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Correct option is B)

Let r and h be the internal radius and height of the cylindrical vessel and let V denote the volume of the material.

Now, $V=πr_{2}h$

$⇒500=πr_{2}h$

$⇒h=πr_{2}500 $ ...(1)

Now, $S=(2πrh+πr_{2})1002 $ (As $1002 $ is thickness)

$=(r1000 +πr_{2}).501 $

For maximum or minimum ,

$drds =(−r_{2}1000 +2πr)501 =0$

$⇒πr_{3}=500$

$∴πr_{3}=500=V=πr_{2}h$

or $r=(π500 )_{1/3}=h$

Clearly$dr_{2}d_{2}S =(r_{3}4V +2π).501 =+ive$

$∴$ For minimum volume, $r=(π500 )_{1/3}=h$

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