Question

11.Air column of 20cmlength in a resonance tube resonates with a certain tuning fork when sounded at its upper open end.The lower end of the tube is closed and adjustable by changing the quantity of mercury filled inside the tube.The temperature of the air is 27⁰C.The change in length of the air column required,if the temperature falls to 7⁰C and the same tuning fork is again sounded at the upper open end is ---

Answer 1

The situation can be considered to be a closed organ pipe.
In closed organ pipes,
Understand that the first fundamental note is in resonance with the air column present in it when it satisfies the condition
$$\begin{gathered} l=\frac{\lambda }{4} \\ l=lengthoftheaircolumn \\ \lambda =wavelength \end{gathered}$$
Given
$l=20cm$
Hence
wavelength is 80 cm.
Thus the natural frequency is
$$\begin{gathered} v=f\lambda \\ v=speedair \\ f=frequency \end{gathered}$$
Speed of air at any given temperature is given by the relation
$$\begin{gathered} V=331.4+0.6T \\ TemperatureexpressedinCelcius \end{gathered}$$
Hence
Speed of air at ${27}^{\circ }C$ is approximately 348 m/s
Hence the frequency of the tuning fork is
$$\begin{gathered} f=\frac{348}{0.80} \\ f=435Hz \end{gathered}$$

When the temperature is reduced by $7^{\circ }\hspace{0.17em}C$ the speed is 343.4 m/s
Hence the wavelength of wave would be
$$\begin{gathered} \lambda =\frac{343.4}{435} \\ \lambda =0.7894m \end{gathered}$$
Hence the length of the air column is
$$\begin{gathered} l=\frac{\lambda }{4} \\ l=0.197m \\ l=19.7cm \end{gathered}$$