Question

A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency $512\text{Hz}$ produces first resonance when the tube is filled with water to a mark $11\text{cm}$ below a reference mark, near the open end of the tube, The experiment is repeated with another fork of frequency $256\text{Hz}$ which produces first resonance when water reaches a mark $27\text{cm}$ below the reference mark. The velocity of sound in air, obtained in the experiment, is close to:

• A. $341{\text{ms}}^{-1}$
• B. $335{\text{ms}}^{-1}$
• C. $328{\text{ms}}^{-1}$
• D. $322{\text{ms}}^{-1}$

Answer 1

The correct option is C $328{\text{ms}}^{-1}$

The length for the first resonance in the closed tube, considering the end correction, is,

$L_1=\frac{{\lambda }_1}{4}+e$

$\Rightarrow L_1=\frac{v}{4f_1}+e---\left(1\right)$

Similarly, for the second tuning fork, the length of the first resonance is given by,

$\Rightarrow L_2=\frac{v}{4f_2}+e---\left(2\right)$

$\left(2\right)-\left(1\right)\Rightarrow L_2-L_1=\frac{v}{4}(\frac{1}{f_2}-\frac{1}{f_1})$

$\Rightarrow 16\times {10}^{-2}=\frac{v}{4}\left(\frac{512-256}{512\times 256}\right)$

$\Rightarrow v=64\times 512\times {10}^{-2}\approx 328{\text{ms}}^{-1}$

Hence, option $\left(D\right)$ is correct.