JEE MAIN - Physics (2022 - 26th July Evening Shift - No. 23)
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
25
Explanation
$${{{F \over A}} \over {{{\Delta L} \over L}}} = Y$$
$$\Delta L = {{FL} \over {AY}} = {{{T_{avg}}L} \over {AY}} = {{MgL} \over {2AY}}$$
$$ = {{20 \times 10 \times 20} \over {2 \times 0.4 \times 2 \times {{10}^{11}}}} = {{4 \times {{10}^3} \times {{10}^{ - 11}}} \over {4 \times 0.4}}$$
$$ = 2.5 \times {10^{ - 8}} = 25 \times {10^{ - 9}}$$
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