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Electricity & Magnetism — Homi Bhabha Class 9 Physics MCQs with Solutions
Free Homi Bhabha Class 9 Physics Electricity & Magnetism MCQs with step-by-step solutions covering Electric Charge & Potential Difference, Ohm's Law, Series & Parallel Resistors, Heating Effects of Current, Domestic Circuits, Magnetic Lines of Force. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Electric Charge & Potential Difference · easy · theory
What is the SI unit of potential difference?
A. Ampere
B. Coulomb
C. Volt ✓ Correct
D. Ohm
Solution: Potential difference is measured in volts. 1 volt = 1 joule of work per coulomb of charge (1 V = 1 J/C).
Q2 — Series & Parallel Resistors · easy · theory
When resistors are connected in series, which quantity is the same through each resistor?
A. the voltage across each resistor
B. the current ✓ Correct
C. the resistance
D. the power dissipated
Solution: In a series circuit there is only one path, so the same current flows through every resistor; the voltage divides among them.
Q3 — Heating Effects of Current · easy · theory
The heat produced in a resistor carrying a current is given by which expression (Joule heating)?
A. H = I²Rt ✓ Correct
B. H = IR²t
C. H = IRt
D. H = I²R/t
Solution: Joule's law of heating states H = I²Rt, where I is the current, R the resistance and t the time.
Q4 — Domestic Circuits · easy · theory
In household wiring, the various appliances are connected to one another in:
A. parallel ✓ Correct
B. a single loop controlled by one switch
C. series
D. no fixed arrangement
Solution: Appliances are wired in parallel so each receives the full mains voltage and can be switched on or off independently.
Q5 — Magnetic Lines of Force · easy · theory
Outside a bar magnet, the magnetic field lines are directed:
A. in no particular direction
B. from the north pole to another north pole
C. from the north pole to the south pole ✓ Correct
D. from the south pole to the north pole
Solution: Outside a magnet the field lines run from the north pole to the south pole; inside the magnet they run from south to north, forming closed loops.
Q6 — Electric Charge & Potential Difference · hard · numerical
If 20 J of work is done in moving 4 C of charge between two points, the potential difference between them is:
A. 80 V
B. 5 V ✓ Correct
C. 24 V
D. 0.2 V
Solution: $V = \dfrac{W}{Q} = \dfrac{20}{4} = 5$ V.
Q7 — Ohm's Law · hard · numerical
A current of 0.5 A flows through a 20 Ω resistor. The potential difference across it is:
A. 10 V ✓ Correct
B. 20.5 V
C. 0.025 V
D. 40 V
Solution: $V = IR = 0.5 \times 20 = 10$ V.
Q8 — Ohm's Law · hard · theory
On a graph of potential difference (V) against current (I) for an ohmic conductor, the slope of the straight line gives the:
A. current
B. resistance ✓ Correct
C. charge
D. power
Solution: Since $V = IR$, the V–I graph is a straight line whose slope equals the resistance R.
Q9 — Series & Parallel Resistors · hard · numerical
Two 6 Ω resistors are connected in parallel. Their equivalent resistance is:
A. 6 Ω
B. 12 Ω
C. 36 Ω
D. 3 Ω ✓ Correct
Solution: For two equal parallel resistors, $R = \dfrac{6}{2} = 3$ Ω.
Q10 — Series & Parallel Resistors · hard · numerical
A 2 Ω and a 3 Ω resistor are connected in parallel. The equivalent resistance is:
A. 1.2 Ω ✓ Correct
B. 2.5 Ω
C. 5 Ω
D. 6 Ω
Solution: $\dfrac{1}{R} = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6} \Rightarrow R = \dfrac{6}{5} = 1.2$ Ω.
Q11 — Series & Parallel Resistors · hard · numerical
Three 2 Ω resistors are connected in series across a 12 V battery. The current in the circuit is:
A. 18 A
B. 6 A
C. 2 A ✓ Correct
D. 0.5 A
Solution: Total resistance $= 2 + 2 + 2 = 6$ Ω; current $= \dfrac{12}{6} = 2$ A.
Q12 — Heating Effects of Current · hard · numerical
An electric device draws a current of 0.5 A at 220 V. Its power consumption is:
A. 110 W ✓ Correct
B. 55 W
C. 220.5 W
D. 440 W
Solution: Power $= V \times I = 220 \times 0.5 = 110$ W.
Q13 — Heating Effects of Current · hard · theory
An electric fuse protects a circuit because, when the current becomes too large, the fuse wire:
A. cools the circuit
B. stores the charge
C. melts and breaks the circuit ✓ Correct
D. increases the current
Solution: Excess current heats the low-melting fuse wire until it melts, breaking the circuit and preventing damage.
Q14 — Domestic Circuits · hard · theory
The earth wire in a domestic circuit is provided mainly for:
A. safety, by carrying leakage current to the ground ✓ Correct
B. supplying current to appliances
C. increasing the voltage
D. reducing the bill
Solution: The earth wire safely diverts any leakage current to the ground, protecting the user from shock.
Q15 — Domestic Circuits · hard · theory
For safety, the fuse (or MCB) in a household circuit is always connected in the:
A. neutral wire
B. appliance body
C. earth wire
D. live wire ✓ Correct
Solution: The fuse is placed in the live wire so that, when it blows, the appliance is disconnected from the live supply.
Q16 — Magnetic Lines of Force · hard · theory
Two magnetic field lines can never intersect each other because:
A. field lines are straight
B. at the point of intersection there would be two directions of the field, which is impossible ✓ Correct
C. they repel each other
D. the field is zero there
Solution: A single point can have only one net field direction, so field lines cannot cross.
Q17 — Motors & Generators · hard · theory
The working of an electric generator is based on the principle of:
A. conservation of charge
B. electromagnetic induction ✓ Correct
C. Ohm's law
D. the heating effect of current
Solution: A changing magnetic flux through the coil induces an EMF — electromagnetic induction (Faraday's law).
Q18 — Electric Charge & Potential Difference · hard · numerical
When a charge of 2 C is moved between two points, 48 J of work is done. If the potential difference is then doubled while the same 2 C charge is moved, what work is done?
A. 48 J
B. 192 J
C. 96 J ✓ Correct
D. 24 J
Solution: Original V = W/Q = 48/2 = 24 V. Doubling gives 48 V. New work W = QV = 2 × 48 = 96 J.
Q19 — Ohm's Law · hard · numerical
The voltage across a fixed resistor is increased from 4 V to 10 V. If the original current was 0.8 A, what is the new current?
A. 1.2 A
B. 2.5 A
C. 0.32 A
D. 2 A ✓ Correct
Solution: The resistance is fixed: R = 4/0.8 = 5 ohm. New current I = V/R = 10/5 = 2 A.
Q20 — Series & Parallel Resistors · hard · numerical
Three equal resistors of 12 ohm each are used. Two are connected in parallel, and that combination is in series with the third. What is the total resistance?
A. 30 ohm
B. 18 ohm ✓ Correct
C. 4 ohm
D. 36 ohm
Solution: Two 12 ohm resistors in parallel give 12/2 = 6 ohm. In series with the third 12 ohm resistor: 6 + 12 = 18 ohm.
Q21 — Heating Effects of Current · hard · numerical
If the current through a heating element is doubled while its resistance and the time stay the same, the heat produced becomes:
A. half
B. 2 times
C. unchanged
D. 4 times ✓ Correct
Solution: Since H = I²Rt, heat is proportional to the square of the current. Doubling I multiplies the heat by 2² = 4.
Q22 — Domestic Circuits · hard · theory
The earth (ground) wire connected to the metal body of an appliance protects the user by:
A. providing a low-resistance path so any leaked current flows to earth and the fuse blows ✓ Correct
B. raising the supply voltage
C. increasing the efficiency of the appliance
D. reducing the current in the live wire during normal operation
Solution: If the live wire touches the metal casing, the earth wire offers a low-resistance path to the ground. The resulting large current blows the fuse and keeps the casing at earth potential, preventing an electric shock.
Q23 — Motors & Generators · hard · theory
A simple electric (DC) motor uses a split-ring commutator. Its main function is to:
A. continuously increase the speed of rotation
B. produce the magnetic field of the motor
C. reduce the current to a safe value
D. reverse the direction of current in the coil every half rotation so the coil keeps turning the same way ✓ Correct
Solution: After every half turn the sides of the coil swap positions between the poles. The split-ring commutator reverses the current in the coil at this instant, so the turning force stays in the same rotational sense and the coil spins continuously.
Q24 — Electric Charge & Potential Difference · medium · theory
The SI unit of electric charge is the:
A. volt
B. ampere
C. ohm
D. coulomb ✓ Correct
Solution: Charge is measured in coulombs (C); 1 C = 1 ampere × 1 second.
Q25 — Electric Charge & Potential Difference · medium · numerical
If a current of 2 A flows through a wire for 5 s, the charge that passes through it is:
A. 10 C ✓ Correct
B. 2.5 C
C. 0.4 C
D. 7 C
Solution: Charge $Q = I \times t = 2 \times 5 = 10$ C.
Q26 — Electric Charge & Potential Difference · medium · theory
The potential difference between two points is defined as the work done in moving a unit charge between them, and its SI unit is the:
A. coulomb
B. watt
C. ampere
D. volt ✓ Correct
Solution: Potential difference $V = \dfrac{W}{Q}$; its unit is the volt (1 V = 1 J/C).
Q27 — Electric Charge & Potential Difference · medium · theory
A material through which electric charge flows easily is called a:
A. insulator
B. conductor ✓ Correct
C. dielectric
D. semiconductor only
Solution: Conductors (e.g. metals) allow charge to flow; insulators (e.g. rubber) resist it.
Q28 — Ohm's Law · medium · theory
Ohm's law states that, at constant temperature, the current through a conductor is:
A. independent of the potential difference
B. inversely proportional to the potential difference
C. proportional to the square of the voltage
D. directly proportional to the potential difference across it ✓ Correct
Solution: $V = IR$, so $I \propto V$ at constant temperature and resistance.
Q29 — Ohm's Law · medium · numerical
A potential difference of 12 V is applied across a 4 Ω resistor. The current through it is:
A. 48 A
B. 0.33 A
C. 3 A ✓ Correct
D. 16 A
Solution: $I = \dfrac{V}{R} = \dfrac{12}{4} = 3$ A.
Q30 — Ohm's Law · medium · numerical
A current of 2 A flows through a resistor when 10 V is applied across it. Its resistance is:
A. 5 Ω ✓ Correct
B. 12 Ω
C. 0.2 Ω
D. 20 Ω
Solution: $R = \dfrac{V}{I} = \dfrac{10}{2} = 5$ Ω.