Redox Reactions And Electrochemistry Question 175
Question: The resistivity of aluminum is $ 2.834\times {10^{-8}}\Omega m. $ Thus, conductance across a piece of aluminum wire, that is 4.0 mm in diameter and 2.00 m long is (assume current=1.25 A)
Options:
A) 111.0 S
B) 1.11 S
C) 222.05 S
D) 1111 S.
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Radius $ =\frac{4.0}{2}=2.0mm=2\times {10^{-3}}m;l=2m $
Area of cross section and resistance (R) is related to the resistivity as, (Specific resistance), the length and area by $ R=Resistivity\times \frac{Length}{Area} $ $ =\frac{2.834\times {10^{-8}}\Omega m\times 2.00m}{\pi {{(2.0\times {10^{-3}}m)}^{2}}} $ $ =4.5\times {10^{-3}}\Omega $
Conductance $ =\frac{1}{R}=\frac{1}{1.5\times {10^{-3}}}=222{{\Omega }^{-1}}(S) $
BETA
