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A uniform wire $20 \mathrm{~m}$ long and weighing $50 \mathrm{~N}$ hangs vertically. The speed of the wave at mid point of the wire is (acceleration due to gravity $=\mathrm{g}=10 \mathrm{~ms}^{-2}$ )
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The correct answer is:
$10 \mathrm{~ms}^{-1}$
$$
\mathrm{m}=\frac{50}{10}=5 \mathrm{~kg}
$$
$$
(\because \mathrm{W}=\mathrm{mg})
$$
Tension in the mid-point of the wire is:
$$
\mathrm{T}=\frac{\mathrm{m}}{2} \mathrm{~g}=\frac{5}{2} \times 10=25 \mathrm{~N}
$$
$\therefore \quad$ Speed of the wave at mid-point of the wire is:
$$
\begin{aligned}
\mathrm{v} & =\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{25}{\left(\frac{5}{20}\right)}} \quad\left(\because \mu=\frac{\mathrm{m}}{\mathrm{L}}\right) \\
\therefore \quad \mathrm{v} & =10 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
\mathrm{m}=\frac{50}{10}=5 \mathrm{~kg}
$$
$$
(\because \mathrm{W}=\mathrm{mg})
$$
Tension in the mid-point of the wire is:
$$
\mathrm{T}=\frac{\mathrm{m}}{2} \mathrm{~g}=\frac{5}{2} \times 10=25 \mathrm{~N}
$$
$\therefore \quad$ Speed of the wave at mid-point of the wire is:
$$
\begin{aligned}
\mathrm{v} & =\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{25}{\left(\frac{5}{20}\right)}} \quad\left(\because \mu=\frac{\mathrm{m}}{\mathrm{L}}\right) \\
\therefore \quad \mathrm{v} & =10 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
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