The Young's modulus of a rope of 10 m length and having diameter of 2 cm is `20xx10^(11)` dyne `cm^(-2)` . If the elongation produced in the rope is 1 cm, the force applied on the rope is

Options

A. `6.28xx10^(5) N`
B. `6.28 xx 10^(4)` N
C. `6.28 xx 10^(4)` dyne
D. `6.28 xx 10^(5)` dyne

Explanation

To solve the problem, we will use the formula for Young's modulus (Y), which is defined as:

\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]

Where:
- \( Y \) = Young's modulus
- \( F \) = force applied
- \( L \) = original length of the rope
- \( A \) = cross-sectional area of the rope
- \( \Delta L \) = elongation produced in the rope

### Step 1: Identify the given values
- Young's modulus, \( Y = 20 \times 10^{11} \, \text{dyne/cm}^2 \)
- Length of the rope, \( L = 10 \, \text{m} = 1000 \, \text{cm} \) (since 1 m = 100 cm)
- Diameter of the rope, \( d = 2 \, \text{cm} \)
- Elongation, \( \Delta L = 1 \, \text{cm} \)

### Step 2: Calculate the radius of the rope
The radius \( r \) can be calculated from the diameter:
\[ r = \frac{d}{2} = \frac{2 \, \text{cm}}{2} = 1 \, \text{cm} \]

### Step 3: Calculate the cross-sectional area \( A \)
The area \( A \) of the rope can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 = \pi (1 \, \text{cm})^2 = \pi \, \text{cm}^2 \]

### Step 4: Rearrange the Young's modulus formula to find the force \( F \)
We need to isolate \( F \) in the Young's modulus formula:
\[ F = \frac{Y \cdot A \cdot \Delta L}{L} \]

### Step 5: Substitute the known values into the formula
Substituting the known values into the equation:
\[ F = \frac{(20 \times 10^{11} \, \text{dyne/cm}^2) \cdot (\pi \, \text{cm}^2) \cdot (1 \, \text{cm})}{1000 \, \text{cm}} \]

### Step 6: Calculate the force \( F \)
Now we can calculate \( F \):
\[ F = \frac{20 \times 10^{11} \cdot \pi \cdot 1}{1000} \]
\[ F = \frac{20 \times 10^{11} \cdot 3.14}{1000} \]
\[ F = \frac{62.8 \times 10^{11}}{1000} \]
\[ F = 6.28 \times 10^{9} \, \text{dyne} \]

### Step 7: Convert dyne to Newton
Since \( 1 \, \text{Newton} = 10^5 \, \text{dyne} \):
\[ F = \frac{6.28 \times 10^{9}}{10^5} \]
\[ F = 6.28 \times 10^{4} \, \text{N} \]

### Final Answer
The force applied on the rope is \( 6.28 \times 10^{4} \, \text{N} \).

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