CSIR UGC NET 2019 (Dec) – Form Correction, Exam Date, Pattern, Syllabus

CSIR UGC NET 2019 – National Testing Agency has started the CSIR UGC NET 2019 application form correction on October 18, 2019 at csirnet.nta.nic.in. Candidates can also remove image discrepancy. It holds CSIR UGC NET Dec 2019 in online mode, on December 15, 2019. CSIR Natioanl Eligibility Test is a national level exam, conducted to determine eligibility for appointment of Lecturer and Junior Research Fellowship. It is a 200 marks exam and is conducted for Life Science, Physical Science, Chemical Science, Mathematical Science, Earth, Atmospheric and Planetary Sciences. You can check CSIR UGC NET 2019 exam dates, registration process, eligibility, pattern, and more details from this page.

Latest: NTA has opened the CSIR NET Form Correction window. Click here to edit and make changes.

CSIR UGC NET 2019

CSIR UGC NET is held twice a year, in the month of June and December. NTA has also released dates for CSIR UGC NET June 2020 exam and you can check it at the end of this page. To know all the dates for the December exam, refer to the table below:

CSIR UGC NET 2019	Important Dates
CSIR NET application form 2019 available from	09 Sep 2019
The closing of online Application	09 Oct 2019 15 Oct 2019
Last date to pay Application fee	10 oct 2019
Application Form Correction from	18 -25 Oct 2019
Issuing of Admit Card	09 Nov 2019
Exam Date	15 Dec 2019
Answer Key releasing date	Dec 2019
Submit objection against question/answer key	Dec 2019
Result declaration date	31 Dec 2019

What is new in CSIR UGC NET December 2019?

The candidates can check below the latest changes in CSIR UGC NET:

• From this session onwards, (National Testing Agency) NTA is responsible to conduct the CSIR UGC NET exam, instead of Council of Scientific and Industrial Research (CSIR).
• The mode of the exam has been changed to Computer Based Test (CBT) from the pen and paper mode (offline).

The candidates can check the official notice by NTA regarding the new changes below:

CSIR UGC NET Application Form 2019

In the application form candidates should enter the valid details. Candidates can use only online mode for doing the registration of CSIR NET Application Form 2019. In the online form, candidates have to fill the Personal, Contact, Educational,  Qualification Details and need to upload the scanned photograph and signature. Application Form has been made available at the official website.

CSIR NET Application Fee 2019

• General category fees: Rs 1000
• OBC NCL fees: Rs 500
• SC / ST / PWD fees: Rs 250

Online Payment – Candidates can pay online through Net Banking/Debit Card/Credit Card/UPI/Paytm Services.

CSIR UGC NET 2019 Eligibility Criteria

Before filling the application form candidates first read the eligibility criteria. The candidates who will found ineligible in any stage of the process those will be not considered for the further process. From below candidates can check the detailed CSIR UGC NET Eligibility Criteria 2019

Nationality: Candidates should hold Indian Citizenship.

Age Limit

• In order to be eligible for JRF NET, candidates must be maximum 28 years old as on 01-01-2019.
• SC/ST/PWD and female applicants will get 5 years age relaxation. OBC candidates will get 3 years age relaxation.
• Lectureship (LS) NET does not have any upper age limit.

Degree: M.Sc or equivalent degree / Integrated BS-MS / BS-4 years / BE / B.Tech / BPharma / MBBS.

Minimum marks in degree: 55% marks for General (UR) and OBC candidates, 50% for SC/ST, Persons with Disability (PWD) candidates.

Those who are enrolled for M.Sc or completed 10+2+3 years of QE as on the last date of the application form are also eligible to apply. They will select option “Result Awaited (RA)”. Condition applicable upon such a candidate would be that he/she must complete the degree with requisite percentage of marks within the validity period of two years to avail the fellowship from the effective date of award. An attestation to this effect, signed by HOD will be submitted.

B.Sc (Hons) or equivalent degree holders or students enrolled in Integrated MS-PhD program with at least 55% marks for General (UR) and OBC candidates; 50% marks for SC/ST, Persons with disability (PWD) candidates.

Can a candidate with a bachelor’s degree appear for CSIR NET?

Yes, bachelor’s degree candidates can also appear for the Joint CSIR-UGC NET exam. They are eligible for CSIR fellowship only after getting registered/enrolled for Ph.D./Integrated Ph.D. program within the validity period of two years.

A candidate with ONLY BACHELOR’S DEGREE will not be eligible for Lectureship.

Eligibility for lectureship of NET qualified candidates

This will be subject to fulfilling the criteria laid down by UGC. Ph.D. degree holders who have passed Master’s degree prior to 19th September 1991, with at least 50% marks are eligible to apply for Lectureship only.

CSIR UGC NET Admit Card 2019

The candidates who submitted the application form on time for those admit card have been issued. The admit card is available at the official website. Admit Card contain the name of candidate, roll number, exam name, center name, father’s name, center code, photograph, signature etc. On the day of the examination, candidates have to carry one valid id proof (aadhar card, pan card, voter id card, passport) and one hard copy of the admit card.

Examination Center Of CSIR UGC NET 2019

Thhe commission has the authority to add, change or dismiss the examination center. The examination will be conducted at 224 cities spread all over India. Candidtes have to select any four cities according to their prefernce for CBT. From below candidates can the Examination Center List Of CSIR UGC NET 2019.

State	City	Code
Arunachal Pradesh	Itanagar/ Naharlagun	AL01
Assam	Dibrugarh	AM01
	Guwahati	AM02
	Jorhat	AM03
	Silchar	AM04
	Tezpur	AM05
Andaman and Nicobar	Port Blair	AN01
Andhra Pradesh	Anantapur	AP01
	Bhimavaram	AP03
	Chirala	AP04
	Chittoor	AP05
	Eluru	AP06
	Guntur	AP07
	Kadapa	AP08
	Kakinada	AP09
	Kurnool	AP10
	Nellore	AP11
	Ongole	AP12
	Rajahmundry	AP13
	Srikakulam	AP14
	Tirupathi	AP16
	Vijayawada	AP17
	Visakhapatnam	AP18
	Vizianagaram	AP19
	Narasaraopet	AP20
	Proddatur	AP21
	Surampalem	AP23
Bihar	Bhagalpur	BR02
	Darbhanga	BR04
	Gaya	BR05
	Muzaffarpur	BR06
	Patna	BR07
	Purnea	BR08
	Arrah	BR09
Chattisgarh	Durg / Bhilai Nagar	CG01
	Bilaspur	CG02
	Raipur	CG03
Chandigarh	Chandigarh Mohali	CH01
Daman & DIU	Daman	DD01
	DIU	DD02
Delhi	Delhi/ New Delhi	DL01
Dadra and Nagar Haveli	Dadra and Nagar Haveli	DN01
Gujarat	Ahmedabad / Gandhinagar	GJ01
	Anand	GJ02
	Bhavnagar	GJ03
	Jamnagar	GJ06
	Junagarh	GJ07
	Mehsana	GJ08
	Rajkot	GJ10
	Surat	GJ11
	Vadodara	GJ12
	Valsad/ Vapi	GJ13
	Himatnagar	GJ14
Goa	Panaji / Madgaon	GO01
Himachal Pradesh	Bilaspur	HP01
	Hamirpur	HP03
	Palampur / Kangra	HP04
	Shimla	HP06
	Solan	HP07
	Mandi	HP08
	Una	HP09
	Kullu	HP10
Haryana	Ambala	HR01
	Faridabad	HR03
	Gurugram	HR04
	Hisar	HR05
	Karnal	HR06
	Kurukshetra	HR07
	Panipat	HR08
	Sonipat	HR09
	Yamuna Nagar	HR10
Jharkhand	Bokaro Steel city	JH01
	Dhanbad	JH02
	Jamshedpur	JH03
	Ranchi	JH04
	Hazaribagh	JH05
Jammu and Kashmir	Baramulla	JK01
	Jammu	JK02
	Samba	JK03
	Srinagar (J & K)	JK04
Karnataka	Belagavi (Belagum)	KK02
	Bellary	KK03
	Bengaluru	KK04
	Bidar	KK05
	Davanagare	KK06
	Gulbarga	KK08
	Hassan	KK09
	Dharwad / Hubli	KK10
	Mangaluru (Mangalore)	KK12
	Mysore	KK14
	Shimoga	KK15
	Tumakuru	KK16
	Manipal / Udupi	KK17
	Mandya	KK18
Kerala	Alappuzha / Chengannur	KL01
	Ernakulam / Angamaly / Moovattupuzha	KL04
	Idukki	KL05
	Kannur	KL06
	Kasaragod	KL07
	Kollam	KL09
	Kottayam	KL11
	Kozhikode	KL12
	Malappuram	KL13
	Palakkad	KL15
	Pathanamthitta	KL16
	Thiruvananthapuram	KL17
	Thrissur	KL18
Lakshadweep	Kavaratti	LD01
Ladakh	Leh	LL01
Meghalaya	Shillong	MG01
Manipur	Imphal	MN01
Madhya Pradesh	Balaghat	MP01
	Betul	MP02
	Bhopal	MP03
	Chhindwara	MP05
	Gwalior	MP06
	Indore	MP07
	Jabalpur	MP08
	Rewa	MP11
	Sagar	MP12
	Satna	MP13
	Ujjain	MP15
Maharasthra	Ahmednagar	MR01
	Akola	MR02
	Amravati	MR03
	Aurangabad	MR04
	Chandrapur	MR09
	Dhule	MR10
	Jalgaon	MR13
	Kolhapur	MR14
	Latur	MR15
	Mumbai / Navi Mumbai	MR16
	Nagpur	MR17
	Nanded	MR18
	Nashik	MR19
	Pune	MR22
	Raigad	MR23
	Ratnagiri	MR24
	Sangli	MR25
	Satara	MR26
	Solapur	MR27
	Thane	MR28
	Wardha	MR29
Mizoram	Aizawl	MZ01
Nagaland	Dimapur	NL01
	Kohima	NL02
Odisha	Balasore	OR02
	Berhampur – Ganjam	OR03
	Bhubaneswar	OR04
	Cuttack	OR05
	Dhenkanal	OR06
	Rourkela	OR08
	Sambalpur	OR09
Punjab	Amritsar	PB01
	Bhatinda	PB02
	Jalandhar	PB04
	Ludhiana	PB05
	Patiala/ Fathegarh Sahib	PB07
	Pathankot	PB08
	Sangrur	PB11
Puducherry	Puducherry	PO01
Rajasthan	Ajmer	RJ01
	Alwar	RJ01
	Bikaner	RJ02
	Jaipur	RJ05
	Jodhpur	RJ06
	Kota	RJ07
	Sikar	RJ08
	Sriganganagar	RJ09
	Udaipur	RJ10
Sikkim	Gangtok	SM01
Tripura	Agartala	TA01
Telangana	Hyderabad / Secunderabad / Ranga reddy	TL01
	Karimnagar	TL02
	Khammam	TL03
	Mahabubnagar	TL04
	Nalgonda	TL05
	Warangal	TL07
Tamil Nadu	Chennai	TN01
	Coimbatore	TN02
	Cuddalore	TN03
	Kanchipuram	TN05
	Kanyakumari / Nagercoil	TN06
	Madurai	TN08
	Namakkal	TN10
	Salem	TN11
	Thanjavur	TN12
	Thoothukudi	TN13
	Tiruchirappalli	TN14
	Tirunelveli	TN15
	Tiruvallur	TN16
	Vellore	TN18
	Virudhunagar	TN20
Uttarakhand	Dehradun	UK01
	Haldwani	UK02
	Haridwar	UK03
	Nainital	UK04
	Pantnagar	UK05
	Roorkee	UK06
Uttar Pradesh	Agra	UP01
	Aligarh	UP02
	Allahabad	UP03
	Bareilly	UP04
	Bulandshahr	UP06
	Faizabad	UP07
	Ghaziabad	UP08
	Gorakhpur	UP09
	Jhansi	UP10
	Kanpur	UP11
	Lucknow	UP12
	Mathura	UP13
	Meerut	UP14
	Moradabad	UP15
	Muzaffarnagar	UP16
	Noida / Greater Noida	UP17
	Sitapur	UP18
	Varanasi	UP19
West Bengal	Asansol	WB01
	Bardhaman	WB02
	Durgapur	WB04
	Hooghly	WB06
	Howrah	WB07
	Kalyani	WB08
	Kolkata	WB10
	Siliguri	WB11

CSIR UGC NET Exam Pattern 2019

Before starting the preparation candidates should know about the exam pattern because this help the candidates to know about the subject name, exam timing, marks, types of questions etc. From below candidates can check the detailed exam pattern of CSIR NET 2019.

• Mode of an exam – Online (Computer Based Test)
• Duration of the exam – 3 hours (180 mins)
• Examination Language – Hindi, EnglishThe selection will be based on the cutoff marks and the marks secured by the candidates in the CSIR UGC NET 2019 Examination.
• Question type – MCQ
• Total No of marks – The test paper carry a maxium of 200 marks for each subject.
• Parts of Exam – The test will consist of three parts (Part A, Part B and Part C)

The subject-wise scheme of examination are as follows:

Pattern for Chemical Science (701):

• Part A: +2 marks for each correct answer and -0.5 for each incorrect answer.
• Part B: +2 marks for each correct answer and -0.5 for each incorrect answer.
• Part C: +4 marks for each correct answer and -1 for each incorrect answer.

No. of Parts	Total no of questions	Max no of Questions to attempt
Part A	20	15
Part B	40	35
Part C	60	25
Total	120	75

Pattern for Earth, Atmospheric, Ocean and Planetary Sciences (702):

• Part A: +2 marks for each correct answer and -0.5 for each incorrect answer.
• Part B: +2 marks for each correct answer and -0.5 for each incorrect answer.
• Part C: +4 marks for each correct answer and -1.32 for each incorrect answer.

No. of Parts	Total no of questions	Max no of Questions to attempt
Part A	20	15
Part B	50	35
Part C	80	25
Total	150	75

Pattern for Life Sciences (703):

• Part A: +2 marks for each correct answer and -0.5 for each incorrect answer.
• Part B: +2 marks for each correct answer and -0.5 for each incorrect answer.
• Part C: +4 marks for each correct answer and -1.32 for each incorrect answer.

No. of Parts	Total no of questions	Max no of Questions to attempt
Part A	20	15
Part B	50	35
Part C	75	25
Total	145	75

Pattern for Mathematical Sciences (704):

• Part A: +2 marks for each correct answer and -0.5 for each incorrect answer.
• Part B: +3 marks for each correct answer and -0.75 for each incorrect answer.
• Part C: +4.75 marks for each correct answer and 0 for each incorrect answer.

No. of Parts	Total no of questions	Max no of Questions to attempt
Part A	20	15
Part B	40	25
Part C	60	20
Total	120	60

Pattern for Physical Sciences (705):

• Part A: +2 marks for each correct answer and -0.5 for each incorrect answer.
• Part B: +3.5 marks for each correct answer and -0.875 for each incorrect answer.
• Part C: +5 marks for each correct answer and -1.25 for each incorrect answer.

No. of Parts	Total no of questions	Max no of Questions to attempt
Part A	20	15
Part B	25	20
Part C	30	20
Total	75	55

Shift and Timings of Examination

Shift	Timings	Duration
Shift 1	09:30 am- 12:30 pm	3 hours (180 mins)
Shift 2	02:30 pm- 5:30 pm

Syllabus Of CSIR UGC NET 2019

From below candidates will get the full detail of CSIR NET Syllabus 2019. Syllabus helps the candidates to know about

• Part A – This is common to all subjects. Questions will be on General Aptitude with emphasis on logical reasoning, graphical analysis, analytical and numerical ability, quantitative comparison, series formation, puzzles.
• Part B – These will be subject-related conventional multiple choice questions.
• Part C – Here there will be higher value questions to test knowledge of scientific concepts and/or application of scientific concepts. To answer such analytical questions, scientific knowledge has to be applied.

Get Here CSIR NET 2019 Complete Syllabus

CSIR UGC NET Answer key 2019

The NTA releases the Provisional Answer Key of the questions on the NTA website, csirnet.nta.nic.in t. Candidtes can use the answer key to cross check the answers and to calculate the probable scores. It also provide an opportunity to the candidates to challenge the Provisional Answer Key of CSIR NET 2019. The Answer Keys is displayed only for two to three days. The Candidates should be given an opportunity to make a challenge online against the Provisional Answer Key on the payment of Rs. 1000/- per question challenged as processing charges.

CSIR UGC NET Result 2019

After releasing of the final CSIR NET Answer Key 2019, the result will be declared at the official website. The result will contain the name and roll number of qualified candidates. The result will be based on the performance of the candidates in the examination and it will be in the form pdf. The candidates whose name will be there in the final result those will finally select for the lectureship or JRF (for which the candidates applied for).

Re-Evaluation/Re-Checking of the result: There shall be no re-evaluation/re-checking of the CSIR UGC NET 2019 result. No correspondence in this regard shall be entertained.

CSIR UGC NET 2020 (June)

The CSIR UGC NET June 2020 exam will also be held by NTA. It will also be held for Lecturer and Junior Research Fellowship. The pattern of CSIR UGC NET 2020 (June) is same as CSIR UGC NET December 2019. CSIR UGC NET 2020 (June) schedule has been relesed by NTA, along with the CSIR UGC NET December 2019 schedule, you can check it from the table provided belowL

CSIR UGC NET 2020 (June)	Dates
Starting of Online registration	16 Mar 2020
Registration ends on	15 Apr 2020
Downloading of Admit Cards	15 May 2020
Date of Examination	21 June 2020
Declartion od Result	05 Jul 2020

Official website: csirnet.nta.nic.in

For more details regarding CSIR UGC NET 2019 (Dec), Download the Information Brochure.

Joint CSIR UGC NET