Let us consider the given diagram when a small element of length $d x$ and of mass $d m$ from the length $L$ of wire at distance is considered at $x$ from the load $(x=0)$.


(a) Let $T(x)$ and $T(x+d x)$ are tensions on the two cross-sections a distance $d x$ apart, then $-T(x+d x)+T(x)=d m g=\mu g d x$ (where $\mu$ is the mass/length).
$$
(\because d m=\mu d x)
$$
$$
d T=\mu g d x[\because d T=T(x+d x)-T(x)]
$$
$\Rightarrow T(x)=\mu g x+C \quad$ (on integrating)
At $x=0, T(0)=M g \Rightarrow C=M g$
Downward force on $d x=w t$ below it $(d x)$
$$
T(x)=\mu g x+M g
$$
Let the length $d x$ at $x$ increase by $d r$, then $d m g$
Young's modulus $Y=\frac{\text { Stress }}{\text { Strain }}$
$$
\frac{T(x) / A}{d r / d x}=Y
$$
where $d r$ is increase in length of wire by $T(x)$.
or, $\frac{d r}{d x}=\frac{1}{Y A} T(x) \Rightarrow d r=\frac{1}{Y A}(\mu g x+M g) d x$
Integrating both sides,
$$
\begin{aligned}
&\Rightarrow \int_0^r d r=\frac{1}{Y A} \int_0^L(\mu g x+M g) d x \\
&\Rightarrow r=\frac{1}{Y A}\left[\frac{\mu g x^2}{2}+M g x\right]_0^L
\end{aligned}
$$
$r$ (change in length $L$ )
$$
=\frac{1}{Y A}\left[\frac{\mu g L^2}{2}+M g L\right] \quad(\because \mu L=m)
$$
( $m$ is the mass of the wire)
$$
\begin{aligned}
&=\frac{1 \times g L}{2 Y A}[\mu L+2 M]=\frac{g L}{2 Y(A)}[m+2 M] \\
&A=\pi \times\left(10^{-3}\right)^2 \mathrm{~m}^2 \\
&Y=200 \times 10^9 \mathrm{Nm}^{-2} \\
&m=\text { volume } \times \text { density } \\
&=\left[\pi \times\left(10^{-3}\right)^2 \times 10\right] \times 7860 \mathrm{~kg} \equiv 0.25 \mathrm{~kg} \\
&r=\frac{10 \times 10(0.25+2 \times 25)}{2 \times 10^{11} \times \pi \times 10^{-6}}=4 \times 10^{-3} \mathrm{~m} \\
&
\end{aligned}
$$
(b) Tension in the wire will be maximum at $x=L$
$$
\begin{aligned}
T=\mu g L+M g=&((\mu L) g+M g) \\
=(m+M) g \quad & {[\because m=\mu L] } \\
\text { The yield force }=& {[\text { Yield strength } Y] \text { area } } \\
&=250 \times 10^6 \times \pi \times\left(10^{-3}\right)^2 \\
&=250 \times \mu N
\end{aligned}
$$
At yield point (Maximum tension) $=$ Yield force
$$
\begin{aligned}
&(m+M) g=250 \times \pi \\
&\therefore[0.25+M] g=250 \times \pi
\end{aligned}
$$
Hence,
$$
\begin{aligned}
&M=\left[\frac{250 \times \pi}{10}-0.25\right]=\frac{250 \times 3.14}{10}-0.25 \\
&M=78.25 \mathrm{~kg}
\end{aligned}
$$