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Q.

A stone is hung in air from a wire which is stretched over a sonometer. The bridges of the sonometer are L cm apart when the wire is in unison with a tuning fork of frequency N. When the stone is completely immersed in water, the length between the bridges is l cm for re-establishing unison, the specific gravity of the material of the stone is

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a

L2L2+l2

b

L2−l2L2

c

L2L2−l2

d

L2+l2L2

answer is C.

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Detailed Solution

Frequency of vib. is stretched string n=12(Length)TmWhen the stone is completely immersed in water, length changes but frequency doesn’t (∵ unison reestablished)Hence length ∝T⇒Ll=TairTwater=VρgV(ρ−1)g(Density of stone = ρ and density of water =1)⇒Ll=ρρ−1⇒ρ=L2L2−l2

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