Question

A spherical balloon is being inflated at the rate of $35 cc / min$. The rate of increase in surface area (in $cm ^2 / min$.) of the balloon when its diameter is $14 cm$, is

• A. 100
• B. $10 \sqrt{10}$
• C. $\sqrt{10}$
• D. 10

Answer 1

Answer: (D)

$ V =\frac{4}{3} \pi r ^3 \Rightarrow \frac{ dV }{ dt }=4 \pi r ^2 \frac{ dr }{ dt } \Rightarrow \frac{ dr }{ dt }=\left(\frac{35}{4 \pi r ^2}\right)...(i)$
Also, $S =4 \pi r ^2 \Rightarrow \frac{ dS }{ dt }=8 \pi r \frac{ dr }{ dt }=(8 \pi r )\left(\frac{35}{4 \pi r ^2}\right)=\frac{70}{ r }$
$\left.\therefore \frac{ dS }{ dt }\right]_{ r =7}=\frac{70}{7}=10$