A hot air balloon is carrying some passengers, and a few sandbags of mass 1 kg each so that its tota

Q: Why do we ignore the mass of the hot air inside the balloon?

We aren't actually ignoring it! The problem states that the 'total mass is 480 kg'. This total mass already includes the mass of the balloon fabric, the basket, the passengers, the sandbags, and the hot air inside. The buoyant force must balance this entire total mass.

Q: Why does the buoyant force change if the volume of the balloon stays the same?

The buoyant force is given by $F_B = V \rho g$. Even though the volume $V$ is constant, the density of the surrounding outside air $\rho$ decreases as the balloon goes higher. Therefore, the buoyant force decreases with altitude.

Q: Is the approximation $e^{-x} \approx 1 - x$ always valid to use?

No, it is only valid when $x$ is very close to zero. In this problem, $x = \frac{1}{120} \approx 0.0083$, which is extremely small. For such small values, the higher-order terms in the Taylor series expansion (like $\frac{x^2}{2}$) become negligible, making the linear approximation highly accurate.

Q: What happens to the acceleration due to gravity, $g$, in the calculations?

Since the balloon is in equilibrium, the upward force equals the downward force: $V \rho g = m g$. Because $g$ appears as a multiplier on both sides of the equation, it perfectly cancels out. The equilibrium height depends on mass and density, not on the local strength of gravity.

Q: Why did the balloon stop exactly at 150 meters?

When the 4 sandbags were thrown out, the balloon became lighter, so the upward buoyant force was temporarily greater than the downward weight. The balloon accelerated upwards. As it rose, the surrounding air density dropped, which reduced the buoyant force. It stopped at 150 meters because at that exact height, the reduced buoyant force perfectly matched the new, lighter weight of the balloon.

Question

A hot air balloon is carrying some passengers, and a few sandbags of mass 1 kg each so that its total mass is 480 kg. Its effective volume giving the balloon its buoyancy is V. The balloon is floating at an equilibrium height of 100 m. When N number of sandbags are thrown out, the balloon rises to a new equilibrium height close to 150 m with its volume V remaining unchanged. If the variation of the density of air with height h from the ground is $ho(h) = ho_0 e^{-\frac{h}{h_0}}$, where $ho_0 = 1.25 	ext{ kg m}^{-3}$ and $h_0 = 6000 	ext{ m}$, the value of N is _________.

Sigma Percentile · Accepted Answer

### The Physics of Flotation

Imagine a hot air balloon floating peacefully in the sky. For it to maintain a constant altitude, it must be in a state of perfect mechanical equilibrium. This means the upward buoyant force exerted by the surrounding air must exactly balance the total downward weight of the balloon, its passengers, and its cargo.

According to Archimedes' principle, the buoyant force $F_B$ is equal to the weight of the displaced fluid. Mathematically, this is expressed as $F_B = V ho g$, where $V$ is the effective volume of the balloon, $ho$ is the density of the surrounding air, and $g$ is the acceleration due to gravity.

### The Initial Equilibrium

Initially, the balloon is floating at a height $h_1 = 100 	ext{ m}$. Let the density of the air at this height be $ho_1$. The total mass of the balloon is given as $480 	ext{ kg}$.

Setting up our force balance equation, we get:

$$V ho_1 g = 480 g$$

We can elegantly cancel out $g$ from both sides, leaving us with a simpler relationship:

$$V ho_1 = 480$$

### Shedding Weight to Rise

To ascend to a higher altitude, the balloon must shed some weight. The problem states that $N$ sandbags, each weighing $1 	ext{ kg}$, are thrown overboard. The new mass of the balloon becomes $(480 - N) 	ext{ kg}$.

The balloon rises and settles at a new equilibrium height $h_2 = 150 	ext{ m}$. At this higher altitude, the air is thinner, meaning the density has decreased to a new value, $ho_2$. The volume $V$ of the balloon is assumed to remain constant. Our new equilibrium equation becomes:

$$V ho_2 g = (480 - N) g$$

Again, canceling $g$, we get:

$$V ho_2 = 480 - N$$

### The Mathematical Bridge

We now have a system of two equations. To eliminate the unknown volume $V$, we can divide the final state equation by the initial state equation:

$$\frac{V ho_2}{V ho_1} = \frac{480 - N}{480}$$

This simplifies beautifully to:

$$\frac{ho_2}{ho_1} = 1 - \frac{N}{480}$$

### The Exponential Atmosphere

The problem provides a model for how air density varies with height: $ho(h) = ho_0 e^{-\frac{h}{h_0}}$. This is a classic exponential decay model, often used in atmospheric physics.

Let's substitute this function into our density ratio:

$$\frac{ho_2}{ho_1} = \frac{ho_0 e^{-\frac{h_2}{h_0}}}{ho_0 e^{-\frac{h_1}{h_0}}}$$

The base density $ho_0$ cancels out. Using the laws of exponents, we can combine the terms:

$$\frac{ho_2}{ho_1} = e^{-\frac{h_2 - h_1}{h_0}}$$

Now, we plug in the given values: $h_1 = 100 	ext{ m}$, $h_2 = 150 	ext{ m}$, and the scale height $h_0 = 6000 	ext{ m}$. The change in height is $h_2 - h_1 = 50 	ext{ m}$.

$$\frac{ho_2}{ho_1} = e^{-\frac{50}{6000}} = e^{-\frac{1}{120}}$$

### The Power of Approximation

Equating our two expressions for the density ratio, we have:

$$1 - \frac{N}{480} = e^{-\frac{1}{120}}$$

Here is where a crucial mathematical tool comes into play. The exponent $-\frac{1}{120}$ is a very small number. For values of $x$ close to zero, we can use the first-order Taylor series expansion: $e^{-x} \approx 1 - x$.

Applying this approximation, our equation transforms into:

$$1 - \frac{N}{480} \approx 1 - \frac{1}{120}$$

The $1$s on both sides cancel out, leaving a straightforward linear equation:

$$\frac{N}{480} = \frac{1}{120}$$

Solving for $N$, we multiply both sides by $480$:

$$N = \frac{480}{120} = 4$$

Thus, exactly **4 sandbags** were thrown out to allow the balloon to reach its new equilibrium height.

Enter Numerical Value:

Visualized Solution

$Equilibrium of the Balloon$

$Initial State (Height h_{1} = 100 m)$

$Final State (Height h_{2} = 150 m)$

$Ratio of Equations$

$Density Variation$

$Substituting Values$

$Approximation$

$Calculating N$

$Food for Thought$

The Sigma Insight: Buoyancy and Archimedes' Principle

The Physics of Flotation

The Initial Equilibrium

Shedding Weight to Rise

The Mathematical Bridge

The Exponential Atmosphere

The Power of Approximation

Similar Questions

A cubical block of side $0.5 m$ floats on water with $30%$ of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [Take, density of water = $1 0^{3} kg / m^{3}$ ]

Consider a solid sphere of radius $R$ and mass density $ρ (r) = ρ_{0} (1 - \frac{r ^{2}}{R ^{2}})$ , $0 < r \leq R$ . The min. density of a liquid in which it will float is

A hollow spherical shell at outer radius $R$ floats just submerged under the water surface. The inner radius of the shell is $r$ . If the specific gravity of the shell material is $\frac{27}{8}$ with respect to water, value of $r$ is

Comprehension Passage

The value of $X$ is _______.

The value of $Y$ is _______.

An air bubble of radius $1 cm$ in water has an upward acceleration $9.8 cm s^{- 2}$ . The density of water is $1 g/cm^{3}$ and water offers negligible drag force on the bubble. The mass of the bubble is (Take, $g = 980 cm/s^{2}$ )

A vessel contains oil (density $= 0.8 g/cm^{3}$ ) over mercury (density $= 13.6 g/cm^{3}$ ). A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of the sphere in $g/cm^{3}$ is

A thin uniform cylindrical shell, closed at both ends, is partially filled with water. It is floating vertically in water in half-submerged state. If $ρ_{c}$ is the relative density of the material of the shell with respect to water, then the correct statement is that the shell is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equilibrium of the Balloon

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial State (Height h1​=100 m)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final State (Height h2​=150 m)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Density Variation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Approximation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating N

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Food for Thought