A metal wire of resistance 3 Ω is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60° at the centre, the equivalent resistance between these two points will be :

Question

A metal wire of resistance 3 Ω is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60° at the centre, the equivalent resistance between these two points will be : (A) (12/5) Ω  (B) (5/3) Ω  (C) (5/2) Ω  (D) (7/2) Ω

Tardigrade Admin · Accepted Answer

R =(ρℓ2/A ℓ D) d = (ρ d ℓ2/m) R propto ℓ2 R = 12Ω (new resistance of wire) R1 = 2 Ω R2 = 10 Ω Req = (10 ×2/10+2) = (5/3) Ω

Q. A metal wire of resistance $3 Ω$ is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle $6 0^{\circ}$ at the centre, the equivalent resistance between these two points will be :

$\frac{12}{5} Ω$

$\frac{5}{3} Ω$

$\frac{5}{2} Ω$

$\frac{7}{2} Ω$

$R = \frac{ρ ℓ ^{2}}{A ℓ D} d = \frac{ρ d ℓ ^{2}}{m}$
$R \propto ℓ^{2}$
$R = 12Ω$ (new resistance of wire)
$R_{1} = 2Ω R_{2} = 10Ω$
$R_{e q} = \frac{10 \times 2}{10 + 2} = \frac{5}{3} Ω$

Q. A metal wire of resistance 3Ω is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60∘ at the centre, the equivalent resistance between these two points will be :

512​Ω

35​Ω

25​Ω

27​Ω

R=AℓDρℓ2​d=mρdℓ2​ R∝ℓ2 R=12Ω (new resistance of wire) R1​=2ΩR2​=10Ω Req​=10+210×2​=35​Ω

$\frac{12}{5} Ω$

$\frac{5}{3} Ω$

$\frac{5}{2} Ω$

$\frac{7}{2} Ω$

$R = \frac{ρ ℓ ^{2}}{A ℓ D} d = \frac{ρ d ℓ ^{2}}{m}$
$R \propto ℓ^{2}$
$R = 12Ω$ (new resistance of wire)
$R_{1} = 2Ω R_{2} = 10Ω$
$R_{e q} = \frac{10 \times 2}{10 + 2} = \frac{5}{3} Ω$