Question

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 :

• A. $\frac{4}{3}$
• B. $\frac{16}{9}$
• C. $\frac{15}{12}$
• D. $\frac{32}{27}$

Answer 1

The correct option is D $\frac{32}{27}$

Frequency of a sonometer is given by
$f=\frac{1}{2l}\sqrt{\frac{Mg}{m}}$, where l is the length an m is mass per unit length of the string.

Frequency of the sonometer when the mass is immersed in water is
$f_w=\frac{1}{2l}\sqrt{\frac{Mg-V{\rho }_{w}g}{m}}$.
Here $V\rho g$ is the buoyancy force.

$f_w=\frac{f}{2}\Rightarrow \frac{1}{4l}\sqrt{\frac{Mg}{m}}=\frac{1}{2l}\sqrt{\frac{Mg-V{\rho }_{w}g}{m}}$.
Here ${\rho }_w=1$.

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