Question

A composite wire consits of a steel wire of length $1.5m$ and a copper wire of length $2.0m$, with a uniform cross-sectional area of $2.5\times {10}^{-5}{m}^2$. It is loaded with a mass of $200kg$. Find the extension produced. Young's modulus of copper is $1.0\times {10}^{11}N{m}^{-2}$ and that of steel is $2.0\times {10}^{11}N{m}^{-2}$.
Take $g=9.8m{s}^{-2}$

Answer 1

The tension due to the load

will oemain same in the composite

wire. So, the tensile stress in

both rods is same.

Tensile stress $=\frac{Load}{Areaofsection}=\frac{200\times 9.8}{2.5\times {10}^{-5}}$

$\Rightarrow$ stress $=7.84\times {10}^{7}N/m^2$

So, let the elongations in the

st and lee wire be $\Delta l_1$ and $\Delta l_2$

Then,

As $r=\frac{T.stress}{strain}$

For st wire

$\Rightarrow 2\times {10}^{11}=\frac{7.84\times {10}^7}{\frac{\Delta l_1}{1.5}}$

$\Rightarrow \Delta l_1=5.88\times {10}^{-4}m$

For Lr wire

$\Rightarrow {10}^{11}=\frac{7.84\times {10}^7}{\frac{\Delta l_2}{2}}$

$\Rightarrow \Delta l_2=1.508\times {10}^{-3}m$

So, extension in composite rod $=\Delta l_1+\Delta l_2$

$=2.156mm$