An aluminum wire of length 1.0 meter has a resistance of 9.0×10^-3 ohm. If the wire were cut into two equal lengths, each length would have a resistance of

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Question:

An aluminum wire of length 1.0 meter has a resistance of 9.0×10^-3 ohm. If the wire were cut into two equal lengths, each length would have a resistance of

Answer:

Each length of wire would have a resistance of 4.5 × 10⁻³ Ω.

Since resistance R = ρl/A where

  • ρ = conductivity of aluminum,
  • l = length and
  • A = cross-sectional area.

Since ρ and A are constant for the aluminum wire, R ∝ l

So. R₂/R₁ = l₂/l₁ where

  • R₁ = initial resistance of wire,
  • R₂ = resistance after wire is cut in two,
  • l₁ = initial length of wire = 1 m and,
  • l₂ = final length of wire = l₁/2

Now, since the length of the wire is cut into two, l₂ = l₁/2

So, R₂/R₁ = l₁/2 ÷ l₁

R₂/R₁ = 1/2

R₂ = R₁/2

R₂ = 9.0 × 10⁻³ Ω/2

R₂ = 4.5 × 10⁻³ Ω

So, each length of wire would have a resistance of 4.5 × 10⁻³ Ω.

Learn more about resistance here:

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