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A spherical balloon is expanding. If the radius is increasing at the rate of 2 centimeters per minute, the rate at which the volume increases (in cubic centimeters per minute) when the radius is 5 , centimetres is
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The correct answer is:
$200 \pi$
Let $\mathrm{r}$ and $\mathrm{V}$ be the respectively radius and volume of the balloon. Let $\mathrm{t}$ represents the time. The rate of increament in radius is $\frac{\mathrm{dr}}{\mathrm{dt}}=2 \mathrm{~cm} / \mathrm{minute} .$ The volume of the balloon is given by
$\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}$
Differentiating w.r. to $\mathrm{t}$, we get
$\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi\left(3 \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}\right)$
Substituting the values of and $\frac{\mathrm{dr}}{\mathrm{dt}}$, we get
$\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi\left(3 \times 5^{2} \times 2\right)=200 \pi \mathrm{cm}^{3} /$ minute
$\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}$
Differentiating w.r. to $\mathrm{t}$, we get
$\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi\left(3 \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}\right)$
Substituting the values of and $\frac{\mathrm{dr}}{\mathrm{dt}}$, we get
$\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi\left(3 \times 5^{2} \times 2\right)=200 \pi \mathrm{cm}^{3} /$ minute
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