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

Answer

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

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Mass of the car, \( m = 1500 \, \text{kg} \)
- Radius ratio of the large piston to the small piston, \( R_L : R_S = 20 : 1 \)

2. **Calculate the Weight of the Car**:
- The weight \( W \) of the car can be calculated using the formula:
\[ W = m \cdot g \]
- Where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \).
- Thus,
\[ W = 1500 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 14715 \, \text{N} \]

3. **Understanding the Hydraulic Lift Principle**:
- In a hydraulic lift, the pressure exerted by the fluid is the same on both pistons. Therefore, we can set up the equation based on pressure:
\[ P_1 = P_2 \]
- Where \( P_1 \) is the pressure on the small piston and \( P_2 \) is the pressure on the large piston.

4. **Express Pressure in Terms of Force and Area**:
- The pressure can be expressed as:
\[ P = \frac{F}{A} \]
- The area \( A \) of a piston is given by \( A = \pi R^2 \).
- Therefore, we can write:
\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \]

5. **Substituting Areas**:
- Let \( F_1 \) be the force (weight) on the small piston and \( F_2 \) be the weight of the car (14715 N).
- The areas can be expressed as:
\[ A_1 = \pi R_S^2 \quad \text{and} \quad A_2 = \pi R_L^2 \]
- Given the radius ratio \( R_L : R_S = 20 : 1 \), we can express \( R_L = 20 R_S \).

6. **Calculating the Areas**:
- Thus, we have:
\[ A_1 = \pi R_S^2 \quad \text{and} \quad A_2 = \pi (20 R_S)^2 = \pi \cdot 400 R_S^2 \]

7. **Setting Up the Pressure Equation**:
- Now, substituting the areas into the pressure equation:
\[ \frac{F_1}{\pi R_S^2} = \frac{14715}{\pi \cdot 400 R_S^2} \]
- The \( \pi R_S^2 \) cancels out:
\[ F_1 = \frac{14715}{400} \]

8. **Calculating the Force on the Small Piston**:
- Now, calculate \( F_1 \):
\[ F_1 = \frac{14715}{400} = 36.7875 \, \text{N} \]

9. **Finding the Equivalent Mass**:
- To find the mass equivalent to this force:
\[ m_1 = \frac{F_1}{g} = \frac{36.7875}{9.81} \approx 3.75 \, \text{kg} \]

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

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