First slide

Stress and strain

Question

A wire of length 2 m is made from 10 cm³ of copper. A force F is applied so that its length increases by 2 mm. Another wire of length 8 m is made from the same volume of copper. If the force F is applied to it, its length will increase by

Moderate

A

0.8 cm
B

1.6 cm
C

2.4 cm
D

3.2 cm

Solution

$I = \frac{FL}{AY} = \frac{{FL}^{2}}{(AL) Y} = \frac{{FL}^{2}}{VY}$
$∴ I \propto L^{2}$ If volume of the wire remains constant
$\begin{array}{l} \frac{I_{2}}{I_{1}} = {(\frac{L_{2}}{L_{1}})}^{2} = {(\frac{8}{2})}^{2} = 16 \\ ∴ I_{2} = 16 \times I_{1} = 16 \times 2 = 32 mm = 3 .2 cm . \end{array}$

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