if an iron wire is drawn to decrease it's radius by 0.5%.then percentage increase in its resistance is
Answer (1)
26th Mar, 2020
Hello,
We know that, R= ρ(L/A)
Resistance, R is inversely proportional to Area A.
Let, R1 be the initial Resistance and R2 be the final resistance.
A1 be the initial area and A2 be the final area.
So, R1 / R2 = A2 / A1
We know that, for a wire, A = π x r^2
Hence, Area is directly proportional to radius square.
So, R1 / R2 = ( r2 / r1 ) ^ 2
Radius is decreased by 0.5 %.
So, r2 = r1 - 0.5% of r1
So, r2 = 0.995 r1
Hence, R1 / R2 = ( 0.995 r1 / r1 ) ^ 2
So, R1 / R2 = 0.99
So, R1 = 0.99 R2
Hence, Percent increase in resistance = ( 100 - 99 )%
So, Percent Increase in resistance = 1%
Best Wishes.
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