MCQ Questions for Class 12 Physics Chapter 13 Nuclei

Nuclei Class 12 MCQs Questions with Answers

Nuclei Class 12 MCQ Chapter 13 Question 1.

The gravitational force between a H – atom and another particle of mass m will be given by Newton’s law:\(\begin{equation}\mathrm{F}=\mathrm{G} \frac{\mathrm{M} \cdot m}{r^{2}}\end{equation}\) , where r is in km and

(A) \(\begin{equation}\mathrm{M}=m_{\text {proton }}+m_{\text {electron }}\end{equation}\)
(B) \(\begin{equation}\mathrm{M}=m_{\text {proton }}+m_{\text {electron }}-\frac{\mathrm{B}}{c^{2}} \quad(\mathrm{~B}=13.6 \mathrm{eV})\end{equation}\)
(C) M is not related to the mass of the hydrogen atom.
(D)\(\begin{equation}\mathrm{M}=m_{\text {proton }}+m_{\text {electron}}-\frac{|\mathrm{V}|}{c^{2}}(|\mathrm{~V}|\end{equation}\)= magnitude of the potential energy of electron in the H-atom).
Answer:
(B) \(\begin{equation}\mathrm{M}=m_{\text {proton }}+m_{\text {electron }}-\frac{\mathrm{B}}{c^{2}} \quad(\mathrm{~B}=13.6 \mathrm{eV})\end{equation}\)

Explanation:
During formation of H – atom, some mass of nucleons convert into energy by the equation
\(\begin{equation}\mathrm{E}=m c^{2}\end{equation}\)
This energy is used to bind the nucleons along with nucleus. So mass of atom becomes slightly less than sum of actual masses of nucleons and electrons.
Actual mass of H – atom
\(\begin{equation}=\mathrm{M}_{p}+\mathrm{M}_{e}-\frac{\text { B.E. }}{c^{2}}\end{equation}\)
\(\begin{equation}\frac{\mathrm{B}}{c^{2}}\end{equation}\) = Binding energy
So, the binding energy of H atoms is 13.6 eV per atom.

MCQ Questions for Class 12 Physics Chapter 13 Nuclei

Nuclear Chapter 13 MCQ Question 2.

Heavy stable nucleus have more neutrons than protons. This is because of the fact that

(A) neutrons are heavier than protons.
(B) electrostatic force between protons are repulsive.
(C) neutrons decay into protons through beta decay.
(D) nuclear forces between neutrons are weaker than that between protons.
Answer:
(B) electrostatic force between protons are repulsive.

Explanation:
Electrostatic force between proton – proton is repulsive which causes theinsfatbility of nucleus. So neutrons are more than the number of protons.

Nuclei Class 12 MCQ Chapter 13 Question 3.

Tritium is an isotope of hydrogen whose nucleus Triton contains 2 neutrons and 1 proton. Free neutrons decay into p + \(\begin{equation}\tilde{e}\end{equation}\) + v. If one of the neutrons in Tritium decays, it would transform into \(\begin{equation}{ }_{2} \mathrm{He}^{3}\end{equation}\) nucleus. This does not happen. This is because

(A) tritium energy is less than that of a \(\begin{equation}\mathrm{He}^{3}\end{equation}\) nucleus.
(B) the electron created in the beta decay process cannot remain in the nucleus.
(C) both the neutrons in triton have to decay simultaneously resulting in a nucleus with 3 protons, which is not a \(\begin{equation}\mathrm{He}^{3}\end{equation}\) nucleus.
(D) because free neutrons decay due to external perturbations which is absent in a triton nucleus.
Answer:
(A) tritium energy is less than that of a \(\begin{equation}\mathrm{He}^{3}\end{equation}\) nucleus.

Explanation:
Tritium \(\begin{equation}
\left({ }_{1}^{3} \mathrm{H}\right)
\end{equation}\) has 1 proton and 2 neutrons.
If a neutron decays as, \(\begin{equation}
n \rightarrow p+\bar{e}+v
\end{equation}\)
then nucleus will have 2 protons and 1 neutron, i.e. triton atom converts in \(\begin{equation}
{ }_{2} \mathrm{He}^{3}
\end{equation}\) (2 proton and 1 neutron). Binding energy of \(\begin{equation}
{ }_{1} \mathrm{He}^{3}
\end{equation}\) is much smaller than \(\begin{equation}
{ }_{2} \mathrm{He}^{3}
\end{equation}\), so transformation is not possible energetically.

MCQ Questions for Class 12 Physics Chapter 13 Nuclei

Nuclear Class 12 Chapter 13 MCQ Question 4.

Nuclear force is a ……….. and ……….. force.

(A) Strong, long – range
(B) Strong, short range
(C) Weak, long – range
(D) Weak, short – range
Answer:
(B) Strong, short range

Explanation:
Nudear force is the strongest short – range force which binds the neutrons and protons in a nudeus.

MCQ On Nuclei Class 12 Chapter 13 Question 5.

Two nuclei have mass number in the ratio 1 : 2. The ratio of their nudear densities is

(A) 1 : 2
(B) 2 : 1
(C) 1 : 1
(D) Cannot be defined from mass number ratio
Answer:
(C) 1 : 1

Explanation:
Nuclear density is same for all nuclei.

Question 6.

The mass of a nucleus in its ground state is

(A) less than the total mass of neutrons and protons.
(B) greater than the total mass of neutrons and protons.
(C) equal to the total mass of neutrons and protons.
(D) equal to the total mass of neutron, protons and electrons.
Answer:
(A) less than the total mass of neutrons and protons.

Explanation:
Protons and neutrons have to come very dose to form a nucleus. In order to achieve this doseness, a lot of energy is required. This energy is provided by the nucleons at the expense of certain portion of their masses. For this reason, the mass of a nucleus in its ground state is always less than the total mass of the constituent neutrons and protons.

Question 7.

……….. has the mass closest to the mass of positron.

(A) Proton
(B) Neutron
(C) Electron
(D) Neutrino
Answer:
(C) Electron

Explanation:
Positron is anti – particle of electron.

Question 8.

X amount of energy is required to remove an electron from its orbit and Y amount of energy is required to remove a nucleon from the nucleus.

(A) X = Y
(B) X > Y
(C) Y > X
(D) X ≥ Y
Answer:
(C) Y > X

Explanation:
Nuclear force is greater than Coulomb force.

MCQ Questions for Class 12 Physics Chapter 13 Nuclei

Question 9.

Suppose we consider a large number of containers each containing initially 10,000 atoms of a radio – active material with a half-life of 1 year. After 1 year,

(A) all the containers will have 5,000 atoms of the material.
(B) all the containers will contain the same number of atoms of the material but that number will only be approximately 5,000.
(C) the containers will in general have different numbers of the atoms of the material but their average will be close to 5,000.
(D)none of the containers can have more than 5000 atoms.
Answer:
(C) the containers will in general have different numbers of the atoms of the material but their average will be close to 5,000.

Explanation:
Half – life time for a radio – active substance is defined as the time in which a radio – active atomic substance remains half of its original value of radio – active atom. Given that, Half – life = 1 year. So, after 1 year means one half – life, that is, average atoms of radioactive substance remain after 1 year in each container is equal to half of 10,0d0 = 5,000 atoms (average).

Question 10.

Consider the following reaction:
\(\begin{equation}{ }_{Z}^{A} X \rightarrow{ }_{Z+1}^{A} Y \rightarrow{ }_{Z-1}^{A-4} Z \rightarrow{ }_{Z-1}^{A-4} Z
\end{equation}\)
Radioactive radiation emitted in the following sequence:

(A) β, α, γ
(B) α, β, γ
(C) γ, β, α
(D) β, γ, α
Answer:
(A) β, α, γ

Explanation:
In β – emission, A remains same and Z increases by 1.
In α – emission, A decreases by 4 and Z decreases by 2.
In γ – emission, there is no change in A and Z.

Question 11.

Which of the following material will be the best moderator for a nuclear power plant ?

(A) Lighter element
(B) Heavier element
(C) Both of the above
(D) None of the above
Answer:
(A) Lighter element

Explanation:
Neutron will be slowed down more if the size of moderator atom is doser to a neutron. For this reason, lighter element will be more efficient moderator.

Assertion And Reason Based MCQs (1 Mark each)

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is true

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Question 1.

Assertion (A): Two atoms of different elements having same mass number but different atomic numbers are called isobars.
Reason (R): Atomic number is the number of protons present and atomic number is the total number of protons and neutrons present in a nucleus.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
Two atoms of different elements having same mass number but different atomic numbers are called isobars. The assertion is true. Atomic number is the number of protons present and atomic number is the total number of protons and neutrons present in a nucleus. The reason is also true. But the reason does not explain the assertion.

MCQ Questions for Class 12 Physics Chapter 13 Nuclei

Question 2.

Assertion (A): Nuclear force is same between neutron – neutron, proton – proton and neutron – proton.
Reason (R): Nuclear force is charge independent.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
Nuclear force acts between nucleons. It is a powerful attractive force and acts in a very short distance.
Neutrons are electrically neutral. So, neutron – neutron is neither attractive nor repulsive force acting. So, Nuclear force binds them together. Protons are having +e charge each. They experience a repulsive force. But for the short distance, the attractive nuclear force is strong enough to overcome this force. So, assertion and reason both are true and the reason explains the assertion.

Question 3.

Assertion (A): Electrons do not experience strong nuclear force.
Reason (R): Strong nuclear force is charge independent.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
Nuclear force is a powerful attractive force acts as long as the distance between particles is within \(\begin{equation}10^{-15} \mathrm{~m}\end{equation}\). This force is charge independent. But as distance increases, the effect of nuclear force rapidly falls.

Electrons are distributed far away. The distance is beyond the range of the nuclear force. Hence nuclear force has no effect on electrons. So, the assertion and reason both are true. But the reason does not explain the assertion.

Question 4.

Assertion (A): The binding energy per nucleon, for nuclei with mass number (A) > 56 decreases with A.
Reason (R): Nuclear force is weak in heavier nuclei.

Answer:
(C) A is true but R is false

Explanation:
From the binding energy per nucleon vs. mass number, we find that binding energy per nucleon is maximum at A = 56. After that, binding energy per nucleon decreases as A increases. So, assertion is true. Nuclear force remains same for all nuclei. Hence the reason is false.

MCQ Questions for Class 12 Physics Chapter 13 Nuclei

Question 5.

Assertion (A): Density of all the nuclei is same.
Reason (R): Radius of nucleus is directly proportional to the cube root of mass number.

Answer:
Assertion (A): Density of all the nuclei is same.

Explanation:
Radius of nucleus = \(\begin{equation}
=\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}
\end{equation}\)
So, Volume of nudeus, \(\begin{equation}
V=\frac{4}{3} \pi R_{0}^{3} A
\end{equation}\)
Considering mass of proton = mass of neutron = m
The mass of the nudeus = M = mA
So, density = \(\begin{equation}
\mathrm{M} / \mathrm{V}=\frac{m \mathrm{~A}}{\frac{4}{3} \pi \mathrm{R}_{0}^{3} \mathrm{~A}}=\frac{m}{\frac{4}{3} \pi \mathrm{R}_{0}^{3}}
\end{equation}\)
So, the mean density is independent of mass number.
So, assertion and reason both are true and the reason properly explains the assertion.

Case-Based MCQs

Attempt any 4 sub – parts out of 5. Each sub – part carries 1 mark.

I. Read the following text and answer the following questions on the basis of the same:

India’s atomic energy programme:
The atomic energy programme in India was launched around the time of independence under the leadership of Homi J. Bhabha (1909 – 1966). An early historic achievement was the design and construction of the first nuclear reactor in India (named Apsara) which went critical on August 4, 1956. India indigenously designed and constructed plutonium plant at Trombay, which ushered in the technology of fuel reprocessing (separating useful fissile and fertile nuclear materials from the spent fuel of a reactor).

Research reactors that have been subsequently commissioned include ZERLINA, PURNIMA (I, II and III), DHRUVA and KAMINI. KAMINI is the country’s first large research reactor that uses \(\begin{equation}
\text { U-233 }\end{equation}\) as fuel. The main objectives of the Indian Atomic Energy programme are to provide safe and reliable electric power for the country’s social and economic progress and to be self reliant in all aspects of nuclear technology.

Exploration of atomic minerals in India undertaken since the early fifties has indicated that India has limited reserves of uranium, but fairly abundant reserves of thorium. Accordingly, our country has adopted a three – stage strategy of nuclear power generation. The first stage involves the use of natural uranium as a fuel, with heavy water as moderator. The Plutonium – 239 obtained from reprocessing of the discharged fuel from the reactors then serves as a fuel for the second stage – the fast breeder reactors.

They are so called because they use fast neutrons for sustaining the chain reaction (hence no moderator is needed) and, besides generating power, also breed more fissile species (plutonium) than they consume. The third stage, most significant in the long term, involves using fast breeder reactors to produce fissile Uranium-233 from Thorium-232 and to build power reactors based on them.

Question 1.

India’s atomic energy programme was launched by:

(A) Shanti Swarup Bhatnagar
(B) Homi J. Bhabha
(C) Meghnad Saha
(D) Daulat Singh Kothari
Answer:
(B) Homi J. Bhabha

Explanation:
The atomic energy programme in India was launched around the time of independence under the leadership of Homi J. Bhabha (1909-1966).

MCQ Questions for Class 12 Physics Chapter 13 Nuclei

Question 2.

First nuclear reactor of India :

(A) APSARA
(B) ZERLINA
(C) DHRUBA
(D) KAMINI
Answer:
(A) APSARA

Explanation: An early historic achievement was the design and construction of the first nuclear reactor in India named APSARA.

Question 3.

Which one of the following is not a nuclear reactor ?

(A) PURNIMA
(B) DHRUVA
(C) KAMINI
(D) ARYABHATTA
Answer:
(D) ARYABHATTA

Explanation:
ARYABHATTA is an Indian artificial satellite.

Question 4.

The main objectives of the Indian Atomic Energy programme:

(A) Development of Nuclear weapons for success in warfare
(B) Generation of safe and reliable electric power
(C) Efficient medical treatment
(D) To breed more fissile species
Answer:
(B) Generation of safe and reliable electric power

Explanation:
The main objectives of the Indian Atomic Energy programme are to provide safe and reliable electric power for the country’s social and economic progress and to be self – reliant in all aspects of nuclear technology.

Question 5.

India has limited reserves of ……….. but fairly abundant reserves of ……….. :

(A) Plutonium, Thorium
(B) Thorium, Uranium
(C) Plutonium, Uranium
(D) Uranium, Thorium
Answer :
(D) Uranium, Thorium

Explanation:
Exploration of atomic minerals in India undertaken since the early fifties has indicated that India has limited reserves of uranium, but fairly abundant reserves of thorium.

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

II. Read the following text and answer the following questions on the basis of the same:

Grand Unification Theory :
There are four fundamental forces in the universe :

  • Gravitational force
  • Electromagnetic force
  • The weak nuclear force
  • The strong nuclear force

The weak and strong forces are effective only over a very short range and dominate only at the level of subatomic particles. Gravitational force and Electromagnetic force have infinite range.

The Four Fundamental Forces and their strengths :

  1. Gravitational Force – Weakest force; but has infinite range.
  2. Weak Nuclear Force – Next weakest; but short range.
  3. Electromagnetic Force – Stronger, with infinite range.
  4. Strong Nuclear Force – Strongest; but short range.

Unification:

  • The weak nuclear force and electromagnetic force have been unified under the Standard Electroweak Theory, (Glashow, Weinberg and Salaam were awarded the Nobel Prize for this in 1979).
  • Grand unification theories attempt to treat both strong nuclear force and electroweak force under the same mathematical structure.
  • Theories that add gravitational force to the mix and try to unify all four fundamental forces into a single force are called Superunified Theories. It has not yet been successful.

MCQ Questions for Class 12 Physics Chapter 13 Nuclei

Question 1.

What are the 4 fundamental forces ?

(A) Gravitational force, electromagnetic force, nuclear force, Tension force
(B) Gravitational force, electromagnetic force, nuclear force, Frictional force
(C) Gravitational force, electromagnetic force, weak nuclear force, strong nuclear force
(D) Frictional force, electric force, nuclear force, magnetic force
Answer:
(C) Gravitational force, electromagnetic force, weak nuclear force, strong nuclear force

Explanation:
There are four fundamental forces in the universe:

  • Gravitational force
  • Electromagnetic force
  • the weak nuclear force
  • the strong nuclear force

Question 2.

Which fundamental force is always attractive ?

(A) Electric force
(B) Magnetic force
(C) Gravitational force
(D) Strong nuclear force
Answer :
(C) Gravitational force

Explanation:
Gravitational force is always attractive. There is no repulsive gravitational force.

Question 3.

Which two fundamental forces have been unified by Standard Electroweak Theory ?

(A) Weak nuclear force and electromagnetic force
(B) Strong nuclear force and electromagnetic force
(C) Gravitational force and electromagnetic force
(D) Weak nuclear force and strong nuclear force
Answer:
(A) Weak nuclear force and electromagnetic force

Explanation:
The weak nuclear force and electromagnetic force have been unified under the Standard Electroweak theory. For this, Glashow, Weinberg and Salaam were awarded the Nobel Prize in 1979.

MCQ Questions for Class 12 Physics Chapter 13 Nuclei

Question 4.

Which one is the weakest force ?

(A) Weak nuclear force
(B) Electromagnetic force
(C) Strong magnetic force
(D) Gravitational force
Answer:
(D) Gravitational force

Explanation:
Gravitational force is the weakest force.

Question 5.

Which of the following forces have infinite ranges ?

(A) Weak nuclear force and strong nuclear force
(B) Gravitational force and Electromagnetic force
(C) Weak nuclear force and Gravitational force
(D) All the forces
Answer:
(B) Gravitational force and Electromagnetic force

Explanation:
Gravitational force and Electromagnetic force are extended upto infinity.

MCQ Questions for Class 12 Physics with Answers

MCQ Questions for Class 12 Physics Chapter 12 Atoms

Atoms Class 12 MCQs Questions with Answers

Atoms Class 12 MCQ Chapter 12 Physics Question 1.

O2 molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms

(A) is not important because nuclear forces are short-ranged.
(B) is as important as electrostatic force for binding the two atoms.
(C) cancels the repulsive electrostatic force between the nuclei.
(D) is not important because oxygen nucleus have equal number of neutrons and protons.
Answer:
(A) is not important because nuclear forces are short-ranged.

Explanation:
As we know that the nuclear forces is too much stronger. Only attractive force as compared to electrostatic repulsive force and nuclear force decreases to zero on increasing distance. So in case of oxygen molecule, the distance between atoms of oxygen is larger as compared to the distances between nucleons in a nucleus. So that, the force between the nuclei of two oxygen atoms is not important as nuclear forces are short-ranged forces.

Atoms MCQ Questions Class 12 Chapter 12 Question 2.

A set of atoms in an excited state decays.

(A) in general, to any of the states with lower energy.
(B) into a lower state only when excited by an external electric field.
(C) all together simultaneously into a lower state.
(D) to emit photons only when they collide.
Answer:
(A) in general, to any of the states with lower energy.

Explanation:
A set of atoms in an excited state decays in general to any of the states with lower energy.

MCQ Questions for Class 12 Physics Chapter 12 Atoms

Physics 12th MCQ Questions Chapter 12 Question 3.

Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced, is

(A) 10.20 eV
(B) 20.40 eV
(C) 13.6 eV
(D) 27.2 eV.
Answer:
(A) 10.20 eV

Explanation:
Total energy of two H-atom in ground state = 2(-13.6) = -27.2 eV The maximum amount by which their combined kinetic energy is reduced when any one H-atom goes into first excited state after the inelastic collision, that is, the total energy of two H-atom after inelastic collision:
E = \(\frac{13.6}{n^{2}}\) M + 13.6
= \(\frac{13.6}{n^{2}}\)+ 13.61 [For excited state (n = 2)]
= 3.4 4- 13.6 = 17.0 eV
So that the loss in kinetic energy due to inelastic collision will be,
= 27.2 -17.0 = 10.2 eV

Physics Class 12 MCQ Questions Chapter 12 Question 4.

Taking the Bohr radius as a0 = 53 pm, the radius of Li++ ion in its ground state, on the basis of Bohr’s model, will be about

(A) 53 pm.
(B) 27 pm.
(C) 18 pm.
(D) 13 pm.
Answer:
(C) 18 pm.

Explanation:
According to Bohr’s model of an atom, radius of an atom in its ground state is
r = \(\frac{r_{0}}{2}\)
where, r0 is Bohr’s radius and Z is atomic number.
As given that,
r0 = 53 pm and atomic number of Lithium atom is 3
so, r = \(\frac{53}{3}\) = 17.67 pm = 18 pm

MCQ Questions for Class 12 Physics Chapter 12 Atoms

Question 5.

The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is B = – \(\frac{m e^{4}}{8 n^{2} \varepsilon_{0}^{2} h^{2}}\) (m = electron mass). If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be Me4
B = – \(\frac{m e^{4}}{8 n^{2} \varepsilon_{0}^{2} h^{2}}\) (M = proton mass) This last expression is not correct because

(A) n would not be integral.
(B) Bohr-quantisation applies only to electron
(C) the frame in which the electron is at rest is not inertial.
(D) the motion of the proton would not be in circular orbits, even approximately.
Answer:
(C) the frame in which the electron is at rest is not inertial.

Explanation:
In a hydrogen atom, electrons revolving around a fixed proton nucleus have some centripetal acceleration. So that, its frame of reference is non-inertial. In the frame of reference, where the electron is at rest, the given expression is not true as it forms the non- inertial frame of reference. As the mass of an electron is negligible as compared to proton, so the centripetal force cannot provide the electrostatic force, So the given expression is not true, as it forms non-inertial frame of reference due to me me< < < mP or centripetal force on Fe< < < FP.

Question 6.

The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because

(A) of the electrons not being subject to a central force.
(B) of the electrons colliding with each other.
(C) of screening effects.
(D) the force between the nucleus and an electron will no longer be given by Coulomb’s law.
Answer:
(A) of the electrons not being subject to a central force.

Explanation:
The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons because when we derive the formula for radius/energy levels, etc., we make the assumption that centripetal force is provided only by electrostatic force of attraction by the nucleus.

So that, this will only work for single electron atoms. In multi-electron atoms, there will also be repulsion due to other electrons. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons.

MCQ Questions for Class 12 Physics Chapter 12 Atoms

Question 7.

For the ground state, the electron in the H-atom has an angular momentum = h, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,

(A) because Bohr model gives incorrect values of angular momentum.
(B) because only one of these would have a minimum energy.
(C) angular momentum must be in the direction of spin of electron.
(D) because electrons go around only in horizontal orbits.
Answer:
(A) because Bohr model gives incorrect values of angular momentum.

Explanation:
According to; Bohr’s second postulate of atomic model, angular momentum of revolving electron must be some integral multiple of \(\frac{h}{2π}\) So the Bohr’s model of atom does not give correct value of angular momentum.

Question 8.

Choose the correct alternative from the clues given at the end of each statement:

(A) The size of the atom in Thomson’s model is ………….. the atomic size in Rutherford’s model. (much greater than/no different from/much less than)
(B) In the ground state of ………….. electrons are in stable equilibrium, while in electrons always experience a net force. (Thomson’s model/Rutherford’s model)
(C) A classical atom based on ………….. is doomed to collapse. (Thomson’s model/Rutherford’s model)
(D) An atom has a nearly continuous mass distribution in ………….. but has a highly non-uniform mass distribution in (Thomson’s model/Rutherford’s model)
(E) The positively charged part of the atom possesses most of the mass in ………….. (Rutherford’s model/both the models)
Answer:
(D) An atom has a nearly continuous mass distribution in ………….. but has a highly non-uniform mass distribution in (Thomson’s model/Rutherford’s model)

Explanation:
(A) The sizes of the atoms taken in Thomson’s model is not different from the atomic size in Rutherford’s model.
(B) In the ground state of Thomson’s model, the electrons are in stable equilibrium, while in Rutherford’s model, electrons always experience a net force.
(C) A classical atom based on Rutherford’s model is doomed to collapse.
(D) An atom has a nearly continuous mass distribution in Thomson’s model, but has a highly non-uniform mass distribution in Rutherford’s model.
(E) The positively charged part of the atom possesses most of the mass in both the models.

MCQ Questions for Class 12 Physics Chapter 12 Atoms

Assertion And Reason Based MCQs (1 Mark each)

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(Q A is true but R is false
(D) A is false and R is true

Question 1.

Assertion (A): Bohr postulated that the electrons in stationary orbits around the nucleus do not radiate.
Reason (R): According to classical Physics, all moving electrons radiate.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
Bohr postulated that electrons in stationary orbits around the nucleus do not radiate. This is true. According to classical Physics, the moving electrons radiate only when they jump from a higher energy orbit to the lower energy orbit. So, the reason is false.

Question 2.

Assertion (A): According to Rutherford, atomic model, the path of electron is parabolic.
Reason (R): Rutherford could not explain the stability of atom.

Answer:
(D) A is false and R is true

Explanation:
According to Rutherford, “the entire positive charge and most of the mass of the atom is concentrated in a small volume called the nucleus, with electrons revolving around the nucleus just as planets revolve around the Sun.” So the assertion is false.
The electron orbiting around the nucleus radiate energy. As a result, the radius of the orbit continuously decreases and the electron fall Into the nucleus. So, stability of atom is not explained. Hence the reason is true.

MCQ Questions for Class 12 Physics Chapter 12 Atoms

Question 3.

Assertion (A): In the a-particle scattering experiment, most of the a-particles pass undeviated.
Reason (R): Most of the space in the atom is empty.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
Most of the a-particles pass roughly in a straight line (within 1°) without deviation. This shows that no force is acting on them. So assertion is true. Most of the space in the atom is empty. Only 0.14% of a-particles are scattered more than 1°.

Question 4.

Assertion (A): Bohr model is not applicable for multi-electron model.
Reason (R): Bohr model cannot account for sublevel (s, p, d,f) orbitals and electron spin.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
Bohr model works well for H and He+ having one electron only. But it does not work for multi-electron atoms, since it cannot account for sublevel (s, p, d, f) orbitals and electron spin. So, assertion and reason both are true and reason explains the assertion.

MCQ Questions for Class 12 Physics Chapter 12 Atoms

Case-Based MCQs

Attempt any 4 sub-parts out of 5. Each sub-part carries 1 mark.

I. Read the following text and answer the following questions on the basis of the same:

Atomic Absorption Spectrometer:
The atomic absorption (AA) spectrometer is used to analyze metals at very low concentrations, typically in the parts per million (ppm) or parts per billion (ppb) ranges. A liquid sample containing dissolved material whose concentration is to be measured is aspirated into a thin, wide AA flame, or is introduced into a small carbon furnace which is heated to a high temperature. Basic Principle of AAS is the measurement of absorption of radiation by free atoms.

The total amount of absorption depends on the number of free atoms present and the degree to which the free atoms absorb the radiation. At the high temperature of the AA flame, the sample is broken down into atoms using an atomizer and it is the concentration of these atoms that is measured. Sample in the form of solution is used. It is broken up into a fine mist with the help of an atomizer. When the mist reaches the flame, the intense heat breaks up the sample into its individual atoms.

Atoms Class 12 MCQ Chapter 12 Physics

When a photon coming out from the hot source hits an atom and the energy of the photon is equal to the gap between two electron energy levels of the atom, then the electron in the lower energy level absorb the photon and jumps up to the higher energy level. If the photon energy does not correspond to the difference between two energy levels, then the photon will not be absorbed (it may be scattered away).

Hence in the spectrum, the wavelength corresponding to the absorbed photons is observed as black lines as shown in the following spectrum of Hydrogen. The dark lines correspond to the frequencies of light those have been absorbed by the sample element. Using this process, a source of photons (generally a white light) of various energies is used to obtain the absorption spectra of different materials and to identify them.

Question 1.

What is the basic principle of Atomic Absorption Spectrophotometer?

(A) Emission of photons when excited electron of an atom comes back to lower energy level.
(B) Absorption of photons when electrons at lower energy level jumps to a higher energy level.
(C) Emission of electrons from an atom at a very high temperature.
(D) Emission of electron when energetic photons bombard an atom.
Answer:
(B) Absorption of photons when electrons at lower energy level jumps to a higher energy level.

Explanation:
Basic principle of AAS is the 1 measurement of absorption of radiation by free atoms. When a photon hits an atom and the
energy of the photon is equal to the gap between two electron energy levels of the atom, the electron in the lower energy level absorb the photon and jumps up to the higher energy level.

MCQ Questions for Class 12 Physics Chapter 12 Atoms

Question 2.

What happens when a photon hits an atom and the energy of the photon is not equal to the gap between two electron energy levels of the atom?

(A) The photon is absorbed and the electron moves to an intermediate energy level.
(B) The photon is absorbed and the electron gets scattered
(C) The photon is not absorbed. It gets scattered.
(D) None of the above
Answer:
(C) The photon is not absorbed. It gets scattered.

Explanation:
When a photon hits an atom and the energy of the photon is equal to the gap between two electron energy levels of the atom, the electron in the lower energy level absorb the photon and jumps up to the higher energy level. If the photon energy does not correspond to the difference between two energy levels, then the photon will not be absorbed (it may be scattered away).

Question 3.

How the corresponding wavelength of the absorbed photon is represented in the absorption spectrum?

(A) By a black line
(B) By a white line
(C) By a black line in the lower wavelength range and by a white line in the higher wavelength range
(D) By a white line in the lower wavelength range and by a black line in the higher wavelength range
Answer:
(A) By a black line

Explanation:
In the spectrum, the wavelength corresponding to the absorbed photons is observed as black lines.

MCQ Questions for Class 12 Physics Chapter 12 Atoms

Question 4.

What should be the concentration of metal for analysis using Atomic Absorption Spectrometer?

(A) Very High concentration
(B) Very Low concentration
(C) Medium concentration
(D) Any concentration
Answer:
(B) Very Low concentration

Explanation:
The atomic absorption (AA) spectrometer is used to analyze metals at very low concentrations, typically in the parts per million (ppm) or parts per billion (ppb) ranges.

Question 5.

How the sample for analysis is driven to atomic state in AAS?

(A) At a very high temperature, the sample is driven to its gaseous state
(B) Using an atomizer and then intense heating.
(C) By rotating the solution of the sample at a very high speed.
(D) None of the above
Answer:
(B) Using an atomizer and then intense heating.

Explanation:
Sample in the form of solution is used. It is broken up into a fine mist with the help of an atomizer. When the mist reaches the flame, the intense heat breaks up the sample into its individual atoms.

II. Read the following text and answer the following questions on the basis of the same:
Spectrum Analysis and Astronomy Each element in the periodic table can appear in gaseous form and produce its own spectrum unique to that element. Hydrogen will not look like Helium, which will not look like carbon which will not look like iron… and soon. Astrophysists can identify what kinds of materials are present in stars from the analysis of star’s spectra. This type of study is called astronomical spectroscopy. The science of spectroscopy is quite sophisticated.

MCQ Questions for Class 12 Physics Chapter 12 Atoms

From spectrum lines analysis astrophysists can determine not only the element, but the temperature and density of that element in the star. The spectral line also can tell us about any magnetic field of the star. The width of the line can tell us how fast the material is moving. We can learn about winds in stars from this.

Temperature (Kelvin)Predominant radiationAstronomical Examples
600 KInfraredPlanets, warm dust
6,000 KOpticalThe photosphere of Sun and other stars
60,000 KUVThe photosphere of very hot stars
600,000 KSoft X-raysThe corona of the Sun
6,000,000 KX-raysThe coronae of active stars

The shifting of spectral lines shift back and forth indicates that the star may be orbiting another star. The following table shows a rough guide for the relationship between the temperature of a star and the electromagnetic spectrum. sight. Edwin Hubble observed that more distant galaxies tended to have more red shifted spectra. This establishes the theory of expansion of the universe.

Question 1.

What is astronomical spectroscopy?

(A) Study of spectrum of star light and to identify its distance from Earth.
(B) Study spectrum of star light and to identify what kinds of elements are present in stars.
(C) Both (A) and (B)
(D) None of the above
Answer:
(B) Study spectrum of star light and to identify what kinds of elements are present in stars.

Explanation:
Astrophysists can identify what kinds of materials are present in stars from the analysis of star’s spectra. This type of study is called astronomical spectroscopy.

Question 2.

From the spectrum analysis the following information of a star can be obtained.

(A) Elements present, temperature
(B) magnetic field, density, mass
(C) distance of the star
(D) Both (A) and (B)
Answer:
(D) Both (A) and (B)

Explanation:
From spectrum lines analysis, astrophysists can determine not only the element, but the temperature and density of that element in the star. The spectral line also can tell us about any magnetic field of the star.

MCQ Questions for Class 12 Physics Chapter 12 Atoms

Question 3.

The lines in a star’s spectrum is found to shift back and forth. What conclusion may be drawn from this observation?

(A) The star may be orbiting another star
(B) There may be a storm in the star
(C) The star may be rotating at a very high speed
(D) None of the above
Answer:
(A) The star may be orbiting another star

Explanation:
The shifting of spectral lines shift I back and forth indicates that the star may be orbiting another star.

Question 4.

What may be the approximate temperature if soft X-rays are found predominantly in the spectrum?

(A) 60000 C
(B) 600000 C
(C) 60000 K
(D) 600000 K
Answer:
(D) 600000 K

Explanation:
From the table, we find that predominant presence of soft X-rays in the spectrum indicated that the temperature is 600000 K.

Question 5.

Which nature of spectrum establishes the theory of the expanding universe?

(A) Red-shift of spectrum
(B) Blue-shift of spectrum
(C) Back and forth movement of spectral lines
(D) None of the above
Answer:
(A) Red-shift of spectrum

Explanation:
If the spectrum of a star is red or blue shifted, then it can be used to infer its velocity along the line of sight. Edwin Hubble observed that more distant galaxies tended to have more red-shifted spectra. This establishes the theory of expansion of the universe.

MCQ Questions for Class 12 Physics with Answers

MCQ Questions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Biotechnology: Principles and Processes Class 12 MCQ Questions With Answers

Biotechnology Principles And Processes MCQ Class 12 Question 1.

Name the enzymes ‘F and ‘Q’ that are involved in the processes given below

Class 12 Biology Chapter 11 MCQ

(A) Enzyme P-Exonuclease and Enzyme Q- Permease
(B) Enzyme P- Exonuclease and Enzyme Q- Ligase
(C) Enzyme P-Endonuclease and Enzyme Q- Permease
(D) Enzyme P-Restriction endonuclease and Enzyme Q- Ligase
Answer:
(C) Enzyme P-Endonuclease and Enzyme Q- Permease

Explanation :
Enzyme P is restriction endonuclease that cuts the DNA into fragments while enzyme join the two fragments.

MCQ Questions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Class 12 Biology Chapter 11 MCQ Question 2.

A biotechnologist wanted to create a colony of E.coli possessing the plasmid pBR322, sensitive to Tetracycline. Which one of the following restriction sites would he use to ligate a foreign DNA?

(A) Sal I
(B) Pvu I
(C) EcoRI
(D) Hind III
Answer:
(A) Sal I

Explanation :
When an alien gene is ligated at the Sal I site of tetracycline resistance gene in the vector pBR322, the recombinant lose tetra-cycline resistance due to insertion of the for-eign DNA.

Chapter 11 Biology Class 12 MCQs Question 3.

What is the criterion for DNA fragments movement on agarase gel during gel electrophoresis?

(A) The larger the fragment size, farther it moves
(B) The smaller the fragment size, farther it moves
(C) Positively charged fragment move to farther end.
(D) Negatively charged fragment do not move.
Answer:
(B) The smaller the fragment size, farther it moves

Explanation :
In agarose gel electrophoresis of DNA molecule smaller molecules travel faster then larger molecules. The movement of DNA fragment is inversely proportional to number of base pairs. Separation of DNA occurs by its length.

MCQ On Biotechnology Principles And Processes Chapter 11 Question 4.

An enzyme catalysing the removal of nucleotides from the ends of DNA is.

(A) endonuclease
(B) exonuclease
(C) DNA ligase.
(D) Hind-II.
Answer:
(B) exonuclease

Explanation:
Exomideases remove nucleotides from the ends of the DNA. Endonucleases make cuts al specific positions within the DNA. DNA ligase (also called genetic gum) is a sealing enyme w’hich is responsible for joining of two individual fragments of DNA, whereas Hind-II is the first discovered restriction endonuclease enzyme.

MCQ Questions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Biotechnology Principles And Processes Class 12 MCQ Chapter 11 Question 5.

The transfer of genetic material from one bacterium to another through the mediation of a vector-like virus is termed as

(A) transduction
(B) conjugation
(C) transformation
(D) translation
Answer:
(A) transduction

Explanation :
Transduction is the process by which genetic material (DNA) is transferred from one bacterium to another through the mediation of a vector, like virus. Bacterial conjugation is the process of transfer of genetic material (plasmid) between bacterial cells by direct cell-to-cell contact or by a bridge-like connection between two cells. Transformation is the genetic alteration of a cell resulting I from the direct uptake and incorporation of exogenous genetic material (exogenous DNA) from its surroundings and taken up through the cell membranes. Translation is the process in which cellular ribosomes create proteins. It is a part of the process of gene expression.

Biotechnology Principles And Processes Class 12 MCQ Questions Question 6.

‘Restriction’ in Restriction enzyme refers to

(A) cleaving of phosphodiester bond in DNA by the enzyme.
(B) cutting of DNA at specific position only.
(C) prevention of the multiplication of bacteriophage in bacteria.
(D) All of the above
Answer:
(B) cutting of DNA at specific position only.

Explanation :
Restriction enzymes (also called molecular scissors) are responsible for cutting DNA. These enzymes belong to a class of enzymes called nucleases and are of two t types (i) Exonuclease which cut DNA at the ends and (ii) endonucleases which make cuts at specific positions within the DNA. The li term ‘restriction’ refers to the function of these enzymes in restricting the propagation of foreign DNA of bacteriophage in host bacterium, that is, cutting of DNA, at specific J position only.

Biotechnology Class 12 MCQ Chapter 11 Question 7.

An antibiotic resistance gene in a vector usually helps in the selection of

(A) competent cells
(B) transformed cells
(C) recombinant cells
(D) None of the above
Answer:
(B) transformed cells

Explanation :
Selectable markers help in identifying and eliminating rum transformants and selectively permitting the growth of the transformants. The normal E. coli cells donot carry resistance against any antibiotic. Competent bacterial cells arc made capable .to take foreign DNA with chemical treatment (e.g, calcium chloride).

MCQ Of Chapter 11 Biology Class 12 Question 8.

Significance of ‘heat shock’ method in bacterial transformation is to facilitate

(A) binding of DNA to the cell wall.
(B) uptake of DNA through membrane transport proteins.
(C) uptake of DNA through transient pores in the bacterial cell wall.
(D) expression of antibiotic resistance gene.
Answer:
(C) uptake of DNA through transient pores in the bacterial cell wall.

Explanation :
In chemical method, the cell is treated with specific concentration of a divalent cation such as calcium which increases the pore size in cell wall. The cells are incubated with recombinant DNA on ice, followed by placing them briefly at 42°C and then putting it back on ice. This is called heat shock method. The bacteria now take up these recombinant DNA.

Biotechnology MCQ Class 12 Chapter 11 Question 9.

While isolating DNA from bacteria, which of the following enzymes is not used?

(A) Lysozyme
(B) Ribonuclease
(C) Deoxyribonuclease
(D) Protease
Answer:
(C) Deoxyribonuclease

Explanation :
Deoxyribonuclease enzyme is not used in the process of isolating DNA from bacteria as this enzyme causes the lysis of DNA molecules

MCQs Of Biology Class 12 Chapter 11 Question 10.

Which of the following bacteria is not a source of restriction endonuclease?

(A) Haemophilus influenzae
(B) Escherichia coli
(C) Agrobacterium tumefaciens
(D) Bacillius amyloli
Answer:
(C) Agrobacterium tumefaciens

Explanation :
Agrobacterium tumefaciens is a pathogen of several dicot plants. It delivers a piece of DNA known as T-DNA in the Ti plas-mid which transforms normal plant cells into tumour cells to produce chemicals required by pathogens. The restriction enzyme Eco Rl, is isolated from E. coli RY13. The first restriction enzymes Hind II was isolated from bacterium Haemophilus influenzas. The restriction enzyme Bam HI is isolated from Bacillus amyloli.

MCQ Questions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Biotechnology Principles And Processes MCQs Question 11.

The correct order of steps in polymerase chain reaction (PCR) is:

(A) Extension, Denaturation, Annealing
(B) Denaturation, Annealing, Extension
(C) Denaturation, Extension, Annealing
(D) Annealing, Extension, Denaturation
Answer:
(B) Denaturation, Annealing, Extension

Explanation :
A single PCR amplification cycle involves three basic steps : denaturation, annealing and extension. PCR stands for polymerase chain reaction in which multiple copies of the gene, for DNA of interest is synthesised in vitro.

MCQ Biotechnology Principles And Processes Question 12.

The process of separation and purification of expressed protein before marketing is called :

(A) Upstream processing
(B) Downstream processing
(C) Bioprocessing
(D) Post-production processing
Answer:
(B) Downstream processing

Explanation :
After completion of biosynthetic stage, the product has to be subjected through a series of processes before it is ready for marketing as a finished product. The processes include separation purification, which are collectively referred to as downstream processing.

MCQ On Biotechnology Class 12 Chapter 11 Question 13.

Stirred-tank bioreactors have been designed for.

(A) ensuring anaerobic conditions in culture vessel
(B) purification of product
(C) addition of preservatives to product
(D) availability of oxygen throughout process
Answer:
(D) availability of oxygen throughout process

Explanation:
The stirred lank bioreactor is well suited for large scale production of micro organism under asepLic condition for a number of days. It can be used easily in research laboratory and main advantage is an oxygen delivery system which provides oxygen without any interruption.

Biotechnology Principles And Processes MCQ Pdf Question 14.

Which of the following has popularised the PCR (polymerase chain reactions)?

(A) Easy availability of DNA template
(B) Availability of synthetic primers
(C) Availability of cheap deoxyribonucleotides
(D) Availability of ‘Thermostable’ DNA polymerase
Answer:
(D) Availability of ‘Thermostable’ DNA polymerase

Explanation :
The polymerase chain reaction (PCR) is a reaction in which amplification of specific DNA sequences is carried out in vitro. Such repeated amplification is achieved by the using thermostable DNA polymerase (isolated from a bacterium, Tlicrmus aquatints) envme which remains active and stable during high temperature and induced denaturation of doublestandard DNA.

Class 12 Biotechnology MCQ Chapter 11 Question 15.

Which of the following steps are catalysed by Taq polymerase in a PCR reaction?

(A) Denaturation of template
(B) Annealing of primers to template DNA
(C) Extension of primer end on the template DNA
(D) All of the above
Answer:
(C) Extension of primer end on the template DNA

Explanation :
In polymerase chain reaction, polymerisation or extension step is catalysed by taq polymerase enzyme. PCR is carried out in the following three steps (i) Denaturation The double-stranded DNA is denatured by applying high temperature of 95 for seconds. Each separated single strand now acts as template for DNA synthesis, (ii) Annealing Two sets of primers are added which anneal to the three ends of each separated strand, Primers act as initiators of replication, (iii)

Extension :
DNA polymerase extends the primers by adding nucleotides complementary to the template provided in the reaction. A thermostable DNA polymerase (Taq DNA polymerase) is used in the reaction which can tolerate the high temperature of the reaction. All these steps are repeated many times to obtain several copies of desired DNA.

Principles And Processes Of Biotechnology MCQ Question 16.

The role of DNA ligase in the construction of a recombinant DNA molecule is

(A) formation of phosphodiester bond between two DNA fragments.
(B) formation of hydrogen bonds between sticky ends of DNA fragments.
(C) ligation of all purine and pyrimidine bases.
(D) None of the above
Answer:
(A) formation of phosphodiester bond between two DNA fragments.

Explanation:
DNA ligase (joining or sealing enzymes) are also called genetic gum. They join two individual fragments of double-stranded DNA by forming phosphodiester bonds between them. Thus, they help in sealing gaps in DNA fragments. Therefore, they act as a moil lecular glue.

MCQ Questions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Biology MCQs Class 12 Chapter 11 Question 17.

Which of the given statement is correct in the context of observing DNA separated by agarose gel electrophoresis?

(A) DNA can be seen in visible light.
(B) DNA can be seen without staining in visible light.
(C) Ethidium bromide stained DNA can be seen in visible light.
(D) Ethidium bromide stained DNA can be seen under exposure to UV light.
Answer:
(D) Ethidium bromide stained DNA can be seen under exposure to UV light.

Explanation :
The separated DNA fragments (by the process of gel electrophoresis) are visu-alised after staining the DNA with ethidium bromide followed by exposure to ultraviolet (UV) radiation. These fragments are seen as orange coloured bands.

Biotechnology MCQ Chapter 11 Question 18.

Who among the following was awarded the Nobel Prize for the development of PCR technique?

(A) Herbert Boyer
(B) Hargovind Khurana
(C) Kary Mullis
(D) Arthur Komberg 0
Answer:
(C) Kary Mullis

12th Biology MCQ Questions Chapter 11 Question 19.

Which of the following statements does not hold true for restriction enzyme?

(A) It recognises a palindromic nucleotide sequence.
(B) It is an endonuclease.
(C) It is isolated from viruses.
(D) It produces the same kind of sticky ends in different DNA Molecules.
Answer:
(C) It is isolated from viruses.

Explanation :
Restriction enzymes are a proloin produced by bacteria that cleaves DNA at specific sites. These are not found in viruses. They are present in bacteria to provide a Lvpe of defence mechanism (called the restriction modification system) against bacterial viruscs. They are of two types endonuclease and exonuclease. They are indispensable tools in recombinant DNA technology and genetic engineering.

Question 20.

Which of the following should be chosen for best yield if one were to produce a recombinant protein in large amounts?

(A) Laboratory flask of largest capacity
(B) A stirred-tank bioreactor without in-lets and outlets
(C) A continuous culture system
(D) Any of the above
Answer:
(C) A continuous culture system

Explanation :
If any protein encoding gene is expressed in a heterologous host, it is called recombinant protein. The cells harbouring cloned genes of interest may be grown on a small scale in the laboratory. The cultures may be used for extracting the desired protein and then purifying it by using different separation techniques. The cells can also be multiplied in a continuous culture system where the used me-dium is drained out from one side while fresh medium is added from the other to maintain the cells in their physiologically most active logexponential phase. This type of culturing method produces a larger biomass leading to higher yields of desired protein.

MCQ Questions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Assertion and Reason based MCQs

Directions: In the following questions a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as :
(A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(C) Assertion (A) is true but reason (R) is false.
(D) Assertion (A) is false but reason (R) is true.

Question 1.

Assertion (A): EcoRl is restriction endonuclease enzyme.
Reason (R): Exonuclease removes nucleotides from the ends of DNA.

Answer:
(B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

Explanation : EcoRI is a restriction endonuclease enzyme. Exonuclease removes nucleotides from Ihe ends of DNA.

Question 2.

Assertion (A): Any fragment of DNA, when linked to the ori region, can be initiated to replicate.
Reason (R): Ori is a genetic sequence that acts as the initiation site for replication of DNA.

Answer:
(A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Explanation :
The process of DNA replication begins at the ori sequence. The presence of ori in a DNA fragment makes it a selfreplicating molecule.

Question 3.

Assertion (A): Restriction enzymes belong to class nucleases.
Reason (R): Nucleases are of two kinds: Exo and endonucleases. Exonucleases remove nucleotides within the DNA.

Answer:
(C) Assertion (A) is true but reason (R) is false.

Explanation :
Nucleases are of two kinds exo and endonucleases, but exonucleases remove nucleotides from the ends of the DNA.

Question 4.

Assertion (A): E. coli having pBR322 with DNA insert at BamH I site cannot grow in medium containing tetracycline.
Reason(R): Recognition site for BamH I is present in TetR region of pBR322.

Answer:
(A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Explanation :
pBR322 carries recognition sites for number of commonly used restriction enzymes. Recognition site for BamHl is present in their region i.e., region responsible for tetra cycling resistance. When an insert is added at the Bam recognition site the gene for tetras cocaine resistance becomes nonfunctional and the recombinant bacteria with plasmid pBR322 that has DNA insert at BamHl lose tetracycline resistance.

Question 5.

Assertion (A): It is essential to have few cloning sites in cloning vector.
Reason(R): It helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.

Answer:
(B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

Explanation:
It is essential to have few cloning sites in cloning vector. It is because, in order to link the alien DNA, the vector needs to have very few recognition sites for the commonly used restriction enzymes. Presence of more than one recognition sites within the vector will generate several fragments, which will complicate the gene cloning. Also, the vector requires a selectable marker, which helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.

MCQ Questions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 6.

Assertion (A): Thermus aquatics, is used in PCR technique.
Reason (R): It is a heat-stable DNA polymerase.

Answer:
(A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Explanation :
Thermus aquaticus, is the source of DNA polymerase because it is a heatstable DNA polymerase.

Question 7.

Assertion (A): Agarose gel electrophoresis is used to check the progression of a restriction enzyme digestion.
Reason (R): Restriction enzyme digestions are performed by incubating purified DNA molecules with restriction enzymes.

Answer:
(B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

Explanation :
In Agarose gel electrophoresis, DNAs fragments cut by restriction enzyme can be arranged according to their sizes.

Question 8.

Assertion (A): DNA is positively charged molecule.
Reason (R): DNA moves towards the positive electrode (anode).

Answer:
(D) Assertion (A) is false but reason (R) is true.

Explanation :
DNA is a negatively charged molecule, hence it moves towards the positive electrode (anode).

Question 9.

Assertion (A): A primer is a small segment of DNA that binds to a complementary strand of DNA.
Reason(R): Primers are necessary to stop the functioning of DNA polymerase enzymes and, therefore, are necessary in polymerase chain reaction.

Answer:
(D) Assertion (A) is false but reason (R) is true.

Explanation :
A primer is a small segment of DNA that binds to a complementary strand of DNA. Primers are necessary to start the functioning of DNA polymerase enzyme and, therefore, are necessary in polymerase chain reaction.

MCQ Questions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 10.

Assertion (A): P-galactosidase coding sequence act as a selectable marker.
Reason (R): Enzyme galactosidase converts the galactose into lactose.

Answer:
(A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Explanation :
Galactosidase is an enzyme that converts the galactose into lactose. This property makes this enzyme to be used as a selectable marker or reporter gene in molecular biology experiments. This property is exploited during the selection of recombinants from the non-recombinants.

Case-Based MCQs

Attempt any four sub-parts from each question, sub-part each carries 1 mark.

I. Read the following and answer questions from Question l. to Question 5. given below:

Restriction endonuclease was isolated for the first time by W. Aber in 1962 in bacteria. Restriction endonucleases cut the DNA duplex at specific points therefore they are also called as molecular scissors or biological scissors. Three types of restriction endemicleases are Type 1 Type 11 and Type HI but only Type II restriction endonucleases are used in recombinant DNA technology. Restriction endonuclease EcoR I recognises the base sequence GAATTC In DNA duplex and cut strands between G and A.

Question 1.

Only type II restriction enzymes are used in gene manipulation because:

(A) ATP is not required for cleaving
(B) it consists of three different subunits
(C) it makes cleavage or cut in both the strands of DNA molecule .
(D) both (A) and (C)
Answer:
(D) both (A) and (C)

Explanation:
Type II enzymes are simpler and don’t require ATP as an energy source, unlike Type I and it makes cleavage or cut in both the strands of DNA molecule.

Question 2.

Which of the following ions are used by restriction endonucleases for restriction?

(A) Mg2+ions
(B) Mn2+ions
(C) Na+ions
(D) K+ions
Answer:
(A) Mg2+ions

Explanation :
The restriction endonuclease binds to magnesium ions. One of these ions,binds to the phosphate grotygOtyhfrrc the cleavage occurs and is requirpftifofi aftlysis.

Question 3.

Restriction endonuclease was,isolated for the first time from a:

(A) plant cell
(B) animal cell
(C) prokaryotic cell
(D) germinal cell
Answer:
(C) prokaryotic cell

Explanation :
Restriction endonuclease was isolated for the first time from the bacterium Haemophilus influenzae (prokaryotic cell).

MCQ Questions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 4.

Restriction endonucleases are also called as molecular or biological scissors because:

(A) they cleave base pairs of DNA only at their terminal ends
(B) they cleave one or both the strands of DNA
(C) they act only on single stranded DNA
(D) none of these
Answer:
(B) they cleave one or both the strands of DNA

Explanation :
Restriction endonucleases naturally targeL double stranded DNA and could cleave one or both strands of the same.

Question 5.

Which type of restriction endonucleases is used most in genetic engineering?

(A) Type I
(B) Type II
(C) Type m
(D) Type IV
Answer:
(B) Type II

Explanation :
Type I and Type III are complex and have only a limited role in genetic engi-neering. Type II restriction endonucleases are used mostly as the culling enzymes in gene cloning.

II. Read the following text and answer the following questions on the basis of the same:
The term biotechnology refers to the use of living organisms or their products to modify human health and their human environment. For example, ‘test- tube’ programme, synthesis of a gene or correcting a defective gene are all part of the biotechnology. The basis of the modern biotechnology are genetic engineering and maintenance of sterile conditions. Genetic engineering is the technique that alter the chemistry of genetic material i.e. DNA and RNA, then this genetic material is introduced into host organisms, which alter the phenotype of the host organism.

Question 1.

Discovery of ——- molecule made genetic engineering possible.

(A) Restriction exonuclease
(B) Restriction exonuclease
(C) Ribozyme
(D) DNA polymerase
Answer:
(B) Restriction exonuclease

Explanation :
Restriction endonucleus act as molecular sensor that gets the DNA from spe-cific nucleotide and gives desired fragment of DNA.

Question 2.

The recognition sequence of the first restriction enzyme isolated was ——– base pair long.

(A) four
(B) six
(C) five
(D) two.
Answer:
(B) six

Explanation :
ECoRI is isolated as first restriction that GAATAC nucleotide base pair.

MCQ Questions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 3.

The specific DNA sequence where EcoRI cuts is:

(A) GATTCG
(B) GAATTC
(C) GTTCAA
(D) TTCCAA.
Answer:
(B) GAATTC

Explanation :
HCoKl recognises GAATAC base pairs that cuts between G and A.

Question 4.

The cutting of DNA at specific locations became possible with the discovery of :

(A) Ligases
(B) Restriction enzymes
(C) Probes
(D) Selectable markers.
Answer:
(B) Restriction enzymes

Explanation :
Restriction enzymes are the molocular scissors that cuts the DNA from specific recognition site.

Question 5.

DNA fragments are :

(A) Positively charged
(B) Negatively charged
(C) Neutral
(D) Either positively or negatively charged depending on their size.
Answer:
(C) Neutral

Explanation :
DNA have PO as negative end So, DNA is negatively charged.

III. A schematic representation of polymerase chain reaction (PCR) up to the extension stage is given below.
Biotechnology Principles And Processes MCQ Class 12

Question 1.

Name the process ‘a’

(A) Termination
(B) Annealing
(C) Denaturation
(D) Extension
Answer:
(C) Denaturation

Explanation :
Denaturating is the first step of PCR which involves the breaking of phos-phate bond belwcent the DNA base pairs at temperature SCPC.

Question 2.

Identify ‘b’

(A) Termination
(B) Annealing primer
(C) Denaturation
(D) Extension
Answer:
(B) Annealing primer

Explanation :

The primer binds on the DNA that initiate the polymerisation with the help of lag polymerase enzyme at 72“C temperature.

MCQ Questions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 3.

Which of the following has popularized the PCR (Polymerase Chain Reaction) ?

(A) Easy availability of DNA template
(B) Availability of synthetic primers
(C) Availability of cheap deoxyribonucleotides
(D) Availability of thermostable DNA polymerase.
Answer:
(D) Availability of thermostable DNA polymerase.

Explanation :
Tag polymerase enzyme help to maintain the stability of DNA to be used repeatedly at high temperature in the PCR.

Question 4.

PCR technique is best for :

(A) DNA synthesis
(B) Protein amplification
(C) DNA amplification
(D) DNA ligation.
Answer:
(C) DNA amplification

Explanation:
PCR techniques is helpful to dctect very minute traces of virus or bacteria DNA and other multiple copies by A amplication.

MCQ Questions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Question 5.

Which among the following is not an application of PCR?

(A) ELISA
(B) Diagnosis of pathogens
(C) DNA fingerprinting
(D) In palaeontology.
Answer:
(A) ELISA

Explanation :
PCR techniques is used to identify traces of DNA but it can not diagnose pathogen or organism.

MCQ Questions for Class 12 Biology with Answers

MCQ Questions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food Production with Answers

Check the below NCERT MCQ Questions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food Production with Answers Pdf free download. MCQ Questions for Class 12 Biology with Answers were prepared based on the latest exam pattern. We have provided Strategies for Enhancement in Food Production Class 12 Biology MCQs Questions with Answers to help students understand the concept very well. https://mcqquestions.guru/mcq-questions-for-class-12-biology-chapter-9/

Strategies for Enhancement in Food Production Class 12 MCQs Questions with Answers

Class 12 Biology Chapter 9 MCQ Questions Question 1.
Zoological name of Indian buffalo is:
(a) Bubalus bubalus
(b) Bos indicus
(c) Bos taurus
(d) Gallusgallus

Answer

Answer: (a) Bubalus bubalus


MCQs Of Biology Class 12 Chapter 9 Question 2.
The best milk breed in the world is:
(a) Chittgong
(b) Deoni
(c) Holstein-Friesian
(d) Sindhi

Answer

Answer: (c) Holstein-Friesian


Food Production Class 12 MCQ Chapter 9 Question 3.
There crops that contribute maximum to global food grain production are :
(a) Wheat, rice and maize
(b) Wheat rice and barley
(c) Wheat, maize and sorghum
(d) Rice, maize and sorghum

Answer

Answer: (a) Wheat, rice and maize


Strategies For Enhancement In Food Production MCQ Class 12 Question 4.
The world’s highly prized wool yielding ‘Pashmina’ breed is :
(a) Sheep
(b) Goat
(c) Goat-sheep cross
(d) Mashmir sheep-Afghan sheep cross

Answer

Answer: (b) Goat


Food Production Class 12 MCQ With Answers Chapter 9 Question 5.
Both in callus and suspension culture commonly used auxin is:
(a) NAA
(b) IEA
(c) 2-4D
(d) Abscisic acid

Answer

Answer: (c) 2-4D


Question 6.
Somatic hybridization can be done by:
(a) By Protoplast fusion
(b) By Haploid anther
(c) By Cell culture
(d) By Totipotency

Answer

Answer: (a) By Protoplast fusion


Question 7.
India’s w heat yield revolution in the 1986s was possible primarily due to:
(a) Mutations resulting in the plant height reduction
(b) Quantitative trait mutations
(c) Hybrid seeds
(d) Increased chlorophyll content

Answer

Answer: (c) Hybrid seeds


Question 8.
One of the following is a disease of poultry:
(a) Anthrax
(b) Pebrine disease
(c) Ranikhet disease
(d) Foot and mouth disease

Answer

Answer: (c) Ranikhet disease


Question 9.
Choose the cat fish from the following:
(a) Katla Katla
(b) Wallage attu
(c) Labeo rohita
(d) Cirrhinus mrigaja

Answer

Answer: (b) Wallage attu


Question 10.
The technique of obtaining large number of plantlets by tissue culture method is called :
(a) Organ culture
(b) Micro-propagation
(c) Macro-propagation
(d) Plantlet culture

Answer

Answer: (b) Micro-propagation


Question 11.
Most cultivated plants are:
(a) Autopolyploid
(b) Allopolyploid
(c) Aneuploids
(d) Haploids

Answer

Answer: (b) Allopolyploid


Question 12.
Triticale first main-made corral crop has been obtained by crossing wheat with
(a) Rye
(b) Pearl mille
(c) Sugarcane
(d) Barley

Answer

Answer: (a) Rye


Question 13.
Most nutritious among the following is:
(a) What
(b) Maize
(c) Pearley
(d) Rice

Answer

Answer: (c) Pearley


Question 14.
Hybrid vagour is induced by:
(a) By Clonal selection
(b) By Crossing of plant
(c) By crossing of two plants
(d) By Species differentiation

Answer

Answer: (c) By crossing of two plants


Question 15.
Norman Borlaugh known as “Father of Green Revolution” has developed new cultivating variety of:
(a) Paddy
(b) Rice
(c) Wheat
(d) Sugarcane

Answer

Answer: (c) Wheat


We hope the given NCERT MCQ Questions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food Production with Answers Pdf free download will help you. If you have any queries regarding Strategies for Enhancement in Food Production CBSE Class 12 Biology MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Dual Nature of Radiation and Matter Class 12 MCQs Questions with Answers

Dual Nature Of Radiation And Matter MCQ Class 12 Question 1.

Consider a beam of electrons (each electron with energy E0) incident on a metal surface kept in an evacuated chamber. Then,

(A) no electrons will be emitted as only photons can emit electrons.
(B) electrons can be emitted but all with an energy, E0.
(C) electrons can be emitted with any energy, with a maximum of E0 – ø (ø is the work function).
(D) electron can be omitted with energy, with a maximum of E0.
Answer:
(D) electron can be omitted with energy, with a maximum of E0.

Eiplanation:
If a beam of electrons of having energy E0 is incident on a metal surface kept in an evacuated chamber.
Dual Nature Of Radiation And Matter MCQ Class 12
The electrons can be emitted with maximum energy E0 (due to elastic collision) and with any energy less than E0, when part of incident energy of electron is used in liberating the electrons from the surface of metal. –

MCQ On Dual Nature Of Matter And Radiation Class 12 Question 2.

The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly

(A) 1.2 nm
(B) 1.2 x 10-3nm
(C) 1.2 x 10-6 nm
(D) 1.2 x 10 nm
Answer:
(B) 1.2 x 10-3nm

Explanation:
Energy of the photon must be equal to the binding energy of proton So, energy of photon = 1 MeV
= 106 x 1.6 x 10-19
λ = \(\frac{h c}{E}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-13}}\) = \(\frac{6.63 \times 3}{1.60}\) x 10-26 + 13
= \(\frac{19.89}{1.60}\) x 10-13 = 12.4 x 10-13 = 1.24 x 10-1 x 10-13
= 1.24 x 10-9 x 10-3 =1.24 x 10-3 nm

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Dual Nature Of Matter And Radiation Class 12 MCQ Chapter 11 Question 3.

The phenomenon which shows quantum nature of electromagnetic radiation is:

(A) photoelectric effect.
(B) Tyndall effect.
(C) interference.
(D) reflection and refraction.
Answer:
(A) photoelectric effect.

Dual Nature Of Radiation And Matter Chapter 11 Class 12 MCQ Question 4.

Kinetic energy of electrons emitted in photoelectric effect is

(A) directly proportional to the intensity of incident light.
(B) inversely proportional to the intensity of incident line.
(C) independent of the intensity of incident light.
(D) independent of the frequency of light.
Answer:
(C) independent of the intensity of incident light.

Explanation:
KE = hv – ø
So, KE is independent of intensity of incident light.

MCQ On Dual Nature Of Matter And Radiation Class 12 Question 5.

Threshold wavelength of a photoelectric emission from a material is 600 nm. Which of the following illuminating source will emit photoelectrons?

(A) 400 W, infrared lamp
(B) 10 W, ultraviolet lamp
(C) 100 W, ultraviolet lamp
(D) Both (B) & (C)
Answer:
(D) Both (B) & (C)

Explanation:
The incident wavelength should be less than threshold wavelength for photoelectric emission. IR has wavelength more than 600 nm. UV has wavelength less than 600 nm. So, photoelectrons emitted when illuminated by UV lamp either 100 W or 10 W

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

MCQ Of Dual Nature Of Radiation And Matter Class 12 Question 6.

Photoelectrons emitted from a metal have

(A) different speeds starting from 0 to certain maximum.
(B) same kinetic energy.
(C) same frequency.
(D) Both (B) & (C)
Answer:
(A) different speeds starting from 0 to certain maximum.

Explanation:
When a photon strikes a metal surface, the surface electrons come out with maximum speed and maximum kinetic energy. But if the electron emission takes place from inner side of metal, then some energy of the electron is lost due to collision with other electrons and so their speed becomes less. So, ultimately the electrons come out with different speeds.

Dual Nature Of Matter And Radiation MCQ Class 12 Chapter 11 Question 7.

At stopping potential, the kinetic energy of emitted photoelectron is

(A) minimum
(B) maximum.
(C) zero
(D) cannot de predicted
Answer:
(C) zero

MCQs On Dual Nature Of Radiation And Matter Class 12 Question 8.

Photons are

(A) electrically neutral and not deflected by electric or magnetic field.
(B) electrically neutral and deflected by magnetic field.
(C) electrically charged and not deflected by electric or magnetic field.
(D) electrically charged and not deflected by electric field.
Answer:
Option (A) is correct

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Class 12 Physics Chapter 11 MCQ Questions Question 9.

A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to

(A) H
(B) H1/2
(C) H0
(D) H-1/2
Answer:
(D) H-1/2

Explanation:
Velocity v, of freely falling body after falling from a height H, will be :
H = υ = \(\sqrt{2gh}\)
We know that de-Broglie wavelength, λ = \(\frac {h}{p}\) ?
λ = \(\frac {h}{mυ}\) = \(\frac{h}{m \sqrt{2 g H}}\)
h, m, and g are constant
∴ \(\frac{h}{m \sqrt{2 g}}\) is constant ⇒λ ∝

Question 10.

A proton, a neutron, an electron and an a-particle have same energy. Then, their de-Broglie wavelengths compare as

(A) λp = λn > λe > λ
(B) λ < λp = λn> λe
(C) λe < λp = λn > λ
(D) λe = λp = λn = λ
Answer:
(B) λ < λp = λn> λe

Explanation:
Matter waves (de-Broglie waves) According to de-Broglie, a moving material particle sometimes acts as a wave and sometimes as a particle.

De- Broglie wavelength:
λd = \(\frac {h}{p}\)
Ed = En = Ee = Ex
K.E. = K = \(\frac {1}{2}\)mυ2
2K = mυ2
2Km = m2υ2
2mK = p2
\(\sqrt{2 m \mathrm{~K}}\) = p
∴ λd = \(\frac {h}{p}\)
λd = \(\frac{h}{\sqrt{2 m \mathrm{~K}}}\)
or λd = \(\frac{h}{\sqrt{2 m \mathrm{~K}}}\)[as h and E(K.E) is const.]
∴ λ ∝ \(\frac{1}{\sqrt{m}}\)
ma >mp = mn >me
∴ λap = λn < λe

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Question 11.

An electron is moving with an initial velocity υ = υ0 \(\hat{\mathbf{i}}\) and is in a magnetic field B = B0\(\hat{\mathbf{j}}\). Then, its de-Broglie wavelength

(A) remains constant.
(B) increases with time.
(C) decreases with time.
(D) increases and decreases periodicalLy.
Answer:
(A) remains constant.

Explanation:
Given, υ = υ0 and B = B0\(\hat{\mathbf{j}}\) Magnetic force on moving electron
= -e[υ0i x B0j] = – eυ0B0 k
MCQ On Dual Nature Of Matter And Radiation Class 12

So, the force is perpendicular to y and B both as the force is perpendicular to the velocity. So, the magnitude will not change V or mv (p = mv momentum) so, the de-Broglie wavelength remains same.

Question 12.

An electron (mass m) with an initial velocity υ υ j (υ0 > 0) is in an electric field E = – E0\(\hat{\mathbf{i}}\) (E0 = constant > 0). Its de-Broglie wavelength at time t is given by

Dual Nature Of Matter And Radiation Class 12 MCQ Chapter 11
(C) decreases with time.
(D) increases and decreases periodicalLy.
Answer:
Option (A) is correct.

Explanation:
The wave associated with moving particle is called matter wave or de-Broglie wave and it propagates in the form of wave packets with group velocity.. According to de-Broglie theory the wavelength of de-Broglie wave is given by
λ = \(\frac{h}{p}=\frac{h}{m v}=\frac{h}{\sqrt{2 m E}}\)
As initial velocity of the electron is υ0\(\hat{\mathbf{i}}\), the initial de-Broglie wavelength of electron,
λ = \(\frac{h}{m v_{0}}\) …………(1)
Electrostatic force on electron in electric field is,
\(\vec{F}\)e = -e\(\vec{E}\) = -e[-E0\(\hat{\mathbf{i}}\)] = eEz0\(\hat{\mathbf{i}}\)

Acceleration of electron, \(\vec{a}\) = \(\frac{\vec{F}}{m}\) = \(\frac{e E_{0} \hat{j}}{m}\)
Velocity of the electron after time t,
Dual Nature Of Radiation And Matter Chapter 11 Class 12 MCQ
de-Broglie wavelength associated with electron at time t is λ = \(\frac{h}{m v}\)
MCQ On Dual Nature Of Matter And Radiation Class 12

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Question 13.

An electron (mass m) with an initial velocity υ = υ0\(\hat{\mathbf{i}}\) is in an electric field E = E0 \(\hat{\mathbf{j}}\) if λ0 = \(\frac{h}{m v_{0}}\) its de-Broglie wavelength at time t is given by

MCQ Of Dual Nature Of Radiation And Matter Class 12

Answer:
Option (C) is correct.

Explanation:
According to the problem de-Broglie wavelength of electron at time
t = 0 is λ0 = \(\frac{h}{m v_{0}}\)
Electrostatic force on electron in electric field is
\(\vec{F}\)e = -e\(\vec{E}\) = -eE0\(\hat{\mathbf{j}}\)
The acceleration of electron, \(\vec{a}\) = \(\frac{\overrightarrow{\mathrm{F}}}{m}\) = \(\frac{e \mathrm{E}_{0}}{m} \hat{j}\)
it is acting along negative y-axis.
The initial velocity of electron along x-axis,
\(v_{x_{0}}\) = v0\(\hat{\mathbf{i}}\)
This component of velocity will remain constant as there is no force on electron in this direction. Now considering y-direction, initia! velodty of electron along y-axis, \(v_{y_{0}}\) = 0 Velocity of electron after time t along y-axis
υy = 0 + \(\left(-\frac{e \mathrm{E}_{0}}{m} \mathrm{~J}\right)\)t = \(-\frac{e \mathrm{E}_{0}}{m} \hat{t} \hat{j}\)
Magnitude of velocity of electron after time t is
Dual Nature Of Matter And Radiation MCQ Class 12 Chapter 11

de-Broglie wavelength, λ0 = \(\frac{h}{m v_{0}}\)
MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter - 9

Question 14.

The ratio of dc-Brogue wavelength associated with two electrons accelerated through 25 V and 36 V is

(A) 25/36
(B) W25
(C) 5/6
(D) 6/5
Answer:
(D) 6/5

Explanation:
λ ∝ 1/ √V
∴λ 12 = √(V2 /V1) = 6/5

Question 15.

Which of the following graphs shows the variation of de-Broglie wavelength with potential through which a particle of charge q and mass m is accelerated?

MCQs On Dual Nature Of Radiation And Matter Class 12
Answer:
Option (B) is correct.

Assertion And Reason Based MCQs (1 Mark each)

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is true

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Question 1.

Assertion (A): The energy (E) and momentum (p) of a photon are related as p = E/c
Reason (R): The photon behaves like a particle.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
Energy= E = hc/λ Momentum = P = h/λ. So, P = E/c
This is true only when photon has a particle nature.
So, assertion and reason both are true and reason properly explains the assertion.

Question 2.

Assertion (A): Photoelectric effect demonstrates the particle nature of light.
Reason (R): The number of photoelectrons is proportional to the frequency of light.

Answer:
(C) A is true but R is false

Explanation:
Photoelectric effect demonstrates the particle nature of light. So assertion is true. Number of emitted photoelectrons depends upon the intensity of light. So reason is false.

Question 3.

Assertion (A): Kinetic energy of photoelectrons emitted by a photosensitive surface depends upon the frequency of incident photon.
Reason (R): The ejection of electrons from metallic surface is possible with frequency of incident photon below the threshold frequency.

Answer:
(C) A is true but R is false

Explanation:
K.E. depends upon the frequency. So assertion is true. Photoelectron emitted when frequency of incident radiation is more than the threshold frequency. So reason is also false.

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Question 4.

Assertion (A): Photosensitivity of a material is high if its work function is low.
Reason (R): Work function = hf0, where /0 is threshold frequency.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Question 5.

Assertion (A): de-Broglie equation is significant for microscopic particles.
Reason (R): de-Broglie wavelength is inversely proportional to the mass of a particle when velocity is kept constant.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
de-Broglie wavelength, λ = h/mυ h and v remaining constant, λ ∝ 1/m So, as the mass of the particle becomes smaller and smaller the de-Broglie wavelength of the particle becomes more and more significant. Hence, assertion and reason both are true and reason explains the assertion properly.

Question 6.

Assertion (A): de-Broglie wavelength of a gas molecule is inversely proportional to the square root of temperature.
Reason (R): The root mean square velocity of gas molecules depends on temperature.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
υRMS = \(\sqrt{\frac{3 R T}{M}}\)
So, υRMS ∝ √T
Again de-Broglie wavelength, λ = h/mυ
So, λ ∝ 1/√T
Hence, assertion and reason both are true and reason explains the assertion

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Question 7.

Assertion (A): Resolving power of electron microscope increases as electron is accelerated through higher voltage.
Reason (R): de-Broglie wavelength of electron decreases as accelerating voltage of electron increases.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
de-Broglie wavelength is given by
λ = \(\frac{h}{\sqrt{2 m e V}}\)
So, as accelerating potential (V) increases, the wavelength decreases.
Resolving power of microscope = \(\frac{2 \mu \sin \beta}{1.22 \lambda}\)
So, as wavelength (X) decreases, resolving power increases. So, assertion and reason both are true and reason explains the assertion.

Case-Based MCQs

Attempt any 4 sub-parts out of 5. Each sub-part carries 1 mark.

I. Read the following text and answer the following questions on the basis of the same:

Photocell:
A photocell is a technological application of the photoelectric effect. It is a device whose electrical properties are affected by light. It is also sometimes called an electric eye. A photocell consists of a semi-cylindrical photo-sensitive metal plate C (emitter) and a wire loop A (collector) supported in an evacuated glass or quartz bulb. It is connected to the external circuit having a high-tension battery B and micro ammeter (pA) as shown in the Figure.
Class 12 Physics Chapter 11 MCQ Questions
Sometimes, instead of the plate C, a thin layer of photosensitive material is pasted on the inside of the bulb. A part of the bulb is left clean for the light to enter it. When light of suitable wavelength falls on the emitter C, photoelectrons are emitted. These photoelectrons are drawn to the collector A. Photocurrent of the order of a few microampere can be normally obtained from a photo cell. A photocell converts a change in intensity of illumination into a change in photocurrent. This current can be used to operate control systems and in light measuring devices.

Question 1.

Photocell is an application of

(A) thermoelectric effect.
(B) photoelectric effect.
(C) photoresistive effect.
(D) None of the above
Answer:
(B) photoelectric effect.

Explanation:
Photocell is a technological application of the photoelectric effect

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Question 2.

Photosensitive material should be connected to

(A) -ve terminal of the battery.
(B) +ve terminal of the battery.
(C) any one of (A) or (B).
(D) connected to ground.
Answer:
(A) -ve terminal of the battery.

Explanation:
Photosensitive material used as emitter should be connected to -ve terminal of the battery so that the emitted electrons are repelled by emitter and collected by collector.

Question 3.

Which of the following statement is true?

(A) The photocell is totally painted black.
(B) A part of the photocell is left clean.
(C) The photocell is completely transparent.
(D) A part of the photocell is made black.
Answer:
(B) A part of the photocell is left clean.

Explanation:
A part of the bulb is left clean for the light to enter in it.

Question 4.

The photocurrent generated is in the order of

(A) ampere
(B) milliampere
(C) microampere
(D) None of the above
Answer:
(C) microampere

Explanation:
Photocurrent of the order of a few microampere can be normally obtained from a photocell.

Question 5.

A photocell converts a change in of incident light into a change in

(A) intensity, photovoltage
(B) wavelength, photovoltage
(C) frequency, photocurrent
(D) intensity, photocurrent
Answer:
(D) intensity, photocurrent

Explanation:
A photocell converts a change in intensity of illumination into a change in photocurrent.

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

II. Read the following text and answer the following questions on the basis of the same:

Electron Microscope
Electron microscopes use electrons to illuminate a sample. In Transmission Electron Microscopy (TEM), electrons pass through the sample and illuminate film or a digital camera. Resolution in microscopy is limited to about half of the wavelength of the illumination source used to image the sample. Using visible light the best resolution that can be achieved by microscopes is about —200 nm. Louis de Broglie showed that every particle or matter propagates like a wave. The wavelength of propagating electrons at a given accelerating voltage can be determined by

l = \(\frac{h}{\sqrt{2 m_{e} v}}\)

Thus, the wavelength of electrons is calculated to be 3.88 pm when the microscope is operated at 100 keV, 2. 74 pm at 200 keV and 2.24 pm at 300 keV. However, because the velocities of electrons in an electron microscope reach about 70% the speed of light with an accelerating voltage of 200 keV, there are relativistic effects on these electrons. Due to this effect, the wavelength at 100 keV, 200 keV and 300 keV in electron microscopes is 3.70 pm, 2.51 pm and 1.96 pm, respectively.

Anyhow, the wavelength of electrons is much smaller than that of photons (2.5 pm at 200 keV). Thus if electron wave is used to illuminate the sample, the resolution of an electron microscope theoretically becomes unlimited. Practically, the resolution is limited to -0.1 nm due to the objective lens system in electron microscopes. Thus, electron microscopy can resolve subcellular structures that could not be visualized using standard fluorescences microscopy.

Question 1.

In electron microscope, electron is used:

(A) to charge the sample.
(B) to. clean the sample.
(C) to illuminate the sample.
(D) All of the above
Answer:
(C) to illuminate the sample.

Explanation:
Electrons as wave is used in electron microscopes to illuminate a sample since it enhances the resolving power.

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Question 2.

Who showed that electron also propagates like a wave?

(A) Louis de Broglie
(B) Albert Einstein
(C) Philipp Lenard
(D) Wilhelm Ludwig Franz Hallwachs
Answer:
(A) Louis de Broglie

Explanation:
Louis de Broglie showed that every particle or matter propagates like a wave.

Question 3.

Why electron as wave is used in electron microscope to illuminate the sample?

(A) The wavelength of electrons as wave is much larger than that of photons, hence resolution is much better.
(B) The wavelength of electrons as wave is much smaller than that of photons, hence resolution is much better.
(C) Electron as wave wave is much brighter than normal light and hence resolution is much better.
(D) Speed of electron as wave wave is greater than the speed of light and hence offers better resolution.
Answer:
(B) The wavelength of electrons as wave is much smaller than that of photons, hence resolution is much better.

Explanation:
Using visible light, the best resolution that can be achieved by microscopes is about – 200 nm. The wavelength of electrons as wave is much smaller than that of photons as wave (2.5 pm at 200 keV). Thus if electron as wave is used to illuminate the sample, the resolution of an electron microscope theoretically becomes unlimited. Practically, the resolution is limited to – 0.1 nm,

Question 4.

As the accelerating voltage increases, the wavelength of electron as wave

(A) decreases.
(B) increases.
(C) remains same.
(D) upto 100 keV increases and then decreases.
Answer:
(A) decreases.

Explanation:
λ = \(\frac{h}{\sqrt{2 m e V}}\)
So, as V increases, λ decreases.

MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Question 5.

Wavelength of electron as wave at accelerating voltage 200 kev is

(A) 2.5 nm
(B) 2.5 mm
(C) 2.5 pm
(D) 2.5 pm
Answer:
(C) 2.5 pm

MCQ Questions for Class 12 Physics with Answers