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 \(\pi=\frac{22}{7}\))

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Answer & Solution

Correct Answer

(D) \(154\)

Step-by-step Solution

Detailed explanation

\(\Delta \ell_{1}+\Delta \ell_{2}=\Delta \ell\)

\(\frac{ F \ell_{1}}{ A _{1} y _{1}}+\frac{ F \ell_{2}}{ A _{2} y _{2}}=\Delta \ell\)

\(F =\frac{\Delta \ell}{\frac{\ell_{1}}{ A _{1} y _{1}}+\frac{\ell_{2}}{ A _{2} y _{2}}}=1.54 \times 10^{2}=154\)

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