A) \[\frac{YA{{x}^{2}}}{L}\]
B) \[\frac{YA{{x}^{2}}}{2L}\]
C) \[\frac{YA{{L}^{2}}}{x}\]
D) \[\frac{YA{{L}^{2}}}{2x}\]
Correct Answer: B
Solution :
In stretching a wire work is done against internal restoring forces. This work is stored in the wire as elastic potential energy or strain energy. If a force F acts along the length L of the wire of cross-section A and stretches it by x, then \[Y=\frac{stress}{strain~}=\frac{F/A}{x/L}=\frac{FL}{Ax}\] \[\Rightarrow \] \[F=\frac{YAx}{L}\] So, the work done for an additional small increase dx in length, \[dW=Fdx=\frac{YA}{L}xdx\] Hence, the total work done in increasing the length by \[l\], \[W=\int_{0}^{x}{dW}=\int_{0}^{x}{Fdx}=\int_{0}^{x}{\frac{YA}{L}xdx=\frac{1}{2}\frac{YA}{L}{{x}^{2}}}\] This work done is stored in the wire. \[\therefore \] Energy stored in wire \[U=\frac{1}{2}\frac{YA{{x}^{2}}}{L}\]
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