The tension of a piano wire is 16 Kg weight. What must be the change in tension so as to produce a tone an octave lower :

Options

A. 8kg weight
B. 32 kg weight
C. 12 kg weight
D. 4 kg weight

Answer

To solve the problem, we need to determine the change in tension required to produce a tone that is an octave lower than the original tone produced by the piano wire.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Initial tension \( T_1 = 16 \, \text{kg} \) (weight)
- We need to find the new tension \( T_2 \) that produces a tone an octave lower.

2. **Understand the Relationship Between Frequency and Tension:**
- The frequency of a vibrating string is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( L \) is the length of the string, \( T \) is the tension, and \( \mu \) is the mass per unit length of the string.

3. **Frequency Change for an Octave:**
- A tone an octave lower means the frequency is halved. Therefore, if the original frequency is \( f_1 \), the new frequency \( f_2 \) will be:
\[ f_2 = \frac{f_1}{2} \]

4. **Set Up the Ratio of Frequencies:**
- From the frequency formula, we can set up the ratio:
\[ \frac{f_1}{f_2} = \frac{\sqrt{T_1}}{\sqrt{T_2}} \]
- Since \( f_2 = \frac{f_1}{2} \), we have:
\[ \frac{f_1}{\frac{f_1}{2}} = 2 \]
- Thus, we can write:
\[ 2 = \frac{\sqrt{T_1}}{\sqrt{T_2}} \]

5. **Square Both Sides:**
- Squaring both sides gives:
\[ 4 = \frac{T_1}{T_2} \]

6. **Substitute Known Values:**
- Substitute \( T_1 = 16 \, \text{kg} \):
\[ 4 = \frac{16}{T_2} \]

7. **Solve for \( T_2 \):**
- Rearranging gives:
\[ T_2 = \frac{16}{4} = 4 \, \text{kg} \]

8. **Calculate Change in Tension:**
- The change in tension \( \Delta T \) is given by:
\[ \Delta T = T_1 - T_2 = 16 \, \text{kg} - 4 \, \text{kg} = 12 \, \text{kg} \]

### Final Answer:
The change in tension required to produce a tone an octave lower is \( 12 \, \text{kg} \).

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