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)

From ideal instruments, current measured is I = 10.0 amp, potential di

1.	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)
Answer» `1.00 xx 10^(-4) Omega - m` `1.0 xx 10^(-4) Omega - m` `1 xx 10^(-4) Omega - m` `1.000 xx 10^(-4) Omega - m` Solution :`rho = (PI D^(2))/(4L) (V)/(i) = ((3.14)(2.00 xx 10^(-3))^(-2))/(4(0.314)) ((100.0)/(10.0))` `rho = 1.00 xx 10^(-4) Omega - m`

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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)

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