Question

Two identical wires are suspended from a roof, but one is of copper and other is of iron. Young’s modulus of iron is thrice that of copper. The weights to be added on copper and iron wires so that the ends are on the same level must be in the ratio of
A. 1:3
B. 2:1
C. 3:1
D. 4:1

Answer 1

As a first step you could recall the expression for young’s modulus. As the given wires are identical we know that their length and area of cross section would be the same and for the wires to be at the same level after adding weights, the elongation in both wires should be the same. So now we will get the force on the wires proportional to their respective young’s modulus whose ratio is already given. And hence, you could find the answer.

Formula used:
Young’s Modulus,
$Y=\dfrac{Fl}{A\Delta l}$

Complete step by step solution:
In the question, we are given two identical wires. One is made of copper and the other of iron and both are suspended from the roof. We are given that the Young’s modulus of iron$\left( {{Y}_{i}} \right)$ is thrice that of copper$\left( {{Y}_{c}} \right)$ , that is,
${{Y}_{i}}=3{{Y}_{c}}$
$\Rightarrow \dfrac{{{Y}_{i}}}{{{Y}_{c}}}=\dfrac{3}{1}$ ……………………………….. (1)
Now we have the definition of Young’s modulus that defines it as the ratio of longitudinal stress $\left( \sigma \right)$ to the longitudinal strain$\left( \varepsilon \right)$, that is,
$Y=\dfrac{\sigma }{\varepsilon }$ …………………………… (2)
But we know that longitudinal stress, just like any other stress is the force acting per unit area, that is,
$\sigma =\dfrac{F}{A}$ ………………………………… (3)
Now, we have the longitudinal strain which is given by the ratio of change in length$\left( \Delta l \right)$ to original length$\left( l \right)$ , that is,
$\varepsilon =\dfrac{\Delta l}{l}$ …………………………………………… (4)
Substituting (3) and (4) in (2),
$Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}$
$\Rightarrow Y=\dfrac{Fl}{A\Delta l}$ ……………………………………… (5)
We are asked to find the ratio of the weights that have to be added on copper and iron wires so that the ends are on the same level. Also, we have already said that the wires are identical in dimension, so, length and area will be the same for iron as well as copper wire and for the ends to be on the same level, the change in length should also be the same. So with all the other quantities constant we could say that,
$Y\propto F$
$\Rightarrow \dfrac{{{Y}_{i}}}{{{Y}_{c}}}=\dfrac{{{F}_{i}}}{{{F}_{c}}}$
From (1),
$\dfrac{{{F}_{i}}}{{{F}_{c}}}=\dfrac{3}{1}$
$\therefore {{F}_{c}}:{{F}_{i}}=1:3$
Therefore, we found the ratio of the weights that have to be added on copper and iron wires so that the ends are on the same level as 1:3.

Hence, option A is the right answer.

Note:
If you are wondering why we have taken the ratio of forces acting on the when we are asked to find the ratio of the weights to be added. In the question we are told that the wires are suspended from the roof. So, in such a situation the only force acting on the wires will be that due to gravity and hence we could consider the force equivalent to weight here.