A rope 1 cm in diameter breaks, if the tension in it exceeds 500 N. The maximum tension that may be given to similar rope of diameter 3 cm is

Options

A. 500 N
B. 3000 N
C. 4500 N
D. 2000 N

Explanation

To solve the problem, we need to determine the maximum tension that a rope with a diameter of 3 cm can withstand, given that a rope with a diameter of 1 cm breaks at a tension of 500 N.

### Step-by-Step Solution:

1. **Understand the relationship between tension and area**:
The breaking stress (σ) of a material is defined as the force (tension, T) applied per unit area (A). It can be expressed as:
\[ \sigma = \frac{T}{A} \]

2. **Calculate the area of the first rope**:
The diameter of the first rope (D1) is 1 cm, so the radius (R1) is:
\[ R1 = \frac{D1}{2} = \frac{1 \, \text{cm}}{2} = 0.5 \, \text{cm} = 0.005 \, \text{m} \]
The cross-sectional area (A1) of the first rope is given by:
\[ A1 = \pi R1^2 = \pi (0.005)^2 = \pi \times 0.000025 = 7.85 \times 10^{-5} \, \text{m}^2 \]

3. **Calculate the breaking stress of the first rope**:
Given that the maximum tension (T1) the first rope can withstand is 500 N, we can calculate the breaking stress (σ) as:
\[ \sigma = \frac{T1}{A1} = \frac{500 \, \text{N}}{7.85 \times 10^{-5} \, \text{m}^2} \approx 6.37 \times 10^6 \, \text{N/m}^2 \]

4. **Calculate the area of the second rope**:
The diameter of the second rope (D2) is 3 cm, so the radius (R2) is:
\[ R2 = \frac{D2}{2} = \frac{3 \, \text{cm}}{2} = 1.5 \, \text{cm} = 0.015 \, \text{m} \]
The cross-sectional area (A2) of the second rope is given by:
\[ A2 = \pi R2^2 = \pi (0.015)^2 = \pi \times 0.000225 = 7.07 \times 10^{-4} \, \text{m}^2 \]

5. **Determine the maximum tension for the second rope**:
Since the breaking stress remains the same for both ropes (as they are made of the same material), we can set up the equation:
\[ \sigma = \frac{T2}{A2} \]
Rearranging gives us:
\[ T2 = \sigma \times A2 \]
Substituting the values we found:
\[ T2 = (6.37 \times 10^6 \, \text{N/m}^2) \times (7.07 \times 10^{-4} \, \text{m}^2) \approx 4500 \, \text{N} \]

### Final Answer:
The maximum tension that may be given to the similar rope of diameter 3 cm is approximately **4500 N**.