A) \[\frac{{{M}^{2}}{{g}^{2}}L}{A({{Y}_{1}}+{{Y}_{2}})}\]
B) \[\frac{2{{M}^{2}}{{g}^{2}}L}{A({{Y}_{1}}+{{Y}_{2}})}\]
C) \[\frac{2{{M}^{2}}{{g}^{2}}L}{A({{Y}_{1}}+{{Y}_{2}})}\]
D) \[\frac{{{M}^{2}}{{g}^{2}}L({{Y}_{1}}+{{Y}_{2}})}{2A{{Y}_{1}}+{{Y}_{2}}}\]
Correct Answer: A
Solution :
Equivalent spring constant a wire is given by \[K=\frac{YA}{L}\] \[{{K}_{aq}}\,={{K}_{1}}+{{K}_{2}}\] \[\frac{Y(2A)}{L}=\frac{{{Y}_{1}}A}{L}+\frac{{{Y}_{2}}A}{L}\] or \[Y=\frac{{{Y}_{1}}+{{Y}_{2}}}{2}\] the system can be replaced as if suspended from a single wire of Young?s modulus Y. \[U=\frac{1}{2}\,\times \] strees \[\times \] strain \[\times \] volume \[=\frac{1}{2}\,\times \,\frac{{{(stree)}^{2}}}{Y}\times AL\] \[=\frac{1}{2}\,{{\left[ \frac{Mg}{A} \right]}^{2}}\,\times \frac{1}{Y}\,\times AL\] \[=\frac{{{M}^{2}}{{g}^{2}}L}{A({{Y}_{1}}+{{Y}_{2}})}\,\,\,\,\,\,\,(\because \,\,2Y={{Y}_{1}}+{{Y}_{2}})\]You need to login to perform this action.
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