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 Mains 2025 January 24 Shift 1 Question Paper with Solutions - JEE Main is a highly anticipated exam for engineering hopefuls, determining access to some of India's leading technical institutions. The January 24 Shift 1 paper of JEE Main 2025, once can provide students with the opportunity to analyze the style of questions, identify trends in topic selection, and their strategies for future attempts. As the exam of other shifts evolve, candidates can expect a thought-provoking and dynamic test that challenges their understanding and application of core concepts.JEE Main answer key is released.
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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.
Aspect |
Details |
Time Duration |
3 hours (180 minutes) |
Number of Questions |
75 questions |
Highest Marks |
300 marks |
Exam Language |
Available in 13 languages |
Mode of Exam |
Computer-Based Test (CBT) |
Question Type |
Multiple Choice Questions (MCQs) and Numerical Value-Based (NVBQs) |
JEE Main 2025 January 22 Shift 1 Question Paper Overview |
The exam consisted of 75 questions spread across three subjects: |
Physics |
25 questions |
Chemistry |
25 questions |
Mathematics |
25 questions |
Q.1. Which of the following is the strongest oxidising agent?
$1 \quad \mathrm{Eu}^{2+}$
$2 \mathrm{Ce}^{2+}$
$3 \mathrm{Ce}^{4+}$
$4 \quad E u^{4+}$
Q.2 If the $5^{\text {th }}, 6^{\text {th }}$ and $7^{\text {th }}$ term of the binomial expansion of $\left(1+x^2\right)^{n+4}$ are in A.P. Then the greatest binomial coefficient in the expansion of $\left(1+x^2\right)^{n+4}$ is
Q.3 The difference in melting point and boiling point of oxygen and sulphur can be explain
1 Electronegativity
2 Electron gain enthalpy
3 Atomicity
4 Ionisation energy
Q.4 The number of 3 digit numbers which is divisible by 2 and 3 but not divisible by 4 and 9 .
Q.5 If $I=I_A \sin \omega t+I_B \cos \omega t$, then find rms value of current
$1 \quad I_{r m s}=I_A+I_B$
$2 \quad I_{\mathrm{rms}}=\sqrt{I_A^2+I_B^2}$
$3 \quad I_{\mathrm{rms}}=\sqrt{\frac{l_{L_2^2+l_B^2}^2}{2}}$
$4 \quad I_{\mathrm{rms}}=\frac{1}{2} \sqrt{I_A^2+I_B^2}$
Q.6. Ribose present in DNA is
(A) It is a pentose sugar
(B) Present in pyranose form
(C) $\alpha$ anomeric carbon is present
(D) Present in D configuration
(E) It is reducing sugar in free form
Choose the correct statements :
1 A, C & E only
2 A, D & E only
3 A, B, C, D & E
4 A & E only
Q.7Difference of M.P & B.P b/w Oxygen & Sulpher can be explained by.
a) EN
b) $\Delta H_{\text {eg }}$
c) Atomicity
d) $I E$
Q.8 If A is $3 \times 3$ matrix such that $\operatorname{det}(\mathrm{A})=2$. Then $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))))$
Q.9 If 10 mol CO and 10 mol of $\mathrm{Fe}_3 \mathrm{O}_4$ reacts according to $\mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{CO} \rightarrow 4 \mathrm{CO}_2+3 \mathrm{Fe}$ What is the Weight of Fe produce?
Q.10 A drop of radius R is split into 27 drops of equal radius, the work done is 10 J . If the same big drop is split into 64 equal drops, the work done is
(A) 10
(B) 15
(C) 20
(D) $75 / 4$
Q.11 What is relative shift of focal length of a lens when optical power is increased from 0.1 D to 2.5 D?
$1 \quad 24 / 25$
2 $213 / 10$
$3 \quad 21 / 25$
$4 \quad 11 / 10$
Q.12 If the $\mathrm{K}_{\text {sp }}$ of $\mathrm{Cr}(\mathrm{OH})_3$ is $1.6 \times 10^{-30} \mathrm{M}^4$. The molar solubility of salt in water is $1.56 \times 10^{-x}$, then value of $x$ is
Q.13 Q) Radius of curvature plano convex lens is 2 cm and refractive index is $\mathbf{1 . 5}$ has focal length $f_1$ in air and $f_{\mathbf{2}}$ in a medium of refractive index 1.2. Calculate $f_1 / f_2$
Q.14 The electric flux through the shaded area of square plate of side $a$ due to point charge placed at distance of $a / 2$ from it as shown in figure, is $\frac{N Q}{48 \varepsilon_0}$. Then $N$ is
Q.15 A solid cylinder of mass $m$ and radius $r$ is released from rest at the top of a rough inclined plane making an angle of $45^{\circ}$ with the horizontal. Assuming the cylinder rolls without slipping, find the accleration of the axis of the cylinder.
(A) $\frac{g}{2}$
(B) $\frac{g}{\sqrt{2}}$
(C) $\frac{2 g}{3 \sqrt{2}}$
(D) $\frac{g}{3 \sqrt{2}}$
Q.16 In a square loop of side length $\frac{1}{\sqrt{2}} \mathrm{~m}, \mathrm{a}$ current of 5 A is flowing. Find magnetic field at its centre in $(\mu \mathrm{T})$.
Q.17 If $\vec{a}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k}, \vec{b}=3 \hat{\imath}+\hat{\jmath}-\hat{k}$ and $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$. Also $\vec{a} \cdot \vec{c}=5$ and $\vec{c}$ is perpendicular to $\vec{b}$. Then $|\vec{c}|$ is
Q.18 $\lim _{x \rightarrow 0} \operatorname{cosec} x$. $\left(\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}\right)$
1) 0
2) 1
3) $\frac{1}{2 \sqrt{5}}$
4) $-\frac{1}{2 \sqrt{5}}$
Q.19 Process is non-spontaneous at freezing point but spontaneous at boiling point, find $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$.
a) Both are Positive
b) Both are Negative
c) $\Delta S$ Positive, $\Delta H$ Negative
d) $\Delta \mathrm{S}$ Negative, $\Delta \mathrm{H}$ Positive
Q.20 Which of the following is most reactive towards nucleophilic addition reaction.
1 Para-nitro benzaldehyde
2 Para-methyl benzaldehyde
3 Benzaldehyde
4 Acetophenone
The JEE Mains 2025 24 Jan Shift 1 Question Paper and Solutions is available on this page as the exam is over. These materials will be highly useful for candidates appearing in later shifts, allowing them to practice and analyze the questions, helping in better preparation for their own exam sessions.
Practicing the previous year’s question papers is an essential part of JEE Main preparation. It allows you to familiarize yourself with the exam structure, question types, and the level of difficulty you may encounter in JEE Main 2025. Analyzing past papers helps build confidence, refine time management skills, and gives you insight into important topics. Be sure to access these valuable resources to strengthen your preparation strategy for the upcoming exam.
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 JEE Main 2025 exam is expected to focus heavily on Calculus and Algebra in Mathematics, particularly Differentiation, Integration, and Quadratic Equations. In Physics, Mechanics, including Laws of Motion and work energy, will be key, along with Electromagnetism and Modern Physics. For Chemistry, Organic Chemistry will be dominant, with significant portions on Physical Chemistry and Inorganic Chemistry. Focus on these key areas for effective preparation.
The difficulty level of the JEE Main 2025 January 24 Shift 1 exam is expected to be moderate, with Mathematics being the most challenging section, followed by Physics and Chemistry. Practicing with the JEE Mains 24 Jan shift 1 question paper and the JEE Mains 2025 question paper pdf shift 1 helps improve accuracy.
The JEE Mains 2025 question paper pdf 24 January and the JEE Mains 2025 24 Jan shift 1 paper highlight commonly asked questions.
The Physics section is predicted to focus on Mechanics (laws of motion, rotational dynamics), Electromagnetism (electric and magnetic fields, circuits), and Modern Physics (photoelectric effect, atomic structure).
The Chemistry section is expected to be the easiest, with a major focus on Organic Chemistry, followed by relatively simpler questions from Physical Chemistry and Inorganic Chemistry.
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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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!
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