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if an iron wire is drawn to decrease it's radius by 0.5%.then percentage increase in its resistance is


dadi srinivas 26th Mar, 2020
Answer (1)
KUNAL LANJEWAR 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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