Two wires have masses in the ratio 3:4. They are made of the same material. Under the same load what is ratio of their elongations ? Explain why?

Answer

To solve the problem of finding the ratio of elongations of two wires made of the same material and under the same load, we can follow these steps:

### Step 1: Understand the relationship between elongation and mass
The elongation (e) of a wire under a load (F) can be described by the formula:
\[ e = \frac{F \cdot L}{A \cdot Y} \]
where:
- \( F \) is the force applied (load),
- \( L \) is the original length of the wire,
- \( A \) is the cross-sectional area,
- \( Y \) is the Young's modulus of the material.

### Step 2: Establish the relationship between mass and dimensions
Since both wires are made of the same material, their Young's modulus (Y) is the same. The mass (m) of a wire can be expressed as:
\[ m = \rho \cdot V = \rho \cdot A \cdot L \]
where \( \rho \) is the density of the material and \( V \) is the volume.

### Step 3: Set up the ratio of elongations
Let the masses of the two wires be \( m_1 \) and \( m_2 \) with a ratio of \( \frac{m_1}{m_2} = \frac{3}{4} \). The elongations of the two wires under the same load can be expressed as:
\[ e_1 = \frac{F \cdot L_1}{A_1 \cdot Y} \]
\[ e_2 = \frac{F \cdot L_2}{A_2 \cdot Y} \]

### Step 4: Relate the lengths and areas of the wires
Since the wires have masses in the ratio \( \frac{3}{4} \), we can express their lengths and areas in terms of their masses. Assuming the wires have the same density, we can write:
\[ \frac{m_1}{m_2} = \frac{\rho \cdot A_1 \cdot L_1}{\rho \cdot A_2 \cdot L_2} \]
This simplifies to:
\[ \frac{A_1 \cdot L_1}{A_2 \cdot L_2} = \frac{3}{4} \]

### Step 5: Substitute into the elongation formula
Now, we can express the ratio of elongations:
\[ \frac{e_1}{e_2} = \frac{F \cdot L_1 / (A_1 \cdot Y)}{F \cdot L_2 / (A_2 \cdot Y)} \]
This simplifies to:
\[ \frac{e_1}{e_2} = \frac{L_1 \cdot A_2}{L_2 \cdot A_1} \]

### Step 6: Use the mass ratio to find the elongation ratio
From the mass ratio and the relationship established earlier, we can express \( A_1 \) and \( A_2 \) in terms of \( L_1 \) and \( L_2 \):
\[ \frac{e_1}{e_2} = \frac{3}{4} \cdot \frac{L_1}{L_2} \]
Assuming the lengths are proportional to the masses, we can set \( L_1 = k \cdot 3 \) and \( L_2 = k \cdot 4 \) for some constant \( k \).

### Step 7: Final calculation
Substituting the lengths into the elongation ratio:
\[ \frac{e_1}{e_2} = \frac{3}{4} \cdot \frac{3}{4} = \frac{9}{16} \]

### Conclusion
Thus, the ratio of elongations \( \frac{e_1}{e_2} \) is \( \frac{9}{16} \).

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