Question 15

Let $h_1$ and $h_2$ be the heights of the liquid in the left and right chambers respectively.

The base area of each chamber is $A = 1 \text{ m} \times 1 \text{ m} = 1 \text{ m}^2$.

Since the total volume of liquid is conserved, $A h_1 + A h_2 = A \times 2$, so $h_1 + h_2 = 2$.

The velocity of efflux through the hole of area $a = \sqrt{10} \times 10^{-4} \text{ m}^2$ is $v = \sqrt{2g(h_1 - h_2)}$.

The rate of decrease of height in the left chamber is given by: $A \frac{dh_1}{dt} = -a \sqrt{2g(h_1 - h_2)}$

Substituting $h_2 = 2 - h_1$:

$\frac{dh_1}{dt} = -\frac{a}{A} \sqrt{2g(2h_1 - 2)} = -\frac{2a\sqrt{g}}{A} \sqrt{h_1 - 1}$

Separating variables and integrating from $t=0$ ($h_1=2$) to $t=500$ ($h_1=h$):

$\int_2^h (h_1 - 1)^{-1/2} dh_1 = -\int_0^{500} \frac{2a\sqrt{g}}{A} dt$

$[2\sqrt{h_1 - 1}]_2^h = -\frac{2a\sqrt{g}}{A} (500)$

$\sqrt{h - 1} - \sqrt{2 - 1} = -\frac{a\sqrt{g}}{A} (500)$

Using $a\sqrt{g} = \sqrt{10} \times 10^{-4} \times \sqrt{10} = 10^{-3}$ and $A = 1$:

$\sqrt{h - 1} - 1 = -10^{-3} \times 500 = -0.5$

$\sqrt{h - 1} = 0.5 \implies h - 1 = 0.25 \implies h = 1.25 \text{ m}$