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a wire of radius r has resistance R if is stretched to the wire of r/2 radius then find new resistance value

Let the initial resistance of the wire be R1 Area A1 Length L1

Given r2 = r1/2

Hence New Area( A2) becomes A1/4 ( Area is proportional to r2 )

thefore the new Length L2 = 4L1 (u may use the relationship A1L1=A2L2 to find out the new length )

resistance R1 is proportional to L1/A1----1

and R2is proportional to 4L1/ (A1/4) .ie, 16L1/A1-----2

on dividing 1/2 : we get the new resistance R2= 16R1

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Formula:- R1=n square of R2 Where R is resistance And n = n times of area, n times of length as u wish And r1=1/2 r2 So area will be A1=1/4A2 ( from πr^2) So R1 = 16 R2
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Two wire A and B are of same metal, have the same area of cross-section and have their lengths in the ratio 2:1. What will be the ratio of currents flowing through them when the same potential difference is applied across the length of each of them? 
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