VIT - VITEEE 2025
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 7th April | NO Further Extensions!
JEE Main 2025 January 29 Shift 2 Question paper with solutions will be updated on this page after the conduction of the exam. We will provide the JEE Main Jan 29 Question paper with solutions for shift 2 based on students’ reactions. JEE Main Jan 29 shift 2 paper has 75 questions that are divided into Physics, Chemistry, and Mathematics sections. The difficulty level for each section may be different. Check this page after the exam for a detailed JEE Main Jan 29 shift 2 Question paper with solutions.
Also Read: JEE Main 2025 Jan 29 Shift 1 Question Paper | JEE Main 2025 Jan 29 Shift 1 Answer Key | JEE Main 2025 Jan 29 Shift 2 Answer Key
JEE Main Session 2 Questions: April 3: Shift 1 | Shift-2 | April 2: Shift 1 | Shift-2 | Overall Analysis
JEE Main 2025: Mock Tests | PYQs | Rank Predictor | College Predictor | Admit Card Link
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The National Testing Agency (NTA) will conduct the Joint Entrance Examination (JEE) Main 2025 April session for the remaining days on April 4, 7, 8 and 9. Candidates can download their exam hall ticket through the official website at jeemain.nta.nic.in.
Total Duration |
3 hours (180 minutes) |
Total Questions |
75 questions |
Maximum Marks |
300 marks |
Medium of Exam |
Available in 13 languages |
Mode of Exam |
Computer Based Test (CBT) |
Question Type |
MCQs and Numerical Value Based |
The JEE Main 2025 January 29 Shift 1 Question Paper and Solutions will be available here post-exam. This will act as a useful resource for students taking further exams, providing a chance to examine the questions and answers thoroughly. It will ultimately help them to familiarize themselves fully with the types of questions and answers, to get an account of the format of the exam for this accurate result and realize the subject for further study on this topic and, finally, to perfect their exam-passing game.
29 Jan shift 2
Q.1 Which of the following form most stable carbocation ?
$1 \quad(\mathrm{Ph})_3 \mathrm{C}-\mathrm{Br}$
$2 \quad \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Br}$
$3 \quad \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3$
$4 \quad \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$
Q.2 Number of $\sigma$ and $\pi$ bonds respectively in hex-1-en-4-yne are
$1 \quad13, 3$
$2 \quad14, 3$
$3 \quad 3,14$
$4 \quad 14,13$
Q.3. If the letters of the word "KANPUR" are arranged in dictionary, then the $440^{\text {th }}$ word is
1 PRKAUN
2 PRKUAN
3 PRKNAU
4 PRKUNA
Q.4 0.41 g of $\mathrm{BaSO}_4$ is obtained from 0.2 g of organic compound in Carius method. What is the percentage of sulphur present in organic compound?
Q.5 A solenoid of radius 10 cm carrying current 0.29 A and having total 200 turns. If the magnetic field inside the solenoid is $2.9 \times 10^{-4} \mathrm{~T}$. Find length of solenoid.
$1 \quad 6 \pi \mathrm{~cm}$
$2 \quad 8 \pi \mathrm{~cm}$
$3 \quad 4.5 \mathrm{~cm}$
$4 \quad 16 \mathrm{~cm}$
Q.6 An equiconvex lens is cut in two ways as shown. If the focal length of the parts are as mentioned in the diagram. Find $\frac{L_1}{L_i}$
$1 \quad 2$
$2 \quad 4$
$3 \quad \frac{1}{2}$
$4 \quad \frac{1}{4}$
Q.7 If $3^{107}$ is divided by 23 , then remainder is
Q.8 Which element in group $\mathbf{1 5}$ has lowest Ionisation Energy
$1 \quad \mathrm{Bi}$
$2 \quad P$
$3 \quad As$
$4 \quad \mathrm{Sb}$
Q.9 Three identical particles, each of mass $m$ move under the influence of mutual attraction forces. Initially, the are on the vertices of an equipotential triangle of side ' $a$ ' and have equal speed $v$ directed towards the adjacent particles as shown. The net angular momentum about the centre just before collision is
$1 \quad \frac{3 m v a}{2}$
$2 \quad \frac{2}{3} m v a$
$3 \quad \frac{\sqrt{3}}{2} m v a$
$4 \quad \frac{2}{\sqrt{3}} m v a$
Q.10 Which of the following formsthe most stable carbocation?
(A) $(\mathrm{Ph})_3 \mathrm{C}-\mathrm{Br}$
B) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Br}$
(C) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3$
b) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$
Q.11 $\mathrm{XeF}_3^{\ominus}$ shape ?
Q.12 Let $a_{i j}=(\sqrt{2})^{i+j}, A=\left[a_{i j}\right]_{3 \times 0}$. If sum of third row of $A^2$ is $\alpha+\beta \sqrt{2}$, then $\alpha+\beta$ is
Q.13 A solenoid of radius 10 cm carrying current 0.29 A and having a total of 200 turns. If the magnetic field inside the solenoid is $2.9 \times 10^{-4} \mathrm{~T}$. Find the length of the solenoid.
(a) $8 \pi \mathrm{~cm}$
(b) $6 \pi \mathrm{~cm}$
(c) 16 cm
(d) 4.5 cm
Q.14 $$
\begin{aligned}
&\text { Match the physical quantities with their corresponding dimensions }\\
&\begin{array}{|l|l|l|l|}
\hline & \text { Column-I } & & \text { Column-II } \\
\hline \text { (A) } & \text { Young's modulus } & \text { (i) } & {\left[\mathrm{AL}^2\right]} \\
\hline \text { (B) } & \text { Magnetic moment } & \text { (ii) } & {\left[\mathrm{ML}^2 T^{-2} \mathrm{~A}^{-1}\right]} \\
\hline \text { (C) } & \text { Magnetic flux } & \text { (iii) } & {\left[\mathrm{AL}^{-1}\right]} \\
\hline \text { (D) } & \text { Magnetic Intensity } & \text { (iv) } & {\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]} \\
\hline
\end{array}
\end{aligned}
$$
$\begin{array}{ll}1 & \text { A-(iii), B-(i), C-(ii), D-(iii) } \\ 2 & \text { A-(iv), B-(ii), C-(i), D-(iii) } \\ 3 & \text { A-(iii), B-(i), C-(ii), D-(iv) } \\ 4 & \text { A-(iii), B-(ii), C-(i), D-(iv) }\end{array}$
Q.15 River is flowing with velocity $9 \mathrm{~km} / \mathrm{hr}$. velocity of boat with respect to river is $27 \mathrm{~km} / \mathrm{hr}$ in the direction of flow of river. Person standing over this boat throw a ball vertically upward with respect to himself with velocity $10 \mathrm{~m} / \mathrm{s}$. Determine the horizontal displacement or range (in cm ) till ball come to same horizontal level from where it was projected.
Q.16 Two particles of same mass are performing SHM vertically with two different springs of spring constants $K_1$ and $K_2$. If the amplitude of both is the same. Find the ratio of the maximum speed of two particles.
Q.17 Q. A physical quantity $Q$ is given as $Q=\frac{a b^4}{c d}$, if the percentage error $\{0, b, c$ and $d$ are $2 \%, 1 \%, 2 \%$ and $1 \%$, the $\%$ error in $Q$ will be
$ 1\quad 5 \%$
$2 \quad15 \%$
$3 \quad9 \%$
$4 \quad2 \%$
Q.18 Q. Let $f(x)=\int_0^x t\left(t^2-3 t+20\right) d t, x \in(1,3)$ and range of $f(x)$ is $(\alpha, \beta)$, then $\alpha+\beta$ is equal to
$1 \quad \frac{185}{4}$
$2 \quad \frac{185}{2}$
$3 \quad \frac{185}{3}$
$4 \quad \frac{37}{4}$
Q.19 The value of the limit
$$
\lim _{x \rightarrow 0}(\operatorname{cosec} x)\left(\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}\right) \text { is }
$$
$1 \quad 0$
$2 \quad 1$
$3 \quad \frac{1}{2 \sqrt{5}}$
$4 \quad-\frac{1}{2 \sqrt{5}}$
Students can download the JEE main Question paper with solutions of 22, 23, and 24 January 2025.
As per the observation of the previous days' analysis, the difficulty level of the JEE Main question paper is overall moderate level. In Chemistry, students would do better with NCERT-only, for all chemistry questions are right out of the NCERT. For Physics, they must practice formula questions, and Mathematics is quite lengthy and time-consuming.
You can check the JEE Main 2025 question paper analysis according to the feedback we got for the last few days’ shifts. Here you can check it:
Mathematics:
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 7th April | NO Further Extensions!
Ranked #42 among Engineering colleges in India by NIRF | Highest CTC 50 LPA , 100% Placements
The paper was manageable for those who had prepared well. The topics covered were a mix of easy questions and moderate-level problems.
Difficult questions were fewer, but those who prepared thoroughly could still tackle them effectively.
Topics like Quadratic Equations, Sequences and Series, and Matrices were frequently asked.
Complex numbers and the Binomial Theorem also had a few questions.
Differentiation and Integration were prominent, with several questions on the Application of Derivatives and Definite Integrals.
Questions on Straight Lines and Conic Sections (especially Circles and Parabolas) were commons
Trigonometric identities and equations were included, though not extensively.
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Most questions were easy to moderate, with only a small portion being tough.
Physics was manageable for most students, especially if they had a solid grasp of basic concepts. Moderate-level questions were expected, and easier questions made the section more approachable.
Topics like Newton's Laws of Motion, Work, Energy & Power, and Rotational Motion appeared frequently.
Gravitation and Kinetic Theory of Gases were also important areas.
Coulomb’s Law, Capacitance, Current Electricity, and Magnetic Effects of Current were commonly tested.
Electromagnetic Induction had a couple of questions as well.
Photoelectric Effect, Atomic Structure, and Radioactivity were recurring topics in the exam.
Chemistry:
Chemistry was slightly more challenging compared to Physics and Mathematics.
The difficult portion was notable, requiring thorough preparation, especially in areas like Physical Chemistry and Inorganic Chemistry. Students should focus more on Chemistry in future shifts as it poses a challenge for many candidates.
Topics like Chemical Kinetics, Thermodynamics, and Mole Concepts were highly featured.
Equilibrium and Electrochemistry appeared in moderate questions.
Reaction Mechanisms, especially Aromatic Substitution and Functional Groups, were frequently tested.
Hydrocarbons and Polymers appeared as well.
Yes, you can appear in multiple shifts, and the best score will be considered.
Solve easier questions first, allocate specific time to each section, and focus on high-scoring topics.
JEE Main answer key is released.
Application Date:03 September,2024 - 16 April,2025
Application Date:14 November,2024 - 25 April,2025
For Class 12th, if you're preparing for JEE while ensuring a strong NCERT foundation, here are some of the best guides with excellent theory explanations:
Physics
1. Concepts of Physics by H.C. Verma (Vol 1 & 2) – Best for conceptual clarity and problem-solving.
2. Understanding Physics by D.C. Pandey (Arihant Series) – Topic-wise coverage with great theory and examples.
3. NCERT Exemplar & Physics NCERT (for basics) – Essential for board exams and JEE base concepts.
Chemistry
1. NCERT Chemistry (Both Physical & Organic) + NCERT Exemplar – Must for theory and direct JEE questions.
2. Modern’s ABC of Chemistry – Great for detailed theory aligned with NCERT.
3. Physical Chemistry by P. Bahadur – For numerical problems and JEE preparation.
4. Organic Chemistry by Morrison & Boyd (Optional) – For deeper conceptual clarity.
5. Inorganic Chemistry by J.D. Lee (adapted version for JEE) – Good for conc
eptual understanding.
Mathematics
1. NCERT Mathematics – Base book for boards and JEE fundamentals.
2. Cengage Mathematics Series – Provides topic-wise theory and a variety of problems.
3. Arihant’s Skills in Mathematics by Amit M Agarwal – Good for understanding advanced concepts.
4. Problems in Calculus of One Variable by I.A. Maron – For strong calculus preparation.
If you're looking for one-stop solutions for JEE + NCERT theory, Arihant’s and Cengage’s books provide excellent detailed explanations.
Hi Aspirant, NTA will conduct JEE entrance exam, here is then link to download the JEE Main 2025 April 2 Shift answer key, visit our site https://engineering.careers360.com/articles/jee-main-april-2-shift-1-answer-key-2025
Hello,
Congratulations on achieving a 90.3 percentile in JEE Main 2025! As an SC category female candidate , you have a favorable position in the admission process for NITs. Let's assess your chances for Computer Science and Engineering (CSE) at NIT Goa and explore other potential NITs.
Understanding Your Rank:
Percentile to Rank Conversion: With a 90.3 percentile , your approximate Common Rank List (CRL) rank would be around 97,000 out of 1,000,000 candidates. For the SC category , this translates to an approximate category rank of 6,000 to 7,000 .
NIT Goa CSE Admission Chances:
Previous Year Cutoffs: In 2024, the SC category closing rank for CSE at NIT Goa was 2,445 .
Analysis: Given your estimated SC category rank of around 6,000 to 7,000 , securing a seat in CSE at NIT Goa may be challenging, as the previous year's cutoff was significantly lower.
Alternative NITs to Consider:
Based on previous years' data, here are some NITs where the SC category closing ranks for CSE were higher, indicating a better chance of admission:
NIT Agartala: Closing rank around 3,682 .
NIT Puducherry: Closing rank around 3,160 .
NIT Sikkim: Closing rank around 3,803 . https://www.careers360.com/
NIT Arunachal Pradesh: Closing rank around 4,295 .
Recommendations:
Participate in JoSAA Counseling: Engage actively in the JoSAA counseling process, listing multiple NITs and branches to maximize your admission chances. https://www.careers360.com/
Consider Related Branches: If CSE is highly competitive, explore related branches like Information Technology (IT) or Electronics and Communication Engineering (ECE) , which may have slightly higher cutoffs.
Stay Updated: Regularly check the official JoSAA website and individual NIT portals for the most current cutoff information and counseling updates.
Please note that cutoff ranks can vary each year based on factors like the number of applicants, seat availability, and exam difficulty. Therefore, while historical data provides a reference, it's essential to stay informed about the current year's trends.
Best wishes for your counseling process and future academic endeavors!
Hello,
Regarding your query there is no single "best" between JEE and NEET the choice depends on individuals understanding, interests, and career aspirations. While JEE s for aspiring engineers NEET is for aspiring doctors.
No, with an NTA score of 52.24 in the General category, you are unlikely to qualify for JEE Advanced as this score is significantly below the required cutoff for the general category. To be eligible for JEE Advanced, you typically need an NTA score well above 90 percentile.
Key points to consider:
Low percentile:
A 52.24 NTA score translates to a very low percentile, which means you rank much lower compared to other general category candidates.
JEE Advanced cutoff:
The minimum required percentile to qualify for JEE Advanced in the general category is usually much higher than 52.24.
What you can do:
Focus on improving your score:
If you really want to attempt JEE Advanced, consider preparing for the next JEE Mains attempt with a goal of achieving a much higher NTA score.
Explore other options:
While you may not qualify for top IITs with this score, you can still explore admission to other good engineering colleges that accept JEE Mains scores based on your percentile
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