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A uniform cylinder rod 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}{\text { 8AY }}


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

\Delta l=\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}

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Riya

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