In a hydraulic lift at a service station, the radii of the large and small piston are in the ratio of 20 : 1. What weight placed on the small piston will be sufficient to lift a car of mass 1200 kg ?

Options

A. 3.75 Kg
B. 37.5 kg
C. 7.5 kg
D. 75 kg

Answer

To solve the problem, we will use the principles of hydraulics, specifically Pascal's law, which states that pressure applied to a confined fluid is transmitted undiminished throughout the fluid.

### Step-by-Step Solution:

1. **Identify the Given Ratios**:
The ratio of the radii of the large piston (R) to the small piston (r) is given as 20:1. This means:
\[ R = 20r \]

2. **Calculate the Area Ratio**:
The area of a circle is given by \( A = \pi R^2 \). Therefore, the areas of the large and small pistons can be expressed as:
\[ A_{large} = \pi R^2 \quad \text{and} \quad A_{small} = \pi r^2 \]
The ratio of the areas is:
\[ \frac{A_{large}}{A_{small}} = \frac{\pi R^2}{\pi r^2} = \frac{R^2}{r^2} = \left(\frac{R}{r}\right)^2 = \left(\frac{20}{1}\right)^2 = 400:1 \]

3. **Apply Pascal's Law**:
According to Pascal's law, the pressure exerted on both pistons must be equal:
\[ \frac{F_{large}}{A_{large}} = \frac{F_{small}}{A_{small}} \]
Where:
- \( F_{large} \) is the weight of the car (mass \( m \) times gravitational acceleration \( g \)).
- \( F_{small} \) is the weight placed on the small piston.

4. **Substituting Values**:
The weight of the car is:
\[ F_{large} = m_{car} \cdot g = 1200 \, \text{kg} \cdot g \]
Since we are looking for the weight \( F_{small} \) that needs to be placed on the small piston, we can express it as:
\[ F_{small} = m_{small} \cdot g \]
Thus, we can rewrite the equation:
\[ \frac{1200g}{A_{large}} = \frac{m_{small}g}{A_{small}} \]

5. **Canceling \( g \)**:
The gravitational acceleration \( g \) cancels out from both sides:
\[ \frac{1200}{A_{large}} = \frac{m_{small}}{A_{small}} \]

6. **Using the Area Ratio**:
From our earlier calculation, we know:
\[ \frac{A_{large}}{A_{small}} = 400 \]
Therefore:
\[ A_{large} = 400 A_{small} \]
Substituting this into the equation gives:
\[ \frac{1200}{400 A_{small}} = \frac{m_{small}}{A_{small}} \]

7. **Solving for \( m_{small} \)**:
Rearranging gives:
\[ m_{small} = \frac{1200}{400} = 3 \, \text{kg} \]

### Conclusion:
The weight that needs to be placed on the small piston to lift the car of mass 1200 kg is **3 kg**.