A force F is needed to break a copper wire having radius R. The force needed to break a copper wire of same length and radius 2R will be

Options

A. `(F)/(2)`
B. `2F`
C. `4F`
D. `(F)/(4)`

Answer

To solve the problem, we need to determine the force required to break a copper wire of radius 2R, given that a force F is needed to break a wire of radius R.

### Step-by-Step Solution:

1. **Understanding Young's Modulus**:
The force required to break a wire can be derived from Young's modulus, which is defined as:
\[ F = Y \cdot A \cdot \frac{\Delta L}{L} \]
where \( F \) is the force, \( Y \) is Young's modulus, \( A \) is the cross-sectional area, \( \Delta L \) is the extension, and \( L \) is the original length of the wire. Since the lengths are the same, we can focus on the area.

2. **Cross-Sectional Area Calculation**:
The cross-sectional area \( A \) of a wire with radius \( r \) is given by:
\[ A = \pi r^2 \]
For the first wire with radius \( R \):
\[ A_1 = \pi R^2 \]
For the second wire with radius \( 2R \):
\[ A_2 = \pi (2R)^2 = \pi \cdot 4R^2 = 4\pi R^2 \]

3. **Relating Forces**:
Since the force is directly proportional to the cross-sectional area, we can express the forces as:
\[ F_1 \propto A_1 \quad \text{and} \quad F_2 \propto A_2 \]
Therefore, we can write:
\[ \frac{F_1}{F_2} = \frac{A_1}{A_2} \]
Substituting the areas:
\[ \frac{F_1}{F_2} = \frac{\pi R^2}{4\pi R^2} = \frac{1}{4} \]

4. **Finding the Relationship Between Forces**:
Rearranging the above equation gives:
\[ F_2 = 4F_1 \]
Since we know \( F_1 = F \), we can substitute:
\[ F_2 = 4F \]

5. **Conclusion**:
The force needed to break a copper wire of radius \( 2R \) is \( 4F \).

### Final Answer:
The force needed to break a copper wire of radius \( 2R \) is \( 4F \).

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