JEE Mains 2026-01-21 Shift 2 Question Paper With Detailed Solutions

JEE Mains 2026 2026-01-21 Shift 2

Chemistry

Consider the following data:

\\ \Delta_{\mathrm{f}} \mathrm{H}^{\Theta}(

methane, g

)=-\mathrm{X} \mathrm{kJ} \mathrm{mol}^{-1}\\

Enthalpy of sublimation of graphite

=\mathrm{Y} \mathrm{kJ} \mathrm{mol}^{-1}\\

Dissociation enthalpy of

\mathrm{H}_{2}=\mathrm{Z} \mathrm{kJ} \mathrm{mol}^{-1}\\

The bond enthalpy of

\mathrm{C}-\mathrm{H}

bond is given by:

A

\frac{X+Y+2 Z}{4}

B

\frac{X+Y+4 Z}{2}

C

X+Y+Z

D

\frac{-X+Y+Z}{4}

Correct Answer:

A

Explanation

\mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g})\\

-\mathrm{x}=\left(\Delta \mathrm{H}_{\text {sub }}\right.

of carbon

)+2 \times(

B.E. of

\mathrm{H}-\mathrm{H})\\

-4 \times(

B.E. of

\mathrm{C}-\mathrm{H})\\

-x=y+2 z-4(

B.E. of

C-H)\\

B.E. of

\mathrm{C}-\mathrm{H}=\frac{\mathrm{y}+2 \mathrm{z}+\mathrm{x}}{4}

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

Match List-I with List-II.

List-I	List-II
Pair of Compounds	Type of Isomers
A. 2-Methylpropene and but-1-ene	I. Stereoisomers
B. Cis-but-2-ene and trans-but-2-ene	II. Position isomers
C. 2-Butanol and diethyl ether	III. Chain isomers
D. But-1-ene and but-2-ene	IV. Functional group isomers

A

A-III, B-I, C-IV, D-II

B

A-III, B-I, C-II, D-IV

C

A-I, B-IV, C-III, D-II

D

A-II, B-I, C-IV, D-III

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

Given below are some of the statements about Mn and

\mathrm{Mn}_{2} \mathrm{O}_{7}

. Identify the correct statements A. Mn forms the oxide

\mathrm{Mn}_{2} \mathrm{O}_{7}

in which Mn is in its highest oxidation state. B. Oxygen stabilizes the Mn in higher oxidation states by forming multiple bonds with Mn C.

\mathrm{Mn}_{2} \mathrm{O}_{7}

is an ionic oxide. D. The structure of

\mathrm{Mn}_{2} \mathrm{O}_{7}

consists of one bridged oxygen. Choose the correct answer from the options given below:

A

A, B, C and D

B

A, B and D Only

C

A, C and D Only

D

A, B and C Only

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

Identify the metal ions among

\mathrm{Co}^{2+}, \mathrm{Ni}^{2+}, \mathrm{Fe}^{2+}, \mathrm{V}^{3+}

and

\mathrm{Ti}^{2+}

having a spin-only magnetic moment value more than 3.0 BM . The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is

Correct Answer:

7

Explanation

\quad \mathrm{V}^{3+}=(\mathrm{Ar})_{18} 3 \mathrm{~d}^{2}\\

\mathrm{Ti}^{2+}=(\mathrm{Ar})_{18} 3 \mathrm{~d}^{2}\\

\mathrm{Ni}^{2+}=(\mathrm{Ar})_{18} 3 \mathrm{~d}^{8}\\

\mathrm{Fe}^{2+}=(\mathrm{Ar})_{18} 3 \mathrm{~d}^{6}\\

\mathrm{Co}^{2+}=(\mathrm{Ar})_{18} 3 \mathrm{~d}^{7}\\

Only for

\mathrm{Fe}^{2+}

and

\mathrm{Co}^{2+} \mu

is more than 3.0 B.M.

∴ Number of unpaired electrons

=4+3=7

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

Consider the above sequence of reactions. The number of bromine atom(s) in the final product (P) will be

A

1

B

6

C

5

D

3

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

Match List-I with List-II.

\\

IAMGE

\\

Choose the correct answer from the options given below:

A

A-II, B-III, C-IV, D-I

B

A-IV, B-III, C-I, D-II

C

A-IV, B-I, C-II, D-III

D

A-II, B-IV, C-I, D-III

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

By usual analysis, 1.00 g of compound

(\mathrm{X})

gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound

(\mathrm{X})

is: (nearest integer)\\ (Given, molar mass in

\mathrm{g} \mathrm{mol}^{-1}: \mathrm{O}=16, \mathrm{Mg}=24

,

\mathrm{P}=31

)

A

50

B

30

C

20

D

40

Correct Answer:

A

Explanation

% of

\mathrm{P}=\frac{\mathrm{n}_{\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}} \times 2 \times 31}{\mathrm{~W}_{\text {(unknown compound) }}} \times 100\\

\begin{aligned} & =\frac{\left(\frac{1.79}{222} \times 2 \times 31\right)}{1} \times 100 \\ & =49.99 \% \approx 50 \% \end{aligned}

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

The correct statements are: A. Activation energy for enzyme catalysed hydrolysis of sucrose is lower than that of acid catalysed hydrolysis. B. During denaturation, secondary and tertiary structures of a protein are destroyed but primary structure remains intact. C. Nucleotides are joined together by glycosidic linkage between

\mathrm{C}_{1}

and

\mathrm{C}_{4}

carbons of the pentose sugar D. Quaternary structure of proteins represents overall folding of the polypeptide chain.\\ Choose the correct answer from the options given below:

A

A, C and D Only

B

A, B and D Only

C

A and B Only

D

B and C Only

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

Given below are two statements : Statement-I: Compound (X), shown below , dissolves in

\mathrm{NaHCO}_{3}

solution and has two chiral

Statement - II : Compound (Y), shown below, has two carbons with

\mathrm{sp}^{3}

hybridization, one carbon with

\mathrm{sp}^{2}

and one carbon with sp hybridization

\\

In the light of the above statements, choose the correct answer from the options given below :

A

Statement I is true but Statement II is false

B

Statement I is false but Statement II is true

C

Both Statement I and Statement II are true

D

Both Statement I and Statement II are false

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

Given below are two statements: Statement-I: The correct order in terms of atomic/ionic radii is

\mathrm{Al}>\mathrm{Mg}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}.\\

Statement-II: The correct order in terms of the magnitude of electron gain enthalpy is

\mathrm{Cl}>\mathrm{Br}>\mathrm{S}>\mathrm{O}.\\

In the light of the above statements, choose the correct answer from the options given below:

A

Both Statement I and Statement II are false

B

Statement I is false but Statement II is true

C

Statement I is true but Statement II is false

D

Both Statement I and Statement II are true

Correct Answer:

B

Explanation

Correct order of size is

\mathrm{Mg}>\mathrm{Al}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}\\

Atomic size depends mainly upon

\mathrm{Z}_{\text {effective }}

and shell number.

\\

Generally on moving down the group electron affinity decreases and on moving across the period electron affinity increase.

\\

In the periodic table Cl has maximum electron affinity. Halogen has higher electron affinity than Chalcogen.

\\ \mathrm{Cl}>\mathrm{Br}>\mathrm{S}>\mathrm{O}

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

The correct order of the rate of the reaction for the following reaction with respect to nucleophiles is:

\\ \mathrm{CH}_{3} \mathrm{Br}+\mathrm{Nu}^{\Theta} \longrightarrow \mathrm{CH}_{3} \mathrm{Nu}+\mathrm{Br}^{\Theta}

A

\mathrm{PhO}^{-}>^{-} \mathrm{OH}>\mathrm{CH}_{3} \mathrm{COO}^{-}>\mathrm{ClO}_{4}^{-}

B

\mathrm{ClO}_{4}^{-}>\mathrm{CH}_{3} \mathrm{COO}^{-}>{ }^{-} \mathrm{OH}>\mathrm{PhO}^{-}

\

C

\mathrm{CH}_{3} \mathrm{COO}^{-}>\mathrm{PhO}^{-}>{ }^{-} \mathrm{OH}>\mathrm{ClO}_{4}^{-}

D

{ }^{-} \mathrm{OH}>\mathrm{PhO}^{-}>\mathrm{CH}_{3} \mathrm{COO}^{-}>\mathrm{ClO}_{4}^{-}

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

Given below are two statements : Statement I: The correct order in terms of bond dissociation enthalpy is

\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{F}_{2}>\mathrm{I}_{2}\\

Statement II : The correct trend in the covalent character of the metal halides is

\left[\mathrm{SnCl}_{4}>\mathrm{SnCl}_{2}\right]

,

\left[\mathrm{PbCl}_{4}>\mathrm{PbCl}_{2}\right]

and

\left[\mathrm{UF}_{4}>\mathrm{UF}_{6}\right]\\

In the light of the above statements, choose the correct answer from the options given below :

A

Statement I is true but Statement II is false

B

Both Statement I and Statement II are true

C

Statement I is false but Statement II is true

D

Both Statement I and Statement II are false

Correct Answer:

A

Explanation

Statement-I : Bond energy order is

\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{F}_{2}>\mathrm{I}_{2}\\

Bond energy increases with increase in bond order.

\\

Statement-II Correct order of covalent character According to the Fajan's rule, higher the charge on cation, greater is the covalent character.

\\ \mathrm{PbCl}_{2}<\mathrm{PbCl}_{4}

,

\\ \mathrm{UF}_{6}>\mathrm{UF}_{4}

,

\\ \mathrm{SnCl}_{4}>\mathrm{SnCl}_{2}

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

The correct order of reactivity of the following benzyl halides towards reaction with KCN is:

A

a>b>c>d

B

b>a>d>c

C

b>a>c>d

D

a>b>d>d

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

A substance ' X ' ( 1.5 g ) dissolved in 150 g of a solvent ' Y ' (molar mass

=300 \mathrm{~g} \mathrm{~mol}^{-1}

) led to an elevation of the boiling point by 0.5 K . The relative lowering in the vapour pressure of the solvent ' Y ' is

\_\_\_\_

\times 10^{-2}

. (Nearest integer)

\\

[Given :

\mathrm{K}_{\mathrm{b}}

of the solvent

=5.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} ]\\

Assume the solution to be dilute and no association or dissociation of X takes place in solution.

Correct Answer:

3

Explanation

\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \times \mathrm{K}_{\mathrm{b}} \times \mathrm{m}\\

0.5=\mathrm{i} \times \mathrm{m} \times 5\\

\mathrm{i} \times \mathrm{m}=\frac{0.5}{5}=0.1\\

\mathrm{i} \times \mathrm{a}=\frac{15}{1000}\\

(where

\mathrm{a}=

moles of solute) Now,

\\ \begin{aligned} & \frac{\mathrm{P}_{o}-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}^{o}}=\mathrm{i} \mathrm{X}_{\text {solute }}=\mathrm{i} \times \frac{\mathrm{a}}{\mathrm{a}+\frac{150}{300}} \\ & =\mathrm{i} \times \frac{\mathrm{a}}{1 / 2} =\frac{15 / 1000}{1 / 2} \\&=\frac{30}{1000}=3 \times 10^{-2}=3 \end{aligned}

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

Aqueous HCl reacts with

\mathrm{MnO}_{2}(\mathrm{~s})

to form

\mathrm{MnCl}_{2}(\mathrm{aq})

,

\mathrm{Cl}_{2}(\mathrm{~g})

and

\mathrm{H}_{2} \mathrm{O}(l)

. What is the weight (in g ) of

\mathrm{Cl}_{2}

liberated when 8.7 g of

\mathrm{MnO}_{2}(\mathrm{~s})

is reacted with excess aqueous HCl solution ? (Given Molar mass in

\mathrm{g} \mathrm{mol}^{-1} \mathrm{Mn}=55, \mathrm{Cl}=35.5, \mathrm{O}=16, \mathrm{H}=1

)

A

7.1

B

71

C

21.3

D

14.2

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

The osmotic pressure of a living cell in 12 atm at 300 K . The strength of sodium chloride solution that is isotonic with the living cell at this temperature is

\_\_\_\_

\mathrm{g} \mathrm{L}^{-1}

. (Nearest integer) Given :

\mathrm{R}=0.08 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\\

Assume complete dissociation of NaCl (Given : Molar mass of Na and Cl are 23 and 35.5

\mathrm{g} \mathrm{mol}^{-1}

respectively.)

Correct Answer:

15

Explanation

\pi=\mathrm{iCRT}\\

12=2 \times \mathrm{C} \times 0.08 \times 300\\

12=2 \times \mathrm{C} \times 24\\

\mathrm{C}=\frac{1}{4} \mathrm{~mole} / \mathrm{L}\\

then strength of NaCl solution

\\=\frac{1}{4} \times 58.5 \mathrm{~g} / \mathrm{L}\\

=14.625 \mathrm{~g} / \mathrm{L}\\

=15 \mathrm{~g} / \mathrm{L}

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

For a closed circuit Daniell cell, which of the following plots is the accurate one at a given temperature?

A

B

C

D

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

Decomposition of A is a first order reaction at

\mathrm{T}(\mathrm{K})

and is given by

\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})

. In a closed 1 L vessel, 1 bar

\mathrm{A}(\mathrm{g})

is allowed to decompose at

\mathrm{T}(\mathrm{K})

. After 100 minutes, the total pressure was 1.5 bar . What is the rate constant (in

\min ^{-1}

) of the reaction ? (

\log 2=0.3)

A

6.9 \times 10^{-1}

B

6.9 \times 10^{-3}

C

6.9 \times 10^{-2}

D

6.9 \times 10^{-4}

Correct Answer:

B

Explanation

\quad \mathrm{A}_{\mathrm{g}} \rightarrow \mathrm{B}_{\mathrm{g}}+\mathrm{C}_{\mathrm{g}}\\

1 - - 1-P P P

\\ \mathrm{P}_{\text {total }}=1+\mathrm{P}\\

1.5=1+\mathrm{P}\\

\mathrm{P}=0.5\\

\mathrm{K}=\frac{1}{100} \ln \frac{1}{0.5}\\

=\frac{0.693}{100}\\

=6.9 \times 10^{-3} \mathrm{~min}^{-1}

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

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

\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}

is greater than that of

\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{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:

A

Both Statement I and Statement II are true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

D

Statement I is false but Statement II is true

Correct Answer:

A

Explanation

\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \Rightarrow

CFSE value is zero because of

\\ \mathrm{d}^{5}

configuraiton with WFL in coordination number 6

\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}.\\

\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \Rightarrow

CFSE value is

-0.6 \Delta_{0}

because of

\mathrm{d}^{4}

configuraiton with WFL in coordination number 6. For :

\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right], \mu=\sqrt{1(1+2)}=\sqrt{3}

B.M.

\\

For :

\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right], \mu=\sqrt{0}

B.M.

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

The correct increasing order of

\mathrm{C}-\mathrm{H}(\mathrm{A})

,

\mathrm{C}-\mathrm{O}(\mathrm{B}), \mathrm{C}=\mathrm{O}(\mathrm{C})

and

\mathrm{C} \equiv \mathrm{N}(\mathrm{D})

bonds in terms of covalent bond length is:

A

<

B

<

C

<

D

B

A

<

D

<

C

<

B

C

D

<

C

<

B

<

A

D

<

C

<

A

<

B

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

Given below are four compounds: (a) n-propyl choride (b) iso-propyl chloride (c) sec-butyl chloride (d) neo-pentyl chloride Percentage of carbon in the one which exhibits optical isomerism is:

A

52

B

56

C

46

D

40

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

Consider the following spectral lines for atomic hydrogen:

\\

A. First line of Paschen series

\\

B. Second line of Balmer series

\\

C. Third line of Paschen series

\\

D. Fourth line of Bracket sereis

\\

The correct arrangement of the above lines in ascending order of energy is:

A

D

<

C

<

A

<

B

A

<

B

<

C

<

D

C

<

D

<

B

<

A

D

<

A

<

C

<

B

Correct Answer:

D

Explanation

\quad \Delta \mathrm{E}=13.6 \mathrm{Z}^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)\\

Series &

\mathrm{n}_{1}

&

\mathrm{n}_{2}

\\

(A) Paschen

\left(1^{\text {st }}\right.

line

)

& 3 & 4

\\

(B) Balmer

\left(2^{\text {nd }}\right.

line

)

& 2 & 4

\\

(C) Paschen

\left(3^{\text {rd }}\right.

line

)

& 3 & 6

\\

(D) Bracket

\left(4^{\text {th }}\right.

line

)

& 4 & 8

\\

So correct ascending order of energy of above lines is :

\\ \mathrm{D}<\mathrm{A}<\mathrm{C}<\mathrm{B}

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

The first and second ionization constants of

\mathrm{H}_{2} \mathrm{X}

are

2.5 \times 10^{-8}

and

1.0 \times 10^{-13}

respectively. The concentration of

\mathrm{X}^{2-}

in

0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{X}

solution is

\_\_\_\_

\times 10^{-15} \mathrm{M}

. (Nearest Integer)

Correct Answer:

100

Explanation

\mathrm{H}_{2} \mathrm{X} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HX}^{-}

\\0.1-x \quad x+y \quad x-y\\

2.5 \times 10^{-8}=\frac{(\mathrm{x}+\mathrm{y})(\mathrm{x}-\mathrm{y})}{0.1-\mathrm{x}}\\

\mathrm{HX}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{2-}

,

\\x-y \quad x+y \quad y\\

1 \times 10^{-13}=\frac{(\mathrm{x}+\mathrm{y})(\mathrm{y})}{\mathrm{x}-\mathrm{y}}\\

Approximate :

\mathrm{K}_{\mathrm{a}_{1}} \gg \mathrm{~K}_{\mathrm{a}_{2}} \Rightarrow

So

\mathrm{x} \gg \mathrm{y}.\\

\mathrm{x}+\mathrm{y} \approx \mathrm{x}, \mathrm{x}-\mathrm{y} \approx \mathrm{x}\\

10^{-13}=\frac{x \cdot y}{x}\\

\mathrm{y}=10^{-13}

\\

\left[\mathrm{X}^{2-}\right]=10^{-13}\\

\left[\mathrm{X}^{2-}\right]=100 \times 10^{-15}

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K

\mathrm{MX}(\mathrm{s}) \\ \rightleftharpoons \mathrm{M}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}) ; \mathrm{K}_{\mathrm{sp}}=10^{-10}\\

If the standard reduction potential for

\mathrm{M}^{+}(\mathrm{aq}) \xrightarrow{+\mathrm{e}^{-}} \mathrm{M}(\mathrm{s})

is

\left(\mathrm{E}_{\mathrm{M}^{+} / \mathrm{M}}^{\Theta}\right)=0.79 \mathrm{~V},\\

then the value of the standard reduction potential for the metal/metal insoluble salt electrode

\mathrm{E}_{\mathrm{X}^{-} / \mathrm{MX}(\mathrm{s}) / \mathrm{M}}^{\Theta}

is

\_\_\_\_

mV. (nearest integer)

\\

[Given:

\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}

]

Correct Answer:

200

Explanation

\mathrm{E}_{\mathrm{X}^{-} / \mathrm{MX}(\mathrm{s}) \mathrm{M}}^{\mathrm{o}}=\mathrm{E}_{\mathrm{M}^{+} / \mathrm{M}}^{\mathrm{o}}+\frac{0.0591}{\mathrm{n}} \log \mathrm{K}_{\mathrm{sp}}\\

=0.79+\frac{0.059}{1} \log 10^{-10}\\

=0.79-0.59\\

=0.20 \mathrm{~V}=200 \mathrm{mV}\\

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JEE Mains 2026 2026-01-21 Shift 2

Chemistry

On heating a mixture of common salt and

\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}

in equal amount along with concentrated

\mathrm{H}_{2} \mathrm{SO}_{4}

in a test tube, a gas is evolved. Formula of the gas evolved and oxidation state of the central metal atom in the gas respectively are :

A

\mathrm{CrO}_{2} \mathrm{Cl}_{2}

and +5

B

\mathrm{CrO}_{2} \mathrm{Cl}_{2}

and +6

C

\mathrm{Cr}_{2} \mathrm{O}_{2} \mathrm{Cl}_{2}$ and +6

D

\mathrm{Cr}_{2} \mathrm{O}_{2} \mathrm{Cl}_{2}

and +3

Correct Answer:

B

Explanation

4 \mathrm{NaCl}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+6 \mathrm{H}_{2} \mathrm{SO}_{4}\\

\longrightarrow 2 \mathrm{KHSO}_{4}+2 \mathrm{CrO}_{2} \mathrm{Cl}_{2}+4 \mathrm{NaHSO}_{4}+3 \mathrm{H}_{2} \mathrm{O}\\

(Chromyl chloride) In Chromyl chloride Cr is in +6 oxidation state.

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

Let

\alpha

and

\beta

be the roots of equation

x^{2}+2 a x+(3 a+10) =0

such that

\alpha<1<\beta

. Then the set of all possible values of

a

is :

A

\left(-\infty, \frac{-11}{5}\right) \cup(5, \infty)

B

(-\infty,-2) \cup(5, \infty)

C

(-\infty,-3)

D

\left(-\infty, \frac{-11}{5}\right)

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

If the system of equations

\\3 \mathrm{x}+\mathrm{y}+4 \mathrm{z}=3\\

2 \mathrm{x}+\alpha \mathrm{y}-\mathrm{z}=-3\\

\mathrm{X}+2 \mathrm{y}+\mathrm{z}=4\\

has no solution, then the value of

\alpha

is equal to:

A

19

B

4

C

13

D

23

Correct Answer:

A

Explanation

for no solution

\Delta=0\\

\left|\begin{array}{ccc}3 & 1 & 4 \\ 2 & \alpha & -1 \\ 1 & 2 & 1\end{array}\right|=0\\

\Rightarrow 3(\mathrm{a}+2)+1(-1-2)+4(4-\alpha)=0\\

\Rightarrow 19-\alpha=0 \Rightarrow \alpha=19\\

\& for

\alpha=19\\

\Delta_{\mathrm{x}}=\left|\begin{array}{ccc}3 & 1 & 4 \\ -3 & 19 & -1 \\ 4 & 2 & 1\end{array}\right|=3(21)+1(-1)+4(-82)\\

\neq 0\\

∴ no solution for

\alpha=19

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

Let [ • denote the greatest integer function and

f(\mathrm{x})=\lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\mathrm{n}^{3}} \sum_{\mathrm{k}=1}^{\mathrm{n}}\left[\frac{\mathrm{k}^{2}}{3^{\mathrm{x}}}\right]

. Then

12 \sum_{\mathrm{j}=1}^{\infty} f(\mathrm{j})

is equal to

\_\_\_\_

Correct Answer:

2

Explanation

\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\frac{\mathrm{k}^{2}}{3^{\mathrm{x}}}-1\right)<\sum_{\mathrm{k}=1}^{\mathrm{n}}\left[\frac{\mathrm{k}^{2}}{3^{\mathrm{x}}}\right] \leq \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{\mathrm{k}^{2}}{3^{\mathrm{x}}}\\

\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6.3^{\mathrm{x}}}<\sum_{\mathrm{k}=1}^{\mathrm{n}}\left[\frac{\mathrm{k}^{2}}{3^{\mathrm{x}}}\right] \leq \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6.3^{\mathrm{x}}} \\

\begin{aligned} & \lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+1)}{6 n^{3} \cdot 3^{x}}<\lim _{n \rightarrow \infty} \frac{1}{n^{3}} \sum_{k=1}^{n}\left[\frac{k^{2}}{3^{x}}\right] \leq \lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+1)}{6 \cdot 3^{x} \cdot n^{3}} \\ & \quad \frac{1}{3^{x+1}}<\lim _{n \rightarrow \infty} \frac{1}{n^{3}} \sum_{k=1}^{n}\left[\frac{k^{2}}{3^{x}}\right] \leq \frac{1}{3^{x+1}} \\ & \Rightarrow f(x)=\frac{1}{3^{x+1}} \\ & \quad \Rightarrow 12 \sum_{j=1}^{\infty} f(j)=12 \sum_{j=1}^{\infty} \frac{1}{3^{j+1}}=12\left[\frac{1}{9}+\frac{1}{27}+--\infty\right] \\ & \quad=12 \cdot\left(\frac{\frac{1}{9}}{1-\frac{1}{3}}\right)=2 \end{aligned}

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

Let

\mathrm{A}=\{2,3,5,7,9\}

. Let R be the relation on A defined by x Ry if and only if

2 \mathrm{x} \leq 3 \mathrm{y}

. Let

\ell

be the number of elements in R , and m be the minimum number of elements required to be added in R to make it a symmetric relation. Then

\ell+\mathrm{m}

is equal to:

A

23

B

25

C

21

D

27

Correct Answer:

B

Explanation

\mathrm{A}=\{2,3,5,7,9\}\\

\mathrm{y} \geq \frac{2 \mathrm{x}}{3}\\

\left.\begin{array}{ll}\mathrm{x}=2, & \mathrm{y}=2,3,5,7,9 \\ \mathrm{x}=3, & \mathrm{y}=2,3,5,7,9 \\ \mathrm{x}=5, & \mathrm{y}=5,7,9 \\ \mathrm{x}=7, & \mathrm{y}=5,7,9 \\ \mathrm{x}=9 & \mathrm{y}=7,9\end{array}\right] \rightarrow \ell=18\\

to make it symmetric elements to be added are

\{(5,2),(7,2),(9,2),(5,3),(7,3),(9,3),(9,5)\}

\\

\mathrm{m}=7\\

\therefore \quad \ell+\mathrm{m}=25

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

Let one end of a focal chord of the parabola

y^{2}=16 x

be (16, 16). If

\mathrm{P}(\alpha, \beta)

divides this focal chord internally in the ratio

5: 2

, then the minimum value of

\alpha+\beta

is equal to :

A

22

B

7

C

5

D

16

Correct Answer:

B

Explanation

\mathrm{y}^{2}=16 \mathrm{x}\\

\\

∵ parameter of point A is

\mathrm{t}=2\\

⇒ Parameter of point

B

is

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

⇒ Coordinates of B is

(1,-4)\\

Case 1:

\\ \alpha=\frac{5+32}{7}=\frac{37}{7}\\

\beta=\frac{-20+32}{7}=\frac{12}{7}\\

\Rightarrow \alpha+\beta=7\\

Case 2 : A(16, 16)

\\ \alpha=\frac{2+80}{7}

,

\\ \beta=\frac{-8+80}{7}\\

\alpha+\beta=22\\

So minimum value of

\alpha+\beta=7

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

Let the maximum value of

\left(\sin ^{-1} \mathrm{x}\right)^{2}+\left(\cos ^{-1} \mathrm{x}\right)^{2}

for

x \in\left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]

be

\frac{m}{n} \pi^{2}

, where

\operatorname{gcd}(m, n)=1.

Then

\mathrm{m}+\mathrm{n}

is equal to

\_\_\_\_

.

Correct Answer:

65

Explanation

\left(\sin ^{-1} \mathrm{x}\right)^{2}+\left(\cos ^{-1} \mathrm{x}\right)^{2}\\

=\left(\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}\right)^{2}-2 \sin ^{-1} \mathrm{x} \cos ^{-1} \mathrm{x}\\

=\frac{\pi^{2}}{4}-2\left(\sin ^{-1} \mathrm{x}\right)\left(\frac{\pi}{2}-\sin ^{-1} \mathrm{x}\right)\\

=2\left(\sin ^{-1} \mathrm{x}-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8} \quad

where

\sin ^{-1} \mathrm{x} \in\left[\frac{-\pi}{3}, \frac{\pi}{4}\right]\\

Then max value occurs at

\sin ^{-1} x=\frac{-\pi}{3}\\

Which is

2\left(\frac{\pi}{3}+\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{8}=\frac{29 \pi^{2}}{36}\\

\Rightarrow \mathrm{m}=29

and

\mathrm{n}=36\\

\therefore \mathrm{m}+\mathrm{n}=65

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

Let

\mathrm{a}_{1}, \frac{\mathrm{a}_{2}}{2}, \frac{\mathrm{a}_{3}}{2^{2}}, \ldots, \frac{\mathrm{a}_{10}}{2^{9}}

be a G.P. of common ratio

\frac{1}{\sqrt{2}}

. If

\mathrm{a}_{1}+\mathrm{a}_{2}+\ldots+\mathrm{a}_{10}=62

, then

\mathrm{a}_{1}

is equal to :

A

2(\sqrt{2}-1)

B

2-\sqrt{2}

C

\sqrt{2}-1

D

2(2-\sqrt{2})

Correct Answer:

A

Explanation

\frac{\mathrm{a}_{2}}{2 \mathrm{a}_{1}}=\frac{\mathrm{a}_{3}}{2 \mathrm{a}_{2}} \\=\frac{\mathrm{a}_{4}}{2 \mathrm{a}_{3}}=\ldots =\frac{\mathrm{a}_{10}}{2 \mathrm{a}_{9}}=\frac{1}{\sqrt{2}}\\

\therefore \mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, ., \mathrm{a}_{10}

are in G.P. with common ratio

\sqrt{2}.\\

\sum_{i=1}^{10} a_{i}=\frac{a_{1}\left((\sqrt{2})^{10}-1\right)}{\sqrt{2}-1}=62\\

\Rightarrow \mathrm{a}_{1}=2(\sqrt{2}-1)

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

In the line

\alpha x+4 y=\sqrt{7}

, where

\alpha \in R

, touches the ellipse

3 x^{2}+4 y^{2}=1

at the point

P

in the first quadrant, then one of the focal distances of P is :

A

\frac{1}{\sqrt{3}}-\frac{1}{2 \sqrt{11}}

B

\frac{1}{\sqrt{3}}+\frac{1}{2 \sqrt{5}}

C

\frac{1}{\sqrt{3}}-\frac{1}{2 \sqrt{5}}

D

\frac{1}{\sqrt{3}}+\frac{1}{2 \sqrt{7}}

Correct Answer:

D

Explanation

\alpha x+4 y-\sqrt{7}=0

touches

3 x^{2}+4 y^{2}=1\\

\therefore \mathrm{c}^{2}=\mathrm{a}^{2} \mathrm{~m}^{2}+\mathrm{b}^{2}\\

\frac{7}{16}=\frac{1}{3} \times \frac{\alpha^{2}}{16}+\frac{1}{4} \Rightarrow \alpha=3,-3\\

Tangent is

3 x+4 y-\sqrt{7}=0\\

Let the point of contact is

\mathrm{P}\left(\mathrm{x}_{1} \mathrm{y}_{1}\right)\\

∴ Tangent is

3 \mathrm{xx}_{1}+4 \mathrm{yy}_{1}=1\\

\therefore \frac{3 \mathrm{x}_{1}}{3}=\frac{4 \mathrm{y}_{1}}{4}=\frac{1}{\sqrt{7}} \quad \therefore \mathrm{P}\left(\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right)\\

\mathrm{e}=\sqrt{1-\frac{3}{4}}=\frac{1}{2}\\

\\ \mathrm{PS}=\mathrm{e}(\mathrm{PM})\\

=e\left(\frac{\mathrm{a}}{\mathrm{e}}-\frac{1}{\sqrt{7}}\right)\\

=\frac{1}{2}\left(\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{7}}\right)=\frac{1}{\sqrt{3}}-\frac{1}{2 \sqrt{7}}\\

\mathrm{PS}^{\prime}=\mathrm{e}\left(\mathrm{PM}^{\prime}\right)=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{e}}+\frac{1}{\sqrt{7}}\right)=\frac{1}{2}\left(\frac{1}{\sqrt{7}}+\frac{2}{\sqrt{3}}\right)\\

=\frac{1}{\sqrt{3}}+\frac{1}{2 \sqrt{7}}

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

For a triangle

A B C

, let

\vec{p}=\overrightarrow{B C}, \vec{q}=\overrightarrow{C A}

and

\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{BA}}

. If

|\overrightarrow{\mathrm{p}}|=2 \sqrt{3},|\overrightarrow{\mathrm{q}}|=2

and

\cos \theta=\frac{1}{\sqrt{3}}

, where

\theta

is the angle between

\vec{P}

and

\vec{q}

, then

|\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}}-3 \overrightarrow{\mathrm{r}})|^{2}+3|\overrightarrow{\mathrm{r}}|^{2}

is equal to:

A

340

B

220

C

410

D

200

Correct Answer:

D

Explanation

\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{r}}\\

\cos (\pi-\theta)=\frac{|\overrightarrow{\mathrm{p}}|^{2}+|\overrightarrow{\mathrm{q}}|^{2}-|\overrightarrow{\mathrm{r}}|^{2}}{2|\overrightarrow{\mathrm{p}} \| \overrightarrow{\mathrm{q}}|}\\

\frac{-1}{\sqrt{3}}=\frac{12+4-|\overrightarrow{\mathrm{r}}|^{2}}{2 \cdot 2 \sqrt{3} \cdot 2}\\

|\overrightarrow{\mathrm{r}}|^{2}=24\\

\therefore|\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}}-3 \overrightarrow{\mathrm{r}})|^{2}+3|\overrightarrow{\mathrm{r}}|^{2}\\

=|\vec{p} \times(\vec{q}-3 \vec{p}-3 \vec{q})|^{2}+72\\

=|\vec{p} \times(-3 \vec{p}-2 \vec{q})|^{2}+72\\

=|-2 \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|^{2}+72\\

=4|\overrightarrow{\mathrm{p}}|^{2}|\overrightarrow{\mathrm{q}}|^{2} \times \sin ^{2} \theta+72\\

=4 \cdot 12.4 \cdot \frac{2}{3}+72\\

=200

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

Let

\mathrm{A}=\left\{\mathrm{x}:\left|\mathrm{x}^{2}-10\right| \leq 6\right\}

and

\mathrm{B}=\{\mathrm{x}:|\mathrm{x}-2|>1\}.

Then

A

\mathrm{A} \cup \mathrm{B}=(-\infty, 1] \cup(2, \infty)

B

\mathrm{A}-\mathrm{B}=[2,3)

C

\mathrm{A} \cap \mathrm{B}=[-4,-2] \cup[3,4]

D

\mathrm{B}-\mathrm{A}=(-\infty,-4) \cup(-2,1) \cup(4, \infty)

Correct Answer:

D

Explanation

\left|\mathrm{x}^{2}-10\right| \leq 6\\

-6 \leq \mathrm{x}^{2}-10 \leq \\

4 \leq \mathrm{x}^{2} \leq 16\\

\mathrm{A}=[-4,-2] \cup[2,4]\\

|x-2|>1\\

\mathrm{B}=(-\infty, 1) \cup(3, \infty)\\

\mathrm{A} \cup \mathrm{B}=(-\infty, 1) \cup[2, \infty)\\

\mathrm{A} \cap \mathrm{B}=[-4,-2] \cup(3,4]\\

\mathrm{A}-\mathrm{B}=[2,3]\\

\mathrm{B}-\mathrm{A}=(-\infty,-4) \cup(-2,1) \cup(4, \infty)

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

For the matrices

\mathrm{A}=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]

and

\mathrm{B}=\left[\begin{array}{ll}-29 & 49 \\ -13 & 18\end{array}\right]

, if

\left(A^{15}+B\right)\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right]

, then among the following which one is true?

A

x=5, y=7

B

x=18, y=11

C

x=11, y=2

D

x=16, y=3

Correct Answer:

C

Explanation

Here

\mathrm{A}^{\mathrm{n}}=\left[\begin{array}{cc}2 \mathrm{n}+1 & -4 \mathrm{n} \\ \mathrm{n} & -2 \mathrm{n}+1\end{array}\right]\\

\Rightarrow \mathrm{A}^{15}=\left[\begin{array}{ll}31 & -60 \\ 15 & -29\end{array}\right]\\

\Rightarrow \mathrm{A}^{15}+\mathrm{B}=\left[\begin{array}{ll}2 & -11 \\ 2 & -11\end{array}\right]\\

Now

\left(A^{15}+B\right)\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right]\\

\Rightarrow\left[\begin{array}{cc}2 & -11 \\ 2 & -11\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y}\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right]\\

\Rightarrow 2 \mathrm{x}-11 \mathrm{y}=0

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

If\\

\left(\frac{1}{{ }^{15} \mathrm{C}_{0}}+\frac{1}{{ }^{15} \mathrm{C}_{1}}\right)\left(\frac{1}{{ }^{15} \mathrm{C}_{1}}+\frac{1}{{ }^{15} \mathrm{C}_{2}}\right) \ldots\left(\frac{1}{{ }^{15} \mathrm{C}_{12}}+\frac{1}{{ }^{15} \mathrm{C}_{13}}\right) =\frac{\alpha^{13}}{{ }^{14} \mathrm{C}_{0}{ }^{14} \mathrm{C}_{1} \ldots{ }^{14} \mathrm{C}_{12}}

, then

30 \alpha

is equal to

\_\_\_\_

Correct Answer:

32

Explanation

\prod_{\mathrm{r}=0}^{12}\left(\frac{1}{{ }^{15} \mathrm{C}_{\mathrm{r}}}+\frac{1}{{ }^{15} \mathrm{C}_{\mathrm{r}+1}}\right) \\=\prod_{\mathrm{r}=0}^{12} \frac{\frac{16}{\mathrm{r}+1} \cdot{ }^{15} \mathrm{C}_{\mathrm{r}}}{{ }^{15} \mathrm{C}_{\mathrm{r}} \cdot{ }^{15} \mathrm{C}_{\mathrm{r}+1}}\\

=\prod_{\mathrm{r}=0}^{12} \frac{16}{(\mathrm{r}+1) \cdot \frac{15}{\mathrm{r}+1} \cdot{ }^{14} \mathrm{C}_{\mathrm{r}}}\\=\prod_{\mathrm{r}=0}^{12} \frac{\left(\frac{16}{15}\right)}{{ }^{14} \mathrm{C}_{\mathrm{r}}}\\

=\frac{\left(\frac{16}{15}\right)^{13}}{{ }^{14} \mathrm{C}_{0} \cdot{ }^{14} \mathrm{C}_{1} \quad{ }^{14} \mathrm{C}_{12}} \\ \Rightarrow \alpha=\frac{16}{15}\\

\Rightarrow 30 \alpha=32

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

Let the line L pass through the point

(-3,5,2)

and make equal angles with the positive coordinate axes. If the distance of L from the point (

-2, \mathrm{r}, 1

) is

\sqrt{\frac{14}{3}}

, then the sum of all possible values of

r

is :

A

12

B

16

C

6

D

10

Correct Answer:

D

Explanation

Equation line is :

\frac{x+3}{1}=\frac{y-5}{1}=\frac{z-2}{1}=\lambda\\

∴ General point R on line is

\mathrm{R}(\lambda-3, \lambda+5, \lambda+2)\\

\\ \overrightarrow{\mathrm{PR}} \equiv(\lambda-1, \lambda+5-\mathrm{r}, \lambda+1)\\

Now

\overrightarrow{\mathrm{PR}} \cdot \overrightarrow{\mathrm{d}}=0\\

\Rightarrow(\lambda-1) 1+(\lambda+5-\mathrm{r}) 1+(\lambda+1) 1=0\\

\Rightarrow 3 \lambda-\mathrm{r}+5=0\\

\Rightarrow \lambda=\frac{\mathrm{r}-5}{3}\\

\therefore \mathrm{R} \equiv\left(\frac{\mathrm{r}-5}{3}-3, \frac{\mathrm{r}-5}{3}+5, \frac{\mathrm{r}-5}{3}+2\right)\\

\mathrm{R} \equiv\left(\frac{\mathrm{r}-14}{3}, \frac{\mathrm{r}+10}{3}, \frac{\mathrm{r}+1}{3}\right)\\

Now

\\ \mathrm{PR}=\sqrt{\frac{14}{3}} \Rightarrow(\mathrm{PR})^{2}=\frac{14}{3}\\

\Rightarrow\left(\frac{\mathrm{r}-14}{3}+2\right)^{2}+\left(\frac{\mathrm{r}+10}{3}-\mathrm{r}\right)^{2}+\left(\frac{\mathrm{r}+1}{3}-1\right)^{2}=\frac{14}{3}\\

\Rightarrow \frac{(\mathrm{r}-8)^{2}}{9}+\frac{(10-2 \mathrm{r})^{2}}{9}+\frac{(\mathrm{r}-2)^{2}}{9}=\frac{14}{3}\\

\Rightarrow\left(r^{2}-16 r+64\right)+\left(100+4 r^{2}-40 r\right)+\left(r^{2}-4 r+4\right)=42\\

\Rightarrow 6 \mathrm{r}^{2}-60 \mathrm{r}+126=0\\

\Rightarrow \mathrm{r}^{2}-10 \mathrm{r}+21=0\\

\Rightarrow \mathrm{r}=3,7\\

sum of possible value of r is =10

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

A random variable

X

takes values

0,1,2,3

with probabilities

\frac{2 \mathrm{a}+1}{30}, \frac{8 \mathrm{a}-1}{30}, \frac{4 \mathrm{a}+1}{30}

, b respectively, where

\mathrm{a}, \mathrm{b} \in \mathbf{R}

. Let

\mu

and

\sigma

respectively be the mean and standard deviation of

X

such that

\sigma^{2}+\mu^{2}=2.\\

Then

\frac{\mathrm{a}}{\mathrm{b}}

is equal to :

A

30

B

3

C

60

D

12

Correct Answer:

C

Explanation

X	0	1	2	3
$p(x)$	$\frac{2a + 1}{30}$	$\frac{8a - 1}{30}$	$\frac{4a + 1}{30}$	b

\\ \sigma^{2}=\sum x_{i}^{2} p\left(x_{i}\right)-\mu^{2} \\

\sigma^{2}+\mu^{2}=\sum x_{i}^{2} p\left(x_{i}\right) \\

=0+1\left(\frac{8 a-1}{30}\right)+4\left(\frac{4 a+1}{30}\right)+9 b \\

\Rightarrow \frac{24 a+270 b+3}{30}=2 \\

24 a+270 b=57 \\

8 a+90 b=19 \ldots(1)

Also

\\ \sum \mathrm{p}(\mathrm{i})=1\\

\frac{2 a+1}{30}+\frac{8 a-1}{30}+\frac{4 a+1}{30}+b=1\\

14 \mathrm{a}+30 \mathrm{~b}=29 \tag......{(2)}\\

Solving (1) & (2)

\\ \mathrm{a}=2, \quad \mathrm{~b}=\frac{1}{30}, \frac{\mathrm{a}}{\mathrm{b}}=60

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

Let

\mathrm{y}^{2}=12 \mathrm{x}

be the parabola with its vertex at O . Let P be a point on the parabola and A be a point on the x -axis such that

\angle \mathrm{OPA}=90^{\circ}

. Then the locus of the centroid of such triangles OPA is :

A

y^{2}-6 x+4=0

B

y^{2}-9 x+6=0

C

y^{2}-2 x+8=0

D

y^{2}-4 x+8=0

Correct Answer:

C

Explanation

\mathrm{m}_{\mathrm{AP}}=\frac{-\mathrm{t}}{2}\\

Equation of AP is

\\y-6 t=\frac{-t}{2}\left(x-3 t^{2}\right)\\

Put

\mathrm{y}=0 \Rightarrow \mathrm{x}=12+3 \mathrm{t}^{2}\\

\Rightarrow \mathrm{A}\left(12+3 \mathrm{t}^{2}, 0\right)\\

Let centroid of

\triangle

OPA be

\mathrm{G}(\mathrm{h}, \mathrm{k})\\

\Rightarrow 3 \mathrm{~h}=0+3 \mathrm{t}^{2}+12+3 \mathrm{t}^{2}\\

3 \mathrm{k}=0+6 \mathrm{t}+0\\

\Rightarrow \mathrm{t}=\frac{\mathrm{k}}{2}, \mathrm{~h}=2 \mathrm{t}^{2}+4\\

\Rightarrow \mathrm{h}=2 \frac{\mathrm{k}^{2}}{4}+4\\

⇒ Locus of

(\mathrm{h}, \mathrm{k})

is

\\y^{2}=2 x-8

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

Let

f: \mathbf{R} \rightarrow \mathbf{R}

be a twice differentiable function such that

f^{\prime \prime}(\mathrm{x})>0

for all

\mathrm{x} \in \mathbf{R}

and

f^{\prime}(\mathrm{a}-1)=0

, where a is real number. Let

\mathrm{g}(\mathrm{x})=f\left(\tan ^{2} \mathrm{x}-2 \tan \mathrm{x}+\mathrm{a}\right)

,

0<\mathrm{x}<\frac{\pi}{2}.\\

Consider the following two statements :

\\

(I) g is increasing in

\left(0, \frac{\pi}{4}\right)\\

(II) g is decreasing in

\left(\frac{\pi}{4}, \frac{\pi}{2}\right)

A

Neither (I) nor (II) is True

B

Only (II) is True

C

Only (I) is True

D

Both (I) and (II) are True

Correct Answer:

A

Explanation

\mathrm{g}(\mathrm{x})=f\left((\tan \mathrm{x}-1)^{2}+\mathrm{a}-1\right)\\

\mathrm{g}^{\prime}(\mathrm{x})=f^{\prime}\left((\tan \mathrm{x}-1)^{2}+\mathrm{a}-1\right) \cdot 2(\tan \mathrm{x}-1) \sec ^{2} \mathrm{x}\\

\because f^{\prime}(\mathrm{a}-1)=0

and

f^{\prime \prime}(\mathrm{x})>0\\

\therefore f^{\prime}\left((\tan \mathrm{x}-1)^{2}+\mathrm{a}-1\right)>0\\

\mathrm{g}^{\prime}(\mathrm{x})>0

if

(\tan \mathrm{x}-1)>0\\

g is increasing in

\mathrm{x} \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\\

\mathrm{g}^{\prime}(\mathrm{x})<0

if

\tan \mathrm{x}-1<0\\

g is decreasing in

\mathrm{x} \in\left(0, \frac{\pi}{4}\right)

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

Let the line

L_{1}

be parallel to the vector

-3 \hat{i}+2 \hat{j}+4 \hat{k}

and pass through the point

(2,6,7)

and the line

L_{2}

be parallel to the vector

2 \hat{i}+\hat{j}+3 \hat{k}

and pass through the point

(4,3,5)

. If the line

\mathrm{L}_{3}

is parallel to the vector

-3 \hat{i}+5 \hat{j}+16 \hat{k}

and intersects the lines

\mathrm{L}_{1}

and

\mathrm{L}_{2}

at the points C and D , respectively, then

|\overrightarrow{\mathrm{CD}}|^{2}

is equal to :

A

171

B

290

C

312

D

89

Correct Answer:

B

Explanation

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

Point C on

\mathrm{L}_{1}:\left(-3 \lambda_{1}+2,2 \lambda_{1}+6,4 \lambda_{1}+7\right)\\

\mathrm{L}_{2}: \frac{\mathrm{x}-4}{2}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-5}{3}\\

Point D on

\mathrm{L}_{2}:\left(2 \lambda_{2}+4, \lambda_{2}+3,3 \lambda_{2}+5\right)\\

Dr's of line

L_{3}

:

\\ \mathrm{L}_{3}: \frac{2 \lambda_{2}+3 \lambda_{1}+2}{-3}=\frac{\lambda_{2}-2 \lambda_{1}-3}{5}=\frac{3 \lambda_{2}-4 \lambda_{1}-2}{16}\\

\lambda_{1}=-3, \lambda_{2}=2\\

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

\\ |\overrightarrow{\mathrm{CD}}|^{2}=3^{2}+5^{2}+16^{2}=290

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

Let

\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{x}^{2} f^{\prime}(1)+2 \mathrm{x} f^{\prime \prime}(2)+f^{\prime \prime \prime}(3)

,

\mathrm{x} \in \mathrm{R}

. Then the value of

f^{\prime}(5)

is :

A

\frac{62}{5}

B

\frac{657}{5}

C

\frac{2}{5}

D

\frac{117}{5}

Correct Answer:

D

Explanation

\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+2 \mathrm{xf}^{\prime}(1)+2 \mathrm{f}^{\prime \prime}(2)\\

\mathrm{f}^{\prime \prime}(\mathrm{x})=6 \mathrm{x}+2 \mathrm{f}^{\prime}(1)\\

\mathrm{f}^{\prime \prime}(2)=12+2 \mathrm{f}^{\prime}(1)\\

\therefore \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+2 \mathrm{xf}^{\prime}(1)+2\left(12+2 \mathrm{f}^{\prime}(1)\right)\\

\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+2(\mathrm{x}+2) \mathrm{f}^{\prime}(1)+24\\

Putting,

\mathrm{x}=1\\

\mathrm{f}^{\prime}(1)=3+6 \mathrm{f}^{\prime}(1)+24\\

-5 f^{\prime}(1)=27 \Rightarrow f^{\prime}(1)=\frac{-27}{5}\\

\therefore \mathrm{f}^{\prime \prime}(2)=12+2\left(\frac{-27}{5}\right)=12-\frac{54}{5}=\frac{6}{5}\\

\therefore \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}-\frac{54}{5} \mathrm{x}+\frac{12}{5}\\

\therefore \mathrm{f}^{\prime}(5)=75-54+\frac{12}{5}=\frac{117}{5}

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

If P is a point on the circle

\mathrm{x}^{2}+\mathrm{y}^{2}=4, \mathrm{Q}

is a point on the straight line

5 \mathrm{x}+\mathrm{y}+2=0

and

\mathrm{x}-\mathrm{y}+1=0

is the perpendicular bisector of PQ , then 13 times the sum of abscissa of all such point P is

\_\_\_\_

.

Correct Answer:

2

Explanation

\\

Mid point of PQ lies on

x-y+1=0\\

\frac{2 \cos \theta+\alpha}{2}-\frac{2 \sin \theta-5 \alpha-2}{2}+1=0\\

2 \cos \theta+\alpha-2 \sin \theta+5 \alpha+2+2=0\\

\cos \theta-\sin \theta+3 \alpha+2=0 \tag........{1}\\

∵ Slope of PQ is -1

\\ \frac{2 \sin \theta+5 \alpha+2}{2 \cos \theta-\alpha}=-1\\

2 \sin \theta+5 \alpha+2=-2 \cos \theta+\alpha\\

\sin \theta+\cos \theta+2 \alpha+1=0 \tag..........{2}\\

eliminate

\alpha

from (1) and (2)

\\ \Rightarrow \cos \theta+5 \sin \theta=1, \theta \in[0,2 \pi]\\

\Rightarrow 5 \times 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}=2 \sin ^{2} \frac{\theta}{2}\\

\therefore \sin \frac{\theta}{2}=0 \Rightarrow \cos \theta=1\\

or

\\ \sin \frac{\theta}{2}=5 \Rightarrow \cos \theta=-\frac{12}{13}\\

Sum of all possible values of abscissa of point P is

\\=2 \times 1+2\left(\frac{-12}{13}\right)=\frac{2}{13}\\

∴ 13 times sum of all possible values of abscissa of point P is 2 .

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

Let

y=y(x)

be the solution of the differential equation

\sec x \frac{\mathrm{dy}}{\mathrm{dx}}-2 \mathrm{y}=2+3 \sin \mathrm{x}, \mathrm{x} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

,

y(0)=-\frac{7}{4}

. Then

y\left(\frac{\pi}{6}\right)

is equal to:

A

-\frac{5}{2}

B

-\frac{5}{4}

C

-3 \sqrt{3}-7

D

-3 \sqrt{2}-7

Correct Answer:

A

Explanation

\frac{d y}{d x}-2 y \cos x=2 \cos x+3 \sin x \cdot \cos x\\

I.F.

=\mathrm{e}^{-2 \sin \mathrm{x}}\\

e^{-2 \sin x} \cdot y=\int e^{-2 \sin x}(3 \sin x \cos x+2 \cos x) d x\\

y . e^{-2 \sin x}=e^{-2 \sin x}\left(-\frac{3}{2} \sin x-\frac{7}{4}\right)+C\\

\Rightarrow \mathrm{y}=-\frac{3}{2} \sin \mathrm{x}-\frac{7}{4}+\mathrm{C} . \mathrm{e}^{2 \sin \mathrm{x}}\\

\because y(0)=-\frac{7}{4} \Rightarrow C=0\\

y\left(\frac{\pi}{6}\right)=\frac{-3}{2} \cdot \frac{1}{2}-\frac{7}{4}=\frac{-5}{2}

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

The positive integer n , for which the solutions of the equation\\

\mathrm{x}(\mathrm{x}+2)+(\mathrm{x}+2)(\mathrm{x}+4)+\ldots .+(\mathrm{x}+2 \mathrm{n}-2)(\mathrm{x}+2 \mathrm{n})

\\

=\frac{8 n}{3}

\\ are two consecutive even integers, is :

A

3

B

6

C

12

D

9

Correct Answer:

A

Explanation

x(x+2)+(x+2)(x+4)+\ldots .+(x+2 n-2)(x+2 n)=\frac{8 n}{3}\\

\Rightarrow \sum_{\mathrm{r}=1}^{\mathrm{n}}(\mathrm{x}+2 \mathrm{r}-2)(\mathrm{x}+2 \mathrm{r})=\frac{8 \mathrm{n}}{3}\\

n x^{2}+2 x \sum_{r=1}^{n}(2 r-1)+4 \sum_{r=1}^{n} r(r-1)=\frac{8 n}{3}\\

n x^{2}+2 x \cdot n^{2}+\frac{4 n\left(n^{2}-1\right)}{3}-\frac{8 n}{3}=0\\

x^{2}+2 n x+\frac{4\left(n^{2}-1\right)}{3}-\frac{8}{3}=0<_{\beta}^{\alpha}\\

\because|\alpha-\beta|=2 \Rightarrow \frac{\sqrt{\mathrm{D}}}{|\mathrm{a}|}=2 \Rightarrow \mathrm{D}=4\\

\Rightarrow 4 \mathrm{n}^{2}-4\left(4 \frac{\left(\mathrm{n}^{2}-1\right)}{3}-\frac{8}{3}\right)=4\\

\Rightarrow \mathrm{n}^{2}-\frac{4 \mathrm{n}^{2}}{3}=-3\\

\Rightarrow \mathrm{n}^{2}=9\\

\Rightarrow \mathrm{n}=3

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

If the area of the region

\left\{(x, y): 1-2 x \leq y \leq 4-x^{2}\right.

,

x \geq 0, y \geq 0\}

is

\frac{\alpha}{\beta}, \alpha, \beta, \in N, \operatorname{gcd}(\alpha, \beta)=1

, then the value of

(\alpha+\beta)

is :

A

73

B

85

C

91

D

67

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

The largest

\mathrm{n} \in \mathrm{N}

, for which

7^{\mathrm{n}}

divides 101!, is :

A

16

B

18

C

15

D

19

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

If

\int_{0}^{1} 4 \cot ^{-1}\left(1-2 \mathrm{x}+4 \mathrm{x}^{2}\right) \mathrm{dx}=\mathrm{a} \tan ^{-1}(2)-\operatorname{blog}_{\mathrm{e}}(5)

, where

\mathrm{a}, \mathrm{b} \in \mathbf{N}

, then

(2 \mathrm{a}+\mathrm{b})

is equal to

\_\_\_\_

.

Correct Answer:

9

Explanation

Let

\mathrm{I}=\int_{0}^{1} \cot ^{-1}\left(1-2 \mathrm{x}+4 \mathrm{x}^{2}\right) \mathrm{dx}\\

I=\int_{0}^{1}\left(\cot ^{-1}(2 x-1)-\cot ^{-1}(2 x)\right) d x \tag{1}\\

Applying king

\\ I=\int_{0}^{1}\left(-\cot ^{-1}(2 x-1)+\cot ^{-1}(2 x-2)\right) d x \tag....{(2)}\\

From (1) & (2)

\\2 \mathrm{I}=\int_{0}^{1}\left(\cot ^{-1}(2 \mathrm{x}-2)-\cot ^{-1}(2 \mathrm{x})\right) \mathrm{dx}\\

=\int_{0}^{1} \cot ^{-1}(2 \mathrm{x}-2) \mathrm{dx}-\int_{0}^{1} \cot ^{-1}(2 \mathrm{x}) \mathrm{dx}\\

Applying King

\\=\int_{0}^{1} \cot ^{-1}(-2 x) d x-\int_{0}^{1} \cot ^{-1}(2 x) d x\\

=\int_{0}^{1}\left(\pi-\cot ^{-1}(2 \mathrm{x})\right) \mathrm{dx}-\int_{0}^{1} \cot ^{-1}(2 \mathrm{x}) \mathrm{dx}\\

=\int_{0}^{1}\left(\pi-2 \cot ^{-1}(2 x)\right) d x\\

=\pi-2 \int_{0}^{1}\left(\cot ^{-1} 2 \mathrm{x}\right) \cdot 1 \mathrm{dx}\\

By parts

\\ \pi - 2 \left[ \left( x \cot^{-1} 2x \right)_0^1 + \int_0^1 \frac{2x}{1 + 4x^2} dx \right] \\

\\

Let

1+4 \mathrm{x}^{2}=\mathrm{t}\\

8 \mathrm{xdx}=\mathrm{dt}\\

=\pi-2\left[\cot ^{-1} 2+\frac{1}{4} \int_{1}^{5} \frac{\mathrm{dt}}{\mathrm{t}}\right]\\

=\pi-2 \cot ^{-1} 2-\frac{1}{2} \ell \mathrm{n} 5\\

2 \mathrm{I}=2 \tan ^{-1} 2-\frac{1}{2} \ln 5\\

\Rightarrow 4 \mathrm{I}=4 \tan ^{-1} 2-\ell \mathrm{n} 5\\

\therefore 2 \mathrm{a}+\mathrm{b}=8+1=9

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JEE Mains 2026 2026-01-21 Shift 2

Mathematics

Let

z

be the complex number satisfying

|z-5| \leq 3

and having maximum positive principal argument. Then

34\left|\frac{5 z-12}{5 i z+16}\right|^{2}

is equal to:

A

16

B

12

C

26

D

20

Correct Answer:

D

Explanation

\\|z-5| \leq 3\\

For

\arg (\mathrm{z})

to be maximum, z lies at P .

\\ \begin{aligned} & z \equiv(4 \cos \theta, 4 \sin \theta) \\ & \equiv\left(4 \cdot\left(\frac{4}{5}\right), 4\left(\frac{3}{5}\right)\right)=\left(\frac{16}{5}, \frac{12}{5}\right) \\=\frac{16}{5}+\frac{12 \mathrm{i}}{5} \end{aligned} \\

Now,

34\left|\frac{5 \mathrm{z}-12}{5 \mathrm{iz}+16}\right|^{2} \\=34\left|\frac{(16+12 \mathrm{i})-12}{(16 \mathrm{i}-12)+16}\right|^{2}\\

=34\left|\frac{4+12 \mathrm{i}}{16 \mathrm{i}+4}\right|^{2}\\

=34\left(\frac{16+144}{256+16}\right) \\=34\left(\frac{160}{272}\right)=20

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JEE Mains 2026 2026-01-21 Shift 2

Physics

The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M . Two blocks of mass of M and m are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its centre. The magnitudes of the acceleration experienced by the blocks is

\_\_\_\_

(assume no slipping of string on pulley.)

\\

A

\frac{(M-m) g}{\left[\left(\frac{13}{6}\right) M+m\right]}

B

\frac{(M-m) g}{M+m}

C

\frac{(M-m) g}{\left[\left(\frac{8}{3}\right) M+m\right]}

D

\frac{(M-m) g}{2 M+m}

Correct Answer:

C

Explanation

\begin{align*} & \mathrm{Mg}-\mathrm{T}_{2}=\mathrm{Ma} \tag....{1}\\ & \mathrm{~T}_{1}-\mathrm{mg}=\mathrm{ma} \tag....{2}\\ & \left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right) \mathrm{r}=\mathrm{I} \frac{\mathrm{a}}{\mathrm{r}} \tag{3}\\ & (1)+(2)+(3) \\ & (\mathrm{M}-\mathrm{m}) \mathrm{g}=\left(\mathrm{M}+\mathrm{m}+\frac{\mathrm{I}}{\mathrm{r}^{2}}\right) \mathrm{a} \end{align*}\\

Here

I=M r^{2}+\frac{M \times(2 r)^{2}}{12} \times 2\\

=\left(1+\frac{2}{3}\right) \mathrm{Mr}^{2}\\

=\frac{5}{3} \mathrm{Mr}^{2}\\

(M-m) g=\left[M+m+\frac{5 M}{3}\right] a\\

a=\frac{(M-m) g}{\left[M+m+\frac{5 M}{3}\right]}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

Two cars A and B each of mass

10^{3} \mathrm{~kg}

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

72 \mathrm{~km} / \mathrm{h}

and

36 \mathrm{~km} / \mathrm{h}

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

\_\_\_\_

J.s.

A

3.6 \times 10^{5}

B

10^{5}

C

3 \times 10^{5}

D

2 \times 10^{5}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

A capacitor C is first charged fully with potential difference of

\mathrm{V}_{0}

and disconnected from the battery. The charged capacitor is connected across an inductor having inductance L . In t s

25 \%

of the initial energy in the capacitor is transferred to the inductor. The value of t is

\_\_\_\_

s.

A

\frac{\pi \sqrt{L C}}{3}

B

\frac{\pi \sqrt{\mathrm{LC}}}{6}

C

\frac{\pi \sqrt{\text { LC }}}{2}

D

\pi \sqrt{\frac{L C}{2}}

Correct Answer:

B

Explanation

\mathrm{U}_{\mathrm{c}_{\mathrm{f}}}=75 \% \mathrm{U}_{\mathrm{c}_{\mathrm{i}}}\\

\mathrm{Q}_{\mathrm{F}}^{2}=\frac{3}{4} \mathrm{Q}_{\mathrm{I}}^{2}\\

\mathrm{Q}_{\mathrm{i}} \cos \omega \mathrm{t}=\frac{\sqrt{3}}{2} \\ \mathrm{Q}_{\mathrm{i}} \Rightarrow \mathrm{t}=\frac{\mathrm{T}}{12}\\

\mathrm{t}=\frac{\pi}{6} \sqrt{\mathrm{LC}}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

In a Young's double slit experiment set up, the two slits are kept

0.4 . \mathrm{mm}

apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness

20 \mu \mathrm{~m}

is introduced in front of one of the slits then centre bring fringe shifts by 20 mm on the screen. The refractive index of transparent sheet is given by

\frac{\alpha}{10}

, where

\alpha

is

\_\_\_\_

.

Correct Answer:

14

Explanation

y_{\text {shift }}=\frac{(\mu-1) t . D}{d}\\

20910^{-3}=\frac{(\mu-1) \times 20 \times 10^{-6} \times 1}{0.4 \times 10^{-3}}\\

(\mu-1) \Rightarrow 0.4\\

\mu \Rightarrow 1.4\\

\frac{\alpha}{10}=1.4, \alpha=14

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JEE Mains 2026 2026-01-21 Shift 2

Physics

Two known resistance of

\mathrm{R} \Omega

and

2 \mathrm{R} \Omega

and and one unknown resistance

\mathrm{X} \Omega

are connected in a circuit as shown in the figure. If the equivalent resistance between points A and B in the circuit is

\mathrm{X} \Omega

, then the value of X is

\_\_\_\_

\Omega.\\

A

(\sqrt{3}-1) \mathrm{R}

B

R

C

2(\sqrt{3}-1) \mathrm{R}

D

(\sqrt{3}+1) \mathrm{R}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

A large drum having radius R is spinning around its axis with angular velocity

\omega

, as shown in figure. The minimum value of

\omega

so that a body of mass M remains stuck to the inner wall of the drum, taking the coefficient of friction between the drum surface and mass M is

\mu

, is :

\\

A

\sqrt{\frac{\mu \mathrm{g}}{\mathrm{R}}}

B

\sqrt{\frac{2 \mathrm{~g}}{\mu \mathrm{R}}}

C

\sqrt{\frac{g}{2 \mu R}}

D

\sqrt{\frac{g}{\mu R}}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

The r.m.s speed of oxygen molecules at

47^{\circ} \mathrm{C}

is equal to that of the hydrogen molecules kept at

\_\_\_\_

{ }^{\circ} \mathrm{C}

. (Mass of oxygen molecule/mass of hydrogen molecule=32/2)

A

-235

B

-100

C

-253

D

-20

Correct Answer:

C

Explanation

\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}\\

\mathrm{V}_{\mathrm{rmsO}_{2}}=\mathrm{V}_{\mathrm{rmsH}_{2}}\\

\mathrm{T}_{\mathrm{O}_{2}}=273+47=320 \mathrm{~K}\\

\sqrt{\frac{3 \mathrm{RT}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{RT}_{\mathrm{H}_{2}}}{\mathrm{M}_{\mathrm{H}_{2}}}}\\

\frac{\mathrm{T}_{2}}{\mathrm{M}_{\mathrm{O}_{2}}}=\frac{\mathrm{T}_{\mathrm{H}_{2}}}{\mathrm{M}_{\mathrm{H}_{2}}}\\

\frac{320}{32}=\frac{\mathrm{T}_{\mathrm{H}_{2}}}{2}\\

\mathrm{T}_{\mathrm{H}_{2}}=20 \mathrm{~K}\\

\mathrm{T}_{\mathrm{H}_{2}}=-253^{\circ} \mathrm{C}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

A spherical body of radius

r

and density

\sigma

falls freely through a viscous liquid having density

\rho

and viscosity

\eta

and attains a terminal velocity

\mathrm{v}_{0}

. Estimated maximum error in the quantity

\eta

is : (Ignore errors associated with

\sigma, \rho

and g , gravitational acceleration)

A

2 \frac{\Delta r}{r}-\frac{\Delta v_{0}}{v_{0}}

B

\frac{2 \Delta r}{r}+\frac{\Delta v_{0}}{v_{0}}

C

2\left[\frac{\Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta \mathrm{v}_{0}}{\mathrm{v}_{0}}\right]

D

2\left[\frac{\Delta \mathrm{r}}{\mathrm{r}}-\frac{\Delta \mathrm{v}_{0}}{\mathrm{v}_{0}}\right]

Correct Answer:

B

Explanation

\mathrm{v}_{0}=\frac{2}{9} \frac{\mathrm{r}^{2} \mathrm{~g}}{\mathrm{n}}\left(\rho_{\mathrm{B}}-\rho_{\mathrm{L}}\right)\\

\mathrm{n}=\frac{2}{9} \frac{\mathrm{r}^{2} \mathrm{~g}}{\mathrm{v}_{0}}\left(\rho_{\mathrm{B}}-\rho_{\mathrm{L}}\right)\\

\frac{\Delta \mathrm{n}}{\mathrm{n}}=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta \mathrm{v}_{0}}{\mathrm{v}_{0}}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

The terminal velocity of a metallic ball of radius 6 mm in a viscous fluid is

20 \mathrm{~cm} / \mathrm{s}

. The terminal velocity of another ball of same material and having radius 3 mm in the same fluid will be

\_\_\_\_

\mathrm{cm} / \mathrm{s}

.

Correct Answer:

5

Explanation

We know : Terminal velocity

\propto(\text { radius })^{2}\\

\frac{\left(\mathrm{v}_{\mathrm{T}}\right)_{1}}{\left(\mathrm{v}_{\mathrm{T}}\right)_{2}}=\left(\frac{6}{3}\right)^{2}\\

\left(\mathrm{v}_{\mathrm{T}}\right)_{2}=\frac{\left(\mathrm{v}_{\mathrm{T}}\right)_{1}}{4}=5 \mathrm{~cm} / \mathrm{sec}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

A battery with EMF E and internal resistance

r

is connected across a resistance

R

. The power consumption in R will be maximum when :

A

R=2 r

B

\mathrm{R}=\frac{\mathrm{r}}{2}

C

R=\sqrt{2} r

D

R=r

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JEE Mains 2026 2026-01-21 Shift 2

Physics

An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of the circular part is r . The magnetic field at the centre O of the circular loop is :

\\

A

\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}}{\mathrm{r}}(\pi+1) \hat{\mathrm{i}}

B

-\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}}{\mathrm{r}}(\pi-1) \hat{\mathrm{i}}

C

\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}}{\mathrm{r}}(\pi-1) \hat{\mathrm{i}}

D

-\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}}{\mathrm{r}}(\pi+1) \hat{\mathrm{i}}

Correct Answer:

B

Explanation

\\ \overrightarrow{\mathrm{B}}_{0}=\overrightarrow{\mathrm{B}}_{\mathrm{AB}}+\overrightarrow{\mathrm{B}}_{\mathrm{DE}}+\overrightarrow{\mathrm{B}}_{\mathrm{BCD}}\\

=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}} \hat{\mathrm{i}}+\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}} \hat{\mathrm{i}}-\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}} \hat{\mathrm{i}}\\

=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}} \hat{\mathrm{i}}-\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}} \hat{\mathrm{i}}\\

=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}(1-\pi) \hat{\mathrm{i}}\\

=-\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}(\pi-1) \hat{\mathrm{i}}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

A body of mass 2 kg is moving along x -direction such that its displacement as function of time is given by

\mathrm{x}(\mathrm{t})=\alpha \mathrm{t}^{2}+\beta \mathrm{t}+\gamma \mathrm{m}

, where

\alpha=1 \mathrm{~m} / \mathrm{s}^{2}

,

\beta=1 \mathrm{~m} / \mathrm{s}

and

\gamma=1 \mathrm{~m}

. The work done on the body during the time interval

\mathrm{t}=2 \mathrm{~s}

to

\mathrm{t}=3 \mathrm{~s}

, is\\

\_\_\_\_

J.

A

49

B

42

C

24

D

12

Correct Answer:

C

Explanation

\mathrm{x}(\mathrm{t})=\mathrm{t}^{2}+\mathrm{t}+1\\

\mathrm{v}(\mathrm{t})=2 \mathrm{t}+1\\

\mathrm{a}(\mathrm{t})=2\\

\mathrm{F}=4 \mathrm{~N}.\\

Displacement

=\mathrm{x}(3)-\mathrm{x}(2)\\

=13-7=6 \mathrm{M}\\

\mathrm{W}=\mathrm{F} . \mathrm{S}=4 \times 6=24 \mathrm{~J}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface.

\\

\\

If refractive index of the material of prism is

\sqrt{2}

, the angle

\theta

of prism is.

A

60^{\circ}

B

75^{\circ}

C

90^{\circ}

D

45^{\circ}

Correct Answer:

A

Explanation

\\

For grazing emergence

\\ \sin \mathrm{r}_{2}=\frac{1}{\mu}\\

By Snell's Law at incident surface

\\1 \times \frac{1}{\sqrt{2}}=\sqrt{2} \sin \mathrm{r}_{1}\\

\mathrm{r}_{1}=30\\

\mathrm{r}_{1}+\mathrm{r}_{2}=\mathrm{A}\\

\mathrm{A}=75\\

75+45+\theta=180^{\circ}\\

\theta=60^{\circ}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

Surface tension of two liquids (having same densities),

\mathrm{T}_{1}

and

\mathrm{T}_{2}

, are measured using capillary rise method utilizing two tubes with inner radii of

r_{1}

and

r_{2}

where

r_{1}>r_{2}

. The measured liquid heights in these tubes are

\mathrm{h}_{1}

and

\mathrm{h}_{2}

respectively. [Ignore the weight of the liquid about the lowest point of miniscus]. The heights

\mathrm{h}_{1}

and

\mathrm{h}_{2}

and surface tensions

\mathrm{T}_{1}

and

\mathrm{T}_{2}

satisfy the relation :

A

h_{1} < h_{2}

and

T_{1} = T_{2}

B

\mathrm{h}_{1}=\mathrm{h}_{2}

and

\mathrm{T}_{1}=\mathrm{T}_{2}

C

\mathrm{h}_{1}>\mathrm{h}_{2}

and

\mathrm{T}_{1}=\mathrm{T}_{2}

D

h_{1} > h_{2}

and

T_{1} < T_{2}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

The total length of potentiometer wire AB is 50 cm in the arrangement as shown in figure. If P is the point where the galvanometer shows zero reading then the length AP is

\_\_\_\_

cm .

\\

A

15

B

30

C

25

D

20

Correct Answer:

B

Explanation

\frac{6}{\mathrm{R}_{\mathrm{AP}}}=\frac{4}{\mathrm{R}_{\mathrm{PB}}}

;

\\ \ell_{\mathrm{AP}}+\ell_{\mathrm{PB}}=50 \tag .....{1}

\\ \begin{aligned} & \frac{\mathrm{R}_{\mathrm{AP}}}{\mathrm{R}_{\mathrm{PB}}}=\frac{\ell_{\mathrm{AP}}}{\ell_{\mathrm{PB}}}=\frac{3}{2} \\ & \ell_{\mathrm{AP}}=\frac{3}{5} \times 50=30 \mathrm{~cm} \end{aligned}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

A particle having electric charge

3 \times 10^{-19} \mathrm{C}

and mass

6 \times 10^{-27} \mathrm{~kg}

is accelerated by applying an electric potential of 1.21 V . Wavelength of the matter wave associated with the particle is

\alpha \times 10^{-12} \mathrm{~m}

. The value of

\alpha

is

\_\_\_\_

. (Take Planck's constant

=6.6 \times 10^{-34} \mathrm{~J} . \mathrm{s}

)

Correct Answer:

10

Explanation

\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}

\\ \begin{aligned} & \lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 18 \times 10^{-46} \times 1.21}} \\ & \lambda=10^{-11} \mathrm{~m}=10 \times 10^{-12} \mathrm{~m} \\ & \alpha=10 \end{aligned}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

The charge stored by the capacitor C in the given circuit in the steady state is

\_\_\_\_

\mu \mathrm{C}.\\

A

12.5

B

10

C

7.5

D

5

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JEE Mains 2026 2026-01-21 Shift 2

Physics

A diatomic gas

(\gamma=1.4)

does 100 J of work when it is expanded isobarically. Then the heat given to the gas

\_\_\_\_

J.

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JEE Mains 2026 2026-01-21 Shift 2

Physics

The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of

176 \mathrm{rad} / \mathrm{s}

. The frequency of this simple harmonic oscillator is\_\_\_ Hz.

\left[\right.

take

\left.\pi=\frac{22}{7}\right]

A

14

B

88

C

28

D

176

Correct Answer:

A

Explanation

\boldsymbol{\omega}=176 \mathrm{rad} / \mathrm{sec}\\

\mathrm{f}_{\mathrm{k}}=\frac{\omega}{2 \pi}=\frac{176}{2 \times 22} \times 7\\

=\frac{176}{44} \times 7\\

=4 \times 7=28 \mathrm{~Hz}\\

So frequency of oscillator

\\=\frac{\mathrm{f}_{\mathrm{k}}}{2}=14 \mathrm{~Hz}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

Consider two identical metallic spheres of radius

R

each having charge

Q

and mass

m

. Their centers have an initial separation of

4 R

. Both the spheres are given an initial speed of

u

towards each other. The minimum value of

u

, so that they can just touch each other is :

\\

(Take

k=\frac{1}{4 \pi \epsilon_{0}}

and assume

k Q^{2}>G m^{2}

where G is the Gravitational constant)

A

\sqrt{\frac{k Q^{2}}{4 m R}\left(1-\frac{G m^{2}}{k Q^{2}}\right)}

B

\sqrt{\frac{k Q^{2}}{4 m R}\left(1+\frac{G m^{2}}{k Q^{2}}\right)}

C

\sqrt{\frac{k Q^{2}}{2 m R}\left(1-\frac{G m^{2}}{k Q^{2}}\right)}

D

\sqrt{\frac{k Q^{2}}{2 m R}\left(1-\frac{G m^{2}}{2 k Q^{2}}\right)}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity,

\sigma=10 \mathrm{mho} / \mathrm{m}

. The ratio of maximum conducting current density to maximum displacement current density is

\_\_\_\_

。\\

\left[\right.

Take

\left.\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right]

Correct Answer:

1800

Explanation

\mathrm{j}_{\mathrm{c}}=\sigma \mathrm{E}\\

\mathrm{E} \Rightarrow \mathrm{E}_{0} \sin (\omega \mathrm{t}-\mathrm{kx})\\

\mathrm{j}_{\mathrm{c}}=\sigma \mathrm{E}_{0} \sin (\omega \mathrm{t}-\mathrm{kx})\\

\Rightarrow\left(\mathrm{j}_{\mathrm{c}}\right)_{\text {max }}=\sigma \mathrm{E}_{0} \tag.......{1}\\

\mathrm{J}_{\mathrm{d}} \Rightarrow \frac{\mathrm{i}_{\mathrm{d}}}{\mathrm{A}}=\frac{1}{\mathrm{~A}} \times \varepsilon_{0} \frac{\mathrm{AdE}}{\mathrm{dt}}\\

\Rightarrow \varepsilon_{0} \times \mathrm{E}_{0} \omega \cos (\omega \mathrm{t}-\mathrm{kx})\\

\left(\mathrm{j}_{\mathrm{d}}\right)_{\text {max }} \Rightarrow \varepsilon_{0} \mathrm{E}_{0} \omega \tag.......{2}\\

(i)/(ii)

\\ \frac{\left(\mathrm{j}_{\mathrm{c}}\right)_{\text {max }}}{\left(\mathrm{j}_{\mathrm{d}}\right)_{\text {max }}}=\frac{\sigma \mathrm{E}_{0}}{\varepsilon_{0} \omega \mathrm{E}_{0}} \Rightarrow \frac{\sigma}{\varepsilon_{0} \omega}\\

\Rightarrow \frac{10 \times 4 \pi \times 9 \times 10^{9}}{2 \pi \times 100 \times 10^{6}}\\

\Rightarrow 1800

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JEE Mains 2026 2026-01-21 Shift 2

Physics

A river of width 200 m is flowing from west to east with a speed of

18 \mathrm{~km} / \mathrm{h}

. A boat, moving with speed of

36 \mathrm{~km} / \mathrm{h}

in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also the displacement along the river bank are

\_\_\_\_

and

\_\_\_\_

respectively.

A

20 s and 100 m

B

40 s and 0 m

C

40 s and 200 m

D

40 s and 100 m

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JEE Mains 2026 2026-01-21 Shift 2

Physics

The energy of an electron in an orbit of the Bohr's atom is

-0.04 \mathrm{E}_{0} \mathrm{eV}

where

\mathrm{E}_{0}

is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then

\frac{2 \pi \mathrm{~L}}{\mathrm{~h}}

is

\_\_\_\_

:

A

2

B

4

C

5

D

6

Correct Answer:

C

Explanation

Angular momentum

\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}\\

\mathrm{n}=\frac{2 \pi \mathrm{~L}}{\mathrm{~h}}\\

Energy

\mathrm{E}=-\frac{13.6}{\mathrm{n}^{2}} . \mathrm{z}^{2}

\\ \begin{aligned} & \mathrm{E} \Rightarrow-\frac{\mathrm{E}_{0}}{\mathrm{n}^{2}}=-0.04 \mathrm{E}_{0} \\ & \mathrm{n}^{2}=25, \mathrm{n}=5 \end{aligned}

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JEE Mains 2026 2026-01-21 Shift 2

Physics

Keeping the significant figures in view, the sum of the physical quantities

52.01 \mathrm{~m}, 153.2 \mathrm{~m}

and 0.123 m is :

A

205 m

B

205.333 m

C

205.33 m

D

205.3 m

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JEE Mains 2026 2026-01-21 Shift 2

Physics

Given below are two statements: Statement I: In a Young's double slit experiment, the angular separation of fringes will increase as the screen is moved away from the plane of the slits

\\

Statement II : In a Young's double slit experiment, the angular separation of fringes will increase when monochromatic source is replaced by another monochromatic source of higher wavelength

\\

In the light of the above statements, choose the correct answer from the options given below:

A

Both Statement I and Statement II are true

B

Both Statement I and Statement II are false

C

Statement I is false but Statement II is true

D

Statement I is true but Statement II is false

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