Given that both bulbs consume the same power at 200V and 300V, their resistances are R₁ = 200²/p and R₂ = 300²/p , so R₁/R₂ = 4/9. Since in series, V₁/V₂ = R₁/R₂, the correct answer is (b) Ratio of ...

In Series: Each wire has resistance R′ = R/4. The total resistance in series is 2R/4 = R/2, increasing power by 2 times. Heating time reduces to 4/2 = 2 min. Correct answer: (b) 2 if wires are in series.

The resistance R is given by Ohm’s law: R = V/I Using error propagation, the percentage error in R is the sum of the percentage errors in V and I. Given 3% error in both, the total error in R ...

Let the three resistances be R₁,R₂,R₃, with R₁ : R₂ = 1 : 2 Given: 1/R₁ + 1/R₂ + 1/R₃ = 1 Solving for integer values, the largest resistance is 6 ohms. Answer: (c) 6.

Let the internal resistance be r and the balance lengths be I₁ = 2m, l₂ = 3m. Using the potentiometer equation: l₁/l₂ = R₁/R₂ Solving for r, we get 1 Ω. Answer: (a) 1 Ω.