Govt. Exams
E/E_std = L/L_std, so E = 1.018 × (60/50) = 1.018 × 1.2 = 1.2216 ≈ 1.221 V
Total EMF = 1.5 + 1.5 = 3V. Total resistance = 2 + 0.5 + 0.5 = 3Ω. Current = 3/3 = 1A. Hmm, that's option C. Let me verify: I = 3V/5Ω = 0.6A
R = ρL/A. New resistance = ρ(L/2)/(A/2) = ρL/A = R. Actually R' = ρL'/A' where same ρ applies. R' = (L/2)/(A/2) × (ρ/A) × L = R. Recalculating: R' = ρ(L/2)/(A/2) = ρL/A × (1/2)/(1/2)⁻¹ = 2R
Terminal voltage V = E - Ir. Since I = E/(R+r), we get V = E - Er/(R+r) = ER/(R+r). Both expressions are equivalent
At null point: X/S = L₁/(100-L₁), so 20/S = 40/60, therefore S = 20 × 60/40 = 30Ω. Wait, let me recalculate: S = 20 × 60/40 = 30Ω. Correction needed - answer should be 30Ω if calculation is 20/S = 40/60
Potential gradient = V/L = 2V / 100cm = 0.02 V/cm
Using I = nAve, where n is number density. A = π(0.5×10⁻³)². Solving gives n = 8.05 × 10²⁸ m⁻³
In meter bridge: R/S = l₁/l₂ where l₁ = 60cm, l₂ = 40cm. R/15 = 60/40 = 3/2 → R = 15 × 3/2 = 22.5Ω.
Resistance R = V²/P = (200)²/1000 = 40Ω (constant). Heat at 100V: P = V²/R = (100)²/40 = 10000/40 = 250W.
As temperature increases, atoms vibrate more vigorously, increasing collision frequency with drifting electrons. This increases mean free path reduction and thus resistance increases.
