written 3.9 years ago by
Ramnath • 5.1k


Given –
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=Stress/Strain
Change in Resistance=$∆R=G_f×Strain$
Poisson^' sratio=$μ=\frac{G_f1}2$
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,
$$∆R=1 ×10^{3}×100=0.1\%$$
Therefore, change in resistance is 0.1%.
Further, Poisson’s ratio is given by,
$$μ=(G_f1)/2 \\
μ=(21)/2 \\
μ=0.5$$
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 1/2.