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Calculate the conductance of a conduit the cross-sectional area of which is 3.0 cm2 and the length of which is 9.0 cm, given that its conductivity is 0.80 ohm-1cm-1.


Sagot :

for resistance we have R=ρ l/a
 thus for conductance we have K=σ a/l
conductance,K=1/R
conductivity,σ =1/ρ

σ = .80 Ω-1 cm-1
l =9 cm
a = 3 cm²
K=.80 ×3/9
  =0.26 Ω-1


Answer: [tex]0.266 ohm^{-1}[/tex]

Explanation: Specific conductance or conductivity is the conductance for unit volume of a solution.

[tex]\kappa=C\times \frac{l}{a}[/tex]

where,

[tex]\kappa[/tex] =Specific conductance  or Conductivity=[tex]0.80ohm^{-1}cm^{-1}[/tex]

C = Conductance = ?

l = length= 9.0 cm

a = area = 3.0[tex]cm^2[/tex]

[tex]\frac{length}{area}[/tex]= cell constant =[tex]\frac{9.0cm}{3.0cm^2}=3cm^{-1}[/tex]

[tex]0.80ohm^{-1}cm^{-1}=C\times 3cm^{-1}[/tex]

[tex]C=0.266 ohm^{-1}[/tex]


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