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?

Correct option is B. 53Ω
R = ρl/A
where ρ is resistivity, l is length and A is area of cross section.
When the length is doubled, the area becomes half (assuming volume remains constant).
Therefore, the new resistance will be:
R' = ρ(2l)/(A/2) = 4ρl/A = 4R = 4 * 3Ω = 12Ω (new resistance of wire)
The wire is bent into a circle. The resistance of the entire circle is 12Ω.
The resistance of an arc subtending an angle θ at the center of a circular wire of resistance R is given by:
R_arc = (θ/360°) * R
In this case, θ = 60°, and the total resistance of the circle is 12Ω. So the resistance of the arc is:
R_arc = (60°/360°) * 12Ω = 2Ω
The remaining portion of the circle has a resistance of 12Ω - 2Ω = 10Ω.
These two resistances (2Ω and 10Ω) are in parallel. The equivalent resistance is given by:
1/Req = 1/R_arc + 1/R_remaining
1/Req = 1/2Ω + 1/10Ω
1/Req = (5 + 1)/10Ω = 6/10Ω
Req = 10/6Ω = 5/3Ω ≈ 1.67Ω
However, there seems to be an error in the given solution. The correct approach is as follows:
The total resistance of the circular wire is 12Ω. The resistance of the arc subtending 60° is (60/360) * 12Ω = 2Ω. The remaining part of the circle has resistance 10Ω. These two resistances are in parallel.
Therefore, the equivalent resistance is:
1/Req = 1/2 + 1/10 = 6/10
Req = 10/6 = 5/3 Ω ≈ 1.67 Ω
This value is not among the options. There must be an error in either the problem statement or the given solution. Let's reconsider the provided solution:
R1 = 2Ω (arc resistance)
R2 = 10Ω (remaining resistance)
Req = (R1 * R2) / (R1 + R2) = (2 * 10) / (2 + 10) = 20/12 = 5/3 Ω
This is approximately 1.67Ω, which is not among the options. The provided solution (53Ω) is incorrect.