# A wire is stretched so that its length becomes 6/5 times its original length. If its original resistance is 25 ohm find its new resistance and resistivity.

2
by garima9cr3

• Brainly User
2014-11-04T20:07:23+05:30
Resistance becomes n² times of length is made n times its original length then answer would be (6/5)²25=36Ω
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Yep
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happy to get help...
2014-11-07T06:02:44+05:30

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Resistance of a conductor of length L and an area of uniform cross section is
R = ρ L / A ,              ρ = resistivity of the material
given     R = 25 Ω

Volume of the material in the conductor = L A.
Let us assume that the density of the material remains the same.  Hence the volume Vol of the conductor remains the same.

Vol = A L = A' L'        => A' = A L / L'

After stretching the wire :    L' = 6 L / 5

A' = 5 A L / 6 L = 5 A / 6

Resistance of stretched wire =  R' = ρ L' / A' = ρ  ( 6 L /5) *  [ 6 / (5 A) ]

R' = 36 ρ L / (25 A)  = 36 R / 25

New resistance = 36 * 25 / 25 Ω = 36 Ω