A glass tube of `1.0 m` length is filled with water . The water can be drained out slowly at the bottom of the tube . If a vibrating tuning fork of frequency `500 Hz` is brought at the upper end of the tube and the velocity of sound is `300 m//s`, then the total number of resonances obtained will be

Options

A. A. 4
B. B. 3
C. C. 2
D. 4. 1

Explanation

To solve the problem of finding the total number of resonances obtained in a glass tube filled with water when a tuning fork is applied, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a glass tube of length \( L = 1.0 \, \text{m} \) filled with water. As the water is drained, the effective length of the air column in the tube will change.

2. **Identify the Frequency and Velocity**:
- The frequency of the tuning fork is given as \( f = 500 \, \text{Hz} \).
- The velocity of sound in air is given as \( v = 300 \, \text{m/s} \).

3. **Calculate the Wavelength**:
- The wavelength \( \lambda \) of the sound can be calculated using the formula:
\[ \lambda = \frac{v}{f} \]
- Substituting the values:
\[ \lambda = \frac{300 \, \text{m/s}}{500 \, \text{Hz}} = 0.6 \, \text{m} \]

4. **Closed Organ Pipe Behavior**:
- The tube behaves like a closed organ pipe (one end closed, one end open). The resonant frequencies for a closed pipe are given by:
\[ f_n = \frac{(2n - 1)v}{4L} \]
- Here, \( n \) is the mode number (1, 2, 3, ...).

5. **Finding the Maximum Resonance Condition**:
- The maximum length of the air column is \( L = 1.0 \, \text{m} \).
- Setting up the inequality for the maximum length:
\[ 1.0 \leq \frac{(2n - 1) \cdot 300}{4 \cdot 500} \]
- Simplifying this:
\[ 1.0 \leq \frac{(2n - 1) \cdot 300}{2000} \]
\[ 1.0 \leq \frac{(2n - 1)}{6.67} \]
\[ 6.67 \leq 2n - 1 \]
\[ 2n \geq 7.67 \quad \Rightarrow \quad n \geq 3.835 \]

6. **Finding Possible Integer Values for n**:
- Since \( n \) must be a positive integer, the smallest integer \( n \) that satisfies this condition is \( n = 4 \).
- However, we need to check the maximum possible \( n \) for which the length of the air column remains less than or equal to 1.0 m.

7. **Counting the Resonances**:
- The possible values of \( n \) are 1, 2, 3, and 4.
- The valid resonances occur at odd integers (1, 3, 5, ...), so we have:
- \( n = 1 \) (1st resonance)
- \( n = 2 \) (2nd resonance)
- \( n = 3 \) (3rd resonance)
- \( n = 4 \) (4th resonance)
- The maximum odd integer \( n \) that satisfies the condition is 3.

8. **Total Number of Resonances**:
- Therefore, the total number of resonances obtained is 2 (for \( n = 1 \) and \( n = 3 \)).

### Final Answer:
The total number of resonances obtained is **2**.