A cone whose height is always equal to its diameter is increasing… - Vidyadip

MCQ

A cone whose height is always equal to its diameter is increasing in volume at the rate of $40 \ cm^3 / \sec$ At what rate is the radius increasing when its circular base area is $1 m^2\ ?$

A
$1mm/ \sec$.
B
$0.001\ cm/ \sec.$
C
$2 mm/ \sec.$
✓
$0.002 \ cm/ \sec.$

✓

Answer

Correct option: D.

$0.002 \ cm/ \sec.$

$\text{V} = \frac{1}{3} \pi \text{r}^{2}\text{h}$
Given that height is equals diamiter.
$\Rightarrow \text{h} = 2\text{r}$
$\text{V} = \frac{1}{3} \pi\text{r}^{2}\text{2r}$
$\text{V} = \frac{2}{3} \pi\text{r}^{3}$
$\Rightarrow \frac{\text{dV}}{\text{dt}} = 2\pi\text{r}^{2} \frac{\text{dr}}{\text{dt}}$
$\Rightarrow \frac{\text{dr}}{\text{dt}} = \frac{1}{2\times10^{4}} \times40 \ \ \begin{pmatrix}\because \pi\text{r}^{2} = 1\text{m}^{2}\\\Rightarrow1\text{m}^{2} = 10^{4} \text{cm}^{2} \end{pmatrix}$
$\Rightarrow \frac{\text{dr}}{\text{dt}} = \frac{1}{2\times10^{4}} \times40 = 0.002\text{cm}/\sec.$

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