Question

A vessel completely filled with water has holes 'A' and 'B' at depths 'h' and '3h' from the top respectively. Hole 'A' is a square of side 'L' and 'B' is circle of radius 'r'. The water flowing out per second from both the holes is same. Then 'L' is equal to

• A. $r^{\frac{1}{2}} \left(\pi\right)^{\frac{1}{2}} \left(3\right)^{\frac{1}{2}}$
• B. $r. \left(\pi\right)^{\frac{1}{4}} \left(3\right)^{\frac{1}{4}}$
• C. $r. \left(\pi\right)^{\frac{1}{2}} \left(3\right)^{\frac{1}{4}}$
• D. $r^{\frac{1}{2}} \left(\pi\right)^{\frac{1}{3}} \left(3\right)^{\frac{1}{2}}$

Answer 1

Answer: (C)

$r. \left(\pi\right)^{\frac{1}{2}} \left(3\right)^{\frac{1}{4}}$

Answer 2

Given, depth of hole A from the top $=h$
Depth of hole $B$ from the top $=3 h$
According to question, we can draw the following diagram

According the continuity equation,
$A_{1} V_{1}=A_{2} v_{2}$
Velocity of flux,
$v_{1}=\sqrt{2 g h}\,\,\,\,\,$ (for square hole)
$v_{2}=\sqrt{2 g(3 h)}=\sqrt{{6g} h} \,\,\,\,$ (for circular hole)
$L^{2} \sqrt{2 g h}=\pi r^{2} \sqrt{6 g h}$
$(\because$ area of square $=L^{2}$, area of circle $=\pi r^{2}$ )
On squaring both sides, we get
$2 L^{4} g h =6 \pi^{2} r^{4} g h \,\,\,\,\, 2 L^{4} =6 \pi^{2} r^{4}$
$L^{4} =3 \pi^{2} r^{4} \,\,\,\,\,\,\,L =\left(3 \pi^{2}\right)^{1 / 4} r$
$L =r \pi^{1 / 2}(3)^{1 / 4}$