MHT CET · Maths · Application of Derivatives
Water is being poured at the rate of \(36 \mathrm{~m}^3 / \mathrm{min}\). into a cylindrical vessel, whose circular base is of radius \(3 \mathrm{~m}\). then the water level in the cylinder is rising at the rate of
- A \(\frac{4}{\pi} \mathrm{m} / \mathrm{min}\)
- B \(4 \pi \mathrm{m} / \mathrm{min}\)
- C \(\frac{1}{4 \pi} \mathrm{m} / \mathrm{min}\)
- D \(\frac{2}{\pi} \mathrm{m} / \mathrm{min}\)
Answer & Solution
Correct Answer
(A) \(\frac{4}{\pi} \mathrm{m} / \mathrm{min}\)
Step-by-step Solution
Detailed explanation
Volume of cylinder \(=\pi \mathrm{r}^2 \mathrm{~h}\)
\(\therefore \frac{\mathrm{dv}}{\mathrm{dt}}=\pi \mathrm{r}^2 \frac{\mathrm{dh}}{\mathrm{dt}}\)
\(\therefore 36=\pi(3)^2 \frac{\mathrm{dh}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dh}}{\mathrm{dt}}=\frac{4}{\pi} \mathrm{m} / \mathrm{min}\)
\(\therefore \frac{\mathrm{dv}}{\mathrm{dt}}=\pi \mathrm{r}^2 \frac{\mathrm{dh}}{\mathrm{dt}}\)
\(\therefore 36=\pi(3)^2 \frac{\mathrm{dh}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dh}}{\mathrm{dt}}=\frac{4}{\pi} \mathrm{m} / \mathrm{min}\)
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