A uniform rod of length L has a mass per unit | vedclass

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(B) Consider a small element of length $dx$ at a distance $x$ from the lower end of the rod.The weight of the portion of the rod below this element is

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$\frac{\lambda g L^2}{2 A Y}$

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(B) Consider a small element of length $dx$ at a distance $x$ from the lower end of the rod.The weight of the portion of the rod below this element is $W(x) = (\lambda x)g$.This weight acts as a tensile force $F = \lambda x g$ on the element $dx$.The elongation $d(\Delta L)$ in the element $dx$ is given by $d(\Delta L) = \frac{F dx}{A Y} = \frac{(\lambda x g) dx}{A Y}$.To find the total elongation $\Delta L$,we integrate from $x = 0$ to $x = L$:$\Delta L = \int_{0}^{L} \frac{\lambda g x}{A Y} dx = \frac{\lambda g}{A Y} \int_{0}^{L} x dx = \frac{\lambda g}{A Y} \left[ \frac{x^2}{2} \right]_{0}^{L} = \frac{\lambda g L^2}{2 A Y}$.

$A$ uniform rod of length $L$ has a mass per unit length $\lambda$ and area of cross-section $A$. If the Young's modulus of the rod is $Y$,then the elongation in the rod due to its own weight is:

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