A steel wire of diameter 2 mm has a breaking strength of `4xx10^(5)`N. What is the breaking strength of similar steel wire of diameter `1.5 mm` ?

Options

A. a.`2.3xx10^(5)N`
B. b.`2.6xx10^(5)N`
C. `c.3xx10^(5)N`
D. d.`1.5xx10^(5)N`

Answer

To find the breaking strength of a steel wire of diameter 1.5 mm, given that a similar wire of diameter 2 mm has a breaking strength of \(4 \times 10^5\) N, we can follow these steps:

### Step 1: Understand the relationship between breaking strength and diameter
The breaking strength (F) of a wire is directly proportional to the cross-sectional area (A) of the wire. For a circular wire, the area can be expressed in terms of the diameter (d):
\[ A = \frac{\pi d^2}{4} \]
Thus, we can express the breaking strength as:
\[ F \propto A \propto d^2 \]

### Step 2: Set up the ratio of breaking strengths
Let \(F_1\) be the breaking strength of the first wire (diameter \(d_1 = 2 \, \text{mm}\)) and \(F_2\) be the breaking strength of the second wire (diameter \(d_2 = 1.5 \, \text{mm}\)). From the proportionality, we have:
\[ \frac{F_1}{F_2} = \frac{d_1^2}{d_2^2} \]

### Step 3: Substitute known values
We know:
- \(F_1 = 4 \times 10^5 \, \text{N}\)
- \(d_1 = 2 \, \text{mm}\)
- \(d_2 = 1.5 \, \text{mm}\)

Substituting these values into the ratio gives:
\[ \frac{4 \times 10^5}{F_2} = \frac{(2)^2}{(1.5)^2} \]

### Step 4: Calculate the squares of the diameters
Calculating the squares:
- \(d_1^2 = 2^2 = 4\)
- \(d_2^2 = (1.5)^2 = 2.25\)

Thus, we can rewrite the ratio:
\[ \frac{4 \times 10^5}{F_2} = \frac{4}{2.25} \]

### Step 5: Solve for \(F_2\)
Cross-multiplying gives:
\[ 4 \times 10^5 \cdot 2.25 = 4 \cdot F_2 \]
\[ F_2 = \frac{4 \times 10^5 \cdot 2.25}{4} \]
\[ F_2 = 2.25 \times 10^5 \, \text{N} \]

### Conclusion
The breaking strength of the steel wire with a diameter of 1.5 mm is:
\[ F_2 = 2.25 \times 10^5 \, \text{N} \]