# Mensuration-44

The height of a solid cylinder is 10 cm and diameter 7 cm. Two equal conical holes of radius 3 cm and height 4 cm
are cut off from both ends of the cylinder. Find the volume and surface area of the remaining solid.

Solution: Radius of the cylinder (R) = 7/2 cm

Height (H) = 10 cm

Therefore, Volume of the cylinder = Π R2 H = 22/7 x (7/2)2 x 10 cm3 = 385 cm3

Total surface of cylinder = 2 Π R (R + H)

= 2 x 22/7 x 7/2 (7/2 + 10) cm2 = 2 x 22/7 x 7/2 x 27/2 cm2

= 297 cm2

Radius of each conical hole (r) = 3 cm, height (h) = 4 cm

Therefore, slant height (l) = √(r2 + h2) = √(32 + 42)

= 5 cm

Volume of the two conical holes = 2 x 1/3 Π r2 h = 2 x 1/3 x 22/7 x 32 x 4 cm3

= 75.43 cm2

Curved surface of the two conical cavities = 2 x Π r l = 2 x 22/7 x 3 x 5 cm2 = 94.29 cm2

Therefore, Volume of the remaining solid = Volume of cylinder - Volume of two cones

= (385 - 75.43) cm3 = 309.57 cm3

Surface area of the remaining solid

= Total surface of the cylinder + curved surface of the two cones - 2(area of the circular base of the cones)

= [297 + 94.29 - 2 x 22/7 x 32] cm2 = (297 + 94.29 - 56.57) cm2

= 334.72 cm2