Question

A copper wire of length $2.4m$ and a steel wire of length $1.6m$, both the diameter $3mm$, are connected end to end. When stretched by a load, the net elongation is found to be $0.7mm$. The load applied is
$(Y_{copper}=1.2\times {10}^{11}N{m}^{-2},Y_{steel}=2\times {10}^{11}N{m}^{-2})$

• A. $1.2\times {10}^{2}N$
• B. $1.8\times {10}^{2}N$
• C. $2.4\times {10}^{2}N$
• D. $3.2\times {10}^{2}N$

Answer 1

Answer: (B)

$1.8\times {10}^{2}N$

Given,

$l_{copper}=2.4m$

$l_{steel}=1.6m$

$D=3mm$

$r=\frac{3}{2}=1.5mm$

$Y_{copper}=1.2\times {10}^{11}N/m^2$

$Y_{steel}=2\times {10}^{11}N/m^2$

$\Delta l=0.7mm$

Stress is same, ${\sigma }_{copper}={\sigma }_{steel}$

$Y_{copper}\frac{l_{copper}}{\Delta l_{copper}}=Y_{steel}\frac{l_{steel}}{\Delta l_{steel}}$

$1.2\times {10}^{11}\times \frac{2.4}{\Delta l_{copper}}=2\times {10}^{11}\times \frac{1.6}{\Delta l_{steel}}$

$\frac{\Delta l_{copper}}{\Delta l_{steel}}=2.5$

$\Delta l_{copper}=2.5\Delta l_{steel}$. . . . . . . (1)

$\Delta l=\Delta l_{copper}+\Delta l_{steel}$

$0.7\times {10}^{-3}=2.5\Delta l_{steel}+\Delta l_{steel}$

$\Delta l_{steel}=\frac{0.7\times {10}^{-3}}{3.5}=0.2mm$

$\Delta l_{copper}=2.5\times 0.2mm=0.5mm$

$Y_{copper}=\frac{F/A}{\frac{\Delta l_{copper}}{l_{copper}}}$

$\frac{F}{A}=Y_{copper}\frac{\Delta l_{copper}}{l_{copper}}$

$\frac{F}{A}=1.2\times {10}^{11}\times \frac{0.5\times {10}^{-3}}{2.4}$

$F=\pi r^2\times 0.25\times {10}^8$

$F=3.14\times 1.5\times 1.5\times {10}^{-6}\times 0.25\times {10}^8$

$F=1.8\times {10}^{2}N$

The correct option is B.