Problem 15 - Master Resource Book in JEE Main Physics

Question

A balloon of volume \(1500 \mathrm{~m}^{3}\) and weighing \(1650 \mathrm{~kg}\) with all its equipment is filled with He (density \(0.2 \mathrm{~kg} \mathrm{~m}^{-3}\) ). If the density of air be \(1.3 \mathrm{kgm}^{-3}\), the pull on the rope tied to the balloon will be (a) \(300 \mathrm{~kg}\) (b) \(1950 \mathrm{~kg}\) (c) \(1650 \mathrm{~kg}\) (d) zero

Step-by-Step Solution

Answer

(d) zero

Step 1: Calculate Weight of Helium

To find the weight of the helium inside the balloon, multiply its density by the balloon's volume. The formula for weight is: \[ \text{Weight of Helium} = \text{Density of Helium} \times \text{Volume of Balloon} \]Substituting the given values:\[ \text{Weight of Helium} = 0.2 \, \text{kg/m}^3 \times 1500 \, \text{m}^3 = 300 \, \text{kg} \]

Step 2: Calculate Weight of Displaced Air

The balloon displaces air, leading to buoyancy. To find the weight of the displaced air, use the formula:\[ \text{Weight of Displaced Air} = \text{Density of Air} \times \text{Volume of Balloon} \]Substituting the given values:\[ \text{Weight of Displaced Air} = 1.3 \, \text{kg/m}^3 \times 1500 \, \text{m}^3 = 1950 \, \text{kg} \]

Step 3: Calculate Net Weight of the Balloon System

The net weight of the balloon system includes the weight of the balloon and its equipment minus the weight of the displaced air. The formula is:\[ \text{Net Weight} = \text{Weight of Balloon System} - \text{Weight of Displaced Air} \]Substitute the weights:\[ \text{Net Weight} = (1650 \, \text{kg} + 300 \, \text{kg}) - 1950 \, \text{kg} = 0 \, \text{kg} \]

Step 4: Determine the Pull on the Rope

Since the net weight of the balloon system is zero, the pull required on the rope to keep the balloon in place is also zero. This means the balloon is neutrally buoyant and does not exert any upward or downward force.

Key Concepts

Archimedes' Principle

Archimedes' Principle is fundamental to understanding buoyancy in physics. It states that a body fully or partially submerged in a fluid experiences an upward force, known as buoyant force, that is equal to the weight of the fluid displaced by the body. In the case of the balloon example, the helium-filled balloon displaces air.
Since the density of helium is less than the density of air, the balloon exerts an upward buoyant force. This principle is what allows balloons to float.

• The balloon's buoyancy is determined by subtracting the weight of the air displaced from the total weight of the balloon system (balloon + helium).
• If the buoyant force is equal to or greater than the weight of the balloon, the balloon rises.
• When the forces are balanced, as in this problem, the balloon remains neutral, resulting in zero net weight on the rope.

Understanding Archimedes' Principle helps in predicting whether an object will float or sink, which is essential for designing lighter-than-air vehicles.

Density Calculation

In physics, density calculation is crucial for determining how substances will interact under various conditions. Density is defined as mass per unit volume and is given by the formula: \[ \rho = \frac{m}{V} \] where \( \rho \) is density, \( m \) is mass, and \( V \) is volume. For our balloon scenario, density plays a key role in understanding why the helium-filled balloon can float.
The densities of both helium and air are given in the problem. By using these values, we calculate the mass each occupies within the volume of the balloon. Despite the helium's lower density, it occupies the same volume as the air it displaces, resulting in the balloon's lift.

• Helium has a density of 0.2 kg/m³, making it lighter than air.
• Air, with a density of 1.3 kg/m³, is heavier, contributing to the upward buoyant force.

Determining densities allows us to predict other properties such as pressure, temperature, and volume changes in balloon physics.

Balloon Physics

Balloon physics involves understanding how various forces and properties of gases affect the behavior of balloons. When dealing with helium balloons, one critical aspect is the interplay between helium's properties and the atmospheric air.
Helium is a lighter-than-air gas, allowing it to rise and displace the denser air outside the balloon. The lift of a balloon depends largely on the balance of weights - the balloon with its gas tries to rise, while gravity pulls it down. In the provided exercise, these principles are assessed by:

• Computing the weight of helium inside the balloon based on its density and volume.
• Determining the weight of the air displaced by the same volume.

The balloon effectively becomes buoyant due to this gas replacement mechanism, lifting the balloon until the forces are in equilibrium.
In real-world applications, factors such as the balloon material, volume, and the altitude at which it is used, all influence its performance.

Weight and Mass Calculations

Weight and mass are often confused, but in physics, they have distinct meanings. Mass is a measure of the amount of matter in an object and is constant regardless of location. Weight, however, is the force exerted by gravity on that mass. It is calculated as: \[ W = m \times g \] where \( W \) is weight, \( m \) is mass, and \( g \) is acceleration due to gravity (approximately 9.81 m/s² on Earth).
In the balloon problem, understanding these concepts helps solve for the forces exerted upon it.

• The system's mass, including the balloon, helium, and equipment, needs careful calculation.
• Buoyant force is the critical factor in offsetting the balloon's weight.
• Achieving a net weight of zero indicates that the upward and downward forces are balanced, leading to neutral buoyancy.

This methodical calculation is essential in engineering tasks, especially those involving buoyant materials or vehicles.