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QJEEM22Ph4109: JEE Main 2022 Physics - Class 11 JEEM22Ph4109 Solution: A uniform heavy rod of mass , cross sectional area and length…

A uniform heavy rod of mass 20 kg20 \mathrm{~kg} , cross sectional area 0.4 m20.4 \mathrm{~m}^{2} and length 20 m20 \mathrm{~m} is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is x×10−9 mx \times 10^{-9} \mathrm{~m} . The value of xx is _______________.

(Given, young modulus Y = 2 ×\times 1011 Nm −- 2 and g = 10 ms −- 2)

Step-by-Step Verified Solution

FAΔLL=Y{{{F \over A}} \over {{{\Delta L} \over L}}} = Y

ΔL=FLAY=TavgLAY=MgL2AY\Delta L = {{FL} \over {AY}} = {{{T_{avg}}L} \over {AY}} = {{MgL} \over {2AY}}

=20×10×202×0.4×2×1011=4×103×10−114×0.4= {{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}}

=2.5×10−8=25×10−9= 2.5 \times {10^{ - 8}} = 25 \times {10^{ - 9}}

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