Question

# A balloon that always remains spherical has a variable radius. The rate at which its volume is increasing is $m\pi c{m}^{2}$ with the radius when the latter is $10cm$. Find $m$.

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Solution

## Solve for the required value:Given: The rate at which the volume of the sphere is increasing is $m\pi c{m}^{2}$, and the radius is $10cm$The volume of the sphere is $V=\frac{4}{3}{\mathrm{\pi r}}^{3}$The rate of change of the volume with respect to radius is $\frac{dV}{dr}$$\frac{\mathrm{dV}}{\mathrm{dr}}=\frac{d}{dr}\left(\frac{4}{3}{\mathrm{\pi r}}^{3}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{dV}}{\mathrm{dr}}=\frac{4}{3}\mathrm{\pi }×3{\mathrm{r}}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{dV}}{\mathrm{dr}}=4{\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{dV}}{\mathrm{dr}}=4\mathrm{\pi }{\left(10\right)}^{2}\left[\because \mathrm{r}=10\right]\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{dV}}{\mathrm{dr}}=400\mathrm{\pi }$Hence, the value of $m$ is $400$

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