1. Use first resonance condition for a closed pipe

For a resonance tube closed at one end by water, the first resonance length is

$L + e = \frac{\lambda}{4}$

where:

• $L$ = measured air column length,
• $e$ = end correction,
• $\lambda$ = wavelength.

Since the tube has a jagged end, $e$ is unknown, but it remains the same in both observations.

2. Write equations for the two tuning forks

Let the reference mark be near the open end.

• For frequency $f_1 = 512\,\text{Hz}$, first resonance occurs at $L_1 = 11\,\text{cm} = 0.11\,\text{m}$

So, $L_1 + e = \frac{v}{4f_1}$ $0.11 + e = \frac{v}{4\times 512}$

• For frequency $f_2 = 256\,\text{Hz}$, first resonance occurs at $L_2 = 27\,\text{cm} = 0.27\,\text{m}$

So, $L_2 + e = \frac{v}{4f_2}$ $0.27 + e = \frac{v}{4\times 256}$

3. Eliminate end correction

Subtract the first equation from the second:

$0.27 - 0.11 = \frac{v}{4\times 256} - \frac{v}{4\times 512}$

$0.16 = \frac{v}{1024}$

Therefore,

$v = 0.16 \times 1024 = 163.84\,\text{m/s}$

This is clearly not a reasonable speed of sound, so let us check carefully.

4. Check the resonance relation properly

For first resonance in a closed pipe:

$L+e = \frac{\lambda}{4} = \frac{v}{4f}$

Thus,

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

Substitute values:

$0.16 = \frac{v}{4}\left(\frac{1}{256}-\frac{1}{512}\right)$

Now,

$\frac{1}{256}-\frac{1}{512}=\frac{2-1}{512}=\frac{1}{512}$

Hence,

$0.16 = \frac{v}{4}\cdot \frac{1}{512} = \frac{v}{2048}$

So,

$v = 0.16 \times 2048 = 327.68\,\text{m/s}$

5. Choose the nearest option

$v \approx 328\,\text{m/s}$

So the correct option is B.

6. Verification with options

• A: $335\,\text{m/s}$ — not closest
• B: $328\,\text{m/s}$ — correct
• C: $341\,\text{m/s}$ — not correct
• D: $322\,\text{m/s}$ — not closest

Therefore, the velocity of sound is $328\,\text{m/s}$.