To break a wire of 1 m length, minimum 40 kg weight is required. Then, the wire of the same material of double radius and 6 m length will require breaking weight.

Options

A. 80 kg weight
B. 240 kg weight
C. 200 kg weight
D. 160 kg weight

Explanation

To solve the problem, we need to understand the relationship between the breaking force, the radius of the wire, and the length of the wire. The breaking force (F) is related to the breaking stress (σ) and the cross-sectional area (A) of the wire.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- For the first wire:
- Length (L1) = 1 m
- Breaking weight (F1) = 40 kg
- For the second wire:
- Length (L2) = 6 m
- Radius (r2) = 2 * r1 (double the radius of the first wire)

2. **Understand Breaking Stress:**
- Breaking stress (σ) is defined as:
\[ \sigma = \frac{F}{A} \]
- Where A is the cross-sectional area of the wire. For a circular wire, the area is given by:
\[ A = \pi r^2 \]

3. **Calculate the Breaking Stress for the First Wire:**
- Using the first wire:
\[ \sigma = \frac{F1}{A1} = \frac{40 \text{ kg}}{\pi r_1^2} \]

4. **Determine the Area of the Second Wire:**
- The radius of the second wire is double that of the first wire:
\[ r2 = 2r1 \]
- Therefore, the area of the second wire (A2) becomes:
\[ A2 = \pi (r2)^2 = \pi (2r1)^2 = 4\pi r1^2 \]

5. **Calculate the Breaking Force for the Second Wire:**
- The breaking stress remains the same since both wires are made of the same material:
\[ \sigma = \frac{F2}{A2} \]
- Substituting the area of the second wire:
\[ \sigma = \frac{F2}{4\pi r1^2} \]
- Setting the two expressions for breaking stress equal:
\[ \frac{40 \text{ kg}}{\pi r1^2} = \frac{F2}{4\pi r1^2} \]
- Canceling \(\pi r1^2\) from both sides gives:
\[ 40 \text{ kg} = \frac{F2}{4} \]
- Solving for F2:
\[ F2 = 40 \text{ kg} \times 4 = 160 \text{ kg} \]

6. **Conclusion:**
- The breaking weight required for the second wire of double radius and 6 m length is **160 kg**.