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Question

A square side 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 h'. It is found that volume flow rate of water through both the holes is founded to be same then :

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

The correct option is A r=a√πAr the same depth 'h' water flows same rate from are direction ar pressure is same at that depth.ar flow rate (=dvdt) is equal, the holes must have equal area ⇒πr2=a2⇒r=a√π (Read)

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