When metal wire is stretched by a load the fectional change in its volume `/_\V/V` is proportional to

Options

A. `(Delta l)/(l)`
B. `((Delta l)/(l))^(2)`
C. `((Delta l)/(l))^(3)`
D. `sqrt((Delta l)/(l))`

Answer

A

Explanation

Let r, l and `sigma` represent the radius, length and Poisson.s ratio, for the material of the wire
As Poisson.s ratio, `sigma = (-Delta r//r)/(Delta l//l) :. (Delta r)/(r ) = - sigma ((Delta l)/(l))` …(i)
Now volume `V= pi r^(2) l`
Taking logarithm on both sides, we get
`log V= log pi + 2 log r + log l`
Differentiating on both sides, we get
`(Delta V)/(V) = 2 (Delta r)/(r ) + (Delta l)/(l)` ...(ii)
Substituting for `(Delta r)/(r )` from equation (i) in equation (ii), we get
`(Delta V)/(V) = 2 (-sigma (Delta l)/(l)) + (Delta l)/(l) = (Delta l)/(l) [1- 2 sigma]`
As Poisson.s ratio is a constant for the given material, hence
`(Delta V)/(V) prop (Delta l)/(l)`