The frequency of a sonometer wire is f, but w | vedclass

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Question

Frequency of the sonometer when the mass is immersed in water is

$f_{w}=\frac{1}{2 l} \sqrt{\frac{M g-V \rho_{w} g}{m}}$

Here $V \rho g$ is the

Vedclass · Accepted Answer

$\frac{32}{27}$

Vedclass · Answer

Frequency of the sonometer when the mass is immersed in water is

$f_{w}=\frac{1}{2 l} \sqrt{\frac{M g-V \rho_{w} g}{m}}$

Here $V \rho g$ is the buoyancy force.

$f_{w}=\frac{f}{2} \Rightarrow \frac{1}{4 l} \sqrt{\frac{M g}{m}}=\frac{1}{2 l} \sqrt{\frac{M g-V \rho_{w g}}{m}}$

Here $\rho_{w}=1$

Squaring both sides of the above equation, we get

$\frac{M}{4}=M-V \Rightarrow V=M-M / 4=\frac{3 M}{4}........(1)$

Frequency of the sonometer when the mass is immersed in the unknown liquid is

$f_{L}=\frac{1}{2 l} \sqrt{\frac{M g-V \rho_{L} g}{m}} \Rightarrow \frac{1}{6 l} \sqrt{\frac{M g}{m}}=\frac{1}{2 l} \sqrt{\frac{M g-V \rho_{L} g}{m}}$

Using the value of $V$ from equation $(1),$ we get

$\Rightarrow \frac{M}{9}=M-V \rho_{L} \Rightarrow \frac{M}{9}=M-\frac{3 M}{4} \rho_{L}$

$\Rightarrow \rho_{L}=\frac{32}{27}$

The frequency of a sonometer wire is $f$, but when the weights producing the tensions are completely immersed in water the frequency becomes $f/2$ and on immersing the weights in a certain liquid the frequency becomes $f/3$. The specific gravity of the liquid is:

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