✅ MCQ (Choose Correct Alternative) — from textbook
Q1. Which of the four bulbs (25W, 40W, 60W, 100W) operating at 230V has the lowest resistance?
Answer: (D) 100 W
R = V²/P. Higher the power, lower the resistance.
R(25W) = 230²/25 = 2116Ω, R(100W) = 230²/100 = 529Ω
So 100W bulb has lowest resistance.
Q2. Which of the following is an ohmic conductor? (A) transistor (B) vacuum tube (C) electrolyte (D) nichrome wire
Answer: (D) Nichrome wire
Ohmic conductors follow V = IR with straight line I-V graph. Nichrome wire is an ohmic (linear) device. Transistors, vacuum tubes and electrolytes are non-ohmic.
Q3. A wire of length L and resistance R is stretched so that its radius of cross-section is halved. What is its new resistance?
Answer: (D) 16R
When radius is halved: Area A' = πr'² = π(r/2)² = A/4
Volume = AL = A'L' → L' = 4L
R' = ρL'/A' = ρ(4L)/(A/4) = 16(ρL/A) = 16R
Q4. The internal resistance of a cell of emf 2V is 0.1Ω connected to 0.9Ω. The voltage across cell will be?
Answer: (B) 1.8V
I = ε/(R+r) = 2/(0.9+0.1) = 2A
V = ε - Ir = 2 - 2×0.1 = 2 - 0.2 = 1.8V
Q5. Five dry cells each of 1.5V, four connected normally and one reversed. What is overall voltage?
Answer: (B) 4.5V
4 cells forward = 4×1.5 = 6V
1 cell reversed = −1.5V
Net = 6 − 1.5 = 4.5V
📝 Short Answer Questions
Q1. Distinguish between Ohmic and Non-Ohmic substances with examples.
Ohmic substances: Those for which I-V graph is a straight line through origin. Resistance is constant. They obey Ohm's law (V = IR). Example: nichrome wire, copper, silver.
Non-Ohmic substances: Those for which I-V graph is NOT a straight line. Resistance varies with V or I. They do not obey Ohm's law. Example: vacuum tubes, junction diodes, thermistors, liquid electrolytes.
Q2. DC current flows in a metal piece of non-uniform cross-section. Which quantity remains constant: current, current density or drift speed?
Current (I) remains constant throughout the conductor.
Current density J = I/A — varies (A changes)
Drift speed Vd = J/ne = I/nAe — also varies
But total current I is same at every cross-section by conservation of charge.
Q3. Prove that current density of metallic conductor is directly proportional to drift speed of electrons.
Consider wire with cross-section A, n = free electrons per unit volume, Vd = drift speed.
Charge crossing section in time t: q = nALe, where L = Vd·t
So q = nAVdte
Current I = q/t = nAVde
Current density J = I/A = nVde
Since n and e are constants: J ∝ Vd ✓
🔢 Numericals (from textbook exercises)
Q1. Find resistance of a railway rail 20km long at 20°C. Cross-section = 25cm², resistivity = 6×10⁻⁸ Ωm.
R = ρl/A = (6×10⁻⁸ × 20×10³) / (25×10⁻⁴)
= (6×10⁻⁸ × 2×10⁴) / (25×10⁻⁴)
= 12×10⁻⁴ / 25×10⁻⁴ = 0.48 Ω
Q2. Battery emf = 24V, internal resistance = 380Ω. Find maximum current. Can it start a car motor?
I_max = ε/r = 24/380 = 0.063 A
No, a car starter motor needs 100–200A. This battery cannot drive it.
Q3. Battery emf = 12V, r = 3Ω, I = 0.5A. Find (a) resistance of resistor (b) terminal voltage.
(a) ε = I(R+r) → 12 = 0.5(R+3) → R+3 = 24 → R = 21Ω
(b) V = ε - Ir = 12 - 0.5×3 = 12 - 1.5 = 10.5V
Q4. Current density in copper = 500 A/cm². Free electrons = 8.47×10²² per cm³. Find drift velocity. (e = 1.6×10⁻¹⁹C)
J = 500 A/cm² = 500×10⁴ A/m²
n = 8.47×10²² per cm³ = 8.47×10²⁸ per m³
Vd = J/ne = (500×10⁴)/(8.47×10²⁸ × 1.6×10⁻¹⁹)
= 5×10⁶/13.55×10⁹ = 3.69×10⁻⁴ m/s
Q5. Three resistors 10Ω, 20Ω, 30Ω in series. (i) Equivalent resistance (ii) PD across each when connected to 12V.
(i) Rs = 10+20+30 = 60Ω
(ii) I = V/Rs = 12/60 = 0.2A
V₁ = 0.2×10 = 2V, V₂ = 0.2×20 = 4V, V₃ = 0.2×30 = 6V
Q6. Two resistors 1kΩ and 2kΩ in parallel with 9V. Find (i) equivalent resistance (ii) current through each.
(i) 1/Rp = 1/1000 + 1/2000 = 3/2000 → Rp = 0.66 kΩ
(ii) I₁ = 9/1000 = 9mA, I₂ = 9/2000 = 4.5mA
Q7. Silver wire: R = 4.2Ω at 27°C, R = 5.4Ω at 100°C. Find temperature coefficient α.
α = (R - R₀)/[R₀(T - T₀)] = (5.4 - 4.2)/[4.2 × (100-27)]
= 1.2/(4.2 × 73) = 1.2/306.6 = 3.91×10⁻³/°C
Q8. Wire: length = 6m, diameter = 0.5mm, resistance = 50Ω. Find resistivity and conductivity.
r = 0.25mm = 0.25×10⁻³m, A = πr² = π(0.25×10⁻³)² = 1.963×10⁻⁷m²
ρ = RA/l = 50×1.963×10⁻⁷/6 = 1.636×10⁻⁶ Ωm
σ = 1/ρ = 6.11×10⁵ (Ωm)⁻¹
Q9. Find resistance colour code values: (1) Blue Green Red Gold (2) Brown Black Red Silver (3) Red Red Orange Gold
(1) Blue=6, Green=5, Red=×10², Gold=±5% → 6.5kΩ ±5%
(2) Brown=1, Black=0, Red=×10², Silver=±10% → 1.0kΩ ±10%
(3) Red=2, Red=2, Orange=×10³, Gold=±5% → 22kΩ ±5%
Q10. A current of 4A flows through an automobile headlight for 2 hours. How many electrons flow through it?
q = I×t = 4 × 2×3600 = 28800 C
Number of electrons = q/e = 28800/(1.6×10⁻¹⁹) = 1.8×10²³ electrons
Q11. Heating element: 230V, 5A, 1 hour. Find heat dissipated in kcal. (J = 4.2 J/cal)
P = VI = 230×5 = 1150W
Energy = P×t = 1150×3600 = 4140000 J
Heat = 4140000/4.2 = 985714 cal = 985.7 kcal
🔌 Circuit Problem (Example 11.8 from textbook)
Network: 15V battery, r=1Ω. A-B: two 4Ω in parallel. B-C: 1Ω. C-D: two 6Ω in parallel. Find (i) equivalent resistance (ii) current in each resistor (iii) voltage drops VAB, VBC, VCD.
(i) RAB = (4×4)/(4+4) = 2Ω, RCD = (6×6)/(6+6) = 3Ω, RBC = 1Ω
RT = 2+1+3 = 6Ω
(ii) I = ε/(RT+r) = 15/(6+1) = 2.1A
Current in each 4Ω = 2.1/2 = 1.05A
Current through 1Ω = 2.1A
Current in each 6Ω = 2.1/2 = 1.05A
(iii) VAB = I×RAB = 2.1×2 = 4.2V
VBC = I×RBC = 2.1×1 = 2.1V
VCD = I×RCD = 2.1×3 = 6.3V
Total = 4.2+2.1+6.3 = 12.6V (loss = 2.4V in internal resistance)