MHT CET · Maths · Application of Derivatives
The radius of right circular cylinder increase at the rate \(0.1 \mathrm{~cm} / \mathrm{min}\) and height decreases at the rate of \(0 \cdot 2 \mathrm{~cm} / \mathrm{min}\) The rate of change of volume of the cylinder in \(\mathrm{cm}^3 / \mathrm{min}\), when the radius is \(2 \mathrm{~cm}\) and height is \(3 \mathrm{~cm}\), is
- A \(-2 \mathrm{pm}^3 / \mathrm{min}\)
- B \(\frac{-3 \pi}{5} \mathrm{~cm}^3 / \mathrm{min}\)
- C \(\frac{-8 \pi}{5} \mathrm{~cm}^3 / \mathrm{min}\)
- D \(\frac{2 \pi}{5} \mathrm{~cm}^3 / \mathrm{min}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 \pi}{5} \mathrm{~cm}^3 / \mathrm{min}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & V=p r^2 h \\ & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=\pi\left\{2 r \cdot \frac{\mathrm{d} r}{\mathrm{~d} t} \cdot h+r^2 \cdot \frac{\mathrm{d} h}{\mathrm{~d} t}\right\} \\ & =\pi\left\{2 \times 2 \times 0 \cdot 1 \times 3+2^2 \times(-0 \cdot 2)\right\} \\ & =\pi\{1 \cdot 2-0 \cdot 8\}=0 \cdot 4 \pi=\frac{2 \pi}{5} \mathrm{~cm}^3 / \mathrm{min}\end{aligned}\)
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