A large open tank has two holes in wall. One is a square hole of side L at a depth y from top and ot

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9-1.Fluid Mechanics

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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 $y$ from the top and the other is a circular hole of radius $R$ at a depth $4y$ 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

$\frac{L}{{2\sqrt \pi }}$

B

$2\pi L$

C

$L$

D

$\frac{L}{{\sqrt {2\pi } }}$

Solution

By principle if continuity: $A_{1} v_{1}=A_{2} v_{2}$

As per quesiton: $A_{1}=L^{2} ; \quad v_{1}=\sqrt{2 g y}$ and

$A_{2}=\pi R^{2}, \quad v_{2}=\sqrt{2 g 4 y}$

$\mathrm{So}$

$L^{2} \sqrt{2 g y}=\pi R^{2} \sqrt{2 g 4 y}$

$\Rightarrow L^{2}=2 \pi R^{2}$

$\Rightarrow R=\frac{L}{\sqrt{2 \pi}}$

Standard 11

Physics

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