A string `25cm` long and having a mass of `2.5 gm` is under tension. A pipe closed at one end is `40cm` long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, `8` beats per second are heard. It is observed that decreasing the tension in the string decreases beat frequency. If the speed of sound in air is `320m//s`, find the tension in the string.

Explanation

To find the tension in the string, we will follow these steps:

### Step 1: Calculate the linear mass density (μ) of the string.
The linear mass density (μ) is given by the formula:
\[ \mu = \frac{m}{L} \]
where:
- \( m = 2.5 \, \text{g} = 0.0025 \, \text{kg} \) (converting grams to kilograms)
- \( L = 25 \, \text{cm} = 0.25 \, \text{m} \) (converting centimeters to meters)

Substituting the values:
\[ \mu = \frac{0.0025 \, \text{kg}}{0.25 \, \text{m}} = 0.01 \, \text{kg/m} \]

### Step 2: Calculate the frequency of the string in its first overtone.
The frequency of a vibrating string in its \( n \)-th overtone is given by:
\[ f_n = n \cdot \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
For the first overtone (\( n = 2 \)):
\[ f_2 = 2 \cdot \frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{1}{L} \sqrt{\frac{T}{\mu}} \]
Substituting \( L = 0.25 \, \text{m} \) and \( \mu = 0.01 \, \text{kg/m} \):
\[ f_2 = \frac{1}{0.25} \sqrt{\frac{T}{0.01}} = 4 \sqrt{\frac{T}{0.01}} = 4 \cdot 10 \sqrt{T} = 40 \sqrt{T} \]

### Step 3: Calculate the frequency of the pipe in its fundamental frequency.
For a pipe closed at one end, the fundamental frequency is given by:
\[ f = \frac{v}{\lambda} \]
where \( v \) is the speed of sound in air and \( \lambda \) is the wavelength. The wavelength for the fundamental frequency in a pipe closed at one end is:
\[ \lambda = 4L \]
Substituting \( L = 0.4 \, \text{m} \):
\[ \lambda = 4 \cdot 0.4 = 1.6 \, \text{m} \]
Now, using \( v = 320 \, \text{m/s} \):
\[ f = \frac{320}{1.6} = 200 \, \text{Hz} \]

### Step 4: Set up the equation for beat frequency.
The beat frequency is given by the difference in frequencies of the two sources:
\[ \text{Beat frequency} = |f_2 - f| = 8 \, \text{Hz} \]
Substituting the expressions for \( f_2 \) and \( f \):
\[ |40 \sqrt{T} - 200| = 8 \]

### Step 5: Solve the equation for T.
This gives us two cases to consider:
1. \( 40 \sqrt{T} - 200 = 8 \)
2. \( 40 \sqrt{T} - 200 = -8 \)

**Case 1:**
\[ 40 \sqrt{T} = 208 \implies \sqrt{T} = \frac{208}{40} = 5.2 \implies T = (5.2)^2 = 27.04 \, \text{N} \]

**Case 2:**
\[ 40 \sqrt{T} = 192 \implies \sqrt{T} = \frac{192}{40} = 4.8 \implies T = (4.8)^2 = 23.04 \, \text{N} \]

### Conclusion:
The tension in the string can be either \( 27.04 \, \text{N} \) or \( 23.04 \, \text{N} \). However, since it is observed that decreasing the tension decreases the beat frequency, we will take the higher value of tension as the correct answer.

Thus, the tension in the string is:
\[ \boxed{27.04 \, \text{N}} \]