What minimum score is required in JEE Mains or JEE Advance to crack an IIT or NIT by a PwD category student?

<p>Those who appear for must check the JEE Main 2025 cutoff for general and other categories. The previous year cutoff marks for JEE Main for the General category are 93.24, 79.67 for OBC, 81.33 for EWS-ALL, 0.0018700 for UR-PWD, 60.0923182 for SC-ALL, and 46.6975840 for ST-ALL, </p><p>In 2022, the required marks to clear JEE Main for the General category were 88.41. For General-PwD candidates, it was 0.003. The cutoff for EWS applicants was 63.11, and for OBC-NCL, it was 67.01. SC category students needed 43.08, and ST candidates required 26.78, </p><p>In 2025, to get a BTech seat in the Indian Institutes of Technology (IITs), the open category candidates had to score a minimum of 8.68% in each subject and a total of 30.34% marks in JEE Advanced. On the other hand, candidates in the OBC category had to score 7.8% in each subject and 27.30% overall.</p><p>&nbsp;</p>

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<p>For a PwD (Persons with Disabilities) category student, the minimum score required to crack an IIT or NIT can vary depending on factors like the difficulty of the exam, the number of applicants, and the specific institute's cutoff for that year. However, in general:</p><ul><li><p><strong>IITs</strong>: To qualify for JEE Advanced and secure a spot in one of the IITs, PwD category candidates typically need to score around 45-50% of the total marks in JEE Mains. After that, a good rank in JEE Advanced is crucial, and the cutoff for PwD candidates can range from around 70-150 marks depending on the year and branch preference.</p></li><li><p><strong>NITs</strong>: For NITs, PwD candidates may need a JEE Mains score in the range of 80-100 marks to be eligible for admission, depending on the NIT and the course.</p></li></ul><p>&nbsp;</p>

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