Question

The volume of spherical ball is increasing at the rate of $4\pi cc/sec$. Then rate of surface area when volume is $288\pi cc$ is

• A. $\frac{3\pi }{4}{cm}^2/sec$
• B. $\frac{4}{3}{cm}^2/sec$
• C. $\frac{4\pi }{3}{cm}^2/sec$
• D. $\frac{3}{4}{cm}^2/sec$

Answer 1

The correct option is C $\frac{4\pi }{3}{cm}^2/sec$

Volume of a spherical ball is given by $V=\frac{4\pi r^3}{3}$, while its surface area is given by $SA=4\pi r^2$

Rate of change of volume is given by

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=4\pi$ cc/sec

$\Rightarrow r^2\frac{dr}{dt}=1$ ...(1)

Also, since the rate of increase of surface area is asked when volume is $288\pi$ cc, i.e. $\frac{4\pi r^3}{3}=288\pi$

$\Rightarrow r=6$ cm

Now,

$\frac{d\left(SA\right)}{dt}=8\pi r\frac{dr}{dt}=\frac{8\pi }{r}$ ..(from equation 1)

Thus,

$\frac{d\left(SA\right)}{dt}=\frac{8\pi }{6}=\frac{4\pi }{3}c{m}^2/sec$