An incompressible liquid of density `rho` is enclosed with two frictionless pistons one of cross-sectional area A and the other of 4 A. When the narrow piston moves out by a distance.

Options

A. h
B. 4 h
C. `(h)/(4)`
D. `(h)/(5)`

Answer

To solve the problem, we need to analyze the movement of the two pistons and the relationship between the volumes displaced by each piston.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have an incompressible liquid with density \( \rho \) enclosed between two frictionless pistons.
- The first piston has a cross-sectional area \( A \) and the second piston has a cross-sectional area \( 4A \).

2. **Displacement of the Narrow Piston**:
- Let the narrow piston (area \( A \)) move out by a distance \( y \).
- The volume displaced by this piston can be calculated as:
\[ V_1 = A \cdot y \]

3. **Displacement of the Larger Piston**:
- The larger piston (area \( 4A \)) will experience a change in volume due to the incompressibility of the liquid.
- If the larger piston moves in by a distance \( h \), the volume displaced by this piston is:
\[ V_2 = 4A \cdot h \]

4. **Volume Conservation**:
- Since the liquid is incompressible, the total volume of the liquid must remain constant. Therefore, the volume displaced by the narrow piston must equal the volume displaced by the larger piston:
\[ V_1 = V_2 \]
- Substituting the expressions for \( V_1 \) and \( V_2 \):
\[ A \cdot y = 4A \cdot h \]

5. **Simplifying the Equation**:
- We can cancel \( A \) from both sides (since \( A \neq 0 \)):
\[ y = 4h \]

6. **Finding the Relationship**:
- Rearranging the equation gives us:
\[ h = \frac{y}{4} \]
- This indicates that for every unit distance \( y \) that the narrow piston moves out, the larger piston moves in by a distance of \( \frac{y}{4} \).

### Final Result:
Thus, the distance \( h \) that the larger piston moves in is:
\[ h = \frac{y}{4} \]