Two wires P and Q are of same material. Their length are in the ratio `4:3` and the diameters are in the ratio `2:3`. If they are pulled by the same force their increase in length will be in the ratio.

Options

A. `3//1`
B. `2//3`
C. `5//4`
D. `3//5`

Explanation

To find the ratio of the increase in length of two wires P and Q, we will use the relationship between stress, strain, and Young's modulus. Here’s a step-by-step solution:

### Step 1: Understand the given information
- Let the lengths of the wires P and Q be \( L_1 \) and \( L_2 \) respectively.
- The ratio of their lengths is given as:
\[ \frac{L_1}{L_2} = \frac{4}{3} \]
- Let the diameters of the wires P and Q be \( D_1 \) and \( D_2 \) respectively.
- The ratio of their diameters is given as:
\[ \frac{D_1}{D_2} = \frac{2}{3} \]

### Step 2: Calculate the areas of the wires
- The area \( A \) of a wire can be calculated using the formula for the area of a circle:
\[ A = \frac{\pi D^2}{4} \]
- Therefore, the areas of wires P and Q are:
\[ A_1 = \frac{\pi D_1^2}{4} \quad \text{and} \quad A_2 = \frac{\pi D_2^2}{4} \]
- The ratio of the areas is:
\[ \frac{A_1}{A_2} = \frac{D_1^2}{D_2^2} \]

### Step 3: Substitute the diameter ratio
- Since \( \frac{D_1}{D_2} = \frac{2}{3} \), we can find the ratio of the areas:
\[ \frac{A_1}{A_2} = \left(\frac{D_1}{D_2}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]

### Step 4: Use the formula for increase in length
- The increase in length \( \Delta L \) can be expressed as:
\[ \Delta L = \frac{F L}{Y A} \]
- For wires P and Q, we can write:
\[ \Delta L_1 = \frac{F L_1}{Y A_1} \quad \text{and} \quad \Delta L_2 = \frac{F L_2}{Y A_2} \]

### Step 5: Find the ratio of the increases in length
- The ratio of the increases in length is:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{F L_1 / (Y A_1)}{F L_2 / (Y A_2)} = \frac{L_1 A_2}{L_2 A_1} \]
- Substituting the ratios we found:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{L_1}{L_2} \cdot \frac{A_2}{A_1} = \frac{4/3}{1} \cdot \frac{9/4}{1} = \frac{4 \cdot 9}{3 \cdot 4} = \frac{3}{1} \]

### Conclusion
The ratio of the increase in length of wires P and Q is:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{3}{1} \]