The frequency of a sonometer wire is f. When | vedclass

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(D) The frequency of a sonometer wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit leng

Vedclass · Accepted Answer

$32/27$

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(D) The frequency of a sonometer wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length. Since $f \propto \sqrt{T}$,we have $f \propto \sqrt{Mg}$.When the mass $M$ is immersed in water,the effective tension is $T_w = Mg - V\rho_w g$,where $V$ is the volume of the mass and $\rho_w$ is the density of water. Given $f_w = f/2$,we have:$\frac{f}{2} = \frac{1}{2l} \sqrt{\frac{Mg - V\rho_w g}{\mu}} \implies \frac{1}{4} = \frac{Mg - V\rho_w g}{Mg} = 1 - \frac{V\rho_w}{M}$.Thus,$\frac{V\rho_w}{M} = 1 - \frac{1}{4} = \frac{3}{4}$,so $V = \frac{3M}{4\rho_w}$.When the mass is immersed in a liquid of density $\rho_L$,the frequency is $f_L = f/3$. The effective tension is $T_L = Mg - V\rho_L g$. Thus:$\frac{f}{3} = \frac{1}{2l} \sqrt{\frac{Mg - V\rho_L g}{\mu}} \implies \frac{1}{9} = \frac{Mg - V\rho_L g}{Mg} = 1 - \frac{V\rho_L}{M}$.Substituting $V = \frac{3M}{4\rho_w}$:$\frac{1}{9} = 1 - \frac{3M}{4\rho_w} \cdot \frac{\rho_L}{M} \implies \frac{3}{4} \cdot \frac{\rho_L}{\rho_w} = 1 - \frac{1}{9} = \frac{8}{9}$.Therefore,the specific gravity $\frac{\rho_L}{\rho_w} = \frac{8}{9} \cdot \frac{4}{3} = \frac{32}{27}$.

The frequency of a sonometer wire is $f$. When the weights producing the tension are completely immersed in water,the frequency becomes $f/2$. When the weights are immersed in a certain liquid,the frequency becomes $f/3$. The specific gravity of the liquid is:

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