A large open tank has two holes in the wall. One is a square hole of side at a depth from the top and the other is a circular hole of radius at a depth from the top. [2000]

Question

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 $4 y$ 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 [2000]

Smart Achievers · Accepted Answer

(1)

$As we know, velocity of efflux V = \sqrt{2 g h}$

$From equation of continuity A_{1} V_{1} = A_{2} V_{2}$

$\sqrt{(2 g y)} \times L^{2} = \sqrt{(2 g \times 4 y)} π R^{2}$

$\Rightarrow L^{2} = 2 π R^{2}$

$∴$ $R = \frac{L}{\sqrt{2 π}}$

Solution

1 $\frac{L}{\sqrt{2 π}}$

2 $2 π L$

3 $L$

4 $\frac{L}{2 π}$

Ans.
(1)

$As we know, velocity of efflux V = \sqrt{2 g h}$

$From equation of continuity A_{1} V_{1} = A_{2} V_{2}$

$\sqrt{(2 g y)} \times L^{2} = \sqrt{(2 g \times 4 y)} π R^{2}$

$\Rightarrow L^{2} = 2 π R^{2}$

$∴$ $R = \frac{L}{\sqrt{2 π}}$

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Solution

1 L2π

2 2πL

3 L

4 L2π