A brass wire 300 cm long when subjected to a force F produces an elongation .a.. Another wire of twice the diameter and of same length and material, when subjected to the same force F, produces an elongation .b.. What is the value of a/b ?

Answer

To solve the problem, we need to use the relationship between stress, strain, and the properties of the materials involved. The elongation of a wire under a tensile force can be described by the formula:

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

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

### Step 1: Identify the parameters for both wires
- For the first wire:
- Length \(L_1 = 300 \, \text{cm}\)
- Elongation \(\Delta L_1 = a\)
- Diameter \(d_1\)
- Radius \(r_1 = \frac{d_1}{2}\)
- Area \(A_1 = \pi r_1^2\)

- For the second wire:
- Length \(L_2 = 300 \, \text{cm}\) (same length)
- Elongation \(\Delta L_2 = b\)
- Diameter \(d_2 = 2d_1\) (twice the diameter)
- Radius \(r_2 = \frac{d_2}{2} = d_1\)
- Area \(A_2 = \pi r_2^2 = \pi (2r_1)^2 = 4\pi r_1^2\)

### Step 2: Write the elongation formulas
Using the elongation formula for both wires:

1. For the first wire:
\[ a = \frac{F \cdot L_1}{A_1 \cdot Y} = \frac{F \cdot 300}{\pi r_1^2 \cdot Y} \]

2. For the second wire:
\[ b = \frac{F \cdot L_2}{A_2 \cdot Y} = \frac{F \cdot 300}{4\pi r_1^2 \cdot Y} \]

### Step 3: Find the ratio \(a/b\)
Now, we can find the ratio of the elongations \(a\) and \(b\):

\[ \frac{a}{b} = \frac{\frac{F \cdot 300}{\pi r_1^2 \cdot Y}}{\frac{F \cdot 300}{4\pi r_1^2 \cdot Y}} = \frac{4\pi r_1^2 \cdot Y}{\pi r_1^2 \cdot Y} \]

This simplifies to:

\[ \frac{a}{b} = 4 \]

### Final Answer
Thus, the value of \(\frac{a}{b} = 4\).

---