Let the emfs be E₁ and E₂. From the potentiometer principle: E₁ + E₂/E₁ – E₂ = 50/10 = 5 Solving, E₁/E₂ = 5/1. Answer: (d) 5:1.

Using P = V²R/(R+r)² , equating power for R = 4Ω and R = 1Ω, solving gives r = 2Ω and E = 12V. Maximum power = 18W. Answer: (c).

The current is given by I = dQ/dt = a – 2bt. It starts at, reaches a maximum at t = 0 and decreases linearly with time. Thus, the correct answer is (a) decreases linearly with time.

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 ...