Consider a solid cube which is subjected to a pressure of `6xx10^(5) Nm^(-2)` . Due to this pressure, the each side of the cube is shortened by 2% . Find out the volumetric strain of the cube.

Explanation

Let L be the initial length of the each side of the cube.
Volumn, `V=LxxLxxL=L^(3)`=Initial volume (`V_(i)` say)
If each side of the cube is shortened by 2% ,then final length of the cube `=L-2%` of L
`=(L-(2L)/(100))=L(1-(2)/(100))`
`therefore` Final volume, `V_(f)=L^(3)(1-(2)/(100))^(3)=V=(1-(2)/(100))^(3)`
Change in volume, `DeltaV=V_(f)-V_(i)`
`=V(1-(2)/(100))^(3)-V=V[(1-(2)/(100))^(3)-1]`
`(DeltaV)/(V)=(1-(2)/(100))^(3)-1~= [1-(2xx3)/(100)]-1`
`" "[because (1-x)^(n)~=1-nx` for `x lt lt 1`]
therefore Volumetric strain, `(DeltaV)/(V)=1-0.06-1=0.06`
`" "`[take positive sign]