In a hydraulic press used for lifting heavy loads, the diameters of smaller and larger pistons are 2 cm and 5 cm, respectively. If the smaller piston is pushed in through 5 cm, then by how much distance will the larger piston move out?

Answer

To solve the problem of how much the larger piston moves out when the smaller piston is pushed in by 5 cm, we can use the principle of conservation of volume in an incompressible fluid. Here’s a step-by-step solution:

### Step 1: Identify the diameters and radii of the pistons
- The diameter of the smaller piston (D1) = 2 cm
- The diameter of the larger piston (D2) = 5 cm
- The radius of the smaller piston (r1) = D1/2 = 1 cm
- The radius of the larger piston (r2) = D2/2 = 2.5 cm

### Step 2: Calculate the volume displaced by the smaller piston
The volume (V1) displaced by the smaller piston when it is pushed in can be calculated using the formula for the volume of a cylinder:
\[ V_1 = \pi r_1^2 h_1 \]
Where:
- \( h_1 = 5 \) cm (the distance the smaller piston is pushed in)

Substituting the values:
\[ V_1 = \pi (1 \text{ cm})^2 (5 \text{ cm}) = \pi \times 1 \times 5 = 5\pi \text{ cm}^3 \]

### Step 3: Set up the volume equation for the larger piston
Let \( h_2 \) be the distance the larger piston moves out. The volume (V2) of the larger piston can be expressed as:
\[ V_2 = \pi r_2^2 h_2 \]

### Step 4: Equate the volumes
Since the fluid is incompressible, the volume displaced by the smaller piston is equal to the volume displaced by the larger piston:
\[ V_1 = V_2 \]
\[ 5\pi = \pi (2.5 \text{ cm})^2 h_2 \]

### Step 5: Simplify the equation
We can cancel \( \pi \) from both sides:
\[ 5 = (2.5)^2 h_2 \]
\[ 5 = 6.25 h_2 \]

### Step 6: Solve for \( h_2 \)
Now, we can solve for \( h_2 \):
\[ h_2 = \frac{5}{6.25} = 0.8 \text{ cm} \]

### Step 7: Convert to mm
To express this in millimeters:
\[ h_2 = 0.8 \text{ cm} = 8 \text{ mm} \]

### Final Answer
The larger piston will move out by **8 mm**.

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