Water leaks from an open tank via a hole of area 2 mm^2 at the...

Let $h$ be the height of water in the tank at any time $t$. The velocity of efflux from the hole at the bottom is given by Torricelli's law:

$v = \sqrt{2gh}$

The rate of decrease of the volume of water in the tank is equal to the rate of flow of water through the hole. Using the equation of continuity:

$-A \dfrac{dh}{dt} = a v = a \sqrt{2gh}$

Rearranging the variables to integrate:

$dt = -\dfrac{A}{a \sqrt{2g}} h^{-1/2} dh$

To find the time $T$ required for half of the water to leak out, we integrate from $h = H$ to $h = H/2$:

$\int_{0}^{T} dt = -\dfrac{A}{a \sqrt{2g}} \int_H^{H/2} h^{-1/2} dh$

$T = \dfrac{A}{a \sqrt{2g}} \left[ 2h^{1/2} \right]_{H/2}^{H}$

$T = \dfrac{A}{a} \sqrt{\dfrac{2}{g}} \left( \sqrt{H} - \sqrt{\dfrac{H}{2}} \right)<br><br>Given values are:<br>$A = 0.4 $\text{ m}^2$
$a = 2$\text{ mm}^2 = 2 $\times$10^{-6} $\text{ m}^2$
$H = 80$\text{ cm} =$0.8$\text{ m}
$g = 10$\text{ m/s}^2$<br><br>Substituting these values into the expression for$T$:<br><br>$T = $\dfrac{0.4}{2 \times 10^{-6}} \sqrt{\dfrac{2}{10}} \left( \sqrt{0.8} - \sqrt{0.4} \right)<br><br>$T = 2 $\times$10^5 \left( \sqrt{0.16} - \sqrt{0.08} \right)

$T = 2$\times$10^5 \left( 0.4 - 0.2828 \right)$

$T = 2 \times 10^5 \times 0.1172 = 23440 \text{ s}$

Converting the time into hours:

$T = \dfrac{23440}{3600} \text{ hours} \approx 6.5 \text{ hours}$