When a piece of aluminium wire of finite length is drawn through a series
of dies to reduce its diameter to half its original value, its resistance will
become (AIIMS 1997.MH-CET 2000,UPSEAT 2001.CBSE PMT 2
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=> 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 .
We know ,
- R= ρ l/A
where,
R is resistance of wire
ρ= resistivity of conductor
l= length of conductor
A= area of cross section
so,
=> R= p V/A²
which implies R is inversely proportional to A²
=> R is inversely proportional to d⁴
=> Rα 1/d⁴
Hence,
d reduces to d/2
and d⁴ reduces to d⁴/16 .
Therefore,
R=16R
Hence R increases 16 times more.
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