A solid weighs 45N and 15N respectively in air and water. Determine the relative density of the solid

Answer Details

The relative density of a substance is defined as the ratio of its density to the density of a reference substance, usually water at 4 degrees Celsius. In this problem, we can use the principle of buoyancy to determine the density of the solid. When an object is submerged in a fluid, it experiences an upward force called the buoyant force, which is equal to the weight of the fluid displaced by the object. If the object is less dense than the fluid, it will float, and if it is more dense, it will sink. We are given that the solid weighs 15 N in water, which means it displaces 15 N of water. The weight of the water displaced is equal to the buoyant force on the solid, which is equal to the weight of the solid when it is completely submerged in water. Therefore, the weight of the solid when it is completely submerged in water is 15 N. We are also given that the weight of the solid in air is 45 N. The difference between the weight of the solid in air and water is equal to the weight of the water displaced, which is 30 N. This means that the volume of water displaced by the solid is 30/9.8 = 3.06 L (since the density of water is 1000 kg/m^3 or 9.8 N/L). The relative density of the solid is equal to its density divided by the density of water. We can find the density of the solid by dividing its weight in air by its volume: Density of solid = Weight of solid in air / Volume of solid Density of solid = 45 N / (45 N - 15 N) [since weight of displaced water is 15N] Density of solid = 45 N / 30 N Density of solid = 1.5 N/L Therefore, the relative density of the solid is: Relative density = Density of solid / Density of water Relative density = 1.5 N/L / 1000 N/L Relative density = 0.0015 So the answer is 0.33 (rounded to two decimal places).