A glass tube of length `1.5 m` is filled completely with water , the water can be drained out slowly at the bottom of the tube . Find the total number of resonance obtained , when a tuning fork of frequency `606 Hz` is put at the upper open end of the tube , `v_(sound) = 340 m//s`.

Options

A. `2`
B. `3`
C. `4`
D. `5`

Explanation

To solve the problem, we need to determine the total number of resonances that can be obtained in a glass tube of length 1.5 m when a tuning fork of frequency 606 Hz is placed at the open end of the tube. The speed of sound in air is given as 340 m/s.

### Step-by-Step Solution:

1. **Understand the Setup**:
- The glass tube is closed at one end (the bottom) and open at the other end (the top).
- The length of the tube (L) is 1.5 m.

2. **Resonance Condition**:
- For a tube closed at one end, the resonance frequencies are given by the formula:
\[ f_n = \frac{(2n + 1)v}{4L} \]
- Here, \( n \) is the mode number (0, 1, 2, ...), \( v \) is the speed of sound, and \( L \) is the length of the tube.

3. **Rearranging the Formula**:
- We can rearrange the formula to find the length corresponding to each resonance:
\[ L_n = \frac{(2n + 1)v}{4f} \]

4. **Substituting Values**:
- Substitute \( v = 340 \, \text{m/s} \) and \( f = 606 \, \text{Hz} \):
\[ L_n = \frac{(2n + 1) \cdot 340}{4 \cdot 606} \]

5. **Calculating Lengths for Different Modes**:
- Calculate \( L_n \) for different values of \( n \):
- For \( n = 0 \):
\[ L_0 = \frac{(2 \cdot 0 + 1) \cdot 340}{4 \cdot 606} = \frac{340}{2424} \approx 0.140 \, \text{m} \, (14 \, \text{cm}) \]
- For \( n = 1 \):
\[ L_1 = \frac{(2 \cdot 1 + 1) \cdot 340}{4 \cdot 606} = \frac{3 \cdot 340}{2424} \approx 0.420 \, \text{m} \, (42 \, \text{cm}) \]
- For \( n = 2 \):
\[ L_2 = \frac{(2 \cdot 2 + 1) \cdot 340}{4 \cdot 606} = \frac{5 \cdot 340}{2424} \approx 0.700 \, \text{m} \, (70 \, \text{cm}) \]
- For \( n = 3 \):
\[ L_3 = \frac{(2 \cdot 3 + 1) \cdot 340}{4 \cdot 606} = \frac{7 \cdot 340}{2424} \approx 0.980 \, \text{m} \, (98 \, \text{cm}) \]
- For \( n = 4 \):
\[ L_4 = \frac{(2 \cdot 4 + 1) \cdot 340}{4 \cdot 606} = \frac{9 \cdot 340}{2424} \approx 1.260 \, \text{m} \, (126 \, \text{cm}) \]
- For \( n = 5 \):
\[ L_5 = \frac{(2 \cdot 5 + 1) \cdot 340}{4 \cdot 606} = \frac{11 \cdot 340}{2424} \approx 1.540 \, \text{m} \, (154 \, \text{cm}) \, \text{(not possible)} \]

6. **Determine the Maximum Number of Resonances**:
- The maximum length \( L \) that can be achieved is 1.5 m. The lengths calculated for \( n = 0, 1, 2, 3, 4 \) are all less than 1.5 m, but \( L_5 \) exceeds 1.5 m.
- Therefore, the total number of resonances is 5.

### Final Answer:
The total number of resonances obtained is **5**.