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The volume of a sphere is increasing at the rate of $4 \pi \mathrm{cc} / \mathrm{sec}$. When its volume is $288 \pi \mathrm{cc}$, the rate of increase (in $\mathrm{cm} / \mathrm{sec}$ ) in its radius is
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The correct answer is:
$\frac{1}{36}$
Given that,
Rate of increase in, $\frac{d V}{d t}=4 \pi \mathrm{cc} / \mathrm{sec}$
Volume, $V=288 \pi \mathrm{cc}$
As we know, volume of a sphere of radius $(r)$,
$V=\frac{4}{3} \pi r^3$ $\ldots(\mathrm{i})$
Differentiating above expression w.r.t. ' $t$ ' on both sides, we get
$\frac{d V}{d t}=\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right)=\frac{4}{3} \pi 3 r^2 \frac{d r}{d t}$
$\frac{d V}{d t}=4 \pi r^2 \frac{d r}{d t}$ $\ldots(\mathrm{ii})$
From Eq. (i) $(288 \pi)=\frac{4}{3} \pi r^3$ or $r=6 \mathrm{~cm}$ Putting this value in Eq. (ii), we get
$4 \pi=4 \pi(6)^2 \frac{d r}{d t} \text { or } \frac{d r}{d t}=\frac{1}{36}$
Rate of increase in, $\frac{d V}{d t}=4 \pi \mathrm{cc} / \mathrm{sec}$
Volume, $V=288 \pi \mathrm{cc}$
As we know, volume of a sphere of radius $(r)$,
$V=\frac{4}{3} \pi r^3$ $\ldots(\mathrm{i})$
Differentiating above expression w.r.t. ' $t$ ' on both sides, we get
$\frac{d V}{d t}=\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right)=\frac{4}{3} \pi 3 r^2 \frac{d r}{d t}$
$\frac{d V}{d t}=4 \pi r^2 \frac{d r}{d t}$ $\ldots(\mathrm{ii})$
From Eq. (i) $(288 \pi)=\frac{4}{3} \pi r^3$ or $r=6 \mathrm{~cm}$ Putting this value in Eq. (ii), we get
$4 \pi=4 \pi(6)^2 \frac{d r}{d t} \text { or } \frac{d r}{d t}=\frac{1}{36}$
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