A solid metallic sphere of radius $12 \; cm$ is melted and recast into a solid cylinder of height $36 \; cm$. Find the radius and the curved surface area of the cylinder. Take $\pi = 3.14$
Radius of sphere $= R = 12 \; cm$
Height of cylinder $= h = 36 \; cm$
Let radius of cylinder $= r \; cm$
Volume of sphere $= V_S = \dfrac{4}{3} \pi R^3$
Volume of cylinder $= V_C = \pi r^2 h$
Since the sphere is melted and recast into a solid cylinder,
Volume of sphere $= $ Volume of cylinder
i.e. $\;$ $\dfrac{4}{3} \pi R^3 = \pi r^2 h$
i.e. $\;$ $r^2 = \dfrac{4 R^3}{3 h} = \dfrac{4 \times 12^3}{3 \times 36} = 64$
$\implies$ $r = \pm 8$
$\therefore \;$ Radius of cylinder $= r = 8 \; cm$ $\;$ [negative value is discarded as radius cannot be negative]
Curved surface area of the cylinder $= CSA = 2 \pi r h$
i.e. $\;$ $CSA = 2 \times 3.14 \times 8 \times 36 = 1808.64 \; cm^2$