When tuning forks A and B are sounded together `5` beats per second are heard. Frequency of A is `250Hz`. On loading A with wax `2` beats per second are produced with B. The frequency of B is

Options

A. `255Hz`
B. `320Hz`
C. `245Hz`
D. `420Hz`

Explanation

To solve the problem, we need to analyze the information given about the tuning forks A and B and the beats produced when they are sounded together.

### Step-by-Step Solution:

1. **Understanding Beats**: When two tuning forks are sounded together, the number of beats per second is equal to the absolute difference in their frequencies. If \( f_A \) is the frequency of fork A and \( f_B \) is the frequency of fork B, then the number of beats \( n \) is given by:
\[ n = |f_A - f_B| \]

2. **Initial Condition**: We know that when forks A and B are sounded together, 5 beats per second are heard. Given that the frequency of fork A (\( f_A \)) is 250 Hz, we can write:
\[ |250 - f_B| = 5 \]
This gives us two possible equations:
\[ 250 - f_B = 5 \quad \text{or} \quad f_B - 250 = 5 \]
From these, we can solve for \( f_B \):
- From \( 250 - f_B = 5 \):
\[ f_B = 250 - 5 = 245 \text{ Hz} \]
- From \( f_B - 250 = 5 \):
\[ f_B = 250 + 5 = 255 \text{ Hz} \]

3. **Loading Fork A**: When fork A is loaded with wax, its frequency decreases. The problem states that now only 2 beats per second are produced with fork B. Let the new frequency of fork A be \( f_A' \):
\[ |f_A' - f_B| = 2 \]
Since \( f_A' < 250 \) (because loading with wax decreases the frequency), we can express this as:
\[ f_B - f_A' = 2 \]

4. **Finding the New Frequency**: We already have two possible values for \( f_B \) from step 2: 245 Hz and 255 Hz. We will check both cases to find the correct one.

- **Case 1**: If \( f_B = 245 \) Hz:
\[ 245 - f_A' = 2 \implies f_A' = 245 - 2 = 243 \text{ Hz} \]
This is a valid solution since \( f_A' < 250 \).

- **Case 2**: If \( f_B = 255 \) Hz:
\[ 255 - f_A' = 2 \implies f_A' = 255 - 2 = 253 \text{ Hz} \]
This is not valid since \( f_A' \) cannot be greater than 250 Hz.

5. **Conclusion**: The only valid case is when \( f_B = 245 \) Hz. Therefore, the frequency of tuning fork B is:
\[ f_B = 245 \text{ Hz} \]

### Final Answer:
The frequency of tuning fork B is **245 Hz**.