A wire of 5 Ohm resistance is stretched to twice its length.What will be its new resistance?

Question

will its diameter become half or its area

Answer 1

Assuming that the wire's cross-sectional area(CSA) remains uniform throughout its length even after stretching it, the CSA will become half its original value. Since Resistance of the wire is inversely proportional to its CSA and is directly proportional to its length, the wire's Resistance will become 4 times its original value after stretching. Hence, the answer is 5*4=20 ohms.

Answer 2

20 Ohm

Answer 3

answer is 20 ohms.

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Answer 4

Resistance = resistivity * Length / cross-sectional area

Let L be the length, d be the diameter, then volume of the wire = pi*d^2*L/4

Now when the wire is stretched to twice of its length, L' =2L, but volume will remain the same

So new diameter d ' = ??

pi*d^2*L/4 = pi* d'^2 * 2L /4

=> d'^2 = d^2/2

=> d' = d /sqrt 2

Now R = ro * L/A => ro = RA/L

So new R' = ro * L'/A' = RA/L *(L'/A')

now L' = 2L,

d' = d /sqrt 2

So A' = pi*d'^2 / 4 =  pi* (d^2 /2 )/4 = (pi*d^2/4)/2 = A/2

So R' = 5 * A/L * 2L /(A/2) = 20 ohms

Answer 5

Previous answers have assumed that volume remains the same. But it is not necessary. In general, volume changes. So, we cannot say that area will become half. How much the area changes will depend on Poisson's ratio, which is not given.

Answer 6

20 ohms. Only the area will be halved, assuming it remains uniform.

Answer 7

20 ohms