A wire of cross-section A is stretched horizontally between two clamps located 2 l metre apart. A weight W kg is suspended from the mid point of the wire. If the mid point sags vertically through a distance x < l the value of stress is:

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$\frac{WI}{2 xA}$

$\frac{WI}{4 xA}$

$\frac{2 WI}{lA}$

$\frac{4 xW}{lA}$

answer is .

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Detailed Solution

From the figure it is clear that 2Tcos $θ$ = W

$\Rightarrow T = \frac{W}{2 cosθ}$

Now $\cos θ = \frac{x}{\sqrt{l^{2} + x^{2}}} = \frac{x}{l [1 + \frac{x^{2}}{2 l^{2}}]}$

If $\frac{x^{2}}{2 l^{2}} is neglected being small, then cosθ = \frac{x}{1}$

T = $\frac{W}{2 x}$

Stress = $\frac{T}{A} = \frac{(\frac{Wl}{2 x})}{A} \frac{Wl}{\frac{2 x}{A}} = \frac{Wl}{2 Ax}$

Correct choice is (a).

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