The resistance of a certain length of wire having a diameter of 6 mm is 5 ohm. The wire is drawn such that the diameter becomes 3mm. The new resistance will be.
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The new resistance will be 80 ohms.
Explanation:
Let A is the original area, A’ is the new area, l is the original length, and l’ is the new length.
A = πr^2 = πd^2/4
A’ = π(d/2)^2/4 = πd^2/16
Now,
Original volume = V = (πd^2/4)l
New volume = V’ = (πd^2/16)l’
As, we know that by changing the diameter and length, volume of the conducting wire remains constant so,
(πd^2/4)l = (πd^2/16)l’
l = l’/4
l’ = 4l
similarly,
A’ = A/4
Now,
R1/ R2 = (A’/A)(l/l’)
R1/ R2 = (1/4) (1/4)
R2 = 16 * 5 = 80 ohm
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