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

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8FL/3AY

3FL/5AY

3FL/8AY

2FL/5AY

answer is D.

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Detailed Solution

$L - \frac{L}{3} = \frac{2 L}{3}, Y = \frac{f}{A} \frac{l}{e}; e = \frac{f l}{AY}$

$e = e_{1} + e_{2}; e = \frac{2 F . \frac{L}{3}}{A . Y} + \frac{3 F . \frac{2 L}{3}}{A . Y} = \frac{8 F . L}{3 A . Y}$

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