Problem 30 - Master Resource Book in JEE Main Physics

Question

A hemispherical bowl just floats without sinking in a liquid of density \(1.2 \times 10^{3} \mathrm{kgm}^{-3}\). If outer diameter and the density of the bowl are \(1 \mathrm{~m}\) and \(2 \times 10^{4} \mathrm{kgm}^{-3}\) respectively, then the inner diameter of the bowl will be (a) \(0.94 \mathrm{~m}\) (b) \(0.96 \mathrm{~m}\) (c) \(0.98 \mathrm{~m}\) (d) \(0.99 \mathrm{~m}\)

Step-by-Step Solution

Answer

The inner diameter of the bowl is 0.96 m.

Step 1: Understand the Problem

We are given a hemispherical bowl that floats in a liquid. The outer diameter is 1 m, and the density of the bowl is \(2 \times 10^{4} \mathrm{kg/m}^{3}\), while the liquid's density is \(1.2 \times 10^{3} \mathrm{kg/m}^{3}\). We need to find the inner diameter of the bowl.

Step 2: Apply Archimedes' Principle

According to Archimedes' principle, the buoyant force is equal to the weight of the displaced liquid. For a bowl floating just right, the volume of the liquid displaced equals the outer hemisphere's volume minus the inner hemisphere's volume.

Step 3: Calculate Volume of Outer Hemisphere

The formula for the volume of a hemisphere is \[ V = \frac{2}{3} \pi r^3 \]. Where the outer radius \( R = 0.5 \mathrm{~m} \). Hence, the outer volume \[ V_o = \frac{2}{3} \pi (0.5)^3 \].

Step 4: Calculate Mass of Bowl

The mass of the bowl is calculated by multiplying its density with the volume difference. The density of the bowl is \(2 \times 10^{4} \mathrm{kg/m}^{3}\).

Step 5: Write Displaced Liquid Weight Equation

The weight of the displaced liquid is given by \(V_o \times \rho_{liquid} \times g\). This should be equal to the weight of the bowl, i.e., its volume difference \((V_o - V_i) \times \rho_{bowl} \times g\).

Step 6: Solve for Inner Volume

Equate the weights to get: \[ V_o \times \rho_{liquid} = (V_o - V_i) \times \rho_{bowl} \]. Solve for \( V_i \), the inner volume.

Step 7: Calculate Inner Diameter

The volume inside, \( V_i \), equates to \( \frac{2}{3} \pi r_{inner}^3 \). Use this to solve for \( r_{inner} \) and subsequently find the inner diameter by doubling \( r_{inner} \).

Key Concepts

Understanding Buoyancy

Buoyancy is a fundamental concept in fluid mechanics that explains why objects float or sink. It is governed by Archimedes' Principle, which states that an object in a fluid experiences an upward force called the buoyant force.
This force is equal to the weight of the fluid displaced by the object. In the context of the floating hemispherical bowl, when the bowl is placed in the liquid, it displaces some of the liquid.
The liquid's weight that has been displaced is equal to the buoyant force acting on the bowl. For the bowl to float perfectly, the buoyant force must equal the gravitational force pulling the bowl down. This means the weight of the displaced fluid is equal to the weight of the bowl.
When these forces balance out, the bowl floats. If the weight of the bowl were more than the weight of the fluid displaced, the bowl would sink. Conversely, if it were less, it would float higher.

Performing Density Calculations

Density is the property that defines how much mass is contained within a specific volume. It's a key factor when considering floating objects like our bowl. The formula for density is:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
In this problem, we have two densities to consider: the liquid and the bowl.

• For the liquid, the density \( \rho_{liquid} \) is given as \(1.2 \times 10^3 \mathrm{kg/m}^3\).
• The bowl's density \( \rho_{bowl} \) is \(2 \times 10^4 \mathrm{kg/m}^3\).

To find out if the bowl will float or sink, we need to balance these densities through their volumes. Recall, the weight of the displaced liquid, described by its volume multiplied by its density and gravity, should match the bowl's weight.
Through appropriate calculations, we use these densities to find out how much of the bowl remains above the liquid surface.

Calculating the Volume of a Hemisphere

When considering hemispheres such as our bowl, calculating volume is crucial. The hemisphere's volume is half of a full sphere's volume, given by:
\[ V = \frac{4}{3} \pi r^3 \]
So the hemisphere's volume becomes:
\[ V = \frac{2}{3} \pi r^3 \]
In this exercise, we need the volume calculations for both the outer and inner hemispheres of the bowl.
For the outer hemisphere, with an outer radius of \(0.5 \mathrm{~m}\), its volume \( V_o \) is:
\[ V_o = \frac{2}{3} \pi (0.5)^3 \]
However, finding the inner hemisphere's radius \(r_{inner}\) requires a reverse calculation of this formula. We find the volume equation for the inner hemisphere \(V_i\), determined by its inner radius, and equate it to the displaced water volume.
With this setup, the inner diameter can be derived by doubling this inner radius once it's calculated.