Two wires A and B are of same material. Their lengths are in the ratio 1:2 and diameters are in the ratio 2:1 when stretched by forces `F_A` and `F_B` respectively they get equal increase in their lengths. Then the ratio `(F_A)/(F_B)` should be

Options

A. `1:2`
B. `1:1`
C. `2:1`
D. `8:1`

Explanation

To find the ratio of the forces \( \frac{F_A}{F_B} \) acting on two wires A and B, we can use the relationship defined by Young's modulus. Here’s a step-by-step solution:

### Step 1: Understand the relationship of Young's modulus
Young's modulus \( Y \) is defined as:
\[ Y = \frac{F}{A \cdot \frac{\Delta L}{L}} \]
where \( F \) is the force applied, \( A \) is the cross-sectional area, \( \Delta L \) is the change in length, and \( L \) is the original length.

### Step 2: Set up the equations for both wires
Since both wires are made of the same material, their Young's moduli are equal:
\[ Y_A = Y_B \]
This gives us:
\[ \frac{F_A}{A_A \cdot \frac{\Delta L}{L_A}} = \frac{F_B}{A_B \cdot \frac{\Delta L}{L_B}} \]
Since the increase in length \( \Delta L \) is the same for both wires, we can simplify this to:
\[ \frac{F_A}{A_A \cdot L_A} = \frac{F_B}{A_B \cdot L_B} \]

### Step 3: Rearranging the equation
Rearranging the equation gives:
\[ \frac{F_A}{F_B} = \frac{A_A \cdot L_A}{A_B \cdot L_B} \]

### Step 4: Find the areas of the wires
The area \( A \) of a wire is given by:
\[ A = \frac{\pi d^2}{4} \]
Given the diameter ratio \( \frac{d_A}{d_B} = \frac{2}{1} \), we can find the areas:
\[ \frac{A_A}{A_B} = \frac{\pi (2r)^2/4}{\pi (r)^2/4} = \frac{4}{1} \]

### Step 5: Find the lengths of the wires
Given the length ratio \( \frac{L_A}{L_B} = \frac{1}{2} \), we can express \( L_B \) in terms of \( L_A \):
\[ L_B = 2L_A \]

### Step 6: Substitute values into the equation
Now substituting the values into the equation:
\[ \frac{F_A}{F_B} = \frac{A_A \cdot L_A}{A_B \cdot L_B} = \frac{4}{1} \cdot \frac{L_A}{2L_A} = \frac{4}{2} = 2 \]

### Final Step: Conclusion
Thus, the ratio of the forces is:
\[ \frac{F_A}{F_B} = 2 \]