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Make Obsession For Excellence Your New Year Resolve Dear Friend, Since you have decided to become an Engineer, I take it for granted that you are ambitious enough to succeed in life. Without an ambition, you cannot reach anywhere. So, you have intelligence, and, by now, you have an ambition too, to become an Engineer. Both are very basic must-haves to chart your own path in life. As renowned Spanish painter of previous century, Salvador Dali (1904-1989), said, "Intelligence without ambition is a bird without wings." With an ambition defining your path of life, you are now ready to fly. But is ambition enough to make your mark? Here comes the passion. Ambition and passion are different constructs, though quite related. Ambition is about individual achievement. It is about reaching your full potential and achieving your personal goals. Passion is about purpose. It has a strong positive connotation. It manifests in persistence and a strong willingness to deal with challenges. Read about the lives of as many successful people as you can and you will find both ambition and passion almost always present in them from the very early stages of their lives. Still, there is a bigger thing, called obsession, which I came across when I was reading about Mr. Pawan Kumar Chandana, the space entrepreneur, who recently launched Vikram-S, India's first-ever private rocket. The co-founder and CEO of Skyroot Aerospace, Indian spaceflight company based at Hyderabad, is a Mechanical Engineering graduate from IIT Kharagpur. Instead of going abroad, like many of his ilk for further studies or enrolling in the high-paying finance or information technology fields, he opted for a job in Indian Space Research Organisation (ISRO). A bonus was the placement at the Vikram Sarabhai Space Centre, Thiruvananthapuram. In six years at ISRO, he learnt enough about rocketry to realise that he could develop "Obsession of a lifetime is the key to it privately like SpaceX and Rocket Lab are doing abroad. Quitting ISRO in 2018, achieve, to excel at what one does. It's Mr. Chandana co-founded Skyroot which just within four years of its existence not passion, it's obsession. Obsession has achieved a historic feat of becoming the first among private players to is what will change the world." test multiple rocket propulsion systems. Funding is no longer a problem and —Mr. Pawan Kumar Chandana, Mr. Chandana has only sky, both literally and proverbially, now to reach. Space entrepreneur What led to such a stupendous success of Mr. Chandana and that too at an early age of 31? "Obsession," Mr. Chandana is quite categorical and emphatic. He had an obsession—yes, obsession—for rockets since his college years. Rockets became his fascination as "they were the fantastic machines that have the power to escape the Earth's gravity and go to space". This obsession for rockets made him shun everything else in life and work progressively to gain command over every aspect of rocketry and reach where he has reached today, with a strong possibility of reaching further newer heights in space and life. That is why his words carry more weight when he says, "Obsession builds the courage and the raw confidence to do anything in life. When you are obsessed, you are hungry 24×7. It's insatiable and it's the only thing in your mind. Obsession is the fuel when it's directed towards the right things." And Mr. Chandana is not alone in believing that you must have an obsession for your chosen field of activity in order to have ultimate success. He draws his inspiration from Mr. Conor McGregor, the Irish professional mixed martial artist, who is the first Ultimate Fighting Championship (UFC) fighter to hold UFC championships in two weight classes of featherweight and lightweight simultaneously. He goes to the extent of discounting talent in favour of obsession when he says, "This is an obsession. Talent doesn't exist, we all are equal as human beings. You could be anyone if you put in the time. I'm not talented, I am obsessed." An obsession, by definition, is something that excessively preoccupies the mind. There is a certain negative connotation attached to it as psychiatry deals with obsession in an entirely different manner. Nonetheless, if you can develop an obsession for excellence in your chosen field, it can result in a very powerful and high-potential positive mental state. New Year is the best opportunity to discover and adopt something new for betterment in your life. Developing an obsession for excellence can be a good New Year resolve for 2023. As Mr. Chandana says, "Obsession of a lifetime is the key to achieve, to excel at what one does. It is a key to navigate the ifs and challenges in life. It is the key to expand the boundaries of what is existing and change the world. So, it's not passion, it's obsession. Obsession is what will change the world." With these words, I wish for you everything that you need to achieve your desired success, with a most-fulfilling New Year 2023. Yours Sincerely

(Surendra Kumar Sachdeva) 4

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

INFORMATION BULLETIN

Joint Entrance Examination (Main) 2023 Conducted by National Testing Agency (NTA) website: www.nta.ac.in, https://jeemain.nta.nic.in

IMPORTANT INFORMATION AT A GLANCE 1. Important Dates : EVENTS

DATES (a) Session-1: JEE (Main) - January 2023 Online Submission of Application Form 15 December 2022 to 12 January 2023 (up to 09.00 P.M.) Last date for successful transaction of prescribed Application Fee 12 January 2023 (up to 11:50 P.M.) Announcement of the City of Examination Second week of January 2023 Downloading Admit Cards from the NTA website Third week of January 2023 Dates of Examination 24, 25, 27, 28, 29, 30, 31 January 2023 Display of Question Paper attempted by the Candidate and To be displayed on the NTA website Answer Keys for inviting challenges Declaration of Result To be displayed on the NTA website (b) Session-2: JEE (Main) - April 2023 Online Submission of Application Form 07 February 2023 to 07 March 2023 (up to 09.00 P.M.) Last date for successful transaction of prescribed Application Fee 07 March 2023 (up to 11.50 P.M.) Announcement of the City of Examination Third week of March 2023 Downloading Admit Cards from the NTA website Last week of March 2023 Dates of Examination 06, 07, 08, 09, 10, 11, 12 April 2023 Display of Question Paper attempted by the Candidate and To be displayed on the NTA website Answer Keys for inviting challenges Declaration of Result To be displayed on the NTA website (c) Duration of Examination for each Session of JEE (Main) - 2023 Paper 1 (B.E./B.Tech) or Paper 2A (B.Arch) or Paper 2B (B.Planning) 3 Hours B.Arch & B.Planning (both) 3 Hours 30 Minutes (d) Timing of Examination for each Session: JEE (Main) - 2023 Duration of Examination First Shift Second Shift For 3 Hours Paper 09.00 a.m. to 12.00 Noon (IST) 03.00 p.m. to 06.00 p.m. (IST) For 3 Hours 30 Minutes Paper

09.00 a.m. to 12.30 p.m. (IST)

03.00 p.m. to 06.30 p.m. (IST)

2. Fee Payable for JEE (Main) 2023 for each session Type of (Through Credit Card/Debit Card/Net Banking/UPI): Candidate General/Gen-EWS/ Male Paper 1 : B.E./B.Tech Female OBC (NCL) or Paper 2A : B.Arch Male SC/ST/PwD or Female Paper 2B : B.Planning Third Gender General/Gen-EWS/ Male Paper 1 : B.E./B.Tech & Paper 2A : B.Arch (or) Female Paper 1 : B.E./B.Tech & Paper 2B : B.Planning (or) OBC (NCL) Paper 1 : B.E./B.Tech, Paper 2A : B.Arch & Male SC/ST/PwD Female Paper 2B : B.Planning (or) Paper 2A : B.Arch & Paper 2B : B.Planning Third Gender

In India Outside India (Fee in Rs.) (Fee in Rs.) 1000 5000 800 4000 500 2500 500 2500 500 3000 2000 10000 1600 8000 1000 5000 1000 5000 1000 5000

Processing charges and Goods & Services Tax (GST) are to be paid by the candidate, as applicable. “Candidates are requested to fill in the Application Form very carefully. No corrections will be permitted once the Application Form is submitted”. Note: Multiple Application Forms submitted by a candidate for the same Session(s) will not be accepted and will lead to the cancellation of his/ her result.. ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

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Joint Entrance Examination (Main) 2023 1. About JEE (Main) - 2023 The Joint Entrance Examination, JEE (Main) comprises two papers. Paper 1 is conducted for admission to Undergraduate Engineering Programs (B.E/B.Tech.) at NITs, IIITs, other Centrally Funded Technical Institutions (CFTIs), and Institutions/ Universities funded/recognized by participating State Governments. JEE (Main) is also an eligibility test for JEE (Advanced), which is conducted for admission to IITs. Paper 2 is conducted for admission to B. Arch and B. Planning courses in the country. The JEE (Main) - 2023 will be conducted in 02 (two) sessions for admissions in the next academic session. The candidates will thus benefit in the following ways: l This will give two opportunities to the candidates to improve their scores in the examination if they are not able to give their best in one attempt. l In the first attempt, the students will get a first-hand experience of taking an examination and will know their mistakes which they can improve while attempting for the second time. l This will reduce the chances of dropping a year and droppers would not have to waste an entire year. l If anyone missed the examination due to reasons beyond control (such as the Board examination), then he/she will not have to wait for one entire year. l A candidate need not appear in both Sessions. However, if a candidate appears in more than one Session then his/her best of the JEE (Main) - 2023 NTA Scores will be considered for preparation of the Merit List/ Ranking. 1. JEE (Main) - 2023 Session 1 for Paper 1 (B.E./B.Tech.) will be held on 24, 25, 27, 28, 29, 30 and 31 January 2023 followed by Session 2 on 06, 07, 08, 09, 10, 11 and 12 April 2023. This is being done to ensure that the JEE (Main) - 2023 does not interfere with the Board examinations, which may be held at different times across the States/UTs. Paper 2A and Paper 2B (B. Arch and B. Planning) will also be held twice a year (January and April 2023). 2. It has been decided to provide choices in one section of each subject of Paper 1 and Part-I of Paper 2A and 2B to cater to the decision of different Boards across the country regarding the reduction of the syllabus. However, the total number of questions to be attempted will remain the same (Physics – 25, Chemistry – 25, and Mathematics – 25), wherever applicable.

2. Examination Scheme 2.1 Mode of Examination JEE (Main) - 2023 will be conducted in the following modes: a) Paper 1 (B.E. /B. Tech.) in “Computer Based Test (CBT)” mode only. b) Paper 2A (B. Arch): Mathematics (Part-I) and Aptitude Test (Part-II) in “Computer Based Test (CBT)” mode only and Drawing Test (Part-III) in pen and paper (offline) mode, to be attempted on drawing sheet of A4 size. c) Paper 2B (B. Planning): Mathematics (Part-I), Aptitude Test (Part-II) and Planning Based Questions (Part-III) in Computer Based Test (CBT) mode only. 2.2 Choice of Medium of Question Papers Medium of the Question Papers: Drawing from the National Education Policy (NEP), the JEE (Main) - 2023 will be conducted in English, Hindi, Assamese, Bengali, Gujarati, Kannada, Malayalam, Marathi, Odia, Punjabi, Tamil, Telugu and Urdu. S. No. Language

Examination Centres

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

English English English English English English English English English English

11. 12. 13.

English and Tamil English and Telugu English and Urdu

and and and and and and and and and

Hindi Assamese Bengali Gujarati Kannada Malayalam Marathi Odia Punjabi

All Examination Centres All Examination Centres in India Examination Centres in Assam Examination Centres in West Bengal, Tripura, and Andaman & Nicobar Islands Examination Centres in Gujarat, Daman & Diu, Dadra & Nagar Haveli Examination Centres in Karnataka Examination Centres in Kerala and Lakshadweep Examination Centres in Maharashtra Examination Centres in Odisha Examination Centres in Punjab, Chandigarh, and Delhi/New Delhi (including Faridabad, Ghaziabad, Gurugram, Meerut, Noida/Greater Noida) Examination Centres in Tamil Nadu, Puducherry, and Andaman & Nicobar Islands Examination Centres in Andhra Pradesh and Telangana All Examination Centres in India

The option of language for Question Paper should be exercised while filling up the Application Form online and it cannot be changed at a later stage. Please note that for the correctness of the questions in all the question papers, the English version will be taken as final. 6

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

2.3 Scheme of Examination Subject combinations for each paper, type of questions in each paper, and mode of examination is given in the table below: PAPER SUBJECTS Mathematics, Paper 1 : B.E./B.Tech. Physics & Chemistry

Mode of Examination “Computer Based Test (CBT)” mode only

Paper 2A : B. Arch

TYPE OF QUESTIONS Objective Type - Multiple Choice Questions (MCQs) & Questions for which answer is a numerical value, with equal weightage to Mathematics, Physics & Chemistry Part-I: Mathematics Objective Type - Multiple Choice Questions (MCQs) & Questions for which answer is a numerical value Part-II: Aptitude Test Objective Type - Multiple Choice Questions (MCQs)

“Computer Based Test (CBT)” mode only

Part-III: Drawing Test Questions to test drawing aptitude

“Pen & Paper Based” (offline) mode to be attempted on Drawing sheet A4 size “Computer Based Test (CBT)” mode only

Paper 2B : B. Planning

Part-I: Mathematics

Objective Type - Multiple Choice Questions (MCQs) & Questions for which answer is a numerical value Part-II: Aptitude Test Objective Type - Multiple Choice Questions (MCQs) Part-III: Planning Based Questions

Objective Type - Multiple Choice Questions (MCQs)

2.4 Pattern of Examination Paper 1 : B.E./B.Tech. in Computer Based Test (CBT) mode: 1. Subject-wise distribution of Questions, Total Number of Questions and Marks

Subject

Section A

Section B

Marks

Mathematics 20* 10* 100 Physics 20* 10* 100 Chemistry 20* 10* 100 Total 90 300 *Each Subject will have two sections. Section A will be of Multiple-Choice Questions (MCQs) and Section B will contain Questions whose answers are to be filled in as a numerical value. In Section B, candidates have to attempt any 05 (five) questions out of 10. There will be negative marking for both Section A and Section B. For each question in Section B, enter the correct integer value of the answer using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. For Section B, the answer should be rounded off to the nearest Integer. 2. Marking Scheme for MCQs Correct Answer or the best Answer Four marks (+4) Incorrect Answer Minus one mark (–1) Unanswered / Marked for Review No mark (0) 3. Marking Scheme for questions for which answer is a Numerical value

Correct Answer Incorrect Answer Minus Unanswered / Marked for Review

4. Method of determining merit

Conversion of the raw score in Mathematics, Physics, Chemistry, and conversion of the total into NTA scores. Overall merit shall be prepared by merging NTA scores of all shifts of all days.

5. Method of resolving ties

Tie between candidates obtaining equal Total NTA scores in Paper 1: B.E./B.Tech will be resolved in the following manner in descending order: 1. NTA score in Mathematics, followed by 2. NTA score in Physics, followed by 3. NTA score in Chemistry, followed by 4. Candidate with less proportion of a number of attempted incorrect answers and correct answers in all the subjects in the Test, followed by 5. Candidate with less proportion of a number of attempted incorrect answers and correct answers in Mathematics in the Test, followed by 6. Candidate with less proportion of a number of attempted incorrect answers and correct answers in Physics in the Test, followed by 7. Candidate with less proportion of a number of attempted incorrect answers and correct answers in Chemistry in the Test followed by 8. Older in Age followed by 9. Application Number in ascending order

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

Four marks (+4) one mark (–1) No mark (0)

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Paper 2A (B. Arch): Mathematics (Part-I) and Aptitude Test (Part-II) in Computer Based Test (CBT) mode only and Drawing Test (Part-III) in Pen & Paper Based (offline) mode, to be attempted on drawing sheet of A4 size. 1. Subject-wise distribution of Questions, Total Number of Questions and Marks

Subject No of Part I : Mathematics 20* (Section A) Part II : Aptitude Test Part III : Drawing Test Total

Questions and 10* (Section B) 50 02 82

Marks 100 200 100 400

*20 questions will be MCQs and 05 (five) questions will have answers to be filled as a numerical value. There will be negative marking for both Section A and Section B. For each question in Section B, a candidate has to enter the correct integer value of the answer using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. The answer should be rounded off to the nearest Integer. 2. Marking Scheme for MCQs

3. (a) Marking Scheme for questions for which answer is a Numerical value (b) Marking Scheme for Drawing Test (Part III)

Correct Answer or the most appropriate Answer Incorrect Answer/Multiple Answer Unanswered /Marked for Review Correct Answer or the most appropriate Answer Incorrect Answer Unanswered / Marked for Review Two questions to be evaluated out of 100 marks.

Four marks (+4) Minus one mark (–1) No mark (0) Four marks (+4) Minus one mark (–1) No mark (0)

4. Method of determining merit

Conversion of the raw score in Mathematics, Aptitude Test, Drawing Test, and Total into NTA Scores. Overall merit shall be prepared by merging NTA Scores of all shifts of all days.

5. Method of resolving ties

Tie between candidates obtaining equal Total NTA scores in Paper 2A : B.Arch will be resolved in the following manner: 1. NTA score in Mathematics, followed by 2. NTA score in Aptitude Test, followed by 3. NTA score in Drawing Test, followed by 4. Candidate with less proportion of a number of attempted incorrect answers and correct answers in all the subjects in the Test, followed by 5. Candidate with less proportion of a number of attempted incorrect answers and correct answers in Mathematics (Part-I) in the Test, followed by 6. Candidate with less proportion of a number of attempted incorrect answers and correct answers in Aptitude Test (Part-II) in the Test followed by 7. Older in Age followed by 8. Application Number in ascending order

Paper 2B (B. Planning) Part-I : Mathematics, Part-II : Aptitude Test and Part-III : Planning-Based Questions in Computer Based Test (CBT) mode only 1. Subject wise distribution of

Subject

No of Questions

Marks

20* (Section A) and 10* (Section B)

100

Questions, Total Number of

Part-I: Mathematics

Questions and Marks

Part-II: Aptitude Test

50

200

Part-III: Planning

25

100

Total

105

400

*20 questions will be MCQs and 05 (five) questions will have answers to be filled as a numerical value. There will be negative marking for both Section A and Section B. For each question in Section B, enter the correct integer value of the answer using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. The answer should be rounded off to the nearest Integer. 2. Marking Scheme for MCQs

3. Marking Scheme for questions for which answer is a Numerical value 4. Method of determining merit

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Correct Answer or the most appropriate Answer Four marks (+4) Incorrect Answer / Multiple Answer Minus one mark (–1) Unanswered / Marked for Review No mark (0) Correct Answer or the appropriate Answer Four marks (+4) Incorrect Answer Minus one mark (–1) Unanswered / Marked for Review No mark (0) Conversion of the raw score in Mathematics, Aptitude Test, Planning Based Test, and Total into NTA Scores. Overall merit shall be prepared by merging NTA Scores of all shifts of all days. ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

5. Method of resolving ties

Tie between candidates obtaining equal Total NTA scores in B.Planning will be resolved in the following manner: 1. NTA score in Mathematics, followed by 2. NTA score in Aptitude Test, followed by 3. NTA score in Planning Based Questions, followed by 4. Candidate with less proportion of a number of attempted incorrect answers and correct answers in all the subjects in the Test, followed by 5. Candidate with less proportion of a number of attempted incorrect answers and correct answers in Mathematics (Part-I) in the Test, followed by 6. Candidate with less proportion of a number of attempted incorrect answers and correct answers in Aptitude Test (Part-II) in the Test, followed by 7. Candidate with less proportion of a number of attempted incorrect answers and correct answers in Planning Based Questions (Part-III) in the Test followed by 8. Older in Age followed by 9. Application Number in ascending order

Important Note: (a) For Multiple Choice Questions: To answer a question, the candidates need to choose one option corresponding to the correct answer or the most appropriate answer. However, if any anomaly or discrepancy is found after the process of challenges of the key verification, it shall be addressed in the following manner: (i) Correct answer or the most appropriate answer: Four marks (+4) (ii) Any incorrect option marked will be given minus one mark (-1). (iii) Unanswered/Marked for Review will be given no mark (0). (iv) If more than one option is found to be correct then Four marks (+4) will be awarded to only those who have marked any of the correct options. (v) If all options are found to be correct then Four marks (+4) will be awarded to all those who have attempted the question. (vi) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then full marks will be awarded to all candidates who have appeared irrespective of the fact whether the question has been attempted or not attempted by the candidate. (b) For Numerical Value Questions: There will be negative marking for Section B. However, if any anomaly or discrepancy is found after the process of challenges of the key verification, it shall be addressed in the following manner: i. Correct Answer: Four marks (+4) ii. Incorrect Answer: Minus one mark (-1) iii. Unanswered/Marked for Review: No mark (0). iv. If a question is found to be incorrect or the Question is dropped then Four marks (+4) will be awarded to all those who have attempted the question. The reason could be due to human error or technical error. v. Candidates are advised to do the calculations with the constants given (if any) in the questions. The answer should be rounded off to the nearest integer.

3. Eligibility And Qualifications 3.1 Age Criteria For appearing in the JEE (Main) - 2023, there is no age limit for the candidates. The candidates who have passed the class 12/equivalent examination in 2021, 2022, or appearing in 2023 irrespective of their age can appear in JEE (Main) - 2023 examination. However, the candidates may be required to fulfill the age criteria of the Institute(s) in which they are desirous of taking admission. 3.2 List of Qualifying Examinations (QE) i. The final examination of the 10+2 system, conducted by any recognized Central/ State Board, such as Central Board of Secondary Education, New Delhi; Council for the Indian School Certificate Examinations, New Delhi; etc. ii. Intermediate or two-year Pre-University examination conducted by a recognized Board/ University. iii. Final examination of the two-year course of the Joint Services Wing of the National Defence Academy iv. Senior Secondary School Exam conducted by the National Institute of Open Schooling with a minimum of five subjects. v. Any Public School/ Board/ University examination in India or any foreign country is recognized as equivalent to the 10+2 system by the Association of Indian Universities (AIU). vi. Higher Secondary Certificate Vocational Examination. vii. A Diploma recognized by AICTE or a State board of technical education of at least 3 years duration. viii. General Certificate Education (GCE) examination (London/Cambridge/Sri Lanka) at the Advanced (A) level. ix. High School Certificate Examination of the Cambridge University or International Baccalaureate Diploma of the International Baccalaureate Office, Geneva. x. Candidates who have completed the Class 12 (or equivalent) examination outside India or from a Board not specified above should produce a certificate from the Association of Indian Universities (AIU) to the effect that the examination they have passed is equivalent to the Class 12 Examination. xi. In case the Class 12 Examination is not a public examination, the candidate must have passed at least one public (Board or Pre-University) examination earlier. 3.3 Year of Appearance in Qualifying Examination Only those candidates who have passed the Class 12/equivalent examination in 2021, 2022, or those who are appearing in Class 12/equivalent examination in 2023, are eligible to appear in JEE (Main) - 2023. Candidates who passed the Class 12/equivalent examination in 2020 or before as well as those who will appear in such examination in 2024 or later are not eligible to appear in JEE (Main) - 2023. ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

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4. State of Eligibility State code of eligibility means the code of the State from where the candidate has passed Class XII (or equivalent) qualifying examination by virtue of which the candidate becomes eligible to appear in JEE (Main) 2023. It is important to note that the State code of eligibility does NOT depend upon the native place or the place of residence of the candidate. For example, if a candidate appears for the Class XII examination in New Delhi and is a resident of Noida, Uttar Pradesh, the candidate’s State code of eligibility will be that of Delhi and NOT that of Uttar Pradesh. If a candidate has passed Class XII (or equivalent) qualifying examination from one State but appeared for improvement from another State, the candidate’s State code of eligibility will be from where the candidate first passed the Class XII (or equivalent) examination and NOT the State from where the candidate has appeared for improvement. Candidate passed/appearing class 12th from NIOS should select the State of Eligibility according to the State in which the study centre is located. For Indian nationals passing the Class XII (or equivalent) examination from Nepal/Bhutan, the State code of eligibility will be determined based on a permanent address in India as given in the passport of the candidate. The State code of eligibility of OCI/PIO passing Class XII (or equivalent) examination in India is at par with Indian nationals. However, OCI/PIO passing the Class XII (or equivalent) examination from an institution abroad are eligible for Other State quota seats or All India quota seats (but NOT for Home State quota seats) in all NITs, IIEST and Other-CFTIs

5. Reservations 5.1 Indian nationals belonging to certain categories are admitted under the seats reserved for them in accordance with the rules prescribed by the Government of India. The categories and the extent of reservation are as follows: l General category belonging to Economically Weaker Section (GEN- EWS) – 10% of seats in every course. The benefit of reservation will be given only to those General category candidates who satisfy the conditions given in the OM No. 20013/ 01/2018-BC-II dated 17 January 2019, issued by the Ministry of Social Justice and Empowerment. The criteria for GENEWS will be as per the prevailing norms and/or notifications of the Government of India. l Other Backward Classes belonging to the Non-Creamy Layer (OBC- NCL) – 27% of seats in every course. i. OBCs should be listed in the current updated central list of OBCs (http://www.ncbc.nic.in). ii. OBCs present in the state list but not covered in the central list of OBCs (as per the list in http://www.ncbc.nic.in) are NOT eligible to claim the reservation. iii. The criteria for OBC-NCL will be as per the notification of the Government of India. iv. Candidates belonging to the creamy layer of OBC are NOT entitled to reservation. Such candidates are treated as belonging to the general (GEN), i.e. unreserved category, and they will be eligible only for the OPEN seats – the seats for which all candidates are eligible. l Scheduled Caste (SC) – 15% of seats in every course. l Scheduled Tribe (ST) – 7.5% of seats in every course. l Persons with Disability (PwD) – 5% seats in each of OPEN, GEN-EWS, OBC-NCL, SC, and ST category seats. i. Candidates with at least 40% impairment irrespective of the type of disability i.e., locomotor, visual or SEVERE dyslexic shall be eligible for the benefits of the PwD category. ii. Leprosy-cured candidates who are otherwise fit to pursue the courses are also included in this category. iii. Candidates with less than 40% disability, but difficulty in writing, as prescribed by the Competent authority, are also included in this category. 5.1.1. For Candidates claiming to the GEN-EWS category GEN-EWS certificate (Annexure-IA) needs to be uploaded in the Online Application Form of JEE (Main) – 2023 which should have been issued on or after 01 April 2022 in consonance with the latest guidelines of the Government of India. If any GEN-EWS candidate fails to submit the GEN-EWS certificate (issued on or after 01 April 2022) at the time of online registration, the candidate has to upload a declaration [Declaration in Lieu of Gen-EWS Certificate at Annexure-IB] to that effect (Reference: No.F.No.20013/01/2018-BC-II). 5.1.2 For Candidates claiming to the OBC-NCL category OBC-NCL certificate (Annexure-IIA) needs to be uploaded in the Online Application Form of JEE (Main) – 2023 which should have been issued on or after 01 April 2022, in consonance with the latest guidelines of the Government of India. If any OBC-NCL candidate fails to upload the OBC- NCL certificate (issued on or after 01 April 2022) at the time of online registration, the candidate has to upload a declaration [Declaration in Lieu of OBC-NCL Certificate as per Annexure-IIB] to that effect. Visit http://www.ncbc.nic.in for the latest guidelines and updates on the Central List of State-wise OBCs. 5.1.3. For Candidates claiming to the SC or ST category Caste (for SC) or tribe (for ST) certificate (Annexure-III) needs to be uploaded in the Online Application Form of JEE (Main) – 2023, as per the latest guidelines of the Government of India. 5.1.4 (a) Guidelines for conducting written examination for Persons with Benchmark Disabilities above 40% vide letter dated 29 August 2018 from Ministry of Social Justice and Empowerment Provisions relating to Persons with Disability (PwD): As per Section 2(t) of the RPwD Act, “Persons with Disability (PwD)” means a person with long-term physical, mental, intellectual, or sensory impairment which, in interaction with barriers, hinders his full and effective participation in society equally with others. According to Section 2(r) of the RPwD Act, 2016, “persons with benchmark disabilities” means a person with not less than forty percent (40%) of a specified disability where specified disability has not been defined in measurable terms and includes a person with disability where specified disability has been defined in measurable terms, as certified by the certifying authority. Facilities for PwD candidates to appear in the exam As per the guidelines issued by the Department of Empowerment of Persons with Disabilities (Divyangjan) under the Ministry of Social Justice and Empowerment issued from time to time on the subject: “Written Examination for Persons with Benchmark Disabilities”, for the candidate with one of the benchmark disabilities [as defined in Section 2(r) of 10

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

RPwD Act, 2016], holding a Disability Certificate in the prescribed format in the Rights of Person with Disabilities Rules, 2017 (link: https://upload.indiacode.nic.in/showfile?actid=AC_CEN_25_54_00002_201649_1517807328299 &type=rule&filename=Rules_notified_15.06.pdf). a. The facility of Scribe, in case he/she has a physical limitation and a scribe is essential to write the examination on his/ her behalf, being so certified in the aforesaid format by a CMO/Civil Surgeon/ Medical Superintendent of a Government Health Care Institution. b. Compensatory time of one hour for examination of three hours duration, whether such candidate uses the facility of Scribe or not. Services of a Scribe As per the office memorandum of the Ministry of Social Justice and Empowerment (Reference: F.No. 34-02/2015-DD-III dated August 29, 2018), the PwD candidates who are visually impaired OR dyslexic (severe) OR have a disability in the upper limbs OR have lost fingers/hands thereby preventing them from properly operating the Computer Based Test platform may avail the services of a scribe (amanuensis). The scribe will help the Candidate in reading the questions and/or keying in the answers as per the directions of the Candidate. A scribe will NEITHER explain the questions NOR suggest any solutions. PwD candidates who desire to avail the services of a scribe need to opt for this during the online registration of JEE (Main) – 2023. If a candidate desires to bring his/her own Scribe, then he/she should submit a Letter of undertaking for using own Scribe as per the format available at Annexure. It is to be noted that the Scribe may be provided by the National Testing Agency (NTA), if requestedin the online Application Form of JEE (Main) - 2023. If it is found at any stage that a candidate has availed the services of a scribe and/or availed the compensatory time, but does not possess the extent of disability that warrants the use of a scribe and/or grant of compensatory time, the candidate will be excluded from the process of evaluation, ranking, counseling, and admission. In case such a candidate has already been admitted to any Institution, the admission of the candidate will be cancelled. The NTA does not guarantee any change in the category or sub-category (PwD status) after the submission of the Online Application Form, and in any case, no change will be entertained by NTA after the declaration of NTA Score for JEE (Main) - 2023. The category/sub-category (PwD status) entered in the JEE (Main) Database by the candidate will be used for the JEE (Advanced). Therefore, the candidates are advised to fill in the category/sub-category column very carefully.

Note: 1. The minimum degree of disability should be 40% (Benchmark Disability) in order to be eligible for availing reservation for persons with specified disability. 2. The extent of “specified disability” in a person shall be assessed in accordance withthe “Guidelines for the purpose of assessing the extent of specified disability in a person included under the Rights of Persons with Disabilities Act, 2016 (49 of 2016)” notified in the Gazette of India by the Ministry of Social Justice and Empowerment [Department of Empowerment of Persons with Disabilities (Divyangjan)] on 4 January 2018. 3. No change in the category will be entertained after the last date specified by NTA for JEE (Main) - 2023 Registration. Candidates must note that the benefit of reservation will be given to them subject to verification of documents. If it is discovered at any stage that a candidate has used a false/fake/incorrect document, or has furnished false, incorrect, or incomplete information, in order to avail the benefit of reservation, then such a candidate shall be excluded from all admission processes. In case such a candidate has already been given admission, the admission shall stand cancelled. 5.2 In the case of the Institutes run/aided/recognized by State Governments, the reservation policy of the respective State Governments shall be applicable. Notes: 1. The benefit of reservation for admission to NITs/IIITs and CFTIs shall be given only to those classes/castes/tribes which are in the respective Central List published by the Government of India from time to time. 2. The benefit of reservation will be given only to those castes and tribes that are mentioned in the respective central list of corresponding states published by the Government of India (websites: http://socialjustice.nic.in and https://ncst.nic.in).

6. Choice of Cities The Cities where the JEE (Main) - 2023 will be conducted are given in Appendix I. While applying, candidates have to select any four cities of their choice. The efforts will be made to allot the city of examination to the candidates in order of choice opted by them in their online Application Form. However, due to administrative/logistic reasons, a different city can be allotted. Choice of Centre Cities will be limited to the State of Permanent Address or State of Present Address only. In case, there are very few candidates from a City, the NTA reserves the right to merge one, two, or more cities. The decision of the NTA regarding the allotment of the City/Centre shall be final. No further correspondence or request shall be entertained in such a case.

7. e-Admit Card The e-Admit Card would be issued provisionally to the candidates through the NTA website: https://jeemain.nta.nic.in/, subject to the fulfillment of the eligibility conditions and receipt of the prescribed application fee by NTA. The candidate has to download the Admit Card from the NTA website. The candidate will appear in the examination at the given Centre on the date and shift/timing as indicated in his/her e-Admit Card. No candidate will be allowed to appear at the examination Centre, on the date and shift/time other than that allotted to him/her in his/her Admit Card. ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

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In case a candidate is unable to download his/her Admit Card from the website, he/she should approach the NTA Help Line Number: 011-40759000 between 10.00 am to 5.00 pm. The candidates are advised to read the instructions on the Admit Card carefully and strictly follow them during the examination. In case of any discrepancy in the particulars of the candidate or his/her photograph and signature shown in the e-Admit Card and Confirmation Page, the candidate may immediately approach the NTA Help Line between 10.00 am to 5.00 pm. In such a case, the candidate would appear in the examination with the already downloaded Admit Card. However, NTA will take necessary action to make corrections in the record later. The Admit Card for Session 1 of the examination may be available for download in the Third week of January 2023. The timing for downloading the Admit Card for subsequent Sessions of the exam will be announced on the NTA website (https://jeemain.nta.nic.in/)

8. Schedule of Examination Dates of Examination

Session 1: 24, 25, 27, 28, 29, 30, 31 January 2023 Session 2: 06, 07, 08, 09, 10, 11, 12 April 2023 (Exact date, shift, and city of examination for Paper 1: B.E./B.Tech and Paper 2A : B. Arch and Paper 2B: B. Planning will be made available to the Candidates on their Admit Cards)

Mode of Examination

Paper 1 : B.E./B.Tech

“Computer Based Test (CBT)” mode only

Paper 2A : B. Arch.

Part-I: Mathematics and Part-II: Aptitude Test in Computer Based Test (CBT) mode only Part-III: Drawing Test in “Pen & Paper” (offline) mode, to be attempted on the Drawing sheet of A4 size.

Paper 2B : B. Planning Timing of Examination

Part-I: Mathematics, Part-II: Aptitude Test and Part-III: Planning related MCQ’s in “Computer Based Test (CBT)” mode only

Duration of Exam

First Shift

Second Shift

9.00 a.m. to 12.00 Noon

3.00 p.m. to 6.00 p.m.

For 3 Hours 30 Min. 09.00 a.m. to 12.30 p.m.

3.00 p.m. to 6.30 p.m.

Entry in the Examination Centre / Hall / Room, frisking, biometric registration/ record of manual attendance by Invigilator, document verification / cross checking of Admit Card, signature, and photo match to be completed by Invigilator, etc.

7.00 a.m. to 8.30 a.m.

1.00 p.m. to 2.30 p.m.

Instructions by the Invigilator(s)

8.30 a.m. to 8.50 a.m.

2.30 p.m. to 2.50 p.m.

For 3 Hours

Candidates log in to read instructions

8.50 a.m.

2.50 p.m.

Test Commences

9.00 a.m.

3.00 p.m.

Note: Those appearing for B.Arch. and B.Planning (both), will have the duration of examination 3½ hrs.

9. Important Instructions for the Candidates 1. Candidates are advised to report at the Examination Center well in time i.e. 2 hours before commencement of the examination. 2. Candidates should take their seats immediately after the opening of the Examination Hall. If the candidates do not report in time due to any reason i.e. traffic jam, train/bus delay, etc, they are likely to miss some of the important instructions to be announced in the Examination Rooms/Halls. The NTA shall not be responsible for any delay. 3. The candidate must show, on-demand, the Admit Card downloaded/printed from the NTA website for admission in the examination room/hall. The Test Centre Staff on duty is authorized to verify the identity of candidates and may take steps to verify and confirm the identity credentials. Candidates are requested to extend their full cooperation. A candidate who does not possess a valid Admit Card and authorized Photo ID shall not be permitted to take the examination under any circumstances by the Centre Superintendent. 4. A seat indicating Roll Number will be allotted to each candidate. Candidates should find and sit in their allocated seats only. In case a candidate ventures to change his/her seat and does not sit on the seat allotted to him/her could face cancellation of candidature. No plea would be entertained in this regard. 5. The candidate should ensure that the Question Paper available on the computer is as per his/her opted subject indicated in the Admit Card. In case, the subject of the Question Paper is other than his/her opted subject, the same may be brought to the notice of the Invigilator concerned. 6. The candidates may approach the Centre Superintendent/Invigilator in the room for any technical assistance, first aid emergency, or any other information during the course of the examination. For any queries or issues regarding 12

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Computer Based Test, the candidates may contact on Helpline Numbers available on Joint Entrance Examination (Main) website. 7. In case a candidate, by furnishing false information, appears in more than one shift/date, his candidature will be cancelled and his result will not be declared. 8. For those who are unable to appear on the scheduled date of test for any reason, re-test shall not be held by the NTA under any circumstances. Note: Candidates shall appear at their own cost at the Centre on the date and shift as indicated in their Admit Card issued by the NTA. Under no circumstances the choice of cities for the Centre and shift provided in the Admit Card shall be changed. Candidates MUST bring the following documents on the day of examination at the test centre. Candidates who will not bring these will not be allowed to sit in the examination. a. Print copy of Admit Card along with Self Declaration (Undertaking) downloaded from the NTA Website (a clear printout on A4 size paper) duly filled in. b. One passport size photograph (same as uploaded on the Online Application Form) for pasting on the specific space in the Attendance Sheet at Centre during the examination. c. Any one of the authorized photo IDs (must be original, valid, and non-expired) – School Identity Card/ PAN card/ Driving License/ Voter ID/ Passport/ Aadhaar Card (With photograph)/E-Aadhaar with photograph/ Ration Card with photograph/ Class 12 Board Admit Card with photograph/ Bank Passbook with Photograph. d. PwD certificate issued by the authorized medical officer, if claiming the relaxation under PwD category (or) PwD Certificate regarding physical limitation in an examination to write as per Annexures given in Information Bulletin, if claiming the relaxation under PwD category. e. A simple transparent Ball Point Pen.

10 . Rough Work All calculations/writing work is to be done only in the Rough Sheet provided at the Test Centre in the Examination Room/ Hall and on completion of the test, candidates must hand over the Rough Sheets to the Invigilator on duty in the Room/Hall.

11. Unfair Means Practices and Breach of Examination Rules 11.1 Definition Unfair Means practice is an activity that allows a candidate to gain an unfair advantage over other candidates. It includes, but is not limited to: a) Being in possession of any item or article which has been prohibited or can be used for unfair practices including any stationery item, communication device, accessories, eatable items, ornaments or any other material or information relevant or not relevant to the examination in the paper concerned; b) Using someone to write examination (impersonation) or preparing material for copying; c) Breaching examination rules or any direction issued by NTA in connection with JEE (MAIN) 2023 examination from time to time; d) Assisting other candidate to engage in malpractices, giving or receiving assistance directly or indirectly of any kind or attempting to do so; e) Contacting or communicating or trying to do so with any person, other than the Examination Staff, during the examination time in the Examination Centre; f) Threatening any of the officials connected with the conduct of the examination or threatening any of the candidates; g) Using or attempting to use any other undesirable method or means in connection with the examination; h) Manipulation and fabrication of online documents viz. Admit Card, Rank Letter, Self-Declaration, etc.; i) Forceful entry in /exit from Examination Centre/Hall; j) Use or attempted use of any electronic device after entering the Examination Centre; k) Affixing/uploading of wrong/morphed photographs/signatures on the Application Form/Admit Card/Proforma; l) Creating obstacles in smooth and fair conduct of examination; m) Any other malpractices declared as Unfair Means by the NTA. n) Any candidate with more than one Application Number (more than one SCORE CARDS) will be treated as UFM, even if found at a later stage, and strict action will be taken against that Candidate. 11.2 Punishment for using Unfairmeans Practices During the course of, before or after the examination if a candidate indulges in any of the above or similar practices, he/she shall be deemed to have used unfair practices and booked under UNFAIRMEANS (U.F.M.) case. The candidate would be debarred for 3 years in future and shall also be liable for criminal action and /or any other action as deemed fit. 11.3 Cancellation of Result The result of JEE (Main) - 2023 of the candidate(s) who indulge in Unfair means Practices will be cancelled and will not be declared. Similarly, the result of those candidates who appear from the Centre other than the one allotted to them or allow another candidate/person to write the exam on his behalf will be cancelled. No plea will be entertained in this regard.

12. Display of Answer Key for Challenge 12.1. Display of Answer Key for Challenges 1. The NTA will display Provisional Answer Key of the questions on the NTA website: https://jeemain.nta.nic.in/, with a Public Notice, issued to this effect on the said website, to provide an opportunity to the candidates to challenge the Provisional Answer Keys with a nonrefundable online payment of Rs. 200/- per question challenged as processing charges. The provisional Answer Keys are likely to be displayed for two to three days. 2. Only paid challenges made during stipulated time through key challenge link will be considered. Challenges without justification/evidence and those filed on any other medium other than the prescribed link will not be considered. ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

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3. The NTA decision on the challenges shall be final and no further communication will be entertained. NTA will not inform the Candidates individually about the outcome of the challenges made. 4. The subject experts will examine all the challenges received and then a final answer key will be displayed and declared. 5. The result will be compiled based on the final answer key declared. No grievance with regard to answer key(s) after the declaration of result/NTA Score of JEE (Main) - 2023 will be entertained. 12.2. Display of Question Paper attempted and Recorded Responses The NTA will display the recorded responses and Question Paper attempted by the candidates on the NTA website https://jeemain.nta.nic.in/ prior to the declaration of the result/NTA Score. The recorded responses are likely to be displayed for two to three days.

13. JEE (Main) NTA Score for B.E./B.Tech, B.Arch, and B.Planning 13.1 JEE (Main) NTA Score for B.E./B.Tech, B.Arch, and B.Planning a) Evaluation of multiple-choice questions of B.E./B.Tech, B.Arch, and B.Planning will be carried out using final answer keys and the raw (actual) marks obtained by a candidate will be considered further for computation of the result of JEE (Main) - 2023. b) For multi-shift papers, raw (actual) marks obtained by the candidates in different shifts/sessions will be converted to NTA Score. c) The detailed procedure for the compilation of the NTA Score is available on NTA Website. 13.2 Compilation and display of Result of Paper-1 (B.E./B.Tech.) of JEE (Main) - 2023: a) Compilation and display of NTA Score for Session 1 of Paper-1 (B.E. /B.Tech.): Since Session 1 of JEE (Main) - 2023 B.E./B.Tech will be conducted in multi-shifts, NTA scores will be calculated corresponding to the raw (actual) marks obtained by a candidate. The calculated NTA scores for all the shifts will be declared on the NTA website(s). This will comprise of the NTA scores for each of the three subjects (Mathematics, Physics, and Chemistry) and the total in B.E./B.Tech of Session 1 of Paper 1 of JEE (Main) – 2023. b) Compilation of NTA Score for Session 2 of Paper 1 (B.E. /B.Tech.): Similarly, Session 2 of JEE (Main) - 2023 B.E./B.Tech will be conducted in multi-shifts, NTA scores will be calculated corresponding to the raw (actual) marks obtained by a candidate. The calculated NTA scores for all the shifts will comprise the two NTA scores for each of the three subjects (Mathematics, Physics, and Chemistry) and the total in B.E./B.Tech of Session 2 of Paper 1 of JEE (Main) - 2023. c) Compilation and display of result including overall Merit List/Ranking of Paper 1 (B.E./B.Tech.): The NTA scores for each of the candidates in Total in B.E./B.Tech for Session 1 as well as for Session 2 of JEE (Main) - 2023 will be merged for compilation of results and preparation of overall Merit List/Ranking. The best of the two NTA Scores (Total) will be considered for further processing for those candidates who appeared in both sessions. In case of a tie, i.e. when two or more candidates obtain equal Total NTA Score in JEE (Main)-2023, inter-se merit of such candidates shall be determined as per “Method of resolving ties”. The NTA scores and rank of B.E./B.Tech of JEE (Main) - 2023 for all candidates who appeared in Session 1 as well as in Session 2 for JEE (Main) - 2023 will be declared on the NTA website(s). This shall comprise of the NTA scores obtained by the candidate in Session 1, NTA scores obtained in Session 2, and NTA scores for those who appeared in both sessions (best of the two Total NTA scores) along with the status of those who qualify for appearing in JEE (Advanced) - 2023 provided and subject to other conditions of eligibility being met. The rank shall comprise All India rank and All India category rank. 13.3 Compilation and display of Result of Paper 2A (B. Arch) or Paper-2B (B. Planning) of JEE (Main)-2023: a) Compilation and display of NTA Score for Session 1 of Paper 2A (B. Arch) or Paper 2B (B. Planning): Since Session 1 of JEE (Main) - 2023 Paper-2A (B. Arch) or Paper-2B (B. Planning) will be conducted in multi-shifts, NTA scores will be calculated corresponding to the raw (actual) marks obtained by a candidate. The calculated NTA scores for all the shifts will be declared on the NTA website(s). This will comprise of the NTA scores for each of the three parts (Mathematics, Aptitude Test and Drawing Test or Planning Based Test) and the total in Paper 2A (B. Arch) or Paper 2B (B. Planning) of Session 1 of JEE (Main) - 2023. b) Compilation of NTA Score for Session 2 of Paper 2A (B. Arch) or Paper 2B (B. Planning): Similarly, Session 2 of JEE (Main) - 2023 Paper 2A (B. Arch) or Paper 2B (B. Planning) will be conducted in multi-shifts, NTA scores will be calculated corresponding to the raw (actual) marks obtained by a candidate. The calculated NTA scores for all the shifts will comprise of the NTA scores for each of the three parts (Mathematics, Aptitude Test, and Drawing Test or Planning Based Test) and the total in Paper 2A (B. Arch) or Paper 2B (B. Planning) of Session 2 of JEE (Main) - 2023. c) Compilation and display of result including overall Merit List/ Ranking of Paper 2A (B. Arch) or Paper 2B (B. Planning): The NTA scores for each of the candidates in Total in Paper 2A (B. Arch) or Paper 2B (B. Planning) for Session 1 as well as for Session 2 of JEE (Main) - 2023 will be merged for compilation of results and preparation of overall Merit List/Ranking. The best of the two NTA Scores (Total) will be considered for further processing for those candidates who appeared in both sessions. In case of a tie, i.e. when two or more candidates obtain equal Total NTA Score in JEE(Main) - 2023, inter-se merit of such candidates shall be determined as per “Method of resolving ties” given in Chapter-3 for Paper-2A (B. Arch) or Paper-2B (B. Planning). The NTA scores and rank of Paper-2A (B. Arch) or Paper-2B (B. Planning) of JEE (Main) - 2023 for all candidates who appeared in Session 1 as well as in Session 2 for JEE (Main) - 2023 will be declared on the NTA website(s). This shall comprise of the NTA scores obtained by the candidate in Session 1, NTA scores obtained in Session 2, and NTA scores for those who appeared in both sessions (best of the two Total NTA scores). The rank shall comprise All India rank and All India category rank. A copy of the Final Scorecard of JEE (Main) – 2023 will be sent to the registered e-mail address of the Candidates.

Note: 1. The All India Rank shall be compiled and declared after conduct of Session 2 of JEE (Main) - 2023 Examination. 2. No Score/Rank Card will be dispatched to the candidates and the candidates are advised to download their Score/Rank Cards from the JEE (Main) website: www.nta.ac.in/, https://jeemain.nta.nic.in/ only. 3. There shall be no provision for re-valuation/re-checking of the Score. No correspondence in this regard shall be entertained. 14

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4. Only the All India Rank (AIR) is used for admissions through Central Seat Allocation Board (CSAB)/Joint Seat Allocation Authority (JoSAA) to NITs/ IIITs/ CFTIs/ SFIs/ Others. 5. All participating Institutions that do not use the Centralized Seat Allocation Process will prepare their own ranking based on the performance in JEE (Main) - 2023 and other criteria as decided by them. 6. The National Testing Agency (NTA) is an examination conducting body and not an admission giving authority. Therefore, NTA does not collect the information regarding the total number of seats available in the institutions, or the eligibility and reservation criteria followed by the institutions. 13.4. Re-Evaluation/Re-Checking of Result There shall be no re-evaluation/re-checking of the result. No correspondence in this regard shall be entertained. 13.5. Use of Scores of JEE (Main) – 2023 by other Organisations The merit list/scores/result of JEE (Main) – 2023 may be utilised by other Entities of Central/State Governments with their eligibility criteria/norms/applicable regulations/guidelines/rules. For the academic year 2023-24, admissions under MEA (Welfare) quota seats shall be based on the Joint Entrance Examination (Main).

14. Admission to NITs, IIITs, CFTIs, SFIs, State Engineering Colleges in the Participating States, and Other Participating Institutions. 14.1 Eligibility for Admission to NITs, IIITs, and CFTIs participating through Central Seat Allocation Board (CSAB): Admission to B.E. / B.Tech / B.Arch / B.Planning. Courses in NITs, IIITs, and CFTIs participating through Central Seat Allocation Board will be based on All India Rank subject to the condition that the candidate should have secured at least 75% marks in the Class 12 examination conducted by the respective Boards. For SC/ST candidates the qualifying marks would be 65% in the Class 12/qualifying examination. The candidate is also required to pass in each of the subjects of Class 12 / qualifying examination. The Admission to NITs, IIITs, and CFTIs will be based on the announced qualifying Percentile by the individual Institutes. For eligibility regarding details, candidates may refer to the website https://csab.nic.in/. The eligibility criteria decided by the Council of Architecture for admission to B.Arch. course other than NITs, IIITs, CFTIs: “No candidate shall be admitted to Architecture course unless she/he has passed an examination at the end of the qualifying examination with 50% marks in Physics, Chemistry, and Mathematics and also 50% marks in aggregate of the qualifying examination”. For B. Planning, the candidates should have passed the qualifying examination with 50 % marks in Mathematics and 50% marks in aggregate of the qualifying examination. Subject combinations required in the qualifying examination for admission to B.E./B.Tech, B. Arch, and B. Planning Courses in NITs, IIITs, and other CFTIs shall be as under. Course

Required Criteria based on Class 12/ Equivalent qualifying Examination

B.E./B.Tech.

Passed qualifying examination with Physics and Mathematics as compulsory subjects along with one of the Chemistry/Biotechnology/Biology/ Technical Vocational subjects.

B.Arch.

Passed qualifying examination with Mathematics, Physics, Chemistry

B.Planning

Passed qualifying examination with Mathematics

The Admission Policy, as announced by the Competent Authority of the admitting institutes shall be followed at the time of admission. The Candidates are advised to satisfy themselves about their Eligibility for Admission from the respective websites of admitting authorities. 14.2 Eligibility for Admission to Other Institutions The above-mentioned policy could also be adopted by other Technical Institutions participating in counseling through JoSAA/CSAB. In case a State opts to admit students to the engineering Colleges affiliated with State Universities, the State may prepare a separate rank list based on criteria decided by them. For all admission-related procedures/queries, the candidates are advised to refer to the website of JoSAA, Central Seat Allocation Board (CSAB), or the concerned State Government/Institute after the declaration of result/ranks of JEE (Main) - 2023. The letters/e-mails/grievances/queries/RTI applications/Court cases pertaining to admission related matters/ procedures will not be entertained by NTA. The same may be addressed to the concerned Counseling/Authorities/ Admitting Institutions. 14.3 Seat Allocation Process and Admission Procedure Candidates shall be offered admission based on their choices and All India Ranks of JEE (Main) - 2023 through a Seat Allocation Process to be announced later CSAB/JoSAA. The authentication/verification of relevant documents in support of identification, date of birth, qualifying examination, State of eligibility, category, and disability (if any) of the qualifying candidates would be done at the time of Seat Allocation/admission process. On failing to produce any of the authentic documents, the candidate may not be considered for admission. Gen-EWS, SC, ST, OBC, and PwD candidates will be required to produce the original certificate as per prescribed formats issued by the competent authority at the time of the Seat Allocation/admission process failing which they will not be considered for admission. 14.4 Eligibility Criteria for Indian Nationals Appearing in JEE (Advanced) - 2023 (Please refer to JEE (Advanced) website https://jeeadv.ac.in/ for the latest information) All the candidates must simultaneously fulfill each one of the following to appear for JEE (Advanced) - 2023. ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

15

Performance in JEE (Main) - 2023: Candidates should be among the top 2,50,000 successful candidates (including all categories) in B.E./B.Tech. Paper of JEE (Main) - 2023. Number of Attempts: A candidate can attempt JEE (Advanced) a maximum of two times in two consecutive years.

15. Test Practice Centres (TPCs) The Ministry of Education has mandated the NTA to set up, establish and create a network of Test Practice Centres for candidates, especially in remote and rural areas to enable them to practice and be comfortable in taking a Computer Based Test (CBT). This facility is completely free of cost. Candidates can register online (on the NTA website) where they are provided a convenient TPC near their location to practice on a given computer node. This facilitates the process of being able to take a Computer Based Test (CBT). The entire experience of using a computer is close to the actual experience of taking a CBT. All efforts are made to provide practice tests and questions so that candidates can familiarize themselves with logging into the system, go through the detailed instructions regarding the test, use the mouse or numeric keyboard on-screen (virtual) for attempting each question, scroll down to the next question, navigate between questions, review and edit their options and submit answer.

16. Common Services Centres/Facilitation Centres Candidates who are not well conversant and submitting face difficulties in the online application due to various constraints can use the services of Common Services Centre, Ministry of Electronics and Information Technology, Government of India under the Digital India initiatives of Hon’ble Prime Minister. The Common Services Centre (CSC) scheme is a part of the ambitious National e-Governance Plan (NeGP) of the Government of India and is managed at each village panchayat level by a Village Level Entrepreneur (VLE). There are more than 1.5 lakhs Common Services Centres (CSC) across the country which will provide the desired support to candidates from urban as well as rural areas in online submission of Application Form and payment of fee through e-wallet. The list of the Common Services Centre is available on the website: www.csc.gov.in. To know the nearest Common Services Centre, please open the link - Find My Csc (https://findmycsc.nic.in/csc/).

17. National Test Abhyas The National Testing Agency (NTA), India’s premier autonomous testing organization for conducting entrance examinations for admission in higher education institutions, launched a mobile app called the “National Test Abhyas” - to enable candidates to take mock tests for various competitive exams such as JEE Main, NEET, UGC-NET, and other exams under the NTA’s purview. The app had been launched to facilitate candidates’ access to high-quality mock tests in the safety and comfort of their homes since NTA’s Test-Practice Centers (TPCs) were closed during the ongoing COVID-19 pandemic. With this facility for the candidates, India has taken the lead in restoring a semblance of normalcy in one more crucial area – education – even as we tackle these unprecedented times that have led to significant changes in life around the world. Candidates across the country can use the App to access high-quality tests, free of cost, in a bid to be fully prepared for the upcoming JEE (Main), NEET (UG), and other competitive exams. The tests can be easily downloaded and taken off-line, thus accommodating candidates with low bandwidth internet connections. The app works on Android-based and iOS smartphones and tablets and can be downloaded from Google Play Store or App Store. Once candidates download the app, they simply need to sign-up or register with some basic details, create a free account, and then start accessing mock tests free of cost for their selected examination(s). In addition to that NTA has developed an extensive support system at http://nta.ac.in/abhyas/help. The National Test Abhyas is available on the website: https://www.nta.ac.in/Abhyas and Play Store: https://play.google.com/ store/apps/details?id=com.abhyas.nta.com

18. Miscellaneous 18.1 Query Redressal System (QRS) National Testing Agency (NTA) has established a Query Redressal System (QRS), an online webenabled system developed by NTA. QRS is the platform based on web technology that primarily aims to enable submission of queries/grievances by the Registered Candidate(s) of JEE (Main) - 2023 Examination with (24×7) facility for speedy and favorable redressal of the queries/grievances. A Unique Registration Number will be generated for tracking the status of the queries/grievances. The Registered Candidate(s) are advised to use the online facility for a speedy response. 18.2 Correspondence with NTA All the correspondence should preferably be addressed by e-mail. The email query shall be addressed only if it is not anonymous and contains the name, postal address, and contact number of the sender. An email containing vague or general queries and other queries as contained in the Information Bulletin shall not be entertained. Queries shall not be entertained from a person claiming to be representatives, associates, or officiates of the applicant candidate. The following information shall not be revealed by phone or email: a. Internal documentation/status. b. The internal decision-making process of NTA. Any claim/counterclaim thereof. c. Dates and venue of internal meetings or name of the staff/officers dealing with it. d. Any information which cannot be revealed in the opinion of NTA. 18.3 Weeding Out Rules The record of Joint Entrance Examination JEE (Main) - 2023 would be preserved up to 90 days from the date of declaration of result. 18.4 Legal Jurisdiction All disputes pertaining to the conduct of JEE (Main) – 2023 Examination including Results shall fall within the jurisdiction of Delhi only. The Director (Admin) of the NTA shall be the official by whose designation the NTA may sue or be sued. 16

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

MOCK TEST FOR JEE (MAIN) 2023 General Instructions: 1.

Duration of Test is 3 hrs.

2.

The Test booklet consists of 90 questions. The maximum marks are 300.

3.

There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each part has two sections. (i) Section-I : This section contains 20 multiple choice questions which have only one correct answer. Each question carries 4 marks for correct answer and –1 mark for wrong answer. (ii) Section-II : This section contains 10 questions. In Section II, attempt any five questions out of 10. The answer to each of the questions is a numerical value. Each question carries 4 marks for correct answer and –1 mark for wrong answer.

PART – A : PHYSICS

.

SECTION - I Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct. 1.

2.

A block A of mass m is placed on another rough plank of mass 4m which in turn is placed on smooth ground. Velocity v0 to be given to block A, so that it just falls off the plank is

(1)

3gl 2

(2)

5 gl 2

(3)

1 5gl 2

(4)

gl 2

A projectile is projected with velocity 20 m/s at an angle of 53° with horizontal. Radius of curvature of trajectory of projectile at the instant when its velocity is perpendicular to its initial velocity

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

3.

(1) 30 m

(2) 22.5 m

(3) 5 m

(4)

225 m 8

A sound source S is moving along a straight track with speed v, and is emitting sound of frequency v0 (see figure). An observer is standing at a finite distance, at the point O, from the track. The time variation of frequency heard by the observer is best represented by (t0 represents the instant when the distance between the source and observer is minimum)

(1)

(2)

17

MOCK TEST (3)

(1) B, E, F (4)

(2) C, G, H (3) B, E, G (4) C, E, G

4.

Velocity (v) of particle of mass (m) varies with time (t) as shown in graph. Power delivered by

7.

A particle is projected with 20 m/s at an angle of 37° with horizontal. At topmost point, it

t the forces acting on the particle at t = 0 is 2

breaks into three identical pieces such that all of them land on ground simultaneously. If x1, x2, x3 are x-coordinates of piece landing on ground, then (x1 + x2 + x3) is

(1)

mv 02 t0

mv 02 (3) 4t0

(2)

mv 02 2t0

(1) 115.20 m (2) 90 m

mv 02 (4) 8t0

(3) 65.80 m (4) 80 m

5.

Kinetic energy of a rod of mass M having velocity of its ends as shown, is

8.

For

a

plane

electromagnetic

wave,

the

magnetic field at a point x and time t is B( x, t ) = 1.2  10−7 sin(0.5  103 x + 1.5  1011t )kˆ  T.

The instantaneous electric field E corresponding to B is (speed of light c = 3 × 108 ms–1)

6.

(1)

13 Mv 02 6

(2)

8 2 Mv 0 5

(1) E ( x, t ) = 36 sin(1 103 x + 1.5  1011t )iˆ 

(3)

15 Mv 02 6

(4)

11 2 Mv 0 5

(2) E ( x, t ) = 36 sin(0.5  103 x + 1.5  1011t )kˆ 

Shape of a wave travelling along positive X-axis

V m

(3) E ( x, t ) =  −36 sin(0.5  103 x + 1.5  1011t ) jˆ  V

m

is shown in figure. Particles moving along negative Y-axis at given instant are 18

V m

(4) E ( x, t ) = 36 sin(1 103 x + 0.5  1011t ) jˆ 

V m

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

MOCK TEST 9.

In line of sight communication, distance

12. If intensity (I) versus wavelength () for three

between receiving and transmitting antenna is

identical black bodies at temperature T1, T2 and

80 km. If height of transmitting antenna is

T3 are shown in figure, then

50 m, then height of receiving antenna should be (1) 100 m (2) 180 m (3) 235 m (4) 140 m 10. The value of the acceleration due to gravity is g1 at a height h =

R (R = radius of the earth) 2

(1) T2 > T3 > T1

(2) T2 > T1 > T3

(3) T3 > T2 > T1

(4) T1 > T2 > T3

from the surface of the earth. It is again equal to g1 at a depth d below the surface of the earth.

13.

d  The ratio   equals R 5 9

In the figure shown, the current in the 10 V

(2)

1 3

(1) 0.36 A from negative to positive terminal

(3)

7 9

(4)

4 9

(1)

battery is close to

(2) 0.42 A from positive to negative terminal (3) 0.71 A from positive to negative terminal (4) 0.21 A from positive to negative terminal 14. Following plots show Magnetization (M) vs

11. In the shown P-V diagram of a cyclic process of an

ideal

gas.

If

U A→B = +400 kJ

and

Magnetising

field

(H)

and

Magnetic

susceptibility  vs Temperature (T) graph:

QB →C = −500 kJ , then heat given to gas in the

process (C → A) is

(a)

(b)

(c)

(d)

Which of the following combination will be represented by a diamagnetic material? (1) –25 kJ

(2) 25 kJ

(1) (a), (d)

(2) (b), (c)

(3) –20 kJ

(4) 20 kJ

(3) (b), (d)

(4) (a), (c)

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

19

MOCK TEST 15. Two large plates each of area A are at separation d and charges given are Q and 2Q as shown. If a dielectric slab of dielectric constant K is filled between slabs, then electric field between plates is

(1)

(3)

Q 2 A0K

(2)

2QK A0

(4)

Q A0K

(1) 8, 8

(2) 4, 8

(3) 8, 16

(4) 8, 4

19. In the arrangement shown in the figure, the distance of final image from the lens is

(1) 40 cm

(2) 50 cm

(3) 25 cm

(4) 70 cm

20. An unpolarized beam of intensity I0 is passed through a polariser. The intensity of emergent beam is

2Q 3 A0

16. A uniformly charged spherical shell of radius R and charge Q is rotating about its diameter with

(1) I0

(2)

I0 4

(4)

constant angular speed . The magnetic (3)

moment of shell is (1)

2QR 3

2

(3)

3QR 5

2

(2)

Q R 3

2

(4)

Q R 2

2

I0 2

I0 2

SECTION - II Numerical Value Type Questions: This section contains 10 questions. In Section II, attempt any five questions out of 10. The answer to each question is

17. The phasor diagram of LCR series circuit is

a NUMERICAL VALUE. For each question, enter the

shown in figure. Phase difference between AC

correct numerical value (in decimal notation,

voltage and AC current is

truncated/rounded‐off to the second decimal place; e.g. 06.25, 07.00, −00.33, −00.30, 30.27, −27.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. 21. A particle A moving with velocity 5 m/s strikes a stationary particle B placed on smooth surface.

(1) 30°

(2) 37°

(3) 45°

(4) 53°

18. In a radioactive decay sequence converted into

206 92

If both of them move with velocity 2 m/s in opposite direction, then coefficient of restitution 238 92

U gets

is

n . Value of n, is 10

X . The number of -particles

and -particles emitted are respectively 20

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

MOCK TEST 22. A cylindrical vessel is filled with liquid up to

26. In a mixture of H atoms and He+ ions, all are

h = 10 m. A small hole has to be punctured on

excited

side wall, so that range of effluxed liquid is

Subsequently H atoms transfer their total

maximum. Maximum value of range, is n m.

excitation energy to He+ ions. Find new

Value of n is

quantum state n of He+.

[g = 10 m/s2]

to

their

first

excited

states.

27. The light source S is over the center of a circular fixed opaque plate of radius r = 1 m at a distance a = 2 m from it. The distance from the plate to the screen is b = 1 m. Rate of increase of diameter of shadow formed on screen if source moves towards the screen with speed 1 m/s, at the given instant, will be k m/s. Find

23. Two point masses m, 2m separated by distance

value of k.

S

2d placed on smooth surface are released from rest. Velocity of mass 2m when separation between them is d is

a

Gm . Value of n, is nd

2r

b screen

24. In the shown part of a circuit current (I) in the shown figure is _____ ampere.

28. Two particles execute SHM with same angular frequency but different amplitudes A and

3 A

along same line with same mean position. If they cross each other in opposite direction at a

3 A from mean position, then phase 2 difference between the two SHM's, in degrees is ___ distance

29. In YDSE arrangement slit separation is d = 3, 25. The diameter of a spherical bob is measured using a vernier callipers. 9 divisions of the main

where  is wavelength of light incident. If distance y of first minima is

scale, in the vernier callipers, are equal to 10 divisions of vernier scale. One main scale

D k

, then find k.

(D>>d)

division is 1 mm. The main scale reading is 10 mm and 8th division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of 0.04 cm, then the radius of the bob is ___ × 10–2 cm. ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

21

MOCK TEST 30. A uniform chain of mass M and length L is held

(assume fallen portion does not form any heap) (M = 1 kg and g = 10 m/s2)

vertical so that its lower end just touches the floor. Now if chain is released from rest, then find force (in N) exerted by chain on the floor

 1 when   th length of chain has already fallen 4

PART – B : CHEMISTRY SECTION - I

34. Which of the following sugar is non-reducing?

Multiple Choice Questions: This section contains

(1) Maltose

(2) Glucose

20 multiple choice questions. Each question has

(3) Sucrose

(4) Lactose

4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.

35. Which among the following oxide gives two oxyacids in water?

31. Most reactive form of phosphorus is (1) White phosphorus (2) Red phosphorus

(1) N2O3

(2) NO

(3) NO2

(4) P4O10

36. pH of resultant solution formed by mixing equal

(3) Black phosphorus

volumes of two solutions having pH = 5.8 and

(4) All allotropes are equally reactive

pH = 12.8 is

32. When copper metal is refined by electrolytic method, incorrect statement is

(1) 5.5 (2) 12.5

(1) Anode is made of pure metal while cathode is made of impure metal

(3) 6.1 (4) 13.1

(2) Anode is made of impure metal while pure metal is cathode (3) During electrolysis impurities collected as anode mud

37.

.

(4) CuSO4 can be used as electrolyte Product formed is 33.

CH3 The product C is

(1)

(1) CH3

22

C

O

C6 H5

C6 H5

(2) (2) C6 H5

(3)

C

(4)

O

CH3

C

C

C6 H5

CH3 ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

MOCK TEST CH3 CH3

(3) C6 H5

C

C

C6 H5

(1) T2 > T1 > T3

(2) T1 > T2 > T3

(3) T3 > T2 > T1

(4) T1 = T2 = T3

OH OH

CH3 (4) C6 H5

C

C

O

CH3

42.

NH2 major product A is

38. A colloidal solution is prepared by following

H

(1)

reaction

D

D

(3)

(excess)

of this solution is

D

D

D

(4)

C

C D

C C Br

H

43. From the following select the correct expression of compressibility factor for H2 at higher pressure

(1) K4[Fe(CN)6]

(2) FeCl3

(3) MgSO4

(4) AlBr3

o Ereduction

H C

H

H C

Br

The most effective electrolyte in the coagulation

39. What is

(2)

C

D

AgNO3 + KI → AgI + KNO3

D

H C

behaving as van der Waals gas (1) 1 +

Pb RT

(2) 1 −

Pb RT

(3) 1 +

a RTVm

(4) 1 −

a RTVm

potential for the half cell

Pt | S2– | CuS| Cu , if E°Cu

2+

/Cu

is 0.34 V and Ksp of

CuS = 10–35? (1) 0.34 V

(2) –0.69 V

(3) 0.96 V

(4) –0.21 V

44. Following graph is not valid for (m = molar conductivity, m = molar conductivity at infinite dilution, c = concentration of electrolyte)

40. Which of the following metal sulphate will give SO2 and SO3 both gaseous products on heating? (1) CuSO4

(2) FeSO4

(3) Fe2(SO4)3

(4) CaSO4

41. Preparation

of

N2 ( g) + 3H2 ( g)

NH3(g)

by

2NH3 (g)

the

reaction

shows

the

following graph of % yield against pressure at different temperature. Then which of the following is correct relation for T1, T2, T3

(1) NaCl (aq)

(2) K2SO4 (aq)

(3) PhCOOH (aq)

(4) CaCl2 (aq)

% yield of NH3

45. The enzyme responsible for conversion of starch T1

into maltose is

T2

(1) Maltase

T3

(2) Amylase (3) Urease

P ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

(4) Pepsin 23

MOCK TEST 46.

An aqueous solution when treated with NH 4OH, it gives a white precipitate. On addition of excess NH4OH, the precipitate gets dissolved. Identify the precipitate which dissolves in NH4OH. (1) Zn(OH)2

(2) Mn(OH)2

(3) Mg(OH)2

(4) Ca(OH)2

50. The compounds B and C are majorly

(1)

47. The correct order of melting point of 14th group elements

(2)

(1) C > Si > Ge > Pb > Sn (2) C < Si < Ge < Sn < Pb (3) Sn > Pb > Ge > Si > C

(3)

(4) Si > C > Ge > Sn > Pb 48.

Select the option in which heat evolved is maximum. Given fHo(CO2, g) = – 75 kcal/mol fHo(CO, g) = – 25 kcal/mol The product will be CO if excess amount of carbon is present and CO2 if excess O2 is present [Consider full oxidation of carbon to decide limiting and excess reagent] (1) 10 moles of carbon and 11 moles of O2 (2) 24 gm of carbon and 65 gm of O2 (3) 4 moles of carbon and 3.5 moles of O2 (4) 32 gm of carbon and 80 gm of O2

49. Racemic mixture is formed by reaction of HCN/KCN with

(4) SECTION - II Numerical Value Type Questions: This section contains 10 questions. In Section II, attempt any five questions out of 10. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded‐off to the second decimal place; e.g. 06.25, 07.00, −00.33, −00.30, 30.27, −27.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. 51. An oxide of vanadium has magnetic moment of 2.83 BM. The oxidation state of vanadium in its given oxide would be

(1)

52. w g of gas (A) of molecular weight 100 and 0.2 g of gas (B) of molecular weight 80 when

(2)

enclosed in one litre container exert a total pressure of 750 mm of Hg. The partial pressure of A is 500 mm Hg. The value of w is_______.

(3)

53. Consider the reaction : Br /H O

2 2 Trans-2-butene ⎯⎯⎯⎯→ A (major product).

(4) Number of chiral carbon in the product A is 24

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

MOCK TEST 57. Consider the following reactions

54. Consider the following reactions: NaCl + K2Cr2O7 + H2SO4 → (A) + side products

(i) CH3MgBr Cu A ⎯⎯⎯⎯⎯→ B ⎯⎯⎯ ⎯ → 2-methyl-2-butene + 573 K (ii) H3O

(Conc.)

(A) + NaOH → (B) + Side products

The mass percentage of carbon in A is ______.

(B) + H2SO4 + H2O2 → (C) + Side products

(Round off answer to the second decimal place)

(dilute)

The sum of the total number of atoms in one molecule each of (A), (B) and (C) is _____. 55. The number of sp2 hybridised carbons present in “Aspartame” is ______. 56. Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve 10 ppm of iron in 100 kg of wheat is_____________ (Report answer to the nearest integer)

58. The number of electrons that are in resonance in histamine is _____. 59. A solution of phenol in chloroform when treated with aqueous NaOH gives compound P as a major product. The number of H-atoms that are more acidic than phenol in P is –11

3

60. If the solubility product of AB2 is 3.20 × 10 M , then the solubility of AB2 in pure water is –4

–1

_____ × 10 mol L . [Assuming that neither kind of ion undergo hydrolysis]

Atomic weight : Fe = 56; S = 32.00; O = 16.00

PART – C : MATHEMATICS SECTION - I Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.

If f ( x ) = x + b x + 3 x + 2 , then x +c x +4 x +3 (2) f(–50) = 501

(3) f(–50) = –1

(4) f(50) = –501 65.

= cos{tan–1x} is (2) 1

1

(

63. If I1 =  1– x 50 0

)

100

(3)

5050 5049

(4)

5049 5050

1 2 1

(

dx and I2 =  1– x 50 0

such that I2 = I1 then  equals ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

(1)

4 2  sq.units 3

3

(3)

1 2  sq.units 3

(2)

3 2  sq.units 4

(4)

5 2  sq.units 3

3

3

62. The value of x for which sin{cot–1(1 + x)}

(4) −

5050 5051

3

(1) f(50) = 1

(3) 0

(2)

y = x2 and y = sec −1  − cos2 x  , where [.] represent greatest integer function, is

x + a x + 2 x +1

1 2

5051 5050

64. The area of the region bounded by the curves

61. Let a – 2b + c = 1.

(1)

(1)

)

101

dx

2 n   1 lim  + + ...... +  is equal to 2 2 n→  1− n 1− n 1− n 2  (1)

−1 2

(2)

1 2

(3) –1 (4) Zero 25

MOCK TEST 66. A bag contains 10 balls, out of which 2 are white, 5 are red and 3 are black balls. If three are drawn at random, then the probability that two balls are of same colour is (1)

79 120

(2)

2 3

(3)

3 4

(4)

1 4

67. The foot of the perpendicular drawn from the point (4, 2, 3) to the line joining the points (1, –2, 3) and (1, 1, 0) lies on the plane (1) x – 2y + z = 1

(2) x + 2y – z = 1

(3) x – y – 2z = 1

(4) 2x + y – z = 1

68. The mean and variance of a binomial distribution are 20 and 4 respectively. The probability of exactly 2 successes is (1)

(3) 69. If

192 5

(2)

23

384 5

(4)

25

a, b and c

are

unit

a + 2b + 3c = 3 + 2 2 ,

384 5

between

such

between



such

that

+ 6 cos is equal to (1) 2 2 + 5

(2) 5 – 2 2

(3) 2 2 – 5

(4) 4 2

)

f ( x ) = min − 1 − ( x + 1) , 1 − ( x − 1) , | x | , then

the number of points where function f is non-

(2) 1 : 2

(3) 2 : 1

(4)

2 :1

73. Let y = y(x) be the solution of the differential dy   equation cos x + 2y sin x = sin2x, x   0,  . dx  2 If y(/3) = 0, then y(/4) is equal to 1 2

−1

(2)

2 −2

(4) 2 + 2 1 128

  ( 7Ci + 7C j )

0i  j 7

(1) 7

(2) 2

(3) 27

(4) 27 – 7

will be

75. The complex number z satisfying the condition  arg {(z –1) / (z + 1)} = lies on 4 (1) A straight line

(2) A parabola

(3) A circle

(4) Pair of straight lines

(1) 66

(2) 33

(3) 132

(4) 165

77. A sequence is obtained by deleting all perfect squares from the set of all natural numbers starting from 1, then (1) The remainder is zero when 2003rd term of the sequence is divided by 2048

differentiable, is

26

(1) 1 : 1

76. The number of integral solutions of the equation 3x + 3y + z = 30, where x  0, y  0 and z  0 is

70. f : [–2, 2] → R, 2

is equal to

(2)

that

  2    ,  , then the maximum value of 4 cos 2 3 

(

 4

74. The value

angle

2

4 − x2

(3) 2 − 2

vectors

b and c is

dx

 2  (3)  (4) 6 72. A closed cylinder has to be made with 100 m2 of plastic. If its volume is maximum, then the ratio of its radius to the height, is (1)

25

a and b is , angle between a and c is  and

angle

x+ 0

(1)

23

192 5

2

71. The value of

(1) Zero

(2) 1

(3) 2

(4) 3

(2) The remainder is 2017 when 2003rd term of the sequence is divided by 2048 ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

MOCK TEST (3) The remainder is 1 when 931st term is divided by 2 (4) The remainder is 2 when 931st term is divided

 −1 1 1    81. Consider a matrix A =  1 −1 1  . If |2 adj(3  1 1 −1 adj(4A–1))| = 2a. 3b, then the value of a – 2b is

by 2 78. If two pairs of tangents drawn from points A and B to the circle

x2

+

y2

– 4x + 6y – 3 = 0 form a

parallelogram then mid-point of AB lies on (2)

(3) 2x – y + 6 = 0

(4) x2 + y2 + 4y – 1 = 0

x2

+

79. A tangent drawn to the hyperbola

– 2x = 0

x2 a2

−

y2 b2

18, 24, 22, 20 is equal to

= 1 at

 3 + 1  84. If A =  3 +    − 6 − 2 

85. The number of non-negative integers n satisfying

the sum of the values of  is

n2 = p + q and n3 = p2 + q2, where p and q are integers is

(2) 5

9 2

(4)

86. If the slope of tangent of the curve y = |x – 1| + |x

17

– 2| + |x – 3| at the point where x = 4, is a, then the value of a2 is

80. The length of common chord of the ellipse

( x − 1)2 ( y − 2 )2 = 1 and the circle (x – 1)2 + + 9 4 (y –

87. The number of solutions of the equation

1 1 1 + = { x } + , is equal to [ x ] [2 x ] 3

= 1, is

(1) Zero (3)

(2) 1 unit

1 unit 4

(4)

(where [] and {} represent greatest integer and fractional part function)

4 unit 9

88. The set A = {(x, y) : x8 + y8 + 6 = 8xy, x, y  R}

SECTION - II

then number of possible subsets of A is

Numerical Value Type Questions: This section

89. If 2sin2x + 3sinx – 2 > 0 and x2 – x – 2 < 0 then

contains 10 questions. In Section II, attempt any five

the exhaustive set of values of x satisfying the two

questions out of 10. The answer to each question is a

  inequalities is  , m then k + m = k 

NUMERICAL VALUE. For each question, enter the correct

6 − 2

square units with coordinate axes. Its

(1) 4

2)2

+6

  2 + 3  , there are 2 − 6  3

two values of  exist for which matrix A is singular,

eccentricity is equal to

(3)



 , forms a triangle 6

a point P with eccentric angle of area

y2

7e2x – 13ex = 6 is 83. Variance of the following data 6, 8, 10, 16, 14, 12,

(1) x + 2y – 3 = 0

3a2

82. Number of real roots of the equation e4x + e3x –

numerical

value

(in

decimal

notation,

truncated/rounded‐off to the second decimal place;

90. If the sum of the series up to infinity

e.g. 06.25, 07.00, −00.33, −00.30, 30.27, −27.30) using the mouse and the on-screen virtual numeric keypad

5 11 17 23 + 2 2 + 2 2 + 2 2 + .......... 2 1 .4 4 .7 7 .10 10 .13

in the place designated to enter the answer.

then 48S is equal to

2

is S

   ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

27

MOCK TEST

MOCK TEST FOR JEE (MAIN)-2023 ANSWERS PHYSICS

28

CHEMISTRY

MATHEMATICS

1.

(2)

31.

(1)

61.

(1)

2.

(4)

32.

(1)

62.

(4)

3.

(1)

33.

(2)

63.

(2)

4.

(2)

34.

(3)

64.

(1)

5.

(1)

35.

(3)

65.

(1)

6.

(2)

36.

(2)

66.

(1)

7.

(1)

37.

(1)

67.

(4)

8.

(3)

38.

(1)

68.

(1)

9.

(3)

39.

(2)

69.

(3)

10.

(1)

40.

(2)

70.

(3)

11.

(1)

41.

(3)

71.

(1)

12.

(4)

42.

(2)

72.

(2)

13.

(4)

43.

(1)

73.

(2)

14.

(4)

44.

(3)

74.

(1)

15.

(1)

45.

(2)

75.

(3)

16.

(2)

46.

(1)

76.

(1)

17.

(1)

47.

(1)

77.

(1)

18.

(3)

48.

(1)

78.

(4)

19.

(1)

49.

(1)

79.

(4)

20.

(2)

50.

(4)

80.

(1)

21.

(08.00)

51.

(03.00)

81.

(11.00)

22.

(10.00)

52.

(00.50)

82.

(01.00)

23.

(03.00)

53.

(02.00)

83.

(33.00)

24.

(09.00)

54.

(18.00)

84.

(00.00)

25.

(52.00)

55.

(09.00)

85.

(03.00)

26.

(04.00)

56.

(05.00)

86.

(09.00)

27.

(00.50)

57.

(66.67)

87.

(03.00)

28.

(90.00)

58.

(06.00)

88.

(04.00)

29.

(35.00)

59.

(01.00)

89.

(08.00)

30.

(07.50)

60.

(02.00)

90.

(16.00)

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

MOCK TEST

ANSWERS AND HINTS PART – A : PHYSICS

P=

1. Answer (2)

arel =

5g 4

v 02 = 2 

5.

mv 02 dK mv 02 = 2 t= dt 2t0 t0

Answer (1) Vcm = 2v0

5g 5g l = l 4 2

2. Answer (4)

=

2v 0 L

K=

1 1 ML2 4v 02 M 4v 02 +  2 2 2 12 L

v cos 37° = 20 cos 53°

= 2Mv 02 + Mv 02 

v = 15 m/s

R=

(15)2 225 = m g cos37 8.0

=

6.

13 Mv 02 6

Answer (2) Vp = −

V y x

At C, G, H

3. Answer (1)

1 6

y 0 x

 Vp < 0 7.

Answer (1)

x= While approaching

x1 + x2 + x3 = 3 x

  c  = 0    c – v cos  

While receding   c  = 0    c + v cos  

4. Answer (2)

P=

dK dt

v v= 0t t0 K=

1 2

mv 02 2 t t02

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

202 sin74 10

8.

=

3  202 3 4  2  10 9 5

=

72  16 = 115.2 10

Answer (3) B = 1.2 × 10–7sin(0.5 × 103x + 1.5 × 1011t) kˆT .

Wave is travelling along –x-axis and B is along +z-axis.

E0 = cB0 = 36

V m

E must be along –y-axis 29

MOCK TEST 9. Answer (3)

11 Eeq 28

Eeq =

d = 2RhT + 2Rhr 80000 = 2  6400  103  50 + 2  64  105  hR

hR = 235 m 10. Answer (1) R GM 4g  g h =  = = 2   3R 2 9   2   

For diamagnetic material  is independent of temperature

For cyclic process U = 0

and magnetisation (M) is directly proportional to

Q=W

 QCA = −25 kJ

30 A 138

14. Answer (4)

11. Answer (1)

UCA = 100 J

150 V 11

150 11 = 150 A I= 28 138 + 10 11

I=

5R 9

WCA = −125 J

20 10 + 7 4

1500 = 10 + 4 I 138

d  4g  g 1 −  =  R 9 d=

=

…(1)

H(M = –CH) 15. Answer (1)

E=

E0 K

q=

Q in capacitor 2

12. Answer (4)

1 mT = b  T    3   2  1

 E=

 T1  T2  T3

13. Answer (4)

16. Answer (2)

=

ql 2m

=

Q  (I ) 2m

=

Q  2    mR 2  2m  3 

=

Q R 2 3

1 1 1 28 = +  Req =  Req 4 7 11 Circuit can be reduced to 30

 Q = K 0 X 2 A0 XK

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

MOCK TEST 17. Answer (1)

tan  =

tan  =

23. Answer (03.00)

(VL − VC ) VR

(3 − 2) 3

=

1 2m 2  Vr = | U | 2 3

m 2 Gm2 Vr = 3 d

1 3

  = 30° 18. Answer (3) In -decay mass number decreases by 4. No. of -particles =

238 − 206 =8 4

No. of -particles = 8  2 = 16 19. Answer (1)

1 1 1 − = v u f

 u1 = (50 − 10) = 40 cm

40  20  v2 = = 40 cm (40 − 20)

24. Answer (09.00)

I=

V R



−3 − V 69 − V 0 – V + + =0 1 2 0.5

 V = 9 volt  I=9A 25. Answer (52.00)

Diameter = Reading – Zero error

I = I0 cos2 

Intensity of unpolarised light becomes half when passed through a polaroid.

I0 2

= 10.4 mm Radius = 5.2 mm 26. Answer (04.00)

 1 1 E = E0 Z 2  2 − 2   n1 n2 

21. Answer (08.00)

Vel. of separation Vel. of approach

Eloss = 10.2 eV by H atom

1 1   10.2 = 13.6  4  − 2  4 n 

4 e = = 0.8 5

 n=4

22. Answer (10.00) y=5m Range =

Vr 3

Reading = 10 + 8 × 0.1 = 10.8 mm

20. Answer (2)

e=

V2m =

Let V is the potential of junction

30  20 v1 = = 60 cm (30 − 20)

I=

3Gm d

Vr =

2gy 

27. Answer (00.50) 2y = 10 m g

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

D x+b = 2r x 31

MOCK TEST 2r ( x + b) x



−2br dx dD = dt x 2 dt



D=

sin  =

1 6

y = D tan  = D 

1 35

30. Answer (07.50)

1 = m/s 2

 Mg  2Mg  3Mg  F = + 4 = 4   4   

x

F=

r

Mg 2Mg 3Mg + = 4 4 4

PART – B : CHEMISTRY D

31. Answer (1)

28. Answer (90.00) Draw the corresponding phase diagram. Asin1 =

3

White phosphorus is most reactive form of phosphorus because of ring strain. 32. Answer (1)

A 2

During electrolytic refining anode is impure metal and pure metal is taken as cathode. 33. Answer (2)

60°

30°

3 A sin(  − 2 ) =

34. Answer (3)

 ( – 2) = 6

1 = 60°, 2 = 150°

NO2 will give HNO2 and HNO3.

29. Answer (35.00)

 x = for minima 2  x = d sin  = 2

32

3 sin  =

Sucrose is non-reducing. Rest all other given carbohydrates are reducing 35. Answer (3)

2 – 1 = 90°



3 A 2

36. Answer (2) pH = 12.8  pOH = 1.2 [OH–] = 10–1.2

 2

pH = 5.8  [H+] = 10–5.8 ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

MOCK TEST 10–5.8 CH3 This

38. Answer (1)

graph C

39. Answer (2)

S

= 0.34 +

0.059 log K sp( CuS ) 2

for

strong

electrolyte.

O

AgI/Ag+ is positively charged colloidal solution

E 2− | CuS | Cu = E°Cu2+ | Cu +

is

OH is weak electrolyte and hence

graph is not valid for PhCOOH. 45. Answer (2)

0.059 log 10−35 2

Amylase converts the starch into maltose

= –0.69 V 46. Answer (1)

40. Answer (2)  2FeSO4 ⎯⎯ → Fe2O3 + SO2 + SO3

ppt of Mn(OH)2, Mg(OH)2, are insoluble in excess NH4OH. Zn forms [Zn(NH3)4]2+ in presence of ammonium salts.

41. Answer (3)

N2 + 3H2

2NH3 ,

H ⎯⎯ → negative

Low temperature favours forward path and

Ca(OH)2 does not give ppts. 47. Answer (1)

hence M.P. of group 14 elements decreases down the group and the M.P. of Sn is less than Pb.

T1 < T2 < T3 42. Answer (2)

Br H

D D

H Br

D NaI Acetone

C H

H C D

(Anti-periplanar elimination)

43. Answer (1) For H2 gas 'a' is negligible

48. Answer (1) C

+ O2

CO2

10

11

–

0

1

10

Heat = –10 × 75 = –750 kcal(max) 49. Answer (1)

 P(Vm – b) = RT  PVm – Pb = RT Pb  Z = 1+ RT ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

33

MOCK TEST 50. Answer (4)

55. Answer (09.00)

Number of sp2 hybridized carbon are ‘9’.

51. Answer (03.00)

n ( n + 2 ) = 2.83

56. Answer (05.00)

n=2

Mass of iron needed in 100 kg wheat =

V+3 → [Ar] 4s0 3d2

10 106

 105

= 1.0 gm Molecular mass of FeSO4.7H2O is 278 52. Answer (00.50)

55.85 gm iron is present in 278 gm of salt 1 gm iron is present in

P(t) = 750 mm of Hg

278.85  5 gm of salt 56

57. Answer (66.67)

w     100 P( A ) =   750 = 500 w 1   +  100 400   w = 0.5 53. Answer (02.00) Me H

CH3 C

CH

Br

CH

OH

Br 2/H2 O

C H

CH3 Me

A is C4H8O

Number of optical centre = 2 % of C =

54. Answer (18.00) K Cr O conc.H2SO4

NaOH

2 2 7 → CrO Cl ⎯⎯⎯⎯ NaCl ⎯⎯⎯⎯⎯⎯ → Na2CrO4 2 2

dil H SO H2O2

(A)

(B)

2 4 → CrO Na2CrO4 ⎯⎯⎯⎯⎯ 5

(B)

= 66.67 58. Answer (06.00)

(C)

Total number of atoms in A, B and C are 18. 34

48  100 72

Histamine : ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

MOCK TEST 59. Answer (01.00)

Here,  =

I2 I1

Now, 1

I2 =  (1 − x 50 )101dx = 0

60. Answer (02.00)

I2 = −

A 2+ (aq) + 2B (aq)

AB2

s

1

0 (1 − x

50 100

)

1

0 (1 − x

50

)(1 − x 50 )100 dx

1

−  x x 49 (1 − x 50 )100 dx 0 I

II

2s

1

3

Ksp = 4s = 3.2 × 10

50 101 1 (1 − x )  x  I2 = I1 +  (1 − x )101  −  dx  5050  0 0 5050

–11

–12

3

 s = 8 × 10

I2 = I1 + 0 −

–4

s = 2 × 10

5051 I2 = I1 5050

PART – C : MATHEMATICS 61. Answer (1)

x +a

x+2

x +1

f (x) = x + b

x +3

x+2

x +c

x+4

x +3



I2 5050 = = I1 5051

64. Answer (1)  0 < cos2x < 1

R1 → R1 + R3 – 2R2

Operate

I2 5050

and

a – 2b + c = 1

 0 > – cos2x > –1  [–cos2x] = –1 or 0

using

But sec–1(0) is not defined  x  (2n + 1)

1

 y = sec–1(–1) = 

0

0

f (x) = x + b

x +3

x+2

x +c

x+4

x +3

Solving y = x2 and y =  we have  = x2, x = ± 

 f(x) = 1  f(50) = 1 62. Answer (4)

  sin(cot −1(1 + x )) = sin  + tan−1 x  2 

  − tan−1(1 + x ) = + tan−1 x 2 2 

 Required area

−1 x= 2



I1 =  (1 − x ) 0

)

0

63. Answer (2) 1

(

= 2   − x 2 dx

50 100

dx, I2 =

1

0 (1 − x

50 101

)

dx

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

 x3  = 2  x −  3   0



35

MOCK TEST     = 2    −  − ( 0 − 0 )   3   

2

3

 4  1 p(x = 2) = 25C2     5 5

4 =  2 sq.units 3

n(n + 1)

n →

2(1 − n ) 2

=−

a + 2b + 3c =



1 2

Required probability 3

192 = 23 5

C2  7C1 + 2C2  8C1 + 5C2  5C1 10

C3

=

79 120

3+2 2



4 cos  + 6 cos  = − 11 + 2 2 − 12 cos 



  2    ,  , then maximum value of 2 3 

66. Answer (1)

=

23

69. Answer (3)

65. Answer (1) lim

4 , n = 25 5

 p=

4 cos  + 6 cos  = 2 2 – 5

70. Answer (3)

67. Answer (4) Equation of line through points (1, –2, 3) and (1, 1, 0) is

Number of points of non-differentiability = 2 x −1 y −1 z − 0 = = ( =  say ) 0 −3 3−0

A point on above line M = (1, – + 1, ) D.Rs of PM = PM ⊥ AB  (–3).0 + (–1 – ) (–1) + ( – 3).1 = 0  =1

71. Answer (1) Put x = 2sin dx = 2cos d  /2

I=

 0

→

cos  d  sin  + cos 

 − 2

 Foot of perpendicular = (1, 0, 1) This point lies on plane 2x + y – z = 1

 /2

 I=

 0

68. Answer (1) np(1 – p) = 4 and np = 20 36

 2I =

sin  d  sin  + cos 

   I= 2 4

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

MOCK TEST 72. Answer (2)

75. Answer (3)

Let x be the radius and y be the height.  2(x2 + xy) = 100

 z – 1  arg  =  z + 1 4 Which represents an arc of circle.



76. Answer (1)

50 −x =y x

3x + 3y + z = 30

 50 x  Volume, V = x2y =   − x3    

x + y = 30 –

z 3

So z can take 0, 3, 6, …30

dV 2 =0  x =5 dx 3

So total number of integral solution = 11 + 10 + 9 + …+ 1 = 66 77. Answer (1)

2 and y = 10 3 

Natural numbers from 1 to 2003 We have perfect squares number (1, 4, 9, ..., 1936(44))

x 1 = y 2

So add 44 to 2003 73. Answer (2) Now from 1 to 2047 we have 2025 which is dy + 2y tan x = 2sin x dx

square of 45 So now sequence is 1 to 2048 where 2048 is 2003rd term

2 tan x dx I.F. = e  = sec 2 x

 y .sec 2 x =  2sin x.sec 2 xdx + c

And for getting 931st term 1 to 931 we have (1, 4, 9, ... 900 (30))

 y .sec 2 x = 2sec x + c

So 1 to 961 we have 961 which is square of 31 So 1 to 962 where 962 is 931st term

 When x = , y = 0; then c = –4 3

78. Answer (4) 2

 y sec x = 2sec x − 4  y =

2sec x − 4 sec 2 x

74. Answer (1) By expanding we get

1 (C0 + C1) + (C0 + C2) + (C0 + C3) 128 ...(C6 + C7) 

1 .7.27 = 7 128

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

Centre (2, –3) is mid-point of AB. 79. Answer (4)     Point P   is  a sec , b tan  6 6 6    

 2a b  i.e., P  ,  3  3

Equation of tangent at P is

x 3a 2

−

y 3b

=1

37

MOCK TEST 1 3a   3b = 3a2 2 2

 Area of triangle =  

b =4 a e2 = 1 +

b

2

a2



x=



2 =

6 + 24 = 15 2

( 92 + 72 + 52 + 32 + 12 ) 2 = 33 10

84. Answer (00.00)

= 17

 3 + 1  A =  3 +    − 6 − 2 

80. Answer (1)



  2 + 3  2 − 6  3

+6 6 − 2

R3 → R3 + 2R1  3 + 1  A =  3 +    

Length of common chord = 0 81. Answer (11.00) |A| = 4 |2 adj(3 adj(4A –1))| = 23 |adj(3 adj(4A–1))| = 23 |3 adj(4A–1)|2 = 23(33|adj 4A–1|)2 = 23 . (33)2 |adj 4A–1|2

Let ex = t

(t + 

–

1)2

6

3  0 2 

85. Answer (03.00) n 2 = p + q and n 3 = p 2 + q 2

4

)

2

 2 p 2 − 2n 2 p + n 4 − n 3 = 0 p =

2n 2  8 x 3 − 4 x 4 4

n 2  n 2n − n 2 2

 Possible value of p be 0,1 and 2 – 13t – 6 = 0

Hence the possible value of n is 0,1,2

(t + 2) (t – 3) = 0

86. Answer (09.00)

ex = –1, –2, 3

87. Answer (03.00) As, x > 0

83. Answer (33.00) Variance,  38

0

|A| = 2(2 – 9)

=

82. Answer (01.00)

+



 n3 = p2 + n 2 − p

 1 = 23 . (33)2 . (43)4   4 = 23 + 24 – 8 . 36 = 219 . 36

7t2

6

 3 + 1  A =  −1   

= 23 . (33)2(43)4 |A–1|4

t3

+6

(

= 23(33)2(43|A–1|)4

  2 + 3  2  3

R2 → R2 – R1 – R3

= 23(33)2(|4A–1|2)2 = 23(33)2|4A–1|4

t4



2

 ( xi =

− x)

N

2

( RHS is positive)

Let x = n + f where n = [x] and f = {x} Now equation reduces to ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

MOCK TEST n=5f1, +ve (2) +ve, >1, –ve (3) –ve, 1, –ve 6. Hydrogen bomb is based on the principle of (1) nuclear fission (2) natural radioactivity (3) nuclear fusion (4) artificial radioactivity 7. The oxidation state of Cr in [Cr(NH3)4Cl2 ] + is (1) +3 (2) +2 (3) +1 (4) 0 8. If D is the degree of dissociation of Na2SO4, the vant Hoff ’s factor (i) used for calculating the molecular mass is (1) 1 + D (2) 1 – D (3) 1+ 2D (4) 1 – 2D 9. Which one of the following species is diamagnetic in nature? (1) He+2 (2) H2 + (3) H 2 (4) H2– 10. Which of the following oxides is amphoteric in character? (1) CaO (2) CO2 (3) SiO2 (4) SnO2 11. The correct acidic order of the following is

1.

112

OH I.

OH

OH II.

III. NO2

CH3 (1) I > II > III (3) II > III > I

(2) III > I > II (4) I > III > II

12. CH3 – CH2 – CH – CH3 obtained by chlorination of | Cl n-butane, will (1) meso form (2) racemic mixture (3) d-form (4) l-form 13. Which alkene on ozonolysis gives CH 3CH2CHO and CH3 C CH3 || O CH 3 (1) CH3CH2CH = C CH 3 (2) CH3CH2 CH = CHCH2 CH3 (3) CH3CH2CH = CHCH3 (4) CH3 — C=CHCH3 . | CH3 14. An organic compound A(C 4H9Cl) on reaction with Na/diethyl ether gives a hydrocarbon which on monochlorination gives only one chloro derivative then, A is (1) t-butyl chloride (2) secondary butyl chloride (3) iso butyl chloride (4) n-butyl chloride. 15. Which of the following is incorrect? (1) FeCl3 is used in detection of phenol. (2) Fehling solution is used in detection of glucose. (3) Tollens reagent is used in detection of unsaturation. (4) NaHSO3 is used in detection of carbonyl compound. 16. Which of the following is not correctly matched? (1) neoprene :

LM – CH – C| =CH – CH MN 2

2

(2) nylon-66 :

LM MN– NH – (CH ) –NH – CO – (CH ) O LM || terylene : – OCH – CH MN 2 6

(3)

O ||

24

2

O ||

2

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

CH LM | PMMA : –M CH – C – MM | COOCH N 3

(4)

2

n

17. The incorrect IUPAC name is (1) CH3 – C – CH – CH3 ; 2 methyl-3-butanone || | O CH3 (2) CH3 – CH – CH – CH3 ; 2, 3-dimethyl pentane | | CH3 CH2CH3 (3) CH3 – C { CCH(CH3)2 ; 4-methyl-2-pentyne (4) CH3 – CH – CH – CH3 ; 3-bromo-2-chloro butane | | Cl Br 18. In preparation of alkene from alcohol using Al2O3 which is the effective factor (1) porosity of Al2O3 (2) temperature (3) concentration (4) surface area of Al2O3 19. Which of following is correct? (1) on reduction, any aldehyde gives secondary alcohol. (2) reaction of vegetable oil with H2SO4 gives glycerine. (3) alcoholic iodine with NaOH gives iodoform. (4) sucrose on reaction with NaCl gives invert sugar. 20. In steam distillation of toluene, the pressure of toluene in vapour is (1) equal to pressure of barometer (2) less than pressure of barometer (3) equal to vapour pressure of toluene in simple distillation (4) more than vapour pressure of toluene in simple distillation 21. A compound of molecular formula C7H16 shows optical isomerism, compound will be (1) 2, 3-dimethyl pentane (2) 2, 2-dimethyl butane (3) 2-methyl hexane (4) None of these 22. Change in enthalpy for reaction, 2H2O2 (l) o 2H2O (l) + O2 (g) if heat of formation of H2O2 (l) and H2O (l) are –188 and –286 kJ/mol respectively, is : (1) –196 kJ/mol (2) +196 kJ/mol (3) +948 kJ/mol (4) –948 kJ/mole 23. When 1 mol of gas is heated at constant volume, temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct (1) q = w = 500 J, 'V = 0 (2) q = 'V = 500 J, w = 0 (3) q = w = 500 J, 'V = 0 (4) 'V = 0, q = w = – 500 J 1 24. Enthalpy of CH4 + O2 o CH3OH is negative. If 2 enthalpy of combustion of CH4 and CH3OH are x and y respectively. Then which relation is correct? (1) x > y (2) x < y (3) x = y (4) x t y 25. A human body required 0.01 m activity of radioactive substance after 24 hours. Half life of radioactive substance ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

is 6 hours. Then injection of maximum activity of radioactive substance that can be injected (1) 0.08 (2) 0.04 (3) 0.16 (4) 0.32 Molarity of liquid HCl, if density of solution is 1.17 mg/cc is : (1) 36.5 (2) 18.25 (3) 32.05 (4) 42.10 Percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (at. wt.= 78.4) then minimum molecular weight of peroxidase anhydrous enzymes is (1) 1.568 × 104 (2) 1.568 × 103 (3) 15.68 (4) 2.136 × 104 Specific volume of cylindrical virus particle is 6.02 × 102 cc/gm whose radius and length are 7 Å and 10 Å respectively. If NA = 6.02 × 1023, find molecular weight of virus. (1) 15.4 kg/mol (2) 1.54 × 104 kg/mol 4 (3) 3.08 × 10 kg/mol (4) 3.08 × 103 kg/mol Standard electrode potentials are Fe2+/Fe; Eº = –0.44 Fe3+/Fe2+; Eº = 0.77 Fe2+, Fe3+ and Fe blocks are kept together, then (1) Fe3+ increases (2) Fe3+ decreases (3) Fe2+/Fe3+ remains unchanged (4) Fe2+ decreases 'G298 < 0 PbO2 o PbO SnO2 o SnO 'G298 > 0 Most probable oxidation state of Pb and Sn will be (1) Pb4+, Sn4+ (2) Pb4+, Sn2+ (3) Pb2+, Sn2+ (4) Pb2+, Sn4+ Which of the following two are isostructural? – (1) XeF2, IF2 (2) NH3 , BF3 –2 (3) CO3 , SO3–2 (4) PCl5, ICl5 Correct order of Ist ionisation potential among following elements Be, B, C, N, O is (1) B < Be < C < O < N (2) B < Be < C < N < O (3) Be < B < C < N < O (4) Be < B < C < O < N Which statement is incorrect? (1) Ni(CO)4 – tetrahedral, paramagnetic (2) Ni(CN)4–2 – square planar, diamagnetic (3) Ni(CO)4 – tetrahedral, diamagnetic (4) [Ni(Cl)4] –2 – tetrahedral, paramagnetic Main axis of a diatomic molecule is z, molecular orbital px and py overlap to form which of the following orbitals. (1) S molecular orbital (2) V molecular orbital (3) G molecular orbital (4) no bond will form Which of the following will exhibit maximum ionic conductivity? (1) K 4 [Fe(CN)6 ] (2) [Co(NH 3 ) 6 ]Cl 3 (3) [Cu(NH 3 ) 4 ]Cl 2 (4) [Ni(CO) 4 ] The following quantum numbers are possible for how many orbitals : n = 3, l = 2, m = ± 2. 113

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

114

(1) 1 (2) 2 (3) 3 (4) 4 – – In HS , I , R — NH2, NH3 order of proton accepting tendency will be (1) I – > NH3 > R — NH2 > HS– (2) NH3 > R — NH2 > HS– > I– (3) R — NH2 > NH3 > HS– > I– (4) HS– > R — NH2 > NH3 > I– The beans are cooked earlier in pressure cooker because (1) boiling point increases with increasing pressure (2) boiling point decreases with increasing pressure (3) extra pressure of pressure cooker softens the beans (4) internal energy is not lost while cooking in pressure cooker Nitrogen forms N2 , but phosphorus does not form P2, however, it converts P4 , reason is (1) triple bond present between phosphorus atom (2) pS – pS bonding is weak (3) pS – pS bonding is strong (4) multiple bonds form easily Ionisation constant of CH3COOH is 1.7 × 10–5 and concentration of H+ ions is 3.4 × 10 – 4. Then find out initial concentration of CH3COOH molecules (1) 3.4 × 10 – 4 (2) 3.4 × 10 – 3 –4 (3) 6.8 × 10 (4) 6.8 × 10 – 3 Solubility of M2S salt is 3.5 × 10 – 6 then find out solubility product : (1) 1.7 × 10 –6 (2) 1.7 × 10 –16 – 18 (3) 1.7 × 10 (4) 1.7 × 10 – 12 b If a a X species emits first a positron, then two D and two E particles and in the last, one D, is also emitted and gets converts to dc Y species. So correct relation is (1) c = b – 12, d = a – 5 (2) c = a – 5, d = d – 1 (3) c = a – 6, d = b – 0 (4) c = a – 4, d = b – 2 K p for equilibrium FeO(s) + CO(g) Fe(s) + CO2(g) at 1000ºC is 0.0403. If CO(g) at pressure of 1.00 atm and excess of FeO(s) are placed in a container at 1000ºC, the pressure of CO(g) at equilibrium would be (1) 0.713 atm (2) 0.813 atm (3) 1.713 atm (4) 0.513 atm Consider the following alcohols (i) CH3 —CH2CH2CH2OH (ii) CH3— CH2—CHOH CH3 (iii) CH3—C—OH—CH3 | CH3 The order of increasing b.p. would be (1) (i), (ii) and (iii) (2) (iii), (ii) and (i) (3) (ii), (iii), and (i) (4) None of the three Ethyl alcohol and methyl alcohol can be distinguished by (1) Lucas test (2) Victor Mayer’s test (3) iodoform test (4) oxidation test When glycerol is heated with excess of HI the final product formed is (1) 2-iodopropane (2) propylene (3) allyl iodide (4) glycerol tri-iodide

47. The equilibrium constant at a certain temperature for the reactions H2 + 12 S2 o H2S and H2+Br2 o 2HBr are K1 and K2 respectively. The value of K for the reaction Br2 + H2S o 2HBr + 12 S2 would be (2) K2 /K 1 (1) K1 /K2 (3) K1 × K2 (4) K1 – K2 48. One mole of O2 combines with 2 moles of SO2 to produce 2 moles of SO3 in a container of volume V. At equilibrium it was observed that 2 moles of SO3 are produced. The value of K for the equilibrium would be (1) x2 V/(1 – x)3 (2) 4x2/(2– x)2(1–x) 3 2 (3) (1 – x) /x V (4) x/(1+ x)V 49. Which one of the following is dibasic acid? (1) metaphosphoric acid (2) orthophosphoric acid (3) pyrophosphoric acid (4) phosphorus acid 50. Which one of the following mixed sulphates is not an alum? (1) K2SO4.Al2(SO4)3.24H2O (2) K2SO4.Cr(SO4)3.24H2O (3) Na2SO4.Al(SO4)3.24H2O (4) CuSO4.Al(SO4)3.24H2O 51. Partial pressure of O2 in the reaction 2Ag2O(s) 4Ag (s) + O2 (g) is (1) KP (2) K P (4) 2KP (3) 3 K P 52. Strength of acidity is in order

53.

54.

55.

56.

57.

OH

OH

(I)

CH3 (II)

OH

NO2 (III)

NO2

OH

NO2 (IV)

(1) II > I > III > IV (2) III > IV > I > II (3) I > IV > III > II (4) IV > III > I > II Metallic lusture is explained by (1) diffusion of metal ions (2) oscillation of loose electrons (3) excitation of free protons (4) existence of b.c.c. lattice Which one is true peroxide? (1) NO2 (2) MnO2 (3) BaO2 (4) SO2 XeF4 on partial hydrolysis produces (1) XeF2 (2) XeOF2 (3) XeOF4 (4) XeO3 Among alkali metal salts, the lithium salts are the poorest conductors of electricity in aqueous solution because of (1) easy diffusion of Li+ ions (2) lower ability of Li+ ions to polarise water molecules (3) lowest charge to radius ratio (4) higher degree of hydration of Li+ ions. In the estimation of an oxalate with KMnO4 solution which of the following is used as indicator? ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

(1) methyl orange (3) starch

(2) phenolphthalein (4) None of these OH |

58.

C6H5 2 6 5 6 5 The above reaction is known as (1) Beckmann rearrangement (2) Benzilic acid rearrangement (3) benzoin condensation (4) addol condensation 59. Increasing order of electronegativity of the hybrid orbitals is:

(1) sp < sp2 < sp3 (2) sp2 < sp < sp3 (3) sp3 < sp2 < sp (4) sp2 < sp3 < sp 60. In the following endothermic reaction BaCO3 BaO + CO2 formation of BaO is favoured by (1) decreasing temperature (2) decreasing pressure (3) more concentration of CO2 (4) all of these will shift the equilibrium towards left

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55.

(1) (1) (1) (3) (3) (1) (3) (1) (4) (2)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56.

(4) (3) (1) (2) (3) (1) (1) (2) (4) (4)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57.

(2) (2) (3) (1) (1) (1) (2) (3) (1) (4)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58.

(2) (4) (3) (1) (1) (1) (4) (1) (2) (1)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59.

(1) (2) (1) (2) (2) (1) (2) (2) (2) (3)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60.

(3) (2) (2) (2) (4) (1) (1) (1) (3) (3)

ANSWERS WITH EXPLANATIONS 1.

(1) For depositing 27 g aluminium, 3 × 96500 C are required. 7 C . C 27 (4) Difluoro acetic acid is highly ionised due to the presence of two electron withdrawing fluorine atom. Hence it would exhibit highest conductivity. (2) Drugs which are used to reduce the body temperature, e.g. fever are called antipyretics. (2) Polymers having amide linkage are called polyamides. n[H2N–(CH2)6–NH2] + n[HOOC–(CH2)4–COOH]

=

2.

3. 4.

Hexamethylene diamine

Adipic acid

o — ( HN–(CH2)6 – [NH–CO](CH2)4 – CO — ) Nylon-66

Amide linkage

5.

(1) For spontaneous process 'G = – ve, K > 1 and Eºcell = +ve

6.

(3) Hydrogen bomb is based on the principle of nuclear fusion. (1) Let the oxidation state of Cr in [Cr(NH3)4Cl2 ]+ = x Now, x + 4 × 0 + 2 × (–1) = +1 or x = +3 (3) Na2SO4 2Na+ + SO42– Initial 1 0 0 At Equi. 1 – D 2D D 1 2 Vant Hoff factor (i) 1 2 1 (2) H2 : V1s2, V*1s2; no unpaired electron; diamagnetic. (1) He2+ : V1s2, V*1s1; one unpaired electron; paramagnetic

7.

8.

9.

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

(3) H2+ : V1s1, V*1s0; one unpaired electron; paramagnetic (4) H2– : V1s2, V*1s1; one unpaired electron; paramagnetic 10. (4) SnO2 is amphoteric, as it reacts both with acids and bases. SnO2 + 4HCl o SnCl4 + 2H2O SnO2 + 2NaOH o Na2SnO3 + H2O (1) is basic, (2) and (3) are acidic. 11. (2) We know that the acidity of phenol is due to the powerful resonance stabilization of the phenoxide ion. Now for the substituents present in phenol, any substituent which stabilizes the phenoxide ion more by dispersal of negative charge will increase the acidity of phenols, conversely, any substituent which destabilize the phenoxide ion by intensifying negative charge will decrease the acidity of phenols. (i) Electron withdrawing groups like —NO2, — CN, —X (halogens) etc. Which stabilize the phenoxide ion by dispersing the negative charge relative to phenol increase the acidic strength of phenols. (ii) Electron donating groups like alkyl, —NH2 , — OR (alkoxy) —(alkyl) etc., which destabilize the phenoxide ion by intensifying the negative charge relative to phenol tend to decrease the acid strength. This explains the following order of acidity. p-nitrophenol > phenol > p-cresol. 12. (2) Chlorination of n-butane takes place through free radical formation. e.g.  Cl2 o  115

 CH3CH2CH2CH3 CH3 C HCH2CH3 + HCl  l attackk sp2 hybrid planar shape intermediate and C from either side to give CH3 C HCH2CH3 + C l H Cl | | o CH3 – C – CH2CH3 + CH3 – C – CH2CH3 | | Cl H

Racemic mixture

13. (1) CH3CH2CH = C

CH 3

+O3 CH 3 O—O | | CH 3 o CH3CH2CH C O CH 3 o CH3CH2CHO + (CH3)2C = O

14. (1) In the given problem : C4 H 9Cl + 2Na + ClC4 H 9 C4 H9 — C4H9 + 2NaCl + Cl2 p mono chloroalkyl derivative

Compound A is t-butyl chloride; in this all —CH3 groups have primary hydrogen only and thus will give only one mono-chloro derivative. (CH3)3C — C(CH3)3 CH2Cl(CH2)2C— C(CH3)3 Mono chloro alkyl derivative

15. (3) Tollens reagent is ammoniacal silver nitrate solution used to detect the presence of aldehydic group. Aldehydes reduces Tollens reagent to metallic silver or silver mirror. CH3CHO + Ag2O    o CH3COOH + 2Ag (from Tollens reagent)

19. (3) C2H5OH + 4I2 + NaOH o CHI3 + NaI + HCOONa + H2O Pale yellow crystals

20. (2) Steam distillation is a distillation with water and is carried out when a solid or liquid is insoluble in water and is volatile with steam, but the impurities are nonvolatile. 21. (1) A compound which shows optical isomerism should have a chiral carbon atom, e.g., a carbon atom to which all the four groups or radicals attached is different. H | CH 3 The structure: CH— — C2H5 | CH 3 CH 3 2, 3-dimethyl has a chiral carbon atim. Others do not have chiral carbon atom. 22. (1) We know 'Hºf = ¦Hºf (product) – ¦Hºf (reactant) For the given reaction : 2H2O2 (l)o2H2O (l) + O2 (g) 'Hºf = 2 × 'Hºf (H2O) – 2 × 'Hºf (H2O2) = 2 × – 286 kJ mol–1 – 2 × (–188) kJ mol –1 = –196 kJ mol –1 23. (2) We know : 'H = 'E + P'V when 'V = 0, W = P'V = 0 ?  'H = 'E we also know from first law that : 'E = q + w ? 'H = q In the present case, 'H = 500 J = 'E = 500 J; q = 500J and w = 0. 24. (2) CH4 (g) + 12 O2(g) o CH3OH (l) 'H = x – y; Given 'H = –ve Hence x – y < 0 25. (3) We know :

Silver mirror

16. (3) Terylene is an example of condensation polymer of methyl terephthalate and glycol CH3COOH—

n

HOCH 2 —CH2 OH glycol

24 4 6 at the end of 24 hours activity (an ) = 0.01 M ?

3

0.01

a0

FG 1 IJ H 2K

4

?

– C —OCH2—CH2—O— C – || || O Terylene or dacironr O

–

n 17. (1) This is the only incorrect IUPAC nomenclature. 18. (2) Alcohols may be dehydrated to the corresponding olefins. The ease of dehydration of alcohols follow the order tertiary > secondary > primary C2H5OH

116

n

no. of half life

—COOCH3

Methyl terephthalate

LM — MM C|| – NO

0

FG 1 IJ H 2K

H 2O

CH2 = CH2

a0 = .01 × 16 = .16 No. of moles 26. (3) molarity (m) per litre of solution 1 117 . 1000 36.5 1 36.5 117 . 1000 = 32.05 27. (1) From the problem it is clear that 0.5 g of Se is present in 100 g of enzyme

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

100 78.4 0.5 = 1.568 × 104 (Because in a molecule of enzyme one Se atom must be present). 28. (1) Given specific volume (volume of 1 gm) of cylindrical virus particle = 6.02 × 10 –2 cm3 g –1 Volume of cylinder = Sr 2l Also given : r = 7Å = 7 × 10 – 8 cm l = 10 Å = 10 × 10 – 8 2 22 8 cc cc . cm ' 7 Volume Wt. of one virus particle ? 78.4 g of Se would be present in

e

j

. 154 10

36. (1) n = 3, l = 2, m = +2 depicts one of the five d-orbitals m= +2, +1, 0, 1, –2 37. (3) Tendency of proton acceptance is known as the basic strength. Basic character decreases as R—NH2 > NH3 > HS– > I– 38. (1) More is the pressure, higher is the boiling point. 39. (2) For strong S-bonding, pS-pS bonding should be strong. In core of phosphorus (P), due to its larger size as compared to nitrogen (N), pS-pSbonding is not so strong. 40. (4) For the ionisation of CH3COOH : CH3COOH CH3COO + H+

6.02 10 2 ? Molecular weight of virus = Wt. of NA particle . 154 10

23 2

29. (2)

30. (4)

31. (1) 32. (1)

6.02 10

CH 3COO H Ka

or [ CH 3 COOH ] Initial conc.

23

6.02 10 = 15400 g mol–1 = 15.4 kg mol–1 The metals having higher negative reduction electrode potential can displace metals having lower negative reduction electrode potential or +ve reduction electrode potential. For a reaction to be spontaneous 'G < O For the reaction : PbO2 o PbO  'G < 0 +4 +2  o favourable For the reaction : SnO2 o SnO; 'G > 0 +4 +2 m favourable – XeF2 and I F2 both are sp3d hybridized. The ionisation potential decreases as the size of the atom decreases. Atoms with fully filled or partially filled orbitals have high ionisation potential.

33. (1) The Ni in Ni(CO)4 is in zero oxidation state 8 2 28Ni : [18Ar] 3d 4s 3 4s2 4p 3d Ni : [18Ar] in hybridised state

[ CH 3COO][ H ] [ CH 3COOH ]

Ka

23

4 CO groups

Hence, Ni(CO)4 is sp3 hybridized, having tetrahedral geometry but diagmagnetic as there are no unpaired electrons. 34. (1) For S-overlap, the lobes of the atomic orbitals are perpendicular to the line joining the nuclei. + + + ... ... S-overlap – – – Hence, only sidewise overlaping takes place. 35. (1) The ionic conductivity increases with increase of number of ions, produced after decomposition. compound (1) 5, (2) 4, (3) 3, (4) 0. ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

(3.4 10

4 2

)

5

1.7 10 = 6.8 × 10 –3 41. (2) For reaction M2S 2M+ + S2– Let s be the solubility in gm mol L–1 ? Ksp (M2S)= (2s)2 × s = 4s3 = 4 × (3.5 × 10– 6 ) 3 = 1.7 × 10 –16 42. (1)

b a

0 1

(

1)

1

2 42

8 5

2 2 0 1

c d

12 5

b 8 ( a 3) X 3

4

4 2

43. (1) FeO(s) + CO(g) Fe(s) + CO(g) (1–x) atm x atm PCO2 0.403 Kp PCO x 0.403 or 1 x = PCO2 =0.287 atm and PCO= 1 – x = 0.713 atm 44. (2) More the branching, lesser is the b.p. 45. (3) Ethyl alcohol with iodine in presence of a base gives a yellow precipitate of iodoform whereas methyl alcohol does not, CH3CH2OH + 4I2 + 3Na2CO3   o CHI3 + 5NaI + HCOONa + 3CO2 + 2H2O 46. (1) When HI is in small quantity : CH2I CH2OH | CH2 | CH2OH

3 H2 O

| CH2I Glycerol tri-iodide

o  CH + I2 | CH2I Alkyl iodide 117

When HI is in larger quantity : CH2 CH3 CH3 CH3 | | | | CH CHI CH CHI M rule M rule | | || | CH2IH CH2I CH2 CH3 2-iodopropane

47. (2) K 1

[H 2 S] [H 2 ][S 2 ]1/2

and K 2

[HBr]2 [H 2 ][Br2 ]

Now, equilibrium constant K for the reaction Br2 + H2S 2HBr+1/2 S2. K 48. (1)

[HBr]2 [S 2 ]1/2 [Br2 ][H 2 S]

K2 K1

2SO2 + O2 o 2SO3 2 mole 1 mole 0 initial 2–2x 1–x 2x moles at equilibrium 2 2x 1 x 2x V V V molar conc. at equilibrium Now K

[SO 3 ]2 [SO 2 ]2 [O 2 ]

LM 2x OP NVQ 2 2x 1 x MNL V OPQ LMN V OPQ 2

2

x2 V (1 x )3

49. (4) The structure of phosphorus acid is O H || OH P | OH

118

50. (4) CuSO4 Al2(SO4)3.24H2O. In this CuSO4 is there in place of sodium or potassium or ammonium sulphate which is a must for alum. 51. (1) As solid has no partial pressure ? [Ag 2O(s) ] = [Ag(s)] = 1 Hence, KP = PO2 52. (2) Electron withdrawing groups increase the acidity while electron-donating groups decrease the acidity of phenols. — CH3 and —NO2 are electron withdrawing groups whereas —OH is electron donating group. Hence the result 53. (2) The presence of mobile electrons or oscillation of loose electrons are responsible for metallic lustre. 54. (3) BaO2 is a true oxide as it contains peroxide ion 22 55. (2) XeF4 on partial hydrolysis produces XeOF2 XeF4 + H2O o XeOF2 + 2HF 56. (4) Because of higher degree of hydration of Li+ ions. 57. (4) KMnO4 is used as self indicator. 58. (1) Benzilic acid rearrangement C 6 H 5 — C — C — C6H5 || || O O C 6 H5

OH C

C 6 H5

BH

COOH

H2 O

OH | (C6H5)2 —C— COOH 59. (3) Greater the s-character of the hybrid orbitals more electronegative is the atom. Thus a sp-hybridised C-atom having hybrid orbitals with 50% s-character is more electronegative than a sp2 hybridised carbon with 33.3% s-character and sp3-hydridised carbon with 25% s-character. 60. (3) This is according to Le Chatelier principle which states as “when a constraint is applied to a system in equilibrium, the system adjusts in a way so as to anull the effect of the constraint.”

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

UNIT 13 : VECTOR ALGEBRA 1. Scalars : The quantities which have only magnitude and are not related to any direction in space are called scalars. Examples : Mass, Time, Volume, etc. 2. Vectors : The quantities which have both magnitude and direction are called vector quantities. Examples : Velocity, Acceleration, Force, etc. G A Vector A is denoted as A . 3. Modulus of a vector : The modulus or magnitude of a vector a is the non-negative number which is the measure of its length and is denoted by a or a. o

Remark : Every vector AB has the following three characteristics : o

o

(a) Length : The length of AB is denoted by AB or AB. (b) Support : The line of unlimited length of which AB o

is a segment is called the support of the vector AB . o

(c) Sense or Direction : The sense of AB is from A to o

B and that of BA is from B to A. Thus,, the sense o of a directed line segment AB is from its initial point A to the terminal point B. o (d) Initial and Terminal Points. For the vector AB , A is called the initial point and B is called the terminal point. 4. Types of Vectors (a) Null vector or Zero vector : If the initial and terminal points of a vector coincide, then it is called a zero vector. It is denoted by 0. Its magnitude is zero and direction is indeterminate. (b) Unit Vector : A vector whose modulus is unity is called a unit vector. The unit vector in the direction of a vector a is denoted by aˆ . ˆ = 1. Thus |a| (c) Reciprocal Vector : A vector, whose direction is same as that of a given vector a but its magnitude is the reciprocal of the magnitude of the given vector a, is called the reciprocal of the vector a and is denoted by a–1. (d) Equal Vectors : Two vectors a and b are said to be equal, written as a = b, if they have the (i) same length (ii) same sense ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

(e) Like and Unlike Vectors : Two vectors are said to be like when they have the same sense of direction and unlike when they have opposite sense of direction. (f) Collinear or Parallel Vectors: Vectors having the same or parallel supports are called collinear vectors. (g) Co-initial Vectors : Vectors having the same initial point are called co-initial vectors. (h) Co-planar Vectors : A system of vectors is said to be coplanar, if their supports are parallel to the same plane. Note that any two vectors are always coplanar. (i) Coterminous Vectors : Vectors having the same terminal points are called coterminous vectors. (j) Negative of a vector : A vector having the same magnitude as that of a given vector a and direction opposite to that of a is called the negative of the vector a and is denoted by – a. o

o

Thus if AB = a, then – a = BA (k) Localised vectors and free vectors : A vector parallel to a given vector through a specified point as the initial point is known as a localised vector. If the initial point of a vector is not specified, it is said to be a free vector. (l) Position vector : The position vector of a point P is the o vector OP , where O is the fixed point, called the origin. 5. Parallelogram Law of Addition : If two vectors a and b are represented in magnitude and direction by the two adjacent sides of a parallelogram, then their sum c is represented by the diagonal of the parallelogram which is co-initial with the given vectors. Choose any point O as origin. Draw two vectors o

o

OA (= a) and OB (= b) such that both have the same initial point.

o

Complete the parallelogram OACB. Then OC is the sum or the resultant of a and b. o

Thus OC

o

o

OA  AC =a+b

o

o

OA  OB

119

Example : If the position vectors of three consecutive vertices of a parallelogram are     3 5  and  then the coordinates of the fourth vertex are (a) (2, 1, 3) (b) (6, 7, 8) (c) (4, 1, 3) (d) (7, 7, 7) (e) (8, 8, 8) Solution: (d) The parallelogram is as shown :



,

C(7, 9, 11)

D

A(1, 1, 1)

B(1, 3, 5)

Point D is A + C – B or 7, 7, 7.

Remarks : (a) For any two non-zero vectors a and b, the magnitude of a + b is not equal to the sum of the magnitudes of a and b. (b) The difference a – b of two vectors is defined as the sum of vectors a and – b. Thus a – b = a + (– b) 6. Properties of Addition of Vectors: (a) Vector addition is commutative, i.e. a+b=b+a for any two vectors a and b. (b) Vector addition is associative, i.e. ab c a bc for any three vectors a, b and c. (c) For every vector a, we have a +0=a=0+a where 0 is the null vector. Thus 0 is the additive identity of vectors. (d) For every vector a, there corresponds a vector –a such that a + (– a) = 0 = (– a) + (a). Thus (– a) is the additive inverse of a. 7. Multiplication of a vector by a scalar : Let a be a given vector and k be any scalar (taken to be real), then ka is another vector whose magnitude is |k| times the magnitude of the vector a and whose direction is either same as that of a, or opposite to that of a, depending upon whether k is positive or negative. If k = 0, then ka will be the zero vector. Thus k > 0 Ÿ ka and a have same sense of direction. k < 0 Ÿ ka and a are in opposite directions. 8. Properties of Multiplication of A Vector by A Scalar : For vectors a, b and scalars m, n, we have (i) m(–a) = (–m)a = –(ma) (ii) (–m)(–a) = ma (iii) m(na) = (mn)a = n(ma) (iv) (m + n)a = ma + na (v) m(a + b) = ma + mb 9. Linear Combination of Vectors : A vector r is said to be a linear combination of vectors a, b, c......, etc. if there exist scalars O, P, J..... etc. such that r = Oa + Pb+ Jc + ........... For example, the vectors r1 = a + 2 b + c and r2 = 3 a + 3 b + c are linear combination of the vectors a, b and c. 120

10. Linear Dependence and Linear Independence of Vectors : The vectors a, b, c are called linearly dependent if there exist scalars O, P, J, not all zero (at least one of them should be non-zero), such that Oa + Pb + Jc = 0 The vectors which are not linearly dependent are called linearly independent. Thus if a, b, c are linearly independent vectors, then Oa + Pb + Jc = 0 ŸO P J 0 For Example : (i) Vectors i + j – k, i – j + k and i are linearly dependent because by choosing O 1, P 1, J 2, we get 1(i + j – k) + 1(i – j + k) – 2 i = 0 (ii) The vectors i + j and 2i + 3j are linearly independent because for any real numbers D and E, D(i + j) + E(2i + 3j) = 0 Ÿ D  2 E 0 and D  3 E 0

ŸD E 0 11. Some Important Results (a) If a and b are non-zero, non-collinear vectors, then they are linearly independent. (b) If a and b are two non-zero, non-collinear vectors, then any vector r in the plane of a and b (since any two vectors are always coplanar) can be expressed uniquely as a linear combination of the vectors a and b, i.e r OaPb (c) Let a, b, c be any three non-coplanar vectors. Then any vector r can be expressed as a linear combination of a, b and c. (d) Any three non-coplanar vectors are linearly independent. (e) If A and B are the points with position vectors a and b respectively and if C is a point dividing AB internally in the ratio m : n, then the position vector of C is given by

n a  mb mn (f) If A and B are two points with position vectors a and b respectively and let C be a point dividing AB externally in the ratio m : n. Then the position vector of C is given by o mb  n a OC m n (g) If C is mid-point of AB, then o

OC

o

OC

ab [ m = n = 1] 2

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

Ÿ or Ÿ ? or

(h) If a, b, c are the position vectors of the vertices of a triangle ABC, then the position vector of the centroid of the triangle is abc 3 o

o

(i) Three points A, B and C are collinear if AB k BC , for some real k. 12. Some Important Results Result 1. If a and b are any two non-collinear vectors, then any vector r which is coplanar with a and b (i.e. in the plane of a and b) can be expressed as r = x a + y b, where x and y are scalars and this representation is unique. xa is called the component of r in the direction of a and yb is called the component of r in the direction of b.

Example : The vector(s) which is/are coplanar with vectors i  j  2 k and i  2 j  k , and perpendicular to the vector i  j  k is/are (A) j  k (C) i  j

Consider a vector a coplanar with the given vectors; it can be written as a O ( i  j  2 k )  P( i  2 j  k ) ( O  P )i  ( O  2P )j  (2O  P )k

can be expressed uniquely as r = xa + yb + zcc where x, y and z are scalars. xa, yb, zc are respectively the components of r in the directions of a, b and c. Remark. Since i, j, k are the unit vectors along the coordinate axes x, y and z, therefore for any point P in space, its position vector r can be expressed as r = xi + yj + zk where x, y, z are scalars, r

2

x y z

2,

Example : The number of distinct real values of O, for which the vectors – O2 i  j  k , i  O2 j  k and i  j  O2 k aree coplanar, is (A) zero (B) One (C) Two (D) Three Solution: (C) The given vectors are coplanar when  O2 1 1

1  O2 1

1 1 O2

0

Ÿ –O2 (O4 – 1) – 1(–O4 – 1) + 1(1 + O2) = 0 ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

It will be perpendicular to the vector ( i  j  k ) when a .( i  j  k ) 0 Ÿ (O  P   O  2P) + (2O  P) Ÿ 4(O  P) = 0 or O  P ? a O(  j  k ) For O = 1, a 1

 j  k j  k

For O = –1, a 1 Thus, (A) and (D) are correct answers.

o

and xi, yj, zk are called the resolved parts of OP along i, j, k respectively. Result 3 (Collinearity of three points). Let A, B, C be three points with position vector a, b and c respectively. Then the necessary and sufficient condition for the points A, B and C to be collinear is that there exist three scalars x, y and z such that xa + yb + zc = 0 where x, y and z are not all zero (i.e. atleast one of them is non-zero) and x + y + z = 0. Result 4 (Coplanarity of four points). The necessary and sufficient condition for four points with position vectors a , b, c and d to be coplanar is that there exist scalars x, y, z and u such that xa + yb + zc + ud = 0 where x, y, z and u are not all zero (atleast one of them is non-zero) and x+y+z+u=0

(B)  i  j (D)  j  k

Solution: (A) and (D)

Result 2. If a, b, c are non-coplanar vectors, then any vector r

2

–O6 + O2 + O2 + 1 + 1 + O2 = 0 O6 – 3O2 – 2 = 0 (1 + O2)2 (O2 – 2) = 0 O2 = 2 O r 2

WORKING RULE What to Prove How to Prove

S. No. 1. Collinearity of two vectors a and b 2.

Collinearity of three points A, B and C

3.

Coplanarity of three vectors a, b and c

Express one vector as some scalar multiple of the other, i.e. there must exist some scalar D such that b = Da o

 2

Example : The vector the vectors

o

Find the vectors AB and AC and then express one of them as some scalar multiple of the other, i.e. there must exist some scalar D such o o that AB D AC Express one of them as a linear combination of the remaining two, i.e. express c as c = xa + yb

  and



 lies in the plane of  and bisects the angle le

between b and c . Then which one of the following gives possible values of D and E ? (a) D = 1, E = 1 (b) D = 2, E = 2 (c) D = 1, E = 2 (d) D = 2, E = 1

Solution: (a) the vector

 2

 lies in the plane of

 , 

 121

and a bisects the angle between b and c. ?

(

 2 Ÿ O = 1, D

)

Thus, when the scalar product of two vectors is zero, then either the two vectors are perpendicular to each other or one of them is a zero vector. Geometrical Interpretation of The Scalar Product. Scalar product in terms of projections Figure given below is self-explanatory.



 2  ) 1, E = 1 b 30º 30º

a c

b c

1

b c

2. 2

cos

1 2

3 12. Product of Two Vectors : The product of two vectors is defined in two ways. (A) Scalar product or Dot product. (B) Vector product or Cross product. A. Scalar product or Dot product : Let a and b be two non-zero vectors inclined at an angle T. Then the scalar product of a and b is denoted by a . b and is defined as the scalar | a |.| b| cosθ .

Here MN is the projection of the vector b on the vector a. Clearly, MN = BP = b cos T ? a . b = ab cos T = (b cos T) a = MN u a = (Projection of b on a) u a Similarly, we can write a . b = (Projection of a on b) u b Scalar Product in Terms of Components. Let a = a1i + a2j + a3k and b = b1i + b2j + b3k Then a . b = a1b1 + a2b2 + a3b3 3

¦ a ibi i 1

Thus the scalar product of two vectors is equal to the sum of the products of their corresponding components.

Example : If a , b and c are unit vectors such thatt Thus a . b = | a |.| b| cosθ

Remark 1. The scalar product of two vectors is a scalar quantity

which can be positive or negative depending upon whether T is acute or obtuse. Remark 2. If a and b are like vectors (i.e. T = 0), then a . b = ab cos 0 = ab and if a and b are unlike vectors (i.e. T = S), then a . b = ab cos S = – ab Also a . a = |a|2 = a2 ( ' T = 0) a . b = b. a a . (b + c) = a . b + a . c In particular, since i, j and k are mutually orthogonal unit vectors, then i. i=j. j= k.k=1 and i. j= j. k= k.i=0

Remark 3. If a or b or both are zero vectors, then T is not defined, as 0 has no direction. In this case, their dot product a . b is defined as the scalar zero. Remark 4. If a and b are perpendicular vectors, then T = 90o and we have a . b = ab cos 90o = 0 122

= (b) 1 (d) 3

0, then (a) –1 (c) –3

Solution: (c) Since a b c Ÿ a Ÿ

b

0

c

Ÿ 1 1 2a. b 1 [' a , b, c are unit vectors] s] Ÿ a. b

1 2

1 2

Similarly, So, 3

FG 1IJ H2K

2

FG 1IJ FG 1IJ = –3 H2K H2K

Example : Two adjacent sides of a parallelogram ABCD are  2  2  . The side 2  10  11  and given by AD is rotated by an acute angle D in the plane of the ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

parallelogram so that AD becomes AD' . If AD' makes a right angle with the side AB, then the cosine of the angle D is given by (A) (C)

8 9

(B)

1 9

(D)

17 9 4 5 9

AD. AB

Solution: (B) cos

AD AB 20 22

2

So, cos (90º – D) 8 9

Ÿ sin Ÿ

C

D T A

B

64 81

(a)

S 3

(b)

S 6

(c)

S 4

(d)

S 2

G G G G G G a  b  c 0 &|a| 3,|b| 5,|c| 7 G G G a  b c G G GG G |a|2 |b|2 2a.b |c|2

GG Ÿ 2a.b 49  9  25 15/ 2 15 1 Ÿ cos T u 3 u 5 2 15

17 9

Angle Between Two Vectors Let a and b be two vectors inclined at an angle T. Then a . b = | a |.| b| cos T

‰

Ÿ cos T

a.b a.b Ÿ T cos 1 | a || b| |a || b|

§

cos

 1¨

a 1b1  a 2 b 2  a 3 b 3

¨¨ ©

a12  a 22  a 32 b12  b 22  b 32

· ¸ ¸¸ ¹

Remark : The condition of perpendicularity in terms of the components is a 1b1  a 2 b 2  a 3 b 3

T cos1(1/ 2)

0

GG Example : If a, b and cG are three non-zero vectors such that GG GG a.b a.c, then G G G (a) a A b and c G G G G G (b) either a A (b – c)or b c

(b) 1

(c) No value of m exist

(d) –1

ˆ Solution: (c) Let aˆ j  ˆj  (m  1)k. G ˆ and cG ˆi  ˆj  mkˆ b j  ˆj  mk.

G G G GGG Vectorsof a, b  c are coplanar if [ a b c] 0 1

1

m 1

1

1

m

1 –1

m

G (d) b

1 –2

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

0

Applying C 2 o C 2  C1 , we get

1 1

Ans. (b)

S 3

coplanar, m = (a) 0

G G G (c) a A (b – c) G c

1 2

ˆ ˆi  ˆj  mk, ˆ ˆi – ˆj  mkˆ are If ˆi  ˆj  (m  1)k,

Example :

If a = a1i + a2j + a3k and b = b1i + b2j + b3k then T

GG a.b G G |a||b]

G Ÿ (3)2  )(5)2  2a.b (7)2

2

1 1

D

D'

8 9

G G G G 0 and|a| 3,|b| 5,|c| 7. Then G G angle between a and b is

G Solution:. (a) Angle between aG & b, cos T

4 100 121 1 8 8 40 8 15 3 9 Since TD= 90º Ÿ T= 90º – D

G G

G

Example : If a  b  c

0 0

m 1 m

0

m

Ÿ 2m(1)  (m  1)( 2) 0 123

Solution : (e) Consider 2( 2 ˆi  ˆj  2kˆ ) 4 ˆi  2 ˆj  4 kˆ

Ÿ 2 0 is not possible.

2 2  Now, a. r 4 2 2 2 4 1 0  2 2  , . 1 4 2 ( 2) 4 2

Hence, for no value of m, the vectors are coplanar.

G G G G G G Example : If i  j  k and 2 i  3 j  k are adjacent sides of

?

a parallelogram, then the lengths of its diagonals are (a)

21, 13

(b)

3, 14

(c)

13, 14

(d)

21, 3

Ans (a)

Example : If the volume of the parallelepiped formed by three G G G non-coplanar vector a, b and c is 4 cubic units, then G G G G G ª¬a u b b u c c u a º¼ (a) 8

(b) 64

(c) 16

(d) 4

Ans. (c)

G G G G G ª¬ a u b b u c c u a º¼

GG G [abc] 4 2

16

Example : A unit vector in the XZ-plane which is perpendicular to the vector (2, 4, – 3) - is ……… 1 1 ( 3, 0 , 3) (a) (b) ( 3, 0 , 2 ) 13 18 1 1 ( 3, 0 , 2 ) (c) (d) (3, 0 , 2 ) 3 13 Solution: (d) Let (a, 0, b) be the unit vector in the xz-plane which is perpendicular to the vector (2, 4, –3) 3b So, 2a – 3b = 0 Ÿ a 2 Also a2 + b2 = 1 9b 2 Ÿ b2 1 4 13 2 b 1 Ÿ 4 2 Ÿ b 13 3 So, a 13 1 So, the required unit vector is ( 3, 0 , 2 ) . 13 Example : Which one of the following vectors is of magnitude 6 and perpendicular to both  2 2  ? 2 2     )  (a) (b) (   ) ) (c) (   (d) (   ) (e) (   124

G r

0

r perpendicular to a and b both. Its magnitude is 2 × 3 = 6

Example : If the scalar product of the vector   2  with the unit vector along  2  3  is equal to 2, then one of the values of m is (a) 3 (b) 4 (c) 5 (d) 6 (e) 7 Solution: (d) According to given condition, we have  2 3  ) (  2  2 m2 4 9 m 2 ? m 2 13 m 8 2 m 2 13 Squaring on both sides, we get, (m + 8)2 = 4(m2 + 13) Ÿ m2 + 16m + 64 – 4m2 – 52 = 0 Ÿ 3m2 – 16m – 12 = 0 Ÿ 3m2 – 18m + 2m – 12 = 0 Ÿ 2 . Ÿ m 6, 3   2, Example : If the vectors     are mutually orthogonal,  2 4 then (O, µ) = (1) (–3, 2) (2) (2, –3) (3) (–2, 3) (4) (3, –2)

Solution: (1) Since a. b where

b. c c. a   2

0

2 4 

 and  



So, a . c and b. c Solving for Oand µ, we get µ = 2 and O = –3 So, (O P) = (–3, 2)

Example : The non-zero vectors a , b and c are related by 8

and

7

. Then the angle between a and c

is : (a) (c)

(d)

4

2 8 , 7 Let T be the angle between a and c

Solution: (a)

2

?

cos

a. c

56 b

a c

8 b 7 b

1

? T S. ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

0

Example : If

| | 3, |

5

7 then

Solution: (e) x

a b

a and b is (a)

(b)

(c)

(d)

x

6 Solution: (b) a b c a b

4

0 a

c

2

is shown in the figure x and y are the diagonals of

2

2

| |

2

a ||gm with sides a and b

2

|

2

( a. b)

|a || b|cos ( 3)(15) cos cos

e'

j

1

3

(a) 190 (c) 300 (e) 192 Solution: (e)

(a) 5 3 (c) 4 3 (e) 6 3

2

is equal to (b) 275 (d) 320

, | | 2 , | | 3 and the

,

angle between a and b is

3

3

, then |

i  j  k . Then the vector b

(

)

3 is

 

(2)



(4)   2  0

0

Ÿ

0

Ÿ 3a 2 b a c



Ÿ 2b

3  

'

0



]

3a a c 3j 3k ( 2 i j k )

LM MM' N

LM i MM0 N1

j 1 1

OP PP Q

k 1 1

2 



OP PP Q

  2

Example : Let the vectors PQ, QR , RS, ST and UP

1 or p = r

2

Example :

Ÿ

2 i 2 j 4 k Ÿ

Example : Vectors a and b are inclined at an angle T = 120º. If |

Solution: (1) Since a b c

)

or .

j  k and c

Example : Let a

, (b) p = 1, q = 1 (d) q = 1, p = r

(a) r = 1, p = q (c) r = 2p, q = 2 (e) q = 1, r = 1 Solution: (d) We have

6 3

satisfying a u b  c 0 and a. b (1) i j 2 k (3)   2 

3

Example :

Ÿ

3 2

2 2 3

B. Vector Product Or Cross Product : The vector product of two vectors, denoted by a u b, is defined as another vector whose magnitude is ab sin T and is in the direction of the normal to the plane of the vectors a and b. Thus, a u b = ab sinT n, where the unit vector n (normal to the plane of a and b) is taken in such a way that a, b, n form a Right-Handed System.

(

|

2|a| | b|cos 60º

15 2 15 2 15 2 1 Ÿ 2

a. b

Ÿ

b

y

| is equal to

(b) 6 (d) 9

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

represent the sides of a regular hexagon, Statement-1 : PQ u ( RS  ST ) z 0. because Statement-2 : PQ u RS 0 and PQ u ST z 0 (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.

Solution: Statement-2 is obviously false since PQ is not parallel to RS. ‰

Condition of Collinearity of Two Vectors : If a and b are collinear vectors, then au b=0 125

Its converse is not true. In fact a u b = 0 implies that the vectors a and b are either collinear or one of them is a zero vector. Therefore, if a and b are non-zero vectors, then a and b are collinear œ a u b = 0 In particular, (i) a u a = 0 (ii) i u i = j u j = k u k = 0 (iii) i u j = k, j u k = i, k u i = j i, j, k are taken in cyclic order

and (b + c) u a = (c) If O and P are (Oa  u Pb   ‰

 2 5

2  3 

, then the value of 14 Solution: (5) Since

S

U

Integer Type Questions

Example : If a and b are vectors in space given by a b

T

bua+cua scalars, then  OPa u b  Pa  u Ob

[ 2a

R

and

is

b ].[ ( a 2 b) ( a b)] '

P

Q

( 2 a b).[(( a 2 b ). b) a (( a 2 b). a ) b ]

(iv) j u i = –k, k u j = –i, i u k = –j ‰

'

=

Geometrical Interpretation of the Vector Product : o

( 2a

LM' N

Choose O as origin and draw co-initial vectors OA (= a) o

and OB (= b). Complete the parallelogram OACB. Clearly,,

b).[( 0 2 )a (1 0 ) b ] 2

2 2 70

LM' N

.

1

( 2a b).( 2 a b) =4+1+0+0

OP Q

2

0

1

( 2 a b).( 2a b) 2

2

0

1

OP Q

1

=5 Area of the ||gm OACB = ab sinT 1 Area of ' OAC = ab sinT 2 ? |a u b| = Area of the ||gm OACB or |a u b| = 2 Area of the ' OAC C ‰

‰

‰

126

Vector Product in Terms of Components : Let a = a1i + a2j + a3k and b = b1i + b2j + b3k be two vectors. Then a u b = (a1i + a2j + a3k) u (b1i + b2j + b3k) = (a2b3 – b2a3)i + (a3b1 – b3a1)j + (a1b2 – b1a2)k i j k a1 a 2 a 3 b1 b 2 b 3 Condition of Collinearity in Terms of Components : Let the vectors a and b be collinear.. Then, a u b = 0 Ÿ (a2b3 – b2a3)i + (a3b1 – b3a1)j + (a1b2 – b1a2)k = 0 Ÿ a2b3 – b2a3 = 0, a3b1 – b3a1 = 0 and a1b2 – b1a2 = 0 a1 a 2 a 3 a a2 a a 3 a 3 a1 Ÿ 2 and 1 Ÿ , b1 b 2 b 3 b 2 b 3 b 3 b1 b1 b 2 Properties of Vector Products : (a) Since a u b = – b u a, therefore the vector product is not commutative. (b) Vector product is distributive. a u (b + c) = a u b + a u c

2

Example : The value of

2 2 2

is

2a b (b) 3/2 (d) 4/3

(a) 1/2 (c) 5/2 2

Solution: (a)

2 2 2

2a b |a b|2

( a . b) 2

2 |a |2 | b|2 2 |a |2 | b|2 2 |a |2 | b|2 1 2 ‰

Applications of Vectors in Mechanics : (a) Work done by a force = Force u Displacement (b) Moment of a force F about a point O is the vector o

M = OP u F, where P is any point on the line of action of the force F. Let r be the position vector of the point P with respect to O. ?r

o

OP Ÿ M

r uF

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

Example : Find the moment about the point i – 2j – k of a force represented by 3i + k acting through the point 2i – j + 3k. Here F = 3i + k, r = (2i – j + 3k) – (i + 2j – k) = (i – 3j + 4k) Moment of the Force F about the point i + 2j – k ? =ruF = (i – 3j + 4k) u (3i + k) = –j + 9k – 3i + 12j = –3i + 11j + 9k

(c) Vector equation of a plane passing through a point a and parallel to two given vectors b and c is r = a + sb + tc, where t and s are arbitrary constants. or [r b c] = [a b c] (d) Vector equation of a plane passing through the points a, b and c is r = (1 – r – t) a + r b + t c or r . (b u c + c u a + a u b) = [a b c] (e) Vector equation of a plane passing through a point a and perpendicular to n is r . n = a . n. Note that the perpendicular distance of the plane from a.n the origin . |n|

(c) Moment of A Couple A system consisting of a pair of equal unlike parallel forces having different points of application is called a couple. The vector sum of the two forces of a couple is always a zero vector.

(f) Perpendicular distance of a point P(r) from a line passing through a and parallel to b is given by 2 1/ 2

ª ª(r  a ) . b º º (r  a)u b «( r  a )2  « PM » » « b ¬ | b | ¼ »¼ ¬ (g) Perpendicular distance of a point P(r) from a plane passing through a and parallel to b and c is r a . buc PM buc (h) Perpendicular distance of a point P(r) from a plane passing through the points a, b and c is given by

The sum of the moments of the forces of the couple about any point O is given by o

M

o

OA u F  OC u  F o · § o ¨ OA  OC ¸ u F © ¹ o o o o CA u F r u F .

o · § o ¨ OA  CO ¸ u F © ¹

? Sum of the moments of the forces of a given couple

is independent of the point O. ?M

o

o

CA u F | F || CA |sin T | F|u CN, Constant

Any line perpendicular to the plane of a couple is called the axis of the couple and thus it is parallel to its moment vector. Thus, the moment of a couple is a vector perpendicular to the plane of the couple and its magnitude is the product of either force with the perpendicular distance between the lines of forces. Example : Find the moment of the couple formed by the forces 3i + k and –3i – k acting at the points i + 2j – k and 2i – j – 3k respectively. Solution : Moment of the couple = (r1 – r2) u F; r1, r2 being the position vectors of the points of application of the forces F and – F. ? r1 – r2 = (i + 2j – k) – (2i – j + 3k) = – i + 3j – 4k F = 3i + k Moment of the couple = (–i + 3j – 4k) u (3i + k) = j – 9k + 3i – 12j = 3i – 11j – 9k Applications of Vectors in Geometry : (a) Vector equation of a straight line passing through a and parallel to b is r = a + t b, where t is an arbitrary constant. (b) Vector equation of a straight line passing through two points a and b is r = a + t (b – a)

‰

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

r a ˜ buc  cua  a ub b uc  cua  a u b

PM

Example : Let P, Q, R and S be the points on the plane with position vectors 2 i j , 4 i , 3i 3j and 3i 2 j respectively.. The quadrilateral PQRS must be a (A) parallelogram, which is neither a rhombus nor a rectangle (B) square (C) rectangle, but not a square (D) rhombus, but not a square

Solution: M

F i jI GH 2 JK

(A) Mid-points of PR and QS are same, i.e.,

( 3 2  )

( 

M

)

F i jI GH 2 JK Q ( 4 i )

( 2  )

6 

Also,

 3

Ÿ PQ || SR and PS || QR Also,

.

0 ,

Hence, PQRS is a parallelogram, but not a rhombus or rectangle. 127

‰ Very Short Answer Type Questions (1 mark) G G 1. If a 3i  2j  k and b 2i  4j  3k G G Then find |a  2b| G 2. Find a vector in the direction of vector a i  2j that has magnitude 7 units. G G G G 3. Find |a u b| if a 2i  j  k and b 3i  5j  2k . 4.

2.

a

G For what value of O are the vectors a 2i  Oj  k and G b i  2j  3k perpendicular to each other? ‰ Short Answer Type Questions (4 marks) G G 6. Show that the vectors a 3i  2j  k , b i  3j  5k and G c

8.

2i  j  4k form a right angled triangle.. G G If a 5i  j  3k and b i  3j  5k . Then show that the G G G G vectors (a  b) and (a  b) are perpendicular.. G G G For any three vectors a, b and c evaluate te G G G G G G G G G a u (b  c)  b u (c  a )  c u (a  b)

Hints And Solutions We have G a 3i  2j  k G and b 2i  4j  3k G ? 2b 2(2i  4j  3k)

4i  8j  6k

3.

86 128

i  2j 5

7  14  i j 5 5

We have G G a 2i  j  k and b

3i  5j  2k

i

j

k

2 1

1

3 5 2

i( 2  5)  j( 4  3)  k(10  3) 7i  7j  7k G G ? |a u b| ( 7)2  72  7 2 7 3 4.

...(i)

...(ii)

§ 1  2 · 7¨ i j¸ 5 ¹ © 5

G G ? aub

5.

G G G G (2a  3b) u (5a  7b) G G G G G G G G 10a u a  14(a u b)  15(b u a )  21(b u b) G G G G 0  14(a u b)  15(a u b)  0 G G G G G G G G (' a u a b u b 0, a u b b u a ) G G (a u b) G G We have a 2i  Oj  k and b i  2j  3k G G G G ' a A b Ÿ a ˜b 0 Ÿ (2i  Oj  k ) ˜ (i  2j  3k ) 0 Ÿ 2  2O  3 0

(3  4)i  ( 2  8)j  (1  6)k

( 1)2  62  72

i  2j 1 4

G 7a

On substracting equation (ii) from equation (i), we get G G a  2b (3i  2j  k)  (4i  8j  6k )

i  6j  7k G G ? |a  2b| | i  6j  7k|

12  ( 2)2

i  2 J

? The vector having magnitude 7 in the direction of G a is

‰ LONG ANSWER TYPE QUESTIONS (6 marks) 9. Find the area of the triangle having the points A (1, 1, 1), B(1, 2, 3) and C(2, 3, 1) as its vertices. G G G 10. Let a i  j  k, b 4i  2j  3k and c i  2j  k and find a vector of magnitude 6 units which is parallel to the G vector 2aG  b  3cG . 1.

G a G |a|

1  2  i j 5 5

G G G G Evaluate (2a  3b) u (5a  7b)

5.

7.

G The unit vector a in the direction of given vector a is

Ÿ O 5.

5 2

We have G G a 3i  2j  k , b i  3j  5k

G and c

2i  j  k

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

G Now |a|

9  4 1

= G |b|

14

12  ( 3)2  5 2

1  9  25 G |c|

JJJG ? OA i  j  k JJJG JJJG OB i  2j  3k and OC 2i  3j  k JJJG JJJG JJJG ? AB OB  OA i  2j  3k  i  j  k

3 2  ( 2)2  12

0i  j  2k JJJG JJJG JJJG AC OC  OA 2i  3j  k  i  j  k

35

2 2  12  ( 4)2

4  1  16

i  2j  0k

21

G G ? |a|2  |c|2 14  21 35 G G G ? |a|2  |c|2 |b|2

JJJG JJJG ? AB u AC

? vectors form a right angled triangle 7.

We have G G a 5i  j  3k , b i  3j  5k G G ? a  b 5i  j  3k  i  3j  5k

24  8  16 0 G G G G ? a  b and (a  b) are perpendicular vectors.

G G G G G G GG G a u (b  c)  b u (c  a)  c(a  b) G G G G G G G G (a u b)  (a u c)  (b u c)  (b u a ) G G G G  (c u a )  (c u b) G G G G G G G G (a u b)  (c u a )  (b u c)  (a u b) G G G G  (c u b)  (b u c) =0

9.

0 1 2 1 2 0

16  4  1

4i  4j  2k G G G G ? (a  b) ˜ (a  b) (6i  2j  8k ) ˜ (4i  4j  2k )

G G G G (a u b), a u c

k

4i  2j  k JJJG JJJG | AB u AC| ( 4 )2  2 2  ( 1)2

G G a  b 5i  j  3k  i  3j  5k

G G [' b u a

j

i(0  4)  j(0  2)  k(0  1)

6i  2j  8k

8.

i

G G G G G G (c u a ), c u b (b u c)]

? Area of 'ABC =

21

1 JJJG JJJG |AB u AC| 2

1 21 sq. units. 2 10. We have G G a i  j  k , b G G G ? 2a  b  3c

G 4i  2j  3k , c

i  2j  k

2(i  j  k)  (4i  2j  3k)  3(i  2j  k ) 2i  2j  2k  4i  2j  3k  3i  6j  3k i  2j  2k . G Unit vector in the direction of vector 2aG  b  3cG is G G G i  2j  2k 2a  b  3c G G G |2a  b  3c| 12  ( 2)2  22

Let O be the origin of reference. i  2j  2k 9

i  2j  2k 3

1 2 2 i  j k 3 3 3 ? Vector of magnitude 6 units parallel to vector G G G 2a  b  3c

§1 2 2 · 6 ¨ i  j  k ¸ ©3 3 3 ¹ ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

2i  4j  4k

129

JJJG JJJG OB i  2j  3k and OC 2i  3j  k JJJG JJJG JJJG ? AB OB  OA i  2j  3k  i  j  k 0i  j  2k JJJG JJJG JJJG AC OC  OA 2i  3j  k  i  j  k

10. We have G G a i  j  k , b G G G ? 2a  b  3c

G 4i  2j  3k , c

i  2j  k

2(i  j  k)  (4i  2j  3k)  3(i  2j  k ) 2i  2j  2k  4i  2j  3k  3i  6j  3k

i  2j  0k

i  2j  2k . JJJG JJJG ? AB u AC

i

j

k

0 1 2 1 2 0

i(0  4)  j(0  2)  k(0  1) 4i  2j  k JJJG JJJG |AB u AC| ( 4 )2  2 2  ( 1)2 16  4  1

21

? Area of 'ABC =

1 JJJG JJJG |AB u AC| 2

1 21 sq. units. 2

130

G Unit vector in the direction of vector 2aG  b  3cG is G G G i  2j  2k 2a  b  3c G G G |2a  b  3c| 12  ( 2)2  22 i  2j  2k 9

i  2j  2k 3

1 2 2 i  j k 3 3 3 ? Vector of magnitude 6 units parallel to vector G G G 2a  b  3c

§1 2 2 · 6 ¨ i  j  k ¸ ©3 3 3 ¹

2i  4j  4k

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

Directory Of Engineering Colleges Continued from ESR January 2023

ANDHRA PRADESH / TELANGANA Nova College of Engineering & Technology President : Dr. M. Vijaya Nirmala Ibrahimpatnam, Krishna District ANDHRA PRADESH Mob : 7675969675 Email : [email protected] Website : www.enggvja.nova.edu.in Osmania University Vice Chancellor : Prof. S. Ramachandram Registrar : Prof. Ch. Gopal Reddy Main Rd, Amberpet, Hyderabad-500007 ANDHRA PRADESH / TELANGANA Tel : 040-27682284, Fax : 27098003, 2798043 Email : [email protected], [email protected], [email protected] Website : www.osmania.ac.in Pragati Engineering College Principal :Dr. K. Satyanarayana 1-378, ADB Road, Surampalem, Near Kakinada, East Godavari District-533437 ANDHRA PRADESH Tel : 08852-252233, 252234, 07330826667 Fax : 08852-252232 Email : [email protected], [email protected] Website : www.pragati.ac.in Prakasam Engineering College Director: Mr. Srikanth Kancharia Principal : Dr. M. Lakshman Rao O.V. Road, Kandukur, Prakasam Dist., ANDHRA PRADESH Tel : 08598-222288, 221300 Fax : 08598-221300 Email : [email protected] Website : www.prakasamec.com Princeton Institute of Engineering and Technology for Women Principal : Prof. Dr. Venkara SheshaGiridhar Chowdaryguda Village, Ghatkesar Mandal, Medchal-Malkajgiri, Tel : 08415 - 200325, 040-27037328, 7989857165 Mob. : 9394544566, 9908224092,9940772472 Email : [email protected], [email protected], [email protected] Website : www.petw.com Pydah College of Engineering & Technology Chairman : Sri Pydah Krishna Prasad, Kakinada - Yanam Road Patavala, Tallarevu Mandal East Godavari-533461, ANDHRA PRADESH

Tel : 0884 2315345 Mb : 7382456539 Email : [email protected] Website : www.pydah.edu.in QIS College of Engineering & Technology Principal : Mr. Dhulipalla Venkata Rao Vengamukkapalem, Ongole-523272 ANDHRA PRADESH Tel : 08592-281023, 222830, 09246419542 Fax : 08592-284524 Email : [email protected], [email protected] Website : www.qiscet.edu.in Quba College of Engineering & Technology Principal : Dr. M.Z.I. Sajid Venkatachalam, Kanpur Bit-I, Nellore-524320, ANDHRA PRADESH Tel : 0861-2383531, 2383019 Fax : 0861-2383532 Email : [email protected], [email protected] Website : www.qubacollge.in Rajeev Gandhi Memorial College of Engineering and Technology Principal : Dr. T. Jaya Chandra Prasad National Highway : 18, Nerawada ‘X’ Roads, Nandyal, Kurnool Dist-518501 ANDHRA PRADESH / TELANGANA Tel : 08514-275203, 04 Fax : 08514-275123 Email : [email protected], [email protected] Website : www.rgmct.edu.in R.V.R. & J.C. College of Engineering Principal : Dr. K. Srinivasu Chandramoulipuram, Chowdavaram, Guntur-522019 ANDHRA PRADESH / TELANGANA Tel : 94910 73317,94910 73318 Email : [email protected], [email protected] Website : rvrjcce.ac.in Raghu Engineering College Director : Sri. S.V.S.S. Ramachandra Raju Dakamarri, Bheemunipatnam Mandal, Visakhapatnam-531162 ANDHRA PRADESH / TELANGANA Tel : 08922-248001, 002 Email : [email protected] Website : www.raghuenggcollege.com Raja Mahendra College of Engineering Chairman : Dr. P. Mahender Reddy Cherlapatelguda, Kalsa, Ibrahimpatnam, R.R. District-501506 TELANGANA / ANDHRA PRADESH Tel : 08414-202109, 9963777785 Email : [email protected] Website : www.rajamahendra.org

ENGINEERING SUCCESS REVIEW, FEBRUARY 2023

Ramappa Engineering College Principal : Dr. G. Subash Chander Shyampet Jagir, Hunter Road, Hanamkonda, Warangal-506001 TELEGANA / ANDHRA PRADESH Tel : 9396579563,9100000362 Fax : 0870-2524111 Email : [email protected] Website : www.ramappa.edu.in Rao & Naidu Engineering College Principal : K. Ravi Kumar NH-5, Near South By-pass Road Junction, Ongole-523001, Prakasam District ANDHRA PRADESH/ TELANGANA Tel : 9246770581, 9246770601 Fax : 08592-226628 Email : [email protected] Website : www.rnec.org R.K. College of Engineering Principal : Dr. K. Rama Krishnaiah Kethanakonda, Ibrahimpatnam (M), Vijayawada ANDHRA PRADESH / TELANGANA Tel : 9515159721, 9701543399 08659-282956, 66, 67, 89 Email : [email protected] Website : www.rkce.in Royal Institute of Technology & Science Vice Chairman : Mhd. R.S. Arif Rizwan Damergidda (V), Chevella (M), R. R. Dist.-501503 ANDHRA PRADESH/ TELANGANA Tel : 040-23313151, Mob. : 08790393627, 9515045798, 9849999121, Fax : 23313161 Email : [email protected] Website : www.rits.ac.in Administrative office : 11-4-641/ A (New Block), Lane Adjacent to Chief Engineers Office, Opp. PTI Building, AC Guards, Hyderabad-500004 ANDHRA PRADESH / TELANGANA Tel : 040-2331347, Mob. : 998922212 Fax : 040-23313161 Website :www.rits.ac.in S R K R Engineering College Sagi Rama Krishnam Raju Engineering College (SRKREC) Principal : Dr. M Jagapathi Raju Chinna Amiram, Bhimavaram-534204 ANDHRA PRADESH Tel : 08816-223332 (Five Lines) Mb: 09848823332, 09848381818 Fax : 08816-224516, 229377 Email : [email protected] Website : www.srkrec.edu.in 131

Sri Venkatesa Perumal College of Engineering & Technology, (SVPCET) Principal : Dr. T.Sunilkumar Reddy RVS Nagar, Chinnaraja Kuppum, K.N. Road, Chittoor (Dist.), Puttur-517583, ANDHRA PRADESH Tel : 9390505457, Fax : 0877-2230199 Email : [email protected] Website : www.svpcet.org S.R. Engineering College Chairman : Sri A. Varadha Reddy Ananthasagar (V), Hasanparthy (M), Warangal-506371, TELANGANA Tel : 0870-2818377, Fax : 0870-2818456 Email : [email protected], Website : www.srecwarangal.ac.in Sri Chaitanya Technical Campus Principal : Dr. S.A. Muzeer,M.E, Ph.D. SHERIGUDA(V), IBRAHIMPATNAM (M), RANGA REDDY DISTRICT, TELANGANA Tel : 08414 – 223222/23 Email : [email protected], [email protected] Website :www.srichaitanya.ac.in Sana Engineering College Principal : Dr. Baddeti Syam Vijayawada Road, Kodad – 508206, Suryapet Dist. Tel : 08683 - 255764 Mb : 9010310400, 09010092224, 09989933786 Email :[email protected] Website : www.sanaeng.in Sankethika Vidya Parishad Engineering College Principal: Dr. M. Ramjee Behind Cricket Stadium, PM Palem, Vizag-530041, ANDHRA PRADESH Tel : 0891-2781375, 2795522 Email : [email protected] Website : www.svpec.edu.in Sasi Institute of Technology & Engineering Director :Dr. K. Bhanu Prasad Tadepalligudem, Near Aerodrome, West Godavari Dist.-534101 ANDHRA PRADESH Tel : 08818-275500, 11, 22, 33 Fax : 08818-244628 Email : [email protected] Website : www.sasi.ac.in Scient Institute of Technology (SIT) Principal : Dr. G.Anil Kumar 3-2-848/9-11, Opp. Hotel Mahaveer, Kachiguda Station Road, Kachiguda, Hyderabad, Telangana - 500027 Tel : 040-65594666 Email : [email protected], [email protected], [email protected] Website : www.scient.ac.in Shaaz College of Engineering & Technology Director: Dr. Duggirala Srinivasa Rao Himayat Nagar Village, Moinabad Mandal, 132

Ranga Reddy Dist.-500075 ANDHRA PRADESH Tel : 9848308313, Fax : 08413-235082 Email : [email protected] Website : www.shaaz.org Shadan College of Engineering & Technology Director :Prof. S A Muneem Peeran Cheru, Himayat Sagar Road, Hyderabad-500086 ANDHRA PRADESH / TELANGANA Tel : 040-29880841 Email : [email protected] Website : www.scet.in Shadan Women’s College of Engineering & Technology Principal : Dr. K. Palani 6-2-980, Raj Bhavan Road, Taj Enclave, Khairatabad, Hyderabad-500004 ANDHRA PRADESH / TELANGANA Tel : 040-23305552, 3, 4, 040 - 23305546 Email : [email protected], [email protected] Website :www.swcet.in Shahjehan Group of Colleges Chairman : Mr. Ahmed Uddin Sy.No 232 & 239 Nyalate Village, Ramanaguda Road, Chevella, R.R. Dist. Hyderabad ANDHRA PRADESH / TELANGANA Tel : 08417-203622, Mob. : 09912253890 Email : [email protected], [email protected] Website : www.sjcolleges.com Shree Kavitha Engineering College Principal : Dr. Koteswara Rao Seelam Karepally (V/M), Khammam (Dist.), TELANGANA-507122 Tel : 08745-246008 Fax : 08745-246017 Email : [email protected], [email protected] Website : www.skec.ac.in

Siddhartha Institute of Engineering & Technology Principal :Dr. Koorapati Eshwara Prasad Vinobha Nagar, Ibrahimpatnam, R.R.district, Hyderabad-501506 ANDHRA PRADESH / TELANGANA Tel : 07893044445 Mb : 09966779182, 7893969789, 09177607659 Email : [email protected], [email protected] Website : www.siddhartha.ac.in Sreenidhi Educational Group (Sreenidhi Institute of Science and Technology) Chairman : Dr. K.T. Mahi Yamnampet, Ghatkesar Mandal, Ranga Reddy District, Hyderabad-501301 TELANGANA Tel : 040-27631236, 27640395 Fax : 040-27640394 Email : [email protected] Website : www.sreenidhi.edu.in Sreenivasa Institute of Technology and Management Studies (SITAMS) Chairman: Sri K. Ranganadham Dr.D.K. Audishsavulu Marg, (Bengaluru-Tirupathi Bye-pass Road), Murukambattu, Chittoor-517127 ANDHRA PRADESH Tel : 08572-246298, 08572-246299 Fax : 08572-246297 Email : [email protected] Website : www.sitams.org Sri Chundi Ranganayakulu Engineering College Principal : Dr. P. Srinivasa Rao Chilakaluripet, Guntur Dist.-522619 ANDHRA PRADESH Tel : 08647-256243, 253412 Fax : 08647-258120 Email : [email protected], [email protected] Website : www.crenggcollege.com Sridevi Women’s Engineering College Principal : Dr. B.L. Malleswari V.N. Pally, Near Gandipet, R.R. Dist.-500075 ANDHRA PRADESH Tel : 08413-234333, 234056 Mb : 09959425729 Email : [email protected], [email protected] Website :www.swec.ac.in

Shree Vishnu Engineering College Chairman : Sri K.V. Vishnu Raju Vishnupur, West Godavari District, Bhimavaram-534202 ANDHRA PRADESH Tel : 08816-250864 Email : [email protected], [email protected] Website : www.svecw.edu.in Shri Vidyaniketan Engineering College Principal : Dr. B.M. Satish Sree Sainath Nagar, A. Rangampet, Near Tirupathi, Chandragiri Mandal, Tirupati, Chittoor District-517102 ANDHRA PRADESH Tel : 0877-2504888 Mb: 09160999977 Email : [email protected] Website : www.vidyanikethan.edu

Sri Indu College Of Engineering & Technology Principal : Dr. G. Suresh Facing Main Road, Sheriguda, Ibrahimpatnam, R.R. Dist.-501510 ANDHRA PRADESH Tel : 08414-202085, 9347067999 Email : [email protected], [email protected] Website : www.sriindu.ac.in

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Sri Krishna Devaraya College of Engineering & Technology Principal : Dr. R. Ramachandra NH-205, Anantpur-Tirupati hwy, Anantpuram-515003, ANDHRA PRADESH Tel : 08554-255714 Mb : 09492473714 Email : [email protected], [email protected] Website : www.skucet.ac.in Sri Sarada Institute of Science & Technology Director/President : Dr. P. Narasimha Reddy Anantharam Village, Bhongir (Mandal), Nalgonda District-508116 ANDHRA PRADESH Tel : 040-27175372, 27844287, 9849160806, 9848023623 Fax : 040-40123287 Website : www.ssist.ueuo.com Sri Sarathi Institute of Engineering & Technology Principal : Mr. S. Sridhar 2-7- 10 A, Nuzvid, Krishna Dist., Gandhi Nagar-521201, ANDHRA PRADESH Tel : 08656-233711, 234001 Fax : 235711 Email : [email protected] Website : www.ssiet.edu.in Sri Vasavi College of Engineering Principal :Dr. Guduru VNSR Ratnakara Rao Pedatadepalli, Tadepalligudem-534101 West Godavari Dist. ANDHRA PRADESH Tel : 08818-284355, 284344 Fax : 08818-284322 Email : [email protected] Website : www.srivasaviengg.ac.in Sri Venkateswara College of Engineering & Technology Principal : Dr. Matam Mohan Babu R.V.S. Nagar, Tirupati Road, Chittoor-517127, ANDHRA PRADESH Tel : 07729999158 Email : [email protected] Website : www.svcetedu.org Sri Venkateswara University College of Engineering Principal : Prof. G N Pradeep Kumar Tirupati-517502, ANDHRA PRADESH Tel : 0877-2289561, 077299 99158 Email : [email protected] Website : www.svuce.edu.in

Pochampally(M), Yadadri Bhuvanagiri(Dt), Greater Hyderabad-508284 ANDHRA PRADESH Tel : 84980-93080 Email : [email protected] Website : www.smecd.com St. Ann’s College of Engineering & Technology Principal : Dr. M. Venu Gopala Rao Nayunipalli, Challareddipalem, Vetapalem, Chirala, Prakasam Dist.-523187 ANDHRA PRADESH Tel : 08594-246100, 247500 Fax : 08594-247300 Email : [email protected], [email protected] Website : www.sacet.ac.in St. John’s College of Engineering & Technology Principal Dr. V. Veeranna Yerrakota, Yemmiganoor, Kurnool-518360 ANDHRA PRADESH Tel : 9394228866, Fax : 08512-228566 Email : [email protected] Website : www.sjcet.ac.in St. Martin’s Engineering College Principal: Dr. P. Santosh Kumar Patra Sy. No. 98 & 100, Dhulapally, Near Kompally, Secunderabad - 500100 TELANGANA Tel :7997267788 Email : [email protected], [email protected] Website : www.smec.ac.in St. Stanley College of Engineering & Technology Principal : Dr. Satya Prakash Lanka H.No. 5-78 to 82, B-1-80 & 5-9-81 Chapel Road, Abids, Hyderabad 500001 ANDHRA PRADESH / TELANGANA Tel : 040-23234880 Email : [email protected], [email protected] Website : www.stanley.edu.in St. Theressa Institute of Engineering & Technology Principal : Dr. V. Joshua Jaya Prasad Garividi (Cheepurupally), Vizayanagaram District ANDHRA PRADESH Tel : 08952-281062, 281061 Email : [email protected] Website : www.avanthi.edu.in

SSJ Engineering College Principal : Dr. M. Ashok Vattinagulapally, Gandipet, Hyderabad-75 ANDHRA PRADESH / TELANGANA Tel : 7675855790, 9010301699 Email : [email protected] Website : www.ssjec.com

Swarna Bharathi Institute of Science & Technology Principal : Dr. Pannala Krishna Murthy Pakabanda Street, Khammam-507002 ANDHRA PRADESH Tel : 8712712331, 8712112331 Fax : 08742-247788 Email : [email protected], [email protected] Website : www.sbit.ac.in

St Marys Engineering College Managing Director: Dr. M. Satyanarayana Rao Near Ramoji Film City, Behind Mount Opera, Deshmukhi(V),

Swarnandhra College of Engineering & Technology Principal : Dr.S.Suresh Kumar Seetharampuram, Narsapur-534280 ANDHRA PRADESH

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Tel : 9346610099, 7989106066 Fax : 08814-240638 Email : [email protected], [email protected] Website : www.swarnandhra.ac.in Thandra Paparaya Institute of Science & Technology Principal : Dr. V. Joshua Jaya Prasad Komatipalli, Bobbili-535558, Vizaianagaram ANDHRA PRADESH Tel : 08944-257501, 257502, 9494150999 Fax : 08944-257503 Email : [email protected], [email protected] Website : www.tpist.org Tirumala Engineering College Principal : Dr. G. Manikandan Bogaram (V), Keesara (M), Medical District-501301 TELANGANA Ph. : 040-30249560, 65, 67, 68 Email : [email protected], [email protected] TKR College of Engineering & Technology Principal : Dr. D.V. Ravi Shankar Survey No. 8/A Medbowli, Meerpet, Saroor Nagar, Hyderabad-500079 ANDHRA PRADESH / TELANGANA Tel : 044-6534 7536, 6558 7536, 9849477550, 9603999946, 9949665436 Fax : 040-24094567 Email : [email protected], [email protected] Website : www.tkrcet.ac.in TRR College of Engineering Principal : Prof. Dr. K. Srinivasa Rao T.R.R. Nagar, Inole Village, Dist. Medak, Patancheru-502301 ANDHRA PRADESH Tel : 08455-288586 Email : [email protected] Website : www.trrinstitutions.edu.in University College of Engineering Principal : Prof. Sriram Venkatesh Osmania University, Hyderabad-500007 ANDHRA PRADESH / TELANGANA Tel : 040-27682384, 27098254, 09440834065, 09848519860 Email : [email protected], [email protected] [email protected] Website : www.uceou.edu V.R. Siddhartha Engineering College Principal : Dr. A.V. Ratna Prasad Kanuru, Krishna Dt, Vijayawada-520007 ANDHRA PRADESH Tel : 0866-2582333, 2584930 Fax : 0866-2582672 Email : [email protected], [email protected], [email protected], [email protected], [email protected] Website : www.vrsiddhartha.ac.in 133

Vaagdevi College of Engineering Chairman : Mr. C. Janga Reddy Bollikunta, Warangal-506005 TELANGANA Tel : 0870-2865182, 83, 86 Mob. : 8886526969, 8886536969, 9849746969 Email : [email protected], [email protected] Website : www.vaagdevieng.ac.in Vardhaman College of Engineering Principal : Dr. S. Sai Satyanarayana Reddy Chairman : Dr. T. Vijendra Reddy Kacharam, Hyderabad, Ranga Reddy (Dt.), Shamshabad-501218 ANDHRA PRADESH / TELANGANA Tel : 08413-253335, 253201, 64582470 Fax : 08413-253482 Email : [email protected], [email protected] Website : www.vardhaman.org Vasavi Engineering College Principal : Dr. S.V. Ramana 9-5-81, Ibrahimbagh, Hyderabad-500031 ANDHRA PRADESH / TELANGANA Tel : 040-23146003, 23146000 Fax : 23146090, 23146080 Email : [email protected], [email protected] Website : www.vce.ac.in Vathsalya Institute of Technology Principal : Dr. E. Janardhana Rao Anantharam Village, Bhongiri Mandal, Nalgonda Dist.-508116 TELANGANA Tel : 08685-245715, 245748, 245749 Email : [email protected], [email protected] Website : www.vistengg.ac.in Vidya Bharathi Institute of Technology Principal : Dr. B. Satyanarayana Pembarthi (V), Warangal Dist., Jangaon-506167, TELANGANA Tel : 08716-288146, 288289, 9948295183 Fax : 08716-288288 Email : [email protected], [email protected] Website : www.vbit.co.in Vidya Jyoti Institute of Technology & Science Director : Dr. P. Venugopal Reddy Aziz Nagar Gate, CB Post, Chilkur Road, Hyderabad-500075 ANDHRA PRADESH / TELANGANA Tel : 08413-235300, 235399, 9849014284 Email : [email protected], [email protected], [email protected] Website : www.vjit.ac.in

VIF College of Engineering & Technology Chairman : Dr. Syed Ainuddin 786, Himayath Nagar, Gandipet ‘X’ Road, Moinabad Mandal, Ranga Reddy Dist.-500075, TELANGANA Tel : 08413-235329, 235045 Fax : 08413-235328 Email : [email protected], [email protected], [email protected] Website : www.vifcollege.org Vignan’s Foundation for Science, Technology & Research (Deemed to be University) Chairman : Dr. L. Rathaiah Vice-Chancellor : Dr. M.Y.S. Prasad Vadlamudi (Chebrolu Mandal) Guntur Dist.-522213 ANDHRA PRADESH Tel : 0863-2344700, 2344701 Fax : 0863-2344707 Email : [email protected] Website : www.vignan.ac.in Vignan Institute of Technology & Science Principal : Dr. G. Durga Sukumar Vignan Hills, Deshmukhi Village, Pochampally Mandal, Nalgonda District-508218 TELANGANA Tel : 08685-226128, 9866399776/861, 9866399861, 9866399776 Fax : 08685-226128 Email : [email protected], [email protected] [email protected] [email protected] Website : www.vignanits.ac.in

Director : Wg. Cdr. I.P.C. Reddy Visvodaya Campus, Udayagiri Road, Kavali, Nellore (Dist.)-524201 ANDHRA PRADESH Tel : 086262-243930, 242422, 240056 Email : [email protected], [email protected] Website : www.vitskavali.in Viswanandha Institute of Technology and Management Principal : Prof. Dr. Matha Govinda Raju Chairman : Sri V. Narasimham Village : Mindivani Palem, Post Office : Sontyam Mandal : Anandapuram, Visakhapatnam-531173 ANDHRA PRADESH Tel : 9440425849, 9949638777, 988560877 Fax : 0891-2561088/2539007 Email : [email protected], [email protected], [email protected] Website : www.vitamenggcollege.com Vivekananda Institute of Technology & Science Chairman : Dr. M. Satchidananda Rao Principal : Dr. D. Ramana Reddy Opp. Housing Board, Bypass Road, Karimnagar-505001, TELANGANA Tel : 0878-2970887, 0878-2950900 9121162057, 9177200699 9866569121, 9866172952 Fax : 0878-2279181 Email : [email protected], [email protected], [email protected] Website : www.vits.ac.in

Vijay Institute of Technology & Science Principal : Prof. Rajani Ganta Markal, Mandal : Sadashivnagar, Kamareddy Revenue Division, Nizamabad-503224, TELANGANA Tel : 08468-326484, 326869, 248764, 248114, 227508, Fax : 08462-227507 Email : [email protected], [email protected] Website : www.vitsedu.in Vijaya Rural Engineering College Principal : Dr. P. Sampath Rao Das Nagar, Makloor, Rochis Valley, Manikbhandar, Nizamabad Dist.-503003, TELANGANA Tel : 08462-280157, 75, Mob. : 9848884300 Email : [email protected], [email protected] Website : www.vrecedu.org Visvodaya Institute of Technology and Science Chairman : Mr. D. Vidyadhar Kumar Reddy

VNR Vignana Jyothi Institute of Engineering & Technology Chairman : Dr. D. Nageswara Rao Principal : Dr. C.D. Naidu Bachupally, Nizampet (SO), Hyderabad-500090, ANDHRA PRADESH Tel : 040-23042758, 59, 60 Fax : 040-3042761 Email : [email protected] Website : www.vnrvjiet.ac.in VRS &YRN College of Technology President : Sri V. Venkata Rama Chandra Vasu Principal : Dr. Y. Venkata Hanumantha Rao Vadarevu Road, Chirala, Prakasam Dist.-523157 ANDHRA PRADESH Tel : 08594-239525, 9849266760 9849266760, 8594239524 Fax : 08594-230435 Email : [email protected], [email protected], [email protected] Website : www.vycet.org (To be continued)

Printed and Published by Surendra Kumar Sachdeva on behalf of Competition Review Pvt. Ltd. and Printed at Paras Offset Pvt. Ltd., Plot No.118-F, Sector-56, Phase-IV, Kundli Indl. Estate, Dist. Sonepat (Haryana) and published from 604, Prabhat Kiran, Rajendra Place, New Delhi-110125. Editor Surendra Kumar Sachdeva. Engineering Success Review is a monthly magazine published from Delhi. All disputes are subject to Delhi jurisdiction only. Disputes, if any will be settled under the jurisdiction of Delhi by sole arbitrator to be appointed by Competition Review Pvt. Ltd. Question Papers being published are based on Memory Retention Contest. Despite every effort on the part of the publisher to ascertain the veracity of the information, if some error or omission creeps in inadvertently, neither the publisher nor any employee of the magazine shall be held responsible. RNI No. DELENG/2009/28848 DL(C) - 01/1351/2021-23 134

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