A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in air. A tuning fork of frequency $512$ Hz produces first resonance when the tube is filled with water to a mark $11$ cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency $256$ Hz which produces first resonance when water reaches a mark $27$ cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to
Answer: (D) $328\ \text{m s}^{-1}$
Let $x$ be the unknown distance from the effective open end (including end correction) to the reference mark. For the first resonance the air column is $\lambda/4$:
$$\frac{v}{4\times512} = 11 + x,\qquad \frac{v}{4\times256} = 27 + x$$
Subtracting removes $x$:
$$\frac{v}{1024} - \frac{v}{2048} = 16 \Rightarrow v = 32768\ \text{cm/s} \approx 328\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics