Question

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

• A. $\frac{1}{2}\frac{YA{x}^2}{g{L}^2}$
• B. $\frac{1}{2}\frac{YA{L}^2}{g{x}^2}$
• C. $\frac{YA{x}^3}{g{L}^3}$
• D. $\frac{YA{L}^3}{g{x}^2}$

Answer 1

The correct option is C $\frac{YA{x}^3}{g{L}^3}$

Let the tension in the string is T and for the equilibrium of mass m ​
$2T\sin \theta =mg\Rightarrow T=\frac{mg}{2\sin \theta }=\frac{mgL}{2x}$ $[\text{As}\theta \text{is small then}\sin \theta =\frac{x}{L}]$
Increment in the length $l=AC-AB=\sqrt{L^2+x^2}-L=(L^1+x^2{)}^{1/2}-L$
$=L\left[(1+{\frac{x^2}{L^2}}^{1/2}-1)\right]=L[1+\frac{1}{2}\frac{x^2}{L^2}-1]=\frac{x^2}{2L}$
As Young's modulus $Y=\frac{T}{A}\frac{L}{l}$ $\therefore T=\frac{YAl}{L}$
Substituting the value of T and l in the above equation we get $\frac{mgL}{2x}=\frac{YA}{L}.\frac{x^2}{2L}$
$\therefore$ $m=\frac{YA{x}^3}{g{L}^3}$