Answer:
It will take 20,000 seconds to get a pressure of 200Pa at the bottom of the tank.
Explanation:
The pressure at the bottom of the tank will be
[tex]P = \dfrac{Mg}{A}[/tex]
where [tex]M[/tex] is the mass of the water, and [tex]A[/tex] is the base area of the tank.
The base area of the tank is
[tex]A = 0.5m* 2m = 1m^2[/tex],
and if we want the pressure at the bottom to be 200pa, then it must be that
[tex]200Pa = \dfrac{M(10m/s^2)}{1m^2}[/tex],
solving for [tex]M[/tex] we get:
[tex]M = 20kg\\[/tex]
which is the required mass of the water in the tank.
Now, the tank fills at a rate of 2 drops per second or
[tex]2 (0.05g)/s = 0.10g/s = 0.0001kg/s[/tex]
since each drop weights 0.05g.
Therefore, the time [tex]t[/tex] it takes to collect 20kg of water will be
[tex]t = 20kg \div \dfrac{0.0001kg}{s}[/tex]
[tex]t = 2*10^5s[/tex]
which is 55.56 hours or 2 days and 7.56 hours.
Thus, it will take 20,000 seconds to get a pressure of 200Pa at the bottom of the tank.
