Question

the radius of a conducting wire is doubled. what will be the ratio of its new specific resistance to the old one?

Answer 1

If specific resistance is P

P = RA/L

P = πr^2R/L

P1/P2 = (r1/r2)^2

P1/P2 = (r/(2r))^2

P1/P2 = 1/4

P2 = 4P1

P2/P1 = 4/1

Ratio of its new specific resistance to the old one is 4 : 1

Answer 2

If specific resistance is P

P = RA/L

P = πr^2R/L

P1/P2 = (r1/r2)^2

P1/P2 = (r/(2r))^2

P1/P2 = 1/4

P2 = 4P1

P2/P1 = 4/1

Ratio of its new specific resistance to the old one is 4 : 1

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