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ID: 3896539 • Letter: #
Question
%3Cp%3E%3Cspan%20class%3D%22c1%22%3EYour%20professor%20gives%20you%20a%20summer%20job%0Aorganizing%20his%20lab.%20On%20the%20first%20day%2C%20you%20come%20across%20three%0Aidentically%20shaped%20cylinders%20of%20metal%2C%20each%20having%20a%20diameter%20of%0A5.19%20mm%20and%20a%20length%20of%204.51%20cm.%20They%20appear%20to%20be%20made%20out%20of%0Adifferent%20metals%2C%20but%20they%20are%20not%20labeled!%20They%20are%2C%20however%2C%0Apackaged%20with%20a%20table%20of%20the%20conductivities%20of%20several%20metals.%20You%0Alocate%20a%20multimeter%20and%20use%20it%20to%20measure%20the%20electrical%20resistance%0Aof%20each%20cylinder.%20Using%20the%20measured%20resistances%20and%20the%20table%20of%0Aconductivities%2C%20identify%20the%20metal%20that%20each%20cylinder%20is%20composed%0Aof.%20%20%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3BSample%201%3A%26nbsp%3BR1%20%3D%2033.8%20%CE%BC%CE%A9%0A%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3BSample%202%3A%26nbsp%3BR2%20%3D%2060.9%20%CE%BC%CE%A9%0A%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3BSample%203%3A%26nbsp%3BR3%20%3D%201180%0A%CE%BC%CE%A9%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cimg%20class%3D%22user-upload%22%20src%3D%0A%22http%3A%2F%2Fmedia.cheggcdn.com%2Fmedia%252Fd22%252Fd22729ec-c2cb-4094-a138-cdb4cbe35bb6%252FphpurQo9c.png%22%0Aheight%3D%22275%22%20width%3D%22624%22%20%2F%3E%3C%2Fp%3E%0AExplanation / Answer
OK, the boring way is to use the formula conductivity is (all in meters) length divided by area and also divided by reisistance (in ohms). Conductivity of first sample is is 0.0451m/(pi*0.002595^2*0.0000338 Ohms) =6.30 x 10^7 or Silver
Switching in Ohms of second sample we get 0.0451m/(pi*0.002595^2*0.0000609 Ohms) = 3.50 x 10^7 or Aluminum
and then the third sample gives 0.0451m/(pi*0.002595^2*0.001180 Ohms) or Titanium
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