Question

Two exactly similar wires of steel and copper are stretched by equal forces. If the total elongation is 2cm, then how much is the elongation in steel and copper wire respectively? Given ${Y_{steel}} = 20 \times {10^{11}}dyne/c{m^2},{Y_{copper}} = 12 \times {10^{11}}dyne/c{m^2}$
$
  (a){\text{ 1}}{\text{.25cm;0}}{\text{.75cm}} \\
  (b){\text{ 0}}{\text{.75cm;1}}{\text{.25cm}} \\
  (c){\text{ 1}}{\text{.15cm;0}}{\text{.85cm}} \\
  (d){\text{ 0}}{\text{.85cm;1}}{\text{.15cm}} \\
 $

Answer 1

- Hint – In this use the concept that since the forces applied onto the two wires is exactly similar thus stress applied onto the two wires will be the same as the two wires are exactly similar, so use the basics of young modulus (Y) that it is defined as the ratio of stress by strain, to calculate the respective elongations in the wire.

Complete step-by-step solution -

Let the elongation in steel wire be $\Delta {l_s}$ and the elongation in copper wire be $\Delta {l_c}$
Now as we know young modulus (Y) is defined as the ratio of stress by strain.
$Y = \dfrac{{{\text{stress}}}}{{{\text{strain}}}}$.
Now as we know that stress is the ratio of force (F) applied per unit area (A).
Therefore, stress = $\dfrac{F}{A}$ dyne/$m^2$.
And strain is the ratio of change in length $\Delta l$ to the original length $l$.
Therefore, strain = $\dfrac{{\Delta l}}{l}$ (it is unit less).
So young modulus (Y) = $\dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta l}}{l}}} = \dfrac{{F \times l}}{{A \times \Delta l}}$ dyne/$m^2$.
So the young modulus of steel (Ys) =$\dfrac{{{F_s} \times {l_s}}}{{{A_s} \times \Delta {l_s}}}$ dyne/$m^2$.
Similarly, the young modulus of copper (Yc) =$\dfrac{{{F_c} \times {l_c}}}{{{A_c} \times \Delta {l_c}}}$ dyne/$m^2$.
Now divide these two equations we have,
$ \Rightarrow \dfrac{{{Y_s}}}{{{Y_c}}} = \dfrac{{\dfrac{{{F_s} \times {l_s}}}{{{A_s} \times \Delta {l_s}}}}}{{\dfrac{{{F_c} \times {l_c}}}{{{A_c} \times \Delta {l_c}}}}}$
Now it is given that the two wires are exactly similar and stretched by equal force therefore,
Fs = Fc, As = Ac and ls = lc, so we have,
$ \Rightarrow \dfrac{{{Y_s}}}{{{Y_c}}} = \dfrac{{\Delta {l_c}}}{{\Delta {l_s}}}$
Now substitute the given values we have,
$ \Rightarrow \dfrac{{20 \times {{10}^{11}}}}{{12 \times {{10}^{11}}}} = \dfrac{{\Delta {l_c}}}{{\Delta {l_s}}}$
Now simplify it we have,
$ \Rightarrow \dfrac{{\Delta {l_c}}}{{\Delta {l_s}}} = \dfrac{5}{3}$
$ \Rightarrow \Delta {l_c} = \Delta {l_s}\dfrac{5}{3}$.............. (1)
Now it is given that the total elongation is 2 cm.
$\Delta {l_c} + \Delta {l_s} = 2$ cm.................... (2)
Now from equation (1) we have,
$ \Rightarrow \Delta {l_s}\dfrac{5}{3} + \Delta {l_s} = 2$
Now simplify it we have,
$ \Rightarrow \Delta {l_s}\left( {\dfrac{{5 + 3}}{3}} \right) = 2$
$ \Rightarrow \Delta {l_s} = \dfrac{6}{8} = 0.75$ cm.
Now from equation (2) we have,
$ \Rightarrow \Delta {l_c} + 0.75 = 2$
$ \Rightarrow \Delta {l_c} = 2 - 0.75 = 1.25$ cm.
So the elongation in steel and copper is 0.75 and 1.25 cm respectively.
So this is the required answer.
Hence option (B) is correct.

Note – Stress is a physical quantity that defines force per unit area applied to any material. The maximum stress of material that it can stand before it breaks is called the breaking point. Strain is simply the measure of how much an object is stretched or deformed when a force is applied to an object. Young’s model is a measure of a solid’s stiffness or resistance to elastic deformation under load.