A uniform heavy rod of weight 10kg ms^-2, cross-sectional area 100 cm²f and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 × 10¹¹ Nm^-2. Neglecting the lateral concentration, find the elongation of rod due to its own weight:

Correct Option - 2


Detailed Solution:

W = 10 kg m/s²

A = 100 cm²

^{$\mathcal{l}=20cm$}

Y = 2 × 10¹¹ N/m²

^{$Y=\frac{F\mathcal{l}}{A\Delta \mathcal{l}}\Rightarrow \Delta \mathcal{l}=\frac{F\mathcal{l}}{AY}$}

For elemental mass of length, dx

Change in its length

$\Delta \mathcal{l}=5\times 10^{-10}m$

 

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