Category: Class 12 Physics Objective Questions

Physics Multiple Choice Questions on “Bohr Model of the Hydrogen Atom”.

1. The radius of the Bohr orbit depends on which of the following?
a) (frac {1}{n })
b) n
c) (frac {1}{n^2})
d) n²
Answer: c
Clarification: The equation is given as:
r_n = (frac {n^2h^2}{4pi^2mkZe^2}).
From this, we can understand that r_n is directly proportional to n².

2. What is the order of the radius of an electron orbit in a hydrogen atom?
a) 10^-8 m
b) 10^-9 m
c) 10^-11 m
d) 10^-13 m
Answer: c
Clarification: The radius of an electron orbit in a hydrogen atom is of the order of 10^-11 m. It is equal to the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state.

3. What will be the longest wavelength in the Balmer series of hydrogen spectrum?
a) 6557 × 10^-10 m
b) 5557 × 10^-10 m
c) 9557 × 10^-10 m
d) 1557 × 10^-10 m
Answer: a
Clarification: (frac {1}{lambda }) = R ( [ frac {1}{n_1^2} – frac {1}{n_2^2} ] )
(frac {1}{lambda }) = 1.098 × 10⁷ ( [ frac {1}{2^2} – frac {1}{3^2} ] )
λ = 36 × (frac {10^{-7}}{5}) × 1.098
λ = 6557 × 10^-10 m.

4. A hydrogen atom in its ground state absorbs 10.2 eV of energy. What is the orbital angular momentum is increased by?
a) 4.22 × 10^-3 Js
b) 2.11 × 10^-34 Js
c) 3.16 × 10^-34 Js
d) 1.05 × 10^-34 Js
Answer: d
Clarification: Increase in angular momentum = (frac {h}{2π}).
(frac {h}{2π}) = 6.6 × (frac {10^{-34}}{2}) × 3.14
(frac {h}{2π}) = 1.05 × 10^-34 Js.

5. Which of the following is true regarding the Bohr model of atoms?
a) Assumes that the angular momentum of electrons is quantized
b) Uses Faraday’s laws
c) Predicts continuous emission spectra for atoms
d) Predicts the same emission spectra for all types of atoms
Answer: a
Clarification: Bohr model of atoms assumes that the angular momentum of electrons is quantized. The atom is held between the nucleus and surroundings by electrostatic forces. The other options are not valid.

6. In terms of Bohr radius a₀, the radius of the second Bohr orbit of a hydrogen atom is given by √2 a₀.
a) True
b) False
Answer: b
Clarification: The equation is given as:
r_n = r₁ n²
r₂ = (2)²r₀ = 4r₀. Thus, in terms of Bohr radius a₀, the radius of the second Bohr orbit of a hydrogen atom is given by 4 a₀.

7. Hydrogen atoms are excited from ground state to the state of principal quantum number 4. Then, what will be the number of spectral lines observed?
a) 3
b) 6
c) 5
d) 2
Answer: d
Clarification: n = 4.
The number of spectral lines emitted = (frac {n(n-1)}{2}).
(frac {n (n-1)}{2} = frac {(4 times 3)}{6}) = 2

8. When a hydrogen atom is in its first excited level, what is the relation of radius and Bohr radius?
a) Twice
b) 4 times
c) Same
d) Half
Answer: b
Clarification: For the first excited level, n = 2.
r₂ = (2)²r₀ = 4r₀.
So, when a hydrogen atom is in its first excited level, its radius is 4 times of the Bohr radius.

Physics Multiple Choice Questions on “Basic Properties of Electric Charge”.

1. A solid sphere will carry more charge than a hollow sphere of the same radius.
a) True
b) False
Answer: b
Clarification: Charge is distributed only on the outer surface of a body. As both the spheres have the same radius and same surface area, they will carry the same amount of charge. But the charge carried by a sphere and a cube will be different as their surface areas are different.

2. +q, +2q, +3q, +4q, ……(up to +20q) charges are situated at coordinates (0,0) , (1,0) ,(2,0) , ….. (Up to 20). What is the total charge stored in the system?
a) +20q
b) +210q
c) +420q
d) +190q
Answer: b
Clarification: As charge is additive, total charge will be (1+2+3+4+…. +20)*q=(frac{20*21}{2})*q= 210q. But if the polarities of the charges are different i.e. some of them are positive and some are negative, then the result will be different. We have to add separately the positive charges and the negative charges.

3. A charged conductor has its charge only on its outer surface. This statement is true for which of the following?
a) For all conductors
b) Only for spherical conductors
c) For hollow conductors
d) For those conductors which don’t have sharp edges
Answer: a
Clarification: Charge remains on the outer surface of a conductor, irrespective of the shape and size of the conductor and also for hollow and solid conductors both. But if there is a sharp edge in the conductor, surface charge density will be more at that point. The surface charge density is uniform in the case of a sphere.

4. What is the unit of surface charge density in the SI unit?
a) C
b) C/m
c) C/m²
d) C/m³
Answer: c
Clarification: Surface charge density means how much charge is stored in the unit surface area of a conductor. So, it’s unit will be=(frac{the , unit , of , the , charge}{the , unit , of , area}). In SI the unit of charge is Coulomb and the unit of area is m². Therefore the required unit will be C/m².

5. What number of electrons will flow in one minute through a conductor that carries 1 Ampere current?
a) 5.2*10²⁰
b) 4.2*10²⁰
c) 3.7*10²⁰
d) 3.7*10¹⁹
Answer: c
Clarification: 1 Ampere means 1 Coulomb per second. So the number of electrons flow per second is ((frac {1}{1.602*10^{-19}})). Therefore number of electron flow in one minute =(frac {1}{1.602*10^{-19}})*60 =3.7*10²⁰.The number will be 60 times more if we have to calculate the number of electron flow in an hour.

6. What is the dimension of volume charge density?
a) [MLAT^-2]
b) [M⁰ L^-3 A T]
c) [M L^-3 A T]
d) [M L^-2 A T]
Answer: b
Clarification: Volume charge density = (frac {charge}{volume}=frac {current*time}{volume}=frac {A*T}{L^3})=[M⁰ A T L^-3]. But in case of surface charge density, the dimension will be [M⁰ L^-2 A T] because surface charge density means an electric charge in a unit area of the surface.

7. Which one of the following is a safe place during lightning?
a) Under a tree
b) Under a light post
c) House with lightning arrester
d) High wall
Answer:c
Clarification: Lightning arrester arrests lightning and allows a safe path of electricity to ground. Thus it is a safe place inside a house that has a lightning arrester on the top of it. Else, electricity finds a high tower or tree or wall and travels through them to ground. Therefore these are not safe places to take shelter during lightning.

8. What should be the shape of a conductor that can hold a charge for long?
a) Cubical
b) Conical
c) Sharp-edged
d) Spherical
Answer: d
Clarification: Sphere has uniform charge distribution over its entire surface. It doesn’t have any sharp edges hence very little chance of charge accumulation at those edges and hence very little chance of discharge of stored charge.

9. A lightning arrester must have the following property.
a) Discontinuity
b) Poor conductivity
c) Needle end
d) Low melting point
Answer: c
Clarification: Needle end can arrest the lightning easily. Besides, the device must provide a continuous path to electricity so that current passes to ground. It must have a high melting point else it will meltdown due to the heat generated during carrying lightning current.

10. Earth is the source of __________
a) An infinite positive and negative charge
b) Positive charge
c) Negative charge
d) Zero charge
Answer: a
Clarification: Earth can be considered as an infinite source of positive and negative charges. This can be justified by the fact that if we connect any positive or negatively charged body to the ground, all of its charges will go to earth.

11. Which one is not the property of charge?
a) Charge is additive
b) Charge is conserved
c) Quantization of charge
d) A charge is self-destructive
Answer: d
Clarification: Electric charge possesses the properties of quantization, conservation of charge. It cannot be destroyed i.e. it is not self-destructive.

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.

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) R_s = R₁ + R₂ + R₃
b) R_s = (frac {1}{(R_1 + R_2 + R_3 )})
c) R_s = (frac {1}{R_1} + frac {1}{R_2} + frac {1}{R_3})
d) R_s = (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 R_s = 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 R_s = 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: R_p = (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, R_s = 2.4 + 1.6 = 4.0.
8 ohms, R_s, and the bottom 8 ohms are in parallel, so, R_p = (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.