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Question

An object of density ρ=500 kg/m3 is hanging from a thin steel wire. The fundamental frequency of transverse standing wave on the wire is 200 Hz. The set up is now immersed in water, so that one third of the object's volume is submerged. The new fundamental frequency (in Hz) is: [Given : Density of water (ρw)=1000 kg/m3]

A
2003
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B
200
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C
2003
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D
100
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Solution

The correct option is C 2003
Density of water (ρw)=1000 kg/m3



Fundamental frequency on a stretched thin wire
f=12lTμ

In air : T=mg=Vρg [ V is the volume of object]
f=12lVρgμ .........(1)

When object is one third immersed in water, then
T=mgupthrust =VρgV3ρwg=V3g(3ρρw)
The new fundamental frequency is
f=12lTμ=12l  Vg3(3ρρw)μ .......(2)

From (1) and (2)

ff=  Vg3(3ρρw)Vρg=(1ρw3ρ)

From the data given,

f=(110003×500)×200=2003 Hz

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