When a tuning fork A of unknown frequency is sounded with another tuning fork B of frequency 256Hz, then 3 beats per second are observed. After that A is loaded with wax and sounded, the again 3 beats per second are observed. The frequency of the tuning fork A is

Options

A. `250Hz`
B. `253Hz`
C. `259Hz`
D. `262Hz`

Explanation

To find the frequency of tuning fork A, we can follow these steps:

### Step 1: Understand the concept of beats
When two tuning forks of slightly different frequencies are sounded together, they produce a phenomenon known as beats. The number of beats per second is equal to the absolute difference between the frequencies of the two tuning forks.

### Step 2: Set up the equations
Let the frequency of tuning fork A be \( f_A \) (unknown) and the frequency of tuning fork B be \( f_B = 256 \, \text{Hz} \). The number of beats observed is 3 beats per second. Therefore, we can write:
\[ |f_A - f_B| = 3 \]

This gives us two possible equations:
1. \( f_A - 256 = 3 \) (when \( f_A > f_B \))
2. \( 256 - f_A = 3 \) (when \( f_A < f_B \))

### Step 3: Solve the equations
**Case 1:** If \( f_A > 256 \):
\[ f_A - 256 = 3 \\ f_A = 256 + 3 = 259 \, \text{Hz} \]

**Case 2:** If \( f_A < 256 \):
\[ 256 - f_A = 3 \\ f_A = 256 - 3 = 253 \, \text{Hz} \]

### Step 4: Consider the effect of loading with wax
When tuning fork A is loaded with wax, its frequency decreases. Since the number of beats remains the same (3 beats per second) after loading with wax, we can conclude that the frequency of tuning fork A must now be less than the frequency of tuning fork B (256 Hz).

Thus, the only valid solution from our cases is:
\[ f_A = 253 \, \text{Hz} \]

### Conclusion
The frequency of tuning fork A is \( 253 \, \text{Hz} \).

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