A steel wire of length $3.2 m \left( Y _{ S }=2.0 \times 10^{11}\,Nm ^{-2}\right)$ and a copper wire of length $4.4\,M$ $\left( Y _{ C }=1.1 \times 10^{11}\,Nm ^{-2}\right)$, both of radius $1.4\,mm$ are connected end to end. When stretched by a load, the net elongation is found to be $1.4\,mm$. The load applied, in Newton, will be. (Given $\pi=\frac{22}{7}$)
A$360$
B$180$
C$1080$
D$154$
JEE MAIN 2022, Medium
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D$154$
d $\Delta \ell_{1}+\Delta \ell_{2}=\Delta \ell$
$\frac{ F \ell_{1}}{ A _{1} y _{1}}+\frac{ F \ell_{2}}{ A _{2} y _{2}}=\Delta \ell$
$F =\frac{\Delta \ell}{\frac{\ell_{1}}{ A _{1} y _{1}}+\frac{\ell_{2}}{ A _{2} y _{2}}}=1.54 \times 10^{2}=154$
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