A wire can be broken by `400kg.wt`. The load required to break the wire of double the thickness of the same material will be (in `kgwt`.)

Options

A. `800`
B. `1600`
C. `3200`
D. `6400`

Explanation

To solve the problem, we need to understand the relationship between the thickness of the wire, the cross-sectional area, and the breaking load.

### Step-by-Step Solution:

1. **Understanding Breaking Load and Stress**:
The breaking load (force) that a wire can withstand before breaking is related to the breaking stress of the material and the cross-sectional area of the wire. The formula for breaking stress (σ) is given by:
\[ \sigma = \frac{F}{A} \]
where \( F \) is the breaking load and \( A \) is the cross-sectional area.

2. **Initial Conditions**:
For the original wire, we know that the breaking load \( F_1 \) is 400 kg.wt.

3. **Cross-sectional Area**:
The cross-sectional area \( A_1 \) of the original wire can be expressed in terms of its thickness \( t \):
\[ A_1 = k \cdot t^2 \]
where \( k \) is a constant that depends on the shape of the wire (for a circular wire, \( k = \frac{\pi}{4} \)).

4. **New Wire Thickness**:
If the thickness of the wire is doubled, the new thickness \( t_2 \) is:
\[ t_2 = 2t \]
The new cross-sectional area \( A_2 \) will be:
\[ A_2 = k \cdot (2t)^2 = k \cdot 4t^2 = 4A_1 \]

5. **Breaking Stress Remains Constant**:
Since the material is the same, the breaking stress \( \sigma \) remains constant. Therefore, we can express the breaking load for the new wire \( F_2 \) as:
\[ \sigma = \frac{F_1}{A_1} = \frac{F_2}{A_2} \]

6. **Relating the Loads**:
Rearranging the equation gives us:
\[ F_2 = \sigma \cdot A_2 \]
Substituting \( A_2 = 4A_1 \):
\[ F_2 = \sigma \cdot 4A_1 \]

7. **Substituting for \( \sigma \)**:
From the original wire, we know \( \sigma = \frac{F_1}{A_1} = \frac{400 \text{ kg.wt}}{A_1} \). Therefore:
\[ F_2 = 4 \cdot \frac{400 \text{ kg.wt}}{A_1} \cdot A_1 = 4 \cdot 400 \text{ kg.wt} = 1600 \text{ kg.wt} \]

### Final Answer:
The load required to break the wire of double the thickness of the same material will be **1600 kg.wt**.