A sonometer wire of 70cm length fixed at one end has a solid mass M, hanging from its other end to produce tension in it. The wire produces a certian frequency. When the same mass 'M' hanging in water, it is found that the length of the wire has to be changed by 5 cm. in order to produce the same frequency. Then the density of the material of mass 'M' is

Options

A. 5 gm /c.c
B. 1 gm/ c.c
C. `(169)/(27)gm// c.c`
D. `(196)/(27)gm//c.c`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Identify the initial conditions
We are given:
- Length of the sonometer wire, \( l = 70 \, \text{cm} = 0.7 \, \text{m} \)
- The mass \( M \) hanging from the wire produces a certain frequency.

### Step 2: Understand the change in conditions
When the mass \( M \) is submerged in water, the length of the wire is changed by 5 cm to maintain the same frequency. Therefore, the new length of the wire is:
\[ l' = 70 \, \text{cm} - 5 \, \text{cm} = 65 \, \text{cm} = 0.65 \, \text{m} \]

### Step 3: Relate frequency to tension and length
The frequency \( f \) of a vibrating wire is given by:
\[ f = \frac{1}{2l} \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire and \( \mu \) is the mass per unit length of the wire.

### Step 4: Determine the tension in both scenarios
1. When the mass is hanging in air:
\[ T = Mg \]

2. When the mass is submerged in water, the effective tension \( T' \) is reduced due to the buoyant force:
\[ T' = Mg - F_b \]
where \( F_b \) is the buoyant force, given by:
\[ F_b = \rho_{water} V = \rho_{water} \frac{M}{\rho_{material}} \]
Here, \( \rho_{water} \) is the density of water, and \( V \) is the volume of the mass.

### Step 5: Set the frequencies equal
Since the frequency remains the same in both cases, we can set the two frequency equations equal:
\[ \frac{1}{2l} \sqrt{\frac{Mg}{\mu}} = \frac{1}{2l'} \sqrt{\frac{T'}{\mu}} \]

### Step 6: Substitute the tensions
Substituting \( T' \) into the frequency equation gives:
\[ \frac{1}{2l} \sqrt{\frac{Mg}{\mu}} = \frac{1}{2l'} \sqrt{\frac{Mg - F_b}{\mu}} \]

### Step 7: Simplify the equation
Canceling out common terms and squaring both sides leads to:
\[ \frac{1}{l^2} (Mg) = \frac{1}{l'^2} (Mg - F_b) \]

### Step 8: Substitute for \( F_b \)
Substituting \( F_b \) into the equation gives:
\[ \frac{1}{l^2} (Mg) = \frac{1}{l'^2} \left(Mg - \rho_{water} \frac{M}{\rho_{material}} g\right) \]

### Step 9: Rearranging and solving for density
After rearranging and simplifying, we can derive an expression for the density of the material:
\[ \frac{1}{\rho_{material}} = \frac{l'^2}{l^2 - l'^2} \]

### Step 10: Plugging in values
Substituting \( l = 0.7 \, \text{m} \) and \( l' = 0.65 \, \text{m} \):
\[ \rho_{material} = \frac{(0.7)^2}{(0.7)^2 - (0.65)^2} \]

Calculating:
1. \( (0.7)^2 = 0.49 \)
2. \( (0.65)^2 = 0.4225 \)
3. \( 0.49 - 0.4225 = 0.0675 \)

Thus:
\[ \rho_{material} = \frac{0.49}{0.0675} \approx 7.26 \, \text{g/cm}^3 \]

### Final Answer
The density of the material of mass \( M \) is approximately \( \frac{196}{27} \, \text{g/cm}^3 \).

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