Question

The frequency of a sonometer wire is f, but when the weight producing the tensions are completely immersed in water the frequency becomes $\dfrac{f}{2}$ and on immersing the weights in a certain liquid the frequency becomes$\dfrac{f}{3}$. The specific gravity of the liquid is
A.)$\dfrac{4}{3}$
B.)$\dfrac{{16}}{9}$
C.)$\dfrac{{15}}{{12}}$
D.)$\dfrac{{32}}{{27}}$

Answer 1

Hint: Specific gravity also known as relative density is the ratio of density of a substance with reference to a certain material. The specific gravity of liquids is usually taken with respect to water. Here in order to find the specific gravity of the liquid we have to find the relative density of the water with respect to weight immersed and also the relative density of the liquid with respect to the weight immersed.

Formula used:

$\rho = \dfrac{M}{V}$, Here $\rho$represents the density, $M$ represents mass $V$ represents volume.

$f = \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg}}{m}}$ .(This is the equation of frequency of a sonometer when a mass $M$ hanged ) represents frequency , represents length , represents mass , represents acceleration due to gravity ,$m$ represents the mass per unit length.

$f = \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg - B}}{m}}$..(This is the equation of frequency of a sonometer when a mass $M$ is immersed in a liquid) represents frequency, represents length , represents mass , represents acceleration due to gravity ,$B$ represents buoyancy force ,$m$ represents the mass per unit length.

$B = V{\rho _L}g$ Here $B$ represents the buoyancy force, $V$ represents the volume, ${\rho _L}$ represents the density of the liquid and represents the acceleration due to gravity.

Complete step by step answer
The frequency of the sonometer when a mass $M$ hanged is given by $f = \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg}}{m}}$.
When we immerse the mass in water, the net buoyant force acting will be $B = V{\rho _w}g$, here ${\rho _w}$ is the density of water.

This frequency of a sonometer when a mass $M$ is immersed in a liquid is given by $f = \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg - B}}{m}}$. Substituting the value of $B$ in the equation we get
$f' = \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg - V{\rho _w}g}}{m}}$
It is already given in the question that this frequency is equal to$f' = \dfrac{f}{2}$.
$\dfrac{f}{2} = \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg - V{\rho _w}g}}{m}}$
Using the value of $f = \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg}}{m}}$ in above equation and squaring the whole equation we get,

$\dfrac{1}{2} \times \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg}}{m}} = \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg - V{\rho _w}g}}{m}} {\text{ }}$
$\dfrac{1}{{16{l^2}}}\dfrac{{Mg}}{m} = \dfrac{1}{{4{l^2}}}\dfrac{{Mg}}{m} - \dfrac{1}{{4{l^2}}}\dfrac{{V{\rho _w}g}}{m}$
$\dfrac{3}{4}M = V{\rho _W}$

Using $M = V\rho$ where the density of the mass connected is $\rho$

$\dfrac{3}{4}\rho = {\rho _W}$

Similarly using $f' = \dfrac{f}{3}$ when the mass was immersed in liquid with density${\rho _L}$

$f' = \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg - V{\rho _L}g}}{m}}$
$\dfrac{f}{3} = \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg - V{\rho _L}g}}{m}}$
Using the value of $f = \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg}}{m}}$ in above equation and squaring the whole equation we get,
$\dfrac{1}{3} \times \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg}}{m}} = \dfrac{1}{{2l}}\sqrt {\dfrac{{Mg - V{\rho _L}g}}{m}}$
$\dfrac{1}{{36{l^2}}}\dfrac{{Mg}}{m} = \dfrac{1}{{4{l^2}}}\dfrac{{Mg}}{m} - \dfrac{1}{{4{l^2}}}\dfrac{{V{\rho _L}g}}{m}$
$\dfrac{8}{9}M = V{\rho _L}$

Using $M = V\rho$ where the density of the mass connected is $\rho$

$\dfrac{8}{9}\rho = {\rho _L}$

Now taking their ratio will give us the specific gravity of the liquid
$\dfrac{{{\rho _L}}}{{{\rho _W}}} = \dfrac{{\dfrac{8}{9}\rho }}{{\dfrac{3}{4}\rho }} = \dfrac{{32}}{{27}}$
The correct option is D.

Note: You the also solve similar question using the following equation when u have the frequency of sonometer with mass immersed in water ${f_w}$ and that of when mass is immersed in liquid ${f_L}$ and you are asked to find the special gravity or in any cases where two of the factors are given and you are asked to find the third.
$\dfrac{{{\rho _L}}}{{{\rho _w}}} = \dfrac{{{f^2} - {{({f_L})}^2}}}{{{f^2} - {{({f_w})}^2}}}$
In this question, by substituting the values we get
$\dfrac{{{\rho _L}}}{{{\rho _w}}} = \dfrac{{{f^2} - {{(\dfrac{f}{3})}^2}}}{{{f^2} - {{(\dfrac{f}{2})}^2}}} = \dfrac{{32}}{{27}}$