A rubber cord of cross sectional area `1mm^(2)` and unstretched length 10cm is stretched to 12cm and then released to project a stone of mass 5 gram.
If Y for rubber `= 5xx10^(8) N//m^(2)`, then the tension in the rubber cord is

Options

A. `25 N`
B. `50 N`
C. `100 N`
D. `200 N`

Explanation

To find the tension in the rubber cord when it is stretched, we can use the relationship defined by Young's modulus. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Cross-sectional area (A) = 1 mm² = \(1 \times 10^{-6}\) m² (conversion to square meters)
- Unstretched length (L) = 10 cm = \(0.1\) m (conversion to meters)
- Stretched length = 12 cm = \(0.12\) m (conversion to meters)
- Change in length (\(\Delta L\)) = Stretched length - Unstretched length = \(0.12 - 0.1 = 0.02\) m
- Young's modulus (Y) for rubber = \(5 \times 10^8\) N/m²
- Mass of the stone = 5 g = \(0.005\) kg (conversion to kilograms)

### Step 2: Calculate the change in length
\[ \Delta L = 0.12 \, \text{m} - 0.1 \, \text{m} = 0.02 \, \text{m} \]

### Step 3: Use Young's modulus formula
Young's modulus (Y) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
Where:
- Stress = \(\frac{F}{A}\)
- Strain = \(\frac{\Delta L}{L}\)

### Step 4: Rearranging the formula to find tension (F)
From the equation of Young's modulus, we can express the force (tension) as:
\[ F = Y \cdot A \cdot \frac{\Delta L}{L} \]

### Step 5: Substitute the values into the equation
Substituting the known values:
\[ F = (5 \times 10^8 \, \text{N/m}^2) \cdot (1 \times 10^{-6} \, \text{m}^2) \cdot \frac{0.02 \, \text{m}}{0.1 \, \text{m}} \]

### Step 6: Simplify the equation
Calculating the fraction:
\[ \frac{0.02}{0.1} = 0.2 \]
Now substituting this back:
\[ F = (5 \times 10^8) \cdot (1 \times 10^{-6}) \cdot 0.2 \]

### Step 7: Calculate the force
\[ F = 5 \times 0.2 \times 10^{8 - 6} = 1 \times 10^2 = 100 \, \text{N} \]

### Final Answer
The tension in the rubber cord is \(100 \, \text{N}\).

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