An open cubical tank of 2 m side is filled with water. If the tank is rotated with an acceleration such that half of the water spills out, then the acceleration is equal to

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This question was previously asked in

JKSSB JE (Civil) Official Paper (Held On: 19 Nov, 2023)

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Answer 1

Concept:

The tank is cubical, side = L
It is rotated about a vertical axis.
The water surface tilts because of centrifugal force.
When half the water spills out, the tilt height h = L. The water will go from depth 2 m at one side to zero at the other.

The tilt height for a rotating open tank: $h=\frac{a.L}g$

Where: $h$ = height difference (tilt); $L$ = side of tank

$a$ = centrifugal acceleration at edge; $g$ = gravity = 9.81 m/s²

Calculation:

L = 2 m

Condition for half spillage:

For half water spillage → the surface tilts from 0 m to 2 m:

So, h = L = 2 m

$2=\frac{a.2}{g}$

$a=g$

For an open cubical tank rotating so that half water spills out → required centrifugal acceleration = g.