Applications Of Derivatives Question 55
Question: 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
Options:
A) $ 162cm^{3}/s $
B) $ 172cm^{3}/s $
C) $ 182cm^{3}/s $
D) $ 192cm^{3}/s $
Show Answer
Answer:
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 ( \frac{h^{2}}{3} )\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. So, $ \frac{dh}{dt}=1cm/s…(2) $ From (1) and (2), $ \frac{dV}{dt}=\frac{\pi h^{2}}{3} $ When $ h=24cm, $ the rate of increase of volume is $ \frac{dV}{dt}=\frac{\pi {{(24)}^{2}}}{3}=192cm^{3}/s $ .
BETA
