# You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrical wire, usin

Question

You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrical wire, using all the metal, with a resistance of 1.3 Ω. How long will your wire be? What will be its diameter? The resistivity of copper is 1.7 x 10-8 Ωm. The mass density of copper is 8.96 g/cm3.

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1 year 2021-09-04T06:25:21+00:00 2 Answers 0 views 0

1. Answer: length=8.55 meters, Diameter= 1.662×10-8meters

Explanation:

Resistivity= 1.7×10-8 ohm meter

Resistance= 1.3 ohm

Density= 8.96g/cm3 or 8960kgm-3

Mass= 1.0g or 1×10^-3kg

To obtain length= mass × resistance/ resistivity × density

= 1×10-3 × 1.3/1.7×10-8 × 8960

= 1.3×10-3/0.152×10-3

= 8.55 meters

Density = 4 × mass/ pi × Length × density

= 4 × 1×10-3/3.142 × 8.55× 8960

= 4×10-3/240702.336= 1.662×10-8 m

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have $$m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}$$ , where:

$$l$$ is the length

$$m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3$$

We divide the first equation by the second equation to get:

$$\frac{m}{R} = \frac{d A^2}{\rho}$$

$$A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2$$

Using this Area, we find the diameter of the wire:

$$D = \sqrt{\frac{4A}{\pi}}$$

$$D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}$$

$$D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm$$

To find the length, we multiply the two equations stated initially:

$$mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}$$

$$l^2 = 8.534\\l = 2.92 \ m$$