A sonometer wire is stretched by hanging a metal bob, the fundamental frequency of the wire is ' $n_1$ '. When the bob is completely immersed in water, the frequency of vibration of wire becomes ' $\mathrm{n}_2$ '. The relative density of the metal of the bob is

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Aicharya · Accepted Answer

$\mathrm{n}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{~T}}{\mu}}$
The sonometer wire is stretched by a metal bob. Hence, we can write, $T=m g=W$ where W is the weight of the metal bob. $\begin{array}{ll} \therefore \quad & \text { From(i), } \\ & \mathrm{n}_1=\frac{1}{2 L} \sqrt{\frac{W_1}{\mu}}, \quad n_2=\frac{1}{2 L} \sqrt{\frac{W_2}{\mu}} \\ \therefore \quad & \frac{W_1}{W_2}=\frac{n_1^2}{n_2^2} \end{array}$ $\text {Relative density }(\sigma)=\frac{\text { weight in air }}{\text { loss of weight in water }}$ $\therefore \quad \frac{\mathrm{W}_1}{\mathrm{~W}_1-\mathrm{W}_2}=\frac{\mathrm{n}_1^2}{\mathrm{n}_1^2-\mathrm{n}_2^2}$

A sonometer wire is stretched by hanging a metal bob, the fundamental frequency of the wire is ' $n_1$ '.…

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