A block of mass 500kg is suspended by wire of length 70 cm. The area of cross-section of wire is 10 `mm^(2)`. When the load is removed, the wire contracts by 0.5cm. The young's modulus of the material of wire will be

Options

A. `10 xx 10^(14)N//m^(2)`
B. `4 xx 10^(14)N//m^(2)`
C. `8xx 10^(11)N//m^(2)`
D. `7xx 10^(10)N//m^(2)`

Explanation

To find the Young's modulus of the material of the wire, we will follow these steps:

### Step 1: Write down the given values
- Mass of the block (m) = 500 kg
- Length of the wire (L) = 70 cm = 0.7 m
- Area of cross-section of the wire (A) = 10 mm² = 10 × 10⁻⁶ m² = 10 × 10⁻⁶ m²
- Change in length of the wire (ΔL) = 0.5 cm = 0.005 m

### Step 2: Calculate the force (F) acting on the wire
The force acting on the wire due to the suspended block can be calculated using the formula:
\[ F = m \cdot g \]
Where \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \).

Substituting the values:
\[ F = 500 \, \text{kg} \times 10 \, \text{m/s}^2 = 5000 \, \text{N} \]

### Step 3: Use the formula for Young's modulus (Y)
The Young's modulus is defined as:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]

### Step 4: Substitute the values into the Young's modulus formula
Substituting the values we have:
- \( F = 5000 \, \text{N} \)
- \( L = 0.7 \, \text{m} \)
- \( A = 10 \times 10^{-6} \, \text{m}^2 \)
- \( \Delta L = 0.005 \, \text{m} \)

So,
\[ Y = \frac{5000 \, \text{N} \cdot 0.7 \, \text{m}}{10 \times 10^{-6} \, \text{m}^2 \cdot 0.005 \, \text{m}} \]

### Step 5: Calculate the Young's modulus
Calculating the numerator:
\[ 5000 \times 0.7 = 3500 \, \text{N m} \]

Calculating the denominator:
\[ 10 \times 10^{-6} \times 0.005 = 5 \times 10^{-8} \, \text{m}^3 \]

Now substituting back into the equation:
\[ Y = \frac{3500}{5 \times 10^{-8}} \]

Calculating:
\[ Y = 7 \times 10^{10} \, \text{N/m}^2 \]

### Final Result
The Young's modulus of the material of the wire is:
\[ Y = 7 \times 10^{10} \, \text{N/m}^2 \]

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