JEE Mains · Maths · STD 12 - 6. Application of derivatives
A water tank has the shape of an inverted right circular cone, whose semi vertical angle is \({\tan ^{ - 1}}\,\left( {\frac{1}{2}} \right)\). Water is poured in at a constant rate of \(5\) cubic meter per minute. Then the rate (in \(m/min\)) at which the level of water is rising at the instant when the depth of water in the tank is \(10\, m\) is
- A \(\frac{2}{\pi }\)
- B \(\frac{1}{{5\pi }}\)
- C \(\frac{1}{{10\pi }}\)
- D \(\frac{1}{{15\pi }}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{{5\pi }}\)
Step-by-step Solution
Detailed explanation
Given \(\theta = {\tan ^{ - 1}}\left( {\frac{1}{2}} \right)\) \( \Rightarrow \tan \theta = \frac{1}{2} = \frac{r}{h}\) \( \Rightarrow r = \frac{h}{2}\) \(V = \frac{1}{3}\pi {r^2}h\) \(V = \frac{1}{3}\pi \frac{{{h^3}}}{4}\)…
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