The side of a cubical block when measured with a vernier callipers is 2.50 cm. The vernier constant is 0.01 cm. The maximum possible error in the area of the side of the block is

Options

A. `pm0.01cm^(2)`
B. `pm0.02cm^(2)`
C. `pm0.05cm^(2)`
D. `pm0.10cm^(2)`

Explanation

To find the maximum possible error in the area of a cubical block given the side length and the vernier constant, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Side length of the cubical block, \( l = 2.50 \, \text{cm} \)
- Vernier constant (least count), \( \Delta l = 0.01 \, \text{cm} \)

2. **Calculate the Area of One Side of the Cube:**
- The area \( A \) of one side of the cube is given by the formula:
\[ A = l^2 \]
- Substituting the value of \( l \):
\[ A = (2.50 \, \text{cm})^2 = 6.25 \, \text{cm}^2 \]

3. **Determine the Maximum Possible Error in Area:**
- The formula for the maximum possible error in area \( \Delta A \) based on the error in length \( \Delta l \) is:
\[ \frac{\Delta A}{A} = 2 \cdot \frac{\Delta l}{l} \]
- Rearranging gives:
\[ \Delta A = A \cdot 2 \cdot \frac{\Delta l}{l} \]

4. **Substituting Values:**
- Substitute \( A = 6.25 \, \text{cm}^2 \), \( \Delta l = 0.01 \, \text{cm} \), and \( l = 2.50 \, \text{cm} \):
\[ \Delta A = 6.25 \cdot 2 \cdot \frac{0.01}{2.50} \]

5. **Calculating the Error:**
- First, calculate \( \frac{0.01}{2.50} \):
\[ \frac{0.01}{2.50} = 0.004 \]
- Now substitute back:
\[ \Delta A = 6.25 \cdot 2 \cdot 0.004 = 6.25 \cdot 0.008 = 0.05 \, \text{cm}^2 \]

6. **Final Result:**
- The maximum possible error in the area of the side of the block is:
\[ \Delta A = 0.05 \, \text{cm}^2 \]

### Conclusion:
The maximum possible error in the area of the side of the block is \( \pm 0.05 \, \text{cm}^2 \).