Category: Class 12 Physics Objective Questions

250+ TOP MCQs on Electrostatics of Conductors | Class12 Physics

Physics Multiple Choice Questions on “Electrostatics of Conductors”.

1. What is the net electrostatic field in the interior of a conductor?
a) Positive
b) Negative
c) Zero
d) Depends on the nature of the conductor
Answer: c
Clarification: Net electrostatic field is zero in the interior of a conductor. When a conductor is placed in an electric field, its free electrons begin to move in the opposite direction. Negative charges are induced on the left end and positive charges on the right end of the conductor. The process continues until the electric field set up by induced charges becomes equal and opposite the external field.

2. What is the electric field in the cavity of a hollow charged conductor?
a) Positive
b) Negative
c) Zero
d) Depends on the nature of the conductor
Answer: c
Clarification: By Gauss’s theorem, the charge enclosed by the gaussian surface is zero. Consequently, the electric field must be zero at every point inside the cavity. Then, the entire excess charge lies on its surface.

3. How is the electric field at the surface of a charged conductor related to the surface charge density?
a) Proportional to each other
b) Indirectly proportional
c) Independent
d) Exponential
Answer: a
Clarification: The electric field at the surface of a charged conductor is proportional to the surface charge density. The electric field is zero inside the conductor and just outside, it is normal to the surface. The contribution to the total flux comes only from its outer cross-section.

4. How is the potential within and on the surface of a conductor?
a) Indirectly proportional
b) Directly proportional
c) Zero
d) Constant
Answer: d
Clarification: Electric field at any point is equal to the negative of the potential gradient. But inside a conductor, the electric field is zero. Hence, the electric potential is constant throughout the volume of a conductor and has the same value on its surface.

5. What is the total work done on moving a test charge on an equipotential surface?
a) Maximum
b) Minimum
c) Constant
d) Zero
Answer: d
Clarification: The potential difference between any two points on an equipotential surface is zero.
Work done=Test charge x potential difference(0)
=Zero .

6. In a thunderstorm accompanied by lightning, it is safest to run near a tree or open ground rather than sitting inside a car.
a) True
b) False
Answer: b
Clarification: In a thunderstorm accompanied by lightning, it is safest to sit inside a car, rather than near a tree or on the open ground. The metallic body of the car becomes an electrostatic shielding from lightning.

7. ‘X’ is the phenomenon of making a region free from any electric field. Identify X.
a) Faraday’s cage
b) Electrostatic shielding
c) Gauss theorem
d) Corona discharge
Answer: b
Clarification: The phenomenon of making a region free from any electric field is called electrostatic shielding. It is based on the fact that the electric field vanishes inside the cavity of a hollow conductor.

8.What type of surface is the surface of a conductor?
a) Equipotential
b)Unipolar
c) Unipotential
d) Bipolar
Answer: a
Clarification: Electric field at any point is equal to the negative of the potential gradient. But inside a conductor, the electric field is zero. Hence, the electric potential is constant throughout the volume of a conductor and has the same value on its surface. Thus the surface of a conductor is equipotential.

250+ TOP MCQs on Combination of Resistors – Series and Parallel | Class12 Physics

Physics Multiple Choice Questions on “Combination of Resistors – Series and Parallel”.

1. In series connection of resistors, what happens to the current across each resistor?
a) Increases
b) Decreases
c) Remain the same
d) Initially increases and then decreases
Answer: c
Clarification: When the resistors are connected in series, and current is passed through them, the current passing through each of the resistor is the same. This is because, the resistors are connected end to the end and, therefore, there is only one path for the current to flow through.

2. What is the equivalent resistance of series combination of 3 resistors?
a) Rs = R1 + R2 + R3
b) Rs = (frac {1}{(R_1 + R_2 + R_3 )})
c) Rs = (frac {1}{R_1} + frac {1}{R_2} + frac {1}{R_3})
d) Rs = (frac {(R_1 + R_2)}{R_3})
Answer: a
Clarification: When three resistors are connected in series, then the equivalent resistance of this combination is Rs = R1 + R2 + R3. So, if 3 resistors having resistances 10, 15, and 20 ohms, are connected in series, then equivalent resistance of this combination is Rs = 10+15+20 = 45 ohms.

3. Identify the combination which is not a series connection.
a) Resistance box
b) Decorative bulbs
c) Fuses
d) Domestic appliances
Answer: d
Clarification: Domestic appliances in a house are connected in parallel combinations, and not in series combinations. This arrangement is done so that each of the appliances can switched on and off independently, which is essential in a house’s wiring.

4. The equivalent overall resistance is smaller than the smallest parallel resistor.
a) True
b) False
Answer: a
Clarification: Yes, the equivalent overall resistance is smaller than the smallest resistor connected in parallel. This is because, the overall equivalent resistance of parallel combination is: Rp = (frac {1}{R_1} + frac {1}{R_2} + frac {1}{R_3}). When the inverse of a resistance value is taken, the value obtained is lesser than the original value. Thus, the sum of inverse values will only provide a lesser value than the initial resistances.

5. Pick out the correct statement from the following about parallel combination of resistors.
a) The current across the resistors are the same
b) The resistance offered by all resistors are the same
c) The potential difference is same across each resistor
d) The equivalent overall resistance is larger than the largest resistor
Answer: c
Clarification: In parallel combination, the resistors are connected together at one end, and are also all connected together at the other end. So, the potential difference across the resistors will not change and thus, remains the same.

6. Calculate resultant resistance between the points A and B in the circuit shown in the figure below.

a) 4 ohms
b) 2 ohms
c) 8 ohms
d) 6 ohms
Answer: b
Clarification: We can simplify the given diagram as follows:

According to the diagram shown above,
(frac {1}{R_1} = frac {1}{4} + frac {1}{8} = frac {4 + 6}{4 times 6} = frac {10}{24}) → R1 = (frac {24}{10}) = 2.4
(frac {1}{R_2} = frac {1}{2} + frac {1}{8} = frac {2 + 8}{2 times 8} = frac {10}{16}) → R2 = (frac {16}{10})= 1.6
Now, R1 and R2 are in series, so, Rs = 2.4 + 1.6 = 4.0.
8 ohms, Rs, and the bottom 8 ohms are in parallel, so, Rp = (frac {1}{8} + frac {1}{4} + frac {1}{8} = frac {1}{2}). Then, when you take the inverse, you will get the equivalent resistance of the given combination, i.e., (frac {1}{R_p} = frac {1}{frac {1}{2}}) = 2 ohms.

7. Two wires of the same material have the same length but their radii are in the ratio of 5:3. They are combined in series, where the resistance of the thicker wire is 12 ohms. Calculate the total resistance of the combination.
a) 40
b) 12
c) 32
d) 20
Answer: c
Clarification: The given ratio of radii = 5:3; (frac {R2}{R1} = frac {5}{3}) → R2 = ((frac {5}{3})) R1
R1 = 12 ohms (given); R2 = (( frac {5}{3} )) × 12 = 20 ohms. So, R1 = 12 ohms and R2 = 20 ohms
Therefore, total resistance (R) = R1 + R2 (since they are combined in series)
= 12 + 20
= 32 ohms
Thus, the total resistance of the combination is 32 ohms.

8. Three resistors each of 5 ohms are connected in the form of a triangle. What is the resistance between the vertices?
a) (frac {3}{10})
b) (frac {10}{3})
c) (frac {15}{50})
d) (frac {2}{5})
Answer: b
Clarification: Equivalent resistance = 5 + 5 + (frac {1}{5}) (since first two are in series, and they are in parallel to the third in case of a triangular arrangement)
(frac {1}{R}) = 10 + (frac {1}{5} = frac {(5 + 10)}{5 times 10} = frac {15}{50} = frac {3}{10})
Thus, R = (frac {1}{frac {3}{10}} = frac {10}{3}) ohms
Therefore, the equivalent resistance is (frac {10}{3}) ohms.

9. Two resistors are connected in parallel, whose resistance values are in the ratio 3:1. Find the ratio of power dissipated.
a) 1:3
b) 3:1
c) 1:2
d) 2:1
Answer: a
Clarification: We can consider the relation that includes power and resistance, i.e. Power = (frac {voltage^2}{resistance}). Since, the resistors are connected in parallel, the voltage across them will be the same. From this relation, power and resistance are inversely proportional to each other.
Thus, (frac {P1}{P2} = frac {R2}{R1} = frac {1}{3})
So, the power dissipated is in the ratio is 1:3.

10. A set up is such that there are three similar resistors, each of 20 ohms resistance. Two of them are connected in parallel, and this combination is connected in series with the third one. The maximum power that can be consumed by each resistor is 30 W. Then, what is the maximum power that can be consumed by the combination of all three resistors?
a) 30
b) 20
c) 35
d) 45
Answer: d
Clarification: The equivalent overall resistance of the parallel combination is:
(frac {1}{R1} = frac {1}{20} + frac {1}{20} = frac {2}{20} = frac {1}{10}) → R1 = 10 ohms.
R1 is in series with R2; So, R3 = R1 + R2 = 10 + 20 = 30 ohms.
Now, we can employ the method of cross-multiplication:
For 20 ohms resistor → 30 W power consumed
For 30 ohms resistor combination → x
20x = 30 × 30
x = (frac {30 times 30}{20})
x = 45
Therefore, the power consumed by the parallel combination is 45 ohms.

250+ TOP MCQs on Moving Coil Galvanometer | Class12 Physics

Physics Multiple Choice Questions on “Moving Coil Galvanometer”.

1. What is moving coil galvanometer used for?
a) Measurement of voltage only
b) Measurement of resistance
c) Measurement of small currents
d) Measurement of electric field
Answer: c
Clarification: Moving Coil Galvanometer is an instrument used for the detection and measurement of current. It is sensitive instrument and can measure current even if it is only a few microamperes. It was invented by Johann Schweigger in the 1800s.

2. Pick out the expression for galvanometer constant from the following?
a) G = (frac {k}{NAB})
b) G = k × NAB
c) G = (frac {NAB}{k})
d) (frac {1}{G} , = , frac {k}{NAB})
Answer: a
Clarification: In a moving coil galvanometer, the current (I) passing through the galvanometer is directly proportional to its deflection (θ), i.e.
I = Gθ
Where G = (frac {k}{NAB}) ➔ Galvanometer constant
N = number of turns in the coil; A = area of coil; B = strength of the magnetic field; k = torsional constant of the spring that means restoring torque per unit twist.

3. Find the true statement.
a) Ammeter is an instrument used to measure potential difference across any element in a circuit
b) Voltmeter is an instrument used to measure current in a circuit
c) Galvanometer constant is dimensionless
d) Current sensitivity is expressed as the exact reverse of the galvanometer constant
Answer: d
Clarification: Current sensitivity is defined as the deflection produced in the galvanometer, when unit current flows through it.
IS = (frac {NAB}{k}) ………………1
The unit of current sensitivity is rad A-1 or div A-1.
From 1, we can understand that current sensitivity is expressed as the exact reverse of galvanometer constant (G = (frac {k}{NAB})).

4. Voltage sensitivity and current sensitivity are related.
a) True
b) False
Answer: a
Clarification: Yes, voltage and current sensitivity are related to each other. Current sensitivity
IS = (frac {NAB}{k}); Voltage sensitivity = (frac {theta }{V} = frac {theta }{IR} = frac {NAB}{Kr})
Therefore, voltage sensitivity ➔ VS = ((frac {1}{R})) x IS.

5. How is a galvanometer converted into an ammeter?
a) By connecting a high resistance shunt in parallel to the galvanometer
b) By connecting a low resistance shunt in parallel to the galvanometer
c) By connecting a high resistance shunt in series with the galvanometer
d) By connecting a low resistance shunt in series with the galvanometer
Answer: b
Clarification: A galvanometer can be converted into an ammeter of given range by connecting suitable low resistance S called shunt in parallel to the given galvanometer, whose value is given by:
S = ( [ frac {I_g}{(I – I_g)} ] )G
Where Ig is the current for full scale deflection of galvanometer, I is the current to be measured by the galvanometer and G is the resistance of galvanometer.

6. How is galvanometer converted into a voltmeter?
a) By connecting a high resistance multiplier in parallel to the galvanometer
b) By connecting a low resistance multiplier in parallel to the galvanometer
c) By connecting a low resistance multiplier in series with the galvanometer
d) By connecting a high resistance multiplier in series with the galvanometer
Answer: d
Clarification: A galvanometer can be converted into voltmeter of given range by connecting a high resistance called multiplier in series with the galvanometer, whose value is given as:
R = ((frac {V}{I_g})) – G
Where V is the voltage to be measured, Ig is the current for full scale deflection of galvanometer and G is the resistance of galvanometer.

7. What should be the value of shunt resistance of the ammeter in order to increase its range?
a) S = (n – 1) G
b) S = (n – 1)/G
c) S = (frac {G}{(n – 1)})
d) S = (frac {G}{(n + 1)})
Answer: c
Clarification: In order to increase the range of an ammeter n times, the value of shunt resistance to be connected in parallel is given by:
S = (frac {G}{(n – 1)})
In order to increase the range of voltmeter n times, the value of resistance to be connected in series with galvanometer is given by:
R = (n – 1) G

8. An ideal ammeter has infinite resistance and an ideal voltmeter has zero resistance.
a) True
b) False
Answer: b
Clarification: An ammeter is a low resistance instrument and it is always connected in series to the circuit. An ideal ammeter has zero resistance. Voltmeter is a high resistance instrument and it is always connected in parallel with the circuit element across which potential difference is to be measured. An ideal voltmeter has infinite resistance.

9. The galvanometer shown below has a resistance (RG) of 50.00 Ω. Now, this galvanometer is converted to an ammeter with the help of a shunt resistance (rS) of 0.05 Ω. Calculate the current passing through the galvanometer in both the cases.

a) 0.0545 A, 0.594 A
b) 0.0545 A, 0.05 A
c) 0.0545 A, 0.06 A
d) 0.07 A, 0.05 A
Answer: a
Clarification: RG = 50.00 Ω; rS = 0.05 Ω; V = 3.00 V; R = 5.00 Ω
First case ➔ Current (I) = (frac {V}{(R_G + R)})
= (frac {3}{(50 + 5)})
= (frac {3}{55})
= 0.0545 A
Second case ➔ Resistance = (frac {(r_S times R_G)}{(r_S + R_G)})
= (frac {(0.05 times 50.00)}{(0.05 + 50.00)})
= 0.0499 = 0.05 Ω
Total resistance = 5 + 0.05 = 5.05 Ω
Therefore, current = (frac {3}{5.05})
= 0.594 A

10. A galvanometer has resistance of 10 ohms and a full scale deflection is produced by 5 milli amperes. What is the value of resistance that should be connected in series with it in order to enable it to read 2V?
a) 20 Ω
b) 10 Ω
c) 40 Ω
d) 30 Ω
Answer: d
Clarification: Given: G = 10 Ω; V = 2 V; Ig = 0.05 A
The required equation ➔ Ig = (frac {V}{(R + G)})
0.05 = (frac {2}{(R + 10)})
(R + 10) = (frac {2}{0.05} = frac {200}{5}) = 40
R = 40 – 10
R = 30 Ω

11. When 0.010A current flows through a moving coil galvanometer, it gives full deflection. Then, it is converted into a voltmeter which gives a reading of 10 V using an external resistance of 950 Ω. What is the resistance of the galvanometer (G)?
a) 40 Ω
b) 50 Ω
c) 60 Ω
d) 70 Ω
Answer: b
Clarification: Given: V = 10 V; Ig = 0.010A; R = 950 Ω
Required equation ➔ R = ((frac {V}{I_g})) – G ➔ G = ((frac {V}{I_g})) – R
G = ((frac {10}{0.010})) – 950
G = 1000 – 950
G = 50 Ω
Therefore, the resistance of galvanometer is 50 Ω.

12. A moving coil galvanometer has the following characteristics – Number of turns of coil = 50; Area of coil = 70 mm2; Resistance of coil = 30 Ω; Flux density of radial field = 0.1 T; Torsional constant of suspension wire = 7 × 10-8 N m/rad. Calculate the current and voltage sensitivity.
a) 10 div/mA, 0.166 div/mV
b) 15 div/mA, 0.115 div/mV
c) 5 div/mA, 0.167 div/mV
d) 20 div/mA, 0.100 div/mV
Answer: c
Clarification: N = 50; A = 70 mm2; B = 0.1 T; k = 7 × 10-8 N m/rad; R = 30 Ω
Current sensitivity (IS) = (frac {NAB}{k}) = 50 × 70 × 10-6 × (frac {0.1}{7}) × 10-8
= 350 × (frac {100}{7}) = 5 × 103 div/amp
= 5 div/mA
Voltage sensitivity (VS) = (frac {NAB}{kR} = frac {I_S}{R})
= (frac {5}{30} = frac {1}{6})
= 0.167 div/mV

13. What is the value of the shunt resistance (S) required if a galvanometer has a resistance f 50 Ω and a maximum of 0.05A current that can be passed through it? The ammeter range is changed to 20 A.
a) 0.120 Ω
b) 0.125 Ω
c) 0.130 Ω
d) 0.145 Ω
Answer: b
Clarification: Given: Ig = 0.05 A; G = 50 Ω; I = 20 A
Required equation ➔ S = Ig × (frac {G}{(I – I_g)})
S = (frac {0.05 times 50}{(20 – 0.05)})
S = (frac {2.5}{19.95})
S = 0.125 Ω
Therefore, the required shunt resistance is 0.125 Ω.

14. The current passing through a galvanometer is 30 mA, resistance of the galvanometer is 50 Ω and a shunt is 1 Ω is connected to the galvanometer. What is the maximum current that can be measured by this ammeter?
a) 1.53 A
b) 15.3 A
c) 0.153 A
d) 153 A
Answer: a
Clarification: Given: Ig = 30 mA = 0.03 A; S = 1 Ω; G = 50 Ω
Required equation ➔ Maximum current (I) = ( [ frac {(S + G)}{S} ] ) × Ig
I = ( [ frac {(50 + 1)}{1} ] ) × 0.03
I = 51 × 0.03
I = 1.53 A

15. In a galvanometer 15% of the total current in the circuit passes through it. If the resistance of the galvanometer is G, then find out the shunt resistance S that is connected to the galvanometer.
a) (frac {17G}{3})
b) (frac {16G}{3})
c) (frac {5G}{17})
d) (frac {3G}{17})
Answer: d
Clarification: Ig = ((frac {15}{100})) × I
S = Ig × (frac {G}{I – I_g})
S = ( ( frac {15}{100})) × I × (frac {G}{(I – 15 times frac {I}{100})})
S = [0.15 × ( frac {I}{(I – 0.15 times I)} ) ] × G
S = [0.15 × ( frac {I}{0.85 times I} ) ] × G
S = ( frac {15}{85} ) × G
S = ( frac {3G}{17} )
Therefore, the shunt resistance required is ( frac {3G}{17} ).

250+ TOP MCQs on Representation of AC Current and Voltage by Rotating Vectors – Phasors | Class12 Physics

Physics Question Bank for Class 12 on “Representation of AC Current and Voltage by Rotating Vectors – Phasors”.

1. Identify the function of a phasor from the following.
a) Phasor is a vector quantity used to represent a sinusoidal signal
b) Phasor is a scalar quantity used to represent a sinusoidal signal
c) Phasor is a vector quantity used to represent a cosine signal
d) Phasor is a scalar quantity used to represent a cosine signal
Answer: a
Clarification: Phasor is a vector quantity used to represent a sinusoidal signal. The vertical component of phasors represents the quantities that are sinusoidally varying for a given equation. The magnitude of the phasors represents the peak value of the voltage and the current.

2. What is the general expression of a sinusoidal signal?
a) A(t) = Am sin⁡(ωt – Φ)
b) A(t) = sin⁡(ωt + Φ)
c) A(t) = Am sin⁡(ωt + Φ)
d) A(t) = 2Am sin⁡(ωt + Φ)
Answer: c
Clarification: A sinusoidal signal is a mathematical curve that describes a smooth periodic oscillation. The general expression of a sinusoidal signal is given as follows:
A(t) = Am sin⁡(ωt + Φ)
Where ➔ Am = Peak Amplitude, ω = Angular Frequency, Φ = Phase Shift.

3. What does a phasor represent?
a) Current and resistance
b) Current and voltage
c) Voltage and resistance
d) Voltage and power
Answer: b
Clarification: Phasors represent alternating current and voltage of same frequency as vectors in a phasor diagram with the phase angle between them. Phasors can also be applied to impedance as well as other related complex quantities that are not dependent on time.

4. Phasors rotate in the clockwise direction.
a) True
b) False
Answer: b
Clarification: No, this statement is false. Phasors rotate in the anti-clockwise direction. Even though phasors are vectors, they represent scalar quantities such as voltage and current. It rotates about the origin with an angular speed ω. With the help of a phasor, we can easily detect whether one of two quantities are in the same phase or not.

5. Find the true statement.
a) Length of the phasor must be greater than the peak value of alternating voltage or alternating current
b) When the current reaches its maximum value after emf becomes maximum, then the current is considered to be leading ahead of emf
c) When the emf reaches its maximum value after current becomes maximum, then the emf is considered to be behind the current
d) Phasors for voltage and current are in the same direction when the phase angle between voltage and current is zero
Answer: d
Clarification: The phasors for voltage and current are in the same direction at all times when the phase angle between voltage and current is zero. All the other statements are not valid. Length of the phasor must be equal to the peak value of alternating voltage or alternating current. Also, when the current reaches its maximum value after emf becomes maximum, then the current is considered to be lagging behind the emf and when the emf reaches its maximum value after current becomes maximum, then the emf is considered to lead the current.

Physics Question Bank for Class 12,

250+ TOP MCQs on Wave Optics – Huygen’s Principle | Class12 Physics

Physics Multiple Choice Questions on “Wave Optics – Huygen’s Principle”.

1. Which among the following isn’t a suitable phenomenon to establish that light is wave motion?
a) Interference
b) Diffraction
c) Reflection
d) Polarization
Answer: c
Clarification: Light undergoes interference, diffraction, and polarization. These phenomena establish that light is a wave motion. Therefore, out of the options, reflection isn’t a suitable phenomenon to establish that light is wave motion.

2. The optical path of monochromatic light is the same if it travels 2 cm thickness of glass or 2.25 cm thickness of water. If the refractive index of water is 1.33, what is the refractive index of glass?
a) 2.5
b) 1.5
c) 3.5
d) 4.5
Answer: b
Clarification: Optical path = μ × Path in medium.
The optical path for glass = Optical path for water.
μg = (frac {(1.33 times 2.25)}{(2.0)})
μg = 1.50.

3. Identify the condition which is not necessary for two light waves to be coherent.
a) The two waves must be continuous
b) The two waves should be of the same frequency or wavelength
c) They should have a constant or zero phases difference
d) They two light sources should be narrow
Answer: d
Clarification: The essential conditions for two light waves to be coherent includes that the two waves must be continuous, the two waves should be of the same frequency or wavelength, and they should have a constant or zero phases difference. So, the unnecessary condition for two light waves to be coherent is the statement that the sources should be narrow.

4. The absolute refractive indices of glass and water are (frac {3}{2}) and (frac {4}{3}). Determine the ratio of the speeds of light in glass and water.
a) 5:7
b) 9:8
c) 7:5
d) 8:9
Answer: d
Clarification: ( frac {v_g}{v_w} = frac {mu_w}{mu_g})
(frac {v_g}{v_w} = frac {(frac {4}{3} )}{(frac {3}{2}) })
(frac {v_g}{v_w}) = 8:9

5. The refractive index of glass is 1.5 and that of water is 1.3, the speed of light in water is 2.25 × 108 m/s. What is the speed of light in glass?
a) 7.95 × 108 m/s
b) 9.95 × 108 m/s
c) 1.95 × 108 m/s
d) 3.95 × 108 m/s
Answer: c
Clarification: ( frac {v_g}{v_w} = frac {mu_w}{mu_g})
Vg = ((frac {mu_w}{mu_g})) × vw
(frac {v_g}{v_w} = ( frac {1.3}{1.5} ) ) × 2.25 × 108
(frac {v_g}{v_w}) = 1.95 × 108 m/s.

6. In Huygens’ theory, light waves are longitudinal and do not require a material medium for their propagation.
a) True
b) False
Answer: b
Clarification: In Huygens’ theory, light waves are longitudinal and require a material medium for their propagation. That is why, Huygens’ assumed the existence of an all-pervading hypothetical medium, called ether.

7. The speed of light in air is 3 × 108 m/s. If the refractive index of glass is 1.5, find the time taken by light to travel a distance of 10 cm in the glass.
a) 0.5 × 10-10 s
b) 5 × 10-10 s
c) 50 × 10-10 s
d) 500 × 10-10 s
Answer: b
Clarification: vg = (frac {c}{mu } = frac {(3 times 10^8)}{(1.5)}) = 2 × 108 m/s.
Time, t = (frac {Displacement}{v_g})
t = (frac {(10 times 10^{-2})}{(2 times 10^8)})
t = 5 × 10-10 s.

8. The speed of yellow light in a certain liquid is 2.4 × 108 m/s. Find the refractive index of the liquid.
a) 1.25
b) 5.55
c) 6.25
d) 12.25
Answer: a
Clarification: μ1 = (frac {c}{v_1})
μ1 = (frac {(3 times 10^8)}{(2.4 times 10^8) })
μ1 = 1.25.

250+ TOP MCQs on The Line Spectra of the Hydrogen Atom | Class12 Physics

Physics Assessment Questions for Class 12 on “The Line Spectra of the Hydrogen Atom”.

1. What causes spectral lines?
a) The transition of electrons between two energy levels
b) The transition of electrons between two wavelength ranges
c) Magnetic and electric field exiting in an atom
d) The transition of electrons from electric to magnetic field
Answer: a
Clarification: The observed spectral lines are caused by the transition of electrons between two energy levels in an atom. The emission spectrum of the hydrogen atom is divided into many spectral series, with wavelengths that are given by Rydberg’s formula.

2. How many spectral lines are there in the hydrogen spectrum?
a) Infinity
b) Zero
c) Multiple
d) One
Answer: c
Clarification: Even though hydrogen has only one electron in its outermost shell, it has multiple spectral lines. This is because hydrogen has many energy levels. When the electron excites from a lower level of energy into a higher level, then it releases a photon, and this photon is the one that appears as spectral lines.

3. Which theory explained that electrons revolved in circular orbits?
a) Einstein theory
b) Bohr theory
c) Rydberg theory
d) de – Broglie theory
Answer: b
Clarification: Niels Bohr explained the line spectrum of the hydrogen atom with the assumption that electrons revolved around an atom in circular orbits and that the orbit closer to the nucleus represented the ground state and the farther orbits represented the higher levels of energy.

4. Bohr’s model only works for hydrogen or helium.
a) True
b) False
Answer: a
Clarification: Yes, this is a true statement. Bohr’s model works only for elements like hydrogen because the model only considers the interactions between one electron and the nucleus. If there are more electrons, then they will repel the one electron, and this, in turn, will change its energy level.

5. When is hydrogen stable?
a) When the electron jumps to higher energy levels
b) When an electric field is introduced
c) When a magnetic field is introduced
d) When the electron is at its ground state
Answer: d
Clarification: The hydrogen atom is stable when the electron rests at the ground level of energy, i.e. when the principal quantum number, n = 1. In other words, when the electron is revolving in the first orbit around the nucleus, the hydrogen atom is stable.

6. Which of the following is found in the UV region of the spectrum?
a) Pfund series
b) Brackett series
c) Lyman series
d) Paschen series
Answer: c
Clarification: The transitions that end at the ground level, where the principal quantum number is one, are called the Lyman series. Here, the energies that are released are so large, and therefore, these spectral lines appear in the ultraviolet region of the spectrum.

7. How many spectral lines does hydrogen have?
a) Four
b) Three
c) Two
d) One
Answer: a
Clarification: Niels Bohr calculated the energies that a hydrogen atom would have in each of its energy levels, based on the wavelength of the spectral lines. Then he found out that there are four spectral lines for hydrogen, namely, Lyman, Balmer, Paschen, and Brackett series. The Lyman series lies in the UV region, whereas the Balmer series lies in the visible region, and the last two lie in the infrared region.

Physics Assessment Questions for Class 12,