TS EAMCET · Maths · Application of Derivatives
The vertical angle of a right circular cone is \(60^{\circ}\). If water is being poured into the cone at the rate of \(\frac{1}{\sqrt{3}} \mathrm{~m}^3 / \mathrm{min}\), then the rate \((\mathrm{m} / \mathrm{min})\) at which the radius of the water level is increasing when the height of the water level is 3 m is
- A \(\frac{1}{3 \sqrt{3 \pi}}\)
- B \(\frac{1}{9 \sqrt{3} \pi}\)
- C \(\frac{1}{9 \pi}\)
- D \(\frac{1}{33}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{9 \pi}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \frac{r}{h}=\tan 30^{\circ} \Rightarrow h=\sqrt{3} r \Rightarrow V=\frac{1}{3} \pi r^2 h=\frac{1}{\sqrt{3}} \pi r^3 \\ & \frac{d V}{d t}=\sqrt{3} \pi r^2 \frac{d r}{d t}=\frac{1}{\sqrt{3}} \Rightarrow \frac{d r}{d t}=\frac{1}{3 \pi r^2}=\frac{1}{\pi h^2} \\ &…
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