Question

3. A wire of length L is drawn such that its diameter is reduced to half of its original

diameter. If the resistance of the wire were 10Ω, its new resistance would be.

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Answer 1

$\bold{AnsweR:}$

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Initial Length of wire = l

Let base area of wire = A

We know, if wire is stretched . Volume of wire remains constant.

∵ V = Al , here A is base area and l is length of wire.

∴ R = ρl/A = ρlA/A² = ρV²/A²

V is constant. ∴ R  1/A²

But we know, base area , A  d² [ actually, Base area = circle area = πd²/4 ∴ A  d²]

∴ R 1/d⁴

we can say

R₁/R₂ = d₂⁴/d₁⁴

Now, according to question, diameter of wire is reduced to half of its original diameter

Means new diameter of wire, d₂ = d/2

Where initial diameter of wire , d₁= d

Initial resistance of wire , R₁ = 10Ω

We have to find R₂ = ?

∴10/R₂ = (d/2)⁴/(d)⁴ = 1/16

R₂ = 160 Ω

Hence , new resistance of wire = 160Ω.

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