Two wires `A` and `B` have equal lengths and aremade of the same material , but diameter of wire `A` is twice that of wire `B`. Then, for a given load,

Options

A. (A) the extension of `B` will be four times that of `A`
B. (B) the extensions of `A` and will be equal
C. (C) the strain in `B` is four times that in `A`
D. (D) the strains in `A` and `B` will be equal

Explanation

To solve the problem, we will analyze the relationship between the extensions in two wires, A and B, which have the same length, are made of the same material, but have different diameters.

### Step-by-Step Solution:

1. **Understanding Young's Modulus**:
Young's modulus (Y) is defined as the ratio of stress to strain. Mathematically, it is given by:
\[ Y = \frac{F/A}{\Delta L/L} \]
where:
- \( F \) is the applied force,
- \( A \) is the cross-sectional area,
- \( \Delta L \) is the extension,
- \( L \) is the original length.

2. **Cross-sectional Area Calculation**:
The cross-sectional area \( A \) of a wire is given by:
\[ A = \frac{\pi d^2}{4} \]
where \( d \) is the diameter of the wire.
Given that the diameter of wire A is twice that of wire B, we can express the areas as:
- For wire A: \( A_A = \frac{\pi (2d_B)^2}{4} = \pi d_B^2 \)
- For wire B: \( A_B = \frac{\pi d_B^2}{4} \)

3. **Setting Up the Equations for Young's Modulus**:
For wire A:
\[ Y_A = \frac{F}{A_A} \cdot \frac{L}{\Delta L_A} \implies Y_A = \frac{F}{\pi d_B^2} \cdot \frac{L}{\Delta L_A} \]
For wire B:
\[ Y_B = \frac{F}{A_B} \cdot \frac{L}{\Delta L_B} \implies Y_B = \frac{F}{\frac{\pi d_B^2}{4}} \cdot \frac{L}{\Delta L_B} = \frac{4F}{\pi d_B^2} \cdot \frac{L}{\Delta L_B} \]

4. **Equating Young's Modulus**:
Since both wires are made of the same material, their Young's moduli are equal:
\[ Y_A = Y_B \]
Thus, we have:
\[ \frac{F}{\pi d_B^2} \cdot \frac{L}{\Delta L_A} = \frac{4F}{\pi d_B^2} \cdot \frac{L}{\Delta L_B} \]

5. **Simplifying the Equation**:
Canceling out common terms (F, \(\pi\), and L) gives:
\[ \frac{1}{\Delta L_A} = \frac{4}{\Delta L_B} \]
Rearranging this, we find:
\[ \Delta L_A = 4 \Delta L_B \]

6. **Conclusion**:
The extension in wire A is four times the extension in wire B:
\[ \Delta L_A = 4 \Delta L_B \]

### Final Answer:
The extension in wire A is 4 times the extension in wire B.