Question

A tuning fork of frequency $300\text{Hz}$ is vibrated on arm-$1$, then the air column vibrates in fundamental mode. If the same tuning fork is vibrated on arm-$2$, the air column vibrates in first overtone. Both the arms are of length $1\text{m}$. If velocity of sound is $300\text{m/s},g=10{\text{m/s}}^2$ and density of water is ${10}^3{\text{kg/m}}^3$ then the density of the unknown liquid is
$($Neglect surface tension and end corrections$)$

• A. $2\times {10}^3{\text{kg/m}}^3$
• B. $4\times {10}^3{\text{kg/m}}^3$
• C. $3\times {10}^3{\text{kg/m}}^3$
• D. $5\times {10}^3{\text{kg/m}}^3$

Answer 1

The correct option is C $3\times {10}^3{\text{kg/m}}^3$

For resonance of tuning fork with arm-$1$,
$f_0=300\text{Hz}=$ frequency of tuning fork.
The respective arm will behave as a closed organ pipe.
$\Rightarrow f_0=\frac{v}{4l_1}$ $($fundamental mode$)$
$\Rightarrow 300=\frac{300}{4l_1}$
$\Rightarrow l_1=0.25\text{m}$
Similarly, for arm-$2$,
$\Rightarrow 300=\frac{3v}{4l_2}$ $($first overtone$)$
$\Rightarrow 300=\frac{3\times 300}{4l_2}$
$\Rightarrow l_2=0.75\text{m}$

Here, $P_0$ is atmospheric pressure.
On applying the Pascal's law in both arms and equating pressure at liquid junction.
$\Rightarrow P_0+{\rho }_{w}g(1-0.25)=P_0+{\rho }_{l}g(1-0.75)$
$\Rightarrow 0.75{\rho }_{w}g=0.25{\rho }_{l}g$
$\Rightarrow {\rho }_l=3{\rho }_w=3\times {10}^3{\text{kg/m}}^3$
Thus, the density of the unknown liquid is $3\times {10}^3{\text{kg/m}}^3$.