JEE Physics - MCQ Practice Questions

V = E - Ir. 9 = 10 - I×r. Current I = V/R = $\frac{9}{90}$ = 0.1A. So 9 = 10 - 0.1r, giving r = 10Ω

In DC circuits, current is continuous (same) through all series elements including capacitors in steady state. However, in AC circuits, current flows through capacitors.

R = ρL/A. If L increases by 10% and volume constant, A decreases by ~9.1%. New R = 1.1R₀/0.91 ≈ 1.21R₀, so 21% increase.

For infinite ladder: Rₑq = 1 + (Rₑq×1)/(Rₑq+1). Solving: Rₑq² - Rₑq - 1 = 0, giving Rₑq = (1+√5)/2 ≈ 1.618Ω (Golden ratio)

R = R₀[1 + α(T-T₀)] = 100[1 + 0.004(120-20)] = 100[1 + 0.4] = 140Ω

Brown=1, Black=0, Red=2 (multiplier=10²=100), Gold=5% tolerance. Resistance = 10×100 = 1000Ω ± 5%

Tungsten filament's resistance increases significantly with temperature due to heating, causing non-linear I-V characteristic (non-ohmic).

P = V²/R. New P = (2V)²/(R/2) = 4V²×2/R = 8(V²/R) = 8P₀

BCS theory explains superconductivity: below critical temperature, electrons form Cooper pairs with no scattering, resulting in zero resistance.

When stretched, length becomes 1.5L. Volume remains constant (LA = A'L'), so A' = A/1.5. New resistance R' = ρ(1.5L)/(A/1.5) = 2.25ρL/A = 2.25R

Resistance of element R = V²/P = 220²/1000 = 48.4Ω (constant). At 110V: P' = V'²/R = 110²/48.4 ≈ 250W. Power varies with square of voltage.

For balanced bridge: P/Q = R/S. Check: $\frac{10}{15}$ = $\frac{20}{30}$ → $\frac{2}{3}$ = $\frac{2}{3}$. The bridge is balanced, so no current flows through galvanometer (zero reading).