JEE Physics - MCQ Practice Questions

Conductivity σ = 1/ρ where ρ is resistivity. Since resistivity is in Ω·m, conductivity is in (Ω·m)⁻¹ = S/m or mho/meter.

Resistivity of aluminum (2.65 × 10⁻⁸ Ω·m) is greater than copper (1.68 × 10⁻⁸ Ω·m). Since P = I²R and R = ρL/A, aluminum has higher resistance and thus higher power dissipation for same current.

Drift velocity vd = I/(nAe) is typically 10⁻⁴ to 10⁻³ m/s for normal currents, much slower than thermal velocity (~10⁶ m/s) but much slower than light speed.

For parallel resistances: 1/Req = 1/R₁ + 1/R₂ = $\frac{1}{4}$ + $\frac{1}{6}$ = $\frac{3}{12}$ + $\frac{2}{12}$ = $\frac{5}{12}$. Therefore Req = $\frac{12}{5}$ = 2.4Ω.

Brown=1, Black=0, Red=2 (multiplier 10²), Gold=5% tolerance. Value = 10 × 10² = 1000Ω ± 5%.

Resistivity ρ is an intrinsic property of material and varies with temperature. It is independent of geometry (length and area) and current/voltage.

1/Req = $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{6}$ = $\frac{3}{6}$ + $\frac{2}{6}$ + $\frac{1}{6}$ = $\frac{6}{6}$ = 1. Therefore Req = 1Ω.

E = V + Ir → 1.5 = 1.2 + 2r → 0.3 = 2r → r = 0.15Ω.

R = ρL/A. R₁/R₂ = (L₁/L₂) × (A₂/A₁) = ($\frac{2}{3}$) × (2²/1²) = ($\frac{2}{3}$) × 4 = $\frac{8}{3}$

At null deflection: P/Q = R/S, which is equivalent to P·S = Q·R. Both are valid balanced bridge conditions

An ideal ammeter must have zero resistance to not affect the circuit current. In practice, ammeters have very low resistance (mΩ range)

Standard color code: 1st band = first digit, 2nd band = second digit, 3rd band = multiplier, 4th band = tolerance

Standard Leclanche cells have internal resistance in the range of 1-2Ω depending on size and condition

Energy = Power × Time = 500W × 4h = 2000Wh = 2 kWh

Resistivity is a material property independent of shape and size. It depends only on the nature of material and temperature

Positive temperature coefficient means R = R₀(1 + αΔT) where α > 0, so resistance increases with increasing temperature

In a purely resistive circuit, voltage and current are in phase, so phase difference is 0°. V = IR is always satisfied

1/R_eq = $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{6}$ = $\frac{3}{6}$ + $\frac{2}{6}$ + $\frac{1}{6}$ = $\frac{6}{6}$ = 1. Therefore R_eq = 1Ω

I = E/(R + r) = 12/(4 + 2) = $\frac{12}{6}$ = 2A

Resistivity is an intrinsic property depending only on the material type and temperature, not on geometry