A circular plate of radius \(5 \mathrm{~cm}\) is heated. Due to expansion, its radius increase at the rate of \(0.05 \mathrm{~cm} / \mathrm{s}\). The rate at which its area is increasing when the radius is \(5.2 \mathrm{~cm}\) is

Let \(r\) be the radius of the circular disc and \(A\) be
the area of the circular disc at any instant of time. we know that
\(\begin{gathered}A=\pi r^2 \\ \frac{d A}{d t}=2 \pi r \frac{d r}{d t}\end{gathered}\)
According to the question, the circular disc expands on heating with the rate of change of radius is \(0.05 \mathrm{~cm} / \mathrm{s}\)
\(\begin{aligned} \frac{d r}{d t} & =0.05 \mathrm{~cm} / \mathrm{s} \\ \frac{d A}{d t} & =2 \times \pi \times 5.2 \times 0.05 \\ \Rightarrow \quad \frac{d A}{d t} & =0.52 \pi \mathrm{cm}^2 / \mathrm{s}\end{aligned}\)
So, the rate at which its area is increasing, when radius is \(5.2 \mathrm{~cm}\) is \(0.52 \pi \mathrm{cm}^2 / \mathrm{s}\).