Class 9 Science Chapter 10 Electric Current and its Effects: 30 MCQs with Detailed Answers

CBSE's 2024–25 rationalized syllabus emphasizes conceptual depth over lengthy answers. MCQs force you to *discriminate* between similar concepts—a skill exams reward. For Chapter 10, this means distinguishing between heating effect (I²Rt law) and magnetic effect (B-field direction by right-hand rule), recognizing circuit symbols instantly, and calculating fuse ratings. Traditional long-answer questions take 10 minutes; MCQs take 1.5 minutes and cover the same ground faster. Board exams now allocate 40–50% marks to objective questions (MCQs, fill-blanks, very-short), making mastery non-negotiable. Moreover, MCQs expose your weak spots immediately: if you choose the wrong option, you know your concept isn't solid. This landing page provides 30 scientifically-graded MCQs (Bloom's taxonomy: remember → understand → apply → analyse) so you progress from basic symbol recognition to complex electromagnet design scenarios. Each question mirrors real exam language and difficulty.

**Q1.** Which symbol represents a fuse in an electric circuit? (A) —⊗— (B) —≡— (C) —/—— (D) ——○—— **Answer: (B)** Reason: A fuse is drawn as a short thick line (≡) between two nodes; it melts when current exceeds its rated value. **Q2.** The heating effect of electric current is given by H = I²Rt. If current doubles, heating effect becomes: (A) 2 times (B) 4 times (C) 8 times (D) 16 times **Answer: (B)** Reason: H ∝ I², so if I → 2I, then H → (2I)² = 4I². **Q3.** An electric bell works on the principle of: (A) Heating effect only (B) Magnetic effect only (C) Both heating and magnetic effects (D) Chemical effect **Answer: (B)** Reason: A bell's electromagnet attracts an armature, striking a gong; no heat is needed. **Q4.** Which device protects a circuit from overcurrent? (A) Voltmeter (B) Ammeter (C) Fuse (D) Rheostat **Answer: (C)** Reason: A fuse melts and breaks the circuit when current exceeds its rating. **Q5.** The magnetic field around a straight conductor is: (A) Radial (pointing outward) (B) Circular (concentric circles) (C) Linear (straight lines) (D) Zero **Answer: (B)** Reason: Using the right-hand rule, thumb along current direction gives concentric field circles perpendicular to the wire. **Q6.** An electromagnet becomes stronger when: (A) Number of coil turns decreases (B) Current decreases (C) Number of coil turns increases (D) Core material changes from iron to copper **Answer: (C)** Reason: More turns amplify the magnetic field; B ∝ N (number of turns). **Q7.** If a fuse rated 5A is used in a circuit with 6A current, it will: (A) Continue working normally (B) Melt and break the circuit (C) Increase the voltage (D) Reduce the current automatically **Answer: (B)** Reason: 6A exceeds the 5A rating; the fuse wire overheats (H = I²Rt) and melts. **Q8.** The heating effect is observed in which of these? (A) Nichrome wire in a heater (B) Copper wire in a circuit (C) Both (A) and (B) (D) Neither **Answer: (C)** Reason: All resistive materials (nichrome R ≈ 1 Ω, copper R ≈ 0.02 Ω) produce heat; nichrome is just more visible due to higher resistance. **Q9.** A device that converts electrical energy to magnetic energy is: (A) Motor (B) Electromagnet (C) Generator (D) Rheostat **Answer: (B)** Reason: An electromagnet uses current to create a magnetic field; it stores no mechanical energy, just magnetic. **Q10.** The direction of the magnetic field around a coil is found using: (A) Left-hand rule (B) Right-hand grip rule (C) Fleming's left-hand rule (D) Ohm's law **Answer: (B)** Reason: Curl right-hand fingers in current direction; thumb points along the magnetic field (north pole inside the coil).

**Q11.** An electric heater has a resistance of 50 Ω and operates at 4A. The heat produced in 60 seconds is: (A) 12,000 J (B) 48,000 J (C) 120,000 J (D) 200,000 J **Answer: (B)** Reason: H = I²Rt = (4)² × 50 × 60 = 16 × 50 × 60 = 48,000 J. **Q12.** Two identical bulbs A and B are connected in series with a 6V battery. If bulb A burns out (open circuit), then: (A) Bulb B still glows at full brightness (B) Bulb B glows dimly (C) Bulb B goes off (D) Total voltage doubles **Answer: (C)** Reason: Series breaks at the open point; no current flows through B either. **Q13.** A solenoid with 500 turns, each of radius 2 cm, carries 2A current. Which factor increases its magnetic field strength? (A) Reducing the current to 1A (B) Decreasing the number of turns to 250 (C) Inserting an iron core (D) Increasing the radius to 4 cm **Answer: (C)** Reason: Iron's high permeability (μ) amplifies B; B = μ₀μᵣNI/L, so μᵣ(iron) >> μᵣ(air). **Q14.** An electric geyser rated 2000W, 250V operates for 30 minutes. The current flowing is: (A) 8 A (B) 10 A (C) 12 A (D) 15 A **Answer: (A)** Reason: P = VI, so I = P/V = 2000/250 = 8 A. **Q15.** Three fuses rated 2A, 5A, and 10A are available. For a circuit drawing 7A, the correct fuse is: (A) 2A (too weak, will blow) (B) 5A (too weak, will blow) (C) 10A (safe, won't blow prematurely) (D) Any fuse works **Answer: (C)** Reason: Fuse rating ≥ circuit current; 10A ≥ 7A allows operation, 5A < 7A triggers false tripping. **Q16.** A wire of length L and cross-sectional area A has resistance R. If the wire is stretched to length 2L (same volume), its new resistance is: (A) R (B) 2R (C) 4R (D) R/2 **Answer: (C)** Reason: R = ρL/A; when stretched, L → 2L and A → A/2 (volume constant), so R' = ρ(2L)/(A/2) = 4ρL/A = 4R. **Q17.** An electromagnet's strength is independent of: (A) Current through the coil (B) Number of turns (C) Shape of the coil (circular vs. square) (D) Nature of the core material **Answer: (C)** Reason: Shape doesn't affect B significantly; current, turns, and core permeability dominate. **Q18.** In an electric bell, the sound is produced by: (A) The electromagnet heating the gong (B) The armature vibrating and striking the gong repeatedly (C) The current flowing through the bell coil (D) The fuse breaking the circuit **Answer: (B)** Reason: AC current alternates, so the electromagnet switches on/off, causing the armature to vibrate and repeatedly strike the gong. **Q19.** A tungsten filament bulb glows because: (A) Tungsten absorbs light (B) High resistance causes intense heating at high current (C) Tungsten is a superconductor (D) Tungsten emits visible light without heating **Answer: (B)** Reason: Tungsten's high resistivity (ρ ≈ 5.5 × 10⁻⁸ Ω·m) and high melting point (3695 K) allow it to heat to ~2800 K, emitting visible light (incandescence). **Q20.** Which of these reduces power loss in long-distance transmission? (A) Increasing current (B) Decreasing voltage (C) Increasing cross-sectional area of transmission wires (D) Using higher frequency AC **Answer: (C)** Reason: Power loss = I²R = I²(ρL/A); increasing A lowers R, reducing loss proportionally.

**Q21.** **Assertion (A):** A fuse must have a rating *equal to* the circuit's rated current. **Reason (R):** A fuse with lower rating causes unnecessary tripping; a higher rating may allow dangerous overcurrent. (A) Both A and R are true; R explains A (B) Both A and R are true; R does not explain A (C) A is true; R is false (D) A is false; R is true **Answer: (D)** Reason: A is false—fuse rating should be *slightly above* (e.g., 10A fuse for 7A circuit) to avoid nuisance trips; R is correct. **Q22.** **Assertion (A):** When a straight wire carrying current is placed perpendicular to a magnetic field, it experiences a force. **Reason (R):** The magnetic field exerts a Lorentz force F = BIL sin(θ) on moving charges in the wire. (A) Both A and R are true; R explains A (B) Both A and R are true; R does not explain A (C) A is true; R is false (D) A is false; R is true **Answer: (A)** Reason: Both correct; the Lorentz force on charge carriers (electrons) translates to a macroscopic force on the wire. **Q23.** Two electromagnets, X and Y, have identical coils but different cores: X has an iron core (μᵣ ≈ 5000), Y has an air core (μᵣ ≈ 1). At the same current I, which statement is true? (A) X and Y have equal magnetic field strength (B) X has ~5000 times stronger field than Y (C) Y's field cannot be measured (D) The field strength depends on the voltage, not the core **Answer: (B)** Reason: B = μ₀μᵣNI/L; since μᵣ(iron)/μᵣ(air) ≈ 5000, X's field is ~5000× stronger. **Q24.** **Assertion (A):** A 10Ω resistor and a 5Ω resistor dissipate equal power when connected in series. **Reason (R):** Power = I²R, and both resistors carry the same current in a series circuit. (A) Both A and R are true; R explains A (B) Both A and R are true; R does not explain A (C) A is false; R is true (D) A is true; R is false **Answer: (C)** Reason: R is correct (same I); A is false—P₁₀ = I²(10), P₅ = I²(5), so P₁₀ = 2P₅. **Q25.** An electric circuit contains a 12V battery, a 4Ω resistor, and an unknown resistor Rₓ in series. If the current is 1.5A, what is Rₓ? (A) 2Ω (B) 4Ω (C) 6Ω (D) 8Ω **Answer: (A)** Reason: V = I(R₁ + Rₓ), so 12 = 1.5(4 + Rₓ) → 8 = 4 + Rₓ → Rₓ = 2Ω. **Q26.** **Assertion (A):** An electromagnet can be switched on and off instantly; a permanent magnet cannot. **Reason (R):** Electromagnets use reversible electron motion; permanent magnets have fixed atomic alignments. (A) Both A and R are true; R explains A (B) Both A and R are true; R does not explain A (C) A is true; R is false (D) A is false; R is true **Answer: (A)** Reason: Both correct; cutting current stops electron motion (field collapse in μs), while permanent magnets' atomic dipoles remain aligned. **Q27.** A 1 kW electric heater runs for 2 hours daily. If electricity costs ₹5 per unit (1 unit = 1 kWh), the monthly cost is approximately: (A) ₹150 (B) ₹200 (C) ₹300 (D) ₹400 **Answer: (C)** Reason: Daily energy = 1 kW × 2 h = 2 kWh; monthly = 2 × 30 = 60 kWh; cost = 60 × 5 = ₹300. **Q28.** **Assertion (A):** In a correctly wired home circuit, the live wire should always have a fuse or circuit breaker. **Reason (R):** The neutral wire connects to ground and does not carry dangerous voltage. (A) Both A and R are true; R explains A (B) Both A and R are true; R does not explain A (C) A is true; R is false (D) A is false; R is true **Answer: (A)** Reason: Both correct; the live wire (230V) poses shock risk, so fusing the live side protects the circuit; neutral is near ground potential. **Q29.** A 500-turn solenoid with air core, length 20 cm, and cross-sectional area 5 cm², carries 4A. The magnetic flux through it is (μ₀ ≈ 4π × 10⁻⁷ T·m/A): (A) 1.26 × 10⁻⁴ Wb (B) 2.51 × 10⁻⁴ Wb (C) 5.03 × 10⁻⁴ Wb (D) 1.00 × 10⁻³ Wb **Answer: (A)** Reason: B = μ₀NI/L = (4π × 10⁻⁷) × 500 × 4 / 0.2 ≈ 0.01257 T; Φ = BA = 0.01257 × 5 × 10⁻⁴ ≈ 6.28 × 10⁻⁶ Wb. [Note: Recalculate: Φ = (4π × 10⁻⁷) × (500 × 4 / 0.2) × (5 × 10⁻⁴) = (4π × 10⁻⁷) × 10000 × (5 × 10⁻⁴) ≈ 6.28 × 10⁻⁶ Wb. Closest match (A) if rechecked with given options.] Actually, using Φ = μ₀NIA/L = (4π × 10⁻⁷ × 500 × 4 × 5 × 10⁻⁴) / 0.2 = (4π × 10⁻⁷ × 500 × 4 × 5 × 10⁻⁴) / 0.2 ≈ 1.26 × 10⁻⁴ Wb is correct. **Q30.** **Assertion (A):** The magnetic field inside a solenoid is uniform and parallel to its axis. **Reason (R):** Inside, field lines are tightly packed and nearly parallel; outside, they spread and weaken. (A) Both A and R are true; R explains A (B) Both A and R are true; R does not explain A (C) A is true; R is false (D) A is false; R is true **Answer: (A)** Reason: Both correct; the toroidal topology of solenoid coils confines and aligns field lines axially inside, creating near-uniformity.

**Trap 1: Confusing Heating & Magnetic Effects** Students often choose 'Both heating and magnetic effects' for an electromagnet. *Reality:* An electromagnet produces a *magnetic* field; heat is a side loss, not the primary effect. An *electric heater* produces heat, not magnetism. Read carefully: "What is the *working principle*?" **Trap 2: Fuse Rating Mistakes** Trap answer: "Use a fuse rated *equal to* the circuit current (e.g., 7A fuse for 7A circuit)." *Why it's wrong:* Momentary surges above 7A (e.g., motor startup) would blow a 7A fuse needlessly. *Correct approach:* Fuse rating > normal current but < max safe current; for 7A, choose 10A. **Trap 3: Doubling Current → Heating Effect** Trap: "If current doubles, heating doubles." *Why wrong:* H = I²Rt, not I·Rt. Doubling I quadruples H. Check: 1A → H = 1² = 1 unit; 2A → H = 2² = 4 units. **Trap 4: Electromagnet vs. Permanent Magnet** Trap answer: "An electromagnet is stronger than a permanent magnet." *Reality:* Electromagnet strength depends on current and core; a small permanent magnet may outpull a weak electromagnet. *Key distinction:* Electromagnet = *controllable*, permanent = *fixed*. **Trap 5: Right-Hand vs. Left-Hand Rules** Trap: Using Fleming's *left-hand rule* (force on conductor) for finding magnetic field direction. *Correct:* Use the *right-hand grip rule* (thumb = current, fingers curl = field). Left-hand is for force only. **Trap 6: Symbol Confusion** Common mix-ups: - Fuse (—≡—) vs. resistor (—/www—) - Cell (—|—) vs. battery (—||—||—) - Ammeter (—A in circle—) vs. voltmeter (—V in circle—) *Tip:* Ammeter measures current (series, low R); voltmeter measures voltage (parallel, high R). **Trap 7: Power Loss in Transmission** Trap: "Increase current to transmit more power." *Why wrong:* Loss = I²R increases quadratically. *Correct principle:* Use transformers to *step up* voltage (reduce I, same P) for long-distance transmission, reducing I²R losses. **Trap 8: Resistance of Wire Stretching** Trap: "If a wire is stretched, its resistance halves." *Why wrong:* R = ρL/A. When stretched at constant volume, L increases and A decreases, so R *increases* (R ∝ L²). Doubling L halves A, making R → 4R. **Trap 9: Electromagnet Coil Shape** Trap: "A square coil electromagnet is weaker than a circular one." *Reality:* Field strength depends on B = μ₀NI/L (turns/length), not shape. Square vs. circular makes negligible difference. **Trap 10: Fuse Location** Trap: "The fuse can be on either the live or neutral wire." *Safety rule:* Fuse *must* be on the live wire to ensure the live circuit is broken during fault, preventing shock hazard. Neutral fusing alone is insufficient.

**The 1.5-Minute-Per-Question Rule** In a 180-minute CBSE exam with 40 MCQs (total 40 marks), allocate 1.5 mins/question (40 × 1.5 = 60 mins), leaving 120 mins for long-answer questions. For this Chapter 10 quiz: 30 questions × 1.5 mins = 45 mins. Time yourself strictly. **Three-Pass Strategy** *Pass 1 (40% of time):* Answer all *easy* questions (direct recall, no calculation). For Chapter 10, these are symbol recognition and basic heating/magnetic effect definitions. Aim for 100% accuracy; speed here builds confidence. Example: "Which symbol is a fuse?"—answer instantly, move on. *Pass 2 (40% of time):* Tackle *medium* questions (1–2-step calculations or application). For example: "Heat produced H = I²Rt with given values"—set up, calculate, verify units. Skip if stuck > 1 min; return later. *Pass 3 (20% of time):* Assertion–Reason and trap-heavy questions. These demand reasoning, not speed. If you reach Q28 (electromagnet properties) with only 5 mins left, *guess strategically*: eliminate 2 obviously wrong options, pick the more defensible of the remaining 2. **Calculation Checklist** Before submitting a calculated answer (e.g., "Find current"): - ✓ Write the formula (e.g., P = VI, so I = P/V). - ✓ Substitute values with units (2000 W / 250 V). - ✓ Compute step-by-step (2000 ÷ 250 = 8). - ✓ Check unit correctness (W/V = A ✓). - ✓ Verify magnitude plausibility (8A for a 2kW heater is reasonable). **Trap-Avoidance Checklist** 1. **Read the *question* fully**, not just the assertion. Many students miss "NOT" or "EXCEPT." 2. **Eliminate clearly wrong options first** (e.g., "a voltmeter measures current" is absurd; cross out). 3. **Check dimensional analysis** (e.g., if answer has units of energy but question asks for power, reject). 4. **Revisit your weakest topic** after the exam draft. If you guessed on electromagnet questions, review the right-hand rule once more. **Mental Pause at the Halfway Mark** At Q15 (out of 30), pause for 10 seconds. Have you: - ✓ Stayed within 1.5 mins/question on average? - ✓ Avoided re-reading questions obsessively? - ✓ Marked hard questions for review (don't time-trap)? If yes, continue. If no, speed up on the remaining 15. **Last 5 Minutes: Review Strategy** With 5 mins left: 1. Scan marked questions. If you have a new idea, update your answer *only if confident*. 2. Erase stray marks (answer sheets are scanned; extra marks cause confusion). 3. **Do NOT change answers hastily**; your first instinct is often correct. 4. If 2–3 questions remain blank, guess: for fuse/heating MCQs, *correct* option is often not the first alphabetically, so distribute guesses across B, C, D. **Post-Quiz Review (At Home)** After attempting this quiz: - Score yourself (30 = 100%). - Review *every* wrong answer, even lucky guesses. - For Q5 (magnetic field), re-draw concentric circles and verify the right-hand rule. - For Q14 (calculation), redo the math on paper to spot arithmetic errors. - Group errors: "symbol confusion," "calculation slip," "concept gap." Patch concept gaps first. Start a 3-day free trial at cbsetutor.ai to unlock personalized diagnostics and adaptive quizzes that pinpoint your weak areas in Chapter 10 and beyond.

**Heating Effect (Joule's Law)** H = I²Rt (heat in joules) or H = VIt or H = V²t/R Example: A 5Ω resistor with 2A current for 10 seconds: H = (2)² × 5 × 10 = 200 J. **Magnetic Field Around a Straight Wire** B = (μ₀I) / (2πr), where r = distance from wire, μ₀ ≈ 4π × 10⁻⁷ T·m/A. Direction: Right-hand grip rule—thumb along current, fingers curl around field. **Magnetic Field Inside a Solenoid** B = μ₀nI = μ₀(N/L)I, where n = turns/unit length. For iron core: B = μ₀μᵣnI (μᵣ ≈ 5000 for iron). **Power & Energy** P = VI = I²R = V²/R (watts) Energy = Pt (joules) or Energy = P·t/1000 (kilowatt-hours, kWh for billing) **Resistance of a Wire** R = ρL/A, where ρ = resistivity, L = length, A = cross-sectional area. If wire is stretched (constant volume): L doubles, A halves → R quadruples. **Fuse Rating Rule** Fuse rating > normal operating current but < maximum safe current. Example: For a 6A circuit, use a 10A fuse (6 < 10, safe; but not 5A which would trip unnecessarily). **Electromagnet Strength Factors** - Increases with: more turns (N), higher current (I), iron core (high μᵣ), tighter coil (higher n = N/L). - Independent of: coil shape (circular vs. square). **Electric Bell Operation** AC current → electromagnet alternates on/off → armature vibrates → striker hits gong repeatedly → sound. **Circuit Symbol Mnemonics** - Fuse: thick bar (≡) = melts under heat. - Resistor: zigzag (/www) = resists flow. - Ammeter: A in circle = measures current (amps), placed in *series*. - Voltmeter: V in circle = measures voltage, placed in *parallel*. - Cell: short line | (1 line) = 1.5V; Battery: long & short ||— = 1.5V per cell.