Water is poured at the rate of \(9\pi\) cm3/s into a vessel in the form of an inverted right circular cone of height 12 cm and base radius 6 cm. The rate at which the water level is rising when the depth of the water is 6 cm is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1 cm/s
Let h be the depth of the water and r the radius of the water surface at time t. The water forms a cone similar to the vessel, so
\( \displaystyle \frac{r}{h} = \frac{6}{12} \), so \( \displaystyle r = \frac{h}{2} \)
The volume of the water is
\( \displaystyle V = \frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi\cdot\frac{h^{2}}{4}\cdot h \)
\( \displaystyle = \frac{\pi h^{3}}{12} \)
Differentiating with respect to t:
\( \displaystyle \frac{dV}{dt} = \frac{\pi h^{2}}{4}\cdot\frac{dh}{dt} \)
With \(\frac{dV}{dt} = 9\pi\) and h = 6:
\( \displaystyle 9\pi = \frac{36\pi}{4}\cdot\frac{dh}{dt} = 9\pi\frac{dh}{dt} \)
\( \displaystyle \frac{dh}{dt} = 1 \) cm/s
Why the other options are incorrect
Option (1) \(\frac{1}{4}\) cm/s — Incorrect. This takes r = h instead of \(r = \frac{h}{2}\): then \(V = \frac{\pi h^{3}}{3}\) and \(36\pi\frac{dh}{dt} = 9\pi\).
Option (2) 3 cm/s — Incorrect. At h = 6 the rate equation is \(9\pi = 9\pi\frac{dh}{dt}\), which gives 1 cm/s, not 3 cm/s.
Option (4) \(\frac{3}{4}\) cm/s — Incorrect. This treats the radius as fixed at 6 cm (\(V = 12\pi h\)); the water surface widens as the level rises, so r = h/2 must be used.