Final answer:
The relative density of a body that weighs 75 N in air and 37.5 N in water is calculated using the loss of weight in water. The weight loss represents the buoyant force equal to the water displaced by the body. By the calculation, the relative density is found to be 2.0.
"the correct option is approximately option D"
Step-by-step explanation:
The student's question is about finding the relative density of a body based on its weight in air and water. The relative density, also known as specific gravity, is a measure of the density of a material compared to the density of water. This concept is studied in the field of physics, particularly when discussing principles like Archimedes' Principle and buoyancy.
Firstly, we assess the weight of the body in air (75 N) and in water (37.5 N). To find the relative density, we use the relation between the loss of weight in water and the weight in air. The weight loss in water is the buoyant force, which is equal to the weight of the water displaced by the body.
Here are the steps to solve the problem:
- Calculate the loss in weight when the body is immersed in water: Weight in air (75 N) - Weight in water (37.5 N) = 37.5 N. This is the weight of the water displaced by the body.
- The relative density (D) is the ratio of the weight in air (Wair) to the loss of weight in the water (Wloss): D = Wair / Wloss.
- Substitute the known values: D = 75 N / 37.5 N = 2.0.
Therefore, the relative density of the body is 2.0, which corresponds to option D.