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 kgf
B. 30 kgf
C. 300kgf
D. 300 kgf

Explanation

To solve the problem of finding the weight that needs to be placed on the small piston of a hydraulic lift to lift a car of mass 1200 kg, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have a hydraulic lift with two pistons. The radius of the large piston (R2) and the small piston (R1) are in the ratio of 20:1. We need to find the weight (F1) that must be placed on the small piston to lift a car weighing 1200 kg (F2).

2. **Identify the Forces**: The force exerted by the car (F2) can be calculated using the formula:
\[ F2 = m \cdot g \]
where \( m = 1200 \, \text{kg} \) and \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity). However, since we are dealing with kgf (kilogram-force), we can consider \( g \) as 1 kgf for simplicity in this context.

3. **Calculate the Force Exerted by the Car**:
\[ F2 = 1200 \, \text{kgf} \]

4. **Use Pascal's Law**: According to Pascal's law, the pressure at both pistons is the same:
\[ \frac{F1}{A1} = \frac{F2}{A2} \]
where \( A1 \) is the area of the small piston and \( A2 \) is the area of the large piston.

5. **Relate the Areas to the Radii**: The area of a circle is given by \( A = \pi R^2 \). Thus:
\[ A1 = \pi R1^2 \quad \text{and} \quad A2 = \pi R2^2 \]

6. **Substitute the Areas into Pascal's Law**:
\[ \frac{F1}{\pi R1^2} = \frac{F2}{\pi R2^2} \]
The \( \pi \) cancels out:
\[ \frac{F1}{R1^2} = \frac{F2}{R2^2} \]

7. **Express F1 in Terms of F2**:
\[ F1 = F2 \cdot \frac{R1^2}{R2^2} \]

8. **Substitute the Ratio of Radii**: Given the ratio \( R2 : R1 = 20 : 1 \), we can express \( R1 \) in terms of \( R2 \):
\[ R1 = \frac{R2}{20} \]
Therefore,
\[ \frac{R1^2}{R2^2} = \left(\frac{1}{20}\right)^2 = \frac{1}{400} \]

9. **Calculate F1**:
\[ F1 = F2 \cdot \frac{1}{400} \]
Substituting \( F2 = 1200 \, \text{kgf} \):
\[ F1 = 1200 \cdot \frac{1}{400} = 3 \, \text{kgf} \]

10. **Conclusion**: The weight that needs to be placed on the small piston to lift the car is **3 kgf**.