$(a)$ $A$ steel wire of mass $\mu$ 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 $\ll$ longitudinal strains,find the extension in the length of the wire. The density of steel is $7860\,kg/m^3$ and Young's modulus $Y = 2 \times 10^{11}\,N/m^2$.
$(b)$ If the yield strength of steel is $2.5 \times 10^8\,N/m^2$,what is the maximum weight that can be hung at the lower end of the wire?

(N/A) Let $L = 10\,m$,$r = 0.1\,cm = 10^{-3}\,m$,$M = 25\,kg$,$\rho = 7860\,kg/m^3$,$Y = 2 \times 10^{11}\,N/m^2$.
The cross-sectional area $A = \pi r^2 = \pi (10^{-3})^2 = \pi \times 10^{-6}\,m^2$.
The mass per unit length $\mu = \rho A = 7860 \times \pi \times 10^{-6} \approx 0.0247\,kg/m$.
The tension at a distance $x$ from the lower end is $T(x) = Mg + \mu gx$.
The extension $d\Delta L$ for an element $dx$ is $d\Delta L = \frac{T(x)dx}{AY} = \frac{(Mg + \mu gx)dx}{AY}$.
Integrating from $x=0$ to $x=L$:
$\Delta L = \int_0^L \frac{(Mg + \mu gx)dx}{AY} = \frac{1}{AY} [MgL + \frac{1}{2}\mu gL^2] = \frac{gL}{AY} (M + \frac{1}{2}\mu L)$.
Substituting values: $\Delta L = \frac{9.8 \times 10}{(\pi \times 10^{-6})(2 \times 10^{11})} (25 + 0.5 \times 0.0247 \times 10) = \frac{98}{2\pi \times 10^5} (25.1235) \approx 3.92 \times 10^{-3}\,m = 3.92\,mm$.
$(b)$ The maximum stress is $\sigma_{max} = \frac{T_{max}}{A} = \text{Yield Strength} = 2.5 \times 10^8\,N/m^2$.
The tension at the top is $T_{max} = Mg + \mu gL$.
$Mg + \mu gL = \sigma_{max} A$.
$Mg = \sigma_{max} A - \mu gL = (2.5 \times 10^8)(\pi \times 10^{-6}) - (0.0247)(9.8)(10) = 785.4 - 2.42 = 782.98\,N$.
The maximum mass $M = \frac{782.98}{9.8} \approx 79.9\,kg$.