Question

A steel wire of length $3.2m(Y_S=2.0\times 10^{11}$ $N{m}^{-2}$ ) 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 $\pi =\frac{22}{7})$

• A. 360
• B. 180
• C. 1080
• D. 154

Answer 1

Answer: (D)

$\Delta {\ell }_1+\Delta {\ell }_2=\Delta \ell$
$\frac{F{\ell }_1}{A_1{y}_1}+\frac{F{\ell }_2}{A_2{y}_2}=\Delta \ell$
$F=\frac{\Delta \ell }{\frac{{\ell }_1}{A_1{y}_1}+\frac{{\ell }_2}{A_2{y}_2}}=1.54\times 10^2=154$