The elastic potential energy of a strained body is
- A$\frac{1}{2} \times \text{stress} \times \text{strain}$
- B$\frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume of the body}$
- C$\frac{1}{2} \times \text{stress} \times \text{strain} \times \text{area of the body}$
- D$\text{stress} \times \text{strain} \times \text{volume of the body}$
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Similar Questions
The work to be done to produce a strain of $10^{-3}$ in a steel wire of mass $2.96 \ kg$ and density $7.4 \ g \ cm^{-3}$ is (Young's modulus of steel $= 2 \times 10^{11} \ Nm^{-2}$)
Two wires of the same diameter and the same material have lengths $l$ and $2l$. If the same force $F$ is applied to each,what is the ratio of the work done in stretching the two wires?
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View Solution$A$ steel uniform rod of length $2L$,cross-sectional area $A$,and mass $M$ is set rotating in a horizontal plane about an axis passing through its center with angular velocity $\omega$. If $Y$ is the Young's modulus for steel,find the total extension in the length of the rod.
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View SolutionThe elastic potential energy stored in a steel wire of length $20 \, m$ stretched by $2 \, cm$ is $80 \, J$. The cross-sectional area of the wire is $......... \, mm^2$ (Given,$Y = 2.0 \times 10^{11} \, N/m^2$).
What is the work done in stretching a uniform metal wire of length $2 \ m$ to $2.004 \ m$ with an area of cross-section $10^{-6} \ m^2$ (in $J$)? [Young's modulus of the wire = $2 \times 10^{11} \ N/m^2$]
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