The length of a rod is 20 cm and area of cross-section 2 cm² . The Young’s modulus of the material of wire is 1.4 x 10¹¹N / m² . If the rod is compressed by 5 kg-wt along its length, then increase in the energy of the rod in joules will be

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22.5 x 10^-4

8.57 x 10^-6

45.0 x 10^-5

9.8 x 10^-5

answer is A.

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Detailed Solution

Energy
$= \frac{1}{2} Fl = \frac{1}{2} \times F \times (\frac{FL}{AY}) = \frac{1}{2} \times \frac{F^{2} L}{AY} = \frac{1}{2} \times \frac{(50)^{2} \times 20 \times 10^{- 2}}{2 \times 10^{- 4} \times 1 .4 \times 10^{11}} = 8 .57 \times 10^{- 6} J$

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