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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.
Solution
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.
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