Question

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.

Answer 1

We have,

the base radius of the cylinder, R=3 cm 

the height of the cylinder, H= 5 cm,

the base radius of the conical hole, r = `3/2` and 

the height of the conical hole, h= `8/9` cm 

Now,

Volume of the cylinder, V = πR²H

= π × 3² × 5

= 45π cm³ 

Also,

Volume of the cone removed from the cylinder, V=πR²H

`= π/3 xx (3/2)^2xx(8/9)`

`=(2π)/3  "cm"^3`

So, the volume of metal left in the cylinder, V' = V - v

`=45pi - (2pi)/3`

`=(133pi)/3  "cm"^3`

`therefore "The required ratio" = (V')/v`

`=(133pi/3)/((2pi)/3)`

`=133/2`

=133:2

So, the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is 133 : 2.