/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 A hemispherical bowl just floats... [FREE SOLUTION] | 91Ó°ÊÓ

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A hemispherical bowl just floats without sinking in a of density \(1.2 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). If the outer diameter a density of the bowl are \(1 \mathrm{~m}\) and \(2 \times 10^{4} \mathrm{~kg} / \mathrm{m}^{3}\), respectively, then the inner diameter of the bowl will be (1) \(0.94 \mathrm{~m}\) (2) \(0.97 \mathrm{~m}\) (3) \(0.98 \mathrm{~m}\) (4) \(0.99 \mathrm{~m}\)

Short Answer

Expert verified
The inner diameter of the bowl is approximately 0.97 m.

Step by step solution

01

Understand the Concept of Buoyancy

When the bowl just floats, the volume of liquid displaced by the bowl equals the volume of the bowl that is immersed in the fluid. This is based on the buoyant force principle from Archimedes' principle, which states that the buoyant force is equal to the weight of the displaced fluid. Since the bowl floats without sinking, the buoyancy must equal the weight of the bowl.
02

Calculate the Mass and Volume of the Bowl

The mass of the hemispherical bowl can be expressed as \(m = \rho_s \times V_s\), where \(\rho_s = 2 \times 10^4 \ \text{kg/m}^3\) is the density of the material of the bowl, and \(V_s\) is the volume of the bowl. Assuming a thin-walled bowl, we consider the volume of the entirety of the material as negligible compared to the entire hemisphere's volume.
03

Equation for Volume of a Hemisphere

The formula for the volume of a hemisphere is given by \(V = \frac{2}{3} \pi R^3\), where \(R\) is the outer radius of the hemispherical bowl. For a diameter of \(1 \ \text{m}\), the outer radius is \(0.5 \ \text{m}\).
04

Establish Equilibrium Condition

For the bowl to just float, the weight of the displaced water must equal the weight of the bowl. This implies, \(\rho_f \times V_f \times g = \rho_s \times V_s \times g\). Simplifying gives us, \(\rho_f \times V_f = \rho_s \times V_s\), with \(\rho_f = 1.2 \times 10^3 \ \text{kg/m}^3\) (density of fluid) and \(V_f\) being the volume of the fluid displaced. Substitute \(V_s = \frac{2}{3} \pi (R_0^3 - r^3)\), where \(r\) is inner radius.
05

Solve for the Inner Diameter

After simplification, we have \(1.2 \times 10^3 \times \frac{2}{3} \pi R_0^3 = 2 \times 10^4 \times \frac{2}{3} \pi (R_0^3 - r^3)\). Solve for \(r\), yielding \(r^3 = R_0^3 - \frac{\rho_f}{\rho_s} \times R_0^3\). Simplifying, we get \(r = (R_0^3 - 0.06 \times R_0^3)^{1/3}\). With numerical calculation, given \(R_0 = 0.5 \text{ m}\), \(r \approx 0.485 \text{ m}\). So, the inner diameter is \(2 \times r \approx 0.97 \text{ m}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Buoyancy
Buoyancy is a concept derived from Archimedes' principle. It explains the upward force that allows objects to float or rise in fluids like water and air. This force is equal to the weight of the fluid that the object displaces. When an object is immersed in a fluid, it encounters a buoyant force that acts in the opposite direction to gravity. If the weight of the object is less than or equal to the weight of the fluid it displaces, it will float. This means the object experiences an upward buoyant force equal to the weight of the displaced fluid, allowing it to hover or stay suspended at a certain level without sinking.
Density
Density is a crucial concept when understanding buoyancy and floating. It is defined as mass per unit volume and typically expressed in kilograms per cubic meter (kg/m³). The formula to calculate density is \( \rho = \frac{m}{V} \), where \( m \) is mass and \( V \) is volume.
Objects with a density higher than the fluid they are placed in will sink, while those with lower density will float. In scenarios involving hemispherical bowls or similar objects, understanding the density of both the object and the fluid is important in predicting whether the object will float or sink. A balance between the densities helps establish floating conditions.
Volume of a Hemisphere
Hemispheres are half of a sphere, which means their volume is half of that of a full sphere. The formula to calculate the volume of a hemisphere is \( V = \frac{2}{3} \pi R^3 \), where \( R \) represents the radius of the hemisphere.
In the context of floating objects like a hemispherical bowl, knowing the volume is essential for computing how much of the bowl will be submerged when it floats. It helps in understanding the volume of the liquid displaced and, consequently, the buoyant force that acts on the object.
Floating Condition
Floating conditions occur when the buoyant force experienced by an object equals its weight. This balance means that the object neither sinks nor rises above the fluid's surface. For a hemispherical object to float, the weight of the fluid displaced must match the object's weight. The equilibrium formula as described by Archimedes' principle is \( \rho_f \times V_f = \rho_s \times V_s \), where \( \rho_f \) is fluid density, \( V_f \) is the volume of fluid displaced, \( \rho_s \) is the object's density, and \( V_s \) the object's volume. Solving this equation with known densities and volumes helps determine conditions such as how much of the object stays above water or floats without being fully submerged.

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