Updated 19 Oct 2026

JEE Main 2025, 7 Apr Shift 2 question paper with solutions

JEE Main 2025, 7 Apr Shift 2 was held on 7 April 2025. The bank holds 73 of its questions (Physics 25, Chemistry 24, Mathematics 24). 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

73 Questions in bank for this shift

Try questions from JEE Main 2025, 7 Apr Shift 2

Q1PhysicsJEE Main 2025, 7 Apr Shift 2Photoelectric effect and Einstein's photoelectric equation

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A photo-emissive substance is illuminated with a radiation of wavelength λ

_{i}

so that it releases electrons with de-Broglie wavelength λ

_{e}

. The longest wavelength of radiation that can emit photoelectron is λ

_{0}

. Expression for de-Broglie wavelength is given by : (m : mass of the electron, h : Planck’s constant and c : speed of light)

Answer key (A) & worked solution

Correct answer: Option A

K.E = E – W

λ_{e} = \frac{h}{2 m K . E}, E = \frac{h c}{λ _{i}}, W = \frac{h c}{λ _{0}}

\frac{h ^{2}}{2 m λ _{e}^{2}} = \frac{h c}{λ _{i}} - \frac{h c}{λ _{0}} λ_{e} = \frac{h}{2 m c ( \frac{1}{λ _{i}} - \frac{1}{λ _{0}} )}

Q2PhysicsJEE Main 2025, 7 Apr Shift 2Conductors and its properties

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The outer body of an air craft is made of metal which protects persons sitting inside from lightning-strikes. Reason (R) : The electric field inside the cavity enclosed by a conductor is zero. In the light of the above statements, chose the most appropriate answer from the options given below :

ABoth (A) and (R) are correct and (R) is the correct explanation of (A)

B(A) is correct but (R) is not correct

CBoth (A) and (R) are correct but (R) is not correct explanation of (A)

D(A) is not correct but (R) is correct

Answer key (A) & worked solution

Correct answer: Option A

Electric field of outside charge is zero inside conductor

Q3ChemistryJEE Main 2025, 7 Apr Shift 2Proteins

Given below are two statements : Statement (I) : On hydrolysis, oligo peptides give rise to fewer number of α-amino acids while proteins give rise to a large number of β-amino acids. Statement (II) : Natural proteins are denatured by acids which convert the water soluble form of fibrous proteins to their water insoluble form. In the light of the above statements, choose the most appropriate answer from the options given below :

ABoth statement I and statement II are correct

BStatement I is incorrect but Statement II is correct

CBoth statement I and statement II are incorrect

DStatement I is correct but Statement II is incorrect

Answer key (C) & worked solution

Correct answer: Option C

(i) Protein does not gives β -amino acid on hydrolysis (ii) Fibrous protein are not water soluble So both statement's are wrong

Q4ChemistryJEE Main 2025, 7 Apr Shift 2VBT in Coordination Compounds

The number of unpaired electrons responsible for the paramagnetic nature of the following complex species are respectively : [Fe(CN)

_{6}

]3–, [FeF

_{6}

]3–, [CoF

_{6}

]3–, [Mn(CN)

_{6}

]3–

A1, 5, 4, 2

B1, 5, 5, 2

C1, 1, 4, 2

D1, 4, 4, 2

Answer key (A) & worked solution

Correct answer: Option A

[F e (C N)_{6}]^{3 -} F e^{3 +} 3 d^{5} t_{2 g}^{2, 2, 1} e_{g}^{0.0} u n p ai r e d e^{-} = 1

[F e F_{6}]^{3 -} F e^{3 +} 3 d^{5} t_{2 g}^{1, 1, 1} u n p ai r e d e^{-} = 5

[C o F_{6}]^{3 -} C o^{3 +} 3 d^{6} t_{2 g}^{2, 1, 1} e_{g}^{1} u n p ai r e d e^{-} = 4

[M n (C N)_{6}]^{3 -} M n^{3 +} 3 d^{4} t_{2 g}^{2, 1, 1} e_{g}^{0, 0} u n p ai r e d e^{-} = 2

Q5MathematicsJEE Main 2025, 7 Apr Shift 2Equation of lines in 3D and vector form

If the equation of the line passing through the point

(0, - \frac{1}{2}, 0)

and perpendicular to the lines

r = λ (i^+ a j^+ b k^) an d

r = (i^- j^- 6 k^) + μ (- b i^+ a j^+ 5 k^)

\frac{x - 1}{- 2} = \frac{y + 4}{d} = \frac{z - c}{- 4},

then a + b+ c+ d is equal to :

A10

B14

C13

D12

Answer key (B) & worked solution

Correct answer: Option B

L in e i s ⊥^{r} t o 2 l in e \Rightarrow

line will be parallel to

(i + a j^+ b k^) \times (- b i^+ a j^+ 5 k^)

Parallel vector along the required line is

i^(5 a - ab) - j^(b^{2} + 5) + k^(a + ab)

Dr’s of required line α (5a –ab), – (b

^{2}

+ 5), (a + ab) Also Dr’s of required line α –2, d, –4

∴ \frac{5 a - ab}{- 2} = \frac{- ( b ^{2} + 5 )}{d} = \frac{a + ab}{- 4} .... (1)

A l so point (0, \frac{- 1}{2}, 0) w i l l l i e o n \frac{x - 1}{- 2} = \frac{y + 4}{d} = \frac{z - c}{- 4}

\frac{0 - 1}{- 2} = \frac{\frac{- 1}{2} + 4}{d} = \frac{0 - c}{- 4} \Rightarrow d = 7, c = 2

F r o m (1) \frac{5 a - ab}{- 2} = \frac{- b ^{2} - 5}{7} = \frac{a + ab}{- 4} \frac{5 a - ab}{- 2} = \frac{a + ab}{- 4}; \frac{- b ^{2} - 5}{7} = \frac{a + ab}{- 4}

- 20 a + 4 ab = - 2 a - 2 ab 4 b^{2} + 20 = 7 a + 7 ab 18 a = 6 ab 36 + 20 = 7 a + 21 a b = 356 = 28 a \Rightarrow a = 2

a + b + c + d = 2 + 3 + 2 + 7 = 14

Q6MathematicsJEE Main 2025, 7 Apr Shift 2Maxima and Minima at a point

Let ƒ : R → R be a polynomial function of degree four having extreme values at x = 4 and x = 5. If

x \to 0 lim \frac{f ( x )}{x ^{2}} = 5,

then ƒ(2) is equal to :

A12

B10

D14

Answer key (B) & worked solution

Correct answer: Option B

x \to 0 lim \frac{f ( x )}{x ^{2}} = 5 x \to 0 lim \frac{a x ^{4} + b x ^{3} + c x ^{2} + d x + e}{x ^{2}} = 5

c = 5 and d = e = 0 f(x) = ax

^{4}

+ bx

^{3}

+ 5x

^{2}

f'(x) = 4ax

^{3}

+ 3bx

^{2}

+ 10x = x (4ax

^{2}

+ 3bx + 10) has extremes at 4 and so f'(4) = 0 & f'(5) = 0

so a = \frac{1}{8} & b = \frac{- 3}{2}

so f (2) = \frac{1}{8} \times 2^{4} - \frac{3}{2} \times 2^{3} + 5 \times 2^{2} = 2 - 12 + 20 = 10