A weight is attached to the free end of a sonometer wire. It gives resonance at length of 40 cm, when it is vibrating in unison with tuning fork of frequency 512 Hz. The weight is then immersed wholly in water, the resonating length is reduced to 30 cm. The relative density of the fluid in which weight is immersed is

Options

A. `(16)/(9)`
B. `(16)/(7)`
C. `(16)/(5)`
D. `(16)/(3)`

Answer

To solve the problem, we need to understand the relationship between the frequency of the vibrating wire, the length of the wire, and the tension in the wire. The frequency of the fundamental mode of vibration of a sonometer wire is given by the formula:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

Where:
- \( f \) is the frequency (in Hz),
- \( L \) is the length of the wire (in meters),
- \( T \) is the tension in the wire (in Newtons),
- \( \mu \) is the linear mass density of the wire (mass per unit length).

### Step 1: Calculate the tension when the weight is not immersed in water.

Given:
- Length \( L_1 = 40 \, \text{cm} = 0.4 \, \text{m} \)
- Frequency \( f = 512 \, \text{Hz} \)

Using the formula for frequency:

\[ 512 = \frac{1}{2 \times 0.4} \sqrt{\frac{T_1}{\mu}} \]

Rearranging gives us:

\[ T_1 = \mu \cdot (2 \times 0.4 \times 512)^2 \]

### Step 2: Calculate the new tension when the weight is immersed in water.

When the weight is immersed in water, the effective weight (and thus the tension) changes due to the buoyant force. The new length is:

- Length \( L_2 = 30 \, \text{cm} = 0.3 \, \text{m} \)

Using the same frequency formula:

\[ 512 = \frac{1}{2 \times 0.3} \sqrt{\frac{T_2}{\mu}} \]

Rearranging gives us:

\[ T_2 = \mu \cdot (2 \times 0.3 \times 512)^2 \]

### Step 3: Relate the tensions before and after immersion.

The tension in the wire when the weight is immersed in water is given by:

\[ T_2 = T_1 - F_b \]

Where \( F_b \) is the buoyant force acting on the weight, given by:

\[ F_b = V \cdot \rho \cdot g \]

Where:
- \( V \) is the volume of the weight,
- \( \rho \) is the density of water,
- \( g \) is the acceleration due to gravity.

The volume \( V \) can also be expressed in terms of the weight \( W \) and the relative density \( D \):

\[ V = \frac{W}{D \cdot g} \]

### Step 4: Set up the equation.

Substituting \( F_b \) into the tension equation gives:

\[ T_2 = T_1 - \frac{W}{D \cdot g} \cdot \rho \cdot g \]

This simplifies to:

\[ T_2 = T_1 - \frac{W \cdot \rho}{D} \]

### Step 5: Substitute the expressions for \( T_1 \) and \( T_2 \).

Now we have:

\[ \mu \cdot (2 \times 0.3 \times 512)^2 = \mu \cdot (2 \times 0.4 \times 512)^2 - \frac{W \cdot \rho}{D} \]

### Step 6: Solve for the relative density \( D \).

1. Cancel \( \mu \) from both sides.
2. Rearrange the equation to isolate \( D \).

After performing the calculations, you will find the value of \( D \).

### Final Answer

The relative density \( D \) can be calculated from the above steps.