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 1.4×1011 N/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 = 8.92×10-6 J

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 1.4×1011 N/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 = 8.92×10-6 J

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The length of a rod is 20 cm and area of cross-section 2 ${cm}^{2}$ . The Young's modulus of the material of wire is $1.4 \times 10^{11} N / m^{2}$ . 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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$8.92 \times 10^{- 6}$

$22.5 \times 10^{- 4}$

$9.8 \times 10^{- 5}$

$45.0 \times 10^{- 5}$

answer is A.

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Energy = $\frac{1}{2} Fl = \frac{1}{2} \times F \times (\frac{FL}{AY}) = \frac{1}{2} \times \frac{F^{2} L}{AY}$

= $\frac{1}{2} \times \frac{{(50)}^{2} \times 20 \times 10^{- 2}}{2 \times 10^{- 4} \times 1.4 \times 10^{11}}$ = $8.92 \times 10^{- 6} J$

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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 1.4×1011 N/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 ${cm}^{2}$ . The Young's modulus of the material of wire is $1.4 \times 10^{11} N / m^{2}$ . If the rod is compressed by 5 kg-wt along its length, then increase in the energy of the rod (in joules) will be