A) \[{{F}_{A}}\times \frac{L}{2AY}\]
B) \[{{F}_{B}}\times \frac{L}{2AY}\]
C) \[\frac{({{F}_{A}}+{{F}_{B}})L}{2AY}\]
D) \[\frac{({{F}_{A}}-{{F}_{B}})L}{2AY}\]
Correct Answer: C
Solution :
Tension at\[x\], \[T={{F}_{A}}-[{{F}_{A}}-{{F}_{B}}]\frac{X}{L}={{F}_{A}}\left( 1-\frac{X}{L} \right)+{{F}_{B}}\left( \frac{X}{L} \right)\] Change in length \[\Delta (dx)=\frac{Tdx}{AY}=\left[ {{F}_{A}}\left( 1-\frac{X}{L} \right)+{{F}_{B}}\left( \frac{X}{L} \right) \right]\frac{dx}{AY}\] Total change in length \[=\frac{1}{AY}\left[ \int_{0}^{L}{{{F}_{A}}(1-x/L)dx+\int_{0}^{L}{{{F}_{B}}\left( \frac{X}{L} \right)dx}} \right]\] \[=\frac{1}{AY}\left[ {{F}_{A}}\left( L-\frac{L}{2} \right)+{{F}_{B}}\times \frac{L}{2} \right]\] \[=\frac{1}{AY}[{{F}_{A}}+{{F}_{B}}]\frac{L}{2}=\frac{L}{2AY}({{F}_{A}}+{{F}_{B}})\]You need to login to perform this action.
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