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

Energy =12Fl=12×F×FLAY=12×F2LAY =12×(50)2×20×10−22×10−4×1.4×1011=8.57×10−6J

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

Energy =12Fl=12×F×FLAY=12×F2LAY =12×(50)2×20×10−22×10−4×1.4×1011=8.57×10−6J

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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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8.57 x 10^-6

22.5 x 10^-4

9.8 x 10^-5

45.0 x 10^-5

answer is A.

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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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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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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