Question

In an experiment, brass and steel wires of length $1\text{m}$ each with areas of cross-section $1{\text{mm}}^2$ are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress requires to produce a net elongation of $0.2\text{mm}$ is
[Take Young’s modulus for steel and brass as $120\times {10}^9{\text{N/m}}^2$ and $60\times {10}^9{\text{N/m}}^2$respectively]

• A. $1.2\times {10}^6{\text{N/m}}^2$
• B. $0.2\times {10}^6{\text{N/m}}^2$
• C. $8\times {10}^6{\text{N/m}}^2$
• D. $4.0\times {10}^6{\text{N/m}}^2$

Answer 1

The correct option is C $8\times {10}^6{\text{N/m}}^2$

In given experiment, a composite wire is stretched by a force $F$.


Net elongation in the wire = elongation in brass wire + elongation in steel wire .. . (i)
Now, Young’s modulus of a wire of cross-section $\left(A\right)$ when some force $(F$ is applied, $Y=\frac{Fl}{A\Delta l}$
We have,
$\Delta l=$ elongation $=\frac{Fl}{AY}$
So, from relation (i), we have
$\Delta l_{\text{net}}=\Delta l_{\text{brass}}+\Delta l_{\text{steel}}$
$\Rightarrow \Delta l_{\text{net}}={\left(\frac{Fl}{AY}\right)}_{\text{brass}}+{\left(\frac{Fl}{AY}\right)}_{\text{steel}}$
As wires are connected in series and they are of same area of cross-section, length and subjected to same force, so
$\Delta l_{\text{net}}=\frac{F}{A}(\frac{l}{Y_{\text{brass}}}+\frac{l}{Y_{\text{steel}}})$
Here,
$\Delta l_{\text{net}}=0.2\text{mm}$
$=0.2\times {10}^{-3}\text{m}$
and $l=1\text{m}$
$Y_{\text{brass}}=60\times {10}^9{\text{Nm}}^{-2}$
$Y_{\text{steel}}=120\times {10}^9{\text{Nm}}^{-2}$
On putting the values, we have
$0.2\times {10}^{-3}=\frac{F}{A}(\frac{1}{60\times {10}^9}+\frac{1}{120\times {10}^9})$
$\Rightarrow$ Stress $=\frac{F}{A}=8\times {10}^6{\text{Nm}}^{-2}$