An aluminum wire of definite length stretched uniformaly sowat it's diameter becomes bait of its mitin diameter. Then what is it's new resistance (assume volume of metai remains constanty: (A) Double (B) Pour times (C) Eight times (D) Sixteen times
Please give me the correct answer with explanation
Then I will mark ur answer as a brainliest
Answers
Answered by
2
When a piece of aluminium wire of finite length is drawn through series of dies to reduce it's diameter to half it's original value, it's resistance will become sixteen times
Explanation:
R= ρ l/A
where R is resistance of wire
ρ= resistivity of conductor
l= length of conductor
A= area of cross section
R= ρ l xA/AxA
R= p V/A²
which implies R is inversely proportional to A²
R is inversely proportional to d⁴
Rα 1/d⁴
where d is the diameter of area of cross sectio n
if d reduces to d/2
then d⁴ reduces to d⁴/16
R=16R
Hence R increases 16 times more.
Similar questions