Question

A stone is hung in the air from a wire which is stretched over a sonometer. The bridge of the sonometer is $40cm$ apart when the wire is in unison with a tuning fork of frequency $256$ . When the stone is completely immersed in water the length between the bridges is $22cm$ for re-establishing (same mode) unison with the same tuning fork. The specific gravity of the material of the stone is
A. $\dfrac{{{{\left( {40} \right)}^2}}}{{{{\left( {40} \right)}^2} + {{\left( {22} \right)}^2}}}$
B. $\dfrac{{{{\left( {40} \right)}^2}}}{{{{\left( {40} \right)}^2} - {{\left( {22} \right)}^2}}}$
C. $256 \times \dfrac{{22}}{{40}}$
D. $256 \times \dfrac{{40}}{{22}}$

Answer 1

Hint: This problem is based on sound waves, we know that frequency of the waves directly varies with tension in the string, but as frequency is unchanged according to the question hence, we will look into other factors such as density (as mediums are different in 2 cases) to identify the steps resulting the solution of the problem.

Formula used:
We know the formula to calculate the frequency($f$) of vibration is given as:
$f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $ or $f.2l = \sqrt {\dfrac{T}{m}} $
Here, $T$ is the tension, $m$ is the mass per unit length and $l$ is the length of the wire.
The expression for specific gravity is,
$\text{Specific Gravity} = \sigma = \dfrac{{{\rho _a}}}{{{\rho _w}}}$
Here, ${\rho _a}$ is the density of air and ${\rho _w}$ is the density of water.

Complete step by step solution:
As, $f$ is constant then $l$ is directly proportional to $\sqrt T $.
Case 1. When a stone is hung in the air, the bridge of the sonometer is $L = 40\,cm$(given)
Case 2. When a stone is completely immersed in water, the bridge of the sonometer becomes$l = 22\,cm$(given)
Considering the above 2 cases, we get
$\dfrac{L}{l} = \dfrac{{\sqrt {{T_{air}}} }}{{\sqrt {{T_{water}}} }}$ … (1)
Also, for a submerged object, the buoyant force is $F = Volume(V) \times density\left( \rho \right) \times g$

Applying this formula in eq. (1), we get
$\dfrac{L}{l} = \dfrac{{\sqrt {V{\rho _a}g} }}{{\sqrt {V{\rho _a}g - V{\rho _w}g} }} = \sqrt {\dfrac{{V{\rho _a}g}}{{V({\rho _a} - {\rho _w})g}}} \\ $
$ \Rightarrow \dfrac{L}{l} = \sqrt {\dfrac{{{\rho _a}}}{{{\rho _w}\left( {\dfrac{{{\rho _a}}}{{{\rho _w}}} - 1} \right)}}} \\ $ … (2)
Here, Specific gravity will be:
$\text{Specific Gravity} = \sigma = \dfrac{{{\rho _a}}}{{{\rho _w}}} \\ $
$ \Rightarrow \dfrac{L}{l} = \sqrt {\dfrac{\sigma }{{\left( {\sigma - 1} \right)}}} $

Squaring both sides, we get
$ \Rightarrow \dfrac{\sigma }{{\left( {\sigma - 1} \right)}} = \dfrac{{{L^2}}}{{{l^2}}}$
$ \therefore \sigma = \dfrac{{{L^2}}}{{{L^2} - {l^2}}} = \dfrac{{{{40}^2}}}{{{{44}^2} - {{22}^2}}}$
Thus, the specific gravity of the material of the stone is $\dfrac{{{{\left( {40} \right)}^2}}}{{{{\left( {40} \right)}^2} - {{\left( {22} \right)}^2}}}$.

Hence, the correct option is B.

Note: Because this is a problem involving sound waves and vibrations, and we know that the nature of sound waves in air and water differs, analyse the given conditions and identify the requirements to calculate the specific gravity as it provides a better understanding of the problem and will aid in obtaining precise results.