Question : A metal wire of resistance 3Ω is elongated to make a uniform wire of double its previous length. This new wire is bent and the ends joined to make a circle. If the two points on this circle make an angle of 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
Doubt by Mohini
Solution :
R = 3Ω
n = 2
R' = n2R
R' = (2)2(3)
R' = 4×3
R' = 12 Ω
Now, we know
R' = 4×3
R' = 12 Ω
Now, we know
Length of Arc = [θ/360°]×2πr
l1 = [60°/360°]×2πr
l1 = πr/3
similarly
l2 = 5πr/3
Also,
l1 = πr/3
similarly
l2 = 5πr/3
Also,
R = ρl/A
R∝l
so
R1/R2 = l1/l2
R1/R2 = (πr/3)/(5πr/3)
R1/R2 = 1/5
R∝l
so
R1/R2 = l1/l2
R1/R2 = (πr/3)/(5πr/3)
R1/R2 = 1/5
R1 : R2 = 1:5
Also, R1+R2 = 12
Hence
R1 = (1/6)×12 = 2 Ω
Also, R1+R2 = 12
Hence
R1 = (1/6)×12 = 2 Ω
R2 = (5/6)×12 = 10 Ω
Clearly, R1 and R2 are connected in parallel.
So equivalent resistance will be given by
Req = R1R2/(R1+R2)
Req = (2×10)/(2+10)
So equivalent resistance will be given by
Req = R1R2/(R1+R2)
Req = (2×10)/(2+10)
Req = 20/12
Req = 5/3 Ω
Req = 5/3 Ω
Hence, b) 5/3 Ω is correct option.
