Ncert Exercises & Solutions

A mild steel wire of length \(\mathrm{2L}\) 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. \( \frac{x^2}{2 L^2} \)
2. \(\frac{x}{\mathrm{~L}} \)
3. \(\frac{x^2}{L}\)
4. \(\frac{x^2}{2L}\)

(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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