A steel wire of length 3.2 m left( Y _{ S }=2.0 × 10^{11},Nm ^{-2}right) and a copper wire of lengt

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8.Mechanical Properties of Solids

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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)

Solution

$\Delta \ell_{1}+\Delta \ell_{2}=\Delta \ell$

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

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

Standard 11

Physics

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