Updated 19 Oct 2026

JEE Main 2025, 29 Jan Shift 1 question paper with solutions

JEE Main 2025, 29 Jan Shift 1 was held on 29 January 2025. The bank holds 62 of its questions (Physics 22, Chemistry 19, Mathematics 21). 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

62 Questions in bank for this shift

Try questions from JEE Main 2025, 29 Jan Shift 1

Q1PhysicsJEE Main 2025, 29 Jan Shift 1Vertical circular motion involving strings and similar cases

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A body of mass 'm' connected to a massless and unstretchable string goes in verticle circle of radius 'R' under gravity g. The other end of the string is fixed at the center of circle. If velocity at top of circular path is

n g R

, where, n ≥ 1, then ratio of kinetic energy of the body at bottom to that at top of the circle is

Answer key (D) & worked solution

Correct answer: Option D

V_{T o p} = n^{2} g R

V_{B o tt o m} = n^{2} g R + 4 g R

Ratio

= \frac{n ^{2} + 4}{n ^{2}}

Q2PhysicsJEE Main 2025, 29 Jan Shift 1Photoelectric effect and Einstein's photoelectric equation

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance. Reason (R) : A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation. In the light of the above statements, choose the most appropriate answer from the options given below:

A(A) is false but (R) is true.

B(A) is true but (R) is false.

CBoth (A) and (R) are true and (R) is the correct explanation of (A).

DBoth (A) and (R) are true but (R) is not the correct explanation of (A).

Answer key (D) & worked solution

Correct answer: Option D

(A) : True (B) : True but not correct explanation

Q3ChemistryJEE Main 2025, 29 Jan Shift 1Electrovalent or Ionic Bond

An element ‘E’ has the ionisation enthalpy value of 374 kJ mol–1. ‘E’ reacts with elements A, B, C and D with electron gain enthalpy values of –328, –349, –325 and –295 kJ mol–1, respectively. The correct order of the products EA, EB, EC and ED in terms of ionic character is :

AEB > EA > EC > ED

BED > EC > EA > EB

CEA > EB > EC > ED

DED > EC > EB > EA

Answer key (A) & worked solution

Correct answer: Option A

Difference between I.E. & E.G.E increases, ionic character increases.

Q4ChemistryJEE Main 2025, 29 Jan Shift 1Stability of Coordination Compounds in Solution

The correct increasing order of stability of the complexes based on value is : (I) [Mn(CN)

_{6}

]3–, (II) [Co(CN)

_{6}

]4– (III) [Fe(CN)

_{6}

]4–, (IV) [Fe(CN)

_{6}

]3–

AII < III < I < IV

BIV < III < II < I

CI < II < IV < III

DIII < II < IV < I

Answer key (C) & worked solution

Correct answer: Option C

(I) [Mn(CN)

_{6}

]3– ⇒ –1.6 ∆

_{o}

(II) [Co(CN)

_{6}

]4– ⇒ –1.8 ∆

_{o}

(III) [Fe(CN)

_{6}

]4– ⇒ –2.4 ∆

_{o}

(IV) [Fe(CN)

_{6}

]3– ⇒ –2.0 ∆

_{o}

I < II < IV < III

Q5MathematicsJEE Main 2025, 29 Jan Shift 1Equation of lines in 3D and vector form

Let

L_{1} : \frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z - 1}{2}

and

L_{2} : \frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z}{1}

be two lines. Let

L_{3}

be a line passing through the point

(α, β, γ)

and be perpendicular to both

L_{1}

and

L_{2} .

L_{3}

intersects

L_{1},

then

∣ 5 α - 11 β - 8 γ ∣

equals:

A18

B16

C25

D20

Answer key (C) & worked solution

Correct answer: Option C

DR’s of

{L_3} = \vec m \times \vec n = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 1&{-1}&2 \\ {-1}&2&1 \end{array}} \right|

= - 5 i^- 3 j^+ k^

L_{3} : \frac{x - α}{- 5} = \frac{y - β}{- 3} = \frac{z - γ}{1} = λ

A (α - 5 λ, β - 3 λ, γ + λ)

L_{1} : \frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z - 1}{2} = k

B (k + 1, - k + 2, 2 k + 1)

Now

α - 5 λ = k + 1 \Rightarrow α = 5 λ + k + 1

β - 3 β = - k + 2 \Rightarrow β = 3 λ - k + 2

γ + λ = 2 k - 1 \Rightarrow γ = - λ + 2 k + 1

∣ 5 α - 11 β - 8 γ ∣ = ∣ - 25 ∣

= 25

Q6MathematicsJEE Main 2025, 29 Jan Shift 1General term of binomial expansion

The least value of n for which the number of integral terms in the Binomial expansion of

(37 + 1211)^{n}

is 183, is

A2184

B2148

C2172

D2196

Answer key (A) & worked solution

Correct answer: Option A

General term

=^{n} C_{r} {7^{\frac{1}{3}}}^{n - r} (11^{\frac{1}{12}})^{r}

=^{n} C_{r} {7}^{\frac{n - r}{3}} (11)^{\frac{r}{12}}

For integral terms, r must be multiple of 12

∴

r = 12k,

k \in W

Total values of r = 183 Hence max r = 12(182) = 2184 Min value of n = 2184