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 1

 
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