $ω = 2 re v s^{- 1} = 2 \times 2 π = 4 π r a d s^{- 1}$ .

$h = \frac{( 4 π ) ^{2} ( 0.05 ) ^{2}}{2 \times 9.8} \approx 0.02$ m.

The final answer is $0.02 m$ .

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