If a metallic circular plate of radius $50 \mathrm{~cm}$ is heated so that its radius increases at the rate…
If a metallic circular plate of radius $50 \mathrm{~cm}$ is heated so that its radius increases at the rate of $1 \mathrm{~mm}$ per hour, then the rate at which, the area of the plate increases (in $\mathrm{cm}^2 /$ hour) is
$5 \pi$
$10 \pi$
$100 \pi$
$50 \pi$
Solution
Let $A=\pi \mathrm{r}^2$ be area of metalic circular plate of $r=50 \mathrm{~cm}$.
Also, given $\frac{d r}{d t}=1 \mathrm{~mm}=\frac{1}{10} \mathrm{~cm}$
$
\begin{aligned}
& \therefore A=\pi r^2 \\
& \Rightarrow \frac{d A}{d t}=2 \pi r \frac{d r}{d t}=2 \pi \cdot 50 \cdot \frac{1}{10}=10 \pi
\end{aligned}
$
Hence, area of plate increases in $10 \pi \mathrm{cm}^2 /$ hour