Question

The radius of a sphere is changing. At an instant of time the rate of change in its volume and its surface area are equal. Then the value of radius at that instant is?

• A. 1
• B. 2
• C. $\frac{3}{2}$
• D. 3

Answer 1

Answer: (B)

Given that, at any instant of time
Rate of change in volume w.r.t. time $=$ rate of change in surface area w.r.t. time
i.e., $\frac{dv}{dt}=\frac{ds}{dt}\dots$ (i)
Volume of sphere of radius $(r),V=\frac{4}{3}\pi r^3$
Differentiating w.r.t, 't', we get
$\frac{dv}{dt}=\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi \frac{d}{dt}\left(r^3\right)$
$\frac{dv}{dt}=\frac{4}{3}\pi \left(3r^2\right)\frac{dr}{dt}$
or $\frac{dv}{dt}=4\pi r^2\frac{dr}{dt}$ ...(ii)
Surface area of sphere of radius $(r)$,
$S=4\pi r^2$
Differentiating w.r.t. $it$, we get
$\frac{ds}{dt}=\frac{d}{dt}\left(4\pi r^2\right)=4\pi \frac{d}{dt}\left(r^2\right)$
$\Rightarrow \frac{ds}{dt}=8\pi r\frac{dr}{dt}\dots$ (iii)
Putting the values from Eqs. (ii) and (iii) in Eq. (i), we get
$4\pi r^2\frac{dr}{dt}=8\pi r\frac{dr}{dt}$ or $r=2$