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(D) Let the inner radius be $r$ and the inner height be $h$. The inner volume is given by $V = \pi r^2 h$,so $h = \frac{V}{\pi r^2}$.The outer radius

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$4$

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(D) Let the inner radius be $r$ and the inner height be $h$. The inner volume is given by $V = \pi r^2 h$,so $h = \frac{V}{\pi r^2}$.The outer radius is $R = r + 2$ and the total height of the container is $H = h + 2$ (since the base is $2 \ mm$ thick and the top is open).The volume of the material $M$ is the difference between the outer volume and the inner volume:$M = \pi (r + 2)^2 (h + 2) - \pi r^2 h$$M = \pi (r^2 + 4r + 4)(h + 2) - \pi r^2 h$$M = \pi (r^2 h + 2r^2 + 4rh + 8r + 4h + 8 - r^2 h)$$M = \pi (2r^2 + 4rh + 8r + 4h + 8)$Substitute $h = \frac{V}{\pi r^2}$ into the expression for $M$:$M(r) = \pi (2r^2 + 4r(\frac{V}{\pi r^2}) + 8r + 4(\frac{V}{\pi r^2}) + 8)$$M(r) = 2\pi r^2 + \frac{4V}{r} + 8\pi r + \frac{4V}{r^2} + 8\pi$To find the minimum volume,differentiate $M$ with respect to $r$ and set it to $0$:$\frac{dM}{dr} = 4\pi r - \frac{4V}{r^2} + 8\pi - \frac{8V}{r^3} = 0$Given that the minimum occurs at $r = 10 \ mm$:$4\pi(10) - \frac{4V}{100} + 8\pi - \frac{8V}{1000} = 0$$40\pi + 8\pi - \frac{40V}{1000} - \frac{8V}{1000} = 0$$48\pi = \frac{48V}{1000}$$V = 1000\pi$Therefore,$\frac{V}{250\pi} = \frac{1000\pi}{250\pi} = 4$.

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