Question

A wire of cross section $A$ is stretched horizontally between two clamps located $2lm$ apart. A weight $Wkg$ is suspended from the mid-point of the wire. If the mid-point sags vertically through a distance $x<<l$, the strain produced is

• A. $\frac{2x^2}{l^2}$
• B. $\frac{x^2}{l^2}$
• C. $\frac{x^2}{2l^2}$
• D. None of these

Answer 1

The correct option is C $\frac{x^2}{2l^2}$



$\Delta l$ = (AC + BC) - AB
Now, AC = BC = $\sqrt{l^2+x^2}$
$\Delta l$ = $2\sqrt{l^2+x^2}$ - $2l$
$\Delta l$ = $2l[\sqrt{1+\frac{x^2}{l^2}}-1]$.....(1)
Now using expansion in (1).....
$(1+x{)}^n=1+nx.....$
$\Delta l=\frac{x^2}{l}$
Strain = $\frac{\Delta l}{2l}=\frac{x^2}{2l^2}$