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Gauge Factor $G_f$=2

Stress =1000 kg/$(cm)^2 $

Modulus of elasticity=ε=2×$(10)^6$ kg/$(cm)^3$

**Formulae –**

Modulus of elasticity=ε=$\frac{Stress}{Strain}$

Change in Resistance=$∆R=G_f$×Strain

Poisson's ratio=$μ=(G_f-1)/2$

**Solution –**

Before we calculate the change in resistance, we have to deduce the value of strain. Strain is given by the ratio of stress and modulus of elasticity,

Strain=$\frac{Stress}ε$

=$\frac{1000}{(2×10^6 )}$

=500×$10^{(-6)}$

Strain=500 μm⁄m

Now, change in resistance is given by,

∆R=$G_f$×Strain

=2×500 μ

∆R=1 ×$(10)^{(-3)}$

Change in resistance in percentage,

Therefore, **change in resistance is 0.1%.**

Further, Poisson’s ratio is given by,

$μ=\frac{(G_f-1)}2$

$μ=\frac{(2-1)}2$

μ=0.5

With this value, it is evident that the metal has a Poisson’s ratio in the upper limit which is 0.5. Generally, Poisson’s ratio always falls between 0 and 0.5. For gold, it is between 0.42 and 0.44; for copper, it is 0.33; and for steel, it is between 0.27 and 0.30.

For the given figures, **the change in resistance is 0.1%** and **the Poisson’s ratio is 0.5** or $\frac{1}2$.