A uniform cylindrical rod of length L and cross-sectional area A having Young's modulus Y is acted upon by forces as shown in figure. [CG PMT 2015]
The elongation produced in the rod is

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

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

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

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

answer is C.

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Detailed Solution

Elongation produced, if the $\frac{2}{3}$ part of length L of rod is stretched by a force of 3F, is given by

$Δ L_{1} = \frac{F L_{1}}{A Y} = \frac{3 F \times \frac{2}{3} L}{A Y} = 2 \frac{F L}{A Y}$

Elongation produced, if the remaining (1/3) part of length L of rod is stretched by a force of 2F, is given by

$Δ L_{2} = \frac{F_{2} L_{2}}{A Y} = \frac{2 F \times \frac{1}{3} L}{A Y} = \frac{2}{3} \frac{F L}{A Y}$

Total elongation produced in the rod,

$Δ L = Δ L_{1} + Δ L_{2} = 2 \frac{F L}{A Y} + \frac{2}{3} \frac{F L}{A Y}$

$∴ Δ L = \frac{8 F L}{3 A Y}$

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