Question
From a solid cylinder whose height is $16 \ cm$ and radius is $12 \ cm,$ a conical cavity of height $8 \ cm$ and of base radius $6 \ cm$ is hollowed out. Find the volume and total surface area of the remaining solid.
✓
Answer
Radius of solid cylinder $(R) = 12 \ cm$
and Height $(H) = 16 \ cm$
$\therefore \text { Volume }=\pi R ^2 H$
$=\frac{22}{7} \times 12 \times 12 \times 16$
$=\frac{50688}{7} \ cm ^3$
Radius of cone $(r) = 6 \ cm,$ and height $(h) = 8 \ cm$.
$\therefore \text { Volume }=\frac{1}{3} \pi r ^2 h$
$=\frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 8$
$=\frac{2112}{7} \ cm ^3$
$(1)$ Volume of remaining solid
$=\frac{50688}{7}-\frac{2112}{7}$
$=\frac{48567}{7}$
$=6939.43^2 \ cm ^3$
$(2)$ Slant height of cone $I =\sqrt{ h ^2+ r ^2}$
$=\sqrt{6^2+8^2}$
$=\sqrt{36+64}$
$=\sqrt{100}$
$=10 \ cm $
Therefore, total surface area of remaining solid $=$ curved surface area of cylinder $+$ curvedsurface area of cone $+$ base area of cylinder $+$ area of circular ring on upper side of cylinder
$=2 \pi RH +\pi rl +\pi R ^2+\pi\left( R ^2- r ^2\right)$
$=\left(2 \times \frac{22}{7} \times 12 \times 16\right)+\left(\frac{22}{7} \times 6 \times 10\right)+\left(\frac{22}{7} \times 12 \times 12\right)+\left(\frac{22}{7}\left(12^2-6^2\right)\right)$
$=\frac{8448}{7}+\frac{1320}{7}+\frac{3168}{7}+\frac{22}{7}(144-36)$
$=\frac{8448}{7}+\frac{1320}{7}+\frac{3168}{7}+\frac{2376}{7}$
$=\frac{15312}{7}$
$=2187.43 \ cm ^2$
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