A tank has a capacity of 100 litres. Initially, the tank contains 10 litres of water thoroughly mixe

Question

A tank has a capacity of 100 litres. Initially, the tank contains 10 litres of water thoroughly mixed with 300 grams of salt. Salt water with a concentration of 5 g/litre is poured into the tank at a constant rate of 2 litres per minute, while the mixture flows out at a constant rate of 1 litre per minute. Let $S$ denote the amount of dissolved salt in the tank (in grams) at time $t$ minutes after salt water is poured into the tank. Show that $\frac{\mathrm{d}S}{\mathrm{d}t}=10-\frac{S}{t+10}$, stating your assumption(s) clearly. [2] By substituting $Q=\left( t+10 \right)S$, solve the differential equation in (i), and find $S$ in terms of $t$. [4] Hence, find the concentration of salt in the tank at the point just before it overflows. [2] Once the volume of salt solution in the tank reaches 100 litres, the pouring stops, and the tank is allowed to drain off. The salt solution drains from the tank at a rate proportional to the volume of solution in the tank. Let $V$ denote the volume of solution in the tank at time $T$ after the tank starts to drain off. If the tanks takes 10 minutes to drain half of its contents, find $V$ in terms of $T$. [4]

Accepted Answer

(i) $\frac{dS}{\mathrm{d}t} = 10 - \frac{S}{10+t}$ (ii) $S = \frac{5t^2 + 100t + 3000}{t + 10}$ (iii) $5.25$ g/L (iv) $V = 100e^{-\frac{T\ln 2}{10}}$

2018 EJC JC2 MYE P2 Q4

2018 EJC JC2 MYE P2 Q4

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