The radius of right circular cylinder increase at the rate $0.1 \mathrm{~cm} / \mathrm{min}$ and height…

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
  1. $-2 \mathrm{pm}^3 / \mathrm{min}$
  2. $\frac{-3 \pi}{5} \mathrm{~cm}^3 / \mathrm{min}$
  3. $\frac{-8 \pi}{5} \mathrm{~cm}^3 / \mathrm{min}$
  4. $\frac{2 \pi}{5} \mathrm{~cm}^3 / \mathrm{min}$

Solution

$\begin{aligned} & V=p r^2 h \\ & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=\pi\left\{2 r \cdot \frac{\mathrm{d} r}{\mathrm{~d} t} \cdot h+r^2 \cdot \frac{\mathrm{d} h}{\mathrm{~d} t}\right\} \\ & =\pi\left\{2 \times 2 \times 0 \cdot 1 \times 3+2^2 \times(-0 \cdot 2)\right\} \\ & =\pi\{1 \cdot 2-0 \cdot 8\}=0 \cdot 4 \pi=\frac{2 \pi}{5} \mathrm{~cm}^3 / \mathrm{min}\end{aligned}$

Asked in: MHT CET 2022 (11 Aug Shift 1)

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