Physics Objective Questions and Answers: Chapter 11 Electricity
Review structured MCQ sets for Class 10 Physics Chapter 11 Electricity. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.
Download Chapter 11 Electricity MCQs with Answers
Access the complete set of multiple-choice questions for Chapter 11 Electricity below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.
Multiple Choice Questions
(a) direct current
(b) electric power
(c) potential difference
(d) alternating current
Show Answer & Explanation
Answer: (d) alternating current
Explanation:
1. Generators at power stations produce alternating current.
2. AC can be stepped up by transformers for long-distance transmission.
Teacher's Note:
1. AC is chosen because transformers reduce transmission losses.
(a) 3 : 5
(b) 5 : 3
(c) 4 : 5
(d) 5 : 4
Show Answer & Explanation
Answer: (a) 3 : 5
Explanation:
1. \( R_A = 220^2/100 = 484\ \Omega \), \( R_B = 220^2/60 \approx 806.7\ \Omega \).
2. Same current, so \( H \propto R \).
3. \( H_A:H_B = 484:806.7 = 3:5 \).
Teacher's Note:
1. For series connection use \( P = I^2R \).
2. The ratio of resistances is inverse to the ratio of the wattage ratings.
(a) only the potential difference across its ends
(b) only the current flowing through it
(c) both the potential difference and the current
(d) neither the potential difference nor the current
Show Answer & Explanation
Answer: (d) neither the potential difference nor the current
Explanation:
1. \( R = \rho l/A \) depends on material, length, area and temperature.
2. \( V/I \) stays constant for an ohmic conductor, so R depends on neither V nor I.
Teacher's Note:
1. V and I change together, their ratio does not.
(a) 0.2 A
(b) 1.0 A
(c) 1.8 A
(d) 1.9 A
Show Answer & Explanation
Answer: (d) 1.9 A
Explanation:
1. Least count = 2/10 = 0.2 A.
2. Readings are multiples of 0.2 A: 0.2, 1.0, 1.8 are possible.
3. 1.9 A is not a multiple of 0.2 A.
Teacher's Note:
1. Least count = range ÷ number of divisions.
(a) thin, long and hot
(b) thick, short and cool
(c) thick, long and hot
(d) thin, short and cool
Show Answer & Explanation
Answer: (b) thick, short and cool
Explanation:
1. R is smaller for larger area (thick), smaller length (short) and lower temperature (cool).
Teacher's Note:
1. Metals' resistance increases with temperature.
(a) 5 V
(b) 2 V
(c) 4 V
(d) 20 V
Show Answer & Explanation
Answer: (a) 5 V
Explanation:
1. Total R = 80 Ω, \( I = 20/80 = 0.25 \) A.
2. Drop across one = 0.25 × 20 = 5 V.
Teacher's Note:
1. Equal resistors in series share the voltage equally.
(a) 1 ampere × 1 second = 1 coulomb
(b) 1 coulomb × 1 joule = 1 volt
(c) 1 volt × 1 coulomb = 1 joule
(d) 1 volt × 1 ampere = 1 joule per second
Show Answer & Explanation
Answer: (b) 1 coulomb × 1 joule = 1 volt
Explanation:
1. \( q=It \) is correct.
2. \( V = W/q \), so 1 V = 1 J/C, not 1 C × 1 J.
3. \( W = qV \) and \( P = VI \) are correct.
Teacher's Note:
1. Check each statement through units.
(a) 200 megawatt hour
(b) 4800 megawatt hour
(c) 4800 megawatt
(d) 4800 joule
Show Answer & Explanation
Answer: (b) 4800 megawatt hour
Explanation:
1. Energy = power × time = 200 MW × 24 h = 4800 MWh.
Teacher's Note:
1. Power is in watts, energy in watt hour or joule.
(a) 25000 Ω
(b) 625 Ω
(c) 25 Ω
(d) 2.5 Ω
Show Answer & Explanation
Answer: (b) 625 Ω
Explanation:
1. \( R = V^2/P = 250^2/100 = 625\ \Omega \).
Teacher's Note:
1. Use the rated V and P to find R.
(a) 90 W
(b) 80 W
(c) 70 W
(d) 75 W
Show Answer & Explanation
Answer: (a) 90 W
Explanation:
1. Series: \( P_s = V^2/(3R) = 10 \) W.
2. Parallel: \( R_p = R/3 \), so \( P_p = 3V^2/R \).
3. \( P_p/P_s = 9 \), so \( P_p = 90 \) W.
Teacher's Note:
1. Parallel combination of n equal resistors dissipates n² times the series power.
(a) high resistance and high melting point
(b) low resistance and high melting point
(c) low resistance and low melting point
(d) high resistance and low melting point
Show Answer & Explanation
Answer: (d) high resistance and low melting point
Explanation:
1. A fuse is a deliberate weak link made of a tin-lead alloy.
2. High resistance gives strong heating, and low melting point makes it melt quickly at excess current.
Teacher's Note:
1. A fuse is always connected in series with the live wire.
(a) not change
(b) increase by 100%
(c) decrease by 400%
(d) increase by 400%
Show Answer & Explanation
Answer: (a) not change
Explanation:
1. \( P = I^2R \).
2. New \( P' = (2I)^2 \times (R/4) = 4I^2 R/4 = I^2R \).
Teacher's Note:
1. Apply the squares: current doubles, power gets 4 times, resistance gives 1/4.
(a) (i) 100 W bulb, (ii) 50 W bulb
(b) (i) 50 W bulb, (ii) 100 W bulb
(c) (i) both equally, (ii) 100 W bulb
(d) none of these
Show Answer & Explanation
Answer: (b) (i) 50 W bulb, (ii) 100 W bulb
Explanation:
1. The 50 W bulb has the larger resistance.
2. Series: same current, \( P = I^2R \), larger R is brighter, so the 50 W bulb.
3. Parallel: same voltage, \( P = V^2/R \), smaller R is brighter, so the 100 W bulb.
Teacher's Note:
1. Series favours high resistance, parallel favours low resistance.
(a) 0.02 A
(b) 0.01 A
(c) 0.2 A
(d) 0.1 A
Show Answer & Explanation
Answer: (d) 0.1 A
Explanation:
1. Least count = 2 A ÷ 20 = 0.1 A.
Teacher's Note:
1. Use only the divisions between the two marks.
(a) 600 V
(b) 500 V
(c) 300 V
(d) 100 V
Show Answer & Explanation
Answer: (c) 300 V
Explanation:
1. \( V = W/q = 6\times10^{-4}/(2\times10^{-6}) = 300 \) V.
Teacher's Note:
1. 1 µC = 10⁻⁶ C.
(a) 40 V
(b) 20 V
(c) 10 V
(d) 4 V
Show Answer & Explanation
Answer: (d) 4 V
Explanation:
1. \( V = IR = 0.2\times20 = 4 \) V.
Teacher's Note:
1. Direct use of Ohm's law.
(a) 4 Ω
(b) 3 Ω
(c) 2 Ω
(d) 1 Ω
Show Answer & Explanation
Answer: (d) 1 Ω
Explanation:
1. \( 1/R = 4\times(1/4) = 1 \), so \( R = 1\ \Omega \).
Teacher's Note:
1. n equal resistors R in parallel give R/n.
(a) double
(b) half
(c) one fourth
(d) four times
Show Answer & Explanation
Answer: (c) one fourth
Explanation:
1. \( H = I^2Rt \); with I halved, H becomes one fourth.
Teacher's Note:
1. Heat is proportional to the square of the current.
(a) 160 × 10⁻¹⁹ erg
(b) 100 J
(c) 160 × 10⁻¹⁹ J
(d) 100 erg
Show Answer & Explanation
Answer: (c) 160 × 10⁻¹⁹ J
Explanation:
1. Gain in KE = eV = 1.6 × 10⁻¹⁹ × 100 = 1.6 × 10⁻¹⁷ J.
2. This is 160 × 10⁻¹⁹ J.
Teacher's Note:
1. 1 erg = 10⁻⁷ J, so the erg options are far too small.
(a) 3R
(b) 2R
(c) R
(d) 2R/3
Show Answer & Explanation
Answer: (d) 2R/3
Explanation:
1. Between two vertices there are two paths: one resistor R and two resistors in series (2R).
2. \( R_{eq} = \dfrac{R\times2R}{R+2R} = \dfrac{2R}{3} \).
Teacher's Note:
1. The two paths are in parallel.
(a) 30 Ω
(b) 5 Ω
(c) 10 Ω
(d) 20 Ω
Show Answer & Explanation
Answer: (d) 20 Ω
Explanation:
1. Rated current \( I = P/V = 500/100 = 5 \) A.
2. Voltage to be dropped \( = 200-100 = 100 \) V.
3. \( R = 100/5 = 20\ \Omega \).
Teacher's Note:
1. Series resistor takes the extra voltage while carrying the rated current.
(a) 200√2 Ω
(b) 800 Ω
(c) 200 Ω
(d) 400 Ω
Show Answer & Explanation
Answer: (d) 400 Ω
Explanation:
1. Volume constant: length ×2 means area ÷2.
2. \( R' = \rho(2l)/(A/2) = 4R = 400\ \Omega \).
Teacher's Note:
1. Recasting keeps volume, so area changes inversely to length.
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Download Chapter MCQs: Class 10 Physics Chapter 11 Electricity
Download Multiple Choice Questions: Chapter 11 Electricity (Class 10 Physics)
Review structured objective questions for Class 10 Physics Chapter 11 Electricity. Built according to official CBSE guidelines, these MCQ sets support daily revision and core concept reinforcement.
Concept Clarification for Chapter 11 Electricity
Each question includes structured solution keys mapped directly to standard CBSE textbooks, helping students evaluate their reasoning and correct mistakes early in their revision.
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FAQs
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