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A metal wire of length 79 cm and diameter 0.31 mm has a resistance of 5.0 Ohm. Find the resistivity of the metal wire.

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3 parts
a

Resistivity ρ=4.8×10−7Ω⋅m\text{Resistivity } \rho = 4.8 \times 10^{-7} \Omega \cdot \text{m}

b

Resistance R=48Ω\text{Resistance } R = 48 \Omega

c

Length L=0.047 m\text{Length } L = 0.047 \text{ m}

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Part a· 3 steps

A metal wire of length 79 cm and diameter 0.31 mm has a resistance of 5.0 Ω\Omega. Find the resistivity of the metal wire.

  1. Convert given values to SI units.

    Length L=79 cm=0.79 mL = 79 \text{ cm} = 0.79 \text{ m}
    Diameter d=0.31 mm=0.31×10−3 md = 0.31 \text{ mm} = 0.31 \times 10^{-3} \text{ m}
    Resistance R=5.0ΩR = 5.0 \Omega

  2. Calculate the cross-sectional area AA of the wire.

    The formula for the area of a circle is A=πd24A = \frac{\pi d^2}{4}.

    A=π(0.31×10−3 m)24A = \frac{\pi (0.31 \times 10^{-3} \text{ m})^2}{4} A=π(9.61×10−8 m2)4A = \frac{\pi (9.61 \times 10^{-8} \text{ m}^2)}{4} A≈7.5498×10−8 m2A \approx 7.5498 \times 10^{-8} \text{ m}^2

  3. Use the resistance formula R=ρLAR = \rho \frac{L}{A} to find resistivity ρ\rho.

    Rearranging the formula for ρ\rho:

    ρ=R⋅AL\rho = \frac{R \cdot A}{L}

    Substitute the values:

    ρ=(5.0Ω)×(7.5498×10−8 m2)0.79 m\rho = \frac{(5.0 \Omega) \times (7.5498 \times 10^{-8} \text{ m}^2)}{0.79 \text{ m}} ρ=3.7749×10−7Ω⋅m0.79\rho = \frac{3.7749 \times 10^{-7} \Omega \cdot \text{m}}{0.79} ρ≈4.778×10−7Ω⋅m\rho \approx 4.778 \times 10^{-7} \Omega \cdot \text{m}

    Rounding to two significant figures:

    Resistivity ρ=4.8×10−7Ω⋅m\boxed{\text{Resistivity } \rho = 4.8 \times 10^{-7} \Omega \cdot \text{m}}

    :

Part b

Resistance R=48Ω\text{Resistance } R = 48 \Omega

Determine the resistance of an aluminium cable of length 2.0×104 m2.0 \times 10^4 \text{ m} and diameter 4.0 mm4.0 \text{ mm}. The resistivity of aluminium is 3.0×10−8Ω⋅m3.0 \times 10^{-8} \Omega \cdot \text{m}.

  1. Convert given values to SI units.

    Length L=2.0×104 mL = 2.0 \times 10^4 \text{ m}
    Diameter d=4.0 mm=4.0×10−3 md = 4.0 \text{ mm} = 4.0 \times 10^{-3} \text{ m}
    Resistivity ρ=3.0×10−8Ω⋅m\rho = 3.0 \times 10^{-8} \Omega \cdot \text{m}

  2. Calculate the cross-sectional area AA of the cable.

    A=πd24A = \frac{\pi d^2}{4} A=π(4.0×10−3 m)24A = \frac{\pi (4.0 \times 10^{-3} \text{ m})^2}{4} A=π(16.0×10−6 m2)4A = \frac{\pi (16.0 \times 10^{-6} \text{ m}^2)}{4} A=π×4.0×10−6 m2A = \pi \times 4.0 \times 10^{-6} \text{ m}^2 A≈1.2566×10−5 m2A \approx 1.2566 \times 10^{-5} \text{ m}^2

  3. Use the resistance formula R=ρLAR = \rho \frac{L}{A}.

    Substitute the values:

    R=(3.0×10−8Ω⋅m)×2.0×104 m1.2566×10−5 m2R = (3.0 \times 10^{-8} \Omega \cdot \text{m}) \times \frac{2.0 \times 10^4 \text{ m}}{1.2566 \times 10^{-5} \text{ m}^2} R=6.0×10−4Ω⋅m21.2566×10−5 m2R = \frac{6.0 \times 10^{-4} \Omega \cdot \text{m}^2}{1.2566 \times 10^{-5} \text{ m}^2} R≈47.74ΩR \approx 47.74 \Omega

    Rounding to two significant figures:

    Resistance R=48Ω\boxed{\text{Resistance } R = 48 \Omega}

    :

Part c

Length L=0.047 m\text{Length } L = 0.047 \text{ m}

A manganin wire of cross-sectional area 4.0×10−7 m24.0 \times 10^{-7} \text{ m}^2 and resistivity 4.3×10−7Ω⋅m4.3 \times 10^{-7} \Omega \cdot \text{m} has a resistance of 0.05Ω0.05 \Omega. Calculate the length of the wire.

  1. Identify the given values.

    Cross-sectional area A=4.0×10−7 m2A = 4.0 \times 10^{-7} \text{ m}^2
    Resistivity ρ=4.3×10−7Ω⋅m\rho = 4.3 \times 10^{-7} \Omega \cdot \text{m}
    Resistance R=0.05ΩR = 0.05 \Omega

  2. Use the resistance formula R=ρLAR = \rho \frac{L}{A} to find length LL.

    Rearranging the formula for LL:

    L=R⋅AρL = \frac{R \cdot A}{\rho}

    Substitute the values:

    L=(0.05Ω)×(4.0×10−7 m2)4.3×10−7Ω⋅mL = \frac{(0.05 \Omega) \times (4.0 \times 10^{-7} \text{ m}^2)}{4.3 \times 10^{-7} \Omega \cdot \text{m}} L=0.2×10−7Ω⋅m24.3×10−7Ω⋅mL = \frac{0.2 \times 10^{-7} \Omega \cdot \text{m}^2}{4.3 \times 10^{-7} \Omega \cdot \text{m}} L=0.24.3 mL = \frac{0.2}{4.3} \text{ m} L≈0.04651 mL \approx 0.04651 \text{ m}

    Rounding to two significant figures:

    Length L=0.047 m\boxed{\text{Length } L = 0.047 \text{ m}}
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