JEE Main 2026 April 2 Shift 2 Physics Question Paper with Solution PDF
JEE Main 2026 April 2 Shift 2 Physics Question Paper with Solution PDF is available — the exam ran on April 2, 2026 from 3:00 PM to 6:00 PM . Candidates who sat for the B.E./B.Tech Physics paper should download the PDF and solutions to check answers and analyse performance.
The Physics paper had 25 questions worth a total of 100 marks . Each correct answer carries +4 marks and each incorrect answer attracts -1 mark . The test was held in computer-based mode and offered in 13 languages .
JEE Main 2026 April 2 Shift 2 Physics Question Paper with Solution PDF — Quick facts
| Detail | Fact |
|---|---|
| Exam date | April 2, 2026 |
| Shift timing | 15:00–18:00 (3:00 PM–6:00 PM) |
| Total questions (Physics) | 25 |
| Total marks (Physics) | 100 |
| Marking scheme | +4 for correct, -1 for wrong |
| Exam mode | Computer-Based Test (CBT) |
| Paper | B.E./B.Tech |
| Languages offered | 13 |
| Question types | MCQs and numerical value questions |
Where to get JEE Main 2026 April 2 Shift 2 Physics Question Paper with Solution PDF
You can download the question paper with solutions from the published PDF accompanying this update. Use the solutions to cross-check your responses and estimate your raw score using the +4/-1 scheme.
Keep in mind the marking rules apply to both MCQs and numerical value questions. If you answered all 25 questions, calculate your raw score as (4 × correct) − (1 × incorrect).
What the available PDF shows: the full set of questions and answer choices, plus answer keys and solution notes for many items. What it does not include: a full question-wise official answer key endorsed by the exam authority, step-by-step workings for every question, or topic-wise weightage and difficulty breakdown.
Use the PDF to do a quick performance check. Do a question-by-question review and flag items you are unsure about for deeper revision.
Quick calculation example
If you get 18 correct and 4 wrong (with 3 unanswered): (18 × 4) − (4 × 1) = 68 marks out of 100.