A long tube contains air pressure of `1 atm` and a temperature of `59^(@) C`. The tube is open at one end and closed at the other by a movable piston . A tuning fork near the open end is vibrating with a frequency of `500 Hz`. Resonance is produced when the piston is at distances `16 cm , 49.2 cm and 82.4 cm` from open end. Molar mass of air is `28.8 g//mol`.
Radius of tube is

Options

A. (a)`1.1 cm`
B. (b)`1 cm`
C. (c)`1.2 cm`
D. (d)`2 cm`

Explanation

To find the radius of the tube, we can follow these steps:

### Step 1: Understand the resonance conditions
In a tube that is open at one end and closed at the other, the resonance occurs at odd multiples of the quarter wavelength. The first resonance occurs at \( \frac{\lambda}{4} \), the second at \( \frac{3\lambda}{4} \), and so on. The distances given (16 cm, 49.2 cm, and 82.4 cm) correspond to these resonances.

### Step 2: Set up the equations for resonance
1. For the first resonance (16 cm):
\[ L_1 = 16 \, \text{cm} = \frac{\lambda}{4} \]
2. For the second resonance (49.2 cm):
\[ L_2 = 49.2 \, \text{cm} = \frac{3\lambda}{4} \]
3. For the third resonance (82.4 cm):
\[ L_3 = 82.4 \, \text{cm} = \frac{5\lambda}{4} \]

### Step 3: Relate the lengths to the wavelength
From the equations above, we can express the wavelength \( \lambda \) in terms of the lengths:
1. From \( L_1 \):
\[ \lambda = 4L_1 = 4 \times 16 = 64 \, \text{cm} \]
2. From \( L_2 \):
\[ \lambda = \frac{4}{3}(L_2 - L_1) = \frac{4}{3}(49.2 - 16) = \frac{4}{3} \times 33.2 = 44.27 \, \text{cm} \]
3. From \( L_3 \):
\[ \lambda = \frac{4}{5}(L_3 - L_1) = \frac{4}{5}(82.4 - 16) = \frac{4}{5} \times 66.4 = 53.12 \, \text{cm} \]

### Step 4: Verify the wavelength consistency
The wavelength calculated from each resonance condition should be consistent. If they are not, we need to check our calculations.

### Step 5: Calculate the speed of sound
Using the ideal gas law and the given conditions (1 atm and 59°C), we can calculate the speed of sound in air:
\[ v = \sqrt{\frac{\gamma R T}{M}} \]
Where:
- \( \gamma \) (ratio of specific heats) for air is approximately \( 1.4 \).
- \( R \) (universal gas constant) is \( 8.314 \, \text{J/(mol K)} \).
- \( T \) (temperature in Kelvin) = \( 59 + 273.15 = 332.15 \, \text{K} \).
- \( M \) (molar mass in kg) = \( 28.8 \, \text{g/mol} = 0.0288 \, \text{kg/mol} \).

### Step 6: Calculate the radius of the tube
Using the wavelength \( \lambda \) and the frequency \( f = 500 \, \text{Hz} \):
\[ v = f \lambda \]
Thus, we can find the radius using the relationship between the diameter and the wavelength in a tube:
\[ d = \frac{\lambda}{0.3} \]
And then the radius \( r \):
\[ r = \frac{d}{2} \]

### Step 7: Final Calculation
Substituting the values, we can find the radius of the tube.