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Q.

An Aluminium wire of length one meter and diameter 2 mm is stretched to increase its length by $\frac{10}{π}$ cm with a force of 10 N. Young’s modulus of the material of the wire is

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a

10⁹ Nm^-2

b

10⁸ Nm^-2

c

10¹⁰ Nm^-2

d

10¹¹ Nm^-2

answer is D.

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Detailed Solution

Young's Modulus of the wire is

$Y = \frac{F / A}{∆ L / L} = \frac{\frac{10}{(π \times {(10^{- 3})}^{2})}}{\frac{(\frac{10}{π} \times 10^{- 2})}{1}} = 10^{8} {Nm}^{- 2}$

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