First slide

Elastic behaviour of solids

Question

The area of cross-section of a wire is $10^{- 5} m^{2}$ when its length is increased by 0.2% a tension of 250N is produced. The Young's modulus of the wire will be ........

Moderate

A

$9.5 \times 10^{9} N m^{- 2}$
B

$11.5 \times 10^{9} N m^{- 2}$
C

$12.5 \times 10^{9} N m^{- 2}$
D

$10.5 \times 10^{9} N m^{- 2}$

Solution

$y = \frac{F L}{A Δ L} = \frac{F / A}{A L / L}$
$y = \frac{250 / 10^{- 5}}{0.2 / 100} = \frac{250}{0.2} \times \frac{100}{10^{- 5}} = 12.5 \times 10^{9}$

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