The length of a strip measured with a meter rod is 10.0cm. Its width measured with a venier calipers is 1.00cm. The least count of the meter rod is 0.1 cm and that of vernier calipers 0.01cm. What will be error in its area?

Options

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

Explanation

To solve the problem of finding the error in the area of a strip given its length and width measurements, we can follow these steps:

### Step 1: Identify the Measurements and Their Errors
- Length (L) of the strip = 10.0 cm
- Width (B) of the strip = 1.00 cm
- Least count of the meter rod (for length) = 0.1 cm
- Least count of the vernier calipers (for width) = 0.01 cm

### Step 2: Determine the Absolute Errors
- The absolute error in length (ΔL) is equal to the least count of the meter rod:
\[ \Delta L = 0.1 \text{ cm} \]
- The absolute error in width (ΔB) is equal to the least count of the vernier calipers:
\[ \Delta B = 0.01 \text{ cm} \]

### Step 3: Calculate the Area of the Strip
- The area (A) of the strip can be calculated using the formula:
\[ A = L \times B \]
- Substituting the values:
\[ A = 10.0 \text{ cm} \times 1.00 \text{ cm} = 10.0 \text{ cm}^2 \]

### Step 4: Use Differentiation to Find the Error in Area
- The error in area (ΔA) can be calculated using the formula:
\[ \Delta A = \Delta L + \Delta B \]
- Substituting the values of ΔL and ΔB:
\[ \Delta A = 0.1 \text{ cm} + 0.01 \text{ cm} = 0.11 \text{ cm}^2 \]

### Step 5: State the Final Result
- Therefore, the error in the area of the strip is:
\[ \Delta A = \pm 0.11 \text{ cm}^2 \]

### Summary
The area of the strip is \(10.0 \text{ cm}^2\) with an error of \(\pm 0.11 \text{ cm}^2\).

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