If the breaking force of a nylon rope of diameter 4cm is `1.6xx10^(5)N`, find the breaking force of the same nylon rope of diameter 2 cm

Answer

To find the breaking force of a nylon rope of diameter 2 cm, given that the breaking force of a nylon rope of diameter 4 cm is \(1.6 \times 10^5 \, \text{N}\), we can use the relationship between breaking force and the radius of the rope.

### Step-by-Step Solution:

1. **Understand the relationship**: The breaking force \(F\) is directly proportional to the square of the radius \(r\) of the rope. This can be expressed as:
\[ F \propto r^2 \]
Therefore, we can write:
\[ \frac{F_1}{F_2} = \frac{r_1^2}{r_2^2} \]

2. **Identify the given values**:
- For the first rope (diameter = 4 cm):
- Diameter \(d_1 = 4 \, \text{cm}\) → Radius \(r_1 = \frac{d_1}{2} = \frac{4}{2} = 2 \, \text{cm}\)
- Breaking force \(F_1 = 1.6 \times 10^5 \, \text{N}\)

- For the second rope (diameter = 2 cm):
- Diameter \(d_2 = 2 \, \text{cm}\) → Radius \(r_2 = \frac{d_2}{2} = \frac{2}{2} = 1 \, \text{cm}\)

3. **Set up the equation**:
Using the relationship established:
\[ \frac{F_1}{F_2} = \frac{r_1^2}{r_2^2} \]
Plugging in the values:
\[ \frac{1.6 \times 10^5}{F_2} = \frac{(2)^2}{(1)^2} \]
This simplifies to:
\[ \frac{1.6 \times 10^5}{F_2} = \frac{4}{1} \]

4. **Cross-multiply to solve for \(F_2\)**:
\[ 1.6 \times 10^5 = 4 \times F_2 \]
Therefore,
\[ F_2 = \frac{1.6 \times 10^5}{4} \]

5. **Calculate \(F_2\)**:
\[ F_2 = 0.4 \times 10^5 = 4 \times 10^4 \, \text{N} \]
Hence,
\[ F_2 = 6.4 \times 10^4 \, \text{N} \]

### Final Answer:
The breaking force of the nylon rope of diameter 2 cm is \(6.4 \times 10^4 \, \text{N}\).