JEE Physics - MCQ Practice Questions

The electric field is related to potential by E = -dV/dr (negative gradient of potential). The negative sign indicates field points toward lower potential.

Distance from each vertex to centroid is a/√3. V = k(q + q - 2q)/(a/√3) = 0. The charges sum to zero, giving zero potential.

All charge elements on the ring are equidistant from the axial point. Distance = √(R² + x²), so V = kQ/√(R² + x²).

For a uniformly charged disc, the field at the center involves integrating contributions from rings. Result: E = σ/(2ε₀) = Q/(2πε₀R²).

Introducing a dielectric increases capacitance by factor κ: C = κε₀A/d = κC₀. This is a fundamental property used in capacitor design.

At midpoint, distance from each charge = 0.5 cm = 0.005 m. Both fields point in same direction (from +q toward -q). E_total = 2 × k × 2×10⁻⁶ / (0.005)² = 7.2 × 10⁷ V/m.

The grounded sphere develops negative charge to maintain V = 0. The charge distribution is non-uniform because the near side accumulates more negative charge.

For an infinite line charge, E = λ/(2πε₀r). Substituting: E = (2 × 10⁻⁸)/(2π × 8.85 × 10⁻¹² × 0.1) = (2 × 10⁻⁸)/(5.57 × 10⁻¹²) ≈ 3.6 × 10³ N/C

Initial force: F₁ = k(Q)(3Q)/r² = 3kQ²/r². When spheres touch, total charge = 4Q, distributed as 2Q each. Final force: F₂ = k(2Q)(2Q)/r² = 4kQ²/r². Ratio: F₂/F₁ = (4kQ²/r²)/(3kQ²/r²) = $\frac{4}{3}$. The force increases by factor $\frac{4}{3}$, or changes by $\frac{4}{3}$ times initial. However, comparing initial to final: change factor = F₂/F₁ = $\frac{4}{3}$. The force becomes ($\frac{4}{3}$) times, meaning it changed by multiplying with $\frac{4}{3}$. If asking reduction: Answer is $\frac{2}{3}$ represents the comparative analysis in different context, but correct ratio of final to initial is $\frac{4}{3}$.

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.

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.

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.

Current I = E/(R + r) = 12/(4 + 2) = 2A. Terminal voltage V = E - Ir = 12 - 2(2) = 8V.

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.

As temperature increases, atoms vibrate more vigorously, increasing collision frequency with drifting electrons. This increases mean free path reduction and thus resistance increases.

Resistance R = V²/P = (200)²/1000 = 40Ω (constant). Heat at 100V: P = V²/R = (100)²/40 = $\frac{10000}{40}$ = 250W.