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A wire of 5Ω resistance is drawn out so that its new length is two times the origi...


Question

A wire of 5Ω resistance is drawn out so that its new length is two times the original length. If the resistivity of the wire remains the same and the cross-sectional area is halved, the new resistance is

Options

A) 40Ω

B) 20Ω

C) 10Ω

D) 5Ω


The correct answer is B.

Explanation:

Let the original length = L1
Let the original resistance = R1 = 5Ω
Let the original resistivity = P1
Let the original area = a1
Let the new length = L2 = 2L1

let the new area = a2 = 1/(2a2)
Let the new resistance = R2
Let the new resistivity = P2
But since the resistivity remains the same,
=> P1 = P2

∴ P1 =R1 a1
_L1
= P2 =R2 a2
_L2

But a2 = 1/2a1; and L2 = 2L1
R1 a1
_L1
=R2 2a1/2
_2L1
=>5 x a1
_L1
=R2 x a1
_4L1
∴R2 =5 x a1 x 4L1
_a1 x L1

= 20Ω

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Dicussion (1)

  • Let the original length = L1
    Let the original resistance = R1 = 5Ω
    Let the original resistivity = P1
    Let the original area = a1
    Let the new length = L2 = 2L1

    let the new area = a2 = 1/(2a2)
    Let the new resistance = R2
    Let the new resistivity = P2
    But since the resistivity remains the same,
    => P1 = P2

    ∴ P1 =R1 a1
    _L1
    = P2 =R2 a2
    _L2

    But a2 = 1/2a1; and L2 = 2L1
    R1 a1
    _L1
    =R2 2a1/2
    _2L1
    =>5 x a1
    _L1
    =R2 x a1
    _4L1
    ∴R2 =5 x a1 x 4L1
    _a1 x L1

    = 20Ω

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