A vertical container (right rectangular prism) with base area measuring 0.15 m by 0.2 m is being filled with identical small stones, each with a volume of 2*10^{-8} m^{3} and a mass of 0.05*10^{-3} kg. Assume that the volume of the empty spaces between the stones is negligible. If the height of the stones in the container increases at the rate of 3*10^{-3} m/s, at what rate does the mass of the stones in the container increase?

###### Solution.

The density of the stone is

ρ = m / V = 0.05*10^{-3} / 2*10^{-8} = 2500 kg/m^{3}.

The total mass of the stones in the container when filled to height h is

M = ρ*V or M = ρ*A*h,

where

A = 0.15*0.2 = 0.03 m^{2}

is the base area of the container that remains constant.

So, the rate of mass change is given by

dM/dt = d(ρ*A*h)/dt or

dM/dt = ρ*A*dh/dt (Since ρ and A are constant).

Then,

ρ*A*dh/dt = 2500 * 0.03 * 3*10^{-3} = 0.225 kg/s.