
A bird weight $2kg$ and is inside a cage of $1kg$ . If it starts flying then the weight of the bird and cage assembly is:
A) $1kg$
B) $2kg$
C) $3kg$
D) $4kg$
Answer
208.2k+ views
Hint: In order to solve this question we need to understand the laws of motion. Also, the concept that helps birds to fly is important. In order to attain stability and fly in the sky every being balances its weight so that it can survive the other forces present in the atmosphere and fly.
Complete step by step answer:
Here we are given that the weight of the bird is $2kg$ .
This bird is inside the cage whose weight is $1kg$ .
As we know that the bird pushes the air down in order to balance its weight and fly.
As the cage is at rest so no pseudo force will be acting on the bird hence weight of bird will be $2kg$ in air.
Hence, the weight of bird and cage assembly will be equal to $2kg$ and $1kg$ that is equal to $3kg$ .
The above expression can be written in the form of an equation as,
${W_{total}} = {W_{bird}} + {W_{cage}}$
Putting the values of all the quantities we have,
${W_{otal}} = 2kg + 1kg$
This gives us the weight of the total assembly as,
${W_{total}} = 3kg$
So, the correct answer is option C, $3kg$.
Note: Here it is important to note that when any bird flies, the upthrust acting on it is equal and opposite to the down thrust on the cage. Due to this reason only the weight of the assembly remains unchanged. The overall forces acting on a body flying in air should be balanced otherwise it would not be possible for it to fly in the air. Here in this question the force acting from upward balances the force acting on the body from down and hence the body is in a stable state.
Complete step by step answer:
Here we are given that the weight of the bird is $2kg$ .
This bird is inside the cage whose weight is $1kg$ .
As we know that the bird pushes the air down in order to balance its weight and fly.
As the cage is at rest so no pseudo force will be acting on the bird hence weight of bird will be $2kg$ in air.
Hence, the weight of bird and cage assembly will be equal to $2kg$ and $1kg$ that is equal to $3kg$ .
The above expression can be written in the form of an equation as,
${W_{total}} = {W_{bird}} + {W_{cage}}$
Putting the values of all the quantities we have,
${W_{otal}} = 2kg + 1kg$
This gives us the weight of the total assembly as,
${W_{total}} = 3kg$
So, the correct answer is option C, $3kg$.
Note: Here it is important to note that when any bird flies, the upthrust acting on it is equal and opposite to the down thrust on the cage. Due to this reason only the weight of the assembly remains unchanged. The overall forces acting on a body flying in air should be balanced otherwise it would not be possible for it to fly in the air. Here in this question the force acting from upward balances the force acting on the body from down and hence the body is in a stable state.
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