A uniform rod of mass M and length L is hanging from point P as shown in the figure. Find the elongation in its length due to the self weight of the rod. The Young's modulus of elasticity of the rod is Y and cross-sectional area of the rod is A.
A
MgL3AY
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B
2MgL3AY
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C
2MgLAY
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D
MgL2AY
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Solution
The correct option is DMgL2AY Since mass M is uniformly distributed along its length.
So, Linear mass density (λ) of the rod =ML
Consider a small section dx of the rod at a distance x from Q. The force acting at the section dx is, W=MgLx=λgx
From Hooke's law, elongation in section dx will be dL=WAYdx=λgxAYdx
So, Total elongation in the rod can be obtained by integrating the above expression from x=0 to x=L