Question

A wire of length l is drawn such that its diameter is reduced to half of its original diameter. if the initial resistance of the wire were 10 ω, its new resistance will be

Answer 1

Given Wire of length l .

Diameter is reduced to half

Area α r²
So Area will become 1/4 th of the original value.

A2=A1/4 ----------(1)

 Before stretching:

Let L1 be the length
A1 be the area
R1= resistance =10ohms

After stretching :

R2 - new resistance
L2=length
A2= area

As mass and density remains constant after stretching.... Volume also remains constant.

L1A1=L2A2

L1A1=L2(A1/4)

L2=4L1-----------equation 2

As we know that R=ρ L/A

R2/R1= L2/L1 (A1/A2)
=4x 4  [ froem equation 1 and 2]

R2/R1= 16
R2= 16xR1=16x10 =160 ohms

Hence resistance of new stretched wire is 160ohms.

Answer 2

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 $\propto$ 1/A²
But we know, base area , A $\propto$ d² [ actually, Base area = circle area = πd²/4 ∴ A $\propto$ d²]

∴ R $\propto$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