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Question

An object of specific gravity ρ is hung from a thin steel wire. The fundamental frequency (in Hz) is 300 Hz. The object is immersed in water with specific gravity ρw, so that one half its volume is submerged. The new fundamental frequency (in Hz) is

A
300(2ρρw2ρ)12
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B
300(2ρ2ρρw)12
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C
300(2ρ2ρρw)
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D
300(2ρρw2ρ)
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Solution

The correct option is A 300(2ρρw2ρ)12

The diagramatic representation of the given problem is shown in figure. The expression of fundamental frequency is v=12lTμ

In air T=mg=(Vρ)g

v=12lvρgμ (1)

When the object is half immersed in water

T=mgupthurst=Vρg(V2)ρwg

=(V2)g(2ρρw)

The new fundamental frequency is

v=12l(Vg2)(2ρpρw)μ (2)

vv=((2ρρw)2ρ)

or v=v(2ρρw2ρ)12=300(2ρρw2ρ)12 Hz


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