Question

A copper wire AB is fused at one end B to an iron wire BC which has rectangularly long shape.AB = 9.00 x 10^2 mm and BC = 4.50 x 10^2 mm. The diameter of the copper wire is 0.50 mm and that of the iron wire is 2.50.mm. The compound wire is stretched so that its length increases by a total amount of 0.1 cm
Calculate:
i) The ratio of the extension of the copper wire to that of the iron wire.
ii) The tension applied to the compound wire.
(NB: Young modulus of copper — 1.3 x 1011 Pa while that of iron = 2.0 x 1011 Pa)

Answer 1

Dimension of wire $AB=9.00 \times 10^2 mm$

$BC = 4.50 \times 10^2 mm.$

Diameter of copper wire $(d)=0.5mm$

Diameter of the iron wire $(D)=2.5mm$

$\Delta l=0.1$

Young modulus of copper(Y1) $= 1.3 \times 10^{11} Pa$ while that of iron(Y2)$= 2.0 \times 10^{11} Pa$

Let the extension in copper wire and the iron wires are $\Delta l_1$ and $\Delta l_2$

$\Delta l= \Delta l_1 +\Delta l_2$

Let the force applied on the compound wire is F, so here force will be same on the both wire,

$Y=\frac{Fl}{A\Delta l}$

$F=\frac{YA \Delta l}{l}$

We know that $F_1= F_2$

Hence, $\frac{Y_1A_1 \Delta l_1}{l_1}=\frac{Y_2A_2 \Delta l_2}{l_2}$

$\Rightarrow \frac{\Delta l_1}{\Delta l_2}=\frac{Y_2A_2 l_1}{Y_1 A_2l_2}$

Now, substituting the values in the above equation,

$\Rightarrow \frac{\Delta l_1}{\Delta l_2}=\frac{ 2.0 \times 10^{11}\times (2.5\times 10^{-3})^2 9\times 10^{2}\times 10^{-3}}{ 1.3 \times 10^{11}\times (0.5\times 10^{-3})^2\times 4.5\times 10^{2}\times 10^{-3}}$

$\Rightarrow \frac{\Delta l_1}{\Delta l_2}=76.92$

b) Now substituting the value of $\Delta l_1$

$\Delta l= \Delta l_1 +\Delta l_2$

$\Rightarrow 0.1=76.92\Delta l_2+\Delta l_2$

$\Rightarrow \Delta l_2=\frac{0.1}{1+76.92}=0.0013cm$

$\Rightarrow F=\frac{Y_2A_2\Delta l_2}{l2}$

$\Rightarrow F=\frac{9\times 10^{11}\times \pi (2.5\times 10^{-3})^2\times 1.3\times 10^{-5}}{0.45\times 4}$

$\Rightarrow F=\frac{229.62}{1.8}N$

Hence, net force applied on the wire will be $F=127.57N$