Q133 / 160 · AP EAMCET 2013 · Physics
A tension of $20 \mathrm{~N}$ is applied to a copper wire of cross sectional area $0.01 \mathrm{~cm}^2$, Young's modulus of copper is $1.1 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$ and Poisson's ratio is 0.32 . The decrease in cross sectional area of the wire is
- Correct answerA$1.16 \times 10^{-6} \mathrm{~cm}^2$
- B$1.16 \times 10^{-5} \mathrm{~m}^2$
- C$1.16 \times 10^{-4} \mathrm{~m}^2$
- D$1.16 \times 10^{-3} \mathrm{~cm}^2$
Solution
Given, $\sigma=0.32, F=20 \mathrm{~N}$
$$
A=0.01 \mathrm{~cm}^2=0.01 \times 10^{-3} \mathrm{~m}
$$
and $Y=1-1 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
We know that
$$
\frac{\Delta l}{l}=\frac{F}{A Y}=\frac{20}{0.01 \times 10^{-3} \times 1.1 \times 10^{11}}=18.1 \times 10^{-7}
$$
and we also known
$$
\begin{gathered}
\sigma=\frac{-\Delta r / r}{\Delta / /} \\
-\frac{\Delta r}{r}=0.32 \times 18.1 \times 10^{-7}=5.79 \times 10^{-7}
\end{gathered}
$$
Hence, decrease in cross reactional area of wire is
$$
\begin{aligned}
\Delta A=2 \frac{\Delta r}{r} \times A & =2 \times 5.79 \times 10^{-7} \times 0.01 \times 10^{-3} \\
& =0.158 \times 10^{-10} \mathrm{~m}^2 \\
& =1.26 \times 10^{-6} \mathrm{~cm}^2
\end{aligned}
$$
Hence, the correct answer is option: 'A'.