SATHEE: Redox Reactions And Electrochemistry Ques 329

Redox Reactions And Electrochemistry Ques 329

Question: The resistivity of aluminum is $2.834 \times 10^{- 8} Ω m .$ 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 $= \frac{4.0}{2} = 2.0 m m = 2 \times 10^{- 3} m; l = 2 m$

Area of cross section and resistance (R) is related to the resistivity as, (Specific resistance), the length and area by $R = R e s i s t i v i t y \times \frac{L e n g t h}{A r e a}$ $= \frac{2.834 \times 10^{- 8} Ω m \times 2.00 m}{π {(2.0 \times 10^{- 3} m)}^{2}}$ $= 4.5 \times 10^{- 3} Ω$

Conductance $= \frac{1}{R} = \frac{1}{1.5 \times 10^{- 3}} = 222 Ω^{- 1} (S)$

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Redox Reactions And Electrochemistry Ques 329

Question: The resistivity of aluminum is 2.834×10−8Ωm. 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:

Answer:

Solution:

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Question: The resistivity of aluminum is $2.834 \times 10^{- 8} Ω m .$ 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)