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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
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B
L2ρg2Y
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C
L2g2Yρ
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D
L2g3Yρ
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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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