You\'ve made the finals of the Science Olympics! As one of your tasks, you\'re g

Question

You've made the finals of the Science Olympics! As one of your tasks, you're given 1.4 g of copper and asked to make a cylindrical wire, using all the metal, with a resistance of 1.0 Omega.

Part A: What length l will you choose for your wire? (answer in m)

Part B: What diameter d will you choose for your wire? (answer in mm)

Explanation / Answer

Given mass of copper = 1.4g we know, density = mass/volume density of copper = 8.96 g/cm3 ==> Volume of copper = 1.4/8.96 = 0.15625 cm3 We also know Resistance(R) of a wire is given by R = ?l/A where '?' is the resistivity of the material 'l' is the length of the wire 'A' is the area of cross-section of wire ==> 1 = 1.67 × 10-6 x (l/A) ==> (l/A) = 1/1.67 x 106 cm-1 ==> (l/A) = 0.595x 106 cm-1 ==> l = (0.595x 106 cm-1)A We also know Volume = length x area of cross-section = l x A = 0.595x 10^6 A2 ==> A = v(0.15625/(0.595x 106 )) ==> A = 5.124x 10^-4 cm2 ==> l =( 0.595x 106 cm-1 )( 5.124x 10-4 cm2) ==> l = 304.878 cm We also know Area of cylindrical wire = pd2/4 where 'd' is the diameter of wire ==> d = v(4A/p) = v(20.496*10^-4 /(p)) = 0.255 mm A) length of wire to be chosen = 304.878 cm B) Diameter of wire = .255mm

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