A sonometer wire is stretched by hanging a metal bob. The fundamental frequency of vibration of wire is '…

A sonometer wire is stretched by hanging a metal bob. The fundamental frequency of vibration of wire is ' $n_1$ '. When the bob is completely immersed in water, the frequency of vibration of wire becomes ' $n_2$ '. The relative density of the metal of the bob is
  1. $\frac{n_1}{n_1-n_2}$
  2. $\frac{\mathrm{n}_2}{\mathrm{n}_1-\mathrm{n}_2}$
  3. $\frac{n_1^2}{n_1^2-n_2^2}$
  4. $\frac{\mathrm{n}_2^2}{\mathrm{n}_1^2-\mathrm{n}_2^2}$

Solution

$\mathrm{n}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{~T}}{\mu}}$
The sonometer wire is stretched by a metal bob, $\mathrm{T}=\mathrm{mg}=\mathrm{W}$ where W is weight of metal bob $\begin{array}{ll} \therefore \quad & \text { From(i), } \\ & \mathrm{n}_1=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{~W}_1}{\mu}}, \quad \mathrm{n}_2=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{~W}_2}{\mu}} \\ \therefore \quad & \frac{\mathrm{~W}_1}{\mathrm{~W}_2}=\frac{\mathrm{n}_1^2}{\mathrm{n}_2^2} \\ & \text { Relative density }(\sigma)=\frac{\text { weight in air }}{\text { loss of weight in water }} \\ \therefore \quad & \frac{\mathrm{W}_1}{\mathrm{~W}_1-\mathrm{W}_2}=\frac{\mathrm{n}_1^2}{\mathrm{n}_1^2-\mathrm{n}_2^2} \end{array}$

Asked in: MHT CET 2024 (02 May Shift 2)

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