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A large open tank has two holes in its wall. One is a square of side $a$ at a depth $x$ from the top and the other is a circular hole of radius $r$ at depth $4 x$ 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 ..........
$2 \pi a$
$a$
$\frac{a}{\sqrt{2 \pi}}$
$\frac{a}{\pi}$
Solution
(c)
Since quantities of water flowing out of both holes is same
$\Rightarrow$ Area of hole $\times$ velocity of efflux $=$ constant
So, $A_1 \times V_1=A_2 \times V_2$
Substituting values.
$a^2 \times \sqrt{2 g x}=\pi r^2 \times \sqrt{8 g x}$ $\left\{\begin{array}{l}A_1=\text { Area of square hole } \\ V_1=\text { Velocity of efflux from square hole }=\sqrt{2 g x} \\ A_2=\text { Area of circular hole } \\ V_2=\text { Velocity of efflux from } \\ \quad \text { circular hole }=\sqrt{2 g(4 x)}\end{array}\right.$
$\Rightarrow a^2=2 \pi r^2$
$\Rightarrow \frac{a}{\sqrt{2 \pi}}=r$
