91Ó°ÊÓ

When we immerse an object in a fluid, it experiences an upward force called the buoyant force. This force is the reason why objects feel lighter in water. The buoyant force occurs because the fluid pressure is greater at the bottom of the object than at the top, effectively pushing the object upward. This concept is governed by Archimedes' Principle, which states that the upward buoyant force is equal to the weight of the fluid displaced by the object.

In the case of our block exercise, we noticed different readings on the scale when the block was submerged in different fluids, like water and oil. For instance, when the block was underwater, the spring scale read 31.1 N, compared to 35.0 N when in air. This difference of 3.9 N is precisely the buoyant force in water. Similarly, in oil, the buoyant force was 3.2 N, demonstrating how different fluids exert different buoyant forces based on their densities.

The volume of a block or any submerged object can be determined using the buoyant force it experiences. In the exercise, we used the buoyant force in water to find out the volume. The formula is derived from the relationship: \[ F_b = \rho V g \]where \( F_b \) is the buoyant force, \( \rho \) is the fluid density, \( V \) is the volume of the object, and \( g \) is the acceleration due to gravity.

To solve for the volume, the equation is rearranged as: \[ V = \frac{F_{b,\text{water}}}{\rho_{\text{water}} \times g} \]Using this, we calculated the block's volume as approximately \( 3.97 \times 10^{-4} \, m^3 \). This calculation is pivotal to further steps, like determining the density of the block and the oil.

Density signifies how compact the mass in a substance is. It is calculated by dividing the mass by the volume and is expressed in kilograms per cubic meter (kg/m³). In the context of this problem, finding the density of the oil is about understanding how much mass is contained in a specific volume of oil.

Given the volume of the block and the buoyant force when submerged in oil, we can derive the density of oil. The relationship used is: \[ \rho_{\text{oil}} = \frac{F_{b,\text{oil}}}{V \times g} \]Here, we find \( \rho_{\text{oil}} \), using the buoyant force of 3.2 N and the known volume of the block. Substituting the values gives us the density of oil as approximately \( 808 \, \text{kg/m}^3 \). This value contrasts with the density of water and highlights the oil's lighter nature.