Question

if the radius of cross section of the conductor is increased by 0.1% keeping volume constant then percentage change in the resistance of the conductor is​

Answer 1

Answer:

#BAL Hi, Here is the answer Resistance = ρ * L /A Area increased by a factor = 1.001² = 1.002001 Length gets reduced by the same factor as the volume is constant = 1.002001 Resistance = ρ * L /A = ρ * (L/1.001002) / A * 1.001002 = [ρ * L /A ] * 1/1.002001² = Original Resistance 'R' * 0.996009 ≈ 0.996 * R The original resistance gets reduced by 0.4 %

Explanation:

Answer 2

The radius of cross section of the conductor is increased by 0.1 % keeping volume constant then the percentage change in the resistance of conductor is 0.4%

- As we know that the formula of resistance = Resistivity× Length/Area.

- Resistance is directly proportional to length and inversely proportional to area and the resistivity depends upon the nature of the material as mentioned in the question the volume of the wire is constant so the resistivity will be constant .

- Here, as per the question there is increase in the radius of the wire by 0.1%

( A ) = π r 2

ΔA/A = 2(Δr/r)

ΔA/A = 2(0.1)%

So, area is increased by 0.2%

- Now, As the volume remain constant there will be decrease in the length of the wire.

- So, the percentage decrease in the length must be 0.2%

- Now, According to the formula of resistance,

R = (ρ)×L/A

ΔR/R = ΔL/L + ΔA/A

So, we have

ΔR/R = -0.2 - 0.2

ΔR/R = - 0.4%

- Length is the percentage decrease so it has negative sign and area on coming up becomes negative from positive sign.