Question

Consider a cylindrical nickel wire 2.1 mm in diameter and 3.2 × 104 mm long. Calculate its elongation when a load of 280 N is applied. Assume that the deformation is totally elastic and that the elastic modulus for nickel is 207 GPa (or 207 × 109 N/m2).

Answer 1

Answer:

Total elongation will be 0.012 m

Explanation:

We have given diameter of the cylinder = 2.1 mm

Length of wire $L=3.2\times 10^4mm$

So radius $r=\frac{d}{2}=\frac{2.1}{2}=1.05mm=1.05\times 10^{-3}m$

Load F = 280 N

Elastic modulus = 207 Gpa

Area of cross section $A=\pi r^2=3.14\times (1.05\times 10^{-3})^2=3.461\times 10^{-6}m^2$

We know that elongation in wire is given by $\delta =\frac{FL}{AE}$, here F is load, L is length, A is area and E is elastic modulus

So $\delta =\frac{FL}{AE}=\frac{280\times 32}{3.461\times 10^{-6}\times 207\times 10^9}=0.012m$