Question

If the surface area of a spherical bubble is increasing at the rate of 4 sq.cm/sec, then the rate of change in its volume (in cubic cm/sec) when its radius is 8 cms is

• A. 8
• B. 12
• C. 15
• D. 16

Hint:

Relate $\frac{dS}{dt}$ and $\frac{dV}{dt}$ using the chain rule with $\frac{dr}{dt}$.

Answer 1

The Correct Option is D

Surface area of a sphere, $S = 4\pi r^2$. Volume of a sphere, $V = \frac{4}{3}\pi r^3$. We are given $\frac{dS}{dt} = 4$ sq.cm/sec. We need to find $\frac{dV}{dt}$ when $r=8$ cm. $\frac{dS}{dt} = 8\pi r \frac{dr}{dt} = 4$, so $\frac{dr}{dt} = \frac{4}{8\pi r} = \frac{1}{2\pi r}$. $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} = 4\pi r^2 (\frac{1}{2\pi r}) = 2r$. When $r=8$, $\frac{dV}{dt} = 2(8) = 16$ cubic cm/sec.