A cylindrical tank having large diameter is filled with water to a height H . A hole of cross-sectional area 5 ~cm ^2 in the tank allows water to drain out. If the water drains out at the rate of 2 10⁻³ ~m ^3 ~s ⁻¹ , then the value of H is (acceleration due to gravity =10 ~ms ⁻² )

Step-by-step solution

The given situation is shown below Velocity of efflux through a hole below depth H of fluid, v= 2 g H Volume flow rate of fluid from hole of area aligned & a=V=a v & V=a 2 g H aligned Here V=2 10⁻³ ~m ^3 / s a=5 ~cm ^2=5 10⁻⁴ ~m ^2 So from eq. (i), we get aligned 2 10⁻³ & =5 10⁻⁴ 2 10 H H & = 8 10 ~m =80 ~cm aligned