JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Fluid Mechanics

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A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of vessel is 5 cm and its rotational speed is 2 rotations per second, then the difference in the heights between the centre and the sides, in cm, will be :

(A)2.0✓ Correct
(B)0.1
(C)0.4
(D)1.2

Answer: (A)

Solution

When the vessel spins, the free surface rises at the rim and dips at the centre, the difference being

y = \frac{ω ^{2} r ^{2}}{2 g}

. Here

r = 0.05

m and the speed is

2

rotations per second, so

ω = 2 \times 2 π = 4 π

rad/s. Then

y = \frac{( 4 π ) ^{2} ( 0.05 ) ^{2}}{2 \times 10} = \frac{16 π ^{2} \times 0.0025}{20}

; taking

π^{2} \approx 10

gives

y = \frac{16 \times 10 \times 0.0025}{20} = 0.02

m, that is

2.0

cm.

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Fluid Mechanics

Solution

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