Question

A resonance tube completely filled with water has small hole at the bottom. Length of the tube is 0.8 m. A vibrating tuning fork of frequency 500 Hz is held near the open end of tube. Water is slowly removed from the bottom. The maximum number of resonances heard will be (neglect end correction. Speed of sound in air = 340 m/s)

• A. 2
• B. 5
• C. 3
• D. 4

Answer 1

Answer: (A)

2

Answer 2

The resonance condition for a tube with one end closed (air column length $L$) is:

$$L = \frac{(2n-1) \lambda}{4} \quad \text{for } n = 1, 2, 3, \dots$$

where the wavelength $\lambda$ is given by:

$$\lambda = \frac{v}{f} = \frac{340}{500} = 0.68\, \text{m}$$

Thus,

$$L = (2n-1) \times \frac{0.68}{4} = (2n-1) \times 0.17\, \text{m}$$

The air column length can vary from 0 (when the tube is completely filled with water) to $0.8\, \text{m}$.

For $n = 1$:

$$L = 1 \times 0.17 = 0.17\, \text{m} \quad (\leq 0.8\, \text{m})$$

For $n = 2$:

$$L = 3 \times 0.17 = 0.51\, \text{m} \quad (\leq 0.8\, \text{m})$$

For $n = 3$:

$$L = 5 \times 0.17 = 0.85\, \text{m} \quad (> 0.8\, \text{m})$$

Thus, the maximum number of resonances heard is when $n = 1$ and $n = 2$, i.e., 2 resonances.