Q116 / 170 · Manipal (MET) 2015 · Mathematics
A cone whose height is always equal to its diameter, is increasing in volume at the rate of $40 \mathrm{~cm}^3 / \mathrm{s}$. At what rate is the radius increasing when its circular base area is $1 \mathrm{~m}^2 ?$
- A$1 \mathrm{~mm} / \mathrm{s}$
- B$0.001 \mathrm{~cm} / \mathrm{s}$
- C$2 \mathrm{~mm} / \mathrm{s}$
- Correct answerD$0.002 \mathrm{~cm} / \mathrm{s}$
Solution
Let $h$ be the height, $r$ be the radius of the base and $V$ be the volume of the cone at time $t$. Then,
$V=\frac{1}{3} \pi r^2 h$
$\Rightarrow \quad V=\frac{2}{3} \pi r^3 \quad[\because h=2 \pi$, given $]$
On differentiating both sides w.r.t. $t$, we get
$\frac{d V}{d t}=2 \pi r^2 \frac{d r}{d t}$
$\Rightarrow \quad 40=2(10)^4 \frac{d r}{d t}$
$\left[\because \pi r^2=1 \mathrm{~m}^2=10^4 \mathrm{~cm}^2\right.$ and $\frac{d V}{d t}=40 \mathrm{~cm}^3 / \mathrm{s}$, given $]$
$\Rightarrow \quad \frac{d r}{d t}=\frac{2}{1000} \mathrm{~cm} / \mathrm{s}=0.002 \mathrm{~cm} / \mathrm{s}$
Hence, the correct answer is option: 'D'.