91Ó°ÊÓ

Density is a fundamental concept used to measure how much mass is contained in a given volume. It is defined as the mass per unit volume and is mathematically represented as \( \rho = \frac{m}{V} \), where \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume. In the context of the hollow spherical shell problem, the density of glass is given as \( 2.60 \times 10^{3} \, \text{kg/m}^3 \). Since the mass of the glass is known to be 1 kilogram, you can calculate the volume of the glass by rearranging the formula to solve for \( V \): \( V = \frac{m}{\rho} \).
This calculation is crucial for determining the volume that the glass occupies when forming the shell.
The understanding of this concept aids in solving problems where you need to find one of the three properties if the other two are known.

Volume calculation is essential when determining the dimensions required for a specific shape or structure.
For the hollow spherical shell, it is crucial to calculate both the outer and inner volumes to find the respective radii. The outer volume \( V_{out} \) corresponds to the volume of water displaced, which is based on the condition for floating. In this problem, it is found using the formula: \( V_{out} = \frac{1}{\rho_w} \), where \( \rho_w \) is the density of water. This gives a volume of \( 10^{-3} \, \text{m}^3 \).
The inner volume \( V_{in} \), on the other hand, is calculated as the difference between the outer volume and the volume of glass (hollow portion): \( V = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3 \); where \( R \) and \( r \) are the outer and inner radii, respectively.Understanding how to calculate these volumes is essential for assessing the spatial requirements of the shell and is foundational in fluid mechanics problems involving buoyancy.

The principle of buoyancy explains how and why objects float. Derived from Archimedes' Principle, it states that for an object to float, the weight of the fluid displaced by the object must be equal to the weight of the object itself. This is the condition required for the hollow spherical shell to float in water.
In mathematical terms, the condition can be expressed as \( \rho_w \times V_{out} \times g = m \times g \). Here, \( \rho_w \) is the density of water, \( V_{out} \) is the volume of the sphere's outer surface (the total volume displaced), and \( \text{g} \) is the acceleration due to gravity.
This results in finding the volume of water displaced as equal to the mass divided by the density of water: \( V_{out} = \frac{m}{\rho_w} = 10^{-3} \, \text{m}^3 \). Understanding this floating condition helps in visualizing the balance of forces that allows the spherical shell to float and further guides in solving equations involving its dimensions.

A hollow spherical shell consists of two concentric spheres: an outer sphere and an inner sphere, with a cavity in between. The difference in volumes between these spheres defines the material volume forming the shell. In the exercise provided, this structure is vital for understanding how the shell floats and holds its shape.A critical aspect of dealing with hollow spherical shells is the radii differencing formula: \( V_{out} - V_{in} = V \) where \( V \) is the volume of the glass forming the shell, calculated from density and mass. The physical condition \( \frac{4}{3} \pi R^3 = 10^{-3} \), derived from floating, sets the equation for the outer radius.The process involves using traditional geometric volume formulas for spheres: \( \frac{4}{3} \pi R^3 \) and \( \frac{4}{3} \pi r^3 \). Solving these simultaneously allows for finding the two radii, helping visualize how the shell forms a balance between material used and space displaced.This understanding of a hollow spherical structure aids in problems across physics and engineering, providing a clear illustration of space and matter interaction in three-dimensional geometric shapes.