What is the resistance of a wire if its diameter is doubled?

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Kevin J
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This was a MCQ question on one of my Cambridge IGCSE Exams, the answer key said that the resistance would decrease by 16 folds (1/16 less then the previous one), I don't know how they got 16 from? (Apparently, I answered 1/4 and it was wrong)
 
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If DC resistance is directly proportional to crossectional area, then 1/4 sounds right.
πR∧2 VERS. π(2R)∧2 (Assuming the length is constant.)
 
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Here's the question, the mark scheme says it's D, how do you get D?
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Chestermiller said:
What happens to the length?
It doesn't say anything, it was a question from CAMBRIDGE IGCSE
 
It says the new cylinder is the same volume as the old one.
 
ChemAir said:
It says the new cylinder is the same volume as the old one.
I have a feeling that the length decreases right?
 
The volume of a cylinder is $$V=\pi\frac{D^2}{4}L$$If D doubles and V remains constant what does L have to do?
 
Kevin J said:
Here's the question, the mark scheme says it's D, how do you get D? View attachment 233716
Kevin J said:
It doesn't say anything, it was a question from CAMBRIDGE IGCSE
Kevin J said:
I have a feeling that the length decreases right?
I does in fact mention the length. It states the volume of the putty remains constant as the diameter is increased by twice. For that to be true, the length must also change. Remember that resistance is approximated from resistivity and is inversely proportional to the cross-sectional area of the putty while also being proportional to the putty's length. If the length of the putty decreases while the area increases, the resistance will decrease by the seemingly "too large" factor shown above. Hope this helps.

The easiest way to handle these problems is to substitute in for some numbers and see what happens when you change the parameters.
 
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