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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:
$\frac{4}{3}$
$\frac{16}{9}$
$\frac{15}{12}$
$\frac{32}{27}$
Solution

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
$\Rightarrow \frac{M}{9}=M-V \rho_{L} \Rightarrow \frac{M}{9}=M-\frac{3 M}{4} \rho_{L}$
$\Rightarrow \rho_{L}=\frac{32}{27}$