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Q.

A brass rod of cross-sectional area 1 cm² and length 0.2 m is compressed lengthwise by a weight of 5 kg. If Young's modulus of elasticity of brass is $1 \times 10^{11} N / m^{2}$ and g = 10 m $s^{- 2}$ then increase in the energy of the rod will be

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a

$10^{- 5} J$

b

$5 \times 10^{- 5} J$

c

$2.5 \times 10^{- 5} J$

d

$2.5 \times 10^{- 4} J$

answer is B.

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Detailed Solution

$\begin{matrix} U = \frac{1}{2} \times \frac{(Stress)^{2}}{Y} \times Volume \\ = \frac{1}{2} \times \frac{F^{2} \times A \times L}{A^{2} \times Y} \\ = \frac{1}{2} \times \frac{F^{2} L}{AY} = \frac{1}{2} \times \frac{(50)^{2} \times 0 .2}{1 \times 10^{- 4} \times 1 \times 10^{11}} = 2 .5 \times 10^{- 5} J \end{matrix}$

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