### Correct Option: B. Decrease #### Detailed Explanation: When a ship moves from salt water into fresh water, the fraction of its volume that is above the water surface decreases. To understand why this happens, we need to consider the concepts of buoyancy and density. 1. **Buoyancy and Archimedes' Principle**: - According to Archimedes' principle, a floating object displaces a volume of fluid equal to the weight of the object. The buoyant force acting on the ship is equal to the weight of the water displaced. - The buoyant force can be calculated using the formula: \[ F_b = \rho_{fluid} \cdot V_{displaced} \cdot g \] where \( F_b \) is the buoyant force, \( \rho_{fluid} \) is the density of the fluid (water in this case), \( V_{displaced} \) is the volume of fluid displaced, and \( g \) is the acceleration due to gravity. 2. **Density of Salt Water vs. Fresh Water**: - Salt water has a higher density than fresh water. The typical density of salt water is about \( 1.025 \, \text{g/cm}^3 \), while the density of fresh water is approximately \( 1.00 \, \text{g/cm}^3 \). - When the ship is in salt water, it displaces a certain volume of salt water that corresponds to its weight. When it moves into fresh water, the same weight of the ship will displace a larger volume of fresh water due to the lower density. 3. **Effect on the Ship's Position**: - As the ship enters fresh water, it will float lower in the water because it needs to displace more volume of the less dense fresh water to balance its weight. - This means that a greater portion of the ship will be submerged in the fresh water compared to when it was in salt water. Consequently, the fraction of the ship's volume that is above the water surface decreases. 4. **Mathematical Representation**: - Let’s denote: - \( W \) = weight of the ship - \( V_s \) = volume of the ship submerged in salt water - \( V_f \) = volume of the ship submerged in fresh water - In salt water: \[ W = \rho_{salt} \cdot V_s \cdot g \] - In fresh water: \[ W = \rho_{fresh} \cdot V_f \cdot g \] - Since \( \rho_{salt} > \rho_{fresh} \), it follows that \( V_f > V_s \). Therefore, the volume of the ship above the water surface decreases when moving from salt water to fresh water. #### Why Other Options Are Incorrect: - **Option A: Increase**: This option suggests that the fraction of the ship's volume above the water surface increases. This is incorrect because, as explained, the ship displaces more volume in fresh water, causing it to sit lower in the water. - **Option C: Remain the same**: This option implies that the fraction of the volume above the water surface does not change. This is also incorrect because the change in water density directly affects how much of the ship is submerged. - **Option D: Increase then decrease**: This option suggests a two-phase change, which is not applicable in this scenario. The transition from salt water to fresh water results in a consistent decrease in the fraction of the volume above the water surface. ### Revision Summary: - A ship floats lower in fresh water than in salt water due to the difference in density. - The buoyant force is determined by the weight of the fluid displaced, which changes with fluid density. - As the ship moves from salt water to fresh water, it displaces more volume, leading to a decrease in the fraction of its volume above the water surface. - Understanding buoyancy and density is crucial for predicting how objects behave in different fluids.