A pipe 20 cm long is open at both ends. Which harmonic of mode of the pipe is resonantly excited by a 1.66 kHz Given : velocity of sound `= 332 ms^(-1)`.

Answer

To determine which harmonic of a pipe that is 20 cm long and open at both ends is resonantly excited by a frequency of 1.66 kHz, we can follow these steps:

### Step 1: Convert the Length of the Pipe to Meters
The length of the pipe is given as 20 cm. We need to convert this to meters for our calculations.

\[ L = 20 \, \text{cm} = 0.2 \, \text{m} \]

### Step 2: Identify the Given Values
We are given:
- Frequency \( f = 1.66 \, \text{kHz} = 1660 \, \text{Hz} \)
- Velocity of sound \( v = 332 \, \text{m/s} \)
- Length of the pipe \( L = 0.2 \, \text{m} \)

### Step 3: Use the Formula for Frequency of the nth Harmonic
For a pipe open at both ends, the frequency of the nth harmonic is given by the formula:

\[ f = n \frac{v}{2L} \]

Where:
- \( f \) = frequency
- \( n \) = harmonic number
- \( v \) = velocity of sound
- \( L \) = length of the pipe

### Step 4: Rearrange the Formula to Solve for n
We can rearrange the formula to solve for \( n \):

\[ n = \frac{f \cdot 2L}{v} \]

### Step 5: Substitute the Known Values into the Equation
Now, we substitute the known values into the equation:

\[ n = \frac{1660 \, \text{Hz} \cdot 2 \cdot 0.2 \, \text{m}}{332 \, \text{m/s}} \]

### Step 6: Calculate the Value of n
Now we perform the calculations:

\[ n = \frac{1660 \cdot 0.4}{332} \]

Calculating the numerator:

\[ 1660 \cdot 0.4 = 664 \]

Now divide by the velocity of sound:

\[ n = \frac{664}{332} = 2 \]

### Conclusion
The harmonic mode of the pipe that is resonantly excited by a frequency of 1.66 kHz is the **2nd harmonic**.

---