Category: Class 12 Physics Objective Questions

Physics Question Papers for Schools on “Electrostatic Potential due to a System of Charges”.

1. Which is the expression for electric potential due to a system of charges from the following?
a) V=(frac {1}{4 pi varepsilon_o} (frac {q_1}{r_1} + frac {q_2}{r_2} + frac {q_3}{r_3} +……. frac {q_n}{r_n}))
b) V=1 χ 4πε_o ( (frac {q_1}{r_1} + frac {q_2}{r_2} + frac {q_3}{r_3} +……. frac {q_n}{r_n}))
c) V=(frac {1}{8 pi varepsilon_o} (frac {q_1}{r_1} + frac {q_2}{r_2} + frac {q_3}{r_3} +……. frac {q_n}{r_n}))
d) V=(frac {1}{4 pi varepsilon_o} (frac {q_1}{r_1} times frac {q_2}{r_2} times frac {q_3}{r_3} times……. frac {q_n}{r_n}))
Answer: a
Clarification: The electric potential at a point due to a system of charges is equal to the algebraic sum of the electric potentials due to individual charges at that point. The expression for electric potential due to a system of charges is given by:
V=(frac {1}{4 pi varepsilon_o} (frac {q_1}{r_1} + frac {q_2}{r_2} + frac {q_3}{r_3} +……. frac {q_n}{r_n}))

2. There are infinite number of charges, each equal to ‘q’, which are placed along the X-axis at points x = 1, x = 4, x = 16, x = 64……..Then, determine the electric potential due to this system of charges at the point x = 0.
a) V=(frac {16q}{3pi varepsilon_o})
b) V=(frac {3q}{4pi varepsilon_o})
c) V=(frac {4q}{8pi varepsilon_o})
d) V=(frac {3q}{16pi varepsilon_o})
Answer: d
Clarification: We know electric potential(V) due to charge ‘q’ is V = (frac {kq}{r}).
Electric potential (V) is a scalar quantity, so, the total potential at x = 0 is the sum of all the individual charges
V=[((frac {kq}{1}) + (frac {kq}{4}) + (frac {kq}{16}) + (frac {kq}{64}) ) + ……..]
V=kq((frac {1}{(1-frac{1}{4})})) → sum of infinite G.P=(frac {a}{(1-r)})
V=((frac {q}{4pi varepsilon_o}) times (frac {3}{4}))
V=(frac {3q}{16pi varepsilon_o})

3. Two equal and opposite charges Q1 = 2 C and Q2 = -2C are placed at a distance of 6m from each other. What is the potential at the midpoint between the two charges?
a) 2 V
b) 0 V
c) 1 V
d) 3 V
Answer: b
Clarification: Electric potential is a scalar quantity. So, the total potential will be the sum of all the individual charges.
V₁ = (frac {kQ}{r} = frac {2k}{3})
V₂ = (frac {kQ}{r} = frac {-2k}{3})
Total potential (V) = V₁ + V₂
V = k [ ((frac {2}{3}) + (frac {-2}{3})) ]
V = 0 V
Therefore, the potential at the midpoint between the two charges is zero volt.

4. Electric field and electric field intensity are related to electric potential.
a) True
b) False
Answer: a
Clarification: Yes, both are related to electric potential. The relationship between electric potential and electric field is a differential → electric field (E) is the gradient of electric potential (V) in the x direction. Thus, as the charge moves in the x direction, the rate of its change in potential is the value of the electric field. Then, electric field can be defined as the negative of the rate of derivative of potential difference.

5. The 3 charges → (frac {q}{2}), -q, (frac {q}{2}) are placed along the x axis at x = 0, x = r, x = 2 respectively. Find the resultant potential at a point A located at a distance y from charge q such that r << y.
a) V=(frac {4qr^2}{4pi varepsilon_o y})
b) V=(frac {qr^3}{4pi varepsilon_o y^3})
c) V=(frac {qr^2}{4pi varepsilon_o y^3})
d) V=(frac {qr^3}{4pi varepsilon_o y^2})
Answer: c
Clarification: Electric potential at point A can be given by:
V=(frac {1}{4pi varepsilon_o}[(frac {(frac {q}{2})}{y+r}) – frac {q}{y} + frac {(frac {q}{2})}{(y-r)}])
V=(frac {(qr^2)}{4pi varepsilon_o y(y^2 – r^2)})
V=(frac {qr^2}{4pi varepsilon_o y^3}) (Since y >> r)

Physics Question Papers for Schools,

Physics Multiple Choice Questions on “Current Electricity – Resistivity of Various Materials”.

1. Identify the type of materials which have resistivities in the range of 10^-8 Ωm to 10^-6 Ωm.
a) Semiconductors
b) Insulators
c) Conductors
d) Thyristors
Answer: c
Clarification: Conductors have low resistivities in the range of 10^-8 Ωm to 10^-6 Ωm. Metals are good conductors. Conductors are objects or types of material that allows the flow of charge in one or more directions, and a result, the resistance offered against the flow of charge will be less.

2. Which of the following is a widely used variety of commercial resistor?
a) Bio amplification resistor
b) Wire-bound resistor
c) Ultrasonic resistor
d) Copper resistor
Answer: b
Clarification: A wire bound resistor is an electrical passive component that limits current. Wire-bound resistors are made by winding the wires of an alloy like manganin on an insulating base. They are relatively insensitive to temperature.

3. Identify the material whose resistivity is more than 10¹⁴ Ωm.
a) Bakelite
b) Copper
c) Aluminum
d) Silicon
Answer: a
Clarification: Insulators are materials which do not conduct electric current, and thereby, offer high resistance to the flow of charges. Insulators like bakelite and hard rubber have very high resistivities in the range of 10¹⁴ to 10¹⁶ Ωm.

4. Identify the wrong statement describing the color code for carbon resistors.
a) The first band indicates the first significant figure
b) The second band indicates the second significant figure
c) The third band indicates the third significant figure
d) The fourth band indicates the possible variation in the percent of the indicated value
Answer: c
Clarification: The third band indicates the power of ten with which the first and second significant figures must be multiplied to get the resistance value in ohms. All the other statements are valid.

5. Find the resistance value of the carbon resistor if the colors of the four bands are red, red, red and silver respectively.
a) 33 × 10²Ω ± 20%
b) 22 × 10²Ω ± 5%
c) 22 × 10²Ω ± 10%
d) 33 × 10³Ω ± 10%
Answer: c
Clarification: A color code is used to indicate the resistance value of a carbon resistor and its percentage accuracy. The corresponding value of resistance for the given color code is 22 × 10² Ω ± 10% (Red-2; Red-2; Red-2; Silver-10%).

6. If the fourth band of the carbon resistor is absent, it implies there is no tolerance.
a) True
b) False
Answer: b
Clarification: The fourth band indicates the tolerance or possible variation in the percent of the indicated value. If the fourth band is absent, it implies the tolerance of the resistor is ±20%.

7. ‘X’ is a type of commercial resistor made from a mixture of clay, carbon black, and resin binder which are pressed and then molded into cylindrical rods by heating. Identify X.
a) Wire-bound resistors
b) Carbon resistors
c) Metal film resistors
d) Surface-mount resistors
Answer: b
Clarification: Carbon resistors are made from a mixture of carbon black, clay and resin binder which are pressed and then molded into cylindrical rods by heating. The rods are enclosed in a ceramic or plastic jacket. They are used in circuits of radio receivers, amplifiers, etc.

8. Find the resistance value of the carbon resistor if the colors of the four bands are yellow, violet, brown and gold respectively.
a) 47 × 10¹Ω ± 5%
b) 47 × 10⁰Ω ± 5%
c) 47 × 10⁶Ω ± 5%
d) 47 × 10²Ω ± 5%
Answer: a
Clarification: A color code is used to indicate the resistance value of a carbon resistor and its percentage accuracy. The corresponding value of resistance for the given color code is 47 × 10¹Ω ± 5% (Yellow-4; Violet-7; Brown-1; Gold-5%).

9. Find the resistance value of the carbon resistor if the colors of bands are green, violet, and red respectively.
a) 57 × 10²Ω ± 5%
b) 57 × 10²Ω ± 10%
c) 57 × 10²Ω ± 20%
d) 57 × 10²Ω
Answer: c
Clarification: A color code is used to indicate the resistance value of a carbon resistor and its percentage accuracy. The corresponding value of resistance for the given color code is 57 × 10²Ω ± 20% (Green-5; Violet-7; Red-2; No fourth band-20%).

10. Identify the material whose resistivity lie between 10^-6 Ωm and 10⁴ Ωm.
a) Silver
b) Mica
c) Copper
d) Silicon
Answer: d
Clarification: The resistivities of semiconductors lie between 10^-6 Ωm and 10⁴ Ωm. A semiconductor material has an electrical conductivity value falling between that of a conductor, such as metallic aluminum, and an insulator, such as wood. Silicon and germanium are typical semiconductors.

Physics Multiple Choice Questions on “Solenoid and Toroid”.

1. Pick out the expression for magnetic field inside a toroid from the following.
a) B = μ₀ NI
b) B = 2μ₀ NI
c) B = (frac {mu_o}{NI})
d) B = (frac {mu_o N}{I})
Answer: a
Clarification: Ampere’s circuital law:
∮ B.dl = ∮ B.dlcos⁡0 = B ∮ dl = B.2πr …………………..1
But, for a toroid:
∮ B.dl = μ₀ × total current threading the toroid
∮ B.dl = μ₀ × total number of turns in the toroid × I
∮ B.dl = μ₀ N2πrI ……………………………2
From 1 and 2
B.2πr = μ₀ N2πrI
Therefore, B = μ₀NI

2. A solenoid of 0.5 m length with 100 turns carries a current of 5 A. A coil of 20 turns and of radius 0.02m carries a current of 0.6 A. What is the torque required to hold the coil with its axis at right angle to that of solenoid in the middle point of it?
a) 1.60 N m
b) 15.89 × 10^-5 N m
c) 1.893 × 10^-5 N m
d) 1.893 × 10^-8 N m
Answer: c
Clarification: Given: μ₀ = 4 × 3.14 × 10^-7; number of turns of solenoid = 100; Current passing through the solenoid = 5 A; Number of turns of coil = 20; Current passing through the coil = 0.6A; Length of the solenoid = 0.5 m; Radius of coil = 0.02 m
Magnetic field of solenoid (B) = μ₀nI = μ₀ × (frac {100}{0.5}) × 5 = 4 × 3.14 × 10^-7 × (frac {100}{0.5}) × 5
Magnetic moment of the coil (M) = I × A × N = 0.6 × (0.02)² × 20
Torque (τ) = MBsin (90^o)
τ = 0.6 × (0.02)² × 20 × 4 × 3.14 × 10^-7 × (frac {100}{0.5}) × 5      [sin (90^o) = 1]
τ = 1.893 × 10^-5 N m

3. Two concentric coils of 20 turns each are situated in the same plane. Their radii are 60 cm and 80 cm and they carry currents 0.4 A and 0.5 A respectively in opposite directions. Calculate the magnetic field at the center.
a) (frac {5mu_0}{4})
b) (frac {5mu_0}{6})
c) (frac {5mu_0}{8})
d) (frac {5mu_0}{10})
Answer: b
Clarification: Since the two coils are concentric and in the same plane, carrying currents in opposite directions, the total magnetic field at the center of the concentric coils is given by:
B = B1 – B2 = (frac {mu_0}{4pi } [ frac {2pi N_1 I_1}{r_1} ] , – , frac {mu_0}{4 pi } [ frac {2 pi N_2 I_2}{r_2} ])
B = (frac {mu_0}{2} [ frac {N_1 I_1}{r_1} , – , frac {N_2 I_2}{r_2} ])
B = (frac {mu_0}{2} big [ )20 × ( frac {0.4}{0.6} ) – 20 × ( frac {0.5}{0.8} big ] )
B = (frac {mu_0}{2} [ frac {40}{3} , – , frac {25}{2} ])
B = (frac {mu_0}{2} [ frac {80-75}{6} ] )
B = (frac {5mu_0}{6})

4. Magnetic field in toroid is stronger than that in solenoid.
a) True
b) False
Answer: a
Clarification: Yes, in the magnetic field at the center in toroid is stronger than that in a solenoid. This is due to its ring structure. The magnetic field inside and outside the toroid is zero. Toroid does not have a uniform magnetic field inside it unlike a solenoid.

5. A long solenoid has 500 turns per cm and carries a current I. The magnetic field at its center is 7.54 × 10^-2 Wb m^-2. Another long solenoid has 300 turns per cm and it carries a current (frac {I}{3}). What is the value of magnetic field at the center?
a) 1.50 × 10^-3 Wb/m²
b) 1.50 × 10^-5 Wb/m²
c) 1.05 × 10^-3 Wb/m²
d) 1.50 × 10^-2 Wb/m²
Answer: d
Clarification: The magnetic field induction at the center of a long solenoid is B = μ_onI→B ∝ nI      n = number of turns per unit length of solenoid. I = current passing through the solenoid.
(frac {B1}{B2} = frac {n1}{n2} [ frac {I1}{I2} ] )
(frac {B1}{B2} = frac {500}{300} , times , frac {I}{(frac {I}{3})})
(frac {B1}{B2}) = 5
B2 = (frac {B1}{5})
B2 = (frac {7.54 , times , 10^{-2}}{5})
B2 = 0.01508 = 1.50 × 10^-2Wb/m²

6. A solenoid has core of a material with relative permeability 200 and its windings carry a current of 2 A. The number of turns of the solenoid is 200 per meter. What is the magnetization of the material?
a) 6 × 10⁴ A/m
b) 7 × 10⁴ A/m
c) 8 × 10⁴ A/m
d) 9 × 10⁴ A/m
Answer: c
Clarification: Given: number of turns (n) = 200 turns per meter; current (I) = 2 A; Relative permeability (µ_r) = 200
Magnetic intensity is given by: H = nI = 200 × 2 = 400 A/m
μ_r = 1 + γ → γ is the magnetic susceptibility of the material
γ = μ_r – 1
Magnetization → M = γ x H
M = (μ_r – 1) × H
M = (200 – 1) × 200
M = 199 × 200
M = 79600
M = 7.96 × 10⁴ A/m ≈ 8 × 10⁴ A/m

7. A rectangular coil, of sides 4 cm and 5 cm respectively, has 50 turns in it. It carries a current of 2 A, and is placed in a uniform magnetic field of 0.5 T in such a manner that its plane makes an angle of 60^o with the field direction. Calculate the torque on the loop.
a) 5 × 10^-4 N m
b) 5 × 10^-2 N m
c) 5 × 10^-3 N m
d) 5 × 10^-5 N m
Answer: b
Clarification: The magnitude of torque experienced by a current carrying coil kept in a uniform magnetic field is given by:
τ = MB sin (θ) = (NIA) × B sin (θ)
M → Magnetic moment of the coil
N → Number of turns
I → Current in coil
A → Area of the coil
θ → Angle between normal to the plane of the coil and the direction of the magnetic field
Given: N = 50; A = 4 × 5 = 20 cm² = 20 × 10^-4 m²; B = 0.5 T; I = 2 A
θ = 90^o – 60^o = 30^o
τ = (NIA) × B sin (θ)
τ = 50 × 2 × 0.5 × 20 × 10^-4 × sin (30^o)
τ = 0.05
τ = 5 × 10^-2 N m

Physics Multiple Choice Questions on “Electromagnetic Induction – Eddy Currents”.

1. Name the current induced in solid metallic masses when the magnetic flux threading through them changes.
a) Ampere currents
b) Faraday currents
c) Eddy currents
d) Solenoidal currents
Answer: c
Clarification: Eddy currents are the currents induced in solid metallic masses when the magnetic flux threading through them changes. These currents look like eddies or whirlpools in water and so they are known as eddy currents.

2. Identify the type of commercial motor which works as a consequence of eddy currents.
a) Compressors
b) Induction motors
c) Turbines
d) Hydropowered motors
Answer: b
Clarification: A rotating magnetic field is produced employing two single-phase currents. A metallic rotor placed inside the rotating magnetic field starts rotating due to large eddy currents produced in it. These motors are commonly used in fans.

3. Which of the following is not an application of eddy currents?
a) Ammeter
b) Dead beat galvanometer
c) Speedometer
d) Energy meters
Answer: a
Clarification: Although eddy currents are undesirable, still they find applications in the following devices: Induction furnace, Electromagnetic damping, Electric brakes, Speedometers, Induction motor, Electromagnetic shielding, and energy meters.

4. What is measured by the eddy currents induced in energy meters?
a) Electric potential
b) Electric induction
c) Electric power
d) Electric energy
Answer: d
Clarification: In energy meters used for measuring electric energy, the eddy currents induced in an aluminum disc are made use of. Therefore, electric energy is measured by the eddy currents induced in energy meters.

5. Eddy currents can be used to heat localized tissues of the human body. What is this branch of science referred to as?
a) Inductology
b) Eddy dynamics
c) Inductothermy
d) Eddy mechanics
Answer: c
Clarification: Eddy currents can be used to heat localized tissues of the human body. This branch is called inductothermy. Inductothermy is a method of electrotherapeutics in which certain parts of the body of the patient are heated by an alternating, predominantly high-frequency electromagnetic field, which induces eddy currents in body tissues.

6. Eddy currents may not be suitable for electrostatic shielding.
a) True
b) False
Answer: b
Clarification: No, Eddy currents can be used for electromagnetic shielding. Electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. The higher the conductivity of the sheet used, the better the shielding of the transient magnetic field.

7. ‘X’ set up in the drum exert a torque on the drum to stop the train. Identify X.
a) Induction
b) Eddy currents
c) Electromagnetic oscillations
d) Air pressure
Answer: b
Clarification: A strong magnetic field is applied to the rotating drum attached to the wheel. The eddy currents set up in the drum exert a torque on the drum to stop the train. That is the reason eddy currents are set up in the drum of the train wheel.

8. Identify the law which is used to find out the direction of eddy currents.
a) Lenz’s law
b) Faraday’s law
c) Ampere circuital law
d) Maxwell’s law
Answer: a
Clarification: Eddy currents are the currents induced in solid masses when the magnetic flux threading through them changes. Eddy currents also oppose the change in magnetic flux, so their direction is given by Lenz’s law.

9. How can electromagnetic damping be increased?
a) Increasing the magnetic density
b) Reduce viscosity
c) Winding the coil on the aluminum frame
d) Increase the temperature of the coil
Answer: c
Clarification: The electromagnetic damping can be increased by winding the coil on a lighter copper or aluminum frame. As the frame moves in the magnetic field, eddy currents are set up in the frame which resists the motion of the coil.

Physics Multiple Choice Questions on “Ray Optics – Refraction through a Prism”.

1. What is the angle between the incident ray and the emergent ray in a prism called?
a) Angle of deviation
b) Angle of refraction
c) Angle of reflection
d) Angle of dispersion
Answer: a
Clarification: A prism is a homogenous, transparent medium enclosed two plane surfaces inclined at an angle. These surfaces are called the refracting surfaces and the angle between the incident ray and emergent ray is known as the angle of deviation.

2. Identify the prism formula from the following.
a) μ=⁡(frac {sin [ frac {A – delta_m}{2} ] }{ sin⁡(frac {A}{2}) } )
b) μ=⁡(frac {sin [ frac {A + delta_m}{4} ] }{ sin⁡(frac {A}{2}) } )
c) μ=⁡(frac {sin [ frac {A + delta_m}{2} ] }{ sin⁡(frac {A}{2}) } )
d) μ=⁡(frac {sin [ frac {A + delta_m}{2} ] }{ cos⁡(frac {A}{2}) } )
Answer: c
Clarification: The refractive index of the material of the prism is given as:
μ=⁡(frac {sin [ frac {A + delta_m}{2} ] }{ sin⁡(frac {A}{2}) } )
Where A ➔ Angle of the prism and δ_m ➔ angle of minimum deviation. This is known as the prism formula.

3. Which of the following is Cauchy’s formula?
a) μ=A+Bλ²+Cλ⁴
b) μ=A+(frac {B}{lambda^2} + frac {C}{lambda^4})
c) μ=A+B+CBλ
d) μ=A+(frac {B}{lambda}+frac {C}{lambda^2})
Answer: b
Clarification: Cauchy’s dispersion formula is an empirical expression that gives an approximate relation between the refractive index of a medium and the wavelength of the light. Cauchy’s formula is given as:
μ=A+(frac {B}{lambda^2} + frac {C}{lambda^4})
Where A, B, and C are the arbitrary constants of the medium.

4. The Refractive index of a material of a prism is different for different colors.
a) True
b) False
Answer: a
Clarification: Yes, this statement is true. The Refractive index is the property of a material. Since δ=(μ-1)A, different colors turn through different angles on passing through the prism. This is the cause of dispersion. Therefore, the refractive index of a material of a prism is different for different colors.

5. What is the difference in a deviation between any two colors called?
a) Linear dispersion
b) Angular dispersion
c) Mean deviation
d) Mean dispersion
Answer: b
Clarification: The difference in a deviation between any two colors is known as angular dispersion. Angular dispersion is given as:
δ_V-δ_R=(μ_V-μ_R)A
Where μ_V and μ_R are the refractive index for violet rays and red rays, respectively. Mean deviation is δ = ( ( frac {delta_V + delta_R}{2} ) ).

6. Pick out the formula for dispersive power from the following.
a) Dispersive power = (frac {mean , deviation}{angular , dispersion })
b) Dispersive power = mean deviation * angular dispersion
c) Dispersive power = mean deviation + angular deviation
d) Dispersive power = (frac {angular , dispersion}{mean , deviation})
Answer: d
Clarification: The formula for dispersive power is given as:
Dispersive power (ω)=(frac {Angular , dispersion (delta_V-delta_R)}{Mean , deviation (delta)})
ω=(frac {mu_V-mu_R}{mu-1})
Where μ=(frac {mu_V+mu_R}{2}) = mean refractive index

7. What is the condition for dispersion without deviation?
a) δ-δ’=0
b) δ+δ’=0
c) δ × δ^‘=0
d) (frac {delta}{delta^{‘}})=0
Answer: b
Clarification: Consider combining two prisms of refracting angles A and A’, and dispersive powers ω and ω’ respectively in such a way that their refracting angles are reversed concerning each other. For no deviation, the condition is:
δ+δ’=0
So, (μ-1)A+(μ’-1)A’=0 or A’=-(frac {(mu -1)A}{(mu^{‘}-1)})

8. The Refractive index and wavelength are directly proportional to each other.
a) True
b) False
Answer: b
Clarification: No, this statement is false. The Refractive index of material and wavelength of color are inversely proportional to each other. For example, the wavelength of the color red is the longest, so the refractive index of the same will be the smallest. Similarly, violet has the greatest refractive index and the shortest wavelength.

9. Calculate the dispersive power of crown glass where μ_V=1.456 and μ_R=1.414.
a) 0.0096
b) 0.45
c) 0.96
d) 0.096
Answer: d
Clarification: Given: The refractive index for violet color = 1.456; Refractive index for red color = 1.414
Required equation ➔
ω=(frac {mu_V – mu_R}{mu – 1})
Also, μ=(frac {(mu_V+mu_R)}{2})
μ=(frac {1.456+1.414}{2})=1.435
Thus, ω=(frac {1.456-1.414}{1.435-1})
ω=(frac {0.042}{0.435})
ω=0.096
Therefore, the dispersive power of crown glass is 0.096.

10. A thin prism with an angle of 3^o and made from glass of refractive index 1.15 is combined with another prism made from glass and has a refractive index of 1.45. If the dispersion were to occur without deviation then what should be the angle of the second prism?
a) 3^o
b) 0^o
c) 1^o
d) 2^o
Answer: c
Clarification: The required equation ➔ δ=(μ-1)A
When two prisms are combined, then:
δ=δ+δ’=(μ-1)A+(μ’-1)A’=0
So, A’=-(frac {(mu-1)A}{mu^{‘}-1})
A’=-(frac {(1.15-1)}{1.45-1}) × 3
A’=-1^o
Therefore, the angle of the other prism is 1^o and opposite of the first prism.

Physics Online Quiz for Class 12 on “Alpha-Particle Scattering and Rutherford’s Nuclear Model of Atom”.

1. The atomic number of silicon is 14. Its ground state electronic configuration is
a) 1s²2s²2p⁶3s²3p⁴
b) 1s²2s²2p⁶3s²3p³
c) 1s²2s²2p⁶3s²3p²
d) 1s²2s²2p⁶3s²3p¹
Answer: c
Clarification: The atomic number of silicon is 14.
Therefore, ¹⁴Si will have the following electronic configuration:
1s²2s²2p⁶3s²3p²

2. What is the valence electron in alkali metal?
a) f-electron
b) p-electron
c) s-electron
d) d-electron
Answer: c
Clarification: The valence electron in an alkali metal is an s-electron. Generally, they make up Group 1 of the periodic table. The different examples that come under this category are lithium, potassium, and francium.

3. Of the following pairs of species which one will have the same electronic configuration for both members?
a) Li⁺ and Na⁺
b) He and Ne⁺
c) H and Li
d) C and N⁺
Answer: d
Clarification: Carbon and the positive ion of nitrogen (N⁺) will have the same electronic configuration.
The electronic configuration of both Carbon and the positive ion of nitrogen is as follows:
1s²2s²2p⁶.

4. Which of the following did Bohr use to explain his theory?
a) Conservation of linear momentum
b) The quantization of angular momentum
c) Conservation of quantum frequency
d) Conservation of mass
Answer: b
Clarification: To explain his theory, Niels Bohr used the quantization of angular momentum. It means the radius of the orbit and the energy will be quantized. The Boundary conditions for the wave function are periodic.

5. According to Bohr’s principle, what is the relation between the principal quantum number and the radius of the orbit?
a) r proportional to (frac {1}{n})
b) r proportional to (frac {1}{n^2})
c) r proportional to n
d) r proportional to n²
Answer: d
Clarification: The equation is given as:
r = (frac {n^2 h^2}{4pi^2 m k Z e^2})
Therefore, we can say that the radius of the orbit is directly proportional to the square of the principal quantum number.

6. The kinetic energy of the α-particle incident on the gold foil is doubled. The distance of closest approach will also be doubled.
a) True
b) False
Answer: b
Clarification: As the distance of the closest approach is inversely proportional to the kinetic energy of the incident α-particle, so the distance of the closest approach is halved when the kinetic energy of α-particle is doubled.

7. Based on the Bohr model, what is the minimum energy required to remove an electron from the ground state of Be atom? (Given: Z = 4)
a) 1.63 eV
b) 15.87 eV
c) 30.9 eV
d) 217.6 eV
Answer: d
Clarification: The equation is given as:
E_n = (frac {-13.6 Z^2}{n^2}) eV
E_n = (frac {-13.6 times 16 }{ 1 })
E_n = -217.6 eV
Hence the ionization energy for an electron in the ground state of Be atom is 217.6 eV.

8. If an α-particle collides head-on with a nucleus, what is its impact parameter?
a) Zero
b) Infinite
c) 10^-10 m
d) 10¹⁰ m
Answer: a
Clarification: The perpendicular distance between the path of a projectile and the center of the potential field is the impact parameter. Therefore, for a head-on collision of the α-particle with a nucleus, the impact parameter is equal to zero.

9. In which of the following system, will the radius of the first orbit (n=1) be minimum?
a) Doubly ionized lithium
b) Singly ionized helium
c) Deuterium atom
d) Hydrogen atom
Answer: a
Clarification: The equation is given as:
r = (frac {h^2}{2pi^2 m k Z e^2})
Hence, of the given atoms/ions, (Z = 3) is maximum for doubly ionized lithium, so the radius of its first orbit is minimum.

10. An α-particle of energy 10 MeV is scattered through 180^o by a fixed uranium nucleus. Calculate the order of distance of the closest approach?
a) 10^-20cm
b) 10^-12cm
c) 10^-11cm
d) 10¹²cm
Answer: b
Clarification: r₀ = (frac {(2Ze^2)}{(4pi varepsilon_0 (frac {1}{2}m v^2))})
r₀ = (frac {9 times 10^9 times 2 times 92 times (1.6 times 10^{-19})^2}{10 times 1 times 10^{-13}}) J
r₀ = 4.239 × 10^-14 m
r₀ = 4.2 × 10^-12 cm.

Physics Online Quiz for Class 12,