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. $\frac{1}{2}\frac{YAx^{2}}{gL^{2}}$
• B. $\frac{1}{2}\frac{YAL^{2}}{gx^{2}}$
• C. $\frac{YAx^{3}}{gL^{3}}$
• D. $\frac{YAL^{3}}{gx^{2}}$

Answer 1

Answer: (C)

$\frac{YAx^{3}}{gL^{3}}$

Answer 2

Let the tension in the string is T and for the equilibrium of mass m

$2T\sin\theta = mg$

$\Rightarrow T = \frac{mg}{2\sin\theta} = \frac{mgL}{2x}$ [As $\theta$ is small then $\sin\theta = \frac{x}{L}$]

Increment in the length $l = AC - AB = \sqrt{L^{2} + x^{2}} - L = (L^{2} + x^{2})^{1/2} - L$

$= L\left\lbrack \left( 1 + \frac{x^{2}}{L^{2}} \right)^{1/2} ⥂ - 1 \right\rbrack = L\left\lbrack 1 + \frac{1}{2}\frac{x^{2}}{L^{2}} - 1 \right\rbrack = \frac{x^{2}}{2L}$

As Young's modulus $Y = \frac{T}{A}\frac{L}{l}$

∴ $T = \frac{YAl}{L}$

Substituting the value of T and l in the above equation we get $\frac{mgL}{2x} = \frac{YA}{L}.\frac{x^{2}}{2L}\therefore m = \frac{YAx^{3}}{gL^{3}}$