Question

A copper wire of length $2.2m$ and a steel wire of length $1.6m$, both of diameter $3.0mm$ are connected end to end. When stretched by a force, the elongation in length $0.50mm$ is produced in the copper wire. The stretching force is $(Y_{Cu}=1.1\times 10^{11}N/m^2$ $Y_{\text{steel }}=2.0\times 10^{11}N/m^2)$

• A. $5.4\times 10^{2}N$
• B. $3.6\times 10^{2}N$
• C. $2.4\times 10^{2}N$
• D. $1.8\times 10^{2}N$

Answer 1

Answer: (D)

For $Cu$ wire, $l_1=2.2m$,
$r_1=1.5mm=1.5\times 10^{-3}m$
$Y_1=1.1\times 10^{11}N/m^2$
For steel wire, $l_2=1.6m$,
$r_2=1.5mm=1.5\times 10^{-3}m$
$Y_2=2.0\times 10^{11}N/m^2$
Let $F$ be the stretching force in both the wires then
For $Cu$ wire, $Y_1=\frac{F}{\pi r_1^2}\times \frac{l_1}{\Delta l_1}$
$\Rightarrow F=\frac{Y_1\pi r_1^2\times \Delta l_1}{l_1}$
$=\frac{1.1\times 10^{11}}{2.2}\times \frac{22}{7}\times {\left(1.5\times 10^{-3}\right)}^2\times 0.5\times 10^{-3}$
$\cong 1.8\times 10^{2}N$