NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields
NCERT Solutions For Class 12 Physics Chapter 1 Electric Charges and Fields: Electrostatics is the study of electric charges at rest. The NCERT Class 12th Physics Electric Charges and Fields deal with the charging of a body, properties of charge, Columbus law, electric field, electric flux, Gauss law and application of Gauss law. Two main laws discussed in physics class 12 chapter 1 are Gauss law and Columbus law. Solutions of NCERT Class 12 physics chapter 1 Electric Charges and Fields explain problems related to these topics.
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roblems in Physics Class 12 chapter 1 are important from the CBSE class 12 exam point of view. Try solving all the questions of NCERT and if any doubt arises then check the CBSE NCERT solutions for Class 12 Physics Chapter 1 Electric Charges and Fields.
The derivations in Physics Class 12 NCERT chapter 1 Electric Charges and Fields of NCERT are very important in CBSE board exam point of view. As the second chapter of NCERT solutions Class 12 is the continuation of the NCERT Class 12 Physics chapter 1 pdf, it is important to remember all points studied in the chapter. The NCERT solutions will help you in a better understanding of the concepts you have studied. While studying NCERT Solutions for Class 12 Physics Chapter 1 formulas can be compared with the chapter Gravitation of Physics NCERT solutions class 11.
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NCERT Exemplar Solutions Class 12 Physics Chapter 1
NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields Exercise
Answer:
Given,
= C
=
r = 30 cm = 0.3 m
We know,
Force between two charged particles, and separated by a distance r.
Since the charges are of the same nature, the force is repulsive.
Answer:
Given,
=
=
F = (Attractive)
We know,
Force between two charged particles, and separated by a distance r.
Therefore, the distance between the two charged spheres is 12 cm.
Answer:
Using Newton's third law, the force exerted by the spheres on each other will be equal in magnitude.
Therefore, the force on the second sphere due to the first = 0.2 N (This force will be attractive since charges are of opposite sign.)
Answer:
Electrostatic force
So the dimension of
..................(1)
The gravitational force between two bodies of mass M and m is
so dimension of
.............(2)
Therefore from (1) and (2)
is dimensionless
or
Here,
K = , where is the permittivity of space.
e = Electric charge ([e] = [C])
G = Gravitational constant. ([G]= )
and are mass of electron and proton ([ ] = [ ] = [Kg])
Substituting these units, we get
Hence, this ratio is dimensionless.
Putting the value of the constants
The given ratio is the ratio of electric force to the gravitational force between an electron and a proton considering the distance between them is constant!
Q 1.4 (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.
Answer:
The given statement "electric charge of a body is quantised" implies that charge on a body can take only integral values. In other words, only the integral number of electrons can be transferred from one body to another and not in fractions.
Therefore, a charged body can only have an integral multiple of the electric charge of an electron.
Answer:
On a macroscopic level, the amount of charge transferred is very large as compared to the charge of a single electron. Therefore, we tend to ignore the quantisation of electric charge in these cases and considered to be continuous in nature.
Answer:
When a glass rod is rubbed with a silk cloth, opposite charges appear on both the rod and the cloth.
The phenomenon of charging bodies by rubbing them against each other is known as charging by friction. Here, electrons are transferred from one body to another giving both the bodies an equal but opposite charge. The number of electrons lost by one body (attains positive charge due to loss of negatively charged electrons) is equal to the number of electrons gained by the other body (attains negative charge). Therefore, the net charge of the system is zero. This is in accordance with the law of conservation of charge
Answer:
Charges at (A, C) and (B, D) are pairwise diametrically opposite and also equal.
Therefore, their force on a point charge at the centre of the square will be equal but opposite in directions.
Now, AC = BD =
AO = BO = CO = DO = r = Half of diagonal =
Force on point charge at centre due to charges at A and C =
Similarly, force on point charge at centre due to charges at B and D =
Net force on point charge =
.
Hence, the charge at the centre experiences no force.
Answer:
A positive point charge experiences a force in an electrostatic field. Since the charge will experience a continuous force and cannot jump from one point to another, the electric field lines must be continuous.
Q 1.7 (b) Explain why two field lines never cross each other at any point?
Answer:
A tangent drawn at any point on a field line gives the direction of force experienced by a unit positive charge due to the electric field on that point. If two lines intersect at a point, then the tangent drawn there will give two directions of force, which is not possible. Hence two field lines cannot cross each other at any point.
Answer:
Given, AB = 20 cm
Since, O is the midpoint of the line AB.
AO = OB = 10 cm = 0.1m
The electric field at a point caused by charge q, is given as,
Where, q is the charge, r is the distance between the charges and the point O
k = 9x10 ^{ 9 } N m ^{ 2 } C ^{ -2 }
Now,
Due to charge at A, electric field at O will be and in the direction AO.
Similarly the electric field at O due to charge at B, also in the direction AO
Since, both the forces are acting in the same direction, we can add their magnitudes to get the net electric field at O:
E' = + = 2E (Since their magnitudes are same)
along the direction AO.
Answer:
Let Q = C
The force experienced by Q when placed at O due to the charges at A and B will be:
where 'E' is the net electric field at point O.
Q being negatively charged will be attracted by positive charge at A and repelled by negative charge at B. Hence the direction of force experienced by it will be in the direction of OA.
Answer:
Given,
and
The total charge of the system =
The system is electrically neutral. (All dipole systems have net charge zero!)
Now, distance between the two charges, d = 15 + 15 = 30 cm = 0.3 m
We know, The electric dipole moment of the system, p = x d = x d (i.e, the magnitude of charge x distance between the two charges)
The direction of a dipole is towards the positive charge . Hence, in the positive z-direction.
Answer:
Given,
Electric dipole moment, p =
E =
We know, the torque acting on a dipole is given by:
Therefore, the magnitude of torque acting on the dipole is
Answer:
Clearly, polyethene being negatively charged implies that it has an excess of electrons(which are negatively charged!). Therefore, electrons were transferred from wool to polyethene.
Given, charge attained by polyethene = C
We know, Charge on 1 electron =
Therefore, the number of electrons transferred to attain a charge of =
electrons.
Answer:
The charge attained by polyethene (and also wool!) is solely due to the transfer of free electrons.
We know, Mass of an electron =
The total mass of electron transferred = number of electrons transferred x mass of an electron
=
Yes, there is a transfer of mass but negligible.
Answer:
Since the radii of the spheres A and B are negligible compared to the distance of separation, we consider them as a point object.
Given,
charge on each of the spheres =
and distance between them, r = 50 cm = 0.5 m
We know,
Therefore, the mutual force of electrostatic repulsion(since they have the same sign of charge)
Answer:
We know, force between two charged particles separated by a distance r is:
Now if
The new value of force:
Therefore, the force increases 16 times!
Answer:
When two spheres of the same size are touched, on attaining the equipotential state, the total charge of the system is equally distributed on both of them.
Therefore, (i) When uncharged third sphere C is touched with A, charge left on A =
and charge attained by C =
(ii) Now, charge on B + charge on C = + =
When touched, charge left on B =
Therefore
Therefore,
Answer:
Charges 1 and 2 are repelled by the negatively charged plate of the system
Hence 1 and 2 are negatively charged .
Similarly, 3 being repelled by positive plate is positively charged.
(charge to the mass ratio: charge per unit mass)
Since 3 is deflected the most, it has the highest charge to mass ratio .
Answer:
Given,
Area of the square =
Since the square is parallel to the yz plane, therefore it's normal is in x-direction.(i.e direction )
therefore, flux through this surface:
Answer:
Now, Since the normal of the square plane makes a angle with the x-axis
therefore, flux through this surface:
Answer:
The net flux of the uniform electric field through a cube oriented so that its faces are parallel to the coordinate planes is z ero .
This is because the number of lines entering the cube is the same as the number of lines leaving the cube.
Alternatively,
using Gauss’s law, we know that the flux of electric field through any closed surface S is times the total charge enclosed by S.
i.e.
where, q = net charge enclosed and = permittivity of free space (constant)
Since there is no charge enclosed in the cube, hence .
Answer:
Using Gauss’s law, we know that the flux of electric field through any closed surface S is times the total charge enclosed by S.
i.e.
where, q = net charge enclosed and = permittivity of free space (constant)
Given,
This is the net charge inside the box.
Answer:
Using Gauss's law, we know that
Since flux is zero, q = 0, but this q is the net charge enclosed by the surface.
Hence, we can conclusively say that net charge is zero, but we cannot conclude that there are no charges inside the box.
Answer:
Let us assume that the charge is at the centre of the cube with edge 10 cm.
Using Gauss's law, we know that the flux of electric field through any closed surface S is times the total charge enclosed by S.
i.e.
where, q = net charge enclosed and = permittivity of free space (constant)
Therefore, flux through the cube:
Due to symmetry, we can conclude that the flux through each side of the cube, , will be equal.
Answer:
Given,
q = net charge inside the cube =
Using Gauss’s law, we know that the flux of electric field through any closed surface S is times the total charge enclosed by S.
i.e.
where, q = net charge enclosed and = permittivity of free space (constant)
(Note: Using Gauss's formula, we see that the electric flux through the cube is independent of the position of the charge and dimension of the cube! )
Q 1.20 (a) A point charge causes an electric flux of to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
Answer:
Given,
Using Gauss’s law, we know that the flux of electric field through any closed surface S is times the total charge enclosed by S.
i.e.
where, q = net charge enclosed and = permittivity of free space (constant)
Therefore, flux does not depend on the radius of the sphere but only on the net charge enclosed. Hence, the flux remains the same although the radius is doubled.
Answer:
Given,
Using Gauss’s law, we know that the flux of electric field through any closed surface S is times the total charge enclosed by S.
i.e.
where, q = net charge enclosed and = permittivity of free space (constant)
Q 1.21 A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is and points radially inward, what is the net charge on the sphere?
Answer:
We know, for determining the electric field at r>R for a conducting sphere, the sphere can be considered as a point charge located at its centre.
Also, electric field intensity at a point P, located at a distance r, due to net charge q is given by,
Given, r = 20 cm = 0.2 m (From the centre, not from the surface!)
Therefore, charge on the conducting sphere is (since flux is inwards)
Answer:
Given,
Surface charge density =
Diameter of sphere = 2.4 m radius of sphere, r = 1.2 m
The charge on the sphere, Q= surface charge density x surface area of the sphere
Q 1.22 (b) A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of . (b) What is the total electric flux leaving the surface of the sphere?
Answer:
Using Gauss's law, we know that :
Answer:
Given,
d = 2 cm = 0.02 m
We know, For an infinite line charge having linear charge density , the electric field at a distance d is:
ï¿¼
The linear charge density is .
Answer:
We know, electric field, E, due to an infinite plate (length>>thickness) having surface charge density = .
(To note: It's independent of distance from the plate!)
In the region outside the first plate,
since both plates have the same surface charge density(in magnitude only), their electric fields are same in magnitude in this region but opposite in direction.
(E due to positive plate away from it and E due to negative plate towards it!)
Hence, the electric field in the outer region of the first plate is zero.
Answer:
We know, electric field, E, due to an infinite plate (length>>thickness) having surface charge density = .
(To note: It's independent of distance from the plate and same everywhere!)
In the region outside the second plate,
Since both plates have the same surface charge density(in magnitude only), their electric fields are same in magnitude in this region but opposite in direction.
(E due to positive plate away from it and E due to negative plate towards it!)
Hence, the electric field in the outer region of the second plate is zero.
Answer:
We know, electric field, E, due to an infinite plate (length>>thickness) having surface charge density = .
(To note: It's independent of distance from the plate!)
Let A and B be the two plates such that:
=
= -
Therefore,
The electric field between the plates, E = =
NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields Additional Exercises
Answer:
The force due to the electric field is balancing the weight of the oil droplet.
weight of the oil drop = density x volume of the droplet x g =
Force due to the electric field = E x q
charge on the droplet, q = No. of excess electrons x charge of an electron =
Balancing forces:
Putting known and calculated values:
Q 1.26 Which among the curves shown in Figure cannot possibly represent electrostatic field lines?
Answer:
(a) Wrong, because field lines must be normal to a conductor.
(b) Wrong, because field lines can only start from a positive charge. It cannot start from a negative charge,
(c) Right;
(d) Wrong, because field lines cannot intersect each other,
(e) Wrong, because electrostatic field lines cannot form closed loops.
Answer:
Force on a charge F=qE
but here E is varying along the Z direction.
Force can be written as,
Torque experienced =0 since both dipole and electric field are in the Z direction. The angle between dipole and the electric field is 180 degrees
Answer:
We know that the electric field inside a conductor is zero.
Using Gauss' law, if we draw any imaginary closed surface inside the solid, net charge must be zero.(Since E= 0 inside)
Hence there cannot be any charge inside the conductor and therefore, all charge must appear on the outer surface of the conductor.
Answer:
We know, electric field inside a conductor is zero.
Now, imagine a Gaussian surface just outside the cavity inside the conductor. Since, E=0 (Using Gauss' law), hence net charge must be zero inside the surface. Therefore, -q charge is induced on the inner side of the cavity(facing conductor B).
Now consider a Gaussian surface just outside the conductor A. The net electric field must be due to charge Q and q. Hence, q charge is induced on the outer surface of conductor A. Therefore, the net charge on the outer surface of A is Q + q.
Answer:
We know that the electric field inside a conductor is zero.
Therefore, a possible way to shield from the strong electrostatic fields in its environment is to enclose the instrument fully by a metallic surface.
Answer:
Let the surface area of the sphere be S.
And assume that the hole is filled. For a point B, just above the hole, considering a gaussian surface passing through B, we have
Now, since the electric field is always perpendicular to the surface of the conductor.
Using Superposition principle, ,
where is due to the hole and is due to the rest of the conductor. (both pointing outwards, i.e away from the centre)
Again, for a point A, just below the hole, Electric field will be zero because of electrostatic shielding.
Using the superposition principle, this will be due to pointing inwards(towards the centre) and due to (Pointing away from the centre)
Using this relation, we get:
Since this is pointing outwards,
is the electric field in the hole.
(Trick: 1. Assume the hole to be filled.
2. Consider 2 points just above and below the hole.
3. Electric fields at these points will be due to the hole and rest of the conductor. Use superposition principle.)
[Hint: Use Coulomb’s law directly and evaluate the necessary integral.]
Answer:
Let AB be a long thin wire of uniform linear charge density λ.
Let us consider the electric field intensity due to AB at point P at a distance h from it as shown in the figure.
The charge on a small length dx on the line AB is q which is given as q = λdx.
So, according to Coulomb’s law, the electric field at P due to this length dx is
But
⇒
This electric field at P can be resolved into two components as dEcosθ and dEsinθ. When the entire length AB is considered, then the dEsinθ components add up to zero due to symmetry. Hence, there is only dEcosθ component.
So, the net electric field at P due to dx is
dE' = dE cosθ
⇒ ………………….(1)
In ΔPOC,
⇒ x = h tan θ
Differentiating both sides w.r.t. θ,
⇒ dx = h sec ^{ 2 } θ dθ …………………….(2)
Also, h ^{ 2 } + x ^{ 2 } = h ^{ 2 } + h ^{ 2 } tan ^{ 2 } θ
⇒ h ^{ 2 } + x ^{ 2 } = h ^{ 2 } (1+ tan ^{ 2 } θ)
⇒ h ^{ 2 } + x ^{ 2 } = h ^{ 2 } sec ^{ 2 } θ ………….(3)
(Using the trigonometric identity, 1+ tan ^{ 2 } θ = sec ^{ 2 } θ)
Using equations (2) and (3) in equation (1),
,
The wire extends from to since it is very long.
Integrating both sides,
This is the net electric field due to a long wire with linear charge density λ at a distance h from it.
In the question linear charge density =E
therefor
Answer:
Given, a proton and a neutron consist of three quarks each.
And, ‘up’ quark is of charge + e, and the ‘down’ quark of charge e
Let the number of 'up' quarks be n. Therefore, the number of 'down' quarks is (3-n).
The net charge =
Now, a proton has a charge +1e
Proton will have 2 u and 1 d, i.e, uud
Similarly, the neutron has a charge 0
Neutron will have 1 u and 2 d, i.e, udd
Answer:
For equilibrium to be stable, there must be a restoring force and hence all the field lines should be directed inwards towards the null point. This implies that there is a net inward flux of the electric field. But this violates Gauss's law, which states that the flux of electric field through a surface not enclosing any charge must be zero. Hence, the equilibrium of the test charge cannot be stable.
Therefore, the equilibrium is necessarily unstable.
Answer:
Two charges of same magnitude and sign are placed at a certain distance apart. The mid-point of the line joining these charges will have E =0.
When a test charged is displaced along the line towards the charges, it experiences a restoring force(which is the condition for stable equilibrium). But if the test charge is displaced along the normal of the line, the net electrostatic force pushes it away from the starting point. Hence, the equilibrium is unstable.
Answer:
Let s be the vertical deflection, t be the time taken by the particle to travel between the plates
Here , u =0 , since initially there was no vertical component of velocity.
The particle in the elecric field will experience a constant force (Since, Electric field is constant.)
F = ma = -qE (Using Newton's Second Law, F = ma)
a = -qE/m (-ve sign implies here in downward direction)
Again, t = Distance covered/ Speed =
(In x-direction, since there is no force, hence component of velocity in x-direction remains constant = .
And, the distance covered in x-direction = length of the plate = L)
Putting these values in our deflection equation,
(S is -ve, which implies it deflect in downwards direction.)
The vertical deflection of the particle at the far edge of the plate is .
This motion is similar to the motion of a projectile in a gravitational field, which is also a constant force. The force acting on the particle in the gravitational field is mg whiles in this case, it is qE. The trajectory will be the same in both cases.
Answer:
The vertical deflection of the particle at the far edge of the plate is
given s= 0.5cm=0.005cm
calculate for L from the above equation
L=1.6 cm
Important Features of NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields
There are two parts of NCERT exercise given in the last part of the NCERT Physics chapter Electric Charges and Fields. The first part is the exercise questions and the second part is the Additional exercise questions. Solutions to both this parts are covered in the Class 12 Physics NCERT solutions. Questions 25 to 34 are the part of additional exercise. Students can download Class 12 NCERT Physics solutions and prepare offline.
Topics Covered in Class 12 NCERT Chapter Electric Charges and Fields:
The main topics covered in the ch 1 physics class 12 are
Charge and its properties
This NCERT Class 12 topic discuss the concept of charge with practical examples of charging through friction, conduction and induction. Also, the concepts of properties like conservation, additivity and quantisation of charge are discussed in the chapter electric charges and fields. Questions based on this are present in electric charges and fields Class 12 solutions.
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Coulomb's law and its applications
Coulomb's law discusses the relation between the force exerted between two charges separated by a distance. For example, questions 1 and 12 of NCERT solutions for Class 12 chapter 1 uses this law to solve it.
Electric field: electric field and properties of electric field lines are discussed in this topic
Electric dipole: The concepts of dipole and field due to dipole at the axial, equatorial and general point and torque due to electric dipole are discussed.
Continuous charge distribution: Linear charge, surface charge and volume charge distributions are covered in this class 12 NCERT Physics chapter 1 topic
Electric flux, Gauss's law and its applications: The concepts of electric flux, the relation between electric charge and flux and application of Gauss law to different charge configurations are discussed in class 12 NCERT.
All the topics mentioned in Physics Class 12 chapter 1 are important and students are advised to go through all the concepts mentioned in the topics. Questions from all the above topics are covered in the electric charges and fields NCERT solutions.
NCERT Exemplar Class 12 Solutions
NCERT Solutions for Class 12 Physics- Chapter Wise
Subject Wise Solutions
- NCERT solutions for class 12 mathematics
- NCERT solutions for class 12 chemistry
- NCERT solutions for class 12 biology
Importance of NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields in Board Exams:
The physics paper for the CBSE board exam is for 70 marks. In 2019 CBSE board exam 9 marks questions were asked from class 12 chapter 1 physics that includes the first two chapters, and this makes solving NCERT class 12 physics chapter 1 quite important. Practising problems from the NCERT class 12 Physics solutions chapter 1 help students to score well in the final exam. Students can also use NCERT class 12 Physics chapter 1 solutions pdf download button to read chapter 1 physics class 12 NCERT solutions offline.
Frequently Asked Question (FAQs) - NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields
Question: What is the weightage of the chapter electric fields and charges for CBSE board exam
Answer:
From the Class 12 Physics Chapter 1 NCERT solutions electric charges and fields 3 to 5 marks questions can be expected for CBSE bord exam. Questions based on the NCERT exercise questions can be expected for the exam. So practising the NCERT solutions for Class 12 Physics chapter 1 is important. The questions may be theory based, numerical or derivations.
Question: Where can I find complete solutions of NCERT class 12 Physics
Answer:
The detailed NCERT Class 12 Physics solutions chapter 1 is provided by careers 360. All the exercise and additional exercise questions of NCERT chapter Electric charges and Fields are solved with necessary explanations.
Question: Whether the unit electrostatics is helpful in higher studies?
Answer:
Yes, it is helpful in engineering streams and for research in the science field. The concepts studied in electrostatics will be useful in electrical and electronics-related branches.
Question: What are the important topics of physics NCERT class 12 chapter 1
Answer:
The two important laws discussed in NCERT solutions for Class 12 Physics Chapter 1 are important. That is the Coulombs law and the Gauss law. The derivations using Gauss laws and numerical related to both laws are important.
Question: What is the weightage of electric charges and fields in JEE Mains
Answer:
In JEE mains around 3.3% of questions are asked from the Physics class 12 chapter 1 questions and answers. That is 1 to 2 questions can be expected.
Question: What is the weightge of NCERT chapter electric charges and fields for neet exam
Answer:
For NEET exam 4% questions can be expected from the NCERT physics chapter 1 Electric Charges and Fields.
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sir board exam 2022-23 me 30% syllabus reduce hoga?
Dear aspirant !
Hope you are doing great ! No, why will it reduce ,there is no covid like situation as in the past years ,so why are you wasting your time by thinking such thoughts,no syllabus will be reduced onwards ,if covid situation arises then there may be possibility to reduce the syllabus , but as see the present situation ,no syllabus will deduct ,and also if syllabus reduced then it is not profitable for you because in neet and jee ,full syllabus comes , so if they reduce the syllabus then study full syllabus !
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I have six subject in cbse class 12th (math ,physics,chemistry, computer science, english, physical education) I fail in chemistry but pass all other subject will cbse consider as pass or fail
Hello ,
If you fail in only one of the six subject of you class 12 boards examination you will be considered to be overall Passed and the subject in which you failed will be considered as the additional subject .You will have the opportunity to attempt the compartment exam if you want to (I would suggest you to attempt it as it might create some problem in future ).
If you fail in two or more subjects you will be considered as fail and will have to repeat class 12 .
Hope this helps.
I bubbled 2 digits of my roll no. and 1 digit of my subject code incorrectly. I feel awful about it. I made a grave mistake and I think my future is ruined. Any chance my board paper wont be rejected?
Hello Aspirant,
See the mistake that you have done is little bit big and it can cause you some trouble because incorrect roll number might misplace your copy and due to which your result might get affected.
So I would recommend you to contact the examination center which has been allotted to you, because they can help you with this issue to some extent and from next time be more careful.
I hope this information can help you upto some extent.
Good Luck!!!
CBSE class 12th English question
You may click on the links given below as desired (to view Class 12 CBSE English Question Papers) -
For Term 1 CBSE English Paper - CBSE Class 12 English Question Paper 2021-22 (Term 1) (careers360.com)
For Term 2 CBSE English Sample Paper - CBSE Class 12 English Core Sample Paper 2022 (Term 2) (careers360.com)
For 2020 CBSE English Paper - CBSE 12th English Core Question Papers 2020 (careers360.com)
For 2019 CBSE English Paper - CBSE Class 12 English Question Paper 2019 (careers360.com)
For 2018 CBSE English Paper - CBSE Class 12th Question Papers 2018 English 1 (careers360.com)
To get complete set of previous year question papers, subject wise, you may visit the link - CBSE Class 12 Previous Year Question Papers With Solutions PDF Download (careers360.com)
Some recommended books to practise questions from, for CBSE English Examination -
- CBSE All In One English Core Class 12 for 2022 Exam (Updated edition for Term 1 and 2)
- Oswaal Term 2 English Core CBSE Class 12 Sample Question Paper Book (For Term-2 2022 Exam) Oswaal Books
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