The radius of conducting wire is doubled. What will be the ratio of new specific resistance to the old one

Dear Student

r1 = R
r2 = 2R

​Specific resistance, ρ = RAL
where, R is resistance, A is cross-sectional area and L is length

So, ρR (πr2)L
Or, ρ1ρ2 = (r1r2)2            ​ρ1ρ2 = ​(R2R)2              ​ρ1ρ2 = 14

So, the ratio is 1:4

Regards

 

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1/4 will the ratio
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