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A bird of mass 1.23 kg is able to hover by imparting a downward velocity 10 m/s uniformly to air of density r kg/m3 over an effective area 0.1 m2. If the acceleration due to gravity is 10 m/ s2, then the magnitude of r in kg/m3 is
- a)0.0123
- b)0.123
- c)1.23
- d)1.32
Correct answer is option 'C'. Can you explain this answer?
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A bird of mass 1.23 kg is able to hover by imparting a downward veloc...
To solve this problem, we can use the principle of conservation of momentum. This principle states that the total momentum before an event is equal to the total momentum after the event, assuming no external forces are acting on the system.
Let's consider the bird as our system. Initially, the bird is at rest, so the total momentum is zero. When the bird imparts a downward velocity to the air, it creates an upward force on itself due to the reaction force from the air. This force allows the bird to hover.
Let's denote the upward force as F and the mass of the bird as m. The momentum imparted to the air is given by the product of the mass of the air (ρA, where ρ is the density of air and A is the effective area) and the velocity imparted to the air (v). Therefore, the momentum imparted to the air is ρAv.
According to the conservation of momentum, the total momentum after the event is also zero. Since the bird is hovering, the only force acting on it is its weight (mg), acting downwards. Therefore, the upward force (F) must be equal in magnitude to the weight of the bird (mg), but in the opposite direction.
Therefore, F = mg.
Now, equating the momentum imparted to the air to the momentum of the bird:
ρAv = mg
We are given the mass of the bird (m = 1.23 kg), the downward velocity imparted to the air (v = 10 m/s), the effective area (A = 0.1 m^2), and the acceleration due to gravity (g = 10 m/s^2).
Substituting these values into the equation, we get:
ρ(0.1)(10) = (1.23)(10)
Simplifying the equation further:
ρ = (1.23)(10) / (0.1)(10)
ρ = 12.3 / 1 = 12.3 kg/m^3
Therefore, the magnitude of ρ is 12.3 kg/m^3.
The correct answer is option C: 1.23.
Let's consider the bird as our system. Initially, the bird is at rest, so the total momentum is zero. When the bird imparts a downward velocity to the air, it creates an upward force on itself due to the reaction force from the air. This force allows the bird to hover.
Let's denote the upward force as F and the mass of the bird as m. The momentum imparted to the air is given by the product of the mass of the air (ρA, where ρ is the density of air and A is the effective area) and the velocity imparted to the air (v). Therefore, the momentum imparted to the air is ρAv.
According to the conservation of momentum, the total momentum after the event is also zero. Since the bird is hovering, the only force acting on it is its weight (mg), acting downwards. Therefore, the upward force (F) must be equal in magnitude to the weight of the bird (mg), but in the opposite direction.
Therefore, F = mg.
Now, equating the momentum imparted to the air to the momentum of the bird:
ρAv = mg
We are given the mass of the bird (m = 1.23 kg), the downward velocity imparted to the air (v = 10 m/s), the effective area (A = 0.1 m^2), and the acceleration due to gravity (g = 10 m/s^2).
Substituting these values into the equation, we get:
ρ(0.1)(10) = (1.23)(10)
Simplifying the equation further:
ρ = (1.23)(10) / (0.1)(10)
ρ = 12.3 / 1 = 12.3 kg/m^3
Therefore, the magnitude of ρ is 12.3 kg/m^3.
The correct answer is option C: 1.23.
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A bird of mass 1.23 kg is able to hover by imparting a downward velocity 10 m/s uniformly to air of density r kg/m3 over an effective area 0.1 m2. If the acceleration due to gravity is 10 m/ s2, then the magnitude of r in kg/m3 isa)0.0123b)0.123c)1.23d)1.32Correct answer is option 'C'. Can you explain this answer?
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