Applications Of Derivatives Question 180
Question: A spherical balloon is being inflated at the rate of 35 cc/min. The rate of increase of the surface area of the balloon when its diameter is 14 cm is
[Karnataka CET 2005]
Options:
A) 7 sq. cm/min
B) 10 sq. cm/min
C) 17.5 sq. cm/min
D) 28 sq. cm/min
Show Answer
Answer:
Correct Answer: B
Solution:
Volume = $ V=\frac{4}{3}\pi r^{3} $
therefore $ \frac{dV}{dt}=4\pi r^{2}.\frac{dr}{dt} $ , at $ r=7 $ cm 35 cc/min = $ 4\pi {{(7)}^{2}}\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=\frac{5}{28\pi } $ Surface area, S = $ 4\pi r^{2} $
$ \frac{dS}{dt}=8\pi r\frac{dr}{dt}=\frac{8\pi .7.5}{28\pi }=10, $ cm2/min.
BETA
