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If the surface area of a sphere of radius $r$ is increasing uniformly at the rate $8\, cm^2/s$, then the rate of change of its volume is :
If the surface area of a sphere of radius $r$ is increasing uniformly at the rate $8\, cm^2/s$, then the rate of change of its volume is :
Updated On: Sep 30, 2024
- constant
- proportional to $\sqrt{r}$
- proportional to $r^2$
- proportional to $r$
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The Correct Option is D
Solution and Explanation
$V = \frac{4}{3}\pi r^{3}\quad\Rightarrow\quad \frac{dV}{dt} = 4\pi r^{2}. \frac{dr}{dt}\quad\quad\ldots\left(i\right)$
$S = 4\pi r^{2} \Rightarrow \frac{dS}{dt} = 8\pi r. \frac{dr}{dt}$
$\Rightarrow 8 = 8\pi \,r \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{\pi r}$
Putting the value of $\frac{dr}{dt}$ in $\left(i\right)$, we get
$\frac{dV}{dt} = 4\pi r^{2} \times\frac{1}{\pi r} = 4r$
$\Rightarrow \quad \frac{dV}{dt}$ is proportional to $r.$
$S = 4\pi r^{2} \Rightarrow \frac{dS}{dt} = 8\pi r. \frac{dr}{dt}$
$\Rightarrow 8 = 8\pi \,r \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{\pi r}$
Putting the value of $\frac{dr}{dt}$ in $\left(i\right)$, we get
$\frac{dV}{dt} = 4\pi r^{2} \times\frac{1}{\pi r} = 4r$
$\Rightarrow \quad \frac{dV}{dt}$ is proportional to $r.$
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