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Thermal Expansion

A uniform rod...

Question

A uniform rod of length L, cress-sectional area A and Young's modulus Y is stretched by applying two force $F_{1}$ and $F_{2}$ as shown in figure. The elongation in the rod will be ( $F_{1}$ > $F_{2}$ )

A

\frac{(F_{1} + F_{2}) L}{A Y}

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B

\frac{(F_{1} - F_{2}) L}{2 A Y}

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C

\frac{(F_{1} + F_{2}) L}{2 A Y}

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D

\frac{F_{1} L}{2 A Y}

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Solution

The correct option is C $\frac{(F_{1} + F_{2}) L}{2 A Y}$
$\begin{matrix} Abplying NLM: 2 law, F_{1} - T = ρ A X . a T - F_{2} = P A \cdot (L - x) a \end{matrix}$
$\frac{(F_{1} - T)}{T - F_{2}} = \frac{x}{L - x}$
$T = F_{1} - \frac{x (F_{1} - F_{2})}{L}$
$Elongation Δ l = \int_{0}^{L} \frac{T d x}{A y} f_{0}^{L} \frac{[F_{1} - \frac{x}{L} (F_{1} - F_{2})] d x}{A Y} = \frac{(F_{1} + F_{2})}{2 A Y}$

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