Question

A tension of 22N is applied to a copper wire of cross-sectional area $0.02c{m^2}$. Young’s modulus of copper $1 \cdot 1 \times {10^{11}}\dfrac{N}{{{m^2}}}$ and Poisson’s ratio is 0.32. The decrease in cross sectional area will be:
A) $1 \cdot 28 \times {10^{ - 6}}c{m^2}$.
B) $1 \cdot 6 \times {10^{ - 6}}c{m^2}$.
C) $2 \cdot 56 \times {10^{ - 6}}c{m^2}$.
D) $0 \cdot 64 \times {10^{ - 6}}c{m^2}$.

Answer 1

Hint:Young’s modulus is defined as the ratio of the longitudinal stress to the strain. The young’s modulus also gives the idea about the strength of any material. Poisson’s ratio is the term which comes when there is expansion of material in the perpendicular direction of the applied force and it is the ratio of the lateral strain to longitudinal strain.

Formula used:The formula used for the Young’s modulus is given by,${\text{Young's modulus}} = \dfrac{{{\text{stress}}}}{{{\text{strain}}}} = \dfrac{{\left( {\dfrac{F}{A}} \right)}}{{\left( {\dfrac{{\Delta l}}{l}} \right)}}$
Also the formula for the Poisson’s ratio is given by${\text{poisson's ratio}}\left( \mu \right){\text{ = }} - \dfrac{{{\text{lateral strain}}}}{{{\text{longitudinal strain}}}} = - \dfrac{{{\varepsilon _{lateral}}}}{{{\varepsilon _{axial}}}}$.

Complete step by step solution:
Here it is given that the Young’s modulus is given by, $Y = \dfrac{{Fl}}{{A\Delta l}}$ and also we know that the Poisson’s ratio is $\mu = \dfrac{{\left( {\dfrac{{\Delta r}}{r}} \right)}}{{\left( {\dfrac{{\Delta l}}{l}} \right)}}$.

Rearranging the terms we get,

$\sigma = \dfrac{{\left( {\dfrac{{\Delta r}}{r}} \right)}}{{\left( {\dfrac{{\Delta l}}{l}} \right)}}$
$\dfrac{{\Delta l}}{l} = \left( {\dfrac{1}{\sigma }} \right) \cdot \left( {\dfrac{{\Delta r}}{r}} \right)$………eq. (1)

Replace the value of equation (1) in the formula of Young’s modulus.

\[ \Rightarrow Y = \dfrac{{\left( {\dfrac{F}{A}} \right) \cdot l}}{{\Delta l}}\]
\[ \Rightarrow Y = \dfrac{\sigma }{{\left( {\dfrac{{\Delta l}}{l}} \right)}}\]

Where $\sigma $ is the stress.

Replace the value of $\dfrac{{\Delta l}}{l} = \left( {\dfrac{1}{\sigma }} \right) \cdot \left( {\dfrac{{\Delta r}}{r}} \right)$.

\[ \Rightarrow Y = \dfrac{\sigma }{{\left( {\dfrac{{\Delta l}}{l}} \right)}}\]
\[ \Rightarrow Y = \dfrac{\sigma }{{\left[ {\left( {\dfrac{1}{\mu }} \right) \cdot \left( {\dfrac{{\Delta r}}{r}} \right)} \right]}}\]
\[ \Rightarrow \dfrac{{\Delta r}}{r} = \dfrac{{\mu \cdot \sigma }}{Y}\]

Put the value force, Poisson’s ratio, area of cross section and Young’s modulus in the above formula.

$ \Rightarrow \dfrac{{\Delta r}}{r} = \dfrac{{\mu \cdot \sigma }}{Y}$
$ \Rightarrow \dfrac{{\Delta r}}{r} = \left[ {\dfrac{{0.32 \times \left( {\dfrac{{22}}{{0.02 \times {{10}^{ - 4}}}}} \right)}}{{1.1 \times {{10}^{11}}}}} \right]$
$ \Rightarrow \dfrac{{\Delta r}}{r} = 32 \times {10^{ - 6}}$………eq. (2)

The area of the cross section is given by $A = \pi {r^2}$ and also$\Delta A = 2\pi r\Delta r$.

Let us calculate$\dfrac{{\Delta A}}{A}$.

$ \Rightarrow \dfrac{{\Delta A}}{A} = \dfrac{{2\pi r\Delta r}}{{\pi {r^2}}}$
$ \Rightarrow \dfrac{{\Delta A}}{A} = \dfrac{{2\Delta r}}{r}$………eq. (3)

Put the value of $\dfrac{{\Delta r}}{r}$ from equation (2) to equation (3).

$ \Rightarrow \dfrac{{\Delta A}}{A} = \dfrac{{2\Delta r}}{r}$
$ \Rightarrow \dfrac{{\Delta A}}{A} = 2 \cdot \left( {32 \times {{10}^{ - 6}}} \right)$
$ \Rightarrow \Delta A = A \cdot \left( {64 \times {{10}^{ - 6}}} \right)$
$ \Rightarrow \Delta A = 0.02 \times 64 \times {10^{ - 6}}$
$ \Rightarrow \Delta A = 1.28 \times {10^{ - 6}}c{m^2}$

So the correct answer for this problem is option A.

Note:The students should remember the formula and concept of the young’s modulus as it can help in solving such problems also should students remember the concept of the Poisson’s ratio because the problem where there is expansion of the material takes place in the perpendicular direction of the applied force the concept of Poisson’s ratio should be used.