MHT CET · Physics · Waves and Sound
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 \(=\mathrm{g}=10 \mathrm{~ms}^{-2}\) )
- A \(4 \mathrm{~ms}^{-1}\)
- B \(10 \sqrt{2} \mathrm{~ms}^{-1}\)
- C \(10 \mathrm{~ms}^{-1}\)
- D Zero \(\mathrm{ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(10 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
\(
\mathrm{m}=\frac{50}{10}=5 \mathrm{~kg}
\)
\(
(\because \mathrm{W}=\mathrm{mg})
\)
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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