question_answer

If water is poured into an inverted hollow cone whose semi-vertical angle is \[30{}^\circ \]. Its depth (measured along the axis) increases at the rate of 1 cm/s. The rate at which the volume of water increases when the depth is 24 cm is

A) \[162\,\,c{{m}^{3}}/s\]

B) \[172\,\,c{{m}^{3}}/s\]

C) \[182\,\,c{{m}^{3}}/s\]

D) \[192\,\,c{{m}^{3}}/s\]

Correct Answer: D

Solution :

[d]

Let A be the vertex and AO the axis of the cone.

Let \[O'A=h\] be the depth of water in the cone.

In \[\Delta AO'C,\]

\[\tan 30{}^\circ =\frac{O'C}{h}\] or \[O'C=\frac{h}{\sqrt{3}}=radius\]

\[V=\] Volume of water in the cone

\[=\frac{1}{3}\pi {{(O'C)}^{2}}\times AO'\]

\[=\frac{1}{3}\pi \left( \frac{{{h}^{2}}}{3} \right)\times h=\frac{\pi }{9}{{h}^{3}}\]

or \[\frac{dV}{dt}=\frac{\pi }{3}{{h}^{2}}\frac{dh}{dt}\]                                 ? (1)

But given that depth of water increases at the rate of 1 cm/s

\[\frac{dh}{dt}=1\,\,cm/s\]                                              ?. (2)

From (1) and (2), \[\frac{dV}{dt}=\frac{\pi {{h}^{2}}}{3}\]

When \[h=24\,\,cm,\] the rate of increase of volume is

\[\frac{dV}{dt}=\frac{\pi {{(24)}^{2}}}{3}=192\,\,c{{m}^{3}}/s\].