MHT CET · Physics · Waves and Sound
A string of mass \(0 \cdot 1 \mathrm{~kg}\) is under a tension \(1 \cdot 6 \mathrm{~N}\). The length of the string is \(1 \mathrm{~m}\). A
transverse wave starts from one end of the string. The time taken by the wave to
reach the other end is
- A \(0.30 \mathrm{~s}\).
- B \(0.50 \mathrm{~s}\).
- C \(0.25 \mathrm{~s}\)
- D \(0.75 \mathrm{~s}\)
Answer & Solution
Correct Answer
(C) \(0.25 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
Mass of string \(=0.1 \mathrm{lgg}\)
Tension \(T=1.6 \mathrm{~N}\)
Length \(\quad \ell=1 \mathrm{~m}\)
Man per unit length \(\mu=\frac{m}{1}=\frac{0.1}{1}=0.1 \mathrm{~kg} / \mathrm{m}\)
The velocity \((v)\) of the transverse wave in the string is given by the relation
\(\begin{aligned}
V &=\sqrt{\frac{T}{4}} \\
&=\sqrt{\frac{1.6}{0.1}} \\
&=4 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
Time taken \(T=\ell / V=1 / 4=0.25 \sec\)
Tension \(T=1.6 \mathrm{~N}\)
Length \(\quad \ell=1 \mathrm{~m}\)
Man per unit length \(\mu=\frac{m}{1}=\frac{0.1}{1}=0.1 \mathrm{~kg} / \mathrm{m}\)
The velocity \((v)\) of the transverse wave in the string is given by the relation
\(\begin{aligned}
V &=\sqrt{\frac{T}{4}} \\
&=\sqrt{\frac{1.6}{0.1}} \\
&=4 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
Time taken \(T=\ell / V=1 / 4=0.25 \sec\)
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