A) \[YxA/2L\]
B) \[Y{{x}^{2}}A/L\]
C) \[Y{{x}^{2}}A/2L\]
D) \[2Y{{x}^{2}}A/L\]
Correct Answer: C
Solution :
Work done in stretching a wire \[w=\frac{1}{2}Fx\] So, Young's modulus \[Y=\frac{F/A}{x/L}=\frac{FL}{Ax}\] or \[F=\frac{YAx}{L}\] So, work done \[W=\frac{1}{2}\frac{YAx}{L}\times x\] \[=\frac{YA{{x}^{2}}}{2L}\]You need to login to perform this action.
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