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JEE Main & Advanced Physics Wave Mechanics Question Bank Mock Test - Waves and Acoustics

  • question_answer
    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

    A) \[\frac{{{L}^{2}}}{{{L}^{2}}+{{l}^{2}}}\]

    B) \[\frac{{{L}^{2}}-{{l}^{2}}}{{{L}^{2}}}\]

    C) \[\frac{{{L}^{2}}}{{{L}^{2}}-{{t}^{2}}}\] 

    D) \[\frac{{{L}^{2}}-{{l}^{2}}}{{{L}^{2}}}\]

    Correct Answer: C

    Solution :

    [c] Frequency of vib. is stretched string \[n=\frac{1}{2(Length)}\sqrt{\frac{T}{m}}\] When the stone is completely immersed in water, length changes but frequency doesn't (\[\therefore \] unison reestablished) Hence length \[\propto \sqrt{T}\Rightarrow \frac{L}{l}=\sqrt{\frac{{{T}_{air}}}{{{T}_{water}}}=\sqrt{\frac{Vpg}{V(p-1)g}}}\] (Density of stone = p and density of water =1) \[\Rightarrow \frac{L}{l}=\sqrt{\frac{\rho }{\rho -1}}\Rightarrow \rho =\frac{{{L}^{2}}}{{{L}^{2}}-{{l}^{2}}}\]


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