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26 August, 21:00

A cylindrical conductor of length 'l' and area of cross section 'A' has a resistance 'R'. Another conductor of length 2.5l and resistance 0.5R of the same material has area of cross section?

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  1. 26 August, 21:06
    0
    R = (ρ x l) / A [Resistance = resisitivity times length divided by area of cross-section]

    0.5R = (ρ x 2.5l) / A'

    0.5R/2.5 = (ρ x l) / A'

    R/5 = (ρ x l) / A'

    R = 5 (ρ x l) / A'

    (ρ x l) / A = 5 (ρ x l) / A'

    A' = 1/5 A

    A' = 0.2 A
  2. 26 August, 21:09
    0
    Answer: A₂ = 5A

    Explanation:

    L1 = L

    A1 = A

    R1 = R

    R2 = 0.5R

    L₂ = 2.5L

    A2 = ?

    from resistance formula,

    R = ρL / A

    but since they are of the same material,

    ρ₁ = ρ₂ = ρ

    R₁ = ρL₁ / A₂ ... equation i

    R₂ = ρL₂ / A₂ ... equation ii

    dividing equation ii by i,

    R₂ / R₁ = (L₂ / A₂) / (L₁ / A₁) ... [ρ cancles out]

    R₂ / R₁ = (L₂ * A₁) / (A₂ * L₁)

    0.5R / R = (2.5L * A) / (A₂*L)

    0.5 = 2.5A / A₂

    A₂ = 2.5A / 0.5

    A₂ = 5A
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