Answer:
[tex]\frac{dP}{P} = 6.25[/tex]
Explanation:
Given data:
Sa = 2.1
[tex]R = \frac{pl}{A}[/tex]
[tex]\frac{dR}{R} =\frac{dP}{P} +\frac{dL}{L} (1_2V)[/tex]
[tex]\frac{dR}{R} =\frac{dP}{P} +\epsilon (1_2V)[/tex]
[tex]Sa = \frac{\frac{dR}{R}}{\epsilon} =\frac{\frac{dP}{P}}{\epsilon} +\frac{\epsilon (1_2V)}{\epsilon}[/tex]
[tex]Sa = (1+2v) + \frac{\frac{dP}{P}}{\epsilon}[/tex]
change in specific resistance is given as [tex]\frac{dP}{P}[/tex]
[tex]\frac{dP}{P} = \frac{Sa -(1-2v)}{\epsilon}[/tex] ........2
where v is elastic range = 0.30
[tex]\epsilon = 0.08[/tex]
[tex]\frac{dP}{P} = \frac{2.1 -(1-2\times 0.30)}{0.08}[/tex]
[tex]\frac{dP}{P} = 6.25[/tex]
