Given that, a man makes complete use of $405$ cc of iron, $783$ cc of aluminium, and $351 cc$ of copper to make a number of solid right circular cylinders of each type of metal.
These cylinders have the same volume and each of these has a radius $3 \;\text{cm}.$
Let the number of cylinder $= x,y,z $
As the number of cylinders is to be kept at a minimum, the volume of each cylinder has to be maximum.
Volume $\uparrow = \dfrac {405}{x \downarrow} = \dfrac{783}{y \downarrow} = \dfrac{351}{z \downarrow}$
The volume of each cylinder should be the highest common factor of $ 405, 783$, and $351.$
$\therefore$ The volume of each cylinder $ = 27$ cc
- Number of iron cylinders $(x) = \frac{405}{27} = 15$
- Number of aluminium cylinders $(y) = \frac {783}{27} = 29$
- Numbers of copper cylinders $ (z) = \frac {351}{27} = 13$
Therefore, the total number of cylinders $ = x+y+z = 15+29+13 =57$
Volume of each cylinder $ = \pi r^{2} h$
$\implies 27 = \pi \times 3\times 3\times h$
$\implies \boxed {h = \frac {3}{\pi}} $
Total surface area (TSA) of all cylinders $=$ total number of cylinders (total surface area of $1$ cylinder $+$ area of top and bottom)
$\qquad = 57 (2 \pi r h + 2 \pi r^{2} )$
$ \qquad = 57 \times 2 (\pi r h + \pi r^{2})$
$ \qquad = 114 (\pi .3.\frac{3}{\pi} + 9\pi )$
$ \qquad = 114 \left( \frac {9\pi}{\pi} + 9\pi \right)$
$ \qquad = 114 \times 9 (1 \times \pi ) $
$ \qquad = 1026 (1 \times \pi)$
Correct Answer $: \text {D}$