Question

Q15. A wire has a resistance of 16 $\Omega$ it is melted and drawn into a wire of half of its length calculate the resistance of the new wire. What is the percentage change in its resistance ?

Answer 1

Dear Student

given,

R₁= 16Ω

l₁ = l

R₂ = ?

l ₂ = 2l₁ = 2l

Now when the wire is melted and recast into a wire of half length, its volume will remain unchanged.

Volume of cylindrical wire = length x Area of cross section of wire

so,

l₁A₁ = l₂A₂

or, l₁A₁ = l₁/2(A₂)

or, A₂ = 2A₁ ..........................(1)

now resistance R = ρl/A

so, R₁ = ρl ₁/A₁

R₂ = ρl₂/A₂

or, R₁/R₂ = (ρl ₁/A₁ )/(ρl₂/A₂) = l₁A₂/l₂A₁

so from (1),

R₁/R₂ = l₁​2A₁/ l₁/2(A₁) = 4

or, R₂ = R₁ /4= 16/ 4 = 4Ω

therefore change in resistance = R₂ - R₁ = 16-4 = 12Ω

% change in resistance = (12/16)x100 = 75%

Regards