Applications Of Derivatives Question 82
Question: If the volume of a spherical balloon is increasing at the rate of 900cm3per sec, then the rate of change of radius of balloon at instant when radius is 15cm
[in cm/sec] [RPET 1996]
Options:
A) $ \frac{22}{7} $
B) 22
C) $ \frac{7}{22} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ V=\frac{4}{3}\pi r^{3} $ Differentiate with respect to t, $ \frac{dV}{dt}=\frac{4}{3}\pi 3r^{2}.\frac{dr}{dt} $
therefore $ \frac{dr}{dt}=\frac{1}{4\pi r^{2}}.\frac{dV}{dt} $
$ \frac{dr}{dt}=\frac{1}{4\times \pi \times 15\times 15}\times 900 $
$ \Rightarrow $ $ \frac{dr}{dt}=\frac{1}{\pi }=\frac{7}{22} $ .
BETA
