Question

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
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Answer 1

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.