A 200-liter tank initially full of water develops a leak at the bottom. Given that 30% of the water leaks out in the first 5 minutes, find the amount of water left in the tank 10 minutes after the leak develops if the water drains off at a rate that is proportional to the amount of water present.

the amount of water that is left in the tank after 10 min is 98 L

Step-by-step explanation:

since the water drains off at rate that is proportional to the water present

(-dV/dt) = k*V, where k = constant

(-dV/V) = k*dt

-∫dV/V) = k*∫dt

-ln V/V₀=k*t

or

V = V₀*e^ (-k*t), where V₀ = initial volume

then since V₁=0.7*V₀ at t₁ = 3 min

-ln V₁/V₀=k*t₁

then for t₂ = 10 min we have

-ln V₂/V₀=k*t₂

dividing both equations

ln (V₂/V₀) / ln (V₁/V₀) = (t₂/t₁)

V₂/V₀ = (V₁/V₀) ^ (t₂/t₁)

V₂=V₀ * (V₁/V₀) ^ (t₂/t₁)

replacing values

V₂=V₀ * (V₁/V₀) ^ (t₂/t₁) = 200 L * (0.7) ^ (10min/5min) = 98 L

then the amount of water that is left in the tank after 10 min is 98 L