question_answer

In the shown figure, length of the rod is\[L\], area of cross-section\[A\], Young's modulus of the material of the rod is\[Y\]. Then, \[B\] and \[A\] is subjected to a tensile force \[{{F}_{A}}\] while force applied at end\[B\], \[{{F}_{B}}\] is lesser than\[{{F}_{A}}\]. Total change in length of the rod will be

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 \[=\frac{1}{AY}\left[ \int_{0}^{L}{{{F}_{A}}(1-x/L)dx+\int_{0}^{L}{{{F}_{B}}\left( \frac{X}{L} \right)dx}} \right]\] \[=\frac{1}{AY}\left[ {{F}_{A}}\left( L-\frac{L}{2} \right)+{{F}_{B}}\times \frac{L}{2} \right]\] \[=\frac{1}{AY}[{{F}_{A}}+{{F}_{B}}]\frac{L}{2}=\frac{L}{2AY}({{F}_{A}}+{{F}_{B}})\]