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A uniform slender rod of length 2L, cross sectional area A and Young’s modulus Y is acted upon the forces shown in the figure.
The elongation of the rod is

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$\frac{5 FL}{3 AY}$

$\frac{3 FL}{AY}$

$\frac{FL}{AY}$

$\frac{11 FL}{3 AY}$

answer is B.

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Detailed Solution

$\begin{array}{l} e = e_{1} + e_{2} \\ = \frac{F}{A} X \frac{(\frac{L}{3})}{Y} + \frac{2 F}{A} X \frac{(2 L - \frac{L}{3})}{Y} \\ = \frac{FL}{3 AY} + \frac{10 FL}{3 AY} = \frac{11 FL}{3 AY} \end{array}$

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