Answer to Question #110391 in Physics for César Rafael Sorgato Santos

Question

A vertical cylinder tank of straight section area 'A1' is open to the air at its top and contains water up to depth 'h0'. A hole of area 'A2' is drilled in the bottom (A1> A2).
(A) Show in detail how to get to the formula below with the data provided.

h (t) = ((√h0) - (A2 / A1). (√g / 2) .t) ²

(B) How long does it take for the tank to empty after the hole is opened? (The answer must be based on A1, A2, h0 and g)

Answer 1

a)

$$dV=A_2vdt=A_2\sqrt{2gh}dt$$

$$dV=A_1dh$$

$$A_2\sqrt{2gh}dt=A_1dh$$

$$\sqrt{h_0}-\sqrt{h}=\frac{A_2}{A_1}\sqrt{\frac{g}{2}}t$$

$$h=\left(\sqrt{h_0}-\frac{A_2}{A_1}\sqrt{\frac{g}{2}}t\right)^2$$

b)

$$h=0\to \sqrt{h_0}=\frac{A_2}{A_1}\sqrt{\frac{g}{2}}t$$

$$t=\sqrt{\frac{2h_0}{g}}\frac{A_1}{A_2}$$