Fundamental frequency of a stretched sonometer wire is `f_(0)`. When its tension is increased by `96%` and length drecreased by `35%`, its fundamental frequency becomes `eta_(1)f_(0)`. When its tension is decreased by `36%` and its length is increased by `30%`, its fundamental frequency becomes `eta_(2)f_(0)`. Then `(eta_(1))/(eta_(2))` is.

Options

A. `4.5`
B. `3.5`
C. `2`
D. `6`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the formula for fundamental frequency
The fundamental frequency \( f \) of a stretched sonometer wire is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( \mu \) is the linear mass density of the wire (constant in this case).

### Step 2: Analyze the first scenario
In the first scenario, the tension is increased by 96% and the length is decreased by 35%.

- The new tension \( T_1 \) can be expressed as:
\[ T_1 = T + 0.96T = 1.96T \]

- The new length \( L_1 \) can be expressed as:
\[ L_1 = L - 0.35L = 0.65L \]

Using these values, we can express the new frequency \( f_1 \):
\[ f_1 = \frac{1}{2L_1} \sqrt{\frac{T_1}{\mu}} = \frac{1}{2(0.65L)} \sqrt{\frac{1.96T}{\mu}} = \frac{1.96}{1.3} \cdot \frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{1.96}{1.3} f_0 \]

### Step 3: Calculate \( \eta_1 \)
From the above expression, we can find \( \eta_1 \):
\[ \eta_1 = \frac{f_1}{f_0} = \frac{1.96}{1.3} = \frac{1.96}{1.3} \approx 1.5077 \]

### Step 4: Analyze the second scenario
In the second scenario, the tension is decreased by 36% and the length is increased by 30%.

- The new tension \( T_2 \) can be expressed as:
\[ T_2 = T - 0.36T = 0.64T \]

- The new length \( L_2 \) can be expressed as:
\[ L_2 = L + 0.3L = 1.3L \]

Using these values, we can express the new frequency \( f_2 \):
\[ f_2 = \frac{1}{2L_2} \sqrt{\frac{T_2}{\mu}} = \frac{1}{2(1.3L)} \sqrt{\frac{0.64T}{\mu}} = \frac{0.8}{2.6} \cdot \frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{0.8}{2.6} f_0 \]

### Step 5: Calculate \( \eta_2 \)
From the above expression, we can find \( \eta_2 \):
\[ \eta_2 = \frac{f_2}{f_0} = \frac{0.8}{2.6} \approx 0.3077 \]

### Step 6: Calculate the ratio \( \frac{\eta_1}{\eta_2} \)
Now we can find the ratio \( \frac{\eta_1}{\eta_2} \):
\[ \frac{\eta_1}{\eta_2} = \frac{1.5077}{0.3077} \approx 4.89 \]

### Conclusion
Thus, the final answer for \( \frac{\eta_1}{\eta_2} \) is approximately \( 4.89 \).