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A sphere increases its volume at the rate of $\pi \mathrm{cc} / \mathrm{s}$. The rate at which its surface area increases when the radius is $1 \mathrm{~cm}$ is
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Verified Answer
The correct answer is:
$2 \pi \mathrm{sq} \mathrm{cm} / \mathrm{s}$
Given rate of increase volume of sphere is
$$
\frac{d V}{d t}=\pi
$$
We know that,
Volume of sphere; $V=\frac{4}{3} \pi r^{3}$
$\Rightarrow \quad \frac{d V}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \cdot \frac{d r}{d t}=4 \pi r^{2} \frac{d r}{d t}$
$\Rightarrow \quad \pi=4 \pi r^{2} \frac{d r}{d t} ; \quad$ [from Eq. (i)]
$\Rightarrow \quad \frac{d r}{d t}=\frac{1}{4 r^{2}}$
5
Also, we know that, surface of sphere,
$$
\begin{aligned}
&S=4 \pi r^{2} \\
&\Rightarrow \quad \frac{d S}{d t}=8 \pi r \frac{d r}{d t}=8 \pi r \cdot \frac{1}{4 r^{2}} \quad[\text { from Eq. (ii)] } \\
&\Rightarrow \quad[\because \text { given } r=1] \\
&\Rightarrow \quad \frac{d S}{d t}=2 \pi
\end{aligned}
$$
So, rate of increase in surface $=2 \pi \mathrm{sq} \mathrm{cm} / \mathrm{s}$
$$
\frac{d V}{d t}=\pi
$$
We know that,
Volume of sphere; $V=\frac{4}{3} \pi r^{3}$
$\Rightarrow \quad \frac{d V}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \cdot \frac{d r}{d t}=4 \pi r^{2} \frac{d r}{d t}$
$\Rightarrow \quad \pi=4 \pi r^{2} \frac{d r}{d t} ; \quad$ [from Eq. (i)]
$\Rightarrow \quad \frac{d r}{d t}=\frac{1}{4 r^{2}}$
5
Also, we know that, surface of sphere,
$$
\begin{aligned}
&S=4 \pi r^{2} \\
&\Rightarrow \quad \frac{d S}{d t}=8 \pi r \frac{d r}{d t}=8 \pi r \cdot \frac{1}{4 r^{2}} \quad[\text { from Eq. (ii)] } \\
&\Rightarrow \quad[\because \text { given } r=1] \\
&\Rightarrow \quad \frac{d S}{d t}=2 \pi
\end{aligned}
$$
So, rate of increase in surface $=2 \pi \mathrm{sq} \mathrm{cm} / \mathrm{s}$
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