Under the same load,wire A having length `5.0m` and cross section `2.5times10^(-5)m^(2)` stretches uniformly by the same amount as another wire `B` of length `6.0m` and a cross section of `3.0 times10^(-5)m^(2)` stretches.The ratio of the Young's modulus of wire `A` to that of wire `B` will be:

Options

A. `1 : 4`
B. `1 : 1`
C. `1 : 10`
D. ` 1 : 2`

Answer

To find the ratio of the Young's modulus of wire A to that of wire B, we will use the formula for Young's modulus, which is given by:

\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]

where:
- \( Y \) is the Young's modulus,
- \( F \) is the force applied (load),
- \( L \) is the original length of the wire,
- \( A \) is the cross-sectional area,
- \( \Delta L \) is the extension (stretch) of the wire.

### Step 1: Write the expressions for Young's modulus for both wires

For wire A:
\[ Y_A = \frac{F \cdot L_A}{A_A \cdot \Delta L} \]

For wire B:
\[ Y_B = \frac{F \cdot L_B}{A_B \cdot \Delta L} \]

### Step 2: Take the ratio of Young's moduli

We need to find the ratio \( \frac{Y_A}{Y_B} \):
\[ \frac{Y_A}{Y_B} = \frac{\frac{F \cdot L_A}{A_A \cdot \Delta L}}{\frac{F \cdot L_B}{A_B \cdot \Delta L}} \]

### Step 3: Simplify the ratio

Since \( F \) and \( \Delta L \) are the same for both wires, they cancel out:
\[ \frac{Y_A}{Y_B} = \frac{L_A \cdot A_B}{L_B \cdot A_A} \]

### Step 4: Substitute the known values

Given:
- Length of wire A, \( L_A = 5.0 \, m \)
- Cross-sectional area of wire A, \( A_A = 2.5 \times 10^{-5} \, m^2 \)
- Length of wire B, \( L_B = 6.0 \, m \)
- Cross-sectional area of wire B, \( A_B = 3.0 \times 10^{-5} \, m^2 \)

Substituting these values into the ratio:
\[ \frac{Y_A}{Y_B} = \frac{5.0 \cdot (3.0 \times 10^{-5})}{6.0 \cdot (2.5 \times 10^{-5})} \]

### Step 5: Calculate the ratio

Calculating the numerator and denominator:
\[ \frac{Y_A}{Y_B} = \frac{5.0 \cdot 3.0}{6.0 \cdot 2.5} \]
\[ = \frac{15.0}{15.0} = 1 \]

### Final Answer

Thus, the ratio of the Young's modulus of wire A to that of wire B is:
\[ \frac{Y_A}{Y_B} = 1 \]