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Edited By Vishal kumar | Updated on Sep 13, 2023 08:46 AM IST | #CBSE Class 12th

**NCERT Solutions for Class 12 Physics Chapter 13 Nuclei** serve as a crucial resource for achieving high scores in both board exams and competitive ones like JEE and NEET. On this NCERT solution page, you'll discover comprehensive and detailed class 12 nuclei ncert solutions to the entire exercise, ranging from questions 13.1 to 13.22 (exercise questions) and 13.23 to 13.31 (additional exercise questions). Additionally, these class 12 physics chapter 13 exercise solutions are conveniently available in PDF format, allowing students to access them offline, free from any internet constraints.

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- NCERT Solutions for Class 12 Physics Chapter 13 – Access and Download Free PDF
- NCERT Solutions for Class 12 Physics Chapter 13 Nuclei
- NCERT solutions for class 12 physics chapter 13 nuclei: Additional Exercise Solution
- NCERT solutions for class 12 physics chapter-wise
- Significance of NCERT solutions for class 12 physics chapter 13 nuclei:
- Key Features of Physics Chapter 13 Class 12 NCERT Solutions
- NCERT solutions subject wise

Do you know that the size of an atom is 10,000 times the size of a nucleus? But the nucleus contains 99.9% of the mass of an atom. We know that an atom has a structure. Does the nucleus also have a structure? If so what are the constituents and how they are arranged? All these questions are answered in Nuclei Class 12 NCERT text book.

Nuclei Class 12 chapter comes under modern Physics and you can expect at least one question for the board exam from NCERT Class 12 Physics chapter 13. Learning the class 12 physics chapter nuclei ncert solutions is important to score well in the board exam. The questions in NCERT Solutions for Class 12 Physics Chapter 13 Nuclei are divided into two parts namely exercise and additional exercise. Nuclei Class 12 NCERT solutions download PDF option is available to read the solution offline.

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Free download **physics chapter 13 class 12 ncert solutions ****pdf** for CBSE exam.

** NCERT solutions for class 12 physics chapter 13 nuclei: Exercise Solution **

** Answer: **

Mass of the two stable isotopes and their respective abundances are and and and .

m=6.940934 u

** Answer: **

The atomic mass of boron is 10.811 u

Mass of the two stable isotopes are and respectively

Let the two isotopes have abundances x% and (100-x)%

Therefore the abundance of is 19.89% and that of is 80.11%

** Answer: **

The atomic masses of the three isotopes are 19.99 u(m _{ 1 } ), 20.99 u(m _{ 2 } ) and 21.99u(m _{ 3 } )

Their respective abundances are 90.51%(p _{ 1 } ), 0.27%(p _{ 2 } ) and 9.22%(p _{ 3 } )

The average atomic mass of neon is 20.1771 u.

** Q. 13.3 ** Obtain the binding energy( in MeV ) of a nitrogen nucleus , given * m *

** Answer: **

m _{ n } = 1.00866 u

m _{ p } = 1.00727 u

Atomic mass of Nitrogen m= 14.00307 u

Mass defect m=7 m _{ n } +7 m _{ p } - m

m=7 1.00866+7 1.00727 - 14.00307

m=0.10844

Now 1u is equivalent to 931.5 MeV

E _{ b } =0.10844 931.5

E _{ b } =101.01186 MeV

Therefore binding energy of a Nitrogen nucleus is 101.01186 MeV.

** Q. 13.4 (i) ** Obtain the binding energy of the nuclei and in units of MeV from the following data:

** Answer: **

m _{ H } = 1.007825 u

m _{ n } = 1.008665 u

The atomic mass of is m=55.934939 u

Mass defect

m=0.528461

Now 1u is equivalent to 931.5 MeV

E _{ b } =0.528461 931.5

E _{ b } =492.2614215 MeV

Therefore the binding energy of a nucleus is 492.2614215 MeV.

Average binding energy

** Q. 13.4 (ii) ** Obtain the binding energy of the nuclei and in units of MeV from the following data:

** Answer: **

m _{ H } = 1.007825 u

m _{ n } = 1.008665 u

The atomic mass of is m=208.980388 u

Mass defect

m=126 1.008665+83 1.007825 - 208.980388

m=1.760877 u

Now 1u is equivalent to 931.5 MeV

E _{ b } =1.760877 931.5

E _{ b } =1640.2569255 MeV

Therefore the binding energy of a nucleus is 1640.2569255 MeV.

** Answer: **

Mass of the coin is w = 3g

Total number of Cu atoms in the coin is n

n=2.871 10 ^{ 22 }

m _{ H } = 1.007825 u

m _{ n } = 1.008665 u

Atomic mass of is m=62.92960 u

Mass defect m=(63-29) m _{ n } +29 m _{ H } - m

m=34 1.008665+29 1.007825 - 62.92960

m=0.591935 u

Now 1u is equivalent to 931.5 MeV

E _{ b } =0.591935 931.5

E _{ b } =551.38745 MeV

Therefore binding energy of a nucleus is 551.38745 MeV.

The nuclear energy that would be required to separate all the neutrons and protons from each other is

n E _{ b } =2.871 10 ^{ 22 } 551.38745

=1.5832 10 ^{ 25 } MeV

=1.5832 10 ^{ 25 } 1.6 10 ^{ -19 } 10 ^{ 6 } J

=2.5331 10 ^{ 9 } kJ

** Q.13.6 (i) ** Write nuclear reaction equations for

** Answer: **

The nuclear reaction equations for the given alpha decay

** Q.13.6 (ii) ** Write nuclear reaction equations for

** Answer: **

The nuclear reaction equations for the given alpha decay is

** Q.13.6 (iii) ** Write nuclear reaction equations for

** Answer: **

The nuclear reaction equations for the given beta minus decay is

** Q.13.6 (iv) ** Write nuclear reaction equations for

** Answer: **

The nuclear reaction equation for the given beta minus decay is

** Q.13.6 (v) ** Write nuclear reaction equations for

** Answer: **

The nuclear reaction for the given beta plus decay will be

** Q.13.6 (vii) ** Write nuclear reaction equations for

Electron capture of

** Answer: **

The nuclear reaction for electron capture of is

** Answer: **

(a) The activity is proportional to the number of radioactive isotopes present

The number of half years in which the number of radioactive isotopes reduces to x% of its original value is n.

In this case

It will take 5T years to reach 3.125% of the original activity.

(b) In this case

It will take 6.64T years to reach 1% of the original activity.

** Answer: **

Since we know that activity is proportional to the number of radioactive isotopes present in the sample.

Also

but

Therefore

The age of the Indus-Valley civilisation calculated using the given specimen is approximately 4217 years.

** Answer: **

Required activity=8.0 mCi

1 Ci=3.7 10 ^{ 10 } decay s ^{ -1 }

8.0 mCi=8 10 ^{ -3 } 3.7 10 ^{ 10 } =2.96 10 ^{ 8 } decay s ^{ -1 }

T _{ 1/2 } =5.3 years

Mass of those many atoms of Cu will be

7.12 10 ^{ -6 } ^{ } ^{ } g of is necessary to provide a radioactive source of 8.0 mCi strength.

** Q. 13.10 ** The half-life of is 28 years. What is the disintegration rate of 15 mg of this isotope?

** Answer: **

T _{ 1/2 } =28 years

The number of atoms in 15 mg of is

N=1.0038 10 ^{ 20 }

The disintegration rate will be

=-1.0038 10 ^{ 20 } 7.85 10 ^{ -10 }

=-7.88 10 ^{ 10 } s ^{ -1 }

The disintegration rate is therefore 7.88 10 ^{ 10 } decay s ^{ -1 } .

** Q.13.11 ** Obtain approximately the ratio of the nuclear radii of the gold isotope and the silver isotope

** Answer: **

The nuclear radii are directly proportional to the cube root of the mass number.

The ratio of the radii of the given isotopes is therefore

** Q.13.12 ** Find the Q-value and the kinetic energy of the emitted -particle in the -decay of

** Answer: **

Mass defect is m

m=226.02540-222.0175-4.002603

m=0.005297 u

1 u = 931.5 MeV/c ^{ 2 }

Q-value= m 931.5

=4.934515 MeV

By using Linear Momentum Conservation and Energy Conservation

The kinetic energy of alpha particle =

=

=4.847 MeV

** Q.13.12 ** ** (b) ** Find the Q-value and the kinetic energy of the emitted -particle in the -decay of

** Answer: **

Mass defect is m

m=220.01137-216.00189-4.002603

m=0.006877 u

1 u = 931.5 MeV/c ^{ 2 }

Q-value= m 931.5

=6.406 MeV

By using Linear Momentum Conservation and Energy Conservation

The kinetic energy of alpha particle =

=

=6.289 MeV

** Q.13.13 ** The radionuclide decays according to

The maximum energy of the emitted positron is .

and

calculate Q and compare it with the maximum energy of the positron emitted.

** Answer: **

If we use atomic masses

Q-value= 0.001033 931.5=0.9622 MeV which is comparable with a maximum energy of the emitted positron.

** Answer: **

The decay equation is

(we did not subtract the mass of the electron as it is cancelled because of the presence of one more electron in the sodium atom)

Q=0.004696 931.5

Q=4.3743 eV

The emitted nucleus is way heavier than the particle and the energy of the antineutrino is also negligible and therefore the maximum energy of the emitted electron is equal to the Q value.

the following

Atomic masses are given to be

** Answer: **

The above negative value of mass defect implies there will be a negative Q value and therefore the reaction is endothermic

Atomic masses are given to be

** Answer: **

The above positive value of mass defect implies Q value would be positive and therefore the reaction is exothermic

** Answer: **

The reaction will be

The mass defect of the reaction will be

Since the mass defect is negative the Q value will also negative and therefore the fission is not energetically possible

** Answer: **

Number of atoms present in 1 kg(w) of =n

Energy per fission (E)=180 MeV

Total Energy released if all the atoms in 1 kg undergo fission = E n

=180 2.52 10 ^{ 24 }

=4.536 10 ^{ 26 } MeV

** Answer: **

The amount of energy liberated on fission of 1 atom is 200 MeV.

The amount of energy liberated on fission of 1g

Total Energy produced in the reactor in 5 years

Mass of which underwent fission, m

=1537.8 kg

The amount present initially in the reactor = 2m

=2 1537.8

=3075.6 kg

** Answer: **

The energy liberated on the fusion of two atoms of deuterium= 3.27 MeV

Number of fusion reactions in 2 kg of deuterium = N _{ A } 500

The energy liberated by fusion of 2.0 kg of deuterium atoms E

Power of lamp (P)= 100 W

Time the lamp would glow using E amount of energy is T=

=4.99 10 ^{ 4 } years

** Answer: **

For a head-on collision of two deuterons, the closest distances between their centres will be d=2 r

d=2 2.0

d=4.0 fm

d=4 10 ^{ -15 } m

charge on each deuteron = charge of one proton=q =1.6 10 ^{ -19 } C

The maximum electrostatic potential energy of the system during the head-on collision will be E

The above basically means to bring two deuterons from infinity to each other would require 360 keV of work to be done or would require 360 keV of energy to be spent.

** Answer: **

Mass of an element with mass number A will be about A u. The density of its nucleus, therefore, would be

As we can see the above density comes out to be independent of mass number A and R _{ 0 } is constant, so matter density is nearly constant

Show that if emission is energetically allowed, electron capture is necessarily allowed but not vice–versa.

** Answer: **

For the electron capture, the reaction would be

The mass defect and q value of the above reaction would be

where m _{ N } and m _{ N } are the nuclear masses of elements X and Y respectively

For positron emission, the reaction would be

The mass defect and q value for the above reaction would be

From the above values, we can see that if Q _{ 2 } is positive Q _{ 1 } will also be positive but Q _{ 1 } being positive does not imply that Q _{ 2 } will also have to positive.

** Answer: **

Let the abundances of and be x and y respectively.

x+y+78.99=100

y=21.01-x

The average atomic mass of Mg is 24.312 u

The abundances of and are 9.3% and 11.71% respectively

** Answer: **

The reaction showing the neutron separation is

But 1u=931.5 MeV/c ^{ 2 }

Therefore E=(0.008978) 931.5

E=8.363007 MeV

Therefore to remove a neutron from the nucleus 8.363007 MeV of energy is required

** Answer: **

The reaction showing the neutron separation is

But 1u=931.5 MeV/c ^{ 2 }

Therefore E=(0.014019) 931.5

E=13.059 MeV

Therefore to remove a neutron from the nucleus 13.059 MeV of energy is required

** Answer: **

Let initially there be N _{ 1 } atoms of and N _{ 2 } atoms of and let their decay constants be and respectively

Since initially the activity of is 1/9 times that of we have

(i)

Let after time t the activity of be 9 times that of

(ii)

Dividing equation (ii) by (i) and taking the natural log of both sides we get

where and

t comes out to be 208.5 days

** Q.13.26 ** Under certain circumstances, a nucleus can decay by emitting a particle more massive than an -particle. Consider the following decay processes:

Calculate the Q-values for these decays and determine that both are energetically allowed.

** Answer: **

1 u = 931.5 MeV/c ^{ 2 }

Q=0.03419 931.5

=31.848 MeV

As the Q value is positive the reaction is energetically allowed

1 u = 931.5 MeV/c ^{ 2 }

Q=0.00642 931.5

=5.98 MeV

As the Q value is positive the reaction is energetically allowed

** Answer: **

The fission reaction given in the question can be written as

The mass defect for the above reaction would be

In the above equation, m _{ N } represents nuclear masses

but 1u =931.5 MeV/c ^{ 2 }

Q=0.247995 931.5

Q=231.007 MeV

Q value of the fission process is 231.007 MeV

** Q.13.28 ** ** (i) ** Consider the D–T reaction (deuterium-tritium fusion)

(a) Calculate the energy released in MeV in this reaction from the data:

** Answer: **

The mass defect of the reaction is

1u = 931.5 MeV/c ^{ 2 }

Q=0.018883 931.5=17.59 MeV

** Q.13.28 (b) ** Consider the D–T reaction (deuterium–tritium fusion)

(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles ; k = Boltzman’s constant, T = absolute temperature.)

** Answer: **

To initiate the reaction both the nuclei would have to come in contact with each other.

Just before the reaction the distance between their centres would be 4.0 fm.

The electrostatic potential energy of the system at that point would be

The same amount of Kinetic Energy K would be required to overcome the electrostatic forces of repulsion to initiate the reaction

It is given that

Therefore the temperature required to initiate the reaction is

** Answer: **

decays from 1.088 MeV to 0 V

Frequency of is

Plank's constant, h=6.62 10 ^{ -34 } Js

Similarly, we can calculate frequencies of and

The energy of the highest level would be equal to the energy released after the decay

Mass defect is

We know 1u = 931.5 MeV/c ^{ 2 }

Q value= 0.001473 931.5=1.3721 MeV

The maximum Kinetic energy of would be 1.3721-1.088=0.2841 MeV

The maximum Kinetic energy of would be 1.3721-0.412=0.9601 MeV

** Answer: **

(a)

The above fusion reaction releases the energy of 26 MeV

Number of Hydrogen atoms in 1.0 kg of Hydrogen is 1000N _{ A }

Therefore 250N _{ A } such reactions would take place

The energy released in the whole process is E _{ 1 }

(b) The energy released in fission of one atom is 200 MeV

Number of atoms present in 1 kg of is N

The energy released on fission of N atoms is E _{ 2 }

** Answer: **

Let the amount of energy to be produced using nuclear power per year in 2020 is E

(Only 10% of the required electrical energy is to be produced by Nuclear power and only 25% of therm-nuclear is successfully converted into electrical energy)

Amount of Uranium required to produce this much energy is M

(N _{ A } =6.023 10 ^{ 23 } , Atomic mass of Uranium is 235 g)

=3.076 10 ^{ 4 } kg

This chapter revolves around the topic of 'Nuclei,' and class 12 physics chapter 13 exercise solutions are vital for several reasons. Firstly, understanding nuclear physics is crucial as it forms the basis for numerous advanced concepts. Secondly, scoring in this chapter can be relatively easier compared to some other physics chapters, provided one grasps the fundamentals. The exercise solutions provide a comprehensive breakdown of each question, aiding students in mastering this essential topic effectively and efficiently, even within a limited time frame.

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Just Study 40% Syllabus and Score upto 100%

Download EBookThe following formulas will be helpful to understand the NCERT solutions for Class 12 Physics chapter 13 Nuclei

- Radii of the nuclei,

Where A is the mass number and

- Mass defect:

Here Z is the atomic number, M is the mass of the nucleus and A is the mass number. This equation tells that the mass of the nucleus is always less than the mass of their constituents.

- Another important relation that helps in NCERT solutions for class 12 is Einstein's mass-energy relation.

Where c is the speed of light.

- Another main concept of NCERT is the law of radioactive decay. This is given by

Where N is the number of nuclei at any time t, is the number of nuclei at any time and lambda is disintegration constant.

- The half-life of a radionuclide is given by

Where lambda is the disintegration constant.

After completing all these topics try to do NCERT class 12 chapter 13 exercises. If you are unable to solve or have any doubts refer to the solutions of NCERT class 12 physics chapter 13 nuclei provided below.

- About 6% of questions are expected from the chapters atoms and NCERT Class 12th Physics Nuclei for CBSE board exams.
- The NCERT Class 12 Physics solutions chapter 13 will help to score well in this chapter.
- The topic of radioactive decay and half-life is important for the board and competitive exams like NEET and JEE Main.
- Sometimes same questions which are discussed in the solutions of NCERT class 12 Physics chapter 13 solutions come in the CBSE12th exam.

**Comprehensive Coverage:**These class 12 nuclei ncert solutions encompass all topics and questions found in Chapter 13, ensuring a thorough understanding of nuclear physics.**Detailed Explanations:**Each nuclei class 12 ncert solutions offers comprehensive, step-by-step explanations, making complex nuclear physics concepts accessible to students.**Clarity and Simplicity:**The class 12 physics chapter nuclei ncert solutions are presented in clear and straightforward language, ensuring ease of understanding.**Practice Questions:**Exercise questions are included for practice and self-assessment, enhancing students' problem-solving skills.**Exam Preparation:**These physics chapter 13 class 12 ncert solutions are essential for board exam preparation and provide valuable support for competitive exams.**Foundation for Advanced Study:**The concepts explored in this chapter serve as the foundation for more advanced studies in nuclear physics and related fields.**Free Access:**These solutions are available for free, ensuring accessibility to all students

- NCERT solutions for class 12 mathematics
- NCERT solutions for class 12 chemistry
- NCERT solutions for class 12 physics
- NCERT solutions for class 12 biology

**NCERT Exemplar Class 12 Solutions**

1. What is the weightage of the chapter nuclei for CBSE board exam

For CBSE board exam from NCERT class, 12 chapters 13 around 4 to 6 marks questions can be expected. All topics of the NCERT syllabus for the chapter Nuclei should be covered for the CBSE board exam.

2. Is the chapter Nuclei important for NEET and JEE Main

Yes the NCERT chapter Nuclei are important for both the exams. Both in NEET and JEE main syllabus the chapter Nuclei is present and 1 or 2 questions from the chapter can be expected for the exams. The questions discussed in the NCERT Solutions for the chapter Nuclei will give a better idea on how to use the formulas and give a better understanding of the concepts discussed.

3. What is the composition of the nucleus according to nuclei ncert solutions?

The nucleus is made up of protons, which are positively charged particles, and neutrons, which are neutral particles.

4. According to nuclei class 12 what is isotopes?

Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons in their nuclei.

5. How nuclei class 12 ncert solutions is important for Board?

nuclei ncert solutions are important for the Board exam as they provide clear explanations, help in solving questions, cover all important topics, provide a structured approach to solving problems, and are designed with the exam pattern in mind, helping in exam-oriented preparation.

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If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

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Thank you and wishing you all the best for your bright future.
**

Hello student,

**
If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:
**

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- Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
- Aim to register before late October to avoid extra fees.
- Schools might not offer classes for private students, so focus on self-study or coaching.

**
Remember
**
, these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

**
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