Birds do not feel electric shock while sitting on current carrying insulated wires because …….
A. The feathers of birds act as a insulator and hence the current does not pass through them
B. The resistance offered by the body of birds is very high
C. The current does not pass through the body as the claws of birds are non-conducting
D. The potential difference between the two claws of the birds is very small
Answer
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Hint: The current flows in the circuit when there is a potential difference between the two points. For the current to flow through the bird, there must be a potential difference across the claws of the bird. Find out the magnitude of the potential difference across the claws of the bird.
Complete answer:
We know that almost every living thing shows a small conducting property. The birds are not exception. The claws too are conducting. We know that the current flows through the circuit when there is a potential difference between the two points. The current flows across the resistance because there is a small voltage drop across the resistor and hence it creates the potential difference.
We also know that the current only flows in the closed circuit. Therefore, the current will flow from one of the claws, pass through the body and will merge at the junction of the other claw and wire. Now, for the current to flow through the bird, there must be a potential difference across the claws of the bird. We know that the electric lines carry a constant voltage supply across the power lines. Therefore, the potential difference across the claws will be negligible or zero. Thus, the current will not flow through the bird and hence they will not be electrocuted.
The option A cannot be true because even if the feathers are non-conducting, the whole body of the bird is conducting. The option B cannot be true because the resistance offered by any living body is not high. The claws too are conducting. Therefore, the option C is also incorrect.
So, the only correct answer is option D.
Note: If the feathers of the birds are very long and the two feathers are suddenly touching the two points on the power line, there will be a small potential difference across the feathers. In this case, the bird will be electrocuted in no time. Also, in other cases, if one of the claws of the bird is touching the tree and another claw is on the power line, the path of the current will be through the bird to the ground through the tree and hence the bird will be electrocuted.
Complete answer:
We know that almost every living thing shows a small conducting property. The birds are not exception. The claws too are conducting. We know that the current flows through the circuit when there is a potential difference between the two points. The current flows across the resistance because there is a small voltage drop across the resistor and hence it creates the potential difference.
We also know that the current only flows in the closed circuit. Therefore, the current will flow from one of the claws, pass through the body and will merge at the junction of the other claw and wire. Now, for the current to flow through the bird, there must be a potential difference across the claws of the bird. We know that the electric lines carry a constant voltage supply across the power lines. Therefore, the potential difference across the claws will be negligible or zero. Thus, the current will not flow through the bird and hence they will not be electrocuted.
The option A cannot be true because even if the feathers are non-conducting, the whole body of the bird is conducting. The option B cannot be true because the resistance offered by any living body is not high. The claws too are conducting. Therefore, the option C is also incorrect.
So, the only correct answer is option D.
Note: If the feathers of the birds are very long and the two feathers are suddenly touching the two points on the power line, there will be a small potential difference across the feathers. In this case, the bird will be electrocuted in no time. Also, in other cases, if one of the claws of the bird is touching the tree and another claw is on the power line, the path of the current will be through the bird to the ground through the tree and hence the bird will be electrocuted.
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