In a meter bridge, the balanced condition gives: R/10 Ω = 3/2 Solving, R= 15Ω. The resistance per unit length = 15Ω/1.5m​ = 10Ω/m. Length of 1Ω = 1/10 = 0.1M = 1.0 × 10⁻¹ m. Answer: (a) 1.0 × ...

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.