A rubber cord has a cross-sectional area `10^(-6) m^(2)` and total unstretched length 0.1 m. It is stretched to 0.125 m and then released to project a particle of mass 5.0 g. The velocity of projection is [Given, Young’s modulus of rubber `Y = 5 xx 10^(8) N//m^(2)` ]

Options

A. 45 m/s
B. 30 m/s
C. 25 m/s
D. 15 m/s

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given data
- Cross-sectional area of the rubber cord, \( A = 10^{-6} \, \text{m}^2 \)
- Unstretched length of the rubber cord, \( L_0 = 0.1 \, \text{m} \)
- Stretched length of the rubber cord, \( L = 0.125 \, \text{m} \)
- Mass of the particle, \( m = 5.0 \, \text{g} = 5.0 \times 10^{-3} \, \text{kg} \)
- Young’s modulus of rubber, \( Y = 5 \times 10^{8} \, \text{N/m}^2 \)

### Step 2: Calculate the extension of the rubber cord
The extension \( \Delta L \) of the rubber cord is given by:
\[ \Delta L = L - L_0 = 0.125 \, \text{m} - 0.1 \, \text{m} = 0.025 \, \text{m} \]

### Step 3: Calculate the stress in the rubber cord
Stress \( \sigma \) is defined as force per unit area. The force \( F \) can be expressed in terms of Young's modulus:
\[ \sigma = \frac{F}{A} = \frac{Y \cdot \Delta L}{L_0} \]
Rearranging gives:
\[ F = \sigma \cdot A = \frac{Y \cdot \Delta L}{L_0} \cdot A \]

### Step 4: Substitute the values to find the force
Substituting the known values:
\[ F = \frac{(5 \times 10^{8} \, \text{N/m}^2) \cdot (0.025 \, \text{m})}{0.1 \, \text{m}} \cdot (10^{-6} \, \text{m}^2) \]
Calculating the force:
\[ F = \frac{(5 \times 10^{8}) \cdot (0.025)}{0.1} \cdot (10^{-6}) = (1.25 \times 10^{8}) \cdot (10^{-6}) = 125 \, \text{N} \]

### Step 5: Calculate the potential energy stored in the rubber cord
The potential energy \( U \) stored in the rubber cord when stretched is given by:
\[ U = \frac{1}{2} F \Delta L \]
Substituting the values:
\[ U = \frac{1}{2} \cdot 125 \, \text{N} \cdot 0.025 \, \text{m} = \frac{1}{2} \cdot 125 \cdot 0.025 = 1.5625 \, \text{J} \]

### Step 6: Set the potential energy equal to the kinetic energy of the particle
When the rubber cord is released, the potential energy converts into kinetic energy \( K \) of the particle:
\[ K = \frac{1}{2} m v^2 \]
Setting \( U = K \):
\[ 1.5625 = \frac{1}{2} (5.0 \times 10^{-3}) v^2 \]

### Step 7: Solve for the velocity \( v \)
Rearranging gives:
\[ v^2 = \frac{2 \cdot 1.5625}{5.0 \times 10^{-3}} = \frac{3.125}{5.0 \times 10^{-3}} = 625 \]
Taking the square root:
\[ v = \sqrt{625} = 25 \, \text{m/s} \]

### Final Answer
The velocity of projection is \( v = 25 \, \text{m/s} \).

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