First slide

Elastic Modulie

Question

A brass rod of cross-sectional area I ${cm}^{2}$ 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 = l0 m/ $\sec^{2}$ , then increase in the energy of the rod will be

Moderate

A

$10^{- 5} J$
B

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

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

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

Solution

$U = \frac{1}{2} \times \frac{{(stress)}^{2}}{Y} \times volume = \frac{1}{2} \times \frac{F^{2} \times A \times L}{A^{2} \times Y}$
= $\frac{1}{2} \times \frac{F^{2} L}{AY} = \frac{1}{2} \times \frac{{(50)}^{2} \times 0.2}{1 \times 10^{- 4} \times 1 \times 10^{11}} = 2.5 \times 10^{- 5} J$

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