Govt. Exams
Entrance Exams
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.
For shunt: S = (Ig × G)/(I - Ig) = (0.001 × 1)/(10 - 0.001) ≈ 0.0001/9.999 ≈ 0.01Ω approximately. More precisely: S/G = Ig/(I-Ig) = 0.0001 gives S ≈ 0.1mΩ. Recalculating: shunt formula gives ~0.1mΩ or ~1/10000 resistance.
Current I = E/(R + r) = 12/(4 + 2) = 2A. Terminal voltage V = E - Ir = 12 - 2(2) = 8V.
For Wheatstone bridge: P/Q = R/S. If P increases by 10%, then Q must also increase by 10% to maintain the ratio and balance condition.
When bent into a square, total resistance is R. Between adjacent corners, one path has R/4 and parallel path has 3R/4. Using parallel formula: (R/4 × 3R/4)/(R/4 + 3R/4) = 3R/16 ÷ 1 = 3R/16. Wait, recalculating: equivalent = (1/4 × 3/4)/(1/4 + 3/4) × R = (3/16)/(1) × R = 3R/16... Actually it's 3R/8 using proper parallel resistance calculation.
