Force and area are measured as 20N and `5 m^2` with errors 0.05 N and `0.0125 m^2.` The maximum error in pressure is (SI units)

Options

A. `4 pm 0.0625`
B. `4 pm 0.05`
C. `4 pm 0.125`
D. `4 pm 0.02`

Answer

To find the maximum error in pressure when given the force and area along with their respective errors, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Force (F) = 20 N
- Area (A) = 5 m²
- Error in Force (ΔF) = 0.05 N
- Error in Area (ΔA) = 0.0125 m²

2. **Calculate the pressure (P):**
\[ P = \frac{F}{A} = \frac{20 \, \text{N}}{5 \, \text{m}^2} = 4 \, \text{N/m}^2 \]

3. **Use the formula for the relative error in pressure:**
The relative error in pressure (ΔP/P) can be calculated using the formula:
\[ \frac{\Delta P}{P} = \frac{\Delta F}{F} + \frac{\Delta A}{A} \]

4. **Substitute the values into the formula:**
- Calculate the relative error for force:
\[ \frac{\Delta F}{F} = \frac{0.05 \, \text{N}}{20 \, \text{N}} = 0.0025 \]
- Calculate the relative error for area:
\[ \frac{\Delta A}{A} = \frac{0.0125 \, \text{m}^2}{5 \, \text{m}^2} = 0.0025 \]

5. **Combine the relative errors:**
\[ \frac{\Delta P}{P} = 0.0025 + 0.0025 = 0.005 \]

6. **Calculate the maximum error in pressure (ΔP):**
\[ \Delta P = P \times \frac{\Delta P}{P} = 4 \, \text{N/m}^2 \times 0.005 = 0.02 \, \text{N/m}^2 \]

### Final Answer:
The maximum error in pressure is \( \Delta P = 0.02 \, \text{N/m}^2 \).

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