Question

The frequency of a sonometer wire is $300\text{Hz}$. The frequency becomes half when the mass producing the tension is completely immersed in water and on immersing the mass in a different liquid, the frequency become one-third. The relative density of the liquid is (upto 2 decimals )

Answer 1

Let ${\rho }_l$ be the density of liquid and ${\rho }_w$ be the density of water.
As we know,
frequency $f\propto \sqrt{T}$ [where $T$ is tension]
$\Rightarrow \frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}}...\left(i\right)$


From figure (I), (II) & (III)
$T_1=\rho Vg$, [where $\rho$ is the density of mass]
$T_2=\rho Vg-{\rho }_{w}Vg$
and $T_3=\rho Vg-{\rho }_{l}Vg$
So, putting all these values in equation (i)
we get, $\frac{f_1}{f_2}=\sqrt{\frac{\rho Vg}{\rho Vg-{\rho }_{w}Vg}}$
$\Rightarrow \frac{300}{150}=\sqrt{\frac{\rho }{\rho -{\rho }_w}}$
$\Rightarrow 4=\frac{\rho }{\rho -{\rho }_w}\Rightarrow 4\rho -4{\rho }_w=\rho$
$\Rightarrow \rho =\frac{4}{3}{\rho }_w.....\left(2\right)$
Again,
$\frac{f_1}{f_3}=\sqrt{\frac{T_1}{T_3}}=\sqrt{\frac{\rho Vg}{\rho Vg-{\rho }_{l}Vg}}$
$\Rightarrow \frac{300}{100}=\sqrt{\frac{\rho }{\rho -{\rho }_l}}\Rightarrow 9=\frac{\rho }{\rho -{\rho }_l}$
$\Rightarrow 9\rho -9{\rho }_l=\rho$
$\Rightarrow \rho =\frac{9}{8}{\rho }_l.....\left(3\right)$
From equation $\left(2\right)$ & $\left(3\right)$,
$\frac{4}{3}{\rho }_w=\frac{9}{8}{\rho }_l$
$\Rightarrow \frac{{\rho }_l}{{\rho }_w}=\frac{4}{3}\times \frac{8}{9}=1.185\approx 1.18$
$[\because \frac{{\rho }_l}{{\rho }_w}=\text{Relative density}]$