Two opposite forces`F_(1) = 120 N` and `F_(2) = 80 N` act on an elastic plank of modulus of elasticity `Y= 2 x 10^(11) N//m^(2)` and length `l=1m` placed over a smooth horizontal surface. The cross-sectional area of the plank is `S = 0.5 m^(2)`. The change in length of the plank is `x xx 10^(-11)m`. Find the value of `x`.

Options

A. A. 100
B. B. 150
C. C. 200
D. B. 1.1

Explanation

To solve the problem, we will use the formula for Young's modulus and the relationship between stress, strain, and change in length. Here are the steps:

### Step 1: Calculate the net force acting on the plank
The net force \( F \) acting on the plank is the difference between the two forces:
\[ F = F_1 - F_2 = 120 \, \text{N} - 80 \, \text{N} = 40 \, \text{N} \]

### Step 2: Calculate the stress on the plank
Stress \( \sigma \) is defined as force per unit area:
\[ \sigma = \frac{F}{S} \]
Substituting the values:
\[ \sigma = \frac{40 \, \text{N}}{0.5 \, \text{m}^2} = 80 \, \text{N/m}^2 \]

### Step 3: Calculate the strain in the plank
Strain \( \epsilon \) is defined as the change in length per unit length:
\[ \epsilon = \frac{\Delta x}{l} \]
Where \( \Delta x \) is the change in length and \( l \) is the original length of the plank.

### Step 4: Relate stress and strain using Young's modulus
Young's modulus \( Y \) is defined as the ratio of stress to strain:
\[ Y = \frac{\sigma}{\epsilon} \]
Rearranging gives:
\[ \epsilon = \frac{\sigma}{Y} \]

### Step 5: Substitute the values into the strain formula
Substituting the values of stress and Young's modulus:
\[ \epsilon = \frac{80 \, \text{N/m}^2}{2 \times 10^{11} \, \text{N/m}^2} = 4 \times 10^{-10} \]

### Step 6: Calculate the change in length \( \Delta x \)
Now we can find \( \Delta x \) using the strain:
\[ \Delta x = \epsilon \cdot l = 4 \times 10^{-10} \cdot 1 \, \text{m} = 4 \times 10^{-10} \, \text{m} \]

### Step 7: Express \( \Delta x \) in the required form
The problem states that the change in length is \( x \times 10^{-11} \, \text{m} \). We can express our result as:
\[ \Delta x = 40 \times 10^{-11} \, \text{m} \]
Thus, \( x = 40 \).

### Final Answer
The value of \( x \) is:
\[ \boxed{40} \]