If the length and beradth of a plate are `(5.0 +- 0.2)cm` and `(4.0 +- 0.1)cm` the the absolue error in measurment of area is...

Options

A. `10cm^(2)`
B. `11cm^(2)`
C. `12cm^(2)`
D. `1.3 cm^(2)`

Explanation

To find the absolute error in the measurement of the area of a plate given its length and breadth with uncertainties, we can follow these steps:

### Step 1: Identify the given values
- Length \( L = 5.0 \, \text{cm} \) with an uncertainty of \( \Delta L = 0.2 \, \text{cm} \)
- Breadth \( B = 4.0 \, \text{cm} \) with an uncertainty of \( \Delta B = 0.1 \, \text{cm} \)

### Step 2: Calculate the area
The area \( A \) of the plate can be calculated using the formula:
\[ A = L \times B \]
Substituting the values:
\[ A = 5.0 \, \text{cm} \times 4.0 \, \text{cm} = 20.0 \, \text{cm}^2 \]

### Step 3: Calculate the relative errors
The relative error in the area can be calculated using the formula:
\[ \frac{\Delta A}{A} = \frac{\Delta L}{L} + \frac{\Delta B}{B} \]
Substituting the values:
\[ \frac{\Delta A}{20.0} = \frac{0.2}{5.0} + \frac{0.1}{4.0} \]

Calculating each term:
- \(\frac{0.2}{5.0} = 0.04\)
- \(\frac{0.1}{4.0} = 0.025\)

Adding these together:
\[ \frac{\Delta A}{20.0} = 0.04 + 0.025 = 0.065 \]

### Step 4: Calculate the absolute error in area
Now, we can find the absolute error \( \Delta A \) using:
\[ \Delta A = A \times \frac{\Delta A}{A} \]
Substituting the values:
\[ \Delta A = 20.0 \, \text{cm}^2 \times 0.065 = 1.3 \, \text{cm}^2 \]

### Final Result
The absolute error in the measurement of the area is:
\[ \Delta A = 1.3 \, \text{cm}^2 \]

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