Each of the two strings of length `51.6 cm` and `49.1 cm` are tensioned separately by `20 N` force. Mass per unit length of both the strings is same and equal to `1 g//m`. When both the strings vibrate simultaneously, the number of beats is

Options

A. 3
B. 5
C. 7
D. 8

Explanation

To solve the problem step by step, we will calculate the frequencies of both strings and then find the number of beats produced when they vibrate simultaneously.

### Step 1: Convert lengths from centimeters to meters
Given lengths:
- \( L_1 = 51.6 \, \text{cm} = 0.516 \, \text{m} \)
- \( L_2 = 49.1 \, \text{cm} = 0.491 \, \text{m} \)

### Step 2: Identify the tension and mass per unit length
Given:
- Tension \( T = 20 \, \text{N} \)
- Mass per unit length \( \mu = 1 \, \text{g/m} = 0.001 \, \text{kg/m} \)

### Step 3: Calculate the frequency of the first string
The formula for the frequency of a vibrating string is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
For the first string:
\[ f_1 = \frac{1}{2 \times 0.516} \sqrt{\frac{20}{0.001}} \]
Calculating \( \sqrt{\frac{20}{0.001}} \):
\[ \sqrt{\frac{20}{0.001}} = \sqrt{20000} \approx 141.42 \]
Now substituting back into the frequency formula:
\[ f_1 = \frac{1}{2 \times 0.516} \times 141.42 \approx \frac{141.42}{1.032} \approx 137.0 \, \text{Hz} \]

### Step 4: Calculate the frequency of the second string
For the second string:
\[ f_2 = \frac{1}{2 \times 0.491} \sqrt{\frac{20}{0.001}} \]
Using the same \( \sqrt{\frac{20}{0.001}} \):
\[ f_2 = \frac{1}{2 \times 0.491} \times 141.42 \approx \frac{141.42}{0.982} \approx 144.0 \, \text{Hz} \]

### Step 5: Calculate the number of beats
The number of beats produced when two frequencies are played together is given by:
\[ \text{Number of beats} = |f_2 - f_1| \]
Substituting the values:
\[ \text{Number of beats} = |144.0 - 137.0| = 7 \]

### Final Answer
The number of beats produced is \( 7 \).

---