Question

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.

(A) 40Ω 

(B) 60Ω 

(C) 120Ω 

(D) 160Ω

Answer 1

 The correct option (D) 160Ω

Explanation:

Diameter i.e. radius is reduced to half.

A ∝ r²

hence area will become (1/4)^th of original.density remains same also mass doesn’t change.

Hence volume is constant

∴ A₁ℓ₁ = A₂ℓ₂

∴ A₁ L₁ = (A₁/4)L₂ as A₂ = (A₁/4)

∴ L₂ = 4 L₁

Now R = (pL/A) i.e. R ∝ (L/A)

∴ (R₂/R₁) = (L₂/L₁) (A₁/A₂)

∴ (R₂ / R₁) = 4 × 4 = 16

∴ R₂ = 16 R₁ = 16 × 10 = 160Ω