Three wires each of length L, cross-sectional area A and resistivity ρ are connected as shown in the figure (two wires L,A connect X to Y in parallel, forming a lens shape, and a third wire L,A connects Y to Z in series). These are to be replaced by another wire of same resistivity such that the resistance between points X and Z does not change. If L1 is the length and A1 is the cross-sectional area of the new wire, then which one among the following is correct?

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L1 = 3L and A1 = 2A. Each of the three wires is identical. Each has resistance R = ρL/A. Two wires connect X to Y in parallel. Their combined resistance is R/2. The third wire connects Y to Z in series. Its resistance is R. So the total resistance from X to Z is R/2 + R, which equals 3R/2. The new wire must give this same resistance. So ρL1/A1 = (3/2) × (ρL/A). This means L1/A1 = 1.5 × (L/A). Check option (a): L1 = 3L and A1 = 2A gives L1/A1 = 3L/2A = 1.5(L/A). This matches. Option (a) is correct.

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