A rubber cord has a cross -sectional area `1 mm^(2)`and total un-stretched length`10.0cm`. It is stretched to `12.0 cm` and then released to project a missile of mass 5.0 g. Taking young's modulus Y for rubber as `5.0xx10^(8) N//m^(2)`.Calculate the velocity of projection .

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert all given units to SI units
- Cross-sectional area \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)
- Original length \( L_0 = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Stretched length \( L = 12 \, \text{cm} = 0.12 \, \text{m} \)
- Change in length \( \Delta L = L - L_0 = 0.12 \, \text{m} - 0.1 \, \text{m} = 0.02 \, \text{m} \)
- Mass of the missile \( m = 5 \, \text{g} = 5 \times 10^{-3} \, \text{kg} \)

### Step 2: Calculate the Young's Modulus
The Young's modulus \( Y \) is given as \( 5.0 \times 10^8 \, \text{N/m}^2 \).

### Step 3: Calculate the force exerted by the rubber cord
Using the formula for stress and Young's modulus:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L_0} \]
Rearranging gives:
\[ F = Y \cdot A \cdot \frac{\Delta L}{L_0} \]
Substituting the values:
\[ F = (5.0 \times 10^8 \, \text{N/m}^2) \cdot (1 \times 10^{-6} \, \text{m}^2) \cdot \frac{0.02 \, \text{m}}{0.1 \, \text{m}} \]
Calculating:
\[ F = (5.0 \times 10^8) \cdot (1 \times 10^{-6}) \cdot 0.2 = 100 \, \text{N} \]

### Step 4: Calculate the elastic potential energy stored in the rubber cord
The elastic potential energy \( U \) is given by:
\[ U = \frac{1}{2} F \Delta L \]
Substituting the values:
\[ U = \frac{1}{2} \cdot 100 \, \text{N} \cdot 0.02 \, \text{m} = 1 \, \text{J} \]

### Step 5: Equate the elastic potential energy to the kinetic energy of the missile
The kinetic energy \( KE \) of the missile is given by:
\[ KE = \frac{1}{2} m v^2 \]
Setting the elastic potential energy equal to the kinetic energy:
\[ 1 \, \text{J} = \frac{1}{2} (5 \times 10^{-3} \, \text{kg}) v^2 \]
Solving for \( v^2 \):
\[ 1 = \frac{1}{2} \cdot 5 \times 10^{-3} \cdot v^2 \]
\[ 2 = 5 \times 10^{-3} \cdot v^2 \]
\[ v^2 = \frac{2}{5 \times 10^{-3}} = 400 \]
\[ v = \sqrt{400} = 20 \, \text{m/s} \]

### Final Answer
The velocity of projection of the missile is \( 20 \, \text{m/s} \).

---