A long glass tube is held vertically , dipping into water, while a tuning fork of frequency `512 H_(Z)` is respeatedly struck and held over the open end. Strong resonance is obtained, when the length of the tube above the surface of water is `50 cm` and again `84 cm`, but not at any intermediate point. Find the speed of sound of sound in air and next length of the air column for resonance.

Explanation

To solve the problem, we need to find the speed of sound in air and the next length of the air column for resonance in a long glass tube dipped in water.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Frequency of the tuning fork, \( f = 512 \, \text{Hz} \)
- Length of the air column for first resonance, \( L_1 = 50 \, \text{cm} = 0.50 \, \text{m} \)
- Length of the air column for second resonance, \( L_2 = 84 \, \text{cm} = 0.84 \, \text{m} \)

2. **Calculate the Difference in Lengths:**
\[ \Delta L = L_2 - L_1 = 0.84 \, \text{m} - 0.50 \, \text{m} = 0.34 \, \text{m} \]

3. **Use the Formula for Speed of Sound:**
The speed of sound in air can be calculated using the formula:
\[ V = 2f \Delta L \]
Substituting the values:
\[ V = 2 \times 512 \, \text{Hz} \times 0.34 \, \text{m} \]

4. **Calculate the Speed of Sound:**
\[ V = 2 \times 512 \times 0.34 = 348.16 \, \text{m/s} \]

5. **Find the Next Length of the Air Column for Resonance:**
To find the next length \( L_3 \) for resonance, we use the relationship:
\[ L_3 - L_2 = \frac{V}{2f} \]
Substituting the values:
\[ L_3 - 0.84 = \frac{348.16}{2 \times 512} \]
Calculate \( \frac{348.16}{1024} \):
\[ L_3 - 0.84 = 0.34 \, \text{m} \]

6. **Calculate \( L_3 \):**
\[ L_3 = 0.84 + 0.34 = 1.18 \, \text{m} \]

### Final Answers:
- Speed of sound in air: \( V \approx 348.16 \, \text{m/s} \)
- Next length of the air column for resonance: \( L_3 = 1.18 \, \text{m} \)