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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.
Solution
Radius of solid cylinder (R) = 12 cm
and Height (H) = 16 cm
∴ Volume = πR2H
= `22/7 xx 12 xx 12 xx 16`
= `50688/7 cm^3`
Radius of cone (r) = 6 cm and height (h) = 8 cm.
∴ Volume = `1/3pir^2h`
= `1/3 xx 22/7 xx 6 xx 6 xx 8`
= `2112/7 cm^3`
(i) Volume of remaining solid
= `50688/7-2112/7`
= `48567/7`
= 6939.43 cm3
(ii) Slant height of cone `l = 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 + curved surface area of cone + base area of cylinder + area of circular ring on upper side of cylinder
= `2piRH + pirl + piR^2 + pi(R^2 - r^2)`
= `(2 xx 22/7 xx 12 xx 16) + (22/7 xx 6 xx 10) + (22/7 xx 12 xx 12) + (22/7(12^2 - 6^2))`
= `8448/7 + 1320/7 + 3168/7 + 22/7(144 - 36)`
= `8448/7 + 1320/7 + 3168/7 + 2376/7`
= `15312/7`
= 2187.43 cm2
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