CBSE Class 10 Physics Electricity MCQs Set 18

Download CBSE MCQs for Class 10 Physics: Chapter 11 Electricity

Explore reliable objective questions for Chapter 11 Electricity tailored for Class 10 learners. Utilizing these Physics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Chapter-wise Objective Questions: Chapter 11 Electricity

View or download the dedicated Chapter 11 Electricity MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Physics evaluations.

Multiple Choice Questions

Question 1: What is the SI unit of electric current?
(a) volt
(b) watt
(c) ampere
(d) joule
Show Answer & Explanation

Answer: (c) ampere

Explanation:
1. Electric current is the rate of flow of charge, I = Q/t.
2. Its SI unit is the ampere (1 A = 1 C s-1).

Teacher's Note:
1. Volt measures potential difference.
2. Watt measures power and joule measures energy.

Question 2: Two resistors A and B in parallel have an effective resistance of 12 Ω. Resistor A has colour bands brown, yellow, black (tolerance neglected). What are the colour bands of B? (Colour code: black 0, brown 1, red 2, orange 3, yellow 4, grey 8)
(a) grey, yellow, black
(b) grey, red, black
(c) grey, orange, black
(d) grey, red, orange
Show Answer & Explanation

Answer: (a) grey, yellow, black

Explanation:
1. A: brown (1), yellow (4), black (×1) gives RA = 14 Ω.
2. 12 = 14RB/(14 + RB) gives RB = 84 Ω.
3. 84 Ω = grey (8), yellow (4), black (×1).

Teacher's Note:
1. The first two bands are digits, the third is the multiplier.
2. Parallel resistance is always less than the smaller resistance.

Question 3: Assertion (A): 1 horse power = 746 watts.
Reason (R): Horse power is a unit of electric potential.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) Both A and R are false.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. One horse power equals 746 W, so A is true.
2. Horse power is a unit of power, not of electric potential, so R is false.

Teacher's Note:
1. Electric potential is measured in volts.
2. Horse power is an old unit of power.

Question 4: Which of the following is correct?
(a) Resistance of a conductor depends on its mass.
(b) Resistance of a conductor is directly proportional to its length.
(c) Resistance of a conductor depends on temperature.
(d) Both (b) and (c).
Show Answer & Explanation

Answer: (d) Both (b) and (c).

Explanation:
1. R = ρℓ/A, so R is directly proportional to length ℓ.
2. The resistivity ρ also changes with temperature, so R depends on temperature.
3. Mass by itself is not a factor, so only (b) and (c) are correct.

Teacher's Note:
1. Resistance depends on length, area, material and temperature.
2. The statements (b) and (c) are both true.

Question 5: It takes 20 minutes to boil water in an electric kettle. By changing the length of the coil, using the same supply, the water is to be boiled in 14 minutes. The coil length must be changed by ______ of the original length.
(a) \( \frac{7}{10} \)th
(b) \( \frac{10}{7} \)th
(c) \( \frac{3}{10} \)th
(d) \( \frac{10}{3} \)rd
Show Answer & Explanation

Answer: (c) \( \frac{3}{10} \)th

Explanation:
1. Heat H = V2t/R is the same, so t ∝ R ∝ ℓ.
2. ℓ2/ℓ1 = 14/20 = 7/10, so the new length is 7ℓ1/10.
3. The change = ℓ1 - 0.7ℓ1 = 3ℓ1/10, i.e. a reduction of 3/10th.

Teacher's Note:
1. For the same heat and voltage, time is proportional to resistance.
2. Resistance is proportional to length.

Question 6: A 200 V battery drives a circuit in which the ammeter reads 5 A. Two parallel branches each consist of two 50 Ω resistors in series, and a copper voltameter is connected across the same two points P and Q (in parallel with both branches). What mass of copper is deposited in 30 min? (e.c.e. of copper = 0.000332 g C-1)
(a) 1.494 kg
(b) 0.5976 g
(c) 1494 kg
(d) 0.5976 kg
Show Answer & Explanation

Answer: (b) 0.5976 g

Explanation:
1. Total resistance = 200/5 = 40 Ω; the two 100 Ω branches in parallel give 50 Ω.
2. 1/40 = 1/50 + 1/Rv gives Rv = 200 Ω, so the branch currents are in the ratio 2 : 2 : 1 and the voltameter current is 5 × 1/5 = 1 A.
3. m = Zit = 0.000332 × 1 × 1800 = 0.5976 g.

Teacher's Note:
1. Current divides inversely with resistance in parallel branches.
2. m = Zit gives the deposit in grams when Z is in g C-1.

Question 7: 200 J of work is done in moving a 10 C charged particle between two points in a uniform electric field of 64 V m-1 along the field. What is the distance between the points?
(a) \( \frac{7}{17} \) m
(b) \( \frac{9}{17} \) m
(c) \( \frac{7}{15} \) m
(d) \( \frac{5}{16} \) m
Show Answer & Explanation

Answer: (d) \( \frac{5}{16} \) m

Explanation:
1. Force F = qE = 10 × 64 = 640 N.
2. W = F × s gives s = 200/640 = 5/16 m.

Teacher's Note:
1. Work = force × displacement along the field.
2. Electric force on a charge is qE.

Question 8: Match Column A with Column B.
(A) Galvanometer (B) Rheostat (C) Voltameter
(P) electrolysis (Q) variable resistor (R) electric current
(a) A → R, B → Q, C → P
(b) A → P, B → R, C → Q
(c) A → Q, B → R, C → P
(d) A → R, B → P, C → Q
Show Answer & Explanation

Answer: (a) A → R, B → Q, C → P

Explanation:
1. A galvanometer detects electric current (R).
2. A rheostat is a variable resistor (Q).
3. A voltameter is used for electrolysis (P).

Teacher's Note:
1. Each instrument is named after its function.
2. A rheostat changes the current by changing resistance.

Question 9: A 20 μC charge placed at a point experiences a force of 20 N. What is the electric field strength at that point?
(a) 103 N C-1
(b) 106 N C-1
(c) 109 N C-1
(d) 1012 N C-1
Show Answer & Explanation

Answer: (b) 106 N C-1

Explanation:
1. E = F/q = 20 / (20 × 10-6) = 106 N C-1.

Teacher's Note:
1. Electric field is force per unit charge.
2. 1 μC = 10-6 C.

Question 10: A total of 1019 electrons flow through a conductor of resistance 20 Ω in one second. What is the potential difference across it? (e = 1.6 × 10-19 C)
(a) 32 V
(b) 64 V
(c) 16 V
(d) 20 V
Show Answer & Explanation

Answer: (a) 32 V

Explanation:
1. Charge = 1019 × 1.6 × 10-19 = 1.6 C in 1 s, so I = 1.6 A.
2. V = IR = 1.6 × 20 = 32 V.

Teacher's Note:
1. Current = charge per second.
2. Then apply Ohm's law.

Question 11: A bulb is connected to a battery through a resistor. If another resistor is connected in parallel with the first resistor, what happens to the brightness of the bulb?
(a) increases
(b) remains the same
(c) decreases
(d) increases or decreases depending upon the resistance of the filament
Show Answer & Explanation

Answer: (a) increases

Explanation:
1. Adding a resistor in parallel with the first resistor lowers the total resistance of the circuit.
2. The battery current, which also flows through the bulb, increases, so the bulb becomes brighter.

Teacher's Note:
1. Parallel addition always reduces effective resistance.
2. Brightness depends on the current in the bulb.

Question 12: Two wires of the same material have lengths in the ratio 1 : 3 and volumes in the ratio 2 : 3. What is the ratio of their resistances?
(a) 1 : 3
(b) 3 : 2
(c) 1 : 4
(d) 1 : 6
Show Answer & Explanation

Answer: (d) 1 : 6

Explanation:
1. For a fixed volume V = Aℓ, R = ρℓ/A = ρℓ2/V.
2. R1/R2 = (ℓ1/ℓ2)2 × (V2/V1) = (1/9) × (3/2) = 1/6.

Teacher's Note:
1. Write R in terms of ℓ and V.
2. Same material means the same ρ.

Question 13: Find the effective resistance between A and B: a parallel pair of 1 Ω and 2 Ω, in series with a 3 Ω resistor, in series with a parallel pair of 4 Ω and 2 Ω.
(a) 5 Ω
(b) 10 Ω
(c) 15 Ω
(d) 20 Ω
Show Answer & Explanation

Answer: (a) 5 Ω

Explanation:
1. 1 Ω and 2 Ω in parallel: (1 × 2)/(1 + 2) = 2/3 Ω.
2. 4 Ω and 2 Ω in parallel: (4 × 2)/(4 + 2) = 4/3 Ω.
3. Total = 2/3 + 3 + 4/3 = 5 Ω.

Teacher's Note:
1. Combine parallel groups first, then add in series.
2. Product over sum works for two resistors.

Question 14: What is the resistance of a bulb rated 100 W - 230 V?
(a) 838 Ω
(b) 434 Ω
(c) 529 Ω
(d) 683 Ω
Show Answer & Explanation

Answer: (c) 529 Ω

Explanation:
1. P = V2/R gives R = 2302/100 = 529 Ω.

Teacher's Note:
1. Rated values give the resistance when the bulb works normally.
2. R = V2/P.

Question 15: In a copper voltameter circuit, a 25 Ω resistor is connected in series with the voltameter and a battery. The heat produced in the resistor in one hour is 3,60,000 J. What mass of copper is deposited at the cathode in one hour? (e.c.e. of copper = 0.0003294 g C-1)
(a) 4.6 g
(b) 0.85 g
(c) 1.32 g
(d) 2.37 g
Show Answer & Explanation

Answer: (d) 2.37 g

Explanation:
1. H = I2Rt gives 360000 = I2 × 25 × 3600, so I2 = 4 and I = 2 A.
2. m = Zit = 0.0003294 × 2 × 3600 = 2.37 g.

Teacher's Note:
1. Series elements carry the same current.
2. Find the current from the heat first.

Question 16: What is the SI unit of potential difference?
(a) volt
(b) watt
(c) ampere
(d) joule
Show Answer & Explanation

Answer: (a) volt

Explanation:
1. Potential difference is work done per unit charge, V = W/Q.
2. Its SI unit is the volt (1 V = 1 J C-1).

Teacher's Note:
1. 1 volt = 1 joule per coulomb.
2. The watt is for power.

Question 17: Resistors A (colour bands blue, yellow, black) and B (red, yellow, black) are connected in parallel. What is their effective resistance? (Colour code: red 2, yellow 4, blue 6, black 0)
(a) 13.5 Ω
(b) 17.5 Ω
(c) 20.5 Ω
(d) 15.7 Ω
Show Answer & Explanation

Answer: (b) 17.5 Ω

Explanation:
1. RA = 64 Ω and RB = 24 Ω.
2. R = (64 × 24)/(64 + 24) = 17.5 Ω.

Teacher's Note:
1. Blue-yellow-black means 6, 4 and ×1.
2. Parallel resistance is less than 24 Ω.

Question 18: Assertion (A): 1 kilowatt = 1000 J s-1.
Reason (R): Watt and J s-1 are both SI units of power.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) Both A and R are false.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. 1 kW = 1000 W and 1 W = 1 J s-1, so A is true.
2. R is true, and it explains why a kilowatt equals 1000 J s-1.

Teacher's Note:
1. 1 watt = 1 joule per second.
2. Kilo means 1000.

Question 19: Which of the following is correct?
(a) The emf of a new dry cell is greater than that of an identical cell used for 3 months.
(b) The internal resistance of a new cell must be greater than that of an old one.
(c) The emf of a new cell is less than that of an old cell and its internal resistance is greater.
(d) The emf is a characteristic of a cell and is different for different cells.
Show Answer & Explanation

Answer: (d) The emf is a characteristic of a cell and is different for different cells.

Explanation:
1. The emf depends on the chemicals of the cell (for example about 1.5 V for a Leclanche cell and 1.08 V for a voltaic cell).
2. So emf is a characteristic of the cell type and differs between different cells; the other statements are not generally true.

Teacher's Note:
1. Emf is set by the chemistry of the cell.
2. Internal resistance of a cell rises as it ages.

Question 20: It takes 5 minutes to boil water in an electric kettle. If the coil length is halved, how long will the same amount of water take with the same supply?
(a) 2.5 min
(b) 5 min
(c) 10 min
(d) 20 min
Show Answer & Explanation

Answer: (a) 2.5 min

Explanation:
1. Halving the length halves the resistance.
2. t = HR/V2 so t is proportional to R; the time becomes 2.5 min.

Teacher's Note:
1. For the same heat and voltage, t ∝ R.
2. Shorter coil means lower resistance and faster heating.

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Chapter 11 Electricity Objective Questions & Solutions for Class 10 Physics

About Chapter 11 Electricity MCQs for Class 10 Physics

Test your conceptual understanding of Chapter 11 Electricity with these targeted multiple-choice questions. Designed in alignment with the latest CBSE curriculum for Class 10 Physics, these problem sets build accuracy and prepare students for objective exams.

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FAQs

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