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 elonation in length 0.50 mm is produced in the copper wire. The stretching force is
$(Y_{cu}=1.1\times {10}^{11}N/m^2,Y_{steel}=2.0\times {10}^{11}N/m^2)$

• 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

The correct option is D $1.8\times {10}^{2}N$

Given that,

Length of the copper wire$l_1=2.2m$

Length of the steel wire $l_2=1.6m$

Elongation in length$△l=\mathrm{0.5.}mm$

$=0.5\times {10}^{-3}m$

Radius of the$Cu$ wire $r_1=1.5\times {10}^{-3}m$

$Y_1=1.1\times {10}^{11}N/m^2$

$Y_2=2.0\times {10}^{11}N/m^2$
Let $F$ be the stretching force in both the wires then,
for $Cu$ wire, $Y_1=\frac{F}{\pi r_1^2}\times \frac{l_1}{△l_1}$
$\Rightarrow F=\frac{Y_1\pi r_1^2\times △l_1}{l_1}$
$=\frac{1.1\times {10}^{-11}}{2.2}\times \frac{22}{7}\times (1.5\times {10}^{-3}{)}^2\times 0.5\times {10}^3$
$=1.8\times {10}^{2}N$

So the streching force in the copper wire be $1.8\times {10}^{2}N$