MHT CET · Maths · Application of Derivatives
If the water is being poured at the rate \(36 \mathrm{~m}^3 / \mathrm{sec}\) in cylindrical vessel of base radius \(3 \mathrm{~m}\), then the rate at which water level is rising, is
- A \(\frac{4}{\pi} \mathrm{m} / \mathrm{sec}\)
- B \(4 \pi \mathrm{m} / \mathrm{sec}\)
- C \(\frac{\pi}{4} \mathrm{~m} / \mathrm{sec}\)
- D \(\frac{3}{\pi} \mathrm{m} / \mathrm{sec}\)
Answer & Solution
Correct Answer
(A) \(\frac{4}{\pi} \mathrm{m} / \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{d} v}{\mathrm{~d} t}=36 \mathrm{~m}^3 / \mathrm{sec}\)
and \(v=\pi r^2 \mathrm{~h}=\pi \times 3^2 \mathrm{~h}=9 \pi \mathrm{h}\)
\(\Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=9 \pi \frac{\mathrm{d} h}{\mathrm{~d} t}\)
\(\Rightarrow 36=9 \pi \frac{\mathrm{d} h}{\mathrm{~d} t}\)
\(\Rightarrow \frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{4}{\pi} \mathrm{m} / \mathrm{sec}\)
and \(v=\pi r^2 \mathrm{~h}=\pi \times 3^2 \mathrm{~h}=9 \pi \mathrm{h}\)
\(\Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=9 \pi \frac{\mathrm{d} h}{\mathrm{~d} t}\)
\(\Rightarrow 36=9 \pi \frac{\mathrm{d} h}{\mathrm{~d} t}\)
\(\Rightarrow \frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{4}{\pi} \mathrm{m} / \mathrm{sec}\)
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