A nylon rope `2 cm` in diameter has a breaking strength of `1.5 xx 10^(5) N`. The breaking strength of a similar rope `1cm` in diameter is

Options

A. `0.375xx10^(5)N`
B. `2xx10^(5)N`
C. `6xx10^(5)N`
D. `9xx10^(4)N`

Explanation

To find the breaking strength of a nylon rope with a diameter of 1 cm, we can use the relationship between breaking strength, area, and breaking stress. Here’s a step-by-step solution:

### Step 1: Understand the relationship
The breaking stress (σ) is defined as the breaking strength (F) divided by the cross-sectional area (A). For two ropes made of the same material, the breaking stress will be the same. Thus, we can write:

\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \]

Where:
- \( F_1 \) = breaking strength of the first rope (2 cm diameter)
- \( A_1 \) = cross-sectional area of the first rope
- \( F_2 \) = breaking strength of the second rope (1 cm diameter)
- \( A_2 \) = cross-sectional area of the second rope

### Step 2: Calculate the areas
The cross-sectional area \( A \) of a circular rope can be calculated using the formula:

\[ A = \frac{\pi}{4} d^2 \]

For the first rope (2 cm diameter):

\[ A_1 = \frac{\pi}{4} (2 \, \text{cm})^2 = \frac{\pi}{4} \times 4 \, \text{cm}^2 = \pi \, \text{cm}^2 \]

For the second rope (1 cm diameter):

\[ A_2 = \frac{\pi}{4} (1 \, \text{cm})^2 = \frac{\pi}{4} \times 1 \, \text{cm}^2 = \frac{\pi}{4} \, \text{cm}^2 \]

### Step 3: Substitute values into the equation
We know that \( F_1 = 1.5 \times 10^5 \, \text{N} \) and we have calculated \( A_1 \) and \( A_2 \). Now we can substitute these into our equation:

\[ \frac{1.5 \times 10^5 \, \text{N}}{\pi \, \text{cm}^2} = \frac{F_2}{\frac{\pi}{4} \, \text{cm}^2} \]

### Step 4: Simplify the equation
Since \( \pi \) appears in both the numerator and denominator, we can cancel it out:

\[ \frac{1.5 \times 10^5}{1} = \frac{F_2}{\frac{1}{4}} \]

This simplifies to:

\[ 1.5 \times 10^5 = 4 F_2 \]

### Step 5: Solve for \( F_2 \)
Now, we can solve for \( F_2 \):

\[ F_2 = \frac{1.5 \times 10^5}{4} = 0.375 \times 10^5 \, \text{N} \]

### Final Answer
Thus, the breaking strength of the nylon rope with a diameter of 1 cm is:

\[ F_2 = 0.375 \times 10^5 \, \text{N} = 3.75 \times 10^4 \, \text{N} \]

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