Question

The frequency of a sonometer wire is $100\text{Hz}$. When the weight producing the tension is completely immersed in water the frequency becomes $80\text{Hz}$ and on immersing the weight in a certain liquid the frequency becomes $60\text{Hz}$. The specific gravity of the liquid is

• A. $1.42$
• B. $1.77$
• C. $1.82$
• D. $1.21$

Answer 1

The correct option is B $1.77$

For sonometer we know that frequency,
$f\propto \sqrt{T}\propto \sqrt{m{g}_{eff}}$

$or$ $f\propto \sqrt{g_{eff}}$

When immersed in water:

Frequency in water, $f_w=0.8f_{\text{air}}$
$\therefore \frac{g_{eff}}{g}=\frac{f_w^2}{f_{air}^2}=(0.8{)}^2=0.64$
Let, ${\rho }_w=$ relative density of water $=1$,
${\rho }_m=$ relative density of mass,
and ${\rho }_L=$ relative density of liquid

$\Rightarrow 1-\frac{{\rho }_w}{{\rho }_m}=0.64$
$\Rightarrow \frac{{\rho }_w}{{\rho }_m}=0.36\dots \dots \left(1\right)$

When immersed in liquid:
Frequency in unknown liquid,
$f_L=0.6f_w$

$\frac{g_{eff}^{\prime }}{g}=(0.6{)}^2=0.36$

$\therefore 1-\frac{{\rho }_L}{{\rho }_m}=0.36$

$\frac{{\rho }_L}{{\rho }_m}=0.64\dots \dots \left(2\right)$

From eqs. $\left(1\right)$ and $\left(2\right)$

$\frac{{\rho }_L}{{\rho }_w}=\frac{0.64}{0.36}$

$\therefore {\rho }_L=1.77$

$\begin{array}{c}\text{Why this question ?}\\ \text{This question introduces the sonometer laws and}\\ \text{its interpretation when the sonometer is immersed}\\ \text{in two different liquids.}\end{array}$