Category: Class 12 Physics Objective Questions

Physics Exam Questions for Schools on “Current Electricity – Cells, Emf and Internal Resistance”.

1. Which of the following is the correct statement regarding electrochemical cell?
a) It converts chemical energy to electrical energy
b) It converts electric energy to chemical energy
c) It converts chemical energy to thermal energy
d) It does not maintain the flow of charge in a circuit
Answer: a
Clarification: An electrochemical cell is a device which converts chemical energy to electric energy, and maintains the flow of charge in a circuit. There are 2 types of electrochemical cells – Galvanic cell and Electrolytic cell.

2. Pick out the dimensional formula of emf from the following.
a) [M¹L²T³A¹]
b) [ML³T³A¹]
c) [M²L²T¹A^-1]
d) [ML²T^-3A^-1]
Answer: d
Clarification: Electromotive force (emf) is defined as the potential difference between the two terminals of a cell in an open circuit. The SI unit of emf is (frac {joule}{coulomb}) or volt. So, its dimensional formula is [ML²T^-3A^-1].

3. Which is the factor that internal resistance does not depend on?
a) Distance between the electrodes
b) Temperature of the electrolyte
c) Nature of electrode and electrolyte
d) Area of the electrode, immersed in the electrolyte
Answer: b
Clarification: Internal resistance is defined as the resistance offered by the electrolyte and electrodes of a cell when the current flows through it. Internal resistance depends on distance between the electrodes, the nature of electrodes and electrolyte, and area of the electrode immersed in the electrolyte. So, that leaves temperature of the electrolyte out, which is the answer.

4. The emf of a cell depends upon concentration of the electrolyte.
a) True
b) False
Answer: a
Clarification: The emf of a cell depends upon the nature of electrodes, nature and concentration of electrolyte used in the cell and its temperature as well. Emf of a cell is inversely proportional to the concentration of the electrolyte.

5. Identify the correct statement from the following about discharging of a cell.
a) The direction of current in the cell is from positive to negative terminal
b) Terminal potential difference is greater than emf of the cell
c) Terminal potential difference is lesser than emf of the cell
d) The current increases and decreases frequently
Answer: c
Clarification: During discharging of a cell terminal potential difference, the terminal potential difference is lesser than the emf of the cell. The direction of current inside the cell is from negative terminal to positive terminal.

6. A current of 3 A passes through an electric circuit for 5 minutes and does a work of 900J. What is the emf of the source?
a) 3V
b) 1V
c) 5V
d) 10V
Answer: b
Clarification: Current = 3 A; Time taken = 5 minutes = 300 seconds
Work done = 900 J; Power = (frac {Work , done}{Time , taken} = frac {900}{300}) = 3 W
Power = Voltage (emf) x Current → Emf = (frac {Power}{Current} = frac {3}{3}) = 1V
Therefore, the emf of the source is 1 volt.

7. The emf of a battery is 86V and internal resistance 1 ohms in the figure shown below. Calculate the current drawn from the battery.

a) 2 A
b) 3 A
c) 5 A
d) 6 A
Answer: d
Clarification: The resistances 8 ohms, 10 ohms, and 2 ohms are in series, so the equivalent resistance is: R_S = 8 + 10 + 2 = 20 ohms; R_S and 4 ohms are in parallel connection to each other, so,
(frac {1}{R_P} = frac {1}{20} + frac {1}{4} = frac {3}{10}) → R_P = (frac {10}{3})
Total resistance (R) = 3 + (frac {10}{3}) + 7 + 1 (internal resistance) = (frac {43}{3})
According to Ohm’s Law → V = IR → I = (frac {V}{R} = frac {86}{frac {34}{3}})
= 3 × 2
= 6 A
Therefore, the current drawn from the battery is 6 amperes.

8. A cell has an emf of 6V, internal resistance of 1 ohms and a current of 0.5 A passing through it. This cell is connected to a resistor. Find out the resistance of the resistor.
a) 10 ohms
b) 11 ohms
c) 12 ohms
d) 13 ohms
Answer: b
Clarification: Emf (e) = 6V; Internal resistance (r) = 1 ohms; Current (I) = 0.5A
Required equation: I = (frac {e}{R}) + r
0.5 = (frac {6}{R}) + 1
R = (frac {(6 – 0.5)}{0.5})
R = 11 ohms
Therefore, the resistance of the resistor is 11 ohms.

9. A group of N cells such its emf E_N = 1.5r_N is shown in the diagram below. What is the current I in the circuit?

a) 51 A
b) 0.015 A
c) 1.5 A
d) 155.1 A
Answer: c
Clarification: The total emf can be calculated as: E = E₁ + E₂ + E₃ + …………… + E_N
(According to the relation given) = 1.5r₁ + 1.5r₂ + 1.5r₃ + ………… + 1.5r_N
= 1.5 (r₁ + r₂ + r₃ + ………… + r_N) → X
Total resistance can be calculated as: R = r₁ + r₂ + r₃ + ……….. + r_N → Y
Substituting Y in X
E = 1.5 × R
We also know that → I = (frac {E}{R})
So, I = (frac {E}{R}) = 1.5 A
Therefore, the current in the circuit = 1.5 ampere

10. Which of the following devices is the more accurate one for the measurement of emf?
a) Meter Bridge
b) Voltmeter
c) Multimeter
d) Potentiometer
Answer: d
Clarification: Potentiometer is the more accurate device to measure emf than the other ones such as multimeter or voltmeter. Potentiometer is highly sensitive and thus, even small emfs can be measured using this device. Moreover, potentiometers do not draw current from the circuit during measurements, like voltmeters.

Physics Exam Questions for Schools,

Physics Multiple Choice Questions on “Bar Magnet”.

1. Which among the following is not attracted by magnets?
a) Iron
b) Cobalt
c) Copper
d) Nickel
Answer: c
Clarification: A magnet is a material that has both attractive and directive properties. It attracts small pieces of iron, nickel, cobalt, etc. This property of attraction is called magnetism. Going by this criteria, copper is not attracted by magnets.

2. Identify the direction in which a thin long piece of magnet comes to rest when suspended freely.
a) East-west
b) North-south
c) Northeast-southeast
d) Northwest-southwest
Answer: b
Clarification: When suspended freely, a thin long piece of magnet comes to rest nearly in the geographical north-south direction. When placed in a non-uniform magnetic field, it tends to move from weaker to stronger magnetic field.

3. Which of the following is not a basic property of magnets.
a) Magnet exists as a monopole
b) Attractive property
c) Directive property
d) Like poles repel and unlike poles attract
Answer: a
Clarification: Attractive property, directive property and the law of magnetic poles are some of the important properties of magnets. So unlike electric charges, magnetic monopoles do not exist. Every magnet exists as a dipole.

4. Give the SI unit of magnetic dipole moment.
a) A^-2m
b) Am^-2
c) A²m
d) Am²
Answer: d
Clarification: The magnetic dipole moment of a magnetic dipole is defined as the product of its pole strength and magnetic length. The SI unit of magnetic dipole moment is ampere metre² (Am²). Magnetic dipole moment is a vector quantity and is directed from south to north pole of the magnet.

5. Which of the following is a natural magnet?
a) Magnetic needle
b) Bar magnet
c) Magnetite
d) Horseshoe magnet
Answer: c
Clarification: Pieces of naturally occurring iron ore, lodestone or magnetite had the property of attracting small pieces of iron. Hence, magnetite is a natural magnet. All natural magnets are permanent magnets, meaning they will never lose their magnetic power.

6. Magnetic lines of force always form open loops. State true or false.
a) True
b) False
Answer: b
Clarification: Magnetic poles always exist in pairs. So magnetic lines of force run from N-pole to S-pole outside the magnet and from S-pole to N-pole inside the magnet. They do not start or end anywhere and always forms closed loops.

7. A magnet can only repel another magnet. So, ‘X’ is a surer test of magnetism. Identify X.
a) Directive
b) Repulsion
c) Remanence
d) Hysteresis
Answer: b
Clarification: A magnet can attract another magnet or a magnetic substance like iron. However, a magnet can repel another magnet only. So repulsion is the surer test of magnetism.

8. A magnetic dipole of length 10 cm has pole strength of 20 Am. Find the magnetic moment of the dipole.
a) 2 Am²
b) 200 Am²
c) 20 Am²
d) 0.2 Am²
Answer: a
Clarification: Magnetic dipole moment = Pole strength × Magnetic length.
Magnetic dipole moment = 20 Am × 0.1 m
Magnetic dipole moment = 2 Am².

9. A magnetized needle of the magnetic moment 20 JT^-1 is placed at 60^o with the direction of the uniform magnetic field of magnitude 9 T. What is the torque acting on the needle?
a) 225 J
b) 625 J
c) 155.8 J
d) 318 J
Answer: c
Clarification: Torque, τ = mBsin (θ).
Given: m = 20 J/T; B =9 T; θ = 60^o
τ = 20 JT^-1 × 9 T × sin(60^o)
τ = 155.8J.

10. A bar magnet of the magnetic moment 5 Am² has poles 20 cm apart. Calculate the pole strength.
a) 250 Am
b) 4 Am
c) 100 Am
d) 25 Am
Answer: d
Clarification: Magnetic dipole moment = Pole strength × Magnetic length.
Pole strength = (frac {Magnetic , dipole , moment}{Magnetic , length})
Pole strength = (frac {5}{0.2})
Pole strength = 25 Am.

Physics Multiple Choice Questions on “AC Voltage Applied to an Inductor”.

1. What is the reactance of an inductor in a dc circuit?
a) Maximum
b) Minimum
c) Zero
d) Indefinite
Answer: c
Clarification: The reactance of an inductor in a dc circuit is zero.
For dc ➔ f = 0
X_L = 2π f L
X_L = 0
Therefore, the reactance of an inductor in a dc circuit is zero.

2. In which type of circuit the value of power factor will be minimum?
a) Resistive
b) Inductive
c) Superconductive
d) Semi conductive
Answer: b
Clarification: For a purely inductive circuit, Φ = ± (frac {pi }{2})
Power factor = cos (±(frac {pi }{2}))
Power factor = 0
Therefore, the power factor will be minimum for an inductive circuit.

3. The frequency of ac is doubled. How does X_L get affected?
a) X_L gets doubled
b) X_L becomes zero
c) X_L is halved
d) X_L is indefinite
Answer: a
Clarification: When the frequency of an ac is doubled ➔ The inductive reactance (X_L) gets doubled.
This is because inductive reactance is directly proportional to the frequency of an alternating current circuit.

4. At what frequency will a coil, which has an inductance of 2.5 H, have a reactance of 3500 Ω?
a) 700Hz
b) 350 Hz
c) 200 Hz
d) 223 Hz
Answer: d
Clarification: f = (frac {X_L}{2pi L})
f = (frac {3500}{(2 times 3.14 times 2.5)})
f = 222.9 Hz ≈ 223 Hz
Therefore, the frequency of a coil having reactance of 3500 Ω is 223 Hz.

5. Why does an inductor offer an easy path to dc and a resistive path to ac?
a) X_L is maximum for dc and infinite for ac
b) X_L is zero for dc and infinite for ac
c) X_L is zero for dc and finite for ac
d) X_L is maximum for dc and finite for ac
Answer: c
Clarification: For dc, f = 0,
X_L = 2π f L
X_L = 0
X_L is zero for dc and has a finite value for ac Hence an inductor offers an easy path to dc and a resistive path to ac.

6. The inductive reactance for an ac circuit containing only an inductor is (frac {E_0}{X_L}).
a) True
b) False
Answer: b
Clarification: No, this statement is false. The inductive reactance for an ac circuit containing only an inductor is X_L = ωL.
X_L = ωL
X_L = 2π f L

7. Determine the peak current if an inductor of inductance 500 mH is connected to an ac source of peak emf 650 V and frequency 100 Hz
a) 1.55 A
b) 2.07 A
c) 7.89 A
d) 9.87 A
Answer: b
Clarification: Peak current (I₀) = (frac {E_0}{X_L})
I₀ = (frac {650}{(2 times 3.14 times 100 times 0.5)})
I₀ = 2.07 A
Therefore, the peak current is calculated as 2.07 A.

8. Find out the rms value of current in the circuit wherein a 35 mH inductor is connected to 200 V, 70 Hz ac supply.
a) 13 A
b) 15 A
c) 20 A
d) 45 A
Answer: a
Clarification: X_L = 2π f L
X_L = 2π × 70 × 35 × 10^-3 Ω.
I_rms = (frac {E_{rms}}{X_L} = frac {200}{(2pi times 70 times 35 times 10^{-3})})
I_rms = 12.99 A ≈ 13 A
Therefore, the rms value of current in the circuit is 13 A.

Physics Objective Questions on “Wave Optics – Refraction and Reflection of Plane Waves Using Huygen’s Principle”.

1. The speed of the yellow light in a certain liquid is 2.4 × 10⁸ m/s. Find the refractive index of the liquid.
a) 6.25
b) 5.73
c) 1.25
d) 9.73
Answer: c
Clarification: μ = (frac {c}{v_t})
μ = 3 × (frac {10^8}{2.4}) × 10⁸
μ = 1.25

2. The wavelength of light coming from a sodium source is 589 nm. What will be its wavelength in the water?
a) 625 nm
b) 443 nm
c) 789 nm
d) 125 nm
Answer: b
Clarification: λ = 589 nm, μ = 1.33.
Wavelength in water, λ_W = (frac {lambda }{mu })
λ_W = (frac {589}{1.33})
λ_W = 443 nm.

3. A light wave enters from air into glass. How will the energy of the wave be affected?
a) Decreases
b) Increases
c) Remains the same
d) Independent
Answer: a
Clarification: When a light wave enters from the air medium into the glass medium, the energy of the wave decreases because a part of the light wave is reflected in the air. Therefore, in this way, energy will be affected.

4. If a wave undergoes refraction, what will be the phase change?
a) 180^o
b) 270^o
c) 90^o
d) 0^o
Answer: d
Clarification: When a wave undergoes refraction, the phase change will be zero. The phase change is what guides the process. Therefore, no phase change occurs to a wave during refraction.

5. When a wave undergoes reflection at a denser medium, what will be the phase change?
a) 2π radian
b) 0
c) π radian
d) 3π radian
Answer: c
Clarification: When a wave is reflected in the rarer medium from the surface of a denser medium, it undergoes a phase change of π radian. Therefore, when a wave undergoes reflection at a denser medium, the phase change will be π radian.

6. Two wave-fronts intersect each other.
a) True
b) False
Answer: b
Clarification: No, this statement is false. If they intersect, then there will be two rays or two directions of propagation of energy at the point of intersection which is not possible. Therefore, two wave-fronts do not intersect with each other.

7. A light wave enters from air into glass. How will the frequency of the wave be affected?
a) Increases
b) Remains unchanged
c) Decreases
d) Insignificant
Answer: b
Clarification: The frequency of the wave remains unchanged. This is because the frequency is independent of the wave dynamics. Therefore, when a light wave enters from the air medium into the glass medium, the frequency will be the same.

Physics Objective Questions and Answers for Class 12,

Physics Problems for Class 12 on “DE Broglie’s Explanation of Bohr’s Second Postulate of Quantisation”.

1. Identify the expression for Bohr’s second postulate.
a) L = (frac {nh}{2p})
b) L = (frac {2p}{nh})
c) L = nh*2p
d) L = (frac {n}{2p})
Answer: a
Clarification: According to Bohr’s second postulate, an electron revolves around the nucleus of an atom in orbits and, therefore, the angular momentum of the revolution is an integral multiple of (frac {h}{2p}), where h is the Planck’s constant. Thus, the expression for angular momentum is given as:
L = (frac {nh}{2p})

2. How did de – Broglie modify Bohr’s postulate?
a) de – Broglie suggested to not take angular momentum into consideration
b) de – Broglie suggested introducing an electric field near the atom
c) de – Broglie suggested that electrons behaved like a wave
d) de – Broglie did not modify Bohr’s second postulate
Answer: c
Clarification: de – Broglie hypothesis did modify Bohr’s second postulate. This postulate of Bohr regarding the quantization of the angular momentum of an electron was further explained by Louis de Broglie. According to de – Broglie, a moving electron in its circular orbit behaves like a particle-wave.

3. Which of the following can be chosen to analogously represent the behavior of a particle?
a) Metal rod
b) String
c) Elastic rubber
d) Glass rod
Answer: b
Clarification: A string is used to represent the behavior of a particle analogously to the waves traveling on it. Particle waves can lead to standing waves held under resonant conditions. When a stationary string is plucked, it causes several wavelengths to be excited. But, we know that only the ones which have nodes at the ends will survive.

4. Distance and wavelengths are proportional in a string when referring to the behavior of a particle.
a) True
b) False
Answer: a
Clarification: Yes, this is a true statement. In a string, standing waves are formed only when the total distance traveled by a wave is an integral number of wavelengths. In this way, distance and wavelength are proportional to each other. Hence, the expression is given as:
2πr_k = kλ

5. How did de – Broglie conclude the modification of Bohr’s II postulate?
a) de – Broglie concluded that electrons cannot be quantized
b) de – Broglie concluded that the wavelength of electrons should be reduced
c) de – Broglie concluded that angular momentum cannot be quantized
d) de – Broglie concluded that wavelengths of matter waves can be quantized
Answer: d
Clarification: de – Broglie concluded his modification of Bohr’s second postulate by stating that the wavelengths of matter waves can be quantized. This implies that the electrons can exist in those orbits which had a complete set of several wavelengths.

Physics Problems for Class 12,

Physics MCQs for Schools on “Forces Between Multiple Charges”.

1. Three equal positive charges are kept at the corners of an equilateral triangle. What will be the vector sum of the forces acting on the particles?
a) Directed towards the center
b) Directed radially outside
c) Acts along one of the sides of the triangle
d) Zero
Answer: d
Clarification: The magnitude of the force acting on each of the charges will be the same and they will differ from each other by 120 degrees. Therefore their resultant force will always be zero.

2. A 1C charge is placed at the origin. Other infinite numbers of unit charges are placed at √2, √4, √8, √16,….(up to infinite) distances from the origin in a straight line. What will be the total force acting on the 1^st charge?
a) 9*10⁹ N
b) 18*10⁹ N
c) Infinite
d) 1 N
Answer: a
Clarification: The force on 1^st charge will be = 9*10^-9*((frac {1}{sqrt{2}^2} + frac {1}{sqrt{4}^2} + frac {1}{sqrt{8}^2}) +⋯) = 9*10^-9*(frac {frac {1}{2}}{1-frac {1}{2}}) = 9*10^-9N [infinite GP formula].

3. Coulomb force is a central force.
a) True
b) False
Answer: a
Clarification: Coulomb force acts along the center-joining-line of the two charges. So this force is called the central force. Gravitational force is also the central force.

4. Four charges are kept at the corner points of a square. The net force on a charge kept at the center of the square is _________
a) Along diagonal
b) Zero
c) Along one side
d) Depends on the nature of the charges
Answer: d
Clarification: Depending on the nature and quantity of charges, the net force on the central charge may vary its magnitude and direction. If all the charges are the same the net force will be zero.

5. Which among the following is false?
a) Coulomb force is a central force
b) The force between two charges depend on the medium between them
c) Coulomb force is a weak force
d) The net force on a charge is the vector sum of the forces acting on it due to several other charges
Answer: c
Clarification: Coulomb force is a strong force. It can be shown that the Coulomb force between two electrons is 10⁴³ times than gravitational force. All the other options are basic properties of Coulomb’s Law.

Physics MCQs for Schools,