JEE Main 2025 Shift 1 Memory Based Question : 22 Jan Shift 1 Questions Available NOW!

This article presents a comprehensive list of JEE Main Paper 1 questions from January 22, 2025. Shift 1 questions are accessible now. Utilize this resource to assess your performance and understand the exam’s challenge level. Expect updates on Shift 2 questions shortly.”

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JEE Main 2025 Paper 1: Access Memory-Based Questions

The JEE Main 2025 exam was successfully conducted on January 22, 2025, without any reported security breaches or irregularities at any examination center. As candidates and experts begin to analyze the exam’s difficulty and assess their performance, a common inquiry arises: What specific questions were asked in Paper 1?

Since the JEE Main exam is administered online, obtaining the exact question paper is not feasible. However, a valuable resource for understanding the exam’s nature and content is through memory-based questions. These questions are compiled by collecting recollections of the exam from candidates who appeared for the test.

Explore these memory-based questions for JEE Main 2025 Paper 1 to:

Gain insights into the exam’s difficulty level.
Understand the topics and concepts covered in the exam.
Analyze the types of questions asked (multiple-choice, numerical, etc.).
Compare your approach to the exam with that of other candidates.

By reviewing these questions, you can gain valuable insights into the JEE Main 2025 exam and better prepare for future attempts or for other competitive exams.

Click here to access the memory-based questions for JEE Main 2025 Paper 1.

Mathematics

Two balls are selected at random one by one without replacement from the bag containing 4 white and 6 black balls. If the probability that the first selected ball is black given that the second selected is also black, is m/n where gcd(m,n) =1, then m + n = ?

Let the triangle PQR be the image of the triangle with vertices (1,3), (3,1), (2,4) in the line x + 2y = 2. If the centroid ofΔPQR is the point (α, β), then 15 (α − β) is equation.

From the English alphabets, 5 letters are chosen and are arranged in alphabetical order. The total number of ways in which the middle letter is M.

A coin tossed three times. Let x denote number of times tail follows a head. If μ & σ² denote the mean and variance of x the value of 64(μ+σ²)

The text states that a₁, a₂, a₃, … are positive terms of a geometric progression (GP). It also provides two equations: a₁ * a₅ = 26 and a₂ + a₆ = 28. Find the value of a₃.

Let the foci of the hyperbola be (1,14) and (1,-11). It passes through the point (1,6) then length of its latus rectum is?

The number of non-empty equivalence relations on the set {1, 2, 3} is…

Using the principle values of the inverse trigonometric functions, the sum of maximum and minimum values of 16[(sec-1 x)²+(cosec-1 x )² ]

Let x = x(y) be the solution of the differential equation y² dx + (x – 1/y) dy = 0. If x(1) = 1 then x(1/2) is equal to
1) 3 + e^2 2) 1/2 + e 3) 3 – e 4) -3/2 + e

Physics

Chemistry

Which of the following lanthanide ion has 7 electrons in the outermost shell?
(1) Eu³⁺ (2) Gd³⁺ (3) Eu²⁺ (4) 4d + 2
Solution:
The correct answer is Eu²⁺
No. of linear compounds?
I3−,NO2,O3,OF2,NO2+,BeCl2,N3−,SO3
Solution:
Let’s determine the number of linear compounds from the given list:
I3−I₃⁻ (Triiodide ion): Linear
NO2NO₂ (Nitrogen dioxide): Bent
O3O₃ (Ozone): Bent
OF2OF₂ (Oxygen difluoride): Bent
NO2+NO₂⁺ (Nitronium ion): Linear
BeCl2BeCl₂ (Beryllium chloride): Linear
N3−N₃⁻ (Azide ion): Linear
SO3SO₃ (Sulfur trioxide): Trigonal planar
So, the linear compounds in the list are:
I3−I₃⁻
NO2+NO₂⁺
BeCl2BeCl₂
N3−N₃⁻
The correct answer is 4 linear compounds.
Given the weight of the organic compound: 180 mg and the weight of the AgCl precipitated: 143.5 g. Calculate the estimation of Cl in ———– %
wt of Cl = 35.5 g wt of Ag = 108 g.

Solution:

Given:

Weight of the organic compound: 180 g

Weight of AgCl precipitated: 143.5 g

Molar mass of AgCl: 108 g/mol (Ag) + 35.5 g/mol (Cl) = 143.5 g/mol

Calculate the moles of AgCl:
Moles of AgCl = Weight of AgCl / Molar mass of AgCl
Moles of AgCl = 143.5 g / 143.5 g/mol = 1 mol

Calculate the moles of Cl:
Moles of Cl = Moles of AgCl
Moles of Cl = 1 mol

Calculate the weight of Cl:
Weight of Cl = Moles of Cl × Atomic mass of Cl
Weight of Cl = 1 mol × 35.5 g/mol = 35.5 g

Calculate the percentage of Cl in the organic compound:
Percentage of Cl = (Weight of Cl / Weight of the organic compound) × 100
Percentage of Cl = (35.5 g / 180 g) × 100 = 19.72%
Electrolysis of which compound gives H₂S₂O₈?
a) Electrolysis of Conc. Na₂SO₄
b) Electrolysis of Dil. Na₂SO₄
c) Electrolysis of Conc. H₂SO₄
d) Electrolysis of Dil. H₂SO₄

Solution: – H₂S₂O₈ is known as peroxodisulfuric acid. – Peroxodisulfuric acid can be formed by the electrolysis of concentrated sulfuric acid (H₂SO₄). Correct Answer:

c) Electrolysis of Conc. H₂SO₄

Question
Which of the given electronegativity orders is incorrect?
(1) Mg < Be < B < N
(2) Al < Si < C < N
(3) S < Cl < O < F
(4) Al < Mg < B < N

Solution:

Electronegativity generally increases across a period from left to right and decreases down a group in the periodic table.

The correct electronegativity order for Group 2 elements is Be > Mg, hence option (1) Mg < Be < B < N is correct.

For Group 13 and Group 14 elements, Si > Al and C > Si. Thus, option (2) Al < Si < C < N is correct.

For Group 16 and Group 17 elements, the electronegativity order is F > O > Cl > S. Therefore, option (3) S < Cl < O < F is incorrect.

For Group 13, Group 2, and Group 15 elements, the order is B > Al > Mg. Hence, option (4) Al < Mg < B < N is correct.

Correct Answer:

(3) S < Cl < O < F
Question:
Which of the following acids is a vitamin?
a) Saccharin acid
b) Aspartic acid
c) Adipic acid
d) Ascorbic acid

Correct Answer:

d) Ascorbic acid

To determine the IUPAC name of the compound in the image, let’s analyze its structure step by step:

Identify the longest carbon chain containing the double bond:
The main chain has four carbon atoms (but-), with a double bond between the second and third carbons (-2-ene).

Number the chain:
Numbering starts from the side closest to the double bond to minimize the position number of the double bond. So, the numbering goes:
CH₃ (C1) – CH(COOH) (C2) – CH=C (C3) – CH(COOCH₃) (C4).

Identify substituents:

A carboxylic acid group (-COOH) is attached to C2.
A methyl ester group (-COOCH₃) is attached to C4.

Name the compound:

The main functional group is the carboxylic acid (-COOH), which takes priority in naming, so the suffix is “-oic acid.”
The ester (-COOCH₃) is a substituent and is named as “methoxycarbonyl.”
The double bond at position 2 makes it “-2-ene.”

IUPAC Name:
2-(Methoxycarbonyl)-2-butenoic acid

Question 8

Reactant: Nitrobenzene (C₆H₅NO₂)

Step i: Br₂ / AcOH (Bromination in the presence of acetic acid)
Nitrobenzene undergoes electrophilic substitution. The nitro group (-NO₂) is a meta-directing deactivator, so bromination occurs at the meta position relative to the nitro group.
Product A: 3-Bromonitrobenzene (m-Bromonitrobenzene)
Step ii: SnCl₂ / HCl (Reduction)
The nitro group (-NO₂) is reduced to an amine group (-NH₂) using SnCl₂ in acidic medium.
Product B: 3-Bromoaniline (m-Bromoaniline)
Step iii: NaNO₂ / HCl (Diazotization)
The amine group (-NH₂) is converted into a diazonium salt (-N₂⁺Cl⁻) when treated with NaNO₂ and HCl at low temperature (0–5°C).
Intermediate: 3-Bromobenzenediazonium chloride
Step iv: EtOH (Ethanol)
In the presence of ethanol, the diazonium salt undergoes decomposition, producing a phenol derivative. Ethanol acts as a reducing agent, replacing the diazonium group with a hydrogen atom.
Final Product: 3-Bromophenol

Final Products:

A: 3-Bromonitrobenzene
B: 3-Bromophenol

Question 9
3 M of NaCl whose density is 1.25 g/mL. Find its Molality.

Given:

Molarity (M) = 3 M
Density (ρ) = 1.25 g/mL = 1250 g/L
Molar mass of NaCl = 58.5 g/mol

Steps:

Mass of 1 Liter of Solution:
Mass = Density × Volume
Mass = 1.25 g/mL × 1000 mL = 1250 g
Mass of NaCl in 1 Liter of Solution:
Moles of NaCl = Molarity × Volume
Moles = 3 mol
Mass of NaCl = Moles × Molar mass
Mass of NaCl = 3 × 58.5 = 175.5 g
Mass of Solvent (Water):
Mass of solvent = Total mass – Mass of solute
Mass of water = 1250 – 175.5 = 1074.5 g = 1.0745 kg
Molality (m):
Molality = Moles of solute / Mass of solvent (kg)
Molality = 3 / 1.0745 ≈ 2.79 mol/kg

Final Answer:
Molality = 2.79 mol/kg

Check: YouTube Channel for solution video for Maths

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JEE Main 2025 Paper 1: Access Memory-Based Questions

Explore these memory-based questions for JEE Main 2025 Paper 1 to:

Gain insights into the exam’s difficulty level.
Understand the topics and concepts covered in the exam.
Analyze the types of questions asked (multiple-choice, numerical, etc.).
Compare your approach to the exam with that of other candidates.

By reviewing these questions, you can gain valuable insights into the JEE Main 2025 exam and better prepare for future attempts or for other competitive exams.

Click here to access the memory-based questions for JEE Main 2025 Paper 1.