A uniform rod of length (L) and area of cross-section (A) is subjected to tensil load (F). If `sigma` be the Poisson's ratio and Y be the Young's modulus of the material of the rod, then find the volumetric strain produced in the rod.

Options

A. `(5)/(AY) (L + 2 sigma)`
B. `(F)/(AY) (L - 2 sigma)`
C. zero
D. None of these

Explanation

Young's modulus i.e., `Y = ("stress")/("Strain") = (F//A)/(Delta //L)`
`:. (Delta L)/(L) = (F)/(AY)`
Now, by definition of poisson's ratio
`sigma = ("Lateral strain")/("Longitudinal strain") = - (Delta r // r)/(Delta L // L)`
By using Eq. (i) , `(Delta r)/(r ) = - (sigma. Delta L)/(L) = - (sigma F)/(AY)`
Now, `(Delta V)/(V) = (2 Delta r)/(r ) + (Delta L)/(L)`
by using Eqs. (i) and (ii) we get
`implies (Delta V)/(V) = 2 (-(sigma F)/(AY)) + (F)/(AY)`
`= (F)/(AY) (1 - 2 sigma)`