Question

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Answer 1

Answer:

CURRENT ELECTRICITY

Explanation:

★ RESISTIVITY ( RHO )

GIVEN ;

★ Length of Wire (L) :- 6 metre

★ Diameter :- 0.5 mm OR 5 × 10-⁴ m

★ [ Radius =0.25mm ]

★ Resistance (R) :- 50

★ Area ( A ) = π × r²

★ RESISTIVITY ( Rho ) = ??

★ By Using Formula ; ←

$r \: = rho \: \frac{l}{a}$

» Rho = ( Resistance × Area ) / length

» Rho = (50 × π × r² ) / 6

» Rho = [ 50 × π × ( 0.25 mm )² ] / 6

» Rho = [ 50 × π × ( 25 × 10^-5 )² ] / 6

» Rho = [ 50 × π × 625 × 10^-10 ] / 6

» Rho = [ (50 × 3.14 × 625 ) × 10^-10 ] / 6

» Rho = [ 98,125 × 10^-10 ] / 6

» Rho = [ 16,354 × 10^-10 ]

★» Rho = 1.63 μ Ω ★«

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★ CONDUCTIVITY ( C ) ★

$c \: = \frac{1}{rho}$

★ C = 1 / 1.63 = 0.613 Ω/m

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ANSWER :- HENCE ; FOR WIRE OF 6m and of Radius 0.25 mm ;

RESISTIVITY = 1.63 μ Ω .... And

CONDUCTIVITY = 0.613 Ω/m.