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A rod of mass M and cross-section area Ais subjected to two forces f and 2 f at both the ends as shown in the figure. If Young’s modulus of its material is y and its length is L , then the total elongation of the rod is $\frac{αfL}{βAy}$ (here $α$ and $β$ are co-primes). Find $(α + β) .$

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Detailed Solution

Tension in the rod at a distance 𝑥 from its
right edge is

$T = F (2 - \frac{x}{L})$

Therefore,
Net extension in the rod,

$δ = \int_{0}^{L} \frac{T}{yA} dx = \frac{3 FL}{2 yA}$

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