Question

# A uniform heavy rod of length $$L$$ and area of cross-section area $$A$$ is hanging from a fixed support. If Young's modulus of the material of the rod is $$Y$$, then the increase in the length of the rod is ($$\rho$$ is a density of the material of the rod) :

A
L2Y2ρg
B
L2ρg2Y
C
L2g2Yρ
D
L2g3Yρ

Solution

## The correct option is D $$\dfrac{L^{2}\rho g}{2Y}$$Consider an element dx at a distance x from the top. Let the wt. of rod is W.The force acting on this because of a part which is below dx is given as $$\dfrac{(L-x)W}{L}$$The elongation in element $$dx$$ is $$=(\dfrac{L-x}{L})\dfrac{Wdx}{AY}$$$$=\dfrac{(L-x)Wdx}{LAY}$$Total elongation, $$\Delta L=\int_{O}^{L}\dfrac{(L-x)Wdx}{LAY}$$$$=\dfrac{WL}{2AY}$$$$W=AL\rho g$$$$\Delta L=\dfrac{AL\rho gL}{2AY}$$$$=\dfrac{L^{2}\rho g}{2Y}$$Physics

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