Question

A 4.0mm-diameter hold is 1.0m below the surface of a 2.0m diameter tank of water. How fast does water squirt out of the hole? (Show Derivation)

Answer 1

let P1 is the pressure at the top of the tank and h1 is the height of tank.

let P2 is the pressure at the hole and h2 is the height of the hole from bottom of the tank.

let rho is the dencity of water.

let velocity of water at the top of the tank, v1 = 0

let v2 = v, is the velocity of the water through the hole.

use Bernoulli's equation,

P1 + (1/2)*rho*v1^2 + rho*g*h1 = P2 + (1/2)*rho*v2^2 + rho*g*h2

PA + 0 + rho*g*h1 = PA + (1/2)*rho*v^2 + rho*g*h2 (here PA is the atmospheric pressure)

(1/2)*rho*v^2 = rho*g*(h1 - h2)

v^2 = 2*g*(h1 - h2)

v = sqrt(2*g*(h1 - h2))

= sqrt(2*9.8*(2 - 1))

= 4.43 m/s <<<<<<<<<<--------------Answer