Science Objective Questions and Answers: Chapter 11 Electricity
Review structured MCQ sets for Class 10 Science 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
Navigate directly to the 50 objective questions for Chapter 11 Electricity using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.
This Class 10 Science Chapter 11 Electricity MCQs set gives 40 multiple choice questions with answers and step-by-step explanations on electric current, charge, potential difference and Ohm's law. The questions cover counting electrons, finding resistance from voltage and current, units of electrical quantities, the use of ammeter and voltmeter with their least count, and six assertion-reason questions.
Multiple Choice Questions
(a) negative to positive potential
(b) positive to negative potential
(c) any of the above two
(d) none of the above two
Show Answer & Explanation
Answer: (b) positive to negative potential
Explanation:
1. By convention the direction of electric current is the direction in which positive charge would flow.
2. Positive charge moves from higher (positive) potential to lower (negative) potential.
Teacher's Note:
1. Electrons actually move in the opposite direction.
2. The convention was set before the electron was discovered.
(a) A to B
(b) B to A
(c) It will not flow
(d) None of these
Show Answer & Explanation
Answer: (b) B to A
Explanation:
1. By convention the direction of electric current is taken opposite to the direction of flow of electrons.
2. Electrons flow from A to B, so the conventional current flows from B to A.
Teacher's Note:
1. Electrons are negatively charged, so they move opposite to the conventional current.
2. Inside a wire it is the electrons that actually move.
(a) A
(b) B
(c) Both have equal potential
(d) Both have lower potential
Show Answer & Explanation
Answer: (a) A
Explanation:
1. Conventional current flows from the point of higher potential to the point of lower potential.
2. Current flows from A to B, so A is at the higher potential.
Teacher's Note:
1. No current flows between points at the same potential.
2. The direction of conventional current is opposite to the direction of flow of electrons.
(a) ions
(b) amperes
(c) electrons
(d) protons
Show Answer & Explanation
Answer: (c) electrons
Explanation:
1. A metallic conductor has a large number of free electrons.
2. When a potential difference is applied, these free electrons drift and constitute the electric current.
Teacher's Note:
1. Ions carry current in electrolytes, not in metals.
2. Ampere is the unit of current and not a particle.
(a) Power
(b) Charge
(c) e.m.f.
(d) Energy
Show Answer & Explanation
Answer: (b) Charge
Explanation:
1. Current I = Q/t, so charge Q = I × t.
2. The product of ampere and second is therefore the unit of charge, which is the coulomb.
Teacher's Note:
1. 1 coulomb = 1 ampere × 1 second.
2. Ampere is a unit of current, and second is a unit of time.
(I) If in an electric circuit, the resistance is doubled, the voltage also gets doubled.
(II) If in an electric circuit, the resistance is halved, the current gets doubled.
(III) Variable resistance is a component used to regulate voltage without changing the current.
(IV) In an electric circuit, a device called rheostat is used to change the resistance in the circuit.
Select the INCORRECT statements:
(a) Both (I) and (II)
(b) Both (I) and (III)
(c) Both (II) and (IV)
(d) (I), (III) and (IV)
Show Answer & Explanation
Answer: (b) Both (I) and (III)
Explanation:
1. Statement (I) is incorrect: the voltage is fixed by the source; if R is doubled, the current becomes half (I = V/R).
2. Statement (II) is correct: for constant V, halving R doubles I.
3. Statement (III) is incorrect: a variable resistance regulates (changes) the current in the circuit.
4. Statement (IV) is correct: a rheostat is a variable resistance. So the incorrect statements are (I) and (III).
Teacher's Note:
1. Ohm's law I = V/R tells how current changes when R changes at constant V.
2. A rheostat is used to control the current, not to change the source voltage.
(a) 1 joule/coulomb
(b) 1 coulomb/joule
(c) 1 joule/coulomb2
(d) 1 joule × coulomb
Show Answer & Explanation
Answer: (a) 1 joule/coulomb
Explanation:
1. Potential difference V = W/Q.
2. So 1 volt = 1 joule of work per 1 coulomb of charge = 1 J/C.
Teacher's Note:
1. Potential difference is work done per unit charge.
2. Its SI unit is volt.
(a) the amount of work done in moving a unit charge by a unit distance
(b) distance between two terminals
(c) length of the connecting wire
(d) the amount of work done in moving a unit charge between the two points
Show Answer & Explanation
Answer: (d) the amount of work done in moving a unit charge between the two points
Explanation:
1. Potential difference between two points is the work done to move a unit positive charge from one point to the other.
2. It is measured in volt (joule per coulomb).
Teacher's Note:
1. It is work per unit charge and not work per unit distance.
2. V = W/Q.
(a) Work done / (Current × Time)
(b) Work done × Charge
(c) (Work done × Time) / Current
(d) Work done × Charge × Time
Show Answer & Explanation
Answer: (a) Work done / (Current × Time)
Explanation:
1. Potential difference (voltage) V = work done/charge.
2. Charge Q = current × time, so V = work done/(current × time).
Teacher's Note:
1. Voltage is the work done per unit charge.
2. Replace charge by current × time to get the required form.
(a) ohmic conductors
(b) non-ohmic conductors
(c) insulators
(d) superconductors
Show Answer & Explanation
Answer: (a) ohmic conductors
Explanation:
1. Materials in which the potential difference is directly proportional to the current (V-I graph is a straight line through the origin) obey Ohm’s law.
2. Such materials are called ohmic conductors.
Teacher's Note:
1. Ohm’s law holds for metallic conductors at constant temperature.
2. Devices such as diodes do not obey Ohm’s law and are called non-ohmic.
(a) potential difference and length
(b) current and length
(c) current and cross-sectional area
(d) current and potential difference
Show Answer & Explanation
Answer: (d) current and potential difference
Explanation:
1. Ohm’s law says V ∝ I, or V = RI.
2. The constant of proportionality R connects the potential difference and the current.
Teacher's Note:
1. Ohm’s law relates only V and I.
2. R = V/I is constant for an ohmic conductor at constant temperature.
(a) only the current is doubled
(b) only the current is halved
(c) only the resistance is doubled
(d) only the resistance is halved
Show Answer & Explanation
Answer: (a) only the current is doubled
Explanation:
1. The resistance of the resistor stays the same.
2. By Ohm’s law I = V/R, so doubling V doubles the current.
Teacher's Note:
1. Resistance does not depend on the applied voltage.
2. Only the current changes.
(a) current is doubled.
(b) current is halved.
(c) current remains same.
(d) current is four times.
Show Answer & Explanation
Answer: (c) current remains same.
Explanation:
1. I = V/R.
2. With V → 2V and R → 2R, I′ = 2V/2R = V/R = I.
Teacher's Note:
1. If the numerator and the denominator are both doubled, the ratio does not change.
2. So the current remains the same.
(a) current will decrease
(b) resistance will increase
(c) resistance will decrease
(d) current will increase
Show Answer & Explanation
Answer: (d) current will increase
Explanation:
1. Resistance of a conductor at constant temperature is fixed.
2. By Ohm’s law I = V/R, so if V increases the current increases in the same ratio.
Teacher's Note:
1. Resistance does not depend on the applied voltage or the current.
2. Current is directly proportional to the voltage across an ohmic conductor.
(a) charge
(b) reciprocal of resistance
(c) resistance
(d) reciprocal of charge
Show Answer & Explanation
Answer: (b) reciprocal of resistance
Explanation:
1. With V on the x-axis and I on the y-axis, slope = I/V.
2. Since R = V/I, I/V = 1/R, which is the reciprocal of resistance.
Teacher's Note:
1. If V is on the y-axis and I on the x-axis, the slope gives the resistance itself.
2. Always check which quantity is on which axis.
(a) Filament of a bulb
(b) LED
(c) Nichrome
(d) Transistor
Show Answer & Explanation
Answer: (c) Nichrome
Explanation:
1. Ohm’s law is obeyed by materials whose resistance stays constant, such as metallic wires and alloys like nichrome.
2. The filament of a bulb changes its resistance as it heats, and an LED and a transistor are non-ohmic devices.
Teacher's Note:
1. Ohm’s law is valid only when the temperature and other physical conditions are constant.
2. A nichrome wire has a nearly constant resistance, so its V-I graph is a straight line.
(a) 2.5 A
(b) 3.5 A
(c) 4.5 A
(d) 5.5 A
Show Answer & Explanation
Answer: (a) 2.5 A
Explanation:
1. Charge Q = 150 C and time t = 1 minute = 60 s.
2. Current I = Q/t = 150/60 = 2.5 A.
Teacher's Note:
1. Always convert time into seconds before using I = Q/t.
2. One ampere means one coulomb of charge flowing per second.
(a) 1200 C
(b) 150 C
(c) 18 C
(d) 1.8 C
Show Answer & Explanation
Answer: (a) 1200 C
Explanation:
1. Time t = 2 minutes = 120 s.
2. Charge Q = It = 10 × 120 = 1200 C.
Teacher's Note:
1. Charge = current × time.
2. Convert minutes to seconds first.
(a) 75 × 1020
(b) 70 × 1015
(c) 60 × 1015
(d) 11 × 1012
Show Answer & Explanation
Answer: (a) 75 × 1020
Explanation:
1. Charge passed Q = It = 10 × 120 = 1200 C.
2. Number of electrons n = Q/e = 1200/(1.6 × 10−19) = 7.5 × 1021 = 75 × 1020.
Teacher's Note:
1. Divide the total charge by the charge of one electron (1.6 × 10⁻¹⁹ C).
2. 7.5 × 10²¹ can be written as 75 × 10²⁰.
(a) 1020
(b) 1016
(c) 1018
(d) 1023
Show Answer & Explanation
Answer: (a) 1020
Explanation:
1. Charge Q = It = 1 × 16 = 16 C.
2. Number of electrons n = Q/e = 16 / (1.6 × 10−19) = 1020.
Teacher's Note:
1. One coulomb of charge contains 6.25 × 1018 electrons.
2. First find the charge using Q = It and then divide by the charge of one electron.
(a) 10
(b) 40
(c) 0.005
(d) 0.05
Show Answer & Explanation
Answer: (b) 40
Explanation:
1. I = Q/t = 20/0.5 = 40 A.
Teacher's Note:
1. Divide the charge by the time in seconds.
2. Dividing by 0.5 is the same as multiplying by 2.
(a) 640 mA
(b) 64 mA
(c) 4.0 mA
(d) 40 mA
Show Answer & Explanation
Answer: (a) 640 mA
Explanation:
1. Charge emitted per second = 4 × 1018 × 1.6 × 10−19 C = 0.64 C.
2. Current I = 0.64 C/1 s = 0.64 A = 640 mA.
Teacher's Note:
1. Current is charge per second, so multiply the number of electrons per second by the electronic charge.
2. 1 A = 1000 mA.
(a) 1800 C
(b) 3600 C
(c) 3000 C
(d) 2400 C
Show Answer & Explanation
Answer: (a) 1800 C
Explanation:
1. Time t = 1 hour = 3600 s.
2. Charge Q = It = 0.5 × 3600 = 1800 C.
Teacher's Note:
1. The power rating (60 W) is not needed to find the charge.
2. Convert the time to seconds.
(a) 0.625 × 10−19
(b) 1.6 × 10−19
(c) 1.6 × 1019
(d) 0.625 × 1019
Show Answer & Explanation
Answer: (d) 0.625 × 1019
Explanation:
1. Charge flowing per second = 1 C.
2. Number of electrons n = 1/(1.6 × 10−19) = 6.25 × 1018 = 0.625 × 1019.
Teacher's Note:
1. 1 coulomb of charge contains about 6.25 × 10¹⁸ electrons.
2. Write the answer in the form given in the options.
(a) 6.25 × 1012
(b) 6.25 × 1015
(c) 6.25 × 1018
(d) 6.25 × 1019
Show Answer & Explanation
Answer: (a) 6.25 × 1012
Explanation:
1. I = 1 μA = 10−6 A, so the charge flowing in one second is 10−6 C.
2. n = 10−6/(1.6 × 10−19) = 6.25 × 1012.
Teacher's Note:
1. 1 μA = 10⁻⁶ A.
2. Number of electrons = charge ÷ charge on one electron.
(a) 1500 volt
(b) 150 volt
(c) 0.15 volt
(d) 0.0015 volt
Show Answer & Explanation
Answer: (a) 1500 volt
Explanation:
1. Potential difference = work done / charge.
2. VB − VA = 15/0.01 = 1500 V.
Teacher's Note:
1. 1 volt means 1 joule of work per coulomb of charge.
2. Divide the work by the charge.
(a) 2 A
(b) 4 A
(c) 10 A
(d) 16 A
Show Answer & Explanation
Answer: (c) 10 A
Explanation:
1. Potential difference V = W/Q = 4000/10 = 400 V.
2. Current I = V/R = 400/40 = 10 A.
Teacher's Note:
1. First find the potential difference from the work done per unit charge.
2. Then use Ohm’s law.
(a) 4 Ω
(b) 40 Ω
(c) 400 Ω
(d) 0.4 Ω
Show Answer & Explanation
Answer: (b) 40 Ω
Explanation:
1. Current I = 100 mA = 0.1 A and V = 4 V.
2. R = V/I = 4/0.1 = 40 Ω.
Teacher's Note:
1. Convert milliampere to ampere before using Ohm's law.
2. 1 mA = 10−3 A.
(a) 90 Ω
(b) 9 Ω
(c) 0.9 Ω
(d) 900 Ω
Show Answer & Explanation
Answer: (a) 90 Ω
Explanation:
1. R = V/I = 9/0.1 = 90 Ω.
Teacher's Note:
1. Use Ohm’s law R = V/I.
2. Dividing by 0.1 is the same as multiplying by 10.
(a) zero
(b) 0.50 A
(c) 2.0 A
(d) 5.0 A
Show Answer & Explanation
Answer: (c) 2.0 A
Explanation:
1. I = V/R = 10/5 = 2.0 A.
Teacher's Note:
1. Use Ohm’s law I = V/R.
2. Check that the units are in volt and ohm.
(a) 20 Ω
(b) 5 Ω
(c) 12 Ω
(d) 8 Ω
Show Answer & Explanation
Answer: (b) 5 Ω
Explanation:
1. R = V/I = 10/2 = 5 Ω.
Teacher's Note:
1. Apply Ohm’s law directly.
2. Resistance is the ratio of voltage to current.
(a) 0.15 ohm
(b) 0.25 ohm
(c) 6.660 ohm
(d) None of these
Show Answer & Explanation
Answer: (a) 0.15 ohm
Explanation:
1. R = V/I = 18/120 = 0.15 Ω.
2. The length and diameter are not needed.
Teacher's Note:
1. Some data in a question may be extra.
2. Ohm’s law gives the resistance directly from V and I.
(a) Ammeter
(b) Voltmeter
(c) Galvanometer
(d) Generator
Show Answer & Explanation
Answer: (a) Ammeter
Explanation:
1. An ammeter is connected in series in a circuit and measures the electric current.
2. A voltmeter measures potential difference.
Teacher's Note:
1. An ammeter has a very low resistance.
2. A galvanometer only detects the presence of current.
(a) 10−6 and 10−9 A respectively
(b) 10−3 and 10−6 A respectively
(c) 10−3 and 10−9 A respectively
(d) 10−6 and 10−3 A respectively
Show Answer & Explanation
Answer: (b) 10−3 and 10−6 A respectively
Explanation:
1. 1 mA (milliampere) = 10−3 A.
2. 1 µA (microampere) = 10−6 A.
Teacher's Note:
1. Milli means one thousandth and micro means one millionth.
2. Convert to ampere before using Ohm’s law.
Assertion (A): When two ends of a metallic wire are connected across the terminals of a cell, then some potential difference is set up between its ends. The direction of electrons is from negative terminal to positive terminal of the cell.
Reason (R): Electrons are flowing through the conductors from its higher potential to its lower potential end.
(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) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. A is true: electrons are negatively charged and move from the negative terminal to the positive terminal of the cell.
2. R is false: the negative terminal is at the lower potential, so electrons move from the lower potential end to the higher potential end.
3. Hence A is true but R is false.
Teacher's Note:
1. Conventional current is opposite to the flow of electrons.
2. Electrons are attracted towards the higher potential (positive) end.
Assertion (A): Voltmeter is always connected in parallel across the circuit while measuring the potential difference.
Reason (R): As the voltage in parallel circuits is measured to be the same.
(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) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. A is true: a voltmeter is connected in parallel across the component whose potential difference is to be measured.
2. R is true: in parallel the potential difference across each branch is the same, so the voltmeter shows the potential difference across the component.
3. R explains A.
Teacher's Note:
1. A voltmeter has a very high resistance, so it takes negligible current.
2. An ammeter, in contrast, is connected in series.
Assertion (A): Electric current flowing through a metallic wire is directly proportional to the potential difference across its ends.
Reason (R): Ohm’s law expression V = IR, where R (resistance) of the wire is always varying.
(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) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. A is true: I ∝ V, which is Ohm’s law.
2. R is false: in Ohm’s law, R is constant for a given wire at a constant temperature, it is not always varying.
3. So A is true but R is false.
Teacher's Note:
1. Ohm’s law is valid only when the resistance is constant.
2. The V-I graph of an ohmic conductor is a straight line through the origin.
Assertion (A): An ammeter is always connected in parallel with the circuit for which current has to be measured.
Reason (R): As the current in a series circuit is same.
(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) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. A is false: an ammeter is always connected in series with the circuit so that the whole current passes through it.
2. R is true: the current is the same everywhere in a series circuit, which is why the ammeter is put in series.
3. So A is false but R is true.
Teacher's Note:
1. Ammeter: low resistance, in series. Voltmeter: high resistance, in parallel.
2. If an ammeter is connected in parallel, it may get damaged.
Assertion (A): Insulators do not allow flow of current through themselves.
Reason (R): They have no free charge carriers.
(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) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. A is true: insulators like glass and rubber do not conduct electricity.
2. R is true: they have almost no free electrons or other charge carriers.
3. R is the correct explanation of A.
Teacher's Note:
1. Conductors have a large number of free electrons.
2. Insulators have very high resistivity, 10¹² to 10¹⁷ Ω m.
Assertion (A): A voltmeter and ammeter can be used together to measure resistance and power.
Reason (R): Power and resistance can be calculated from voltage and current.
(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) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. A is true: the voltmeter gives V and the ammeter gives I of a resistor.
2. R is true: R = V/I and P = VI.
3. R is the correct explanation of A.
Teacher's Note:
1. The experiment to verify Ohm’s law uses an ammeter and a voltmeter.
2. Both resistance and power can be found from the same readings.
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Multiple Choice Questions (MCQs) for Class 10 Science Chapter 11 Electricity
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