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The first semester of the B.E. Aerospace Engineering programme at Chandigarh University is scheduled to commence on 1st August 2025, as per the university's academic calendar for the session July–December 2025. The university will also conduct Orientation and Induction Programs for all undergraduate and postgraduate programs, including B.E. Aerospace Engineering, starting from 2nd August 2025 and continuing through the weekends until 25th August/September 2025. For detailed information on the course curriculum, you can refer to the official website:

The CUET PG exam is conducted annually, and generally takes place in March or April. The specific date of the exam is provided to candidates (by the NTA, on their website) months in advance.

When the NTA issues the notification for the CUET PG examination, it will include everything candidates should know regarding the exam including exam eligibility, application details, exam format, syllabus and timetable.

Candidates should regularly check the official NTA website for any recent updates concerning the CUET PG exam.

For a good preparation, the students can refer to the popularly used books for WBJEE. Along with the books, the students can also attempt mock tests and sample papers regularly. The previous years' question papers are also available for the students to solve and practice.

Since the syllabus level of WBJEE is similar to the board exams, the students are advised to refer to the NCERT books for preparation. However, for a better understanding of the concepts and topics of WBJEE syllabus, the students should also check other important reference books.

Students can check the previous years' WBJEE question papers to check the topics and questions that are repeatedly present and have a high weightage. These topics can be picked out as important topics and the students can focus on them during the preparation process. However, the students should ensure that the entire syllabus is also covered.

Sphalarite             ZnS

Calamine              ZnCO₃

Galena                  PbS

Siderite                FeCO₃

The total marks of GATE AR question paper are 100. General aptitude section consists of 15 marks and part A is of 60 marks. On the other hand, part B consists of 25 marks.

The GATE pattern and structure for data Science and artificial intelligence can be checked below -

• The exam mode is online and the question paper will be available in English language.
• The general aptitude section carries 15 marks and the remining 85 marks are reserved for the subject-based questions. 
• The time duration of GATE DA paper is 3 hours. 
• However, PwD or dyslexic candidates will get one hour compensatory time to attempt the GATE exam. 

Acidic oxide Þ Cl₂O₇

Neutral oxide Þ N₂O, NO

Basic oxide Þ Na₂O

Amphoteric oxide Þ As₂O₃

NAT questions in the GATE exam are the Numerical Answer Type questions, where the candidate has to enter a numerical value for the question using virtual keypad. The NAT question has no negative marking and no multiple options. To answer the GATE NAT questions, candidates must know how to efficienty answer using the virtual keyboard. 

$T_{90\mathrm{\%}}=\frac{2.303}{k}log\frac{100}{10}$

$T_{50\mathrm{\%}}=\frac{2.303}{k}log\frac{100}{50}$

$\frac{T_{90\mathrm{\%}}}{T_{50\mathrm{\%}}}=\frac{log10}{log2}=\frac{1}{0.301}=3.32$

WBJEE is divided into two papers – Paper 1 (Maths) and Paper 2 (Physics and Chemistry). For each paper, the students will have two hours to complete the exam. Overall, the duration of the WBJEE paper is four hours. No extra time will be given to complete the exam.

$2C\left(s\right)+2H_2\left(g\right)\text{?}\text{?}\to C_2{H}_6\left(g\right)\Delta H=\Delta H_f\left(C_2{H}_6\right)$

$\Delta H_f\left(C_2{H}_6\right)=\left[2\times \left(-394\right)\right]+\left[3\times \left(-286\right)\right]-\left[1\times \left(-1560\right)\right]KJ/mole$

= (-788 – 858 + 1560) KJ/mole = -86 kJ/mole

Yes, the WBJEE exam will be held as a multiple-choice paper. Every question will have multiple options, and the students will have to select the correct answer. In the OMR sheet, the correct answer has to be marked and submitted.

$I_{S_4}=4I_0$

$I_{S_3}=I_0$

$\therefore \frac{I_{max}}{I_{min}}=\frac{{\left(\sqrt[]{I_{S_4}}+\sqrt[]{I_{S_3}}\right)}^2}{{\left(\sqrt[]{I_{S_4}}-\sqrt[]{I_{S_3}}\right)}^2}={\left(\frac{2+1}{2-1}\right)}^2=\frac{9}{1}$

WBJEE question paper will consist of questions from Physics, Chemistry and Mathematics. Both Physics and Chemistry section will have 40 questions each. A total of 75 questions have to be attempted in the Mathematics section. Each subject will have different number of questions in each category.

WBJEE question paper will have three categories. Category 1 and 2 questions will have only one correct option. Category 3 questions may have one or more correct options.

${\lambda }_1=\frac{h}{\sqrt[]{2m{E}_1}}={\lambda }_2=\frac{h}{\sqrt[]{2m{E}_2}}$

$=E_2=\frac{4}{9}{E}_1=4eV$

$E_2=E_1-e{V}_0=V_0=5V$

Bond order of $C_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-4}{2}=3$  

 Bond order of $N_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-6}{2}=2$   

Bond order of  $O_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-8}{2}=1$  

N_B = No. of electrons in bonding molecular orbitals.

N_A = No. of electron is Anti bonding molecular orbitals

Energy of 1 photon = $\frac{hc}{\lambda }$  

Energy of 1 mole of photon =  $N_A\times \frac{hc}{\lambda }$  

$=\frac{6.022\times 10^{23}\times 6.63\times 10^{-34}\times 3\times 10^8}{300\times 10^{-9}}$  

=0.399 × 10⁶ J = 399KJ