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Question: From some instruments, current measured is I = 10.0 Amp., potential different measured is V = 100.0 ...

From some instruments, current measured is I = 10.0 Amp., potential different measured is V = 100.0 V, length of wire is 31.4 cm, and diameter of wire is 2.00 mm (all in correct significant figure). The resistivity of wire (in correct significant figure) will be: (Use π = 3.14)

A

1.00×10-4Ω-m

B

1.0×10-4Ω-m

C

1×10-4 Ω-m

D

1.000×10-4 Ω-m

Answer

1.00×10-4Ω-m

Explanation

Solution

The resistance of the wire is calculated using Ohm's law: R=VIR = \frac{V}{I}. Given V=100.0V = 100.0 V and I=10.0I = 10.0 Amp., R=100.0 V10.0 Amp=10.0 ΩR = \frac{100.0 \text{ V}}{10.0 \text{ Amp}} = 10.0 \text{ } \Omega.

The wire is cylindrical, so its cross-sectional area is A=πr2A = \pi r^2. The diameter d=2.00d = 2.00 mm, so the radius r=d2=2.00 mm2=1.00r = \frac{d}{2} = \frac{2.00 \text{ mm}}{2} = 1.00 mm. Converting radius to meters: r=1.00×10−3r = 1.00 \times 10^{-3} m. The cross-sectional area is: A=πr2=π(1.00×10−3 m)2=π×1.00×10−6 m2A = \pi r^2 = \pi (1.00 \times 10^{-3} \text{ m})^2 = \pi \times 1.00 \times 10^{-6} \text{ m}^2. Using π=3.14\pi = 3.14, A=3.14×1.00×10−6 m2A = 3.14 \times 1.00 \times 10^{-6} \text{ m}^2.

The length of the wire L=31.4L = 31.4 cm. Converting length to meters: L=31.4×10−2L = 31.4 \times 10^{-2} m =0.314= 0.314 m.

The resistivity (ρ\rho) is related to resistance (R), length (L), and area (A) by the formula: R=ρLAR = \rho \frac{L}{A} Rearranging for resistivity: ρ=R×AL\rho = \frac{R \times A}{L}

Substitute the values: ρ=10.0 Ω×(3.14×1.00×10−6 m2)0.314 m\rho = \frac{10.0 \text{ } \Omega \times (3.14 \times 1.00 \times 10^{-6} \text{ m}^2)}{0.314 \text{ m}} ρ=31.4×10−6 Ω⋅m20.314 m\rho = \frac{31.4 \times 10^{-6} \text{ } \Omega \cdot \text{m}^2}{0.314 \text{ m}} ρ=100×10−6 Ω⋅m\rho = 100 \times 10^{-6} \text{ } \Omega \cdot \text{m} ρ=1.00×10−4 Ω⋅m\rho = 1.00 \times 10^{-4} \text{ } \Omega \cdot \text{m}

Considering significant figures: I=10.0I = 10.0 A (3 s.f.) V=100.0V = 100.0 V (4 s.f.) R=V/I=100.0/10.0=10.0R = V/I = 100.0/10.0 = 10.0 Ω\Omega (3 s.f., limited by I) L=31.4L = 31.4 cm (3 s.f.) d=2.00d = 2.00 mm (3 s.f.)   ⟹  r=1.00\implies r = 1.00 mm (3 s.f.) A=πr2A = \pi r^2. Using π=3.14\pi = 3.14 (3 s.f.). A=3.14×(1.00×10−3)2=3.14×10−6A = 3.14 \times (1.00 \times 10^{-3})^2 = 3.14 \times 10^{-6} m2^2 (3 s.f.). ρ=R×AL=10.0×3.14×10−60.314\rho = \frac{R \times A}{L} = \frac{10.0 \times 3.14 \times 10^{-6}}{0.314}. ρ=31.4×10−60.314=100×10−6=1.00×10−4Ω⋅m\rho = \frac{31.4 \times 10^{-6}}{0.314} = 100 \times 10^{-6} = 1.00 \times 10^{-4} \Omega \cdot \text{m}. All calculations maintain 3 significant figures.