A segment of wire vibrates with a fundamental frequency of 450 Hz under a tension of 9Kg-wt.Then, tension at which the fundamental frequency of the same wire becomes 900 Hz is

Options

A. 36 kg - wt
B. 27 kg - wt
C. 18 kg - wt
D. 72 kg - wt

Answer

To solve the problem, we will use the relationship between the fundamental frequency of a vibrating wire and the tension in the wire. The fundamental frequency (f) of a wire is given by the formula:

\[ f = \frac{1}{2\pi} \sqrt{\frac{T}{m}} \]

where:
- \( f \) is the fundamental frequency,
- \( T \) is the tension in the wire,
- \( m \) is the mass per unit length of the wire.

Given:
- Initial frequency \( f_1 = 450 \, \text{Hz} \)
- Final frequency \( f_2 = 900 \, \text{Hz} \)
- Initial tension \( T_1 = 9 \, \text{kg-wt} \)

We need to find the new tension \( T_2 \) when the frequency is \( 900 \, \text{Hz} \).

### Step-by-Step Solution:

1. **Write the relationship between frequencies and tensions:**
Since the frequency is proportional to the square root of the tension, we can write:
\[ \frac{f_1}{f_2} = \sqrt{\frac{T_1}{T_2}} \]

2. **Substitute the known values:**
Substitute \( f_1 = 450 \, \text{Hz} \), \( f_2 = 900 \, \text{Hz} \), and \( T_1 = 9 \, \text{kg-wt} \):
\[ \frac{450}{900} = \sqrt{\frac{9}{T_2}} \]

3. **Simplify the left side:**
\[ \frac{1}{2} = \sqrt{\frac{9}{T_2}} \]

4. **Square both sides to eliminate the square root:**
\[ \left(\frac{1}{2}\right)^2 = \frac{9}{T_2} \]
\[ \frac{1}{4} = \frac{9}{T_2} \]

5. **Cross-multiply to solve for \( T_2 \):**
\[ T_2 = 9 \times 4 \]
\[ T_2 = 36 \, \text{kg-wt} \]

### Final Answer:
The tension at which the fundamental frequency of the same wire becomes \( 900 \, \text{Hz} \) is \( 36 \, \text{kg-wt} \).