In an experiment, current measured is, `1 = 10.0 A`, potential difference measured is `V = 100.0 C`, length of the wire is `31.4 cm` and the diameter of the wire ` 2.00 mm` (all in correct significant figures). Find resistivity of the wire in correct significant figures. [Take `pi = 3.14`, exact]. (a) (b) (c) (d) .
Text Solution
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The correct Answer is:
A, D
`rho = (pid^2V)/(4lI) =((3.14)(2.00 xx 10^-3)(100.0))/((4)(31.4)(10.0)xx10^-2` `1.00 xx 10^-4 Omega - m` (to three significant figures).
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Knowledge Check
From some instruments current measured is I=10.0amp , potential different measured is V=100.0 V , length of wire is 31.4 cm , and diameter of wire is 2.00mm (all in correct significant figure). The resistivity of wire (in correct significant figures)will be (use pi=3.14 )
A
`1.00xx10^(-4)Omega-m`
B
`1.0xx10^(-4)Omega-m`
C
`1xx10^(-4)Omega-m`
D
`1.000xx10^(-4)Omega-m`
From ideal instruments, current measured is I = 10.0 amp, potential difference measured is V = 100.0 volt, length of wire is 31.4 cm , and diameter of wire is 2.00 mm , the resistivity of wire will be (in correct significant figure) (pi = 3.14)
A
`1.00 xx 10^(-4) Omega - m`
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`1.0 xx 10^(-4) Omega - m`
C
`1 xx 10^(-4) Omega - m`
D
`1.000 xx 10^(-4) Omega - m`
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