Q68 / 160 · AP EAMCET 2012 · Mathematics

If the volume of a sphere increases at the rate of $2 \pi \mathrm{cm}^3 / \mathrm{s}$, then the rate of increase of its radius (in $\mathrm{cm} / \mathrm{s}$ ), when the volume is $288 \pi \mathrm{cm}^3$, is

  • A
    $\frac{1}{36}$
  • Correct answer
    B
    $\frac{1}{72}$
  • C
    $\frac{1}{18}$
  • D
    $\frac{1}{9}$

Solution

Given $\frac{d V}{d t}=2 \pi \mathrm{cm}^3 / \mathrm{s}$ $\because$ Volume of sphere, $V=\frac{4}{3} \pi r^3$ On differentiating w.r.t. $t$, we get $\begin{aligned} & \frac{d V}{d t}=\frac{4}{3} \pi \times 3 r^2 \frac{d r}{d t} \\ & \Rightarrow \quad 2 \pi=4 \pi r^2 \frac{d r}{d t} \\ & \Rightarrow \quad \frac{d r}{d t}=\frac{1}{2 r^2}=\frac{1}{2 \times 6^2}=\frac{1}{72} \mathrm{~cm} / \mathrm{s} \\ & \end{aligned}$ $\left[\because V=288 \pi=\frac{4}{3} \pi r^3 \Rightarrow 216=r^3 \Rightarrow r=6\right]$

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

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