A large open tank has two small holes in its vertical wall as shown in figure. One is a square hole of side 'L' at a depth ' 4 y' from the top and the other is a circular hole of radius ' R ' at a depth ' 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 :

Question

A large open tank has two small holes in its vertical wall as shown in figure. One is a square hole of side 'L' at a depth ' 4 y' from the top and the other is a circular hole of radius ' R ' at a depth ' 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 : (A) (L/√2 π) (B) 2 π L (C) √(2/π) L (D) (L/2 π)

Tardigrade Admin · Accepted Answer

u 1=√2 gy water from rate =π R 2 √2 gy = A 1 u 1 u 2=√8 gy water from rate = L 2 √2 gy = A 2 u 2 given A 1 u 1= A 2 u 2 R = L √(2/π)

$\frac{L}{2 π}$

$2 π L$

$\frac{2}{π} L$

$\frac{L}{2 π}$

$u_{1} = 2 g y$
water from rate $= π R^{2} 2 g y = A_{1} u_{1}$
$u_{2} = 8 g y$
water from rate $= L^{2} 2 g y = A_{2} u_{2}$
given $A_{1} u_{1} = A_{2} u_{2}$
$R = L \frac{2}{π}$

2π​L​

2πL

π2​​L

2πL​

u1​=2gy​ water from rate =πR22gy​=A1​u1​ u2​=8gy​ water from rate =L22gy​=A2​u2​ given A1​u1​=A2​u2​ R=Lπ2​​

$\frac{L}{2 π}$

$2 π L$

$\frac{2}{π} L$

$\frac{L}{2 π}$

$u_{1} = 2 g y$
water from rate $= π R^{2} 2 g y = A_{1} u_{1}$
$u_{2} = 8 g y$
water from rate $= L^{2} 2 g y = A_{2} u_{2}$
given $A_{1} u_{1} = A_{2} u_{2}$
$R = L \frac{2}{π}$