1.

A wire has a resistance of 32 Ω. It is melted and drawn into a wire of half of its original length. Calculate the resistance of the new wire. What is the percentage change in resistance?

Answer»

Given,

R1= 32Ω,l1= l ,l2= l/2 ,,R2= ?

Now when the wire is melted its volume will remain unchanged.

Volume of cylindrical wire = length x Area of cross section of wire

so,

l1A1= l2A2

or,lA1= (l/2)A2

or, A2= 2A1..........................(1)

Now,resistance R = ρl/A

so, R1= ρl1/A1

R2=ρl2/A2

or, R1/R2= (ρl1/A1)/(ρl2/A2) = l1A2/l2A1

so from (1),

R1/R2=(l x2A1)/(l/2)A1

=4

or, R2=32/4 = 8Ω


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