Question

A spherical balloon is being inflated at the rate of $35cc$ per minute, When its radius is $7cm$, its surface area increases at the rate of

• A. 10 sq. cm/min
• B. 15 sq. cm/min
• C. 20 sq. cm/min
• D. 25 sq. cm/min

Answer 1

Answer: (A)

Given , $\frac{dV}{dt}=35,$
where $V$ is volume of spherical balloon.
Also, $V=\frac{4}{3}\pi r^3$
$\Rightarrow \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=35$
$\Rightarrow \frac{4}{3}\pi \times 3r^2\frac{dr}{dt}=35$
$\Rightarrow \frac{dr}{dt}=\frac{35\times 3}{4\pi \times 3r^2}$
Let $S$ be surface area of sphere then $S=4\pi r^2$
Taking derivatives w.r.t. ${}^{\prime }{t}^{\prime }$
$\Rightarrow \frac{dS}{dt}=8\pi \times r\frac{dr}{dt}=8\pi \times r\times \frac{35\times 3}{4\pi \times 3r^2}$
Substituter r = 7
$\frac{dS}{dt}=\frac{2\times 35\times 3}{3\times 7}=10c{m}^2/min$
$=2{\log }_{e}a$