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CSIR NET Final Answer Key 2024 Released, Response Sheet PDF

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CSIR NET Final Answer Key 2024 for Joint CSIR-UGC NET July 2024 has been released by the National Testing Agency (NTA) on 12th September 2024. The final answer key with correct answers is now available on the official website https://csirnet.nta.ac.in/. All those candidates who appeared in the written examination can check and download the answer key from the direct link provided in the article.

CSIR NET Answer Key 2024 Released

National Testing Agency (NTA) has successfully conducted CSIR NET Exam 2024 on 25th, 26th and 27th July 2024 at a total of 348 exam centers located in 187 cities across the country. According to officials, 2,25,335 candidates have appeared in the eligibility test. CSIR NET Final Answer Key 2024 has been published online, which helps candidates to evaluate their answers and estimate their scores.

CSIR UGC NET Answer Key 2024

The Council of Scientific and Industrial Research National Eligibility Test (CSIR NET) 2024 is conducted online and has a duration of 3 hours. It consists of five subjects: Earth, Atmospheric, Ocean, and Planetary Sciences; Chemical Sciences; Life Sciences; Mathematical Sciences; and Physical Sciences. The exam includes a negative marking scheme where 25% marks are deducted for each wrong answer.

CSIR NET Final Answer Key 2024 – Key Highlights
Organization National Testing Agency(NTA)
Name of the Exam CSIR UGC NET July 2024
post name Junior Fellowship (JRF) and Lectureship/Assistant Professor
Social class answer key
Situation Issued
CSIR NET Exam Date 2024 25, 26 and 27 July 2024
CSIR NET Final Answer Key 2024 12 September 2024
CSIR NET Result 2024 By the end of August 2024
official website https://csirnet.nta.ac.in/

CSIR UGC NET Final Answer Key 2024 Link

The download link for CSIR NET Final Answer Key 2024 has been made available on 12th September 2024 at https://csirnet.nta.ac.in/. The answer key will not be sent directly to the candidates’ email address. The answer key has been published in PDF format, no login details are required to check the answer key. Once the official download link is activated, we have provided the direct link here to access the CSIR UGC NET Answer Key 2024, which covers both Junior Fellowship (JRF) and Lectureship/Assistant Professor categories.

Steps to Download CSIR UGC NET Final Answer Key 2024

Candidates don’t need to be worried about downloading the CSIR UGC NET Final Answer Key 2024. Just follow the below steps to view and download the answer key.

  1. Open your web browser and visit the official website of National Testing Agency (NTA) https://csirnet.nta.ac.in/.
  2. Once you are on the homepage of the NTA website, look for the section called “Latest News.”
  3. In the Latest News section, find a link that specifically mentions “Joint CSIR UGC NET July 2024 – Click here to download Provisional Answer Key”.
  4. Click on the link you found in the latest news.
  5. This will usually open a PDF file in a new tab. The PDF will contain the questions and their final answer keys for the CSIR UGC NET July 2024 exam.
  6. To save the PDF for future reference, click the download icon (usually a downward-pointing arrow) in your browser.
  7. Open the downloaded PDF and carefully compare your answers with the provided provisional answer key.

CSIR UGC NET Marking Scheme

The CSIR UGC NET exam consists of multiple-choice questions. The exam is for a total of 200 marks for each paper. This exam has different subjects and the number of questions varies. However, there is a penalty for wrong answers: 25% of the marks will be deducted for each wrong question, and this penalty can be up to 33% in some sections. This system is designed to discourage guessing and ensure that only accurate answers contribute to your final score.

CSIR UGC NET Marking Scheme
Subject and parts Marks allotted to each question Negative Marking
Chemical Science
A + 2 – 0.5
B + 2 – 0.5
C + 4 – 1
Earth Sciences
A + 2 – 0.5
B + 2 – 0.5
C + 4 – 1.32
Life Sciences
A + 2 – 0.5
B + 2 – 0.5
C + 4 – 1
Mathematical Sciences
A + 2 – 0.5
B + 3 – 0.75
C + 4.75 0
Physics
A + 2 – 0.5
B + 3.5 – 0.875
C + 5 – 1.75
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Hello friends, I am Ashok Nayak, the Author & Founder of this website blog, I have completed my post-graduation (M.sc mathematics) in 2022 from Madhya Pradesh. I enjoy learning and teaching things related to new education and technology. I request you to keep supporting us like this and we will keep providing new information for you. #We Support DIGITAL INDIA.

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