MHT CET · Physics · Waves and Sound
A uniform rope of length \(12 \mathrm{~m}\) and mass \(6 \mathrm{~kg}\) hangs vertically from the rigid support. A block of mass \(2 \mathrm{~kg}\) is attached to the free end of the rope. A transverse pulse of wavelength \(0.06 \mathrm{~m}\) is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is
- A 0.12 m
- B 0.4 m
- C 0.8 m
- D 0.16 m
Answer & Solution
Correct Answer
(A) 0.12 m
Step-by-step Solution
Detailed explanation
Speed of a wave in a string is given by
\(
\begin{aligned}
& \mathrm{V}=\mathrm{f} \lambda=\sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\
& \therefore \lambda=\frac{1}{\mathrm{f}} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\
& \therefore \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}
\end{aligned}
\)
Tension at the bottom of the rope \(=\mathrm{T}_1=2 \mathrm{~kg}\) Tension at the top of the rope \(\mathrm{T}_2=2+6=8 \mathrm{~kg}\)
\(
\begin{aligned}
& \therefore \lambda_2=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \cdot \lambda_1=\sqrt{\frac{8}{2}} \times 0.06 \\
& =2 \times 0.06=0.12 \mathrm{~m}
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{V}=\mathrm{f} \lambda=\sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\
& \therefore \lambda=\frac{1}{\mathrm{f}} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\
& \therefore \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}
\end{aligned}
\)
Tension at the bottom of the rope \(=\mathrm{T}_1=2 \mathrm{~kg}\) Tension at the top of the rope \(\mathrm{T}_2=2+6=8 \mathrm{~kg}\)
\(
\begin{aligned}
& \therefore \lambda_2=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \cdot \lambda_1=\sqrt{\frac{8}{2}} \times 0.06 \\
& =2 \times 0.06=0.12 \mathrm{~m}
\end{aligned}
\)
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