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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, students should check popularly used books for WBJEE. Along with books, students can also attempt mock tests and sample papers regularly. Previous years' question papers are also available for students to solve and practice.

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

Students can check previous years' WBJEE question papers to check  topics and questions that are repeatedly present and have high weightage. These topics can be picked out as important topics and students can focus on them during preparation. Students should make sure that 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 are Numerical Answer Type questions, where candidates have to enter a numerical value using GATE virtual calculator. There is no negative marking for GATE NAT questions.

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

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

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

Total duration of WBJEE is four hours. WBJEE is divided into two papers. 2 hours each will be given for Paper 1 (Mathematics) and Paper 2 (Physics and Chemistry). No extra time will be given to complete 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

WBJEE will be held as a multiple choice examination. Every question will have multiple options, and students will have to select correct answer.

$I_{S_4}=4I_0$

$I_{S_3}=I_0$

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

WBJEE is held as an offline test. Mathematics will have 75 questions and Physics and Chemistry will have 80 questions each. Total number of questions in WBJEE is 155.

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