A cylindrical vessel of base radius 42 cm contains water to a certain level. A solid cylinder of height 18 cm is dropped into it and is completely submerged in water. If the level of the water in the vessel rises by 8 cm, what is the total surface area (in cm²) of the solid cylinder? (Use $\pi = \frac{22}{7}$ ) (Assuming no overflow.)

A fully submerged solid pushes up water equal to its own volume.

Volume of the solid cylinder $=$ volume of the water rise.

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Let the radius of the solid cylinder be $r$ cm.

$\pi r^2 \times 18 = \pi \times 42^2 \times 8$

$r^2 = \dfrac{1764 \times 8}{18}$

$= 784$

$r = 28$ cm

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Total surface area of a cylinder $= 2\pi r(r + h)$

$= 2 \times \dfrac{22}{7} \times 28 \times (28 + 18)$

$= 176 \times 46$

$= 8096$

Therefore, the total surface area is $8096$ cm$^2$.