A steal wire of cross-section area `3xx10^(-6) m^(2)` can withstand a maximum strain of `10^(-3)` .Young's modulus of steel is `2xx10^(11) Nm^(-2)` .The maximum mass this wire can hold is

Options

A. 40kg
B. 60kg
C. 80kg
D. 100kg

Explanation

To find the maximum mass that a steel wire can hold, we will use the relationship between stress, strain, and Young's modulus. Here are the steps to solve the problem:

### Step 1: Identify the given values
- Cross-sectional area of the wire, \( A = 3 \times 10^{-6} \, \text{m}^2 \)
- Maximum strain, \( \epsilon = 10^{-3} \)
- Young's modulus of steel, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)

### Step 2: Use the relationship between stress, strain, and Young's modulus
The relationship is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Where stress is defined as:
\[ \text{Stress} = \frac{F}{A} \]
Thus, we can rewrite the equation as:
\[ Y = \frac{F/A}{\epsilon} \]
Rearranging gives:
\[ F = Y \cdot \epsilon \cdot A \]

### Step 3: Substitute the known values into the equation
Substituting the values we have:
\[ F = (2 \times 10^{11} \, \text{N/m}^2) \cdot (10^{-3}) \cdot (3 \times 10^{-6} \, \text{m}^2) \]

### Step 4: Calculate the force \( F \)
Calculating the right-hand side:
\[ F = 2 \times 10^{11} \cdot 10^{-3} \cdot 3 \times 10^{-6} \]
\[ F = 2 \cdot 3 \cdot 10^{11 - 3 - 6} = 6 \times 10^{2} \, \text{N} \]
\[ F = 600 \, \text{N} \]

### Step 5: Relate the force to mass
The force \( F \) acting on the wire is equal to the weight of the mass \( m \) it can hold:
\[ F = mg \]
Where \( g \) is the acceleration due to gravity, approximately \( 10 \, \text{m/s}^2 \). Therefore:
\[ m = \frac{F}{g} = \frac{600 \, \text{N}}{10 \, \text{m/s}^2} \]
\[ m = 60 \, \text{kg} \]

### Final Answer
The maximum mass that the wire can hold is \( \boxed{60 \, \text{kg}} \).