Respuesta :
Answer:
The length of the copper wire is 0.946 m.
Explanation:
Resistance ;
- Resistance of a wire is directly proportional to its length.
- Resistance of a wire is inversely proportional to the magnitude of its cross section area.
[tex]R\propto \frac{l}{A}[/tex]
[tex]\therefore R=\frac{\rho l}{A}[/tex]
[tex]\rho[/tex] is the resistivity of the wire, [tex]l[/tex] is the length of the wire and A is the cross section area of the wire.
The magnitude of resistivity of copper wire is 1.68×10⁻⁸Ωm.
The magnitude of resistivity of silver wire is 1.59×10⁻⁸Ωm.
The length of silver wire is 1.00 m
The resistance of the copper wire is
[tex]R_{Cu}=\frac{(1.68\times 10^{-8}\Omega m)\times l_{Cu} \ m}{A_{Cu}}[/tex]
The resistance of the silver wire is
[tex]R_{Ag}=\frac{(1.59\times 10^{-8}\Omega m)\times 1.00 \ m}{A_{Ag}}[/tex]
Given that,
[tex]R_{Cu}=R_{Ag}[/tex], [tex]A_{Cu}=A_{Ag}[/tex]
[tex]\therefore \frac{(1.68\times 10^{-8}\Omega m)\times l_{Cu}\ m}{A_{Cu}}=\frac{(1.59\times 10^{-8}\Omega m)\times1.00\ m}{A_{Ag}}[/tex]
[tex]\Rightarrow {(1.68\times 10^{-8}\Omega m)\times l_{Cu}\ m}={(1.59\times 10^{-8}\Omega m)\times 1.00 \ m}[/tex] [[tex]A_{Cu}=A_{Ag}[/tex]]
[tex]\Rightarrow l_{Cu}=\frac{(1.59\times 10^{-8}\Omega m)\times 1.00 \ m} {(1.68\times 10^{-8}\Omega m)}[/tex]
[tex]\Rightarrow l_{Cu}=0.946[/tex] m
The length of the copper wire is 0.946 m.
The length of the copper wire will be "0.946 m".
Resistance
According to the question,
Resistivity of copper wire's magnitude = 1.68 × 10⁻⁸ Ωm
Resistivity of silver wire's magnitude = 1.59 × 10⁻⁸ Ωm
Silver wire's length = 1.00 m
We know that,
[tex]R_{Cu} = R_{Ag}[/tex],
[tex]A_{Cu} = A_{Ag}[/tex]
As we know the relation,
→ R ∝ [tex]\frac{l}{A}[/tex]
∴ R = [tex]\frac{\rho l}{A}[/tex]
here, ρ = Resistivity,
l = length, and
A = Cross section area
Now,
The copper wire's resistance will be:
R = [tex]\frac{1.68\times 10^{-8}\times l_{Cu}}{A_{Cu}}[/tex]
The silver wire's resistance will be:
R = [tex]\frac{1.59\times 10^{-8}\times 1.00}{A_{Ag}}[/tex]
According to the given information,
→ [tex]\frac{1.68\times 10^{-8}\times l_{Cu}}{A_{Cu}} = \frac{1.59\times 10^{-8}\times 1.00}{A_{Ag}}[/tex]
By applying cross-multiplication, we get
1.68 × 10⁻⁸ × [tex]l_{Cu}[/tex] = 1.59 × 10⁻⁸ × 1.00
[tex]l_{Cu}[/tex] = [tex]\frac{1.59\times 10^{-8}\times 1.00}{1.68\times 10^{-8}}[/tex]
= 0.946 m
Thus the above response is correct.
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