Question

The ratio of diameters of two wires of the same material is $n:1$. The length of each wire is $4m$. On applying the same load, the increases in the length of the thin wire will be $\left( {n > 1} \right)$
(A) ${n^2}$ times
(B) $n$ times
(C) $2n$ times
(D) $\left( {2n + 1} \right)$ times

Answer 1

Hint
We need to use the formula for the Young’s modulus to find the increase in the length of the wires for the 2 cases. Then by substituting the given values and then taking the ratio of the 2 cases, we will get the increase in length of the thin wire in the second case with respect to the first case.
In this solution we will be using the following formula,
$\Rightarrow Y = \dfrac{{{F \mathord{\left/
 {\vphantom {F A}} \right.
} A}}}{{{{\Delta L} \mathord{\left/
 {\vphantom {{\Delta L} L}} \right.
} L}}}$
Where $Y$ is the young’s modulus of the wire
$F$ is the force due to the loads on the wire
$A$ is the area of cross-section of the wire
$L$ is the length of the wire and $\Delta L$ is the increase in length.

Complete step by step answer
In this problem we are given 2 wires. Now let us consider the first wire to calculate the change in its length. The young’s modulus of the wire is given by the formula,
$\Rightarrow Y = \dfrac{{{F \mathord{\left/
 {\vphantom {F A}} \right.
} A}}}{{{{\Delta L} \mathord{\left/
 {\vphantom {{\Delta L} L}} \right.
} L}}}$
We can simplify it as,
$\Rightarrow Y = \dfrac{{FL}}{{A\Delta L}}$
We can take the $Y$ to the RHS and bring $\Delta L$ to the LHS as,
$\Rightarrow \Delta L = \dfrac{{FL}}{{AY}}$
Now in the case of both the wires, the Young’s modulus is the same. This is because young’s modulus is the same in both the cases as both the wires are made of the same material. The load on both the wires and the length of both the wires is also the same as in the question. So therefore, we can write this as,
$\Rightarrow \Delta L \propto \dfrac{1}{A}$
Now we can write the area of cross-section of the wire as,
$\Rightarrow A = \pi {\left( {\dfrac{d}{2}} \right)^2}$
where $d$ is the diameter of the wire. The diameter of the wires are given to be in the ratio $n:1$, that is,
$\Rightarrow \dfrac{{{d_1}}}{{{d_2}}} = n$
Or, we can write,
$\Rightarrow {d_1} = n{d_2}$
So the area of cross-section of the first wire can be written as,
$\Rightarrow {A_1} = \pi {\left( {\dfrac{{n{d_2}}}{2}} \right)^2}$
Hence we get,
$\Rightarrow {A_1} = {n^2}\pi {\left( {\dfrac{{{d_2}}}{2}} \right)^2}$
and ${A_2} = \pi {\left( {\dfrac{{{d_2}}}{2}} \right)^2}$
So if the change in the length of the wires are $\Delta {L_1}$ and $\Delta {L_2}$. And the area of cross-sections be ${A_1}$ and ${A_2}$ respectively.
So we can take the ratio of the change in lengths,
$\Rightarrow \dfrac{{\Delta {L_1}}}{{\Delta {L_2}}} = \dfrac{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
 {\vphantom {1 {{A_1}}}}\right.}
\!\lower0.7ex\hbox{${{A_1}}$}}}}{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/
 {\vphantom {1 {{A_2}}}}\right.}
\!\lower0.7ex\hbox{${{A_2}}$}}}}$
So we can write this as,
$\Rightarrow \dfrac{{\Delta {L_1}}}{{\Delta {L_2}}} = \dfrac{{{A_2}}}{{{A_1}}}$
$\Rightarrow \dfrac{{\Delta {L_1}}}{{\Delta {L_2}}} = \dfrac{{\pi {{\left( {\dfrac{{{d_2}}}{2}} \right)}^2}}}{{{n^2}\pi {{\left( {\dfrac{{{d_2}}}{2}} \right)}^2}}}$
Now we can cancel $\pi {\left( {\dfrac{{{d_2}}}{2}} \right)^2}$ from the numerator and denominator. Hence we get,
$\Rightarrow \dfrac{{\Delta {L_1}}}{{\Delta {L_2}}} = \dfrac{1}{{{n^2}}}$
Therefore, arranging this equation we get,
$\Rightarrow \Delta {L_2} = {n^2}\Delta {L_1}$
So the length in the second case is ${n^2}$ times the first case.
Therefore the correct answer is option (A); ${n^2}$.

Note
The Young’s modulus is also termed as the modulus of elasticity. It is a mechanical property of a solid and is given by the ration of the stress to the strain. It can be described as the elastic deformation that a solid undergoes when some load is applied on it.