Question 48 of 118

A steel wire of length $3.2m\left({Y}_{S}=2.0\times {10}^{11}N{m}^{-2}\right)$ and a copper wire of length $4.4m\left({Y}_{C}=1.1\times {10}^{11}N{m}^{-2}\right)$ , both of radius $1.4mm$ are connected end to end. When stretched by a load, the net elongation is found to be $1.4mm$. The load applied, in Newton, will be: (Given $\pi =\dfrac{22}{7}$)

A$360$B$180$C$1080$D$154$

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Answer: (D) $154$

When two wires are connected end to end, the tension in them will be same.
Change in length using Hooke's law can be found as,
$Y=\dfrac{Fl}{A∆l}\Rightarrow \Delta l=\dfrac{Fl}{AY}$
Total change in length can be written as,
$\Delta l=\Delta {l}_{1}+\Delta {l}_{2}$
$\Rightarrow \Delta l=\dfrac{F{l}_{1}}{{A}_{1}{Y}_{1}}+\dfrac{F{l}_{2}}{{A}_{2}{Y}_{2}}$
$\Rightarrow F=\dfrac{\Delta l}{\dfrac{{l}_{1}}{{A}_{1}{Y}_{1}}+\dfrac{{l}_{2}}{{A}_{2}{Y}_{2}}}=\dfrac{1.4\times {10}^{-3}}{\dfrac{3.2}{\pi {\left(1.4\times {10}^{-3}\right)}^{2}\times 2.0\times {10}^{11}}+\dfrac{4.4}{\pi {\left(1.4\times {10}^{-3}\right)}^{2}\times 1.1\times {10}^{11}}} \Rightarrow F=1.54\times {10}^{2}=154N$