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PhysicsSound WaveMiscellaneousHard2 minPYQ_2020

PhysicsHardnumerical

A uniform rope of mass $6 kg$ and length $6 m$ hangs vertically from a rigid support. A block of mass $2 kg$ is attached to the free end of the rope. A transverse pulse of wavelength $0.075 m$ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is $K cm$ where $K$ is

Answer:

15.00

Solution:

$T_{1} = 2 \times 10 = 20 N$
$T_{2} = (2 + 6) 10 = 80 N$
$\frac{V_{2}}{V_{1}} = \sqrt{\frac{80}{20}} = 2$
Frequency remains unchanged
So $\frac{λ_{2}}{λ_{1}} = 2$
$λ_{2} = 2 \times 0.075$
$= 0.15 m$
$K = 15 c m$

Stream:JEESubject:PhysicsTopic:Sound WaveSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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