Height of a water tower, using Pascal's Law [closed]

Question

This is a problem on Brilliant.org which asks, if you want to receive 400kPa of water pressure to a house that is at the foot of a water tower, how tall must the water tower be? It provides you with the formula for Pascal's Law:

$\Delta P = \rho q\Delta h$

The result to the problem is 40.8m which you can figure out with intuition that was explained earlier, but the calculation provided in the solution was that if you substitute the values so that you have:

$400 kPa = (9.8 m/s^2)(1000kg/m^3)(\Delta h)$

It would equal 40.8m. This doesn't make sense to me for two reasons, first, the $s^2$ doesn't cancel out so how can the answer be in meters. Second, when working out for these values, the answer I got was 0.04081. I worked it out using $40 000kg/m^2$ but the answer is then 4.08m, and the $s^2$ still doesn't cancel. If you times this by 10 you get the right answer so I assume that I am just missing something.

For those who want more context, this is on the 3rd page in Physics of the Everyday, In the House, Water Towers. I believe to access this you need a paid subscription. What am I missing? I am not familiar with Pacal's Law and it has not yet been explained mathematically.

Answer 1

First, to see why you are having problems with the units, let's perform dimensional analysis:

$$ [P] = Pa = M L^{-1} T^{-2} $$

$$ [\rho] = M L^{-3} $$

$$ [g] = L T^{-2} $$

So in the expression of Pascal's law, rearranging for $\Delta h$ we have:

$$[\Delta h] = \frac{[P]}{[\rho][g]} = \frac{M L^3 T^2}{L T^2 M L} = L$$

As required.

Then to get the numerical value, I will first convert all units to base units.

$$ 400 kPa = 400 \times 10^3 Pa$$

Then, solving for $\Delta h$:

$$ 400 \times 10^3 Pa = 9.8 m/s^2 \times 1000 kg / m^3 \Delta h$$ $$ \Delta h = (400 / 9.8) m \approx 40.8 m$$