the frequency of a sonometer wire is 100 Hz. When the weight producing th tensions are completely immersed in water the frequency becomes 80 Hz and on immersing the weight in a certain liquid the frequency becomes 60 Hz. The specific gravity of the liquid is

Options

A. 1.42
B. 1.77
C. 1.82
D. 1.21

Answer

To solve the problem step by step, we will use the relationship between frequency, tension, and mass per unit length of the sonometer wire.

### Step 1: Understanding the relationship between frequency and tension
The fundamental frequency \( f \) of a vibrating string (or 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 mass per unit length of the wire.

### Step 2: Expressing frequency in terms of tension
Since the tension \( T \) is equal to the weight \( mg \) (where \( m \) is the mass and \( g \) is the acceleration due to gravity), we can rewrite the frequency as:
\[ f \propto \sqrt{T} \propto \sqrt{mg} \]
This means that the frequency is proportional to the square root of the tension.

### Step 3: Setting up the ratios for different conditions
Let:
- \( f_1 = 100 \, \text{Hz} \) (frequency in air),
- \( f_2 = 80 \, \text{Hz} \) (frequency in water),
- \( f_3 = 60 \, \text{Hz} \) (frequency in a certain liquid).

From the relationship established, we can set up the following ratios:
\[ \frac{f_2^2}{f_1^2} = \frac{g_{\text{water}}}{g_{\text{air}}} \]
Substituting the values:
\[ \frac{80^2}{100^2} = \frac{g_{\text{water}}}{g_{\text{air}}} \]
Calculating this gives:
\[ \frac{6400}{10000} = 0.64 \]

### Step 4: Relating the specific gravities
Since the specific gravity \( S \) of a liquid is defined as the ratio of the density of the liquid to the density of water, we can express the frequency in the liquid as:
\[ \frac{f_3^2}{f_1^2} = \frac{g_{\text{liquid}}}{g_{\text{air}}} \]
Substituting the values:
\[ \frac{60^2}{100^2} = \frac{g_{\text{liquid}}}{g_{\text{air}}} \]
Calculating this gives:
\[ \frac{3600}{10000} = 0.36 \]

### Step 5: Setting up the equations for specific gravity
From the previous steps, we have:
1. For water: \( g_{\text{water}} = 0.64 g_{\text{air}} \)
2. For the liquid: \( g_{\text{liquid}} = 0.36 g_{\text{air}} \)

The specific gravity \( S \) of the liquid can be expressed as:
\[ S = \frac{g_{\text{liquid}}}{g_{\text{water}}} \]
Substituting the values:
\[ S = \frac{0.36 g_{\text{air}}}{0.64 g_{\text{air}}} = \frac{0.36}{0.64} \]
Calculating this gives:
\[ S = 0.5625 \]

### Step 6: Final calculation of specific gravity
To find the specific gravity, we can also express it as:
\[ S = \frac{\text{Density of liquid}}{\text{Density of water}} = \frac{0.64}{0.36} \approx 1.777 \]

Thus, the specific gravity of the liquid is approximately **1.777**.

### Final Answer:
The specific gravity of the liquid is **1.777**.