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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 4 y from the top. W

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 the holes are the same. Then R is equal to

Option: 1

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


Option: 2

\mathrm{2 \pi L}


Option: 3

\mathrm{L}


Option: 4

\mathrm{\frac{L}{2 \pi}}


Answers (1)

best_answer

The ratio of volume of water flowing out per second is given by

               \mathrm{ \frac{V_1}{V_2}=\frac{v_1 a_1}{v_2 a_2}=\frac{v_1(L)^2}{v_2\left(\pi R^2\right)} }                                    (1)

The velocities of water flowing out are given by

                  \mathrm{ \begin{aligned} & v_1=\sqrt{2 g y} \text { and } v_2=\sqrt{2 g(4 y)} \\\\ \therefore \quad \frac{v_1}{v_2} & =\frac{\sqrt{2 g y}}{\sqrt{8 g y}}=\frac{1}{2} \end{aligned} }                 (2)

Using (2) in (1), we have

                           \mathrm{ \frac{V_1}{V_2}=\frac{1}{2} \frac{L^2}{\pi R^2} }

Given \mathrm{\quad V_1=V_2}.

Therefore \mathrm{1=\frac{1}{2} \frac{L^2}{\pi R^2}}

or
                        \mathrm{R=\frac{L}{\sqrt{2 \pi}}}

               which is choice (a).

Posted by

Ajit Kumar Dubey

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