Q108 / 160 · TS EAMCET 2005 · Mathematics

The radius of a circular plate is increasing at the rate of $0.01 \mathrm{~cm} / \mathrm{s}$ when the radius is $12 \mathrm{~cm}$. Then, the rate at which the area increases, is

  • Correct answer
    A
    $0.24 \pi \mathrm{sq} \mathrm{cm} / \mathrm{s}$
  • B
    $60 \pi \mathrm{sq} \mathrm{cm} / \mathrm{s}$
  • C
    $24 \pi \mathrm{sq} \mathrm{cm} / \mathrm{s}$
  • D
    $1.2 \pi \mathrm{sq} \mathrm{cm} / \mathrm{s}$

Solution

The area of circular plate is $A=\pi r^2$ On differentiating w.r.t. $t$, we get $\begin{aligned} & \frac{d A}{d t}=2 \pi r \frac{d r}{d t} \\ & \frac{d A}{d t}=2 \pi(12)(0.01) \\ & \Rightarrow \quad {\left[\because \text { given } \frac{d r}{d t}=0.01 \Rightarrow r=12\right] } \\ & \frac{d A}{d t}= 0.24 \pi \mathrm{sq} \mathrm{cm} / \mathrm{sec} \end{aligned}$

Hence, the correct answer is option: 'A'.

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