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A spherical balloon is being inflated at the rate of $35 \mathrm{~cm} / \mathrm{min}$. When its radius is $7 \mathrm{~cm}$, its surface area increases at the rate of
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Verified Answer
The correct answer is:
$10 \mathrm{~cm}^{2} / \mathrm{min}$
Given, $\frac{d V}{d t}=35$
where, $V$ is volume of spherical balloon.
$$
\begin{aligned}
&\text { Also, } V=\frac{4}{3} \pi r^{3} \\
&\Rightarrow \frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right)=35 \Rightarrow \frac{4}{3} \pi \times 3 r^{2} \frac{d r}{d t}=35 \\
&\Rightarrow \quad \frac{d r}{d t}=\frac{35 \times 3}{4 \pi \times 3 r^{2}}
\end{aligned}
$$
Let $S$ be surface area of sphere, then $S=4 \pi r^{2}$
Taking derivative w.r.t. $r$
$$
\frac{d S}{d t}=8 \pi \times r \frac{d r}{d t}=8 \pi \times r \times \frac{35 \times 3}{4 \pi \times 3 r^{2}}
$$
Substitute, $r=7$
$$
\frac{d S}{d t}=\frac{2 \times 35 \times 3}{3 \times 7}=10 \mathrm{~cm}^{2} / \mathrm{min}
$$
where, $V$ is volume of spherical balloon.
$$
\begin{aligned}
&\text { Also, } V=\frac{4}{3} \pi r^{3} \\
&\Rightarrow \frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right)=35 \Rightarrow \frac{4}{3} \pi \times 3 r^{2} \frac{d r}{d t}=35 \\
&\Rightarrow \quad \frac{d r}{d t}=\frac{35 \times 3}{4 \pi \times 3 r^{2}}
\end{aligned}
$$
Let $S$ be surface area of sphere, then $S=4 \pi r^{2}$
Taking derivative w.r.t. $r$
$$
\frac{d S}{d t}=8 \pi \times r \frac{d r}{d t}=8 \pi \times r \times \frac{35 \times 3}{4 \pi \times 3 r^{2}}
$$
Substitute, $r=7$
$$
\frac{d S}{d t}=\frac{2 \times 35 \times 3}{3 \times 7}=10 \mathrm{~cm}^{2} / \mathrm{min}
$$
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