What force is applied on a piston of area of cross section `2cm^(2)` to obtain a force 150 N on the piston of area of cross section `12 cm^(2)` in a hydraulic machine ?

Answer

To solve the problem, we will use Pascal's law, which states that the pressure applied to a confined fluid is transmitted undiminished in all directions throughout the fluid. This principle can be applied to hydraulic machines.

### Step-by-step Solution:

1. **Understand the Given Data:**
- Area of piston 1 (A1) = 2 cm²
- Area of piston 2 (A2) = 12 cm²
- Force on piston 2 (F2) = 150 N
- We need to find the force applied on piston 1 (F1).

2. **Use Pascal's Law:**
According to Pascal's law, the pressure on both pistons is equal:
\[ P_1 = P_2 \]
Where:
\[ P_1 = \frac{F_1}{A_1} \quad \text{and} \quad P_2 = \frac{F_2}{A_2} \]

3. **Set Up the Equation:**
From the equality of pressures, we have:
\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \]

4. **Substitute the Known Values:**
Substitute the values into the equation:
\[ \frac{F_1}{2} = \frac{150}{12} \]

5. **Cross-Multiply to Solve for F1:**
Cross-multiplying gives:
\[ F_1 \cdot 12 = 150 \cdot 2 \]
Simplifying the right side:
\[ F_1 \cdot 12 = 300 \]

6. **Solve for F1:**
Now, divide both sides by 12 to find F1:
\[ F_1 = \frac{300}{12} = 25 \text{ N} \]

7. **Conclusion:**
The force that needs to be applied on the piston of area of cross-section 2 cm² is **25 N**.