Height of a hollow right circular cylinder, open at both ends, is `2*8` metres. If length of inner diameter of the cylinder is `4*6` dcm and the cylinder is made up of `84*48` cubic-dcm of iron, then calculate the length of outer diameter of the cylinder.

Explanation

Let the outer diameter be d dcm.
`:.` outer radius `= d/2` dcm.
As per question , `pi {(d/2)^(2)-(4.6/2)^(2)}xx28=84.48[ :' 2.8 m = 29 dcm ]`
or, `(22)/(7) {(d/2)^(2)-(2.3)^(2)}xx28=84.48`
or, `{(d/2)^(2)-5.29}xx28=84.48xx(7)/(22)` or, `(d/2)^(2)-5.29=(84.48xx7)/(28xx22)`
or, `(d/2)^(2)-5.29=0.96` or, `(d/2)^(2)=0.96+5.29`
or, `(d/2)^(2) = 6.25 = (2.5)^(2) rArr (d)/(2) 2.5` or, `d=2.5xx2=5`
Hence the length of diameter of the cylinder = 5 dcm.