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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
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.
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