A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is

Solution

Physics sir · 3 steps

Across any one edge you get two paths in parallel: that edge alone, and the rest of the loop.

1. Split the wire. Resistance is proportional to length, so the wire divides its R equally among the sides. Each triangle side is R/3 and each square side is R/4.
2. Two parallel paths. Between the ends of one edge, current can flow through that edge, or the long way round through all the other sides. These two paths are in parallel, and together they add up to R.

   R_△ = (R/3 × 2R/3)/R = 2R/9, R_□ = (R/4 × 3R/4)/R = 3R/16

3. So. The ratio is (2R/9)/(3R/16) = (2 × 16)/(9 × 3) = 32/27, which is option (c).

The idea to keep. For a uniform wire bent into a closed loop, the resistance between two points is the two arcs in parallel: (R₁ R₂)/(R₁ + R₂), where R₁ + R₂ is the whole wire's R.

Answer from the NTA final answer key. Solution by PrepWiser.