Ncert Exercises & Solutions

A mild steel wire of length $2L$ and cross-sectional area $A$ is stretched, well within the elastic limit, horizontally between two pillars (figure). A mass $m$ is suspended from the mid-point of the wire. Strain in the wire is:

1.	$\dfrac{x^2}{2 L^2}$	2.	$\dfrac{x}{\mathrm{~L}}$
3.	$\dfrac{x^2}{L}$	4.	$\dfrac{x^2}{2L}$

(1) Hint: Use Pythagoras theorem to find the change in length.

Step 1: Find the change in length.

Consider the diagram below

Hence, change in length

$∆ L = BO + OC - (BD + DC)$

$= 2 BO - 2 BD (∴ BO = OC, BD = DC)$

$= 2 [BO - BD]$

$= 2 [{(x^{2} + L^{2})}^{\frac{1}{2}} - L]$

$= 2 L [{(1 + \frac{x^{2}}{L^{2}})}^{\frac{1}{2}} - 1]$

$= 2 L [(1 + \frac{x^{2}}{2 L^{2}}) - 1] = \frac{x^{2}}{L}$

Step 2: Find the strain.

$Strain = \frac{∆ L}{L} = \frac{x^{2}}{2 L^{2}}$

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