The volume of sphere is increasing at the rate of 1200 $\mathrm{cu} \mathrm{cm} / \mathrm{s}$. The rate of…
The volume of sphere is increasing at the rate of 1200 $\mathrm{cu} \mathrm{cm} / \mathrm{s}$. The rate of increase in its surface area when the radius is $10 \mathrm{~cm}$ is
$120 \mathrm{sq} \mathrm{cm} / \mathrm{s}$
$240 \mathrm{sq} \mathrm{cm} / \mathrm{s}$
$200 \mathrm{sq} \mathrm{cm} / \mathrm{s}$
$100 \mathrm{sq} \mathrm{cm} / \mathrm{s}$
Solution
Let $V$ be the volume, $S$ be the surface area and $r$ be the radius of the sphere.
It is given that, $\frac{d V}{d t}=1200 \mathrm{cu} \mathrm{cm} / \mathrm{s}$ and $r=10 \mathrm{~cm}$
Now, Volume of sphere $=V=\frac{4}{3} \pi r^3$
$\begin{aligned}
& \therefore \quad \frac{d V}{d t}=\frac{4}{3} \pi 3 r^2 \frac{d r}{d t} \\
& 1200=4 \pi(10)^2 \frac{d r}{d t} \\
& \frac{d r}{d t}=\frac{3}{\pi}
\end{aligned}$
and Surface Area of sphere,
$\begin{aligned}
& S=4 \pi r^2=\frac{d S}{d t}=4 \pi \cdot 2 r \frac{d r}{d t} \\
& \Rightarrow \quad \frac{d S}{d t}=4 \pi \cdot 2(10) \frac{d r}{d t} \\
& \Rightarrow \quad \frac{d S}{d t}=4 \pi \cdot 2 \times 10\left(\frac{3}{\pi}\right) \\
& \therefore \quad \frac{d S}{d t}=240 \mathrm{sq} \mathrm{cm} / \mathrm{s}
\end{aligned}$