A wire of diameter `1mm` breaks under a tension of 1000 N. Another wire, of same material as that of the first one, but of diameter `2mm` breaks under a tension of

Options

A. 500 N
B. 1000 N
C. 10000 N
D. 4000 N

Answer

To solve the problem, we need to understand the relationship between the diameter of the wire, the area of cross-section, and the breaking tension.

### Step-by-Step Solution:

1. **Understand the relationship between diameter and area**:
The area \( A \) of a wire with diameter \( d \) is given by the formula:
\[ A = \frac{\pi d^2}{4} \]

2. **Calculate the area for the first wire**:
For the first wire with a diameter \( d_1 = 1 \, \text{mm} = 0.001 \, \text{m} \):
\[ A_1 = \frac{\pi (0.001)^2}{4} = \frac{\pi \times 0.000001}{4} = \frac{\pi}{4000000} \, \text{m}^2 \]

3. **Determine the breaking tension for the first wire**:
The breaking tension \( F_1 \) for the first wire is given as \( 1000 \, \text{N} \).

4. **Calculate the area for the second wire**:
The second wire has a diameter \( d_2 = 2 \, \text{mm} = 0.002 \, \text{m} \):
\[ A_2 = \frac{\pi (0.002)^2}{4} = \frac{\pi \times 0.000004}{4} = \frac{\pi}{1000000} \, \text{m}^2 \]

5. **Relate the breaking tensions of both wires**:
Since both wires are made of the same material, the breaking tension is proportional to the cross-sectional area. Therefore, we can write:
\[ \frac{F_2}{F_1} = \frac{A_2}{A_1} \]

6. **Substituting the areas**:
Substituting the areas calculated:
\[ \frac{F_2}{1000} = \frac{\frac{\pi}{1000000}}{\frac{\pi}{4000000}} \]
Simplifying the right side:
\[ \frac{F_2}{1000} = \frac{4000000}{1000000} = 4 \]

7. **Calculate \( F_2 \)**:
Therefore:
\[ F_2 = 4 \times 1000 = 4000 \, \text{N} \]

### Final Answer:
The second wire, with a diameter of \( 2 \, \text{mm} \), breaks under a tension of \( 4000 \, \text{N} \).