Question

A vessel completely filled with water has holes 'A' h'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. $r^{\frac{1}{2}}\left(\pi {)}^{\frac{1}{2}}\right(3{)}^{\frac{1}{2}}$
• B. $r.\left(\pi {)}^{\frac{1}{4}}\right(3{)}^{\frac{1}{4}}$
• C. $r.\left(\pi {)}^{\frac{1}{2}}\right(3{)}^{\frac{1}{4}}$
• D. $r^{\frac{1}{2}}\left(\pi {)}^{\frac{1}{3}}\right(3{)}^{\frac{1}{2}}$

Answer 1

Answer: (C)

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

We know that the velocity of efflux at depth $d$ is given by $v=\sqrt{2gd}$

Now, $v_A=\sqrt{2gh}$= velocity of fluid flow at hole A

$v_B=\sqrt{2g\left(3h\right)}=\sqrt{6gh}$=velocity of fluid flow at hole B

volume rate of fluid flow at hole A=$V_A=A_A{v}_A=L^2\sqrt{2gh}$

volume rate of fluid flow at hole B=$V_B=A_B{v}_B=\pi r^2\sqrt{6gh}$

Since, $V_A=V_B$

$⟹L^2\sqrt{2gh}=\pi r^2\sqrt{6gh}$

$⟹L^2=\sqrt{3}\pi r^2$

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

Hence, option (C) is correct.