Question

A $5m$ aluminium wire $\left( {Y = 7 \times {{10}^{10}}N{m^{ - 2}}} \right)$ of diameter $3mm$ supports a $40kg$ mass. In order to have same elongation in a copper wire $\left( {Y = 12 \times {{10}^{10}}N{m^{ - 2}}} \right)$ of same length under same weight, the diameter should be (in $mm$)
(A) $2.0$
(B) $2.3$
(C) $1.75$
(D) $5.0$

Answer 1

Hint:Here, you are supposed to compare quantities for aluminium and copper. Given $Y$ is the Young’s modulus. You need to consider the definition of the Young’s modulus. It gives the relation between stress and strain when a particular substance is under tension and tends to elongate.

Complete step by step answer:
The Young’s modulus is the representation of the rigidity of a particular material. If this material is under tension and it tends to elongate, that is, it is under stress as well as strain, then the Young’s modulus is defined as the ratio of stress to strain.Stress is defined as force divided by the cross section. Strain is defined as the change in dimension occurred due to the acting force divided by the original dimension.Mathematically,
$stress = \dfrac{F}{A}$ and $strain = \dfrac{{\Delta l}}{l}$

So, the Young’s modulus can be given as,
$Y = \dfrac{{stress}}{{strain}}
\Rightarrow Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta l}}{l}}}
\Rightarrow Y = \dfrac{{Fl}}{{A\Delta l}}$
Let us come back to our question. For aluminium,
${Y_{al}} = \dfrac{{{F_{al}}{l_{al}}}}{{{A_{al}}\Delta {l_{al}}}}$.
Since it’s a wire, $A = \dfrac{{\pi {d^2}}}{4}$
${Y_{al}} = \dfrac{{4{F_{al}}{l_{al}}}}{{\pi {d^2}\Delta {l_{al}}}}$, $d$ is the diameter. For copper, ${Y_{cu}} = \dfrac{{4{F_{cu}}{l_{cu}}}}{{\pi {d^2}\Delta {l_{cu}}}}$. In the given question, except the Young’s modulus and the diameter of the wires, rest quantities are equal. If you compare, you will get the diameter of copper as follows
$
{Y_{al}}{d_{al}}^2 = {Y_{cu}}{d_{cu}}^2 \\
\Rightarrow{d_{cu}} = {d_{al}}\sqrt {\dfrac{{{Y_{al}}}}{{{Y_{cu}}}}} \\ $
Let us substitute the given values,
${d_{cu}} = \left( 3 \right)\sqrt {\dfrac{{\left( {7 \times {{10}^{10}}} \right)}}{{\left( {12 \times {{10}^{10}}} \right)}}} \\
\therefore{d_{cu}} = 2.29 \approx 2.3mm$
Hence, in order to have the same elongation in a copper wire of same length under same weight, the diameter should be $2.3mm$.

Hence, option B is correct.

Note: Remember all the formulae that are used in this question in order to find the diameter of the copper wire. Keep in mind the Young’s modulus is defined as the ratio of stress to strain. Also remember that the stress is equal to force divided by cross section area and the strain is defined as change in length divided by original length.