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

For a resonance tube closed at the water surface and open at the top, the first (fundamental) resonance condition is stated by the formula

$$L \;=\;\frac{\lambda}{4},$$

where $$L$$ is the effective length of the vibrating air column and $$\lambda$$ is the wavelength of the sound in air. The effective length is always a little greater than the measured air-column length because of the end correction at the open end.

Let us denote by $$e$$ the end correction produced by the open end and by $$d$$ the unknown extra length between the jagged tip of the tube and the clearly visible reference mark. Both $$e$$ and $$d$$ are constant for every observation on this particular tube.

When the 512 Hz fork is used, the water is 11 cm below the reference mark. The measured length of the air column is therefore 11 cm, so the effective length becomes

$$L_1 \;=\;11\;\text{cm}+d+e.$$

When the 256 Hz fork is used, the water is 27 cm below the reference mark, giving

$$L_2 \;=\;27\;\text{cm}+d+e.$$

For each fork the fundamental condition $$L=\lambda/4$$ must be satisfied. We use the relation $$\lambda=v/f,$$ where $$v$$ is the velocity of sound and $$f$$ is the frequency. Substituting for each fork we have

$$11+d+e \;=\;\frac{v}{4\,f_1},\qquad f_1=512\;\text{Hz},$$

$$27+d+e \;=\;\frac{v}{4\,f_2},\qquad f_2=256\;\text{Hz}.$$

It is convenient to combine the two unknown constants $$d$$ and $$e$$ into a single constant

$$a \;=\;d+e.$$

Writing the two equations with this shorthand gives

$$11+a \;=\;\frac{v}{4\times512},$$

$$27+a \;=\;\frac{v}{4\times256}.$$

From the first equation we have

$$v \;=\;4\times512\,(11+a)\;=\;2048\,(11+a).$$

From the second equation we have

$$v \;=\;4\times256\,(27+a)\;=\;1024\,(27+a).$$

Because both right-hand sides equal the same velocity $$v$$, we equate them:

$$2048\,(11+a) \;=\;1024\,(27+a).$$

Dividing every term by 1024 simplifies the equation to

$$2\,(11+a) \;=\;27+a.$$

Now expanding and gathering like terms,

$$22+2a \;=\;27+a,$$

$$2a-a \;=\;27-22,$$

$$a \;=\;5\;\text{cm}.$$

Substituting $$a=5\;\text{cm}$$ into the first velocity expression gives

$$v \;=\;2048\,(11+5)\;=\;2048\times16\;\text{cm s}^{-1}.$$

Calculating the product,

$$2048\times16=32768\;\text{cm s}^{-1}.$$

Converting centimetres per second to metres per second (1 m = 100 cm),

$$v \;=\;\frac{32768}{100}\;\text{m s}^{-1}\;=\;327.68\;\text{m s}^{-1}.$$

Among the given options, 328 m s−1 is the closest to this calculated value.

Hence, the correct answer is Option D.