A large open tank has two holes on the wall. One is a square hole of side L at a depth y from the top and other is a circular hole of radius R at a depth 4y from the top when the tank is completely filled with water flowing out per second from the two holes are the same then, the value of R is
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Equating the rate of flow, we have
(2gy)−−−−√×L2=(2g×4y)−−−−−−−−√πR2
[Flow=(area)xx(velocity), velocity=2gx−−−√]
where x=height from top
⇒L2=2πR2⇒R=L2π
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