Updated 19 Oct 2026

JEE Main 2025, 28 Jan Shift 1 question paper with solutions

JEE Main 2025, 28 Jan Shift 1 was held on 28 January 2025. The bank holds 68 of its questions (Physics 24, Chemistry 21, Mathematics 23). A sample with answer keys and worked solutions is below, or sit the complete shift paper in the official NTA CBT interface (180 minutes, 300 marks, +4 / −1). Real PYQ papers are never filtered by your covered syllabus.

Sit this paper in the exam screen All JEE Main 2025 shifts Marks vs percentile table

2025 Exam session year

300 Maximum marks (+4 / −1)

180 min Official duration

68 Questions in bank for this shift

Try questions from JEE Main 2025, 28 Jan Shift 1

Q1PhysicsJEE Main 2025, 28 Jan Shift 1Matter waves

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A proton of mass ‘m

_{p}

’ has same energy as that of a photon of wavelength ‘λ’. If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.

Answer key (C) & worked solution

Correct answer: Option C

E is missing in the question but considering E as energy, the solution will be

E_{p h o t o n} = \frac{h c}{λ} = E; E_{p r o t o n} = \frac{1}{2} m_{p} v^{2} = E

\frac{λ _{p r o t o n}}{λ _{p h o t o n}} = \frac{h / p}{h c / E} = \frac{h / 2 m _{p} E}{h c / E}

= \frac{E}{c 2 m _{p} E}

\frac{λ _{p r o t o n}}{λ _{p h o t o n}} = \frac{1}{c} \frac{E}{2 m _{p}}

Q2PhysicsJEE Main 2025, 28 Jan Shift 1Intensity of EM waves

Due to presence of an em-wave whose electric component is given by E = 100 sin(ωt–kx) NC–1, a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

25 sin (ω t - k x) N C^{- 1}

200 sin (ω t - k x) N C^{- 1}

400 sin (ω t - k x) N C^{- 1}

50 sin (ω t - k x) N C^{- 1}

Answer key (B) & worked solution

Correct answer: Option B

Energy density

= \frac{1}{2} ε_{0} E^{2} \times c

Energy

= \frac{1}{2} ε_{0} E^{2} \times c \times v o l u m e

(Energy)

_{1}

= (Energy)

_{2}

(Given)

= \frac{1}{2} ε_{0} E_{1}^{2} c π R_{1}^{2} \times L_{1} = \frac{1}{2} ε_{0} E_{2}^{2} c π R_{2}^{2} \times L_{2}

E_{1}^{2} R_{1}^{2} = E_{2}^{2} R_{2}^{2}

E_{1} R_{1} = E_{2} R_{2}

100 \times R_{1} = E_{2} \times \frac{R _{1}}{2}

E_{2} = 200 N / C

Q3ChemistryJEE Main 2025, 28 Jan Shift 1Carbohydrates

Given below are two statements : Statement I: D-glucose pentaacetate reacts with 2, 4-dinitrophenylhydrazine. Statement II: Starch, on heating with concentrated sulfuric acid at 100°C and 2-3 atmosphere pressure produces glucose. In the light of the above statements, choose the correct answer from the options given below

ABoth Statement I and Statement II are false

BStatement I is false but Statement II is true

CStatement I is true but Statement II is false

DBoth Statement I and Statement II are true

Answer key (B) & worked solution

Correct answer: Option B

Statement I is false but Statement II is true

Q4ChemistryJEE Main 2025, 28 Jan Shift 1Order and Molecularity of Reactions

[A]

_{0}

/molL–1t1/2/min0.1002000.025100 For a given reaction R → P, t1/2 is related to [A]

_{0}

as given in table : Given : log2 = 0.30 Which of the following is true ? A. The order of the reaction is

\frac{1}{2}

. B. If [A]

_{0}

is 1M, then t1/2 is

20010

min C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M. D. t1/2 is 800 min for [A]

_{0}

= 1.6 M Choose the correct answer from the options given below :

AA and C only

BA and B only

CA, B and D only

DC and D only

Answer key (C) & worked solution

Correct answer: Option C

t_{1/2} \propto \frac{1}{A _{0}^{n - 1}}

\frac{( t _{1/2} ) _{1}}{( t _{1/2} ) _{1}} = \frac{( A _{0} ) _{0}^{n - 1}}{( A _{0} ) _{1}^{n - 1}}

\frac{200}{100} = (\frac{0.025}{0.100})^{n - 1}

2 = (\frac{1}{4})^{n - 1}

n - 1 = \frac{1}{2}

n = \frac{1}{2}

(order)

\Rightarrow t_{1/2} \propto A_{0}

\frac{200}{t _{1/2}} = \frac{( 0.1 ) ^{1/2}}{( 1 ) ^{1/2}}

, When A

_{0}

= 1M

t_{1/2} = 20010

min * Ist order kinetics have t1/2 independent of their concentration. So upon changing the concentration t1/2 should not change for first order reaction.

\frac{200}{t _{1/2}} = \frac{( 0.1 ) ^{1/2}}{( 1.6 ) ^{1/2}}

, When A

_{0}

= 1.6 M t1/2 = 800 min

Q5MathematicsJEE Main 2025, 28 Jan Shift 1Concept of DC and DR

Let A(x, y, z) be a point in xy-plane, which is equidistant from three points (0, 3, 2), (2, 0, 3) and (0, 0, 1). Let B = (1, 4, –1) and C = (2, 0, –2). Then among the statements (S1) : ∆ABC is an isosceles right angled triangle and (S2) : the area of ∆ABC is

\frac{9 2}{2} .

Aboth are true

Bonly (S1) is true

Conly (S2) is true

Dboth are false

Answer key (B) & worked solution

Correct answer: Option B

A(x,y,z) Let P (0,3,2), Q(2,0,3), R(0,0,1) AP = AQ = AR x

^{2}

+ (y–3)

^{2}

+ (z – 2)

^{2}

= (x – 2)

^{2}

+ y

^{2}

+ (z – 3)

^{2}

= x

^{2}

+ y

^{2}

+ (z – 1)

^{2}

In xy plane z = 0 So, x

^{2}

– 4x + 4 + y

^{2}

+ 9 = x

^{2}

+ y

^{2}

+ 1 x = 3 9 + y

^{2}

– 6y + 9 + 4 = x

^{2}

+ y

^{2}

+ 1 So, A(3,2,0) also B(1,4,–1) & C(2,0,–2) Now

A B = 4 + 4 + 1 = 3

A C = 1 + 4 + 4 = 3

B C = 1 + 16 + 1 = 18

AB = AC isosceles ∆ & AB

^{2}

+ AC

^{2}

= BC

^{2}

right angle ∆ Area of ∆ ABC =

\frac{1}{2}

× base. Height

\frac{1}{2} \times 3 \times 3 = \frac{9}{2}

So only S

_{1}

is true

Q6MathematicsJEE Main 2025, 28 Jan Shift 1Basics of modulus functions

The sum, of the squares of all the roots of the equation x

^{2}

+ |2x – 3| – 4 = 0, is:

3 (3 - 2)

6 (3 - 2)

6 (2 - 2)

3 (2 - 2)

Answer key (C) & worked solution

Correct answer: Option C

x^{2} + ∣2 x - 3∣ - 4 = 0

Case I :

x ⩾ \frac{3}{2}

x^{2} + 2 x - 3 - 4 = 0

x^{2} + 2 x - 7 = 0

x = 22 - 1

Case II :

x < \frac{3}{2}

x^{2} + 3 - 2 x - 4 = 0

x^{2} - 2 x - 1 = 0

x = 1 - 2

Sum of squares =

(22 - 1)^{2} + (1 - 2)^{2}

= 8 - 42 + 1 + 1 - 22 + 2

= 6 (2 - 2)

∴ (3)