Category: Class 12 Physics Objective Questions

Physics Multiple Choice Questions on “Earth’s Magnetism”.

1. How many quantities are required to specify the magnetic field of the earth?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: Three quantities are needed to specify the magnetic field of the earth on its surface, namely, the horizontal component, the magnetic declination and the magnetic dip. These are known as elements of the earth’s magnetic field or magnetic elements.

2. Which of the following is the definition for magnetic meridian of Earth?
a) Vertical plane passing through the axis of a freely suspended or pivoted magnet
b) Horizontal plane passing through the axis of a freely suspended or pivoted magnet
c) Vertical plane passing through the geographical North Pole and South Pole at a given place
d) Horizontal plane passing through the geographical North Pole and South Pole at a given place
Answer: a
Clarification: A vertical plane passing through the axis of a freely suspended or pivoted magnet is known as magnetic meridian of Earth. The vertical plane passing through the geographical North Pole and South Pole at a given place is known as the geographical meridian of that place.

3. Which among the following is denoted by δ?
a) Horizontal component
b) Magnetic meridian
c) Magnetic declination
d) Magnetic inclination
Answer: d
Clarification: Magnetic dip or inclination at a place is defined as the angle made by the Earth’s magnetic field with the horizontal in the magnetic meridian. It is denoted by δ.
Magnetic declination at a place is defined as the angle between the geographic meridian and magnetic meridian.

4. The Earth always have both horizontal and vertical components everywhere.
a) True
b) False
Answer: b
Clarification: No, this statement is false. The Earth always has a vertical component except at the equator. Similarly, the earth always has a horizontal component except at the poles.

5. Identify the expression for horizontal component from the following.
a) B_H=Bcosδ
b) B=B_H cosδ
c) B_H=(frac {B}{cos delta })
d) B_H=2Bcosδ
Answer: a
Clarification: Horizontal component is a component of Earth’s magnetic field along the horizontal direction in the magnetic meridian. It is denoted by B_H. If B is the intensity of Earth’s total magnetic field, then the horizontal component of Earth’s magnetic field is given by:
B_H=Bcosδ

6. At the magnetic North Pole of the Earth, what is the value of the angle of dip?
a) Zero
b) Minimum
c) Infinity
d) Maximum
Answer: d
Clarification: Angle of dip is 90^o at geographical North Pole because attraction on the North Pole of needle is very strong and the needle remains in vertical plane. At the magnetic equator, the needle will point horizontally, i.e. dip angle is 0^o. As you move from the magnetic equator towards the magnetic pole, the angle increases in the northern hemisphere. So, the angle of dip is maximum.

7. At a given place on the Earth’s surface, the horizontal component of Earth’s magnetic field is 9 × 10^-5 T and the resultant magnetic field is 180 × 10^-6. Calculate the angle of dip at this place.
a) 45^o
b) 0^o
c) 60^o
d) 30^o
Answer: c
Clarification: Given: H = 9 × 10^-5; R = 180 × 10^-6 = 18 × 10^-5
The required equation ➔ H=Rcosδ
cosδ = (frac {H}{R}=frac {9 times 10^{-5}}{18 times 10^{-5}}=frac {1}{2})
Therefore, δ = cos^-1(frac {1}{2})=60^o
Thus, the angle of dip is 60^o.

8. When is the angle of dip at a place equal to 45^o?
a) When the vertical and horizontal components of earth’s magnetic field are equal
b) When the vertical component is twice the horizontal component of earth’s magnetic field
c) When the vertical component is half the horizontal component of earth’s magnetic field
d) When either the vertical component or the horizontal components of earth’s magnetic field is equal to zero
Answer: a
Clarification: Yes, the angle of dip at a place is equal to 45^o, when the vertical and horizontal components of earth’s magnetic field are equal. In this case ➔ δ=45^o
tanδ=( frac {B_V}{B_H})=1
So, B_V=B_H.

9. The angle of dip at a certain place on Earth is 30^o and the magnitude of Earth’s horizontal component of magnetic field is 0.35 G. Find the magnetic field at that place on Earth.
a) 0.35 G
b) 0.40 G
c) 0.45 G
d) 0.50 G
Answer: b
Clarification: Given: Horizontal component (H) = 0.35 G; Angle of dip (δ) = 30^o
Required equation ➔ cosδ=(frac {H}{B})
B=(frac {H}{cosdelta } = frac {0.35}{cos30^o} = frac {0.35}{frac {sqrt {3}}{2}} =frac {0.35times 2}{1.732})=0.40
Therefore, the magnitude of the magnetic field at that place is 0.40 G.

10. What is the relation between angle of dip and magnetic latitude?
a) tanδ=4tanλ
b) tanδ=(frac {4}{tan lambda })
c) tanδ=(frac {2}{tan lambda })
d) tanδ=2tanλ
Answer: d
Clarification: If a magnet is freely suspended or inclined or dipped, the angle between the horizontal and the angle of inclination of the magnet that is inclined only by the earth’s magnetic field will indicate the magnetic latitude. The relation is as follows:
tanδ=2tanλ

Physics Multiple Choice Questions on “Power in AC Circuit : The Power Factor”.

1. How many types of power can be defined in an AC circuit?
a) 3
b) 2
c) 1
d) 5
Answer: a
Clarification: In an AC circuit, we can define three types of power, namely, Instantaneous power, Average power, and Apparent power. A circuit element produces or dissipates power according to the equation ➔ P = IV, where I is the current through the element and V is the voltage across it.

2. Which among the following varies in both magnitude and sign over a cycle?
a) Apparent power
b) Effective power
c) Instantaneous power
d) Average power
Answer: c
Clarification: Instantaneous power is defined as the power in an AC circuit at any instant of time. It is equal to the product of values of alternating voltage and alternating current at that time. And, because instantaneous power varies in both magnitude and sign over a cycle, it seldom has any practical importance.

3. Identify the expression for average power.
a) P_av=(frac {V_o I_o}{2})sinΦ
b) P_av=(frac {V_o I_o}{2})cosΦ
c) P_av=(frac {V_o I_o}{4})cos⁡Φ
d) P_av=2V_oI_osin⁡Φ
Answer: b
Clarification: Average power can be defined as the power averaged over one full cycle of alternating current. It is also known as true power. The expression for average power is given by:
P_av=V_rmsI_rmscos⁡Φ=(frac {V_o I_o}{2})cosΦ

4. Apparent power is also known virtual power.
a) True
b) False
Answer: a
Clarification: Yes, apparent power is also known as virtual power. Apparent power is defined as the product of virtual voltage (V_rms) and virtual current (I_rms). It is the combination of reactive power and true power and is measured in the unit of volt-amps (VA).
P_v=V_rmsI_rms=(frac {V_o I_o}{2})

5. Which of the following is true about power factor?
a) sin⁡Φ=(frac {True , power}{Apparent , power})
b) cosΦ=(frac {True , power}{Apparent , power})
c) sin⁡Φ=(frac {Apparent , power}{True , power})
d) cosΦ=(frac {Apparent , power}{True , power})
Answer: b
Clarification: The power factor of an AC electrical power system is defined as the ratio of the real or true power absorbed by the load to the apparent power flowing in the circuit, and in the closed interval of −1 to 1. The expression for power factor is given as:
cos⁡Φ=(frac {True , power}{Apparent , power})

6. What is the power factor in a pure resistive circuit?
a) 0
b) -1
c) Infinity
d) 1
Answer: d
Clarification: A circuit containing only a pure resistance in an AC circuit is known as a pure resistive AC Circuit. For a purely resistive circuit, the power factor is one, because the reactive power equals zero, i.e.
Φ = 0 ➔ Power factor (cos⁡0)=1

7. What is the power factor in a pure inductive or capacitive circuit?
a) -1
b) 0
c) 1
d) Infinity
Answer: b
Clarification: A circuit which contains only inductance is called a pure inductive circuit and a circuit containing only a pure capacitor is known as a pure capacitive circuit. In a pure inductive circuit or a pure capacitive circuit, the current is lagging or ahead by 90 degrees from the voltage. The power factor is the cosine of the angle between the voltage and the current. Therefore:
Φ = (frac {pi}{2}) ➔ Power factor (cos⁡(frac {pi}{2}))=0

8. Power factor has a unit of Watts.
a) True
b) False
Answer: b
Clarification: No, this statement is false. Power factor is also defined as the ratio of the resistance to the impedance of an AC circuit. Impedance is the effective resistance of an electric circuit or component to alternating current. So, both are resistance quantities. Therefore, the ratio between them will be 1, and hence power factor is a unit less and dimensionless quantity.

9. What is the power factor for a series LCR circuit at resonance?
a) Infinity
b) -1
c) 0
d) 1
Answer: d
Clarification: The impedance for a series LCR circuit is given as:
Z=(sqrt {R^2+(X_L-X_C)^2}) and cos⁡Φ=(frac {R}{Z})
At resonance, X_L=X_C ➔ Z = R;
So, the total impedance of the circuit is only due to resistor. The phase angle between voltage and current is zero. Therefore,
Φ=0 ➔ Power factor (cosΦ)=1

10. The power in an AC circuit contains an inductor of 30 mH, a capacitor of 300 μF, a resistor of 70 Ω, and an AC source of 24 V, 60 Hz. Calculate the energy dissipated in the circuit in 1000 s.
a) 8.22 J
b) 8.22 × 10² J
c) 8.22 × 10³ J
d) 82.2 × 10³ J
Answer: c
Clarification: We know that, P_av=V_rms I_rms cos⁡Φ ………1 and cosΦ;=(frac {R}{Z}) …………………2
In LCR, cosΦ=(frac {V_R}{V}=frac {IR}{IZ}) and I_rms = ( frac {V_{rms}}{Z}) ………………….3
Substituting 2 and 3 in 1
P_av=(frac {V_{rms} big ( frac {V_rms}{Z} big )R}{Z})
P_av=(frac {V_{rms}^2R}{Z^2})
Given: V_rms = 24 V; Resistance (R) = 70 Ω; Inductance (I) = 30 mH = 20 × 10^-3 H; Capacitance (C) = 300 μF
X_L=2πυL=2π (60)(30 × 10^-3)
X_L=11.304 Ω
X_C=(frac {1}{2pi υC}=frac {1}{2pi (60)(300 times 10^{-6})})
X_C=8.846 Ω
For series LCR circuit➔Z=(sqrt {R^2+(X_L-X_C)^2})
Z=(sqrt {70^2+(11.304-8.846)^2})
Z=70.04 ≈ 70 Ω
So, the energy used in 1000 seconds is ➔ P_avt=((frac {V_{rms}^2R}{Z^2}))t
P_avt=(frac {24^2 times 70}{70^2}) × 1000 = 8.22 × 10³J
Therefore, the energy dissipated in the circuit at 1000 seconds is 8.22 × 10³J.

Physics Multiple Choice Questions on “Wave Optics – Interference of Light Waves and Young’s Experiment”.

1. In Young’s double-slit experiment with monochromatic light, how is fringe width affected, if the screen is moved closer to the slits?
a) Independent
b) Remains the same
c) Increases
d) Decreases
Answer: c
Clarification: Fringe width is given as:
β = (frac {Dlambda }{d})
As the screen is moved closer to the two slits, the distance denoted by D decreases and so fringe width β increase.

2. In Young’s double-slit experiment, lights of green, yellow, and orange colors are successively used. Write the fringe widths for the three colors in increasing order.
a) β_G < β_Y < β_O
b) β_O < β_Y < β_G
c) β_O < β_G < β_Y
d) β_Y < β_G < β_O
Answer: b
Clarification: Fringe width is expressed as:
β = (frac {Dlambda }{d}).
Since the wavelength are related as shown ➔ λ_G < λ_Y < λ_O, we can say that ➔ β_G < β_Y < β_O.

3. In Young’s double-slit experiment, the two parallel slits are made one millimeter apart and a screen is placed one meter away. What is the fringe separation when blue-green light of wavelength 500 nm is used?
a) 0.5 mm
b) 50 mm
c) 0.25 mm
d) 25 mm
Answer: a
Clarification: Fringe width is given as:
β = (frac {Dlambda }{d}).
So β = (frac { (1 times 500 times 10^{-9} ) }{(10^{-3})})
β = 0.5 mm
Therefore, the fringe separation when blue-green light of wavelength 500 nm is used is 0.5 mm.

4. What would be the resultant intensity at a point of destructive interference, if there are two identical coherent waves of intensity I₀ producing an interference pattern?
a) 5 I₀
b) 2 I₀
c) I₀
d) zero
Answer: d
Clarification: Resultant intensity at the point of destructive interference will be as follows:
I = I₀ + I₀ + 2√I₀ I₀ cos 180^o
I = 0
Therefore, the value of the resultant intensity at a point of destructive interference is zero.

5. What happens to the interference pattern if the phase difference between the two sources varies continuously?
a) Brightens
b) No change
c) Disappears
d) Monochromatic pattern
Answer: c
Clarification: The positions of bright and dark fringes will change rapidly. Such rapid changes cannot be detected by our eyes. Uniform illumination is seen on the screen i.e., the interference pattern disappears.

6. Two independent light sources act as coherent sources’.
a) True
b) False
Answer: b
Clarification: Two independent sources of light cannot be coherent. This is because the light is emitted by individual atoms when they return to the ground state. Even the smallest source of light contains billions of atoms which obviously cannot emit light waves in the same phase.

7. What will be the effect on the fringes formed in Young’s double-slit experiment if the apparatus is immersed in water?
a) Increases
b) Decreases
c) Remains the same
d) Independent
Answer: b
Clarification: The wavelength of light in water (λ’ = (frac {lambda }{mu })) is less than that in air. When the apparatus is immersed in water, the fringe width decreases. This is the impact experienced by the apparatus when immersed in water.

8. What would be the resultant intensity at a point of constructive interference, if there are two identical coherent waves of intensity I’ producing an interference pattern?
a) 4 I’
b) 0
c) I’
d) 2 I’
Answer: a
Clarification: Resultant intensity at the point of constructive interference is given as follows:
I = I’ + I’ + 2√I’I’ cos 0^o
I = 4 I^‘
Therefore, the value of the resultant intensity at a point of constructive interference is 4 I’.

9. In Young’s double-slit experiment if the distance between two slits is halved and distance between the slits and the screen is doubled, then what will be the effect on fringe width?
a) Doubled
b) Decreases four times
c) Increases four times
d) Halved
Answer: c
Clarification: Original fringe width is given as ➔ β = (frac {Dlambda }{d}).
So, the new fringe width is ➔ β’ = (frac {lambda.2D}{ ( frac {d}{2} ) })
β’ = 4β
Therefore, the new fringe width is 4 β.

10. If the separation between the two slits is decreased in Young’s double-slit experiment keeping the screen position fixed, what will happen to the width of the fringe?
a) Decreases
b) Increases
c) Remains the same
d) Independent
Answer: b
Clarification: Fringe width is expressed as:
β = (frac {Dlambda }{d}).
So, as the separation (d) between the two slits decreases, the fringe width increases. They are inversely proportional to each other.

Physics Multiple Choice Questions on “Nuclear Force”.

1. Which of the following best define nuclear forces?
a) The attraction between protons and neutrons
b) Repulsion between protons and neutrons
c) The attraction between protons and electrons
d) The attraction between electrons and neutrons
Answer: a
Clarification: The protons and neutrons are held together by the strong attractive forces inside the nucleus of an atom. These forces are known as nuclear forces. The other statements are not valid and do not define nuclear forces.

2. Find the true statement.
a) Nuclear charge is dependent on the charge
b) The nuclear force is weaker than the electromagnetic force
c) The nuclear force is independent of charge
d) The nuclear force is weaker than the gravitational force
Answer: c
Clarification: Nuclear force is independent of charge. The nuclear force between two protons is the same as that between two neutrons or between a neutron and proton. This is known as charge independent character of nuclear force. Also, nuclear force is stronger than the electromagnetic and gravitational force.

3. What is the energy released in a nuclear reaction called?
a) R-value
b) Q value
c) P-value
d) Nuclear energy
Answer: b
Clarification: In a nuclear reaction, the sum of masses before the reaction is greater than the sum of masses after the reaction. The difference in masses appears in the form of energy following the law of inter-conversion of mass and energy. Thus, the energy released in a nuclear reaction is called a Q value of a reaction.

4. The nuclear force is short-ranged.
a) True
b) False
Answer: a
Clarification: Yes, this is a true statement. The nuclear force exists in small regions, i.e. the diameter of 10^-15 m = 1 fm. The nuclear force between two nucleons decreases rapidly as the separation between them increases and becomes negligible at separation more than 10 fm.

5. Which of the following is the main result of nuclear fission?
a) Helium
b) Strontium
c) Krypton
d) Barium
Answer: d
Clarification: The breaking of a heavy nucleus into two or more fragments of comparable masses, with the release of a tremendous amount of energy is called nuclear fission. The most typical fission reaction occurs when slow-moving neutrons strike Uranium 235 to form the main product barium. The nuclear reaction is given as:
( _{92}^{235})U + (_0^1)n ➔ ( _{56}^{141})Ba + ( _{36}^{92})Kr + 3( _{0}^{1})n + 200 MeV

6. Uranium 235 mass should be greater than X, then it is capable of continuous fission by itself. Identify X.
a) Critical size
b) Threshold point
c) Critical shape
d) Specific size
Answer: a
Clarification: If more than one of the neutrons produced in the fission of uranium 235 is capable of inducing a fission reaction, then the number of fissions taking place at successive stages goes increasing at a very brisk rate and this generates a series of fission reactions. This is known as a chain reaction. If the mass of the Uranium235 sample is greater than a certain size called the critical size then it is capable of continuous fission by itself. So, X is critical size.

7. Which of the following forms the basis of a nuclear reactor?
a) Uncontrolled chain reaction
b) Fast nuclear reaction
c) Controlled chain reaction
d) Catalyst controlled nuclear reaction
Answer: c
Clarification: In a chain reaction, the fast-moving neutrons are absorbed by certain substances known as moderators such as heavy water, then the number of fissions can be controlled and the chain reaction in such a case is known as a controlled chain reaction. This forms the basis of a nuclear reactor.

8. Controlled chain reactions form the basis of an atomic bomb.
a) True
b) False
Answer: b
Clarification: No, this is a false statement. Uncontrolled chain reactions form the basis of an atomic bomb. If the number of fissions in a given interval of time goes on increasing continuously, then a condition of an explosion is created, and this is known as uncontrolled chain reactions.

9. Why is a fusion reaction difficult to perform?
a) The nuclei are set up far from each other
b) The attraction between the nuclei
c) Sun’s energy is not sufficient
d) Repulsion between the nuclei
Answer: d
Clarification: For the fusion reaction to occur, the light nuclei are brought closer to each other, at a distance of about 10^-14 m. This is only possible at a very high temperature to counter the repulsive force between the two nuclei. Due to this reason, the fusion reaction is difficult to perform.

10. What is the energy released in the fission of 2 kg of Uranium 235? (Given: energy per fission = 200 MeV)
a) 1.64 × 10¹⁴ J
b) 1.64 × 10¹⁵ J
c) 2.64 × 10¹⁴ J
d) 1.64 × 10²⁰ J
Answer: a
Clarification: Given: energy per fission = 200 MeV; For Uranium 235:
The number of atoms in 235 g = 6.023 × 10²³
So, the number of atoms in 2 kg of Uranium 235 = ( ( frac {2000}{235} )) × 6.023 × 10²³
N = 5.125 × 10²⁴
Therefore, the energy released in fission of 2 kg of Uranium 235 is:
E = 200 MeV × N × 1.6 × 10^-19 J
E = 200 × 10⁶ × 5.125 × 10²⁴ × 1.6 × 10^-19 J
E = 1.64 × 10¹⁴ J

Physics Multiple Choice Questions on “Effect of Dielectric on Capacitance”.

1. What is the dielectric constant of the medium if the capacitance of a parallel plate capacitor increases from 40F to 80F on introducing a dielectric medium between the plates?
a) 20
b) 0.5
c) 2
d) 5
Answer: c
Clarification: Capacitance without dielectric = 40 F.
Capacitance with dielectric = 80 F.
k = (frac {80}{40})
k = 2.

2. How does the potential difference change with the effect of the dielectric when the battery is kept disconnected from the capacitor?
a) Increases
b) Decreases
c) Remains constant
d) Becomes zero
Answer: b
Clarification: When the dielectric slab is introduced between the plates, the induced surface charge on the dielectric reduces the electric field. The reduction in the electric field results in a decrease in potential difference.
V = Ed = (frac {E_0 d}{k}=frac {V_0}{k}).

3. How does the potential difference change with the effect of the dielectric when the battery remains connected across the capacitor?
a) Increases
b) Decreases
c) Remains constant
d) Becomes zero
Answer: c
Clarification: As the battery remains connected across the capacitor, so the potential difference remains constant at V₀ even after the introduction of the dielectric slab. In this way, dielectric has an effect on potential difference.

4. How does the capacitance change with the effect of the dielectric when the battery remains connected across the capacitor?
a) Increases
b) Decreases
c) Zero
d) Remains constant
Answer: a
Clarification: When a dielectric is introduced, and the battery remains connected across the capacitor, the capacitance increases from C₀ to C.
C = kC₀.

5. How does the electric field change with the effect of the dielectric when the battery remains connected across the capacitor?
a) Increases
b) Decreases
c) Remains unchanged
d) Zero
Answer: c
Clarification: As the potential difference remains unchanged, so the electric field E₀ between the capacitor plates remain unchanged.
E = (frac {V}{d}) = (frac {V_0}{d}) = E₀.

6. The charge on the capacitor plates decreases from Q₀ to Q with the effect of the dielectric when the battery remains connected across the capacitor.
a) True
b) False
Answer: b
Clarification: When a dielectric is introduced, and he battery remains connected across the capacitor, the charge on the capacitor plates increases from Q₀ to Q.
Q = CV = kC₀.V₀ = kQ₀.

7. How does the capacitance change with the effect of the dielectric when the battery is kept disconnected from the capacitor?
a) Increases
b) Decreases
c) Remains constant
d) Zero
Answer: a
Clarification: When the battery is disconnected, and a dielectric is introduced, there will be a decrease in potential difference and as a result, the capacitance increases k times.
C = (frac {Q_0}{V})
C = ( [ frac {Q_0}{(frac {V_0}{k})} ] )
C = (frac {kQ_0}{V_0})
C = kC₀.

8. In a parallel plate capacitor, the capacitance increases from 100 F to 800 F, on introducing a dielectric medium between the plates. What is the dielectric constant of the medium?
a) 0.125
b) 125
c) 80
d) 8
Answer: d
Clarification: Capacitance with dielectric = 800 F
Capacitance without dielectric = 100 F
Dielectric constant = ( ( frac {Capacitance , with , dielectric}{Capacitance , without , dielectric} ) )
k = (( frac {800 F}{100 F} ))
k = 8.
Therefore, the dielectric constant is calculated as 8.