2025 RVHS Promo Q1

2025 RVHS Promo Q1

Junior College 1
4 marks

A conical water tank is being filled with water at a constant rate of \(0.2\,\mathrm{m}^3\mathrm{s}^{-1}\). The tank has a height of \(6\,\mathrm{m}\) and a radius at the top of \(3\,\mathrm{m}\).

  1. Show that the volume of the water in the tank, \(V\), can be expressed as \(\frac{2}{3}\pi r^3\), where \(r\) is the radius of the top water surface.[1]
  2. Find the exact rate of change of \(r\) when the height of the water is \(4\,\mathrm{m}\).[3]

[The volume of a cone of base radius \(r\) and height \(h\) is given by \(V=\frac{1}{3}\pi r^2h\).]

Solution:

Solution locked

Sign in to view the step-by-step solution

Similar questions are unavailable for this question.
Answer:\(\dfrac{dr}{dt}=\dfrac1{40\pi}\text{ m s}^{-1}\)

Need help? Join our JC Math tuition classes.

Learn more