This question is the combination of cylinder and conical shape problem. Last time this question was asked in 2015 cbse board exam. Question number 25 from RS Aggarwal book page number 288.

Deepak BoraNewbie

# A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weight, a conical hole is drilled in the cylinder. The conical hole has a radius of 3/2 cm and its depth is 8/9 cm. Calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape.

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Given data,

Radius of the cylinder = 3cm

Height of the cylinder = 5cm

volume of a cylinder = π r² h

= 3.14 × (3)² × 5

= 3.14 × 9 × 5

= 141.3 cm³

∴ volume of a cylinder = 141.3 cm³

Radius of the conical hole = 3/2 cm

Depth of the conical hole = 8/9 cm

volume of the conical drill = [1/3] π r² h

= [1/3] × [3.14] × (3/2)² × 8/9

= 2.09 cm³

volume of the conical drill = 2.09 cm³

∴ volume of the metal left = 141.3 cm³ – 2.09 cm³

= 139.21 cm³

∴ volume of the metal left = 139.21 cm³

Now,

Ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape =

= volume of the metal left / volume of the conical drill

= 139.21 cm³/2.09 cm³

∴ Ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is 139:2 (Approximately)