**Question:**

**(a) A steel wire of mass μ per unit length with a circular cross-section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs **

**from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find **

**the extension in the length of the wire. The density of steel is 7860 kg/m3.**

**(b) If the yield strength of steel is 2.5 × 108 N/m2, what is the maximum weight that can be hung at the lower end of the wire?**

**Solution:**

Let dx be the small element

Let dm be the mass

Let L be the length of the wire

Let x be the distance from the end where it is hung

Let μ be the mass per unit length

Therefore,

dm = μ.dx

r = 0.1 cm

A = π(10)-3

M = 25 kg

L = 10 m

a) The downward force is used for calculating the mass, m = 0.25 kg

b) The maximum weight that can be hung at the lower end of the wire is calculated using the tension in the wire which is the maximum at x = L

Therefore, the mass is = 78.25 kg