Electricity — Set 5

Correct Answer: C. 9×10^9 N m^2/C^2

• **9×10^9 N m^2/C^2** = Coulomb's constant k = 1/(4πε₀) ≈ 9×10⁹ N m²/C². • **k ≈ 9×10⁹ N m²/C²; used in F = kq₁q₂/r²** — It links force, charges, and distance in electrostatics. • 💡 Wrong-option analysis: 3×10⁸ m/s: speed of light in vacuum; 8.85×10⁻¹² C²/(N m²): that is permittivity ε₀, not k; 9×10⁻⁹: incorrect power of ten.

Correct Answer: B. 1 C/s

• **1 C/s** = One ampere is defined as the flow of one coulomb of charge per second. • **1 A = 1 C/s** — About 6.24×10¹⁸ electrons pass a cross-section each second when 1 A flows. • 💡 Wrong-option analysis: 1 V/m: unit of electric field strength; 1 N/C: also unit of electric field (equivalent to V/m); 1 J/C: unit of electric potential (volt).

Correct Answer: A. 3.6×10^6 J

• **3.6×10^6 J** = 1 kWh = 1000 W × 3600 s = 3.6×10⁶ J. • **1 kWh = 3.6 MJ** — This is the domestic unit on electricity bills; 1 unit = 1 kWh. • 💡 Wrong-option analysis: 3.6×10⁵ J: off by factor 10, forgets the 1000 in kW; 3.6×10³ J: confuses kW with W; 360 J: uses minutes instead of seconds.

Correct Answer: A. Brown-Black-Red

• **Brown-Black-Red** = Brown = 1, Black = 0, Red = multiplier 10² = 100; so 10 × 100 = 1000 Ω = 1 kΩ. • **1(Brown) 0(Black) × 10²(Red) = 1000 Ω** — The standard colour code uses BBROYGBVGW for digits 0–9. • 💡 Wrong-option analysis: Brown-Red-Black: gives 12×1 = 12 Ω; Red-Black-Brown: gives 20×10 = 200 Ω; Black-Brown-Red: gives 01×100 = 100 Ω.

Correct Answer: A. P/Q = R/S

• **P/Q = R/S** = A Wheatstone bridge is balanced when P/Q = R/S, making the galvanometer current zero. • **P/Q = R/S → zero galvanometer current** — Used to find an unknown resistance with high accuracy. • 💡 Wrong-option analysis: PR = QS: algebraic rearrangement of the correct condition but non-standard form; P+Q = R+S: sum equality does not ensure balance; P/Q = S/R: inverted second ratio, a different and incorrect condition.

Correct Answer: A. 1.0 V

• **1.0 V** = Terminal voltage V = E − Ir = 1.5 − 1×0.5 = 1.0 V. • **V = E − Ir = 1.5 − 0.5 = 1.0 V** — Higher current draws more voltage across internal resistance, lowering terminal voltage. • 💡 Wrong-option analysis: 2.0 V: uses E+Ir for a charging cell; 0.5 V: only the internal drop Ir, not the terminal voltage; 1.5 V: open-circuit emf when no current flows.

Correct Answer: C. 1/Ceq = 1/C1 + 1/C2

• **1/Ceq = 1/C1 + 1/C2** = In series, capacitors share the same charge so their reciprocals add. • **1/Ceq = 1/C1 + 1/C2; equal C: Ceq = C/2** — Series capacitance is always less than the smallest individual capacitor. • 💡 Wrong-option analysis: Ceq = C1+C2: this is the parallel rule; Ceq = C1C2: dimensionally gives F², not F; Ceq = (C1+C2)/2: arithmetic mean, no circuit basis.

Correct Answer: D. 1×10^-4 J

• **1×10^-4 J** = Using U = ½CV² = 0.5 × 2×10⁻⁶ × 100 = 1×10⁻⁴ J. • **U = ½CV² = 0.5 × 2×10⁻⁶ × 10² = 10⁻⁴ J = 0.1 mJ** — Doubling voltage quadruples stored energy. • 💡 Wrong-option analysis: 1×10⁻² J: forgets ½ and misplaces power of ten; 1×10⁻⁵ J: exponent arithmetic error; 1×10⁻³ J: omits the ½ factor.

Correct Answer: C. σ = 1/ρ

• **σ = 1/ρ** = Conductivity is the reciprocal of resistivity; a better conductor has higher σ and lower ρ. • **σ = 1/ρ (unit S/m); copper σ ≈ 6×10⁷ S/m** — Silicon σ ≈ 10⁻³ S/m (semiconductor). • 💡 Wrong-option analysis: σ = ρ: better conductors would have higher resistivity, the opposite of reality; σ = ρ²: dimensionally incorrect (S/m ≠ Ω²m²); σ = 1/ρ²: also dimensionally wrong.

Correct Answer: C. 12 Ω

• **12 Ω** = Using R = R₀(1 + αΔT) = 10(1 + 0.004×50) = 10 × 1.2 = 12 Ω. • **R = 10(1 + 0.2) = 12 Ω** — For most metals α > 0, so resistance increases with temperature. • 💡 Wrong-option analysis: 14 Ω: uses ΔT = 100 instead of 50; 10.2 Ω: uses ΔT = 5 by mistake; 20 Ω: simply doubles R₀ without using the formula.