$(a)$ Consider an element $d x$ at a distance $x$ from the load $(x=0)$. If $\mathrm{T}(x)$ and $\mathrm{T}(x+d x)$ are tensions on the two cross sections a distance $d x$ apart then,

$\therefore \mathrm{T}(x+d x)-\mathrm{T}(x)=g d m$

$\quad=\mu g d x$

(where $d m=$ mass of wire $d x$ and $\mu=$ is the mass per unit length $=\frac{d m}{d x}$ )

$\therefore d \mathrm{~T}=\mu g d x \quad[\because \mathrm{T}(x+d x)-\mathrm{T}(x)=d \mathrm{~T}]$

Integrating on both side,

$\therefore \mathrm{T}(x)=\mu g x+\mathrm{C}$

but at $x=0$, tension $\mathrm{T}(0)=0+\mathrm{C}$

$\therefore \mathrm{Mg}=\mathrm{C}$ where $\mathrm{M}$ is suspended mass

$\therefore \mathrm{T}(x)=\mu g x+\mathrm{Mg}$

If element $d x$, increases by length $d r$ then strain $=\frac{d r}{d x}$

Young modulus $\mathrm{Y}=\frac{\frac{\mathrm{T}(x)}{\mathrm{A}}}{\frac{d r}{d x}}$

$\frac{d r}{d x}=\frac{\mathrm{T}(x)}{\mathrm{YA}}$