Question

The volume of a sphere is increasing the rate of $1200c.cm/sec$. The rate of increase in its surface area when the radius is $10cm$ is

• A. $120sq.cm/sec$
• B. $240sq.cm/sec$
• C. $200sq.cm/sec$
• D. $100sq.cm/sec$

Answer 1

The correct option is B $240sq.cm/sec$

$V=\frac{4}{3}\pi r^3$

$\frac{dV}{dt}=\frac{4}{3}\pi 3r^2\frac{dr}{dt}$

$1200=4\pi r^2\frac{dr}{dt}$

So, $\frac{dr}{dt}=\frac{300}{\pi r^2}$ ...(1)

Now, $S=4\pi r^2$

$\frac{dS}{dt}=4\pi 2r\frac{dr}{dt}$

$\frac{dS}{dt}=8\pi r\times \frac{300}{\pi r^2}$ ...From (1)

$\frac{dS}{dt}=\frac{2400}{r}=\frac{2400}{10}=240$ sq. cm / sec.