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Question

A closed tube filled with water is rotating about an axis as shown in figure. The pressure difference PAPC is


A
12×103 N/m2
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B
24×103 N/m2
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C
6×103 N/m2
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D
20×103 N/m2
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Solution

The correct option is B 24×103 N/m2
Pressure at any point, x distance away from the axis in a rotating fluid is given by:
P=P0+12ρω2x2
where, P0 Pressure at the point on the axis at the same level i.e at point B in this case

At point A,
PA=P0+12×1000×(2)2×22
ω=2 rad/s, ρ=1000 kg/m3, x=2 m
PA=P0+8×103 ...(1)
Similarly at point C,
PC=P0+12×1000×22×42
x=4 m
PC=P0+32×103 ...(2)
On subtracting Eq.(2) from Eq.(1)
PAPC=24×103 N/m2
The pressure difference between point A and C is negative, hence PC>PA.

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