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Question

A square hole of side a is made at a depth h below water surface and to the side of a water container another circular hole of radius r is made to the same container at a depth of 4h. It is found that volume flow rate of water through both the holes is found to be same then:

A
r=a2π
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B
a=r2π
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C
a=r2π
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D
r=a2π
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Solution

The correct option is D r=a2π
ρ1A1v1=ρ2A2v2 as volume rate is constant
velocity by toricellis therom is 2gh
a2(2gh)=πr2(2g(4h))
2πr2=a2
r=a2π

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