What should be the diameter of a copper wire \(\left(Y=12 \times 10^{10} \mathrm{Nm}^{-2}\right)\) of length…
What should be the diameter of a copper wire \(\left(Y=12 \times 10^{10} \mathrm{Nm}^{-2}\right)\) of length \(5 \mathrm{~m}\) to produce the same elongation produced by a \(5 \mathrm{~m}\) long aluminium wire \(\left(Y=7 \times 10^{10} \mathrm{Nm}^{-2}\right)\) of diameter \(3 \mathrm{~mm}\) with the same \(40 \mathrm{~kg}\) mass ?
\(1.5 \mathrm{~mm}\)
\(5 \mathrm{~mm}\)
\(2.3 \mathrm{~mm}\)
\(10 \mathrm{~mm}\)
Solution
For copper wire, \(Y=12 \times 10^{10} \mathrm{Nm}^{-2}\)
Length, \(\quad l=5 \mathrm{~m}\),
\(F=m g=40 \times 10=400 \mathrm{~N}\)
We know that, \(Y=\frac{F l}{A \Delta l}\)
\(\begin{aligned}
\Rightarrow \quad \Delta l & =\frac{F l}{A Y}=\frac{400 \times 5}{\pi r^2 \times 12 \times 10^{10}} \\
\Delta l & =\frac{2000}{12 \pi r^2 \times 10^{10}} \quad \ldots (i)
\end{aligned}\)
For aluminium wire, \(Y=7 \times 10^{10} \mathrm{Nm}^{-2}\)
Diameter, \(d=3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}\)
\(\therefore \quad r=\frac{d}{2}=\frac{3 \times 10^{-3}}{2} \mathrm{~m}=1.5 \times 10^{-3} \mathrm{~m}\)
Similarly, \(\quad Y=\frac{F l}{A \Delta l}\)
\(\Rightarrow \quad \Delta l=\frac{F l}{A Y}=\frac{400 \times 5}{\pi\left(1.5 \times 10^{-3}\right)^2 \times 7 \times 10^{10}}\)
\(\Delta l=\frac{2000}{15.75 \pi \times 10^4}\) ...(ii)
According to given condition, \(\Delta l\) is same in both cases.
From Eqs. (i) and (ii), we get
\(\begin{array}{rlrl}
\therefore & \frac{2000}{12 \pi r^2 \times 10^{10}} & =\frac{2000}{15.75 \pi \times 10^4} \\
\Rightarrow & 12 r^2 & =15.75 \times 10^{-6} \\
r & = & 1.15 \times 10^{-3} \mathrm{~m}=1.15 \mathrm{~mm}
\end{array}\)
\(\therefore\) Diameter, \(d=2 r=2 \times 1.15=2.3 \mathrm{~mm}\)