If `L = (20 +- 0.01) m` and `B = (10 +- 0.02)m` then `L//B` is

Options

A. `(2 +- 0.03)m`
B. `(2 +- 0.15)m`
C. `(2 +- 0.01)m`
D. `(2 +- 0.05)m`

Explanation

To solve the problem of finding \( \frac{L}{B} \) given \( L = (20 \pm 0.01) \, \text{m} \) and \( B = (10 \pm 0.02) \, \text{m} \), we will follow these steps:

### Step 1: Identify the values and uncertainties
We have:
- \( L = 20 \, \text{m} \) with an uncertainty \( \Delta L = 0.01 \, \text{m} \)
- \( B = 10 \, \text{m} \) with an uncertainty \( \Delta B = 0.02 \, \text{m} \)

### Step 2: Calculate the value of \( \frac{L}{B} \)
Using the values of \( L \) and \( B \):
\[ \frac{L}{B} = \frac{20 \, \text{m}}{10 \, \text{m}} = 2 \]

### Step 3: Calculate the relative uncertainties
To find the uncertainty in \( \frac{L}{B} \), we use the formula for the propagation of uncertainty for division:
\[ \frac{\Delta x}{x} = \frac{\Delta L}{L} + \frac{\Delta B}{B} \]
Where \( x = \frac{L}{B} \).

Calculating the relative uncertainties:
1. For \( L \):
\[ \frac{\Delta L}{L} = \frac{0.01}{20} = 0.0005 \]
2. For \( B \):
\[ \frac{\Delta B}{B} = \frac{0.02}{10} = 0.002 \]

### Step 4: Add the relative uncertainties
Now, we add the relative uncertainties:
\[ \frac{\Delta x}{x} = 0.0005 + 0.002 = 0.0025 \]

### Step 5: Calculate the absolute uncertainty in \( \frac{L}{B} \)
Now, we can find the absolute uncertainty \( \Delta x \):
\[ \Delta x = x \times \frac{\Delta x}{x} = 2 \times 0.0025 = 0.005 \]

### Step 6: Write the final result
Thus, the final result for \( \frac{L}{B} \) with its uncertainty is:
\[ \frac{L}{B} = 2 \pm 0.005 \]

### Final Answer:
\[ \frac{L}{B} = 2 \pm 0.005 \]