A uniform heavy rod of mass $20 \mathrm{~kg}$, cross sectional area $0.4 \mathrm{~m}^{2}$ and length $20 \mathrm{~m}$ is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is $x \times 10^{-9} \mathrm{~m}$. The value of $x$ is _______________.
(Given, young modulus Y = 2 $\times$ 1011 Nm$-$2 and g = 10 ms$-$2)
Answer (integer)
25
Solution
<p>${{{F \over A}} \over {{{\Delta L} \over L}}} = Y$</p>
<p>$\Delta L = {{FL} \over {AY}} = {{{T_{avg}}L} \over {AY}} = {{MgL} \over {2AY}}$</p>
<p>$$ = {{20 \times 10 \times 20} \over {2 \times 0.4 \times 2 \times {{10}^{11}}}} = {{4 \times {{10}^3} \times {{10}^{ - 11}}} \over {4 \times 0.4}}$$</p>
<p>$= 2.5 \times {10^{ - 8}} = 25 \times {10^{ - 9}}$</p>
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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