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 A uniform cylinder red of length L, cross-sectional area A and Young's modulus Y is acted upon by the forces shown in the figure.

The elongation of the rod is-

Option: 1

\frac{3 F L}{5 A Y}


Option: 2

\frac{2 F L}{5 A Y}


Option: 3

\frac{3 F L}{8 A Y}


Option: 4

\frac{8F L}{3 A Y}


Answers (1)

best_answer

The free body diagrams of two parts are-


Both parts are stretched. Therefore, total elongation -
\begin{aligned} & \Delta e=\Delta l_1+\Delta l_2 \\ & \Delta l=\frac{3 F\left(\frac{2 L}{3}\right)}{A Y}+\frac{2 F\left(\frac{L}{3}\right)}{A Y} \\ & \Delta l=\frac{8 F L}{3 A Y} \end{aligned}
 

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Rishabh

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