The following four wires are made of the same material. If same tension is applied to each, the wire having…
The following four wires are made of the same material. If same tension is applied to each, the wire having largest extension is
- length 0.5 m, diameter 0.5 mm.
- length 1 m, diameter 1 mm.
- length 2 m, diameter 2 mm.
- length 3 m, diameter 3 mm.
Solution
Given, that material is same, so Young's modulus $=$ constant and tension $=$ constant . Hence, the extension,
$
\begin{aligned}
\Delta L & =\frac{F L}{Y A} \\
\text { So, } \quad \Delta L & =\frac{4 F}{Y \pi} \times \frac{L}{D^2} \quad\left(\because A=\frac{\pi D^2}{4}\right) \\
\Rightarrow \quad \Delta L & \propto \frac{L}{D^2}
\end{aligned}
$
So, now checking the options,
(a)
$
\begin{aligned}
& L=0.5 \mathrm{~m}, D=0.5 \mathrm{~mm}=0.5 \times 10^{-3} \mathrm{~m} \\
& \Delta L=\frac{4 F}{Y \pi} \times \frac{0.5}{\left(0.5 \times 10^{-3}\right)^2}=2 \times 10^6 \times \frac{4 F}{Y \pi}
\end{aligned}
$
(b)
$
\begin{aligned}
L & =1 \mathrm{~m}, D=1 \mathrm{~mm} \\
\Delta L & =\frac{4 F}{Y \pi} \times 1.0 \times 10^6
\end{aligned}
$
(c)
$
\begin{aligned}
L & =2 \mathrm{~m}, D=2 \mathrm{~mm} \\
\Delta L & =\frac{4 F}{Y 4 \pi} \times 0.5 \times 10^6
\end{aligned}
$
(d)
$
\begin{aligned}
L & =3 \mathrm{~m}, D=3 \mathrm{~mm} \\
\Delta L & =\frac{4 F}{Y \pi} \times 0.33 \times 10^6
\end{aligned}
$
Hence, $\frac{4 F}{Y \pi}$ is a constant in all quantities. Now after checking the multiplier of $\frac{4 F}{Y \pi}$. The highest value of $\Delta L$ is in option (a)
Asked in: AP EAMCET 2019 (21 Apr Shift 1)
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