Category: Class 12 Physics Objective Questions

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,

Physics Multiple Choice Questions on “Digital Electronics and Logic Gates”.

1. Which of the following is correct about logic gates?
a) Logic gates have one or more input signals and only one output signal
b) Logic gates have only one input and output signal
c) Logic gates are analogous circuits
d) Logic gates have only one input and many output signals
Answer: a
Clarification: Logic gates are the basic building blocks of any digital system. A digital circuit with one or more input signals but only one output signal is known as a logic gate. All the other statements are not valid.

2. How many basic logic gates are there?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: The logic gates are considered to be the building blocks of a digital system. Each logic gate follows a certain logical relationship between input and output voltage. There are 3 basic logic gates, namely, OR gate, AND gate, and NOT gate.

3. Which of the following gates can have only one input?
a) OR gate
b) NOT gate
c) AND gate
d) NAND gate
Answer: b
Clarification: The NOT gate is the simplest of all logic gates. It has only one input and one output signal. NOT gate is also called as the inverter because t inverts the input signal. All the other gates can have one or more input signals.

4. The truth table shows both the input and output signals.
a) True
b) False
Answer: a
Clarification: Yes, this is a true statement. The truth table is a table that shows all possible input combinations and the corresponding output combinations for a particular logic gate. These are used in connection with Boolean algebra, Boolean functions, and propositional calculus.

5. Which gate will a NAND gate be equivalent to when two inputs of NAND gates are shorted?
a) AND gate
b) OR gate
c) NAND gate
d) NOT gate
Answer: d
Clarification: When two inputs of a NAND gate are shorted, then the Boolean expression for it becomes:
Y=not(A.A)
Y=not(A)
Therefore, the NAND gate will be equivalent to the NOT gate.

6. What should be done to obtain an OR gate from a NAND gate?
a) We need only 3 NAND gates
b) We need two NOT gates obtained from NAND gates and one NAND gate
c) We need 3 NOT gates obtained from NAND gates and 3 NAND gates
d) We need 2 NAND gates and 4 AND gates obtained from NAND gates.
Answer: b
Clarification: We need OR gate (Y = A + B) from NAND gate (Y = not (A.B))
Y = not (not (A.B))
Y = not (not (A)) + not (not (B))        [Using the Boolean identify ➔ not (A.B) = not (A) + not (B)]
So, Y = A + B [Since not (not (A)) = A and not (not (B)) = B]
Therefore, to obtain an OR gate from a NAND gate, we need two NOT gates obtained from NAND gates and one NAND gate.

7. In Boolean algebra, what will not (A + not (B)).C) be equal to?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: not (A + not (B)).C) = not (A + not (B)) + not (C)     [From De Morgan’s theorem]
not (A + not (B)).C) = not (A).not (B) + not(C)     [From De Morgan’s theorem]
not (A + not (B)).C) = not (A).B + not (C)

Physics Multiple Choice Questions on “Current Electricity – Temperature Dependence of Resistivity”.

1. Identify the material which is suitable for making standard resistors.
a) Silver
b) Copper
c) Constantan
d) Germanium
Answer: c
Clarification: Alloys like constantan or manganin are used for making standard resistance coils due to their high resistivity values and very small temperature coefficient.

2. What is the unit of the temperature coefficient of resistance (α)?
a) ^oC
b) ^oC^-1
c) (frac {Omega}{ ^{circ} C})
d) (frac { ^{circ} C}{Omega})
Answer: b
Clarification: The temperature coefficient of resistivity is defined as the increase in resistivity per unit resistivity per degree rise in temperature.
The unit of the temperature coefficient of resistance is ^oC^-1.

3. Which of the following relation is significant for metals when the temperature increases?
a) Resistivity increases and conductivity decreases
b) Resistivity decreases and conductivity decreases
c) Resistivity and conductivity do not change with temperature
d) Temperature dependence is non-linear
Answer: a
Clarification: The resistivity of a metal increases and the conductivity decreases with the increase in temperature. With an increase in temperature, the free electrons collide more frequently with the metal ions. The mean collision time also decreases.

4. Identify the type of material based on the T-ρ graph given below.

a) Silicon
b) Polymer
c) Nichrome
d) Copper
Answer: d
Clarification: For metals, the temperature coefficient of resistivity is positive. At lower temperatures, the resistivity of a pure metal increases as a higher power of temperature. So, the answer is copper, which is a metal.

5. Which among the following has weak temperature dependence values with resistivity?
a) Silver
b) Copper
c) Nichrome
d) Germanium
Answer: c
Clarification: Alloys have high resistivity. The resistivity of nichrome has weak temperature dependence. At absolute zero, a pure metal has negligibly small resistivity while an alloy like nichrome has some residual resistivity.

6. The resistivity of semiconductors and insulators decreases linearly with the increase of temperature.
a) True
b) False
Answer: b
Clarification: The resistivity of semiconductors and insulators decreases exponentially with the increase in temperature. This is because the number density of free electrons increases exponentially with the increase in temperature.

7. The resistivity of ‘X’ decreases with temperature and its coefficient of resistivity is negative. Identify X.
a) Silver
b) Silicon
c) Copper
d) Nichrome
Answer: b
Clarification: The coefficient of resistivity is negative for semiconductors and their resistivity decreases with temperature. The relaxation time does not change with temperature but the number density of free electrons increases exponentially with the increase in temperature. Consequently, the resistivity decreases exponentially with the increase in temperature.

8. A wire has a resistance of 5.5 Ω at 19^oC and 21.5 Ω at 200^oC. Find the temperature coefficient of resistivity(α) of the material.
a) 0.016 ^oC^-1
b) 0.160 ^oC^-1
c) 1.600 ^oC^-1
d) 16.00 ^oC^-1
Answer: a
Clarification: Temperature coefficient α=(frac {(R_2-R_1)}{R_2(T_2-T_1)}).
α = (frac {(21.5-5.5)}{(5.5(200-19))})
= 0.01607 ^oC^-1.

9. Which of the following is not a valid reason for using alloys to make standard resistors?
a) Alloys have a high value of resistivity
b) They are least affected by air and moisture
c) Alloys have a large temperature coefficient
d) Their contact potential with copper is small
Answer: c
Clarification: Alloys have a high value of resistivity. They have a very small temperature coefficient. So their resistance does not change appreciably even for several degrees rise of temperature. That leaves the answer – alloys have a large temperature coefficient.

10. Identify the temperature at which the resistance of copper would be double of its resistance at ^oC. Given α (temperature coefficient of resistivity) for copper=3.9 x 10^-3oC^-1.
a) 125^oC
b) 256^oC
c) 1080^oC
d) 273^oC
Answer: b
Clarification: α=(frac {(R_2-R_1)}{R_2(T_2-T_1)}).
α=(frac {(2R_0-R_0)}{R_0(T-0)} = frac {1}{T}).
T=(frac {1}{alpha})
= (frac {1}{3.9}) x 10^-3oC^-1
= 256^oC.
Therefore, the required temperature is 256^oC.