A steel wire of diameter d, area of cross-section A and length 2L is clamped firmly at two points A and B which are 2L metre apart and in the same plane. A body of mass m is hung from the middle point of wire such that the middle point sags by x lower from original position. If Young’s modulus is Y then m is given by

Let the steel wire of length $2L$ be clamped between two fixed points $A$ and $B$ separated by a distance of $2L$. The midpoint of the wire is $C$, so $AC=CB=L$ initially.
When a mass $m$ is hung from the midpoint $C$, it sags vertically downwards by a distance $x$ to a new position $C^{\prime }$.

The new length of each half of the wire is given by:
$A{C}^{\prime }=B{C}^{\prime }=\sqrt{L^2+x^2}=L{\left(1+\frac{x^2}{L^2}\right)}^{1/2}$

Since the sag $x$ is very small compared to the length $L$ ($x\ll L$), we can use the binomial approximation:
$A{C}^{\prime }\approx L\left(1+\frac{x^2}{2L^2}\right)=L+\frac{x^2}{2L}$

The increase in length (elongation) of each half of the wire is:
$\Delta L=A{C}^{\prime }-L\approx \frac{x^2}{2L}$

The longitudinal strain in the wire is:
$\text{Strain}=\frac{\Delta L}{L}=\frac{x^2}{2L^2}$

Let $T$ be the tension developed in the wire. The longitudinal stress is:
$\text{Stress}=\frac{T}{A}$
where $A$ is the cross-sectional area of the wire.

Using Young's modulus $Y$:
$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{T/A}{x^2/2L^2}$

Solving for the tension $T$:
$T=\frac{YA{x}^2}{2L^2}$

At the point $C^{\prime }$, the mass $m$ is in equilibrium under the action of the downward gravitational force $mg$ and the upward components of the tension $T$ from both segments of the wire.
If $\theta$ is the angle made by each segment of the wire with the horizontal, then:
$2Tsin\theta =mg$

From the geometry of the triangle, $sin\theta =\frac{x}{\sqrt{L^2+x^2}}\approx \frac{x}{L}$ (since $x\ll L$).
Substituting $sin\theta$ into the equilibrium condition:
$2T\left(\frac{x}{L}\right)=mg$

Substitute the value of tension $T$ in the equation above:
$2\left(\frac{YA{x}^2}{2L^2}\right)\left(\frac{x}{L}\right)=mg$

Simplifying the equation gives:
$\frac{YA{x}^3}{L^3}=mg$

Thus, the mass $m$ is given by:
$m=\frac{YA{x}^3}{g{L}^3}$