Question

a metallic cylinder has radius 3cm and height 5cm.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 8/9cm. calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape.

Answer 1

Hey Friend ☺

Given

For Cylinder

Radius = 3cm , Height = 5cm

For Conical hole

Radius = 3/2 cm , Height = 8/9 cm

To Find

The ratio of the volume of metal left in cylinder to the volume of metal taken out
Solution

Firstly let us find the volume of cyclinder

We know that

Volume of cyclinder = pie × r^2 × h

= 3.14 × 3 × 3 × 5

= 28.26 × 5

= 141.30 cm^3

So the volume of the cyclinder is 141.40 cm^3

The conical hole is drilled in the cylinder

Volume of conical hole = 1/3 × pie × r^2 × h

= 1/3 × 3.14 × 3/2 × 3/2 × 8/9

= 2.09 ( approx )

Volume of metal left in cyclinder = 141.30 - 2.09

= 139.21

Ratio = 139.21/2.09

= 13921/209

Hope it helps you ..!!

✌

Answer 2

SMASH ❕ BASH ❕

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. THE ANS IS IT S


. Radius of the metallic cylinder = 3 cm

Height of metallic cylinder = 5 cm

Radius of conical drill = 3/2

Depth of conical drill = 8/9


Volume of cylinder = pr²h
. 22/7*3*3*5
. 45p cm ³

Volume of conical drill = 1/3pr²h
. 1/3*p*9/3*8/9
. 2p/3 cm ³


Volume of metal left in the cylinder == volume of cylinder - volume of conical drill

. ==== 45 p - 2p/3
. 135p - 2p / 3
. 133p/3


Volume of metal left in the cylinder / volume of conical drill
====== > 133p/3 /2p/3
. 133p/2p
. 133/2
. 66.5 cm

. The ratio is 133 : 2



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