The length of a metallic sheet is measured as 10.0 cm using a metre scale of L.C. 0.1cm and its breadth is, measured as 1.00 cm using a vernier callipers of L.C. 0.01 cm, the error in area is

Options

A. `pm 0.01 cm^(2)`
B. `pm 0.1 cm^(2)`
C. `pm 0.11 cm^(2)`
D. `pm 0.2 cm^(2)`

Answer

To solve the problem of finding the error in the area of a metallic sheet given its length and breadth measurements, we will follow these steps:

### Step 1: Identify the Measurements and Their Uncertainties
- Length (L) = 10.0 cm with a least count (L.C.) of 0.1 cm
- Breadth (B) = 1.00 cm with a least count (L.C.) of 0.01 cm

### Step 2: Determine the Absolute Errors
- The absolute error in length (ΔL) = L.C. of the meter scale = 0.1 cm
- The absolute error in breadth (ΔB) = L.C. of the vernier caliper = 0.01 cm

### Step 3: Calculate the Area
The area (A) of the metallic sheet is given by:
\[ A = L \times B \]
Substituting the values:
\[ A = 10.0 \, \text{cm} \times 1.00 \, \text{cm} = 10.0 \, \text{cm}^2 \]

### Step 4: Calculate the Fractional Errors
The fractional error in area (ΔA/A) can be calculated using the formula:
\[ \frac{\Delta A}{A} = \frac{\Delta L}{L} + \frac{\Delta B}{B} \]

Substituting the values:
- \(\Delta L = 0.1 \, \text{cm}\)
- \(L = 10.0 \, \text{cm}\)
- \(\Delta B = 0.01 \, \text{cm}\)
- \(B = 1.00 \, \text{cm}\)

Calculating each term:
\[ \frac{\Delta L}{L} = \frac{0.1}{10.0} = 0.01 \]
\[ \frac{\Delta B}{B} = \frac{0.01}{1.00} = 0.01 \]

Adding these fractional errors:
\[ \frac{\Delta A}{A} = 0.01 + 0.01 = 0.02 \]

### Step 5: Calculate the Absolute Error in Area
Now, we can find the absolute error in area (ΔA):
\[ \Delta A = \frac{\Delta A}{A} \times A \]
Substituting the values:
\[ \Delta A = 0.02 \times 10.0 \, \text{cm}^2 = 0.2 \, \text{cm}^2 \]

### Step 6: Conclusion
The error in the area of the metallic sheet is:
\[ \Delta A = \pm 0.2 \, \text{cm}^2 \]

### Final Answer
The error in area is \( \pm 0.2 \, \text{cm}^2 \).

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