JEE Mains 2026-01-21 Shift 1 Question Paper With Detailed Solutions

JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Given below are two statements:

\\

Statement I: Among

\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2+},\left[\mathrm{Ni}(\mathrm{en})_{3}\right)\right]^{2+}

,

\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}

and

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

, has the maximum number of unpaired electrons.

\\

Statement II: The number of pairs among

\left\{\left[\mathrm{NiCl}_{4}\right]^{2-},\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\right\},\left\{\left[\mathrm{NiCl}_{4}\right]^{2-},\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\right.

and

\left\{\left[\mathrm{Ni}(\mathrm{CO})_{4}\right], \quad\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\right\}

that contain only diamagnetic species is two.

\\

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

A

Statement I is false but Statement II is true

B

Both Statement I and Statement II are true

C

Both Statement I and Statement II are false

D

Statement I is true but Statement II is false

Explanation

\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} \Rightarrow \mathrm{d}^{9}, \mathrm{dsp}^{2}

one unpaired electron

\\ \left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+} \Rightarrow \mathrm{d}^{8}, \mathrm{sp}^{3} \mathrm{~d}^{2}

two unpaired electrons

\\ \left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+} \\ \Rightarrow \mathrm{d}^{8}, \mathrm{sp}^{3} \mathrm{~d}^{2}

two unpaired electrons

\\ \left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \\ \Rightarrow \mathrm{d}^{5}, \mathrm{sp}^{3} \mathrm{~d}^{2}

five unpaired electrons

\\ \left[\mathrm{Ni}(\mathrm{CO})_{4}\right]

(diamagnetic)

\\ \left[\mathrm{NiCl}_{4}\right]^{-2}

(paramagnetic)

\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{-2}

(diamagnetic)

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by

20 \mathrm{~kJ} \mathrm{~mol}^{-1}

. If

\mathrm{k}_{1}

and

\mathrm{k}_{2}

are the rate constants of first and second reaction respectively at 300 K , then In

\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}

will be (nearest integer)

\left[\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]

Explanation

\mathrm{A} \xrightarrow{\mathrm{RX}^{\mathrm{n}}(1)}

product

\mathrm{E}_{1}\\

\mathrm{B} \xrightarrow{\mathrm{RX}^{\mathrm{n}}(2)}

product

\mathrm{E}_{2}\\

Assuming 'A' same for both reaction.

\\ \ln \mathrm{k}_{1}=\ln \mathrm{A}-\frac{\mathrm{E}_{1}}{300 \mathrm{R}}\\

\ln \mathrm{k}_{1}=\ln \mathrm{A}-\frac{\mathrm{E}_{2}}{300 \mathrm{R}}\\

\ln \left(\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}\right)=\frac{\mathrm{E}_{1}-\mathrm{E}_{2}}{300 \mathrm{R}}=\frac{20 \times 1000}{300 \mathrm{R}}\\

=8.032

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Consider the following reactions.

\\ \mathrm{PbCl}_{2}+\mathrm{K}_{2} \mathrm{CrO}_{4} \rightarrow \mathrm{~A}+2 \mathrm{KCl}

\\ (Hot solution)

\\ \mathrm{A}+\mathrm{NaOH} \rightleftharpoons \mathrm{B}+\mathrm{Na}_{2} \mathrm{CrO}_{4}

\\

\mathrm{PbSO}_{4}+4 \mathrm{CH}_{3} \mathrm{COONH}_{4} \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}+\mathrm{X}\\

In the above reactions,

\mathrm{A}, \mathrm{B}

and X are respectively.

A

\mathrm{Na}_{2}\left[\mathrm{Pb}(\mathrm{OH})_{2}\right], \mathrm{PbCrO}_{4} and \left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{Pb}\left(\mathrm{CH}_{3}\,\mathrm{COO}\right)_{4}\right]

B

\mathrm{PbCrO}_{4}, \mathrm{Na}_{2}\left[\mathrm{Pb}(\mathrm{OH})_{4}\right] and \left[\mathrm{Pb}\left(\mathrm{NH}_{3}\right)_{4}\right] \\mathrm{SO}_{4}

C

\mathrm{Na}_{2}\left[\mathrm{Pb}(\mathrm{OH})_{2}\right], \mathrm{PbCrO}_{4} and \left[\mathrm{Pb}\left(\mathrm{NH}_{3}\right)_{4}\right] \\mathrm{SO}_{4}

D

\mathrm{PbCrO}_{4}, \mathrm{Na}_{2}\left[\mathrm{Pb}(\mathrm{OH})_{4}\right] and \left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{Pb}\left(\mathrm{CH}_{3}\,\mathrm{COO}\right)_{4}\right]

Explanation

\underset{\text { (Hot solution) }}{\mathrm{PbCl}_{2}}+\mathrm{K}_{2} \mathrm{CrO}_{4} \rightarrow \underset{\text { (A) }}{\mathrm{PbCrO}_{4}}+2 \mathrm{KCl}\\

\begin{aligned} \mathrm{PbCrO}_{4}+4 \mathrm{NaOH} & \text { (excess) } \\ & \rightarrow \mathrm{Na}_{2}\left[\mathrm{~Pb}(\mathrm{OH})_{4}\right]+\mathrm{Na}_{2} \mathrm{CrO}_{4} \end{aligned}\\

\begin{aligned} \mathrm{PbSO}_{4} & +4 \mathrm{CH}_{3} \mathrm{COONH}_{4} \\ & \rightarrow\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{~Pb}\left(\underset{(\mathrm{X})}{\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{4}}\right]+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\right. \end{aligned}

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

An organic compound "P" of molecular formula

\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{3}

gives positive lodoform test but negative Tollen's test. When "P" is treated with dilute acid, it produces "Q". "Q" gives positive Tollen's test and also iodoform test. The structure of " P " is :

A

B

C

D

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Which of the following represents the correct trend for the mentioned property?

\\

A.

\mathrm{F}>\mathrm{P}>\mathrm{S}>\mathrm{B}-

Frist Ionization Energy

\\

B.

\mathrm{Cl}>\mathrm{F}>\mathrm{S}>\mathrm{P}-

Electron Affinity

\\

C.

\mathrm{K}>\mathrm{Al}>\mathrm{Mg}>\mathrm{B}-

Metallic character

\\

D.

\mathrm{K}_{2} \mathrm{O}>\mathrm{Na}_{2} \mathrm{O}>\mathrm{MgO}>\mathrm{Al}_{2} \mathrm{O}_{3}

- Basic character

\\

Choose the correct answer from the option given below.

A

A, B and D only

B

A, B, C and D

C

A and B only

D

B and C only

Explanation

On moving left to right in a period IE increases and from top to bottom in a group IE decreases.

\\ \mathrm{F}>\mathrm{P}>\mathrm{S}>\mathrm{B}

(IE order)

\\

⇒ On moving left to right in a period metallic and basic character decreases.

\\

\mathrm{K}>\mathrm{Mg}>\mathrm{Al}>\mathrm{B}

(Metallic character order)

\\

⇒ On moving top to bottom in a group metallic and basic character increases.

\\

\mathrm{K}_{2} \mathrm{O}>\mathrm{Na}_{2} \mathrm{O}>\mathrm{MgO}>\mathrm{Al}_{2} \mathrm{O}_{3}\\

⇒ EA : Group

17>

Group

16>

Group 15

\\ \mathrm{Cl}>\mathrm{F}>\mathrm{S}>\mathrm{P}

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Consider the following reaction sequence

\\

\\

The percentage of nitrogen in product ' T ' formed is

\_\_\_\_

\%. (Nearest integer)\\ (Given molar mass in

\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14, \mathrm{O}: 16

)

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Use the following data : TABLE One mole each of

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

and

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

are taken in a 1L closed flask and allowed to establish the equilibrium at 500 K .

\\ \mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})\\

The value of

\mathrm{x}\left(\mathrm{in} \mathrm{kJ} \mathrm{mol}^{-1}\right)

is

\_\_\_\_

(Nearest integer)

\\

(Given:

\log \mathrm{K}=2.2 \mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}

)

Explanation

\mathrm{A}_{2}+\mathrm{B}_{2} \xrightarrow{500 \mathrm{~K}} 2 \mathrm{AB} \log \mathrm{K}=2.2\\

\Delta \mathrm{H}^{\circ}=(2 \times 32)-(6+\mathrm{x})=(58-\mathrm{x}) \mathrm{kJ}\\

\Delta \mathrm{S}^{\circ}=(2 \times 222)-(146+280)=18

Joule

\\ \Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K}\\

\Delta \mathrm{G}^{\circ}=-\frac{8.314 \times 500 \times 2.2 \times 2.303}{1000}\\

\Delta \mathrm{G}^{\circ}=-21.06\\

\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}=-21.06\\

58-\mathrm{x}-500\left(\frac{18}{1000}\right)=-21.06\\

\mathrm{x}=70.06 \mathrm{KJ} / \mathrm{mol}

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

From the following, the least stable structure is :

A

B

C

D

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and

\mathrm{PQ}_{2}

. When 1 g of PQ is dissolved in 50 g of solvent ' A '.

\Delta \mathrm{T}_{\mathrm{b}}

was 1.176 K while when 1 g of

\mathrm{PQ}_{2}

is dissolved in 50 g of solvent ' A ',

\Delta \mathrm{T}_{\mathrm{b}}

was 0.689 K . (

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

of ' A '

=5 \mathrm{~K} \mathrm{kg} \mathrm{mol}^{-1}

). The molar masses of elements P and Q (in

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

) respectively, are :

A

70,110

B

65, 145

C

60, 25

D

25, 60

Explanation

\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{PQ}}=\mathrm{K}_{\mathrm{b}} \mathrm{m}\\

1.176=5 \times \frac{1}{\mathrm{M}_{1}} \times \frac{1000}{50}\\

\mathrm{M}_{1}=85.03\\

\left(\Delta \mathrm{T}_{\mathrm{b}}\right)_{\mathrm{PQ}_{2}}=5 \times \frac{1}{\mathrm{M}_{2}} \times \frac{1000}{50}=0.68\\

\mathrm{M}_{2}=145.13\\

Let molar mass of

\mathrm{P} \& \mathrm{Q}

are

\mathrm{M}_{\mathrm{P}}

and

\mathrm{M}_{\mathrm{Q}}

respectively

\\ \mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}=85.03\\

\mathrm{M}_{\mathrm{P}}+2 \mathrm{M}_{\mathrm{Q}}=145.13\\

\mathrm{M}_{\mathrm{p}}=24.93 \approx 25\\

\mathrm{M}_{\mathrm{Q}}=60.1 \approx 60

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Identify the correct statements.

\\

A. Arginine and Tryptophan are essential amino acids.

\\

B. Histidine does not contain heterocyclic ring in its structure.

\\

C. Proline is a six membered cyclic ring amino acid.

\\

D. Glycine does not have chiral centre.

\\

E. Cysteine has characteristic feature of side chain as

\mathrm{MeS}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-.\\

Choose the correct answer from the options given below:

A

C and E Only

B

B and E Only

C

C and D Only

D

A and D Only

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Consider the following reactions:

\\ \mathrm{NaCl}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{~A}+\mathrm{KHSO}_{4}+\mathrm{NaHSO}_{4}+\mathrm{H}_{2} \mathrm{O}\\

\mathrm{A}+\mathrm{NaOH} \rightarrow \mathrm{B}+\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}\\

\mathrm{B}+\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{C}+\mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\\

In the product ' C ', ' X ' is the number of

\mathrm{O}_{2}^{2-}

units, ' Y ' is the total number oxygen atoms present and '

Z

' is the oxidation state of Cr .

\\

The value of

\mathrm{X}+\mathrm{Y}+\mathrm{Z}

is

\_\_\_\_

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

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

32 \mathrm{~g} \mathrm{~mol}^{-1}.\\

Molar mass of barium sulphate is

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

.

A

4.55 \%

B

10.30 \%

C

21.97 \%

D

16.48 \%

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

The pH and conductance of a weak acid (HX) was found to be 5 and

4 \times 10^{-5} \mathrm{~S}

, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of

1 \mathrm{~cm}^{2}

were at a distance of 15 cm apart. The value of the limiting molar conductivity is

\_\_\_\_

\mathrm{S} \mathrm{m}^{2} \mathrm{~mol}^{-1}

. (nearest integer)

\\

(Given: degree of dissociation of the weak acid (a) << 1)

Explanation

\mathrm{pH}=5\\

\left[\mathrm{H}^{+}\right]=10^{-5}=[\mathrm{HX}] . \alpha\\

=[\mathrm{HX}] \cdot \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\infty}}\\

\Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{[\mathrm{HX}]}\\

\mathrm{K}=\mathrm{G} . \mathrm{G}^{*} \\=4 \times 10^{-5} \times \frac{15}{1}=6 \times 10^{-4} \mathrm{~S} . \mathrm{cm}^{-1}\\

\left[\mathrm{H}^{+}\right]=10^{-5} \\=[\mathrm{HX}] \times \frac{6 \times 10^{-4} \times 1000}{\Lambda_{\mathrm{m}}^{\infty} \times[\mathrm{HX}]}\\

\Lambda_{\mathrm{m}}^{\infty} =60000 \mathrm{S.cm}^{2} \mathrm{~mol}^{-1}\\

\Lambda_{\mathrm{m}}^{\infty} =6 \mathrm{~S} . \mathrm{m}^{2} \mathrm{~mol}^{-1}

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Which of the following graphs between pressure 'P' versus volume 'V' represent the maximum work done?

A

B

C

D

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Given below are two statements :

\\

Statement I: The number of pairs among [

\mathrm{SiO}_{2}, \mathrm{CO}_{2}

],

\left[\mathrm{SnO}, \mathrm{SnO}_{2}\right],\left[\mathrm{PbO}, \mathrm{PbO}_{2}\right]

and

\left[\mathrm{GeO}, \mathrm{GeO}_{2}\right]

, which contain oxides that are both amphoteric is 2.

\\

Statement II :

\mathrm{BF}_{3}

is an electron deficient molecule can act as a lewis acid, forms adduct with

\mathrm{NH}_{3}

and has a trigonal planar geometry.

\\

In the light of the above statement, choose the correct answer from the option 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 Statement II is true

Explanation

\Rightarrow \mathrm{SiO}_{2}, \mathrm{CO}_{2}, \mathrm{GeO}, \mathrm{GeO}_{2}

are acidic in nature.

\\ \mathrm{SnO}, \mathrm{SnO}_{2}, \mathrm{PbO}, \mathrm{PbO}_{2}

are amphoteric in nature.

\\ \Rightarrow \mathrm{BF}_{3}

is lewis acid acoording to lewic octet theory and has

\mathrm{sp}^{2}

hybridization with trigonal planar geometry and it can accept lone pair form ammonia to form adduct.

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Identify correct statement from the following :

\\

A. Propanal and propanone are functional isomers

\\

B. Ethoxyethane and methoxypropane are metamers.

\\

C. But-2-ene shows optical isomerism.

\\

D. But-1-ene and but-2-ene are functional isomers.

\\

E. Pentane and 2, 2-dimethyl propane are chain isomers.

\\

Choose the correct answer from the options given below :

A

B, C and D only

B

A, B and C only

C

A, B and E only

D

C, D and E only

Explanation

\\ \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3} \& \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}\\

Are Positional isomer

\\ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{3} \& \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{3}\\

are Metamer

\\ \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}

does not have optical isomer

\\

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Given below are two statements:

\\

Statement I: The number of species among

\mathrm{SF}_{4}

,

\mathrm{NH}_{4}^{+}, \quad\left[\mathrm{NiCl}_{4}\right], \quad \mathrm{XeF}_{4}, \quad\left[\mathrm{PtCl}_{4}\right]^{2-}, \quad \mathrm{SeF}_{4} \quad

and

\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}

, that have tetrahedral geometry is 3 .

\\

Statement II: In the set

\left[\mathrm{NO}_{2}, \mathrm{BeH}_{2}, \mathrm{BF}_{3}, \mathrm{AIC1}_{3}\right]

, all the molecules have incomplete octet around central atom. 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 false

C

Statement I is false but Statement II is true

D

Both Statement I and Statement II are true

Explanation

Statement-I

\\

\mathrm{SF}_{4}

(See-saw)

\\

\mathrm{XeF}_{4}

(square planar),

\\

\left[\mathrm{PtCl}_{4}\right]^{2-}

(square planar),

\\

\left[\mathrm{NiCl}_{4}\right]^{2-}

(Tetrahedral),

\\

\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}

(square planar),

\\

\mathrm{SeF}_{4}

(See-saw)

\\

\mathrm{NH}_{4}^{+}

(Tetrahedral)

\\

Statement-II

\\

\mathrm{NO}_{2}

(seven electrons on N )

\\

\mathrm{BeH}_{2}

(four electrons on Be )

\\

\mathrm{BF}_{3}

(six electrons on B )

\\

\mathrm{AlCl}_{3}

(six electrons on Al )

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

For the reaction,

\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}

, graph is plotted as shown below. Identify correct statements.

\\

A. Standard free energy change for the reaction is

-5.40 \mathrm{~kJ} \mathrm{~mol}^{-1}.\\

B. As

\mathrm{AG}^{\ominus}

in graph is positive,

\mathrm{N}_{2} \mathrm{O}_{4}

will not dissociate into

\mathrm{NO}_{2}

at all.

\\

C. Reverse reaction will go to completion.

\\

D. When 1 mole of

\mathrm{N}_{2} \mathrm{O}_{4}

changes into equilibrium mixture, value of

\Delta \mathrm{G}^{\ominus}=-0.84 \mathrm{~kJ} \mathrm{~mol}^{-1}\\

E. When 2 mole of

\mathrm{NO}_{2}

, changes into equilibrium mixture,

\Delta \mathrm{G}^{\ominus}

for equilibrium mixture is

-6.24 \mathrm{~kJ} \mathrm{~mol}^{-1}.\\

E. When 2 mole of

\mathrm{NO}_{2}

, changes into equilibrium mixture,

\Delta \mathrm{G}^{\ominus}

for equilibrium mixture is

-6.24 \mathrm{~kJ} \mathrm{~mol}^{-1}

Choose the correct answer from the options given below :

A

D and E only

B

C and E only

C

A and D only

D

B and C only

Explanation

\\

(A)

\Delta_{\mathrm{r}} \mathrm{G}^{\circ}=\mathrm{G}^{\circ}{ }_{\mathrm{B}}-\mathrm{G}^{\circ}{ }_{\mathrm{A}}=+\mathrm{ve}\\

(B)

\Delta_{\mathrm{r}} \mathrm{G}^{\circ}=+\mathrm{ve}, \mathrm{N}_{2} \mathrm{O}_{4}

will partially dissociates into

\mathrm{NO}_{2}.\\

(C) For reverse reaction It is partially completed as there is equilibrium at E .

\\

(D) For 1 mole

\mathrm{N}_{2} \mathrm{O}_{4} ; \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-0.84 \mathrm{~kJ} \mathrm{~mol}^{-1}\\

(E) For 2 mole

\mathrm{NO}_{2} ; \Delta \mathrm{G}^{\circ}=-5.4-0.84\\

=-6.24 \mathrm{~kJ} \mathrm{~mol}^{-1 .}

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

A hydrocarbon ' P '

\left(\mathrm{C}_{4} \mathrm{H}_{8}\right)

on reaction with HCl gives an optically active compound ' Q '

\left(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}\right)

which on reaction with one mole of ammonia gives compound ' R '

\left(\mathrm{C}_{4} \mathrm{H}_{11} \mathrm{~N}\right)

. ' R ' on diazotization followed by hydrolysis gives ' S '. Identify

\mathrm{P}, \mathrm{Q}, \mathrm{R}

and S .

A

B

C

D

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL . When the system is treated with KOH solution, the volume decreases to 64 mL . The formula of the hydrocarbon is :

A

\mathrm{C}_{2}\,\mathrm{H}_{4}

B

\mathrm{C}_{4}\,\mathrm{H}_{10}

C

\mathrm{C}_{2}\,\mathrm{H}_{2}

D

\mathrm{C}_{2}\,\mathrm{H}_{6}

Explanation

\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}(\mathrm{~g})}+\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right) \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{xCO}_{2(\mathrm{~g})}+\frac{\mathrm{y}}{2} \mathrm{H}_{2} \mathrm{O}_{(\ell)}\\

\begin{array}{llccl}\mathrm{t}=0 & 80 & 264 & 0 & - \\ \mathrm{t}=\mathrm{t}_{\text {final }} & - & 264-80\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right) & 80 \mathrm{x} & -\end{array}\\

264-80\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right)+80 \mathrm{x}=224\\

264-\frac{80 \mathrm{y}}{4}=224\\

40=\frac{80 \mathrm{y}}{4} \Rightarrow \mathrm{y}=2\\

264-80\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right)=64\\

264-80\left(\mathrm{x}+\frac{1}{2}\right)=64\\

264-80 \mathrm{x}-40=64\\

\mathrm{x}=2

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

An organic compound

(\mathrm{P})

on treatment with aqueous ammonia under hot condition forms compound

(\mathrm{Q})

which on heating with

\mathrm{Br}_{2}

and KOH forms compound (R) having molecular formula

\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{~N}

. Names of

\mathrm{P}, \mathrm{Q}

and R respectively are.

A

Benzoic acid, benzamide, aniline

B

Toluic acid, methylbenzamide, 2-methylaniline

C

Benzoic acid,4-methylbenzamide,4-methylaniline

D

Phenylethanoic acid, phenylethanamide, benzamine

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

14.0 g of calcium metal is allowed to react with excess HCl at 1.0 atm pressure and 273 K .

\\

Which of the following statements is incorrect?

\\

[Given : Molar mass in

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

of

\mathrm{Ca}-40, \mathrm{Cl}-35.5, \mathrm{H}-1

]

A

0.35\,\mathrm{~mol}\,\mathrm{O}_{2}\,\text{ gas is evolved.}

B

7.84\,\mathrm{~L}^{\circ}\,\text{ of }\mathrm{H}_{2}\,\text{ gas is evolved.}

C

33.3\,\mathrm{g}\,\text{ of }\mathrm{CaCl}_{2}\,\text{ is produced.}

D

The limiting reagent is calcium metal.

Explanation

\underset{\frac{14}{40}}{\mathrm{Ca}_{(\mathrm{s})}}+\underset{\text { excess }}{2 \mathrm{HCl}_{(\mathrm{g})}} \longrightarrow \underset{-}{\mathrm{CaCl}_{2(\mathrm{~s})}}+\underset{-}{\mathrm{H}_{2(\mathrm{~g})}}\\

=0.35

mole

\quad 0.35

mole 0.35 mole

\\

Volume of

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

evolved

\\=0.35 \times 22.4=7.84 \mathrm{~L}\\

(3) is wrong because weight of

\mathrm{CaCl}_{2}=.35 \times 111=

38.85 gm

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

\mathrm{MnO}_{4}^{2-}

, in acidic medium, disproportionates to :

A

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

and

\mathrm{MnO}_{2}

B

\mathrm{MnO}_{4}^{-}

and MnO

C

\mathrm{MnO}_{4}^{-}

and

\mathrm{MnO}_{2}

D

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

and MnO

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Identify A in the following reaction.

A

B

C

D

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JEE Mains 2026 2026-01-21 Shift 1

Chemistry

Given below are two statements:

\\

Statement I: When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the energetically excited hydrogen atoms produce electromagnetic radiation of discrete frequencies.

\\

Statement II: The frequency of second line of Balmer series obtained from

\mathrm{He}^{+}

is equal to that of first line of Lyman series obtained from hydrogen atom.

\\

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

Explanation

\frac{1}{\lambda}=R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\\

For Ist line of lyman series in H -atom

\\ \frac{1}{\lambda}=\mathrm{R}(1)^{2}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)

\\ \frac{1}{\lambda}=\frac{3 R}{4}\\

for

2^{\text {nd }}

line of Balmer series of

\mathrm{He}^{+}\\

\frac{1}{\lambda^{\prime}}=R(2)^{2}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)\\

\frac{1}{\lambda^{\prime}}=\frac{3 \mathrm{R}}{4}\\

As

\lambda

and

\lambda^{\prime}

is equal so frequency of these lines will be also equal.

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

Let O be the vertex of the parabola

x^{2}=4 y

and Q be any point on it. Let the locus of the point P , which divides the line segment OQ internally in the ratio 2: 3 be the conic C . Then the equation of the chord of C , which is bisected at the point

(1,2)

, is:

A

5 x-y-3=0

B

4 x-5 y+6=0

C

x-2 y+3=0

D

5 x-4 y+3=0

Explanation

h=\frac{4 t}{5}\\

\\

\mathrm{k}=\frac{2 \mathrm{t}^{2}}{5}=\frac{2}{5} \cdot\left(\frac{5 \mathrm{~h}}{4}\right)^{2}\\

8 \mathrm{k}=5 \mathrm{~h}^{2}\\

\Rightarrow 5 \mathrm{x}^{2}=8 \mathrm{y}\\

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

5\left(\mathrm{xx}_{1}\right)-4\left(\mathrm{y}+\mathrm{y}_{1}\right)=5 \mathrm{x}_{1}{ }^{2}-8 \mathrm{y}_{1}\\

5(\mathrm{x})-4(\mathrm{y}+2)=5-8.2\\

5 x-4 y+3=0

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

The number of strictly increasing functions f from the set

\{1,2,3,4,5,6\}

to the set

(1,2,3, \ldots, 9)

such that

\mathrm{f}(\mathrm{i}) \neq \mathrm{i}

for

1 \leq \mathrm{i} \leq 6

, is equal to :

A

21

B

27

C

22

D

28

Explanation

\mathrm{f}(\mathrm{i}) \neq \mathrm{i}, \mathrm{f}(\mathrm{x})

is strictly increasing function

\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}

, where

\mathrm{A}\{1,2,3 \ldots \ldots .6\}\\

B

\{1,2,3, \ldots \ldots .9\}

, then number of function

\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}

is equal to

\\

\\ \mathrm{f}(\mathrm{i}) \neq \mathrm{i}

Case -

\mathrm{i} \mathrm{f}(1)=2 \Rightarrow{ }^{7} \mathrm{C}_{5}=21\\

Case- ii

\mathrm{f}(1)=3 \Rightarrow{ }^{6} \mathrm{C}_{5}=6\\

Case- iii

\mathrm{f}(1)=4 \Rightarrow{ }^{5} \mathrm{C}_{5}=1\\

No of function A to

\mathrm{B}=21+6+1=28

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

The sum of all the roots of the equation

(\mathrm{x}-1)^{2}-5|\mathrm{x}-1|+6=0

, is:

A

4

B

3

C

1

D

5

Explanation

Let

|\mathrm{x}-1|=\mathrm{t}\\

\begin{aligned} & \mathrm{t}^{2}-5 \mathrm{t}+6=0 \\ & \mathrm{t}=2 \& \mathrm{t}=3 \\ & |\mathrm{x}-1|=2 \&|\mathrm{x}-1|=3 \\ & \mathrm{x}-1 \pm 2 \& \mathrm{x}-1= \pm 3 \\ & \mathrm{x}=1 \pm 2 \& \mathrm{x}=1 \pm 3 \\ & \therefore \text { root }=3,-1,4,-2 \\ & \therefore \text { Sum of root }=3+(-1)+4+(-2)=4 \end{aligned}

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

Let

\overrightarrow{\mathrm{c}}

and

\overrightarrow{\mathrm{d}}

be vectors such that

|\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|=\sqrt{29}

and

\overrightarrow{\mathrm{c}} \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times \overrightarrow{\mathrm{d}}

. If

\lambda_{1}, \lambda_{2}\left(\lambda_{1}>\lambda_{2}\right)

are the possible values of

(\vec{c}+\vec{d}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})

, then the equation

\\ \mathrm{K}^{2} \mathrm{x}^{2}+\left(\mathrm{K}^{2}-5 \mathrm{~K}+\lambda_{1}\right) \mathrm{xy}+\left(3 \mathrm{~K}+\frac{\lambda_{2}}{2}\right) \mathrm{y}^{2}-8 \mathrm{x}+12 \mathrm{y}+\lambda_{2}=0

represents a circle, for k equal to:

A

4

B

1

C

-1

D

2

Explanation

|\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|=\sqrt{29}\\

\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}=\lambda((2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})\\

\lambda= \pm 1\\

\lambda(-14+6+12)=4 \lambda, \lambda_{1}=4, \lambda_{2}=-4\\

k^{2} x^{2}+\left(k^{2}-5 k+4\right) x y+(3 k-2) y^{2}-8 x+12 y-4 =0\\

is circle

\mathrm{k}^{2}-5 \mathrm{k}+4=0 \Rightarrow \mathrm{k}=1,4\\

\mathrm{k}^{2}=3 \mathrm{k}-2 \Rightarrow \mathrm{k}=1,2\\

\mathrm{k}=1

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

The value of

\int_{-\pi / 6}^{\pi / 6}\left(\frac{\pi+4 x^{11}}{1-\sin (|x|+\pi / 6)}\right) d x

is equal to

A

2 \pi

B

4 \pi

C

8 \pi

D

6 \pi

Explanation

=2 \pi \int_{0}^{\pi / 6} \frac{1}{1-\sin \left(x+\frac{\pi}{6}\right)} d x\\

let

x+\frac{x}{6}=t d x=d t\\

=2 \pi \int_{\pi / 6}^{\pi / 3} \frac{\mathrm{dt}}{1-\sin \mathrm{t}} \\=2 \pi \int_{\pi / 6}^{\pi / 3} \frac{1+\sin \mathrm{t}}{\cos ^{2} \mathrm{t}} \mathrm{dt}\\

=2 \pi\left[\int_{\pi / 6}^{\pi / 3} \sec ^{2} \mathrm{tdt}+\int_{\pi / 6}^{\pi / 3} \sec \mathrm{t} \tan \mathrm{tdt}\right]\\

=2 \pi\left[(\tan \mathrm{t})_{\pi / 6}^{\pi / 3}+(\sec \mathrm{t})_{\pi / 6}^{\pi / 3}\right]\\

=2 \pi\left[\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)+\left(2-\frac{2}{\sqrt{3}}\right)\right]\\

=2 \pi[\sqrt{3}+2-\sqrt{3}]=4 \pi

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

If the domain of the function

f(x)=\cos ^{-1}\left(\frac{2 x-5}{11-3 x}\right)+\sin ^{-1}\left(2 x^{2}-3 x+1\right)

is the interval

[\alpha, \beta]

, then

\alpha+2 \beta

is equal to :

A

1

B

3

C

5

D

2

Explanation

f(x)=\cos ^{-1}\left(\frac{2 x-5}{11-3 x}\right)+\sin ^{-1}\left(2 x^{2}-3 x+1\right)\\

-1 \leq \frac{2 \mathrm{x}-5}{11-3 \mathrm{x}} \leq 1\\

-1 \leq 2 \mathrm{x}^{2}-3 \mathrm{x}+1 \leq 1\\

2 \mathrm{x}^{2}-3 \mathrm{x}+2 \geq 0,2 \mathrm{x}^{2}-3 \mathrm{x} \leq 0\\

\mathrm{x} \in\left[0, \frac{3}{2}\right] \tag .......{(1)}\\

\frac{2 \mathrm{x}-5}{11-3 \mathrm{x}}+1 \geq 0\\

\frac{2 \mathrm{x}-5}{11-3 \mathrm{x}}-1 \leq 0\\

\frac{2 \mathrm{x}-5+11-3 \mathrm{x}}{11-3 \mathrm{x}} \geq 0\\

\frac{5 \mathrm{x}-16}{11-3 \mathrm{x}} \leq 0\\

\\ \frac{6-x}{11-3 x} \geq 0\\

x \in\left(-\infty, \frac{16}{5}\right] \cup\left(\frac{11}{3}, \infty\right) \\

\mathrm{x} \in\left(-\infty, \frac{11}{3}\right) \cup[6, \infty)

intersection

\\ \mathrm{x} \in\left(-\infty, \frac{16}{5}\right] \cup[6, \infty) \tag.......{(2)}\\

Intersection of (i) & (ii)

\mathrm{x} \in\left[0, \frac{3}{2}\right]\\

\alpha=0, \beta=\frac{3}{2} \Rightarrow \alpha+2 \beta=3

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

Let

f: \mathrm{R} \rightarrow(0, \infty)

be a twice differentiable function such that

\mathrm{f}(3)=18, \mathrm{f}^{\prime}(3)=0

and

\mathrm{f}^{\prime \prime}(3)=4

.\\ Then

\lim _{x \rightarrow 1}\left(\log _{e}\left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^{2}}}\right)

is equal to :

A

1

B

9

C

2

D

18

Explanation

Let

T=\lim _{x \rightarrow 1}\left(\frac{f(x+2)}{f(3)}\right)^{\frac{18}{(x-1)^{2}}} ; 1^{\infty}

form

\\ \Rightarrow \mathrm{T}=\mathrm{e}^{\lim _{\mathrm{x} \rightarrow 1} \frac{18}{(\mathrm{x}-1)^{2}}\left(\frac{\mathrm{f}(\mathrm{x}+2)-\mathrm{f}(3)}{\mathrm{f}(3)}\right)}\\

\Rightarrow \mathrm{T}=\mathrm{e}^{\lim _{\mathrm{x} \rightarrow 1} \frac{18}{(\mathrm{x}-1)^{2}}\left(\frac{\mathrm{f}(\mathrm{x}+2)-\mathrm{f}(3)}{18}\right)}\\

\Rightarrow \mathrm{T}=\mathrm{e}^{\lim _{\mathrm{x} \rightarrow 1}\left(\frac{\mathrm{f}(\mathrm{x}+2)-\mathrm{f}(3)}{(\mathrm{x}-1)^{2}}\right) \frac{0}{0} \text { form }}

apply L' pital

\\ \Rightarrow \mathrm{T}=\mathrm{e}^{\lim _{\mathrm{x} \rightarrow 1} \frac{\mathrm{f}^{\prime}(\mathrm{x}+2)}{2(\mathrm{x}-1)} ; \frac{0}{0} \text { form }} \quad

apply L' pital

\\ \Rightarrow \mathrm{T}=\mathrm{e}^{\lim _{\mathrm{x} \rightarrow 1} \frac{\mathrm{f}^{\prime \prime}(\mathrm{x}+2)}{2}} ;=\mathrm{e}^{\frac{4}{2}}=\mathrm{e}^{2}\\

\Rightarrow \log _{\mathrm{e}}(\mathrm{T})=2

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

Let

\vec{a}=-\hat{i}+2 \hat{j}+2 \hat{k}, \vec{b}=8 \hat{i}+7 \hat{j}-3 \hat{k}

and

\vec{c}

be a vector such that

\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}

. If

\overrightarrow{\mathrm{c}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=4

, then

A

33

B

30

C

35

D

27

Explanation

\vec{a}=-\hat{i}+2 \hat{j}+2 \hat{k}

\\ \begin{aligned} & \vec{b}=8 \hat{i}+7 \hat{j}-3 \hat{k} \\ & \vec{c}=c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k} \\ & \vec{a} \times \vec{c}=\vec{b} \Rightarrow\left(2 c_{3}-2 c_{2}\right) \hat{i}+\left(c_{3}+2 c_{1}\right) \hat{j}- \\ & \left(c_{2}+2 c_{1}\right) \hat{k}=8 \hat{i}+7 \hat{j}-3 \hat{k} \\ & 2 c_{3}-2 c_{2}=8, c_{3}+2 c_{1}=7, c_{2}+2 c_{1}=3 \\ & \left(c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}\right) \cdot(\hat{i}+\hat{j}+\hat{k})=4 \\ & \Rightarrow c_{1}+c_{2}+c_{3}=4, c_{1}=2, c_{2}=-1, c_{3}=3 \\ & |\vec{a}+\vec{c}|^{2}=|\hat{i}+\hat{j}+5 \hat{k}|^{2}=27 \end{aligned}

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

If the coefficient of

x

in the expansion of

\left(a x^{2}+b x+c\right)(1-2 x)^{26}

is -56 and the coefficients of

\mathrm{x}^{2}

and

\mathrm{x}^{3}

are both zero, then

\mathrm{a}+\mathrm{b}+\mathrm{c}

is equal to

A

1300

B

1500

C

1403

D

1483

Explanation

\left(a x^{2}+b c+c\right) \sum_{\mathrm{r}=0}^{26}{ }^{26} \mathrm{C}_{\mathrm{r}}(-2 \mathrm{x})^{\mathrm{r}}\\

Coeff. of

\mathrm{x}^{2}: \mathrm{a} .{ }^{26} \mathrm{C}_{0}(-2)^{0}+\mathrm{b} .{ }^{26} \mathrm{C}_{1}(-2)+

c.

{ }^{26} \mathrm{C}_{2}(-2)^{2}=0\\

\Rightarrow a-52 b+1300 c=0 \tag......{(1)}\\

Coeff. of

\mathrm{x}^{3}: \mathrm{a} .{ }^{26} \mathrm{C}_{1}(-2)+\mathrm{b} .{ }^{26} \mathrm{C}_{2}(-2)^{2}+

c.

{ }^{26} \mathrm{C}_{3}(-2)^{3}=0\\

\Rightarrow-52 a+1300 b-20800 c=0 \tag.....{(2)}\\

Coeff. of

x=-56\\

\Rightarrow \mathrm{b} .{ }^{26} \mathrm{C}_{0}(-2)^{0}+\mathrm{c} .{ }^{26} \mathrm{C}_{1}(-2)^{1}=-56\\

b-52 c=-56 \tag.....{(3)}\\

After solving (1), (2) & (3)

\\ \mathrm{a}=1300, \mathrm{~b}=100, \mathrm{c}=3\\

\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=1403

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

Let

f: R \rightarrow R

be a twice differentiable function such that the quadratic equation

\mathrm{f}(\mathrm{x}) \mathrm{m}^{2}-2 \mathrm{f}^{\prime}(\mathrm{x}) \mathrm{m}+\mathrm{f}^{\prime}(\mathrm{x})=0

in m , has two equal roots for every

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

. If

\mathrm{f}(0)=1

,

\mathrm{f}^{\prime}(0)=2

and

(\alpha, \beta)

is the largest interval in which the function

f\left(\log _{e} x-x\right)

is increasing, then

\alpha+\beta

is equal to

Explanation

Given quadratic equation has equal roots, thus

\\ \mathrm{D}=0 \Rightarrow\left(\mathrm{f}^{\prime}(\mathrm{x})\right)^{2}=\mathrm{f}^{\prime \prime}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{x})\\

\frac{f^{\prime}(x)}{f(x)}=\frac{f^{\prime \prime}(x)}{f^{\prime}(x)}\\

Integrate

\\ \ell \mathrm{n}(\mathrm{f}(\mathrm{x}))=\ell \mathrm{n}\left(\mathrm{f}^{\prime}(\mathrm{x})\right)+\ell \mathrm{nC} \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{c} . \mathrm{f}^{\prime}(\mathrm{x})\\

Put

=\mathrm{x}=0\\

1=\mathrm{c} .2 \Rightarrow \mathrm{c}=\frac{1}{2}\\

Now

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

\Rightarrow \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}=2\\

Integrate

\\ \ell \mathrm{n}(\mathrm{f}(\mathrm{x}))=2 \mathrm{x}+\mathrm{d}\\

\Rightarrow \mathrm{d}=0\\

\Rightarrow \ln (\mathrm{f}(\mathrm{x}))=2 \mathrm{x} \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{2 \mathrm{x}}\\

Now let

\mathrm{g}(\mathrm{x})=\mathrm{f}(\ln \mathrm{x}-\mathrm{x})=\mathrm{e}^{2(\operatorname{nx} \mathrm{x})}\\

\therefore \mathrm{g}^{\prime}(\mathrm{x})=2 \mathrm{e}^{2(\ell \mathrm{n} \mathrm{x}-\mathrm{x})}\left(\frac{1}{\mathrm{x}}-1\right) \geq 3\\

\Rightarrow \frac{1-x}{x} \geq 0\\

\Rightarrow \mathrm{x} \in(0,1]\\

\Rightarrow \alpha=0, \beta=\\

\alpha+\beta=1

.

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

The value of

\operatorname{cosec} 10^{\circ}-\sqrt{3} \sec 10^{\circ}

is equal to:

A

2

B

4

C

8

D

6

Explanation

=\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}\\

=\frac{\cos 10^{\circ}-\sqrt{3} \sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}\\

=4\left[\frac{\frac{1}{2} \cos 10^{\circ}-\frac{\sqrt{3}}{2} \sin 10^{\circ}}{2 \sin 10^{\circ} \cos 10^{\circ}}\right]\\

=4\left[\frac{\sin \left(30^{\circ}-10^{\circ}\right)}{\sin 20^{\circ}}\right]\\

=4

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

Let the mean and variance of 7 observations

2,4,10, x, 12,14, y, x>y

, be 8 and 16 respectively. Two numbers are chosen from

\{1,2,3, \mathrm{x}-4, \mathrm{y}, 5\}

one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4 , is:

A

rac{3}{5}

B

rac{4}{5}

C

rac{2}{5}

D

rac{1}{3}

Explanation

Mean

(\overline{\mathrm{x}})=8

(Given)

\\ \Rightarrow \frac{2+4+10+x+12+14+y}{7}=8\\

\Rightarrow x+y=14 \ldots(1) \ldots.\\

Variance

\left(\sigma^{2}\right)=16

(Given)

\\ \Rightarrow 16=\frac{2^{2}+4^{2}+10^{2}+x^{2}+12^{2}+14^{2}+y^{2}}{7}-8^{2}\\

\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=100 \quad \ldots \ldots(2) \ldots.\\

\because(x+y)^{2}=x^{2}+y^{2}+2 x y\\

\Rightarrow \mathrm{xy}=48

(sum is 14 product is 48 )

\\

Since problem states

x>y\\

\therefore \mathrm{x}=8

and

\mathrm{y}=6\\

Now set

X=\{1,2,3,4,6,5\}\\

Now we choose two numbers one after author without replacement total outcomes

=6 \times 5=30\\

We want the prob. That the smaller number among the two is less than 4

\\ \mathrm{P}(

smaller

<4)=1-\mathrm{P}(

smaller

\geq 4)\\

=1-\frac{6}{30}=\frac{4}{5}

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

Let PQ and MN be two straight lines touching the circle

x^{2}+y^{2}-4 x-6 y-3=0

at the points A and B respectively. Let O be the centre of the circle and

\angle \mathrm{AOB}=\pi / 3

. Then the locus of the point of intersection of the lines PQ and MN is:

A

3\left(x^{2}+y^{2} ight)-18 x-12 y+25=0

B

x^{2}+y^{2}-12 x-18 y-25=0

C

x^{2}+y^{2}-18 x-12 y-25=0

D

3\left(x^{2}+y^{2} ight)-12 x-18 y-25=0

Explanation

Given circle

\\

x^{2}+y^{2}-4 x-6 y-3=0\\

C(2,3) \& r=4\\

\cos 30^{\circ}=\frac{\mathrm{r}}{\mathrm{OR}}=\frac{4}{\mathrm{OR}}\\

\Rightarrow \mathrm{OR}=\frac{8}{\sqrt{3}}\\

Now

\\ \mathrm{OR}^{2}=(\mathrm{h}-2)^{2}+(\mathrm{k}-3)^{2}\\

\Rightarrow 3\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)-12 \mathrm{x}-18 \mathrm{y}-25=0

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

If

x^{2}+x+1=0

, then the value of

\left(x+\frac{1}{x}\right)^{4}+\left(x^{2}+\frac{1}{x^{2}}\right)^{4}+\left(x^{3}+\frac{1}{x^{3}}\right)^{4}+\ldots+\left(x^{25}+\frac{1}{x^{25}}\right)^{4}

is :

A

128

B

162

C

175

D

145

Explanation

x^{2}+x+1=0\\

\Rightarrow \mathrm{x}=\omega

or

\omega^{2}\\

\therefore \alpha=\omega \& \beta=\omega^{2}\\

=\left(\omega+\omega^{2}\right)^{4}+\left(\omega^{2}+\omega^{4}\right)^{4}+\left(\omega^{3}+\omega^{6}\right)^{4}+\ldots+ \left(\omega^{25}+\omega^{50}\right)^{4}\\

=\left[\left(\omega+\omega^{2}\right)^{4}+\left(\omega^{2}+\omega^{4}\right)^{4}+\left(\omega^{4}+\omega^{8}\right)^{4}+\ldots .+\right. \left.\left(\omega^{25}+\omega^{50}\right)^{4}\right]+\left[\left(\omega^{3}+\omega^{6}\right)^{4}+\left(\omega^{6}+\omega^{12}\right)^{4}+\right. \left.\left(\omega^{9}+\omega^{18}\right)^{4}+\ldots+\left(\omega^{24}+\omega^{48}\right)^{4}\right]\\

=\underbrace{[1+1+1 \ldots \ldots+1]}_{17 \text { times }}+\underbrace{\left[(1+1)^{4}+(1+1)^{4}+\ldots . .(1+1)^{4}\right]}_{8 \text { times }}\\

=17+128\\

=145

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

Let the foci of hyperbola coincide with the foci of the ellipse

\frac{x^{2}}{36}+\frac{y^{2}}{16}=1

. If the eccentricity of the hyperbola is 5 , then the length of its latus rectum is:

A

12

B

16

C

\frac{96}{\sqrt{5}}

D

24 \sqrt{5}

Explanation

Let

\mathrm{e}_{1}

be eccentricity of ellipse

\\ \Rightarrow e_{1}=\sqrt{1-\frac{16}{36}}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}\\

So

\mathrm{ae}_{1}=6 \cdot \frac{\sqrt{5}}{3}=2 \sqrt{5}\\

Now H :

\frac{x^{2}}{p^{2}}-\frac{y^{2}}{q^{2}}=1\\

p.e

=\mathrm{ae}_{1}\\

p.

5=2 \sqrt{5}\\

\mathrm{p}=\frac{2}{\sqrt{5}} \Rightarrow \mathrm{e}^{2} \\=1+\frac{\mathrm{q}^{2}}{\mathrm{p}^{2}} \Rightarrow 25=1+\frac{5 \mathrm{q}^{2}}{4} \\ \Rightarrow \mathrm{q}^{2} =\frac{96}{5}\\

So length of

L R \frac{2 q^{2}}{p}=\frac{96}{\sqrt{5}}

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

Let

y=y(x)

be the solution curve of the differential equation

\left(1+x^{2}\right) d y+\left(y-\tan ^{-1} x\right) d x=0

,

y(0)=1

. Then the value of

y(1)

is :

A

\frac{2}{e^{\frac{\pi}{4}}}+\frac{\pi}{4}-1

B

\frac{2}{e^{\frac{\pi}{4}}}-\frac{\pi}{4}-1

C

\frac{4}{e^{\frac{\pi}{4}}}+\frac{\pi}{2}-1

D

\frac{4}{e^{\frac{\pi}{4}}}-\frac{\pi}{2}-1

Explanation

\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x}^{2}+1}=\frac{\tan ^{-1} \mathrm{x}}{\mathrm{x}^{2}+1}\\

I.f.

=e^{\tan ^{-1} x}\\

y \times e^{\tan ^{-1} x}=\int e^{\tan ^{-1} x} \cdot \frac{\tan ^{-1} x}{1+x^{2}} d x\\

y \times e^{\tan ^{-1} x}=\tan ^{-1} x\left(e^{\tan ^{-1} x}\right)-e^{\tan ^{-1} x}+c\\

y(o)=1 \Rightarrow c=2\\

\mathrm{y}(1)=\frac{2}{\mathrm{e}^{\pi / 4}}+\frac{\pi}{4}-1

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

6 \int_{0}^{\pi}|(\sin 3 \mathrm{x}+\sin 2 \mathrm{x}+\sin \mathrm{x})| \mathrm{dx}

is equal to.......

Explanation

\quad 6 \int_{0}^{\pi}|2 \sin 2 \mathrm{x} \cos \mathrm{x}+\sin 2 \mathrm{x}| \mathrm{dx}\\

6 \int_{0}^{\pi}\left|4 \sin x \cos ^{2} x+2 \sin x \cos x\right| d x\\

I=12 \int_{0}^{\pi} \sin x\left|2 \cos ^{2} x+\cos x\right|\\

Put

\cos \mathrm{x}=\mathrm{t},-\sin \mathrm{x} \mathrm{dx}=\mathrm{dt}\\

\begin{aligned} & I=-12 \int_{1}^{-1}\left|2 t^{2}+t\right| d t \\ & I= 12\left(\int_{-1}^{-1 / 2}\left(2 t^2+t\right) d t+\int_{-1 / 2}^0-\left(2 t^2+t\right) d t+\int_0^1\left(2 t^2+t\right) d t\right) \end{aligned}

\\ \mathrm{I}=17

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

Let

(\alpha, \beta, \gamma)

be the co-ordinates of the foot of the perpendicular drawn from the point

(5,4,2)

on the line

\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})

Then the length of the projection of the vector

\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}

on the vector

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

is

A

rac{15}{7}

B

4

C

rac{18}{7}

D

3

Explanation

\overline{\mathrm{r}}=(-\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})\\

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

Any general point P on the line is

\\(2 \lambda-1,3 \lambda+3,-\lambda+1)\\

Let the given point is

\mathrm{A}(5,4,2)\\

\overline{\mathrm{AP}}(2 \lambda-6) \hat{\mathrm{i}}+(3 \lambda-1) \hat{\mathrm{j}}+(-\lambda-1) \hat{\mathrm{k}}\\

\because \overline{\mathrm{AP}} \perp^{\mathrm{r}}

Line

(\mathrm{L})\\

\therefore \overline{\mathrm{AP}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})=0\\

2(2 \lambda-6)+3(3 \lambda-1)-1(-\lambda-1)=0\\

\Rightarrow \lambda=1\\

\therefore \alpha=1\\

\beta=6\\

\gamma=0\\

Let the vector

\overline{\mathrm{u}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}

\overline{\mathrm{u}}=\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+\mathrm{O} \hat{\mathrm{k}}

\& \overline{\mathrm{w}} \\=6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\\

So projection

=\frac{|\overline{\mathrm{u}} \cdot \overline{\mathrm{w}}|}{|\overline{\mathrm{w}}|}=\frac{18}{7}

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

The area of the region, inside the ellipse

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

and outside the region bounded by the curves

\mathrm{y}=|\mathrm{x}|-1

and

\mathrm{y}=1-|\mathrm{x}|

, is :

A

2(\pi-1)

B

2 \pi- rac{1}{2}

C

3(\pi-1)

D

2 \pi-1

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

For some

\alpha, \beta \in R

, let

A=\left[\begin{array}{ll}\alpha & 2 \\ 1 & 2\end{array}\right]

and

B=\left[\begin{array}{ll}1 & 1 \\ 1 & \beta\end{array}\right]

be such that

A^{2}-4 A+2 I=B^{2}-3 B+I=O

. Then

\left(\operatorname{det}\left(\operatorname{adj}\left(\mathrm{A}^{3}-\mathrm{B}^{3}\right)\right)\right)^{2}

is equal to

\_\_\_\_

Explanation

\operatorname{Tr}(\mathrm{A})=4 \Rightarrow \alpha+2=4 \Rightarrow \alpha=2\\

\operatorname{Tr}(\mathrm{B})=3 \Rightarrow \beta+1=3 \Rightarrow \beta=2\\

A^{2}-4 A+2 I=O\\

A^{3}=4 A^{2}-2 A=16 A-8 I-2 A\\

14A-8I

\\=\left[\begin{array}{cc}28 & 28 \\ 14 & 28\end{array}\right]+\left[\begin{array}{cc}-8 & 0 \\ 0 & -8\end{array}\right]\\

=\mathrm{A}^{3}=\left[\begin{array}{ll}20 & 28 \\ 14 & 20\end{array}\right]\\

\mathrm{B}^{2}-3 \mathrm{~B}+\mathrm{I}=\mathrm{O}\\

\mathrm{B}^{2}=3 \mathrm{~B}-\mathrm{I}\\

\Rightarrow \mathrm{B}^{3}=3 \mathrm{~B}^{2}-\mathrm{B}=3(3 \mathrm{~B}-\mathrm{I})-\mathrm{B}=8 \mathrm{~B}-3 \mathrm{I}\\

\begin{aligned} & \mathrm{B}^{3}=\left(\begin{array}{cc} 8 & 8 \\ 8 & 16 \end{array}\right)+\left(\begin{array}{cc} -3 & 0 \\ 0 & -3 \end{array}\right)=\left(\begin{array}{cc} 5 & 8 \\ 8 & 13 \end{array}\right) \\ & \mathrm{A}^{3}-\mathrm{B}^{3}=\left(\begin{array}{ll} 20 & 28 \\ 14 & 20 \end{array}\right)-\left(\begin{array}{cc} 5 & 8 \\ 8 & 13 \end{array}\right)=\left(\begin{array}{cc} 15 & 20 \\ 6 & 7 \end{array}\right) \\ & \Rightarrow\left|\mathrm{A}^{3}-\mathrm{B}^{3}\right|=105-120=-15 \\ & \Rightarrow\left|\operatorname{adj}\left(\mathrm{~A}^{3}-\mathrm{B}^{3}\right)\right|=\left|\mathrm{A}^{3}-\mathrm{B}^{3}\right|=-15 \\ & \Rightarrow\left|\operatorname{adj}\left(\mathrm{~A}^{3}-\mathrm{B}^{3}\right)\right|^{2}=225 \end{aligned}

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

Let

\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots

. be a G.P. of increasing positive terms such that

a_{2} \cdot a_{3} \cdot a_{4}=64

and

a_{1}+a_{3}+a_{5}=\frac{813}{7}

. Then

\mathrm{a}_{3}+\mathrm{a}_{5}+\mathrm{a}_{7}

is equal to :

A

3256

B

3252

C

3244

D

3248

Explanation

ar

. \mathrm{ar}^{2} . \mathrm{ar}^{3}=64\\

\mathrm{a}^{3} \mathrm{r}^{6}=64 \quad \Rightarrow \mathrm{ar}^{2}=4\\

\mathrm{a}+\mathrm{ar}^{2}+\mathrm{ar}^{4}=\frac{813}{7}\\

\mathrm{r}^{2}=28\\

a r^{2}+a r^{4}+a r^{6}=

?

\\ \operatorname{ar}^{2}\left(1+\mathrm{r}^{2}+\mathrm{r}^{4}\right)=4(1+28+784)=3252

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

Let a point A lie between the parallel lines

\mathrm{L}_{1}

and

\mathrm{L}_{2}

such that its distances from

\mathrm{L}_{1}

and

\mathrm{L}_{2}

are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC , where the points B and C lie on the lines

\mathrm{L}_{1}

and

\mathrm{L}_{2}

respectively, is :

A

15 \sqrt{6}

B

27

C

21 \sqrt{3}

D

12 \sqrt{2}

Explanation

\sin \theta=\frac{3}{\mathrm{a}}\\

\sin \left(60^{\circ}+\theta\right)=\frac{9}{\mathrm{a}}\\

\frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta=\frac{9}{\mathrm{a}}\\

\sqrt{3} \sqrt{1-\frac{9}{\mathrm{a}^{2}}}+\frac{3}{\mathrm{a}}=\frac{18}{\mathrm{a}}\\

\mathrm{a}=\sqrt{84}\\

Area of

\triangle \mathrm{ABC}=\frac{\sqrt{3}}{4} \mathrm{a}^{2} \\=\frac{\sqrt{3}}{4} \times 84=21 \sqrt{3}

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

The number of relations, defined on the set

\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\}

, which are both reflexive and symmetric, is equal to:

A

256

B

16

C

1024

D

64

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

Let

\mathrm{S}=\{(\mathrm{m}, \mathrm{n}): \mathrm{m}, \mathrm{n} \in\{1,2,3, \ldots . ., 50\}\}

. If the number of elements (

\mathrm{m}, \mathrm{n}

) in S such that

6^{\mathrm{m}}+9^{\mathrm{n}}

is a multiple of 5 is p and the number of elements

(\mathrm{m}, \mathrm{n})

in S such that

\mathrm{m}+\mathrm{n}

is a square of a prime number is q , then

\mathrm{p}+\mathrm{q}

is equal to

\_\_\_\_

Explanation

\mathrm{S}=\{1,2,3, \ldots .50\}\\

\mathrm{p}=\left(6^{\mathrm{m}}+9^{\mathrm{n}}\right)

is divisible by 5 No. of ways

\\6^{\mathrm{m}}=(5 \lambda+1)^{\mathrm{m}}=5 \mathrm{k}+1\\

9^{\mathrm{n}}=(10-1)^{\mathrm{n}}=10 \mu-1

if n is odd

\\

⇒ n must be odd

\\ 10 \mu+1

if n is even

\\

⇒ No. of ways

=50 \times 25=1250\\

\mathrm{q} \Rightarrow(\mathrm{m}+\mathrm{n})

is square of a prime

\\ TABLE

\\ \mathrm{q}=3+8+24+48=83\\

=\mathrm{p}+\mathrm{q}=1250+83=1333

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JEE Mains 2026 2026-01-21 Shift 1

Mathematics

Let

\mathrm{a}_{1}=1

and for

\mathrm{n} \geq 1, \mathrm{a}_{\mathrm{n}+1}

\\

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

. Then

\left|\sum_{n=1}^{\infty}\left(a_{n}-\frac{2}{n^{2}}\right)\right|

is equal to

Explanation

a_{n+1}-\frac{1}{2} a_{n}=\frac{n^{2}-2 n-1}{n^{2}(n+1)^{2}}=\frac{2 n^{2}-(n+1)^{3}}{n^{2}(n+1)^{2}}\\

\Rightarrow \mathrm{a}_{\mathrm{n}+1}-\frac{1}{2} \mathrm{a}_{\mathrm{n}}=\frac{2}{(\mathrm{n}+1)^{2}}-\frac{1}{\mathrm{n}^{2}}\\

\mathrm{n}=1 \mathrm{a}_{2}-\frac{1}{2} \mathrm{a}_{1}=\frac{2}{2^{2}}-\frac{1}{1^{2}}\\

2\left[\mathrm{a}_{3}-\frac{1}{2} \mathrm{a}_{2}=\frac{2}{3^{2}}-\frac{1}{2^{2}}\right]\\

2^{2}\left[a_{4}-\frac{1}{2} a_{3}=\frac{2}{4^{2}}-\frac{1}{3^{2}}\right]\\

⋮

\\2^{n-2}\left[a_{n}-\frac{1}{2} a_{n-1}=\frac{2}{n^{2}}-\frac{1}{(n-1)^{2}}\right]\\

2^{n-1}\left[a_{n+1}-\frac{1}{2} a_{n}=\frac{2}{(n+1)^{2}}-\frac{1}{n^{2}}\right]\\

Adding

\\ \mathrm{a}_{\mathrm{n}+1}=\frac{2}{(\mathrm{n}+1)^{2}}-\frac{1}{2^{\mathrm{n}}} \\ \Rightarrow \mathrm{a}_{\mathrm{n}}=\frac{2}{\mathrm{n}^{2}}-\frac{1}{2^{\mathrm{n}-1}}\\

\Rightarrow\left|\sum_{n=1}^{\infty}\left(a_{n}-\frac{2}{n^{2}}\right)\right| \\ \Rightarrow\left|\sum_{n=1}^{\infty}-\frac{1}{2^{n-1}}\right|=2

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JEE Mains 2026 2026-01-21 Shift 1

Physics

Initially a satellite of

100 \text{ kg}

is in a circular orbit of radius

1.5R_E

. This satellite can be moved to a circular orbit of radius

3R_E

by supplying

\alpha \times 10^8 \text{ J}

of energy. The value of

\alpha

is ___________. (Take Radius of Earth

R_E = 6 \times 10^6

m and

g = 10

m/s²)

A

150

B

500

C

100

D

1000

Explanation

E = \frac{-GM_Em}{2r}; r = \text{radius of circular orbit}

Required energy to be supplied =

\Delta E = E_f - E_i

\Delta E = \left( \frac{-GM_Em}{2(3R_E)} \right) - \left( \frac{-GM_Em}{2(1.5R_E)} \right)

= \frac{GM_Em}{6R_E}

Now,

g = \frac{GM_E}{R_E^2} \Rightarrow \frac{GM_E}{R_E} = gR_E

\therefore \Delta E = \frac{1}{6} gmR_E

= \frac{1}{6} \times 10 \times 100 \times 6 \times 10^6

= 1000 \times 10^6

\alpha = 1000

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JEE Mains 2026 2026-01-21 Shift 1

Physics

10 mole of oxygen is heated at constant volume from 30°C to 40°C. The change in the internal energy of the gas is ______ cal. (The molecular specific heat of oxygen at constant pressure,

C_p = 7 \text{ cal./mol } ^\circ\text{C}

and

R = 2 \text{ cal./mol } ^\circ\text{C}

.)

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JEE Mains 2026 2026-01-21 Shift 1

Physics

Two identical thin rods of mass M kg and length L m are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point P and perpendicular to the plane of the rods is

\frac{x}{12} ML^2 \text{ kg m}^2

. The value of

x

is ___________.

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JEE Mains 2026 2026-01-21 Shift 1

Physics

A

1\text{ m}

long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of

0.10\text{ T}

. If the resistance of the total circuit is

2\Omega

then the force needed to move the rod towards right with constant speed (v) of

1.5\text{ m/s}

is ________ N.

A

7.5 imes 10^{-2}

B

5.7 imes 10^{-3}

C

5.7 imes 10^{-2}

D

7.5 imes 10^{-3}

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JEE Mains 2026 2026-01-21 Shift 1

Physics

A conducting circular loop of area

1.0 m^2

is placed perpendicular to a magnetic field which varies as

B = sin(100 t)

Tesla. If the resistance of the loop is 100

\omega

, then the average thermal energy dissipated in the loop in one period is ______J.

A

\frac{\pi}{2}

B

2\pi

C

\pi

D

\pi^2

Explanation

Area of the loop

= 1 \text{ m}^2

B = \sin(100t)

\therefore \phi = BA = \sin(100t)

\therefore \frac{d\phi}{dt} = 100 \cos(100t)

\therefore P = \frac{V^2}{R} = \frac{10^4 \cos^2(100t)}{100}

\therefore

Thermal energy dissipated in 1 time period

= \int_{0}^{T} P dt = \int_{0}^{T} 100 \cos^2(100t) dt

T = \frac{2\pi}{100} = \frac{\pi}{50} \text{ sec}

\therefore Q = 100 \int_{0}^{\pi/50} \cos^2(100t) dt

= 100 \int_{0}^{\pi/50} \frac{1 + \cos 200t}{2} dt

= 100 \left[ \frac{\pi}{100} \right] = \pi

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JEE Mains 2026 2026-01-21 Shift 1

Physics

An aluminium and steel rods having same lengths and cross-sections are joined to make total length of

120\text{ cm}

at

30^\circ\text{C}

. The coefficient of linear expansion of aluminium and steel are

24 \times 10^{-6}/^\circ\text{C}

and

1.2 \times 10^{-5}/^\circ\text{C}

, respectively. The length of this composite rod when its temperature is raised to

100^\circ\text{C}

, is _____

\text{cm}

.

A

120.20

B

120.15

C

120.03

D

120.06

Explanation

\ell_{\text{final}} = \ell_0(1 + \alpha_{\text{A}}\Delta\text{T}) + \ell_0(1 + \alpha_{\text{B}}\Delta\text{T})

= \ell_0[2 + (\alpha_{\text{A}} + \alpha_{\text{B}})\Delta\text{T}]

= 60[2 + (36 \times 10^{-6}) \times 70]

= 60[2 + 0.0025]

= 120.15 \text{ cm}

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JEE Mains 2026 2026-01-21 Shift 1

Physics

The given circuit works as :

A

AND gate

B

NOR gate

C

NAND gate

D

OR gate

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JEE Mains 2026 2026-01-21 Shift 1

Physics

A current carrying is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy :

A

a = g

B

a > g

C

a = 0

D

0 < a < g

Explanation

Since the solenoid is placed vertically, the magnetic field inside the solenoid will be either along - y or +y axis.

\Rightarrow

Particle will gain velocity along

-y

axis.

\Rightarrow \vec{F}_B = q(\vec{v} \times \vec{B})

\Rightarrow \vec{F}_B = 0

\Rightarrow \vec{F}_{\text{net}} = m\vec{g}

\Rightarrow a_{\text{net}} = g

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JEE Mains 2026 2026-01-21 Shift 1

Physics

The electric field of a plane electromagnetic wave is given by:

E_y = 69 \sin[0.6 \times 10^3x - 1.8 \times 10^{11}t]\text{ V/m}.

The expression for magnetic field associated with this electromagnetic wave is ______ T.

A

B_z = 2.3 \times 10^{-7} \, \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t]

B

B_z = 2.3 \times 10^{-7} \, \sin[0.6 imes 10^3 x + 1.8 imes 10^{11} t]

C

B_y = 69 \, \sin[0.6 imes 10^3 x + 1.8 imes 10^{11} t]

D

B_y = 2.3 imes 10^{-7} \, \sin[0.6 imes 10^3 x - 1.8 imes 10^{11} t]

Explanation

\hat{B} = \hat{c} \times \hat{E}

\Rightarrow \hat{c} = \hat{i}

because phase of electric field is function of

x

.

\Rightarrow \hat{E} = \hat{j}

(given)

\Rightarrow \hat{B} = \hat{i} \times \hat{j} = \hat{k}

|B| = \frac{|E|}{c} = \frac{69 \times 0.6 \times 10^3}{1.8 \times 10^{11}} = \frac{69}{3 \times 10^8}

|B| = 2.3 \times 10^{-7}

\vec{B}_z = 2.3 \times 10^{-7} \sin(0.6 \times 10^3x - 1.8 \times 10^{11}t)

(phase is same as that of electric field)

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JEE Mains 2026 2026-01-21 Shift 1

Physics

A gas based geyser heats water flowing at the rate of

5.0

litres per minute from

27^\circ\text{C} \text{ to } 87^\circ\text{C}

. The rate of consumption of the gas is ____ g/s. (Take heat of combustion of gas =

5.0 × 104J/g

) specific heat capacity of water =

4200 J/kg. °C

A

2.1

B

4.2

C

0.42

D

0.21

Explanation

Water flow rate

= 5\text{ l/min} = \frac{5}{60}\text{ kg/s}

\therefore \text{Power of heater} = \frac{dm}{dt}S\Delta T = \frac{1}{12} \times 4200 \times 60\text{ W}

\therefore

Let the rate of consumption of gas be

x\text{ g/s}

.

\therefore x \times 5.0 \times 10^4 = \frac{1}{12} \times 4200 \times 60

\Rightarrow x = 4200 \times 10^{-4} = 0.42\text{ g/s}

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JEE Mains 2026 2026-01-21 Shift 1

Physics

Consider a modified Bernoulli equation.

\left( P + \frac{A}{Bt^2} \right) + \rho g(h + Bt) + \frac{1}{2}\rho V^2 = \text{constant}

If

t

has the dimension of time then the dimensions of

A

and

B

are ______, ______ respectively.

A

[ML^0T^{-1}]

and

[M^0LT]

B

[ML^0T^{-1}]

and

[M^0LT^{-1}]

C

[ML^0T^{-2}]

and

[M^0LT^{-2}]

D

[ML^0T^{-2}]

and

[M^0LT^{-1}]

Explanation

\Rightarrow [P] = \left[ \frac{A}{Bt^2} \right] \dots (1)

\Rightarrow [h] = [Bt] \dots (2)

\Rightarrow [B] = \left[ \frac{h}{t} \right] = \left[ \frac{L}{T} \right] = [LT^{-1}]

Putting

B

in equation (1)

[ML^{-1}T^{-2}] = \left[ \frac{A}{LT^{-1} \times T^2} \right]

[A] = [ML^0T^{-1}]

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JEE Mains 2026 2026-01-21 Shift 1

Physics

A collimated beam of light of diameter

2\text{ mm}

is propagating along x-axis. The beam is required to be expanded in a collimated beam of diameter

14\text{ mm}

using a system of two convex lenses. If first lens has focal length

40\text{ mm}

, then the focal length of second lens is __________

\text{mm}

.

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JEE Mains 2026 2026-01-21 Shift 1

Physics

In a microscope the objective is having focal length

f_o = 2 \text{ cm}

and eye-piece is having focal length

f_e = 4 \text{ cm}

. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is __________.

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JEE Mains 2026 2026-01-21 Shift 1

Physics

The heat generated in

1

minute between points

A

and

B

in the given circuit, when a battery of

9\text{V}

with internal resistance of

1\text{ }\Omega

is connected across these points is __________

\text{J}

.

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JEE Mains 2026 2026-01-21 Shift 1

Physics

A uniform rod of mass

m

and length

l

suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is ________. (g acceleration due to gravity)

A

mg/2

B

mg/4

C

mg/3

D

mg

Explanation

mg \frac{l}{2} = \frac{ml^2}{3} \alpha

\alpha = \frac{3g}{2l} \dots(1)

mg - T = ma_c

T = mg - ma_c

= mg - m \left( \frac{l}{2} \alpha \right)

= mg - m \left( \frac{l}{2} \cdot \frac{3g}{2l} \right)

T = \frac{mg}{4}

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JEE Mains 2026 2026-01-21 Shift 1

Physics

In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index n(= 1.5), the central fringe shifts by 0.2 cm. The value of t is ____ cm.

A

8 imes 10^{-4}

B

6.0 imes 10^{-3}

C

5.6 imes 10^{-4}

D

5.0 imes 10^{-3}

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JEE Mains 2026 2026-01-21 Shift 1

Physics

A

4\text{ kg}

mass moves under the influence of a force

\vec{F} = (4t^3\hat{i} - 3t\hat{j})\text{ N}

where

t

is the time in second. If mass starts from origin at

t = 0

, the velocity and position after

t = 2\text{s}

will be :

A

\vec{v}=3\hat{i}+\frac{3}{2}\hat{j} \quad \vec{r}=\frac{6}{5}\hat{i}+\hat{j}

B

\vec{v}=4\hat{i}-\frac{3}{2}\hat{j} \quad \vec{r}=\frac{8}{5}\hat{i}-\hat{j}

C

\vec{v}=4\hat{i}+\frac{5}{2}\hat{j} \quad \vec{r}=\frac{8}{5}\hat{i}+2\hat{j}

D

\vec{v}=4\hat{i}-\frac{3}{2}\hat{j} \quad \vec{r}=\frac{6}{5}\hat{i}-\hat{j}

Explanation

\vec{F} = 4t^3\hat{i} - 3t\hat{j}

\vec{a} = \frac{\vec{F}}{m} = t^3\hat{i} - \frac{3}{4}t\hat{j}

x-componenty-component

a_x = t^3

a_y = -\frac{3}{4}t

\frac{dv_x}{dt} = t^3

\frac{dv_y}{dt} = -\frac{3}{4}t

\int_{v_x=0}^{v_{x_2}} dv_x = \int_{t=0}^{t=2} t^3 \, dt

\int_{0}^{v_{y_2}} dv_y = \int_{0}^{2} -\frac{3}{4}t \, dt

v_{x_2} - 0 = \left[ \frac{t^4}{4} \right]_0^2

v_{y_2} = -\frac{3}{4} \left[ \frac{t^2}{2} \right]_0^2

v_{x_2} = 4

v_{y_2} = -\frac{3}{2}

\vec{v}_2 = 4\hat{i} - \frac{3}{2}\hat{j}

x-componenty-component

v_x = \frac{t^4}{4}

v_y = -\frac{3}{8}t^2

\int_0^{x_2} dx = \int_0^2 \frac{t^4}{4} \, dt

\int_0^{y_2} dy = \int_0^2 -\frac{3}{8}t^2 \, dt

x_2 - 0 = \left[ \frac{t^5}{20} \right]_0^2

y_2 - 0 = -\frac{3}{8} \left[ \frac{t^3}{3} \right]_0^2

x_2 = \frac{8}{5}

y_2 = -1

\vec{r} = \frac{8}{5}\hat{i} - \hat{j}

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JEE Mains 2026 2026-01-21 Shift 1

Physics

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from the nucleus it can reach is ________ m. (Atomic number of gold = 79 and

\frac{1}{4\pi\epsilon_0} = 9 \times 10^9

in SI units)

A

2.95 imes 10^{-14}

B

2.95 imes 10^{-16}

C

3.85 imes 10^{-16}

D

3.85 imes 10^{-14}

Explanation

Energy conservation

K_i + U_i = K_f + U_f

7.7 \times 10^6 \times 1.6 \times 10^{-19} + 0

= 0 + \frac{9 \times 10^9 (1.6 \times 10^{-19}) (79 \times 1.6 \times 10^{-19})}{r}

r = 2.95 \times 10^{-14}

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JEE Mains 2026 2026-01-21 Shift 1

Physics

A point charge of

10^{-8} \text{ C}

is placed at origin. The work done in moving a point charge

2 \text{ \mu C}

from point

A(4, 4, 2) \text{ m}

to

B(2, 2, 1) \text{ m}

is ________

J

.

(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ in SI units})

A

45 imes 10^{-6}

B

0

C

30 imes 10^{-6}

D

15 imes 10^{-6}

Explanation

Work done by external agent :

W_{ext} = \Delta U;

\Delta U \rightarrow

Change in potential energy in taking the charge from initial to final configuration

\Rightarrow W_{ext} = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r_f} - \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r_i}

Now,

r_f = \sqrt{(2-0)^2 + (2-0)^2 + (1-0)^2} = 3 \text{ m}

r_i = \sqrt{(4-0)^2 + (4-0)^2 + (2-0)^2} = 6 \text{ m}

\therefore W_{\text{ext}} = (9 \times 10^9) \times (10^{-8} \times 2 \times 10^{-6}) \left[ \frac{1}{3} - \frac{1}{6} \right]

= 3 \times 10^{-5}

= 30 \times 10^{-6} \text{J}

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JEE Mains 2026 2026-01-21 Shift 1

Physics

Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is

5 cm

and the area of cross-sections at A and B are

6 cm^2

and

3 cm^2

respectively. The rate of flow will be_____

cm^3/s.

(take g = 1

0 m/s^2

)

A

\frac{200}{\sqrt{3}}

B

200 \sqrt{6}

C

200 \sqrt{3}

D

100 \sqrt{3}

Explanation

From continuity equation

A_A V_A = A_B V_B \Rightarrow 6V_A = 3V_B \Rightarrow V_B = 2V_A

Applying Bernoullis equation between A & B,

P_A + \frac{1}{2} \rho V_A^2 = P_B + \frac{1}{2} \rho V_B^2

\Rightarrow \rho g \times 0.05 = \frac{1}{2} \rho [V_B^2 - V_A^2] = \frac{1}{2} \rho (3V_A^2)

\Rightarrow V_A = \sqrt{\frac{2g \times 0.05}{3}} \text{ m/s} = \frac{1}{\sqrt{3}} \text{ m/s} = \frac{100}{\sqrt{3}} \text{ cm/s}

\Rightarrow \text{Volume flow rate} = A_A V_A = \frac{6 \times 100}{\sqrt{3}} \text{ cm}^3\text{/sec}

= 200 \sqrt{3} \text{ cm}^3\text{/sec}

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JEE Mains 2026 2026-01-21 Shift 1

Physics

Two strings (A, B) having linear densities

\mu_A = 2 \times 10^{-4} \text{ kg/m}

and

\mu_B = 4 \times 10^{-4} \text{ kg/m}

and lengths

L_A = 2.5 \text{ m}

and

L_B = 1.5 \text{ m}

respectively are joined. Free ends of A and B are tied to two rigid supports C and D, respectively creating a tension of

500 \text{ N}

in the wire. Two identical pulses, sent from C and D ends, take time

t_1

and

t_2

, respectively, to reach the joint. The ratio

t_1/t_2

is:

A

1.08

B

1.90

C

1.67

D

1.18

Explanation

Given

L_A = 2.5 \text{ m}

,

L_B = 1.5 \text{ m}

,

T = 500 \text{ N}

v_A = \sqrt{\frac{T}{\mu_A}} = \sqrt{\frac{500}{2 \times 10^{-4}}} = 5\sqrt{10} \times 10^2 \text{ m/s}

v_B = \sqrt{\frac{T}{\mu_B}} = \sqrt{\frac{500}{4 \times 10^{-4}}} = 5\sqrt{5} \times 10^2 \text{ m/s}

t_1 = \frac{L_A}{v_A} = \frac{2.5}{5\sqrt{10}} \times 10^{-2} \text{ s}

t_2 = \frac{L_B}{v_B} = \frac{1.5}{5\sqrt{5}} \times 10^{-2} \text{ s}

\therefore \frac{t_1}{t_2} = \frac{2.5}{5\sqrt{10}} \times \frac{5\sqrt{5}}{1.5} = \frac{5}{3} \times \frac{1}{\sqrt{2}} = \frac{1.66}{1.41} = 1.18

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JEE Mains 2026 2026-01-21 Shift 1

Physics

A light wave described by

E = 60[\sin(3 \times 10^{15})t + \sin(12 \times 10^{15})t]

(in SI units) falls on a metal surface of work function

2.8 \text{ eV}

. The maximum kinetic energy of ejected photoelectron is (approximately) ________

\text{eV}

. (

h = 6.6 \times 10^{-34} \text{ J-s}

and

e = 1.6 \times 10^{-19} \text{C}

)

A

5.1

B

3.8

C

6.0

D

7.8

Explanation

\omega_1 = 3 \times 10^{15} \text{ rad/sec}

\omega_2 = 12 \times 10^{15} \text{ rad/sec}

\because \nu = \frac{\omega}{2\pi}

E_{\text{photon}} = h\nu = 6.6 \times 10^{-34} \times 1.91 \times 10^{15}

= 1.26 \times 10^{-18} \text{ J}

E_{\text{max}} = \frac{1.26 \times 10^{-18}}{1.6 \times 10^{-19}} \simeq 7.9 \text{ eV}

K_{\text{max}} = E_{\text{max}} - \phi_0

= 7.9 - 2.8

K_{\text{max}} = 5.1 \text{ eV}

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JEE Mains 2026 2026-01-21 Shift 1

Physics

In an experiment the values of two spring constants were measured as k1= (10 0.2) N/m and k2 = ( 20 0.3) N/m. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :

A

2.67%

B

2.33%

C

1.33%

D

1.67%

Explanation

For parallel combination of spring:

K_{eq} = K_1 + K_2 = 30 \text{ N/m}

\Delta K_{eq} = \Delta K_1 + \Delta K_2 = 0.2 + 0.3 = 0.5 \text{ N/m}

\therefore \text{ \%Error in } K = \frac{0.5}{30} \times 100 = 1.67\%

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JEE Mains 2026 2026-01-21 Shift 1

Physics

A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness

(\frac{1}{3})^{rd}

of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :

A

rac{3KC}{2K + 1}

B

rac{CK}{2 + K}

C

rac{3CK^2}{(2K + 1)^2}

D

rac{4KC}{3K - 1}

Explanation

C_1 = \frac{3A \epsilon_0}{2d}

C_2 = \frac{3A \epsilon_0 \times K}{d}

C_1 = \frac{3}{2}C

C_2 = 3KC

C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{\frac{3}{2}C \times 3KC}{\frac{3}{2}C + 3KC}

C_{eq} = \frac{\frac{9}{2}KC^2}{\frac{3}{2}C(2K + 1)} = \frac{3KC}{2K + 1}

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JEE Mains 2026 2026-01-21 Shift 1

Physics

Potential energy (V) versus distance (x) is given by the graph, Rank various regions as per the magnitudes of the force (F) acting on a particle from high to low.

A

ext{F}_{ ext{BC}} > ext{F}_{ ext{CD}} > ext{F}_{ ext{DE}} > ext{F}_{ ext{AB}}

B

ext{F}_{ ext{CD}} > ext{F}_{ ext{AB}} > ext{F}_{ ext{BC}} > ext{F}_{ ext{DE}}

C

ext{F}_{ ext{CD}} > ext{F}_{ ext{DE}} > ext{F}_{ ext{AB}} > ext{F}_{ ext{BC}}

D

ext{F}_{ ext{BC}} > ext{F}_{ ext{AB}} > ext{F}_{ ext{DE}} > ext{F}_{ ext{CD}}

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