A thick rope of density rho and length L is h | vedclass

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(B) Consider a small element of length $dx$ at a distance $x$ from the free end of the rope.The weight of the portion of the rope below this element i

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$(1 / 2) \rho g L^2 / Y$

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(B) Consider a small element of length $dx$ at a distance $x$ from the free end of the rope.The weight of the portion of the rope below this element is $F = (A x \rho) g$,where $A$ is the cross-sectional area.The extension $d(\Delta L)$ in this element $dx$ is given by $d(\Delta L) = \frac{F dx}{AY} = \frac{(A x \rho g) dx}{AY} = \frac{\rho g}{Y} x dx$.To find the total increase in length $\Delta L$,we integrate this expression from $x = 0$ to $x = L$:$\Delta L = \int_{0}^{L} \frac{\rho g}{Y} x dx = \frac{\rho g}{Y} \left[ \frac{x^2}{2} \right]_{0}^{L} = \frac{1}{2} \frac{\rho g L^2}{Y}$.

$A$ thick rope of density $\rho$ and length $L$ is hung from a rigid support. The Young's modulus of the material of the rope is $Y$. The increase in length of the rope due to its own weight is

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