Mechanical Properties of Solids

JEE Main level, single correct, medium

39 of 118

Question 39 of 118

Under the same load, wire A having length 5.0m5.0m and cross section 2.5×10−5m22.5\times {10}^{-5}{m}^{2} stretches uniformly by the same amount as another wire BB of length 6.0m6.0m and a cross section of 3.0×10−5m23.0\times {10}^{-5}{m}^{2} stretches. The ratio of the Young's modulus of wire AA to that of wire BB will be:

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Answer: (B) 1:11:1

Elongation in uniform wire under load is given by, Δl=FlSY,\Delta l=\dfrac{Fl}{SY}, where SS is area of cross-section.
As FF is same for both wire and Δl\Delta l is also same, we can write
ΔlF=lSY⇒lASAYA=lBSBYB\dfrac{\Delta l}{F}=\dfrac{l}{SY} \Rightarrow \dfrac{{l}_{A}}{{S}_{A}{Y}_{A}}=\dfrac{{l}_{B}}{{S}_{B}{Y}_{B}}
⇒5(2.5×10−5)×YA=6(3×10−5)×YB\Rightarrow \dfrac{5}{\left(2.5\times {10}^{-5}\right)\times {Y}_{A}}=\dfrac{6}{\left(3\times {10}^{-5}\right)\times {Y}_{B}}
⇒YAYB=1\Rightarrow \dfrac{{Y}_{A}}{{Y}_{B}}=1

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