Question

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

• 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)

$1.8\times 10^2 \,N$

Answer 2

The correct answer is(D): 1.8×102 N.

For $Cu$ wire, $l_{1}=2.2 \,m$,
$r_{1}=1.5 \,mm =1.5 \times 10^{-3} m$
$Y_{1}=1.1 \times 10^{11} \,N / m ^{2}$
For steel wire, $l_{2}=1.6\,m$,
$r_{2}=1.5 \, mm =1.5 \times 10^{-3} m$
$Y_{2}=2.0 \times 10^{11} \, N / m ^{2}$
Let $F$ be the stretching force in both the wires then
For $Cu$ wire, $Y_{1}=\frac{F}{\pi r_{1}^{2}} \times \frac{l_{1}}{\Delta l_{1}}$
$\Rightarrow F=\frac{Y_{1} \pi r_{1}^{2} \times \Delta l_{1}}{l_{1}}$
$=\frac{1.1 \times 10^{11}}{2.2} \times \frac{22}{7} \times\left(1.5 \times 10^{-3}\right)^{2} \times 0.5 \times 10^{-3}$
$\cong 1.8 \times 10^{2} N$