JEE Advance - Mathematics (2015 - Paper 1 Offline - No. 13)
If the volume of the material used to make the container is minimum when the inner radius of the container is $$10 $$ mm,
then the value of $${V \over {250\pi }}$$ is
then the value of $${V \over {250\pi }}$$ is
Answer
4
Explanation
Given: The inner volume of cylinder $=\mathrm{Vmm}^3$
Thickness of wall $=2 \mathrm{~mm}$
Thickness of bottom circular disc $=2 \mathrm{~mm}$
Let the inner radius of cylinder $=r \mathrm{~mm}$ and height of the inner cylinder $=h \mathrm{~mm}$.
$$ \Rightarrow \mathrm{V}=\pi r^2 h $$
Now, volume of the material used $=$ volume of outer cylinder - volume of the inner cylinder + Volume of the circular disc
$$ \begin{gathered} \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi(r+2)^2 h-\pi r^2 h+\pi(r+2)^2 2 \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi h\left\{(r+2)^2-r^2\right\}+2 \pi(r+2)^2 \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi h(4 r+4)+2 \pi(r+2)^2 \\\\ \left\{\because(a+b)^2=a^2+b^2+2 a b\right\} \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=2 \pi\left\{2 h(r+1)+(r+2)^2\right\} \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi r^2}(r+1)+(r+2)^2\right\} \\\\ \left\{\because \mathrm{V}=\pi r^2 h\right\} \end{gathered} $$
For $\mathrm{V}_{\mathrm{m}}$ to be minimum, $\frac{d v_m}{d r}=0$
Differentiating the above equation w.r.t.r,
$$ \begin{aligned} \quad & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \frac{d}{d r}\left\{\frac{r+1}{r^2}\right\}+\frac{d}{d r}(r+2)^2\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \cdot \frac{r^2(1)-(r+1)(2 r)}{r^4}+2(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \cdot \frac{r^2-2 r^2-2 r}{r^4}+2(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=4 \pi\left\{\frac{\mathrm{V}}{\pi r^3}(-r-2)+(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=4 \pi(r+2)\left(-\frac{\mathrm{V}}{\pi r^3}+1\right) \\\\ \because & \frac{d \mathrm{~V}_m}{d r}=0 \\\\ \Rightarrow & 4 \pi(r+2)\left(\frac{-\mathrm{V}}{\pi r^3}+1\right)=0 \end{aligned} $$
Also, given that $\mathrm{V}_{\mathrm{m}}$ is minimum at $r=10 \mathrm{~mm}$
$$ \begin{array}{ll} \Rightarrow & 4 \pi(10+2)\left(\frac{-\mathrm{V}}{10^3 \pi}+1\right)=0 \\\\ \Rightarrow & 48 \pi\left(\frac{-\mathrm{V}}{10^3 \pi}+1\right)=0 \\\\ \Rightarrow & \frac{\mathrm{V}}{10^3 \pi}=1 \\\\ \Rightarrow & \frac{\mathrm{V}}{250 \pi}=4 \end{array} $$
Thickness of wall $=2 \mathrm{~mm}$
Thickness of bottom circular disc $=2 \mathrm{~mm}$
Let the inner radius of cylinder $=r \mathrm{~mm}$ and height of the inner cylinder $=h \mathrm{~mm}$.
$$ \Rightarrow \mathrm{V}=\pi r^2 h $$
Now, volume of the material used $=$ volume of outer cylinder - volume of the inner cylinder + Volume of the circular disc
$$ \begin{gathered} \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi(r+2)^2 h-\pi r^2 h+\pi(r+2)^2 2 \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi h\left\{(r+2)^2-r^2\right\}+2 \pi(r+2)^2 \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=\pi h(4 r+4)+2 \pi(r+2)^2 \\\\ \left\{\because(a+b)^2=a^2+b^2+2 a b\right\} \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=2 \pi\left\{2 h(r+1)+(r+2)^2\right\} \\\\ \Rightarrow \mathrm{V}_{\mathrm{m}}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi r^2}(r+1)+(r+2)^2\right\} \\\\ \left\{\because \mathrm{V}=\pi r^2 h\right\} \end{gathered} $$
For $\mathrm{V}_{\mathrm{m}}$ to be minimum, $\frac{d v_m}{d r}=0$
Differentiating the above equation w.r.t.r,
$$ \begin{aligned} \quad & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \frac{d}{d r}\left\{\frac{r+1}{r^2}\right\}+\frac{d}{d r}(r+2)^2\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \cdot \frac{r^2(1)-(r+1)(2 r)}{r^4}+2(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=2 \pi\left\{\frac{2 \mathrm{~V}}{\pi} \cdot \frac{r^2-2 r^2-2 r}{r^4}+2(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=4 \pi\left\{\frac{\mathrm{V}}{\pi r^3}(-r-2)+(r+2)\right\} \\\\ \Rightarrow & \frac{d \mathrm{~V}_m}{d r}=4 \pi(r+2)\left(-\frac{\mathrm{V}}{\pi r^3}+1\right) \\\\ \because & \frac{d \mathrm{~V}_m}{d r}=0 \\\\ \Rightarrow & 4 \pi(r+2)\left(\frac{-\mathrm{V}}{\pi r^3}+1\right)=0 \end{aligned} $$
Also, given that $\mathrm{V}_{\mathrm{m}}$ is minimum at $r=10 \mathrm{~mm}$
$$ \begin{array}{ll} \Rightarrow & 4 \pi(10+2)\left(\frac{-\mathrm{V}}{10^3 \pi}+1\right)=0 \\\\ \Rightarrow & 48 \pi\left(\frac{-\mathrm{V}}{10^3 \pi}+1\right)=0 \\\\ \Rightarrow & \frac{\mathrm{V}}{10^3 \pi}=1 \\\\ \Rightarrow & \frac{\mathrm{V}}{250 \pi}=4 \end{array} $$
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