Question

From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the volume of the remaining solid. Also, find the total surface area of the remaining solid. $[Take\pi =\mathrm{3.14.}]$

Answer 1

Volume of the remaining solid = Volume of cylinder - volume of the cone

$\text{Volume of the remaining solid}=\pi r^{2}h-\frac{1}{3}\pi r^{2}h=\frac{2}{3}\pi r^{2}h=\frac{2}{3}\times 3.14\times (6{)}^2\times 8=602.88c{m}^3$

Slant height of cone, $AB=\sqrt{B{C}^2+A{C}^2}=\sqrt{8^2+6^2}=\sqrt{64+36}=10cm$

Total surface area of the remaining solid = curved surface area of cylinder + area of base of cylinder + curved surface area of cone
$=2\pi rh+\pi r^2+\pi rl=2\pi \times 6\times 8+\pi \times 6^2+\pi \times 6\times 10=96\pi +36\pi +60\pi =192\pi =192\times 3.14=602.88c{m}^2$