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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
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Verified Answer
The correct answer is:
0.12 m
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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