A hydraulic automobile lift is designed to lift cars with a maximum mass of `3000Kg`. The area of cross section of the piston carrying the load is `425cm^(2)`. What maximum pressures would the smaller piston have to bear?

Explanation

To solve the problem of determining the maximum pressure that the smaller piston of a hydraulic lift must bear, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Mass of the car, \( m = 3000 \, \text{kg} \)
- Area of the larger piston, \( A = 425 \, \text{cm}^2 \)

2. **Convert Area to Square Meters**:
- Since pressure is typically expressed in Pascals (N/m²), we need to convert the area from cm² to m²:
\[ A = 425 \, \text{cm}^2 = 425 \times 10^{-4} \, \text{m}^2 = 0.0425 \, \text{m}^2 \]

3. **Calculate the Force Exerted by the Car**:
- The force exerted by the car due to gravity can be calculated using the formula:
\[ F = m \cdot g \]
- Where \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \):
\[ F = 3000 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 29400 \, \text{N} \]

4. **Calculate the Pressure on the Larger Piston**:
- The pressure \( P \) exerted on the larger piston can be calculated using the formula:
\[ P = \frac{F}{A} \]
- Substituting the values we have:
\[ P = \frac{29400 \, \text{N}}{0.0425 \, \text{m}^2} \]

5. **Perform the Calculation**:
- Now, calculate the pressure:
\[ P = \frac{29400}{0.0425} \approx 691764.71 \, \text{N/m}^2 \]
- Rounding this value gives:
\[ P \approx 6.92 \times 10^5 \, \text{N/m}^2 \]

6. **Conclusion**:
- The maximum pressure that the smaller piston would have to bear is approximately \( 6.92 \times 10^5 \, \text{N/m}^2 \) or Pascal.