Answer:
The resistance of the copper rod = 4.39×10⁻¹⁵ Ohms.
Explanation:
Resistance: This is the opposition to current flow in a circuit.
The S.I unit of resistance is Ohms.
R = Aσ/L ........................... Equation 1
Where σ = resistivity of the copper, R = resistance of the copper rod, A = cross sectional area of the copper, L = length of the copper rod.
Given:
σ = 1.756×10⁻⁸ ohm-meters, L = 2 meters.
But A = πd²/4
Where d = 8 mm = (8/1000) m = 0.0008 m, π = constant = 3.143
Therefore,
A = 3.143(0.0008)²/4
A = 0.0000005 m²
Substituting these values into equation 1,
R = 5×10⁻⁷×1.756×10⁻⁸/2
R = (8.78 ×10⁻¹⁵)/2
R = 4.39×10⁻¹⁵ Ohms.
Thus the resistance of the copper rod = 4.39×10⁻¹⁵ Ohms.
Resistance is directly proportional to the voltage. The resistance of the copper rod is [tex]4.39\times 10^{-15} \Omega[/tex].
Resistance:
It is defined as the degree of opposition to the current flow in a circuit. The S.I unit of resistance is Ohms.
[tex]R = \dfrac {A\sigma}{L}[/tex] ...........................1
Where,
σ = resistivity of the copper = 1.756×10⁻⁸ ohm-meters
[tex]R[/tex] = resistance of the copper rod
[tex]A[/tex] = cross sectional area of the copper
[tex]L[/tex]= length of the copper rod = 2 m
The cross-sectional area of the copper rod can be calculated by the formula,
[tex]A = \dfrac {\pi d^2} 4[/tex]
Where,
d = density = 8 mm = 0.0008 m,
So,
[tex]A = \dfrac {3.143\times (0.0008)^2}4\\\\A = 5 \times 10^7\rm \ m^2[/tex]
Put the values into equation 1,
[tex]R =\dfrac { 5\times 10^{-7}\times 1.756\times 10^{-8}}2\\\\R = \dfrac {8.78 \times 10^{-15}}2\\\\R = 4.39\times 10^{-15} \Omega[/tex]
Therefore, the resistance of the copper rod is [tex]4.39\times 10^{-15} \Omega[/tex].
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