JEE Main CBT Mock Test Interface (Playable 6-Q Demo)

Practise the JEE Main CBT interface (6-question demo)

In the real JEE Main computer-based test, clicking an option radio button and moving to another question does not save your answer: your choice counts only after you click SAVE & NEXT or SAVE & MARK FOR REVIEW. Try the playable 6-question NTA CBT demo below with real past-paper questions, then check your score immediately.

The five NTA palette states explained

Not Visited (Grey square #E3E3E3): You have not opened this question yet.
Not Answered (Orange-red badge #E24401): You visited the question (or even clicked a radio button) and left without clicking SAVE & NEXT.
Answered (Green badge #2FAF00): You selected an option and clicked SAVE & NEXT. It will be evaluated.
Marked for Review (Purple circle #6B41AE): You clicked MARK FOR REVIEW & NEXT without saving an answer. It counts as unattempted (0 marks) if left as is.
Answered & Marked for Review (Purple circle with green ✓): You selected an option and clicked SAVE & MARK FOR REVIEW. NTA evaluates this response for full marks unless you clear it later.

Read the worked solutions for the 6 demo PYQs

Q1PhysicsJEE Main 2026, 22 Jan Shift 2Properties of images formed by a mirror

In parallax method for the determination of focal length of a concave mirror, the object should always be placed :

Abetween the focus (F) and the centre of curvature (C) of the mirror ONLY

Bat any point beyond the focus

(F)

of the mirror

Cbeyond the centre of the curvature ( C ) of the mirror ONLY

Dbetween the pole ( P ) and the focus ( F ) of the concave mirror ONLY

Answer key (B) & worked solution

Correct answer: Option B

Image should be real. So object should be placed beyond focus.

Q2PhysicsJEE Main 2026, 21 Jan Shift 2Angular momentum of a point particle

Two cars A and B each of mass

1 0^{3} kg

are moving on parallel tracks separated by a distance of 10 m , in same direction with speeds

72 km / h

and

36 km / h

. The magnitude of angular momentum of car A with respect to car B is

____

J.s.

3.6 \times 1 0^{5}

1 0^{5}

3 \times 1 0^{5}

2 \times 1 0^{5}

Answer key (B) & worked solution

Correct answer: Option B

L = m . V_{rel} r_{⊥}

= 1000 \times (36 \times \frac{5}{18}) \times 10 = 1 0^{5} kg m^{2} / s

Q3ChemistryJEE Main 2026, 21 Jan Shift 2Crystal Field Theory

Given below are two statements: Statement I: Crystal Field Stabilization Energy (CFSE) of

[Cr (H_{2} O)_{6}]^{2 +}

is greater than that of

[Mn (H_{2} O)_{6}]^{2 +}

. Statement II: Potassium ferricyanide has a greater spin-only magnetic moment than sodium Ferrocyanide. In the light of the above statements, choose the correct answer from the options given below:

ABoth Statement I and Statement II are true

BBoth Statement I and Statement II are false

CStatement I is true but Statement II is false

DStatement I is false but Statement II is true

Answer key (A) & worked solution

Correct answer: Option A

[Mn (H_{2} O)_{6}]^{2 +} \Rightarrow

CFSE value is zero because of

d^{5}

configuraiton with WFL in coordination number 6

[Cr (H_{2} O)_{6}]^{2 +}

[Cr (H_{2} O)_{6}]^{2 +} \Rightarrow

CFSE value is

- 0.6 Δ_{0}

because of

d^{4}

configuraiton with WFL in coordination number 6. For :

K_{3} [Fe (CN)_{6}], μ = 1 (1 + 2) = 3

B.M. For :

Na_{4} [Fe (CN)_{6}], μ = 0

B.M.

Q4ChemistryJEE Main 2026, 21 Jan Shift 1Quantitative analysis of elements

In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass

32 g mol^{- 1}

). Molar mass of barium sulphate is

233 g mol^{- 1}

4.55%

10.30%

21.97%

16.48%

Answer key (C) & worked solution

Correct answer: Option C

\frac{n _{BaSO_{4}} \times 32}{W _{(unknown comp.)}} \times 100

= \frac{1.2 \times 32}{233} \times \frac{100}{0.75} = 21.97%

Q5MathematicsJEE Main 2026, 21 Jan Shift 2Shortest distance between two lines

Let the line

L_{1}

be parallel to the vector

- 3 i^+ 2 j^+ 4 k^

and pass through the point

(2, 6, 7)

and the line

L_{2}

be parallel to the vector

2 i^+ j^+ 3 k^

and pass through the point

(4, 3, 5)

. If the line

L_{3}

is parallel to the vector

- 3 i^+ 5 j^+ 16 k^

and intersects the lines

L_{1}

and

L_{2}

at the points C and D , respectively, then

∣ CD ∣^{2}

is equal to :

A171

B290

C312

D89

Answer key (B) & worked solution

Correct answer: Option B

L_{1} : \frac{x - 2}{- 3} = \frac{y - 6}{2} = \frac{z - 7}{4}

Point C on

L_{1} : (- 3 λ_{1} + 2, 2 λ_{1} + 6, 4 λ_{1} + 7)

L_{2} : \frac{x - 4}{2} = \frac{y - 3}{1} = \frac{z - 5}{3}

Point D on

L_{2} : (2 λ_{2} + 4, λ_{2} + 3, 3 λ_{2} + 5)

Dr's of line

L_{3}

L_{3} : \frac{2 λ _{2} + 3 λ _{1} + 2}{- 3} = \frac{λ _{2} - 2 λ _{1} - 3}{5} = \frac{3 λ _{2} - 4 λ _{1} - 2}{16}

λ_{1} = - 3, λ_{2} = 2

C (11, 0, -5) D (8, 5, 11)

∣ CD ∣^{2} = 3^{2} + 5^{2} + 1 6^{2} = 290

Q6MathematicsJEE Main 2026, 21 Jan Shift 2Characteristic Equation & Cayley Hamilton Theorem

For the matrices

A = [31 - 4 - 1]

and

B = [- 29 - 13 4918]

, if

(A^{15} + B) [x y] = [00]

, then among the following which one is true?

x = 5, y = 7

x = 18, y = 11

x = 11, y = 2

x = 16, y = 3

Answer key (C) & worked solution

Correct answer: Option C

Here

A^{n} = [2 n + 1 n - 4 n - 2 n + 1]

\Rightarrow A^{15} = [3115 - 60 - 29]

\Rightarrow A^{15} + B = [22 - 11 - 11]

Now

(A^{15} + B) [x y] = [00]

\Rightarrow [22 - 11 - 11] [x y] = [00]

\Rightarrow 2 x - 11 y = 0