Physics › Mechanical Properties of Solids › Young’s Modulus and Breaking Stres
A copper wire of cross-sectional area $0.01 \mathrm{~cm}^2$ is under a tension of $22 \mathrm{~N}$. The…
A copper wire of cross-sectional area $0.01 \mathrm{~cm}^2$ is under a tension of $22 \mathrm{~N}$. The decrease in the cross-sectional area is
(Young modulus $=1.1 \times 10^{11} \mathrm{Nm}^{-2}$, Poisson's ratio $=0.32$ )
$0.128 \times 10^{-6} \mathrm{~cm}^2$ $128 \times 10^{-6} \mathrm{~cm}^2$ $12.8 \times 10^{-6} \mathrm{~cm}^2$ $1.28 \times 10^{-6} \mathrm{~cm}^2$ Solution
Young's modulus,
$
\begin{aligned}
Y & =\frac{F / A}{\Delta l / l} \\
\frac{\Delta l}{l} & =\frac{F}{Y A}
\end{aligned}
$
where, $\frac{\Delta l}{l}=$ longitudinal strain
Given, $F=22 \mathrm{~N}, Y=1.1 \times 10^{11} \mathrm{~N}-\mathrm{m}^2$,
$
\begin{aligned}
A & =0.01 \mathrm{~cm}^2=10^{-6} \mathrm{~m}^2 \\
\frac{\Delta l}{l} & =\frac{22}{1.1 \times 10^{11} \times 10^{-6}}=2 \times 10^{-4}
\end{aligned}
$
Now, Poisson ratio
$
\begin{aligned}
\sigma & =\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=\frac{\Delta d / d}{\Delta l / l} \\
\frac{\Delta d}{d} & =\sigma \cdot \frac{\Delta l}{l}=0.32 \times 2 \times 10^{-4}
\end{aligned}
$
Change in diameter, $\frac{\Delta d}{d}=6.4 \times 10^{-5}$
or change (decrease) in radius, $\frac{\Delta r}{r}=6.4 \times 10^{-5}$
Area,
$
A=\pi r^2
$
Fractional change in area,
$
\frac{\Delta A}{A}=2 \cdot \frac{\Delta r}{r}
$
$
\begin{aligned}
& \frac{\Delta A}{A}=2 \times 6.4 \times 10^{-5} \\
& \Delta A=\left(12.8 \times 10^{-5}\right) \mathrm{A}
\end{aligned}
$
Decrease in area,
$
\begin{gathered}
\Delta A=\left(12.8 \times 10^{-5}\right) \times(0.01) \mathrm{cm}^2 \\
\Delta A=1.28 \times 10^{-6} \mathrm{~cm}^2
\end{gathered}
$
Asked in: AP EAMCET 2018 (24 Apr Shift 1)
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