When two closed organ pipes at `0^@C` are sounded together 20 beats are heard in 4 seconds. How many beats will be heard per second when the temperature is `100^@C.` The increase in length of the pipes due to rise in temperature may be neglected.

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the Beat Frequency
When two sound waves of slightly different frequencies interfere, they produce a phenomenon known as beats. The beat frequency is given by the absolute difference between the two frequencies.

Given that 20 beats are heard in 4 seconds, we can calculate the beat frequency (B) at 0°C:
\[ B = \frac{20 \text{ beats}}{4 \text{ seconds}} = 5 \text{ beats per second} \]

### Step 2: Relate Frequencies to Lengths of Pipes
For closed organ pipes, the frequency of the sound produced is inversely proportional to the length of the pipe. The frequencies of the two pipes at 0°C can be expressed as:
\[ f_1 = \frac{v_0}{2L_1} \quad \text{and} \quad f_2 = \frac{v_0}{2L_2} \]
where \(v_0\) is the speed of sound at 0°C.

The beat frequency can be expressed as:
\[ f_1 - f_2 = 5 \text{ beats per second} \]
Substituting the expressions for \(f_1\) and \(f_2\):
\[ \frac{v_0}{2L_1} - \frac{v_0}{2L_2} = 5 \]
Factoring out \(\frac{v_0}{2}\):
\[ \frac{v_0}{2} \left( \frac{1}{L_1} - \frac{1}{L_2} \right) = 5 \]
Let’s denote this equation as (1).

### Step 3: Analyze Frequencies at 100°C
At 100°C, the speed of sound increases. The new frequencies can be expressed as:
\[ f_1' = \frac{v_{100}}{2L_1} \quad \text{and} \quad f_2' = \frac{v_{100}}{2L_2} \]
where \(v_{100}\) is the speed of sound at 100°C.

The new beat frequency is:
\[ f_1' - f_2' = B' \]
Substituting the expressions for \(f_1'\) and \(f_2'\):
\[ \frac{v_{100}}{2L_1} - \frac{v_{100}}{2L_2} = B' \]
Factoring out \(\frac{v_{100}}{2}\):
\[ \frac{v_{100}}{2} \left( \frac{1}{L_1} - \frac{1}{L_2} \right) = B' \]
Let’s denote this equation as (2).

### Step 4: Divide Equations (1) and (2)
Dividing equation (1) by equation (2):
\[ \frac{\frac{v_0}{2} \left( \frac{1}{L_1} - \frac{1}{L_2} \right)}{\frac{v_{100}}{2} \left( \frac{1}{L_1} - \frac{1}{L_2} \right)} = \frac{5}{B'} \]
The terms \(\left( \frac{1}{L_1} - \frac{1}{L_2} \right)\) cancel out:
\[ \frac{v_0}{v_{100}} = \frac{5}{B'} \]

### Step 5: Relate Speeds to Temperature
The speed of sound is directly proportional to the square root of the absolute temperature:
\[ \frac{v_0}{v_{100}} = \sqrt{\frac{T_0}{T_{100}}} \]
Where \(T_0 = 273 \text{ K}\) (0°C) and \(T_{100} = 373 \text{ K}\) (100°C):
\[ \frac{v_0}{v_{100}} = \sqrt{\frac{273}{373}} \]

### Step 6: Substitute and Solve for \(B'\)
Substituting this into the equation:
\[ \sqrt{\frac{273}{373}} = \frac{5}{B'} \]
Rearranging gives:
\[ B' = 5 \cdot \sqrt{\frac{373}{273}} \]

### Step 7: Calculate \(B'\)
Calculating the numerical value:
\[ B' \approx 5 \cdot \sqrt{1.366} \approx 5 \cdot 1.167 \approx 5.835 \text{ beats per second} \]

### Final Answer
Thus, the number of beats heard per second when the temperature is 100°C is approximately:
\[ B' \approx 6 \text{ beats per second} \]