If the surface area of a spherical balloon of radius $6 \mathrm{~cm}$ is increasing at the rate $2 \mathrm{~cm}^2 / \mathrm{sec}$, then the rate of increase in its volume in $\mathrm{cm}^3 / \mathrm{sec}$ is

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Aicharya · Accepted Answer

Surface area, $S=4 \pi r^2$ $\begin{aligned} & 	herefore \quad \frac{\mathrm{dS}}{\mathrm{dt}}=8 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}} \ & \Rightarrow 2=8 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}} \ & \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{4 \pi \mathrm{r}}...(i) \ & 	ext { Volume, } V=\frac{4}{3} \pi r^3 \ & 	herefore \quad \frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} 	imes 3 \pi \mathrm{r}^2 	imes \frac{\mathrm{dr}}{\mathrm{dt}} \ & =4 \pi r^2 	imes \frac{1}{4 \pi r}... [From (i)] \ & =\mathrm{r} \ & =6 \mathrm{~cm}^3 / \mathrm{sec} \ & \end{aligned}$

If the surface area of a spherical balloon of radius $6 \mathrm{~cm}$ is increasing at the rate $2…

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