In a hydraulic lift, used at a service station the radius of the large and small piston are in the ration of `20 : 1`. What weight placed on the small piston will be sufficient to lift a car of mass `1500 kg` ?

Options

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

Explanation

To solve the problem of determining the weight that must be placed on the small piston of a hydraulic lift to lift a car of mass 1500 kg, we can follow these steps:

### Step 1: Understand the Principle of Hydraulic Lifts
Hydraulic lifts operate based on Pascal's principle, which states that pressure applied to a confined fluid is transmitted undiminished in all directions. This means that the pressure exerted on the small piston will be equal to the pressure exerted on the large piston.

### Step 2: Define the Areas of the Pistons
Given the ratio of the radii of the large piston (R1) to the small piston (R2) is 20:1, we can express the areas of the pistons as follows:
- The area of the large piston (A1) is proportional to \( R1^2 \).
- The area of the small piston (A2) is proportional to \( R2^2 \).

Since \( R1 = 20R2 \), we can calculate the areas:
- \( A1 = \pi R1^2 = \pi (20R2)^2 = 400\pi R2^2 \)
- \( A2 = \pi R2^2 \)

Thus, the ratio of the areas is:
\[ \frac{A1}{A2} = \frac{400\pi R2^2}{\pi R2^2} = 400 \]

### Step 3: Apply the Pressure Equation
The pressure exerted by the weight of the car on the large piston is given by:
\[ P1 = \frac{F1}{A1} = \frac{m_{\text{large}} \cdot g}{A1} \]
Where \( m_{\text{large}} = 1500 \, \text{kg} \) (mass of the car) and \( g \) is the acceleration due to gravity.

The pressure exerted by the weight on the small piston is given by:
\[ P2 = \frac{F2}{A2} = \frac{m_{\text{small}} \cdot g}{A2} \]
Where \( m_{\text{small}} \) is the mass we need to find.

### Step 4: Set the Pressures Equal
Since the pressures are equal (P1 = P2), we can write:
\[ \frac{m_{\text{large}} \cdot g}{A1} = \frac{m_{\text{small}} \cdot g}{A2} \]
The \( g \) cancels out:
\[ \frac{m_{\text{large}}}{A1} = \frac{m_{\text{small}}}{A2} \]

### Step 5: Substitute the Area Ratio
Substituting the area ratio:
\[ m_{\text{small}} = m_{\text{large}} \cdot \frac{A2}{A1} \]
Using the area ratio we found earlier:
\[ \frac{A2}{A1} = \frac{1}{400} \]
Thus:
\[ m_{\text{small}} = 1500 \cdot \frac{1}{400} \]

### Step 6: Calculate the Mass on the Small Piston
Calculating the mass:
\[ m_{\text{small}} = \frac{1500}{400} = 3.75 \, \text{kg} \]

### Final Answer
The weight that must be placed on the small piston to lift the car is **3.75 kg**.

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