An expansible ballon filled with air floats on the surface of a lake with 2/3 of its volume submerged .How deep must it be sunk in the water so that it is just inequilibrium neither sinking further nor rising? It is assumed that the temperature of the water is constant and that the height of water barometer is 9 meters.
To find the depth at which the balloon is in equilibrium, we need to consider the forces acting on the balloon.
The buoyant force exerted on an object immersed in a fluid is given by Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object.
In this case, the balloon is filled with air, and the portion that is submerged displaces water. The buoyant force on the balloon is given by the weight of the water displaced.
Let's assume the volume of the balloon is V. Since 2/3 of its volume is submerged, the volume of water displaced is (2/3)V. The weight of the water displaced is equal to the weight of the column of water above the balloon.
The pressure exerted by a column of water depends on the height of the column and the density of water. The pressure at the surface of the lake is atmospheric pressure, which we can assume is equal to the height of the water barometer, 9 meters.
So, the pressure difference between the top and bottom of the balloon (due to the water column) is 9 meters of water.
Using the formula P = ρgh, where P is pressure, ρ is the density of water, g is the acceleration due to gravity, and h is the height of the water column, we can calculate the pressure difference.
Assuming the density of water is approximately 1000 kg/m³ and the acceleration due to gravity is approximately 9.8 m/s², the pressure difference is:
ΔP = ρgh = 1000 kg/m³ * 9.8 m/s² * 9 m = 88200 Pa
Now, the buoyant force acting on the balloon is equal to the weight of the water displaced, which is given by the formula F = ρVg, where F is the force, ρ is the density of water, V is the volume of water displaced, and g is the acceleration due to gravity.
The weight of the balloon itself is balanced by the upward buoyant force when the balloon is fully submerged. Therefore, for the balloon to be in equilibrium and neither sinking nor rising, the buoyant force must be equal to the weight of the balloon.
Let's assume the mass of the balloon is m. The weight of the balloon is given by the formula W = mg, where W is the weight and g is the acceleration due to gravity.
Setting the buoyant force equal to the weight of the balloon:
F = mg
ρVg = mg
V = m/ρ
Now we can substitute this V into the equation for the buoyant force:
F = ρVg
F = (m/ρ) * ρ * g
F = mg
Therefore, the depth at which the balloon is in equilibrium is directly proportional to the mass of the balloon.
There is not enough information given in the problem to determine the mass of the balloon or the depth at which it will be in equilibrium.
To find the depth at which the balloon will be in equilibrium, we need to consider the buoyant force acting on the balloon and the weight of the balloon.
Let's denote the density of air as ρ_air and the density of water as ρ_water.
Given that 2/3 of the volume of the balloon is submerged, it means that the remaining 1/3 of the volume is above the water's surface.
Step 1: Determine the relationship between densities of air and water.
Since the balloon is floating, the densities of air and water are related by the following equation:
ρ_water × V_submerged = ρ_air × V_balloon
where V_submerged is the volume of the balloon submerged in water, and V_balloon is the total volume of the balloon.
Step 2: Calculate the volume of the balloon submerged.
Given that 2/3 of the volume of the balloon is submerged, we have:
V_submerged = (2/3) × V_balloon
Step 3: Determine the density of air.
The density of air can be calculated using the Ideal Gas Law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since the temperature of the water is constant, we can assume the temperature inside the balloon is the same. Therefore, the density of air can be taken as the same as the density of the outside air.
Step 4: Calculate the weight of the balloon.
The weight of the balloon can be calculated using:
Weight = ρ_air × g × V_balloon
where g is the acceleration due to gravity.
Step 5: Calculate the buoyant force.
The buoyant force is given by:
Buoyant force = ρ_water × g × V_submerged
Step 6: Set up equilibrium condition.
To be in equilibrium, the buoyant force acting on the balloon must be equal to the weight of the balloon. So, we have:
Buoyant force = Weight
Step 7: Substitute the values and solve the equation.
Substituting the values from the given information and equations, we can solve for the depth at which the balloon is in equilibrium.
For example, let's say the volume of the balloon (V_balloon) is 0.1 m^3 and the density of water (ρ_water) is 1000 kg/m^3. We can assume the density of air (ρ_air) to be approximately 1.2 kg/m^3.
V_submerged = (2/3) × V_balloon = (2/3) × 0.1 = 0.0667 m^3
Weight = ρ_air × g × V_balloon = 1.2 × 9.8 × 0.1 = 1.176 N
Buoyant force = ρ_water × g × V_submerged = 1000 × 9.8 × 0.0667 = 652.86 N
Setting up equilibrium: Buoyant force = Weight
652.86 = 1.176
Simplifying the equation, we find:
652.86 = 1.176
652.86 - 1.176 = 0
651.684 = 0
There seems to be an error in the calculations, as it appears that there is no depth at which the balloon can be in equilibrium. Double-check the given information and calculations to ensure accuracy.