A boy's catapult is made of rubber cord which is `42cm` long, with `6mm` diameter of cross-section and of negligible mass. The boy keeps a stone weighing `0.02kg` on it and stretches the cond by `20cm` by applying a constant force. When released. The stone flies off with a velocity of `20ms^(-1)`. Neglect the change in the area of cross-section of the cord while stretched. The Young's modulus of rubber is closet to:

Options

A. `10^(6)Nm^(-2)`
B. `10^(4)Nm^(-2)`
C. `10^(8)Nm^(-2)`
D. `10^(3)Nm^(-2)`

Answer

To find the Young's modulus of the rubber cord used in the catapult, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the rubber cord (L) = 42 cm = 0.42 m
- Diameter of the rubber cord (D) = 6 mm = 0.006 m
- Radius of the rubber cord (r) = D/2 = 0.003 m
- Mass of the stone (m) = 0.02 kg
- Stretching of the cord (ΔL) = 20 cm = 0.20 m
- Final velocity of the stone (v) = 20 m/s

2. **Calculate the Cross-Sectional Area (A):**
The cross-sectional area of the rubber cord can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 = \pi (0.003)^2 = \pi (9 \times 10^{-6}) \approx 2.827 \times 10^{-5} \, \text{m}^2 \]

3. **Calculate the Strain (ε):**
Strain is defined as the change in length divided by the original length:
\[ \text{Strain} (\epsilon) = \frac{\Delta L}{L} = \frac{0.20}{0.42} \approx 0.4762 \]

4. **Calculate the Stress (σ):**
Stress is defined as the force applied per unit area. The force (F) exerted by the stone can be calculated using its weight:
\[ F = mg = 0.02 \times 9.81 \approx 0.1962 \, \text{N} \]
Then, stress can be calculated as:
\[ \text{Stress} (\sigma) = \frac{F}{A} = \frac{0.1962}{2.827 \times 10^{-5}} \approx 6947.5 \, \text{N/m}^2 \]

5. **Relate Young's Modulus (Y) to Stress and Strain:**
Young's modulus is defined as the ratio of stress to strain:
\[ Y = \frac{\sigma}{\epsilon} = \frac{6947.5}{0.4762} \approx 14580.5 \, \text{N/m}^2 \]

6. **Equate Energy:**
The elastic potential energy stored in the stretched rubber cord is equal to the kinetic energy of the stone when it is released:
\[ \frac{1}{2} mv^2 = \frac{1}{2} Y \epsilon^2 A L \]
Rearranging gives:
\[ Y = \frac{mv^2}{\epsilon^2 A L} \]

7. **Substituting Values:**
\[ Y = \frac{0.02 \times (20)^2}{(0.4762)^2 \times (2.827 \times 10^{-5}) \times 0.42} \]
Calculate the above expression to find Y.

8. **Final Calculation:**
After substituting the values and simplifying, we find:
\[ Y \approx 3 \times 10^6 \, \text{N/m}^2 \]

### Conclusion:
The Young's modulus of the rubber is approximately \(3 \times 10^6 \, \text{N/m}^2\).