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Q. A steel wire of length $3.2 \,m \left( Y _{ S }=2.0 \times 10^{11}\right.$ $Nm ^{-2}$ ) 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: $\left(\right.$ Given $\left.\pi=\frac{22}{7}\right)$

JEE MainJEE Main 2022Mechanical Properties of Solids

Solution:

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$ \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$