Resistance of a piece of metal

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Reducing both the length and diameter of a metal piece to half their original values results in a significant change in resistance. The resistance formula R = ρL/A indicates that resistance is directly proportional to length and inversely proportional to cross-sectional area. When the diameter is halved, the area decreases by a factor of four, while the length is halved, effectively doubling the resistance. Consequently, the new resistance is four times the original value. This calculation confirms that the resistance increases significantly due to the changes in dimensions.
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Homework Statement




How does the resistance of a piece of metal change if both its length and its diameter are reduced to one half their original values?
Question 1 answers
It is increased to 4 times the original value
It doubles
It remains unchanged
It is halved
It is reduced to 1/4 of the original value

Homework Equations



R = pL/A where P is the resistivity constant

The Attempt at a Solution



I am not sure how to relate the area to change in area
 
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05holtel said:

The Attempt at a Solution



I am not sure how to relate the area to change in area

I think you can assume it's a round wire.
 


they call one dimension a "diameter" , which becomes half
... how does the new Area compare to the old Area ?
(it's not about _change_ , it is about new = "some number" times old)
 


lightgrav said:
they call one dimension a "diameter" , which becomes half
... how does the new Area compare to the old Area ?
(it's not about _change_ , it is about new = "some number" times old)

Assume it's a round wire, with diameter d, the new diameter will be d/2.
old length is L, new length is 2L

Calculate old and new resistance, compare.
 

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