When two resistance R₁ and R₂ connected in series and introduced into the left gap of a meter bridge and a resistance of 10 Ω is introduced into the right gap, a null point is found at 60 cm from left side. When R₁ and R₂ are connected in parallel and introduced into the left gap, a resistance of 3 Ω is introduced into the right gap to get null point at 40 cm from left end. The product of R₁ R₂ is ____________Ω²

Solution

Physics sir · 4 steps

Balance the bridge twice: once you get the sum, once the parallel value, and their product falls out.

1. Balance condition. At the null point, the left resistance divided by the right resistance equals the left length divided by the right length. The wire is 100 cm, so a null point at l cm leaves 100 - l cm on the right.

   R_(left)/R_(right) = l/(100 - l)

2. Series setup. With the null point at 60 cm, (R₁ + R₂)/10 = 60/40. So R₁ + R₂ = 15 Ω.
3. Parallel setup. With the null point at 40 cm, Rₚ/3 = 40/60. So Rₚ = 2 Ω, where Rₚ = (R₁ R₂)/(R₁ + R₂).
4. So. R₁ R₂ = Rₚ × (R₁ + R₂) = 2 × 15 = 30 Ω².

The common slip: 2. Taking the parallel value Rₚ = 2 Ω as the product gives 2. You still have to multiply it by R₁ + R₂ = 15 Ω.

The idea to keep. The series combination gives you R₁ + R₂ and the parallel combination gives you (R₁ R₂)/(R₁ + R₂). Multiply the two and you get R₁ R₂ without finding either resistance.

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