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# A large open tank has two holes in the wall. One is a square hole of side L at a depth γ from the top and the other is a circular hole of radius R at a depth 4 γ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, R is equal to (a) L / √2π(b) 2πL(c) L(d) L / 2π

Navjyot Kalra
6 years ago
(a) Equating the rate of flow, we have √ (2gy) x L2 = √(2g x 4 y) πR2
[Flow = (area) x (velocity), velocity = √2gx]
Where x = height from top
⇒ L2 = 2 πR2 ⇒ R = L / √2π
9 months ago
Dear student,

Equating the rate of flow, we have √ (2gy) x L2 = √(2g x 4 y) πR2
[Flow = (area) x (velocity), velocity = √2gx]
Where x = height from top
⇒ L2 = 2 πR2 ⇒ R = L / √2π
Hence option (a) is correct.

Thanks and regards,
Kushagra