Category: Class 12 Physics Objective Questions

Physics Multiple Choice Questions on “Current Electricity – Cells in Series and in Parallel”.

1. What will be the grouping of cells when the current in the circuit is (frac {ne}{(R + nr)})?
a) Parallel grouping
b) Series grouping
c) Mixed grouping
d) When there is no grouping
Answer: b
Clarification: When n identical cells, each of emf ‘e’ and internal resistance ‘r’ are connected to the external resistance ‘R’ in series, its called series grouping. In series grouping e_eq = ne and r_eq = nr Therefore, current in the circuit (I) = (frac {ne}{(R + nr)}).

2. Which of the following is correct when one cell is wrongly connected in series circuit?
a) The total emf reduces by e
b) The total emf increases by e
c) The total emf increases by 2e
d) The total emf decreases by 2e
Answer: c
Clarification: When one cell is wrongly connected in series of n identical cells, each of emf e, it will reduce the total emf by 2e. So, effective emf is calculated as e_EFF = ne – 2e. This happens in case of mixed grouping.

3. Calculate the number of dry cells, each of emf 2V and internal resistance 1V that is joined in series with a resistance of 30 ohms so that a current of 0.8A passes through it.
a) 20
b) 10
c) 30
d) 40
Answer: a
Clarification: Emf = 2V; r = 1 ohm; I = 0.8A; R = 30 ohms
The required equation: I = (frac {(n times e)}{ [(n times r) + R] })
0.8 = (frac {(n times 2)}{ [(n times 1) + 30] })
0.8n + 24 = 2n
1.2n = 24
n = (frac {24}{1.2}) = 20
Therefore, the number of dry cells required are 20.

4. In parallel grouping of cells, we obtain more current.
a) True
b) False
Answer: a
Clarification: Yes, in parallel grouping of cells, we obtain more current. In parallel combination, the voltage remains the same and the resistance offered is minimum. As more cells are added parallel to each other, then the resistance will keep reducing. As a result, more current can be obtained.

5. There are 4 resistors, each having the same resistance of 4 ohms. These are first connected in series with a cell of internal resistance 2 ohms. Then, they are connected in parallel to the same cell. Find the ratio of the respective currents in the two cases.
a) 1:8
b) 1:7
c) 1:6
d) 6:1
Answer: c
Clarification: When the resistors are connected in series:
R_S = 4 + 4 + 4 + 4 = 16 ohms
They are connected to a cell of internal resistance of 2 ohms, so current (I₁) = (frac {e}{R_S}) + r
= (frac {e}{6}) + 2
= (frac {e}{18})
When the resistors are connected in parallel:
(frac {1}{R_P} = frac {1}{4} + frac {1}{4} + frac {1}{4} + frac {1}{4}) = 1 → R_P = 1
The current through the circuit (I₂) = (frac {e}{1+2})
= (frac {e}{3})
Ratio of both the currents = (frac {I_1}{I_2} = frac {frac {e}{18}}{frac {e}{3}} = frac {1}{6}) → 1:6

6. 36 cells, each of emf 4V are connected in series and kept in a box. The combination shows an emf of 88V on the outside. Calculate the number of cells reversed.
a) 2
b) 5
c) 10
d) 7
Answer: d
Clarification: Number of cells (n) = 36; Emf of each cell (e) = 4V; Total emf (E) = 88V;
Let the number of reversed cells be ‘y’
The required equation: E_EFF = n × e – 2y × e
88 = 36 × 4 – 2y × 4
88 = 144 – 8y
8y = 56
y = 7
Therefore, there are 7 reversed cells.

7. ‘n’ cells have emf ‘e’ and internal resistance ‘r’ and connected to an external resistance ‘R’. They pass the same current whether the cells are connected in series or in parallel to each other. Then which of the following conditions are true?
a) R = r
b) r = nR
c) R = nr
d) R = n²r
Answer: a
Clarification: Current passed through the external resistance when the cells are connected in series:
I₁ = (frac {ne}{(R + nr)})
Current passed through the external resistance when the cells are connected in parallel:
I₂ = (frac {ne}{(nR + r)})
Given: I₁ = I₂
(frac {ne}{(R + nr)} = frac {ne}{(nR + r)})
R + nr = nR + r
R – nR = r – nr
R (1 – n) = r (1 – n)
R = r

8. When cells are connected incorrectly in series, total internal resistance is also affected.
a) True
b) False
Answer: b
Clarification: When ‘n’ cells, each of internal resistance ‘r’, are incorrectly connected in series, the total internal resistance of cells still remains nr, i.e. there is no effect on the total internal resistance of the cells.

9. A battery of emf 10V has an internal resistance of 1 ohms and is charged by a 150V dc supply using a series resistance of 19 ohms. What is the terminal voltage of the battery?
a) 15V
b) 20V
c) 17V
d) 25V
Answer: c
Clarification: Emf (e) = 10; Internal resistance (r) = 1 ohm; DC supply given = 150V; Resistance(R) = 19 ohms E_EFF = 150 – 10 = 140V; R_TOT = R + r = 19 + 1 = 20 ohms.
I = (frac {E_{EFF}}{(R + r)} = frac {140}{20}) = 7A
Terminal voltage = emf of battery + voltage drop across battery
= 10 + Ir
= 10 + (7 x 1)
= 10 + 7 = 17V
Therefore, the terminal voltage of battery is 17V.

10. A cell has an emf ‘e’ and internal resistance ‘r’ and is connected across a variable external resistance R. Identify the correct plot from the following of potential difference across resistance R when R is increased.
a)

b)

c)

d)

Answer: b
Clarification: Current passing in the circuit (I) = (frac {e}{(R + r)})
Potential difference (V) = IR = [ (frac {e}{(R + r)}) ] × R
V = (frac {e}{(1 + frac {r}{R})})
Therefore, when R = 0 → V = 0
R = infinity → V = e
Thus, an upward curve will only be obtained.

Physics Multiple Choice Questions on “Magnetism and Gauss’s Law”.

1. What is the net magnetic flux through a closed surface?
a) Positive
b) Negative
c) Zero
d) Depends on the nature of the surface
Answer: c
Clarification: Gauss’s law indicates that there are no sources or sinks of the magnetic field inside a closed surface. A “closed surface” is a surface that completely encloses a volume with no holes. Therefore, the net magnetic flux through a closed surface is zero.

2. What is the torque exerted by a bar magnet on itself due to its field?
a) Maximum
b) Zero
c) Minimum
d) Depends on the direction of the magnetic field
Answer: b
Clarification: A bar magnet does not exert a force or torque on itself due to its field. But an element of a current-carrying conductor experiences forces due to another element of the conductor. So, the torque exerted by a bar magnet on itself is zero.

3. When does a magnetic dipole possess maximum potential energy inside a magnetic field?
a) Magnetic moment and magnetic field are antiparallel
b) Magnetic moment and magnetic field are parallel
c) The magnetic moment is zero
d) The magnetic field is zero
Answer: a
Clarification: A magnetic dipole possess maximum potential energy when its magnetic moment and the magnetic field are antiparallel. When the magnetic dipole is aligned along the magnetic field, i.e. when θ = 180⁰, it is in unstable equilibrium having maximum potential energy.

4. Calculate the surface integral of a magnetic field over a surface.
a) Maximum
b) Minimum
c) Zero
d) Equal to its magnetic flux through that surface
Answer: d
Clarification: The Gauss’s law states that the surface integral of a magnetic field over a closed surface is always zero. But the surface integral of a magnetic field over a surface gives magnetic flux through that surface.

5. Which of the following is not a consequence of Gauss’s law?
a) The magnetic poles always exist as unlike pairs of equal strength
b) If several magnetic lines of force enter a closed surface, then an equal number of lines of force must leave that surface
c) There are abundant sources or sinks of the magnetic field inside a closed surface
d) Isolated magnetic poles do not exist
Answer: c
Clarification: Gauss‘s law indicates that there are no sources or sinks of the magnetic field inside a closed surface. In other words, there are no free magnetic charges. Hence, magnetic monopoles do not exist.

6. Isolated magnetic poles exist.
a) True
b) False
Answer: b
Clarification: Gauss’s law indicates that there are no sources or sinks of the magnetic field inside a closed surface. So there is no point at which the field lines start or there is no point at which the field lines terminate. Hence, isolated magnetic poles do not exist.

7. Which among the following is the source of the magnetic field (magnetism)?
a) Mechanical origin
b) Electrical origin
c) Chemical origin
d) Potential origin
Answer: b
Clarification: Magnetism is of electrical origin. The electrons revolving in an atom behave as tiny current loops and these current loops give rise to magnetism. An electric current produces a magnetic field. This magnetic field can be visualized as a pattern of circular field lines circling a wire.

8. The line of force in a magnetic field represents the direction at each point that a magnetic needle placed at the point takes up. Do they also represent the direction of the force on a moving charge at each point?
a) Not possible
b) Represent circular motion
c) Represent tangential motion
d) Represents translatory motion
Answer: a
Clarification: No. The force on a charge is perpendicular to the direction of the magnetic field at each point.
F = q (v × B).
It is inappropriate to call magnetic field lines as lines of force.

Physics Multiple Choice Questions on “AC Voltage Applied to a Capacitor”.

1. A 1.5 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance in the circuit.
a) 2120 Ω
b) 21.2 Ω
c) 212 Ω
d) 2.12 Ω
Answer: c
Clarification: X_c = (frac {1}{2πfC})
X_c = (frac {1}{(2 times 3.14 times 50 times 1.50 times 10^{-6})})
X_c = 212 Ω

2. A capacitor of capacitance 10 μF is connected to an oscillator giving an output voltage, E = 10 sin ωt volt. If ω = 10 rad s^-1, find the peak current in the circuit.
a) 197 mA
b) 1 mA
c) 179 mA
d) 5 mA
Answer: b
Clarification: I₀ = (frac {E_0}{(frac {1}{omega C})})
I₀ = ωCE₀.
I₀ = 10 × 10 × 10^-6 × 10
I₀ = 10^-3A = 1 mA.

3. What is the capacitive reactance of a 5 μF capacitor when it is a part of a circuit whose frequency is 50 Hz?
a) 636.6 Ω
b) 1636.6 Ω
c) 2636.6 Ω
d) 4636.6 Ω
Answer: a
Clarification: X_c = (frac {1}{2πfC})
X_c = (frac {1}{(2 times 3.14 times 50 times 5 times 10^{-6})})
X_c = 636.6 Ω.

4. How will the capacitive reactance be affected if the frequency is doubled?
a) Doubled
b) Insignificant
c) Remains the same
d) Halved
Answer: d
Clarification: If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled. This is because, when capacitive reactance, being a resistance, is lowered, the current can flow more easily through the circuit.

5. Give the SI unit of capacitive reactance.
a) Am
b) Ω
c) Ωm
d) A
Answer: c
Clarification: Capacitive Reactance is the complex impedance of a capacitor whose value changes with respect to the applied frequency. The SI unit of capacitive reactance is the ohm (Ω). Capacitive reactance is denoted by X_c.

6. The capacitive reactance varies directly with the frequency.
a) True
b) False
Answer: b
Clarification: X_c = (frac {1}{omega_C})
X_c = (frac {1}{2πfC}).
Thus the capacitive reactance varies inversely with the frequency. Therefore as frequency increases, the capacitive reactance decreases.

7. Calculate the rms value of current in the circuit wherein an 80 μF capacitor is connected to a 100 V, 80 Hz ac supply.
a) 4 A
b) 2 A
c) 7 A
d) 50 A
Answer: a
Clarification: X_c = (frac {1}{2πfC})
X_c = (frac {1}{(2 times 3.14 times 80 times 80 times 10^{-6})})
X_c = 24.8 Ω.
I_rms = (frac {E_{rms}}{X_C})
I_rms = (frac {100}{24.8})
I_rms = 4.03 A ≈ 4 A.

8. What will be the reactance of a capacitor at 150 Hz, if it has a reactance of 200 Ω at 50 Hz?
a) 67 Ω
b) 40 Ω
c) 150 Ω
d) 60 Ω
Answer: a
Clarification: (frac {X_C^{’}}{X_C} =frac {f}{f^{’}})
X_c = (frac {(50 times 200)}{(150)})
X_c = 66.7 Ω ≈ 67 Ω
Therefore, the reactance at 150 Hz will be 67 Ω.

9. A 1.50 μF capacitor is connected to a 220 V, 50 Hz source. If the frequency is doubled, what happens to the capacitive reactance?
a) Remains the same
b) Doubled
c) Halved
d) Becomes zero
Answer: c
Clarification: When the frequency is doubled, the capacitive reactance will be halved. This is because the capacitive reactance (X_c) is inversely proportional to the frequency (f). This is the effect of capacitive reactance on frequency.

10. Which among the following is the correct expression for finding capacitive reactance for an ac circuit containing capacitor only?
a) X_c = 2πf
b) X_c = (frac {1}{2πfC})
c) X_c = 2πfC
d) X_c = (frac {2πf}{C})
Answer: b
Clarification: For an ac circuit containing capacitor only,
X_c = (frac {1}{omega_C})
X_c = (frac {1}{2πfC}).

Physics Multiple Choice Questions on “Coherent and Incoherent Addition of Waves”.

1. Which of the following is a form of light whose photons share the same frequency and whose wavelengths are in phase with one another?
a) Coherent sources
b) Incoherent sources
c) Electromagnetic waves
d) Sunlight
Answer: a
Clarification: Coherent light is a form of light whose photons share the same frequency and whose wavelengths are in phase with one another. The phase difference between the waves should be constant in case of coherent sources.

2. Which among the following is an example of coherent sources?
a) Fluorescent tubes
b) LED light
c) LASER
d) Tungsten filament lamps
Answer: c
Clarification: LASER is the short form for Laser Amplification by Stimulated Emission Radiation. The amplified light beam coming out of a LASER is essentially due to the emission of electrons stimulated by incident radiation consisting of photons. As a result, this causes the coherent behavior of the LASER beam.

3. Pick the odd one out.
a) LASER
b) LED
c) Sound waves
d) Radio transmitters
Answer: b
Clarification: LED is the odd one out. LED is short for light-emitting diode. LED is not a coherent source, whereas, others are examples of coherent sources. The light emitted from an LED is neither spectrally coherent nor even highly monochromatic.

4. Scattering of waves can be coherent and incoherent.
a) True
b) False
Answer: a
Clarification: Yes, the scattering of waves can be coherent and incoherent. The scattering of a wave is coherent and constructive if the phase delay is the same for all the scattered waves. If it varies randomly, then it is considered to be incoherent.

5. Identify the factor is not the same for coherent waves.
a) Frequency
b) Phase difference constant
c) Amplitude
d) Wavelength in phase with each other
Answer: c
Clarification: Coherent waves are the waves with the same frequency and the wavelength of the waves are in phase as well. Therefore, the phase difference is constant. But the coherent waves do not have the same amplitude. Since the amplitude is different, there will be no complete constructive interference where they meet, so they will contribute poorly to an interference pattern.

6. Which of the following is the formula for calculating coherence time?
a) Τ_c = (frac {lambda^3}{(cDelta lambda)})
b) Τ_c = (frac {lambda}{(cDelta lambda)})
c) Τ_c = (frac {lambda^2}{(cDelta lambda)})
d) Τ_c = (frac {lambda^2}{(cDelta lambda)})
Answer: d
Clarification: The formula for calculating coherence time is given as:
Τ_c = (frac {lambda^2}{(cDelta lambda)})
Where Τ_c is the coherence time, λ is the wavelength, ∆λ is the spectral width of the source, and c is the speed of light in a vacuum (i.e. 3 × 10⁸ m/s).

7. When is the wave interference strong?
a) When the paths taken by all of the interfering waves are greater than the coherence length
b) When the paths taken by all of the interfering waves are lesser than the coherence length
c) When the paths taken by all of the interfering waves are equal than the coherence length
d) When the paths taken by all of the interfering waves are independent of the coherence length
Answer: b
Clarification: Coherence length is defined as the propagation distance over which a coherent wave maintains a specified degree of coherence. Wave interference is strong when the paths taken by all of the interfering waves are lesser than the coherence length.

Physics Online Quiz for IIT JEE Exam on “Size of the Nucleus”.

1. Identify the expression for the nuclear radius from the following.
a) R = R₀ ∛A
b) R = R₀ √A
c) R = R₀ A³
d) R = R₀ A²
Answer: a
Clarification: Most of the nuclei of atoms are spherical in structure. The expression for the nuclear radius is given by:
R = R₀ ∛A
Where R₀ is a constant and A is the mass number. The nuclear radius is measured in Fermi meter.
1 fm = 10^-15 m.

2. Who measured the size of the nucleus first?
a) Bohr
b) Einstein
c) Rutherford
d) Geiger and Marsden
Answer: c
Clarification: Rutherford was the form to measure the size of the nuclei of an atom by 1911. He discovered that these were about 10⁴ times smaller than the atoms that constituted them and this was due to the fascinating way that the alpha particles scattered from metal foils.

3. Which of the following is a stable nucleus?
a) The nucleus with even protons and odd electrons
b) The nucleus with even number of protons and neutrons
c) The nucleus with even neutrons and odd protons
d) The nucleus with odd protons and neutrons
Answer: b
Clarification: The nuclei of atoms having even numbers of both protons and neutrons are the most stable ones and this also means that they are less radioactive than nuclides containing even numbers of protons and odd numbers of neutrons.

4. If the internal energy of a nucleus is high, then it is radioactive.
a) True
b) False
Answer: a
Clarification: Yes, this is a true statement. An atom is said to be unstable, or in other words, radioactive when the forces that it is subjected to it are unbalanced. Therefore, when there is an excess of internal energy, instability of an atom’s nucleus occurs and hence they can become radioactive. This may also result from an excess of either neutrons or protons in the nucleus.

5. A nucleus at rest splits into two nuclear parts having radii in the ratio of 1:3. Find the ratio of their velocities.
a) 1:9
b) 3:1
c) 1:27
d) 27:1
Answer: d
Clarification: Given: R1 = R₀ (A1)^1/3 and R2 = R₀ (A2)^1/3
(frac {R1}{R2} = (frac {A1}{A2} ) ^{frac {1}{3}} ) or ( frac {A1}{A2} = (frac {R1}{R2} )^3 = (frac {1}{3} )^3 )
( frac {A1}{A2} = frac { 1}{27})
So, the ratio of their masses is given as:
( frac {m1}{m2} = frac { 1}{27})
According to the principle of conservation of momentum, the magnitude of p1 = magnitude of p2
m1v1 = m2v2
( frac {v1}{v2} = frac {m2}{m1} = frac {27}{1}) = 27:1.

Physics Online Quiz for IIT JEE Exam,