2022 VJC P1 Q11

2022 VJC P1 Q11

13 marks
Prelims

A tank contains \(500\) litres of water in which \(100\) g of a poisonous chemical called Prokrastenate is dissolved. A solution containing \(0.1\) g of Prokrastenate per litre is pumped into the tank at a rate of \(5\) litres per minute, and the well-mixed solution is pumped out at the same rate. By letting \(x\) grams be the amount of Prokrastenate in the tank after \(t\) minutes, show that

\(\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{k-x}{100}\),

where \(k\) is a constant to be determined.[2]

Find \(x\) in terms of \(t\) and find the value of \(t\) when \(x=75\).[5]

The well-mixed solution that is pumped out flows into an empty container, in the form of an open inverted cone with a height of \(60\) cm and base radius \(30\) cm, at the same rate (see diagram).

Given that \(1\) litre \(1000\) cm\(^{3}\),

  1. Show that the volume of the well-mixed solution in the container, \(V\) cm\(^{3}\) can be expressed as \(V=\frac{\pi {{h}^{3}}}{12}\), where \(h\) cm is the depth of the solution at that instant.[2]
  2. Hence, or otherwise, find the rate of change of the depth of solution after \(5\) minutes.[4]
    [The volume of a cone of base radius \(r\) and height \(h\) is given by \(V=\frac{1}{3}\pi {{r}^{2}}h\).]

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Answer:\(k=50\) and \(x=50+50e^{-t/100}\). When \(x=75\), \(t=100\ln2\approx69.3\) minutes. (i) \(V=\frac{\pi h^3}{12}\). (ii) \(\frac{\mathrm{d}h}{\mathrm{d}t}\approx3.05\text{ cm per minute}\) after \(5\) minutes.

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