A glass tube `1.5 m` long and open at both ends, is immersed vertically in a water tank completely. A tuning fork of 660 Hz is vibrated and kept at the upper end of the tube and the tube is gradually raised out of water the total number of resonances heard before the tube comes out of water taking velocity of sound air `330m//s` is

Options

A. 12
B. 6
C. 8
D. 4

Answer

To solve the problem, we need to find the total number of resonances heard as the glass tube is raised out of the water. The tube is open at both ends, and we can use the formula for the frequency of a tube open at both ends to find the number of resonances.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the tube, \( L = 1.5 \, \text{m} \)
- Frequency of the tuning fork, \( f = 660 \, \text{Hz} \)
- Velocity of sound in air, \( v = 330 \, \text{m/s} \)

2. **Understand the Resonance Condition:**
For a tube open at both ends, the resonance frequencies are given by:
\[ f_n = \frac{n \cdot v}{2L} \]
where \( n \) is the harmonic number (1, 2, 3,...).

3. **Rearranging the Formula:**
We can rearrange the formula to solve for \( n \):
\[ n = \frac{2Lf}{v} \]

4. **Substituting the Values:**
Now, substitute the known values into the equation:
\[ n = \frac{2 \cdot 1.5 \, \text{m} \cdot 660 \, \text{Hz}}{330 \, \text{m/s}} \]

5. **Calculating the Value of \( n \):**
- Calculate the numerator:
\[ 2 \cdot 1.5 \cdot 660 = 1980 \]
- Now divide by the velocity:
\[ n = \frac{1980}{330} = 6 \]

6. **Conclusion:**
The total number of resonances heard before the tube comes out of the water is \( n = 6 \).

### Final Answer:
The total number of resonances heard is **6**.

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