If the radius of a circle increases at the rate of $7 \mathrm{~cm} / \mathrm{sec}$, then the rate of…
If the radius of a circle increases at the rate of $7 \mathrm{~cm} / \mathrm{sec}$, then the rate of increase of
its area after 10 minutes is
- $1,84,800 \mathrm{~cm}^{2} / \mathrm{sec}$
- $1,64,800 \mathrm{~cm}^{2} / \mathrm{sec}$
- $1,88,400 \mathrm{~cm}^{2} / \mathrm{sec}$
- $1,68,400 \mathrm{~cm}^{2} / \mathrm{sec}$
Solution
$\begin{aligned} \frac{d \varepsilon}{d t} &=7 \mathrm{~cm} / \mathrm{sec} . \\ A &=\pi \varepsilon^{2} \\ \frac{d A}{d t} &=2 \pi \varepsilon \frac{d \varepsilon}{d t}-(1) \\ \text { Radius after } 10 \text { minutes } ; & \varepsilon=70 \mathrm{~cm} \times 60 \mathrm{~cm} \\ \frac{d A}{d t} &=2 \times \frac{22}{7} \times 70 \times 7 \times 60[\mathrm{Put} \text { in (1)... }\\ &=1,84,800 \mathrm{~cm}^{2} / \mathrm{sec} . \end{aligned}$
Asked in: MHT CET 2020 (14 Oct Shift 1)
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