Let u and v be unit vectors inclined at an acute angle such that |u × v| = √(3) 2 . If A = λ u + v + (u × v) , then λ is equal to:

(4)/(3)(A · u) - (2)/(3)(A · v)

The integral ∫_0^1 ^-1 (1 + x + x^2 ) dx is equal to:

2 ^-1 2 - (1)/(2) _e ((5)/(4) ) - (π)/(2)

A magnetic field vector in an electromagnetic wave is represented by B = B_0 (2πν t - (2π x)/(λ) )j . Its associated electric field vector is _____ .

E = -νλ B_0 (2πν t - (2π x)/(λ) )k

JEE Main 2026 April 04, Shift 2 Question Paper with Solutions

All 75 questions from the JEE Main 2026 (April 04, Shift 2) shift — Physics (25), Chemistry (25) and Mathematics (25) — with the correct answer and a step-by-step solution for every question.

Physics PYQs 2026 Chemistry PYQs 2026 Mathematics PYQs 2026

Physics25 questions

Q26Single correctUnits and Measurements

Match the LIST-I with LIST-II

List-I	List-II
A. Planck's constant	I. $ML^{2}T^{-2}$
B. Stopping potential	II. $T^{-1}$
C. Work function	III. $ML^{2}T^{-1}$
D. Threshold frequency	IV. $ML^{2}T^{-3}A^{-1}$

(A)

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

(B)

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

(C)

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

(D)

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

Q27Single correctKinematics

Two cars A and B are moving in the same direction along a straight line with speeds 100km/h and 80km/h, respectively such that car A is moving ahead of car B. A person in car B throws a stone with a speed v so that it hits the car A with a speed of 5m/s. The value of v is ____ km/h.

(A)

(B)

(C)

(D)

Q28Single correctLaws of Motion

At t = 0, a body of mass 100 g starts moving under the influence of a force

(5\hat{i} + 10\hat{j})\,\text{N}

from the origin. After 2 s its position is

(2x\hat{i} + 5y\hat{j})\,\text{m}

. The ratio x : y is ____ .

(A)

1 : 2

(B)

2 : 5

(C)

5 : 2

(D)

5 : 4

Q29Single correctKinematics

If x and y coordinates of a projectile as a function of time (t) are given as

24t

and

43.6t - 4.9t^{2}

, respectively, then the angle (in degrees) made by the projectile with horizontal when t = 2s is ____ .

(A)

(B)

(C)

(D)

Q30Single correctGravitation

The height in terms of radius of the earth (R), at which the acceleration due to gravity becomes

\dfrac{g}{9}

, where g is acceleration due to gravity on earth's surface, is

(A)

\displaystyle \sqrt{3}R

(B)

\displaystyle 2\sqrt{2}R

(C)

\displaystyle 2R

(D)

\displaystyle \dfrac{4}{9}R

Q31Single correctProperties of Solids and Liquids

A metal string A is suspended from a rigid support and its free end is attached to a block of mass M. Second block having mass 2M is suspended at the bottom of the first block using a string B. The area of cross sections of strings A and B are same. The ratio of lengths of strings of A to B is 2 and the ratio of their Young's moduli (

Y_A/Y_B

) is 0.5. The ratio of elongations in A to B is ____ .

(A)

(B)

(C)

(D)

Q32Single correctProperties of Solids and Liquids

A water spray gun is attached to a hose of cross sectional area

30\,\text{cm}^{2}

. The gun comprises of 10 perforations each of cross sectional area

15\,\text{mm}^{2}

. If the water flows in the hose with the speed of

50\,\text{cm/s}

, calculate the speed at which the water flows out from each perforation. (Neglect any edge effects)

(A)

100 m/s

(B)

10 m/s

(C)

1000 m/s

(D)

\displaystyle 15 \times 10^{2}

m/s

Q33Single correctKinetic Theory of Gases

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion A: If the average kinetic energy of

H_{2}

and

O_{2}

molecules, kept in two different sized containers are same, then their temperatures will be same.

Reason R: The r.m.s. speed of

H_{2}

and

O_{2}

molecules are same at same temperature.

Choose the correct answer from the options given below

(A)

Both A and R are true and R is the correct explanation of A

(B)

Both A and R are true but R is NOT the correct explanation of A

(C)

A is true but R is false

(D)

A is false but R is true

Q34Single correctProperties of Solids and Liquids

The temperature of a metal strip having coefficient of linear expansion

\alpha

is increased from

T_{1}

T_{2}

resulting in increase of its length by

\Delta L_{1}

. The temperature is further increased from

T_{2}

T_{3}

such that the increase in its length is

\Delta L_{2}

. Given

T_{3} + T_{1} = 2T_{2}

and

T_{2} - T_{1} = \Delta T

, the value of

\Delta L_{2}

(A)

\displaystyle \Delta L_{1}\left[1 + 2\alpha^{2}(\Delta T)^{2}\right]

(B)

\displaystyle \Delta L_{1}\left[1 + \alpha^{2}(\Delta T)^{2}\right]

(C)

\displaystyle \Delta L_{1}\left[1 + 2\alpha \Delta T\right]

(D)

\displaystyle \Delta L_{1}\left[1 + \alpha \Delta T\right]

Q35Single correctOscillations and Waves

A uniform disc of radius R and mass M is free to oscillate about the axis A as shown in the figure. For small oscillations the time period is ____ . (g is acceleration due to gravity)

A uniform disc of radius R and mass M with a horizontal axis A passing through a point on the topmost circumference of the disc (tangent in the plane of the disc). The center of the disc is labeled R below A and the disc is labeled M. A curved arrow at A indicates rotation about this tangential horizontal axis.

(A)

\displaystyle 2\pi\sqrt{\dfrac{5R}{4g}}

(B)

\displaystyle 2\pi\sqrt{\dfrac{2R}{3g}}

(C)

\displaystyle 2\pi\sqrt{\dfrac{3R}{2g}}

(D)

\displaystyle 2\pi\sqrt{\dfrac{3R}{g}}

Q36Single correctElectrostatics

A rigid dipole undergoes a simple harmonic motion about its centre in the presence of an electric field

\vec{E}_1 = E_0\hat{x}

. If another electric field

\vec{E}_2 = 2E_0(\hat{y} + \hat{z})

is introduced to the system, what will be the percentage change in the frequency of the oscillation (approximate)?

(A)

73%

(B)

63%

(C)

83%

(D)

53%

Q37Single correctElectrostatics

From the circuit given below, the capacitance between terminals A and B shown in the circuit is _____

\mu F

. (take

C_1 = C_2 = C_3 = 1\,\mu F

and

C_4 = 2\,\mu F

)

Capacitor network between terminals A and B: $C_1$ and $C_2$ are in series in the upper branch, $C_3$ forms a parallel branch across the combination of $C_1$ and $C_2$, and $C_4$ forms another parallel branch across A-B.

(A)

(B)

7/2

(C)

7/3

(D)

5/2

Q38Single correctElectrostatics

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: In electrostatics, a conductor does not store any net charge inside.
Reason R: Inside the capacitor (with no dielectric medium), the free charge carriers, if placed between the plates of capacitor, experience force and drift.
Choose the correct answer from the options given below

(A)

Both A and R are true and R is the correct explantion of A

(B)

Both A and R are true but R is NOT the correct explanation of A

(C)

A is true but R is false

(D)

A is false but R is true

Q39Single correctMagnetic Effects of Current and Magnetism

A solenoid has a core made of material with relative permeability 400. The magnetic field produced in the interior of solenoid is 1.0 T. The magnetic intensity in SI units is

\alpha \times 10^5

. The value of

\alpha

is ___. (Free space permeability

\mu_0 = 4\pi \times 10^{-7}

SI units.)

(A)

\displaystyle \dfrac{25}{\pi}

(B)

\displaystyle \dfrac{1}{16\pi}

(C)

\displaystyle \dfrac{1}{\pi}

(D)

\displaystyle \dfrac{1}{4\pi}

Q40Single correctElectromagnetic Waves

A magnetic field vector in an electromagnetic wave is represented by

\vec{B} = B_0 \sin\left(2\pi\nu t - \dfrac{2\pi x}{\lambda}\right)\hat{j}

. Its associated electric field vector is _____ .

(A)

\displaystyle \vec{E} = -\nu\lambda B_0 \sin\left(2\pi\nu t - \dfrac{2\pi x}{\lambda}\right)\hat{k}

(B)

\displaystyle \vec{E} = -\nu\lambda B_0 \sin\left(2\pi\nu t - \dfrac{2\pi x}{\lambda}\right)\hat{i}

(C)

\displaystyle \vec{E} = \nu\lambda B_0 \sin\left(2\pi\nu t - \dfrac{2\pi x}{\lambda}\right)\hat{k}

(D)

\displaystyle \vec{E} = \nu\lambda B_0 \sin\left(2\pi\nu t - \dfrac{2\pi x}{\lambda}\right)\hat{i}

SolutionAnswer: Option 1

\vec{E} = -\nu\lambda B_0 \sin\left(2\pi\nu t - \dfrac{2\pi x}{\lambda}\right)\hat{k}

Approach:

Identify the propagation direction from the wave's phase, use

E_0 = cB_0 = \nu\lambda B_0

for the amplitude, and fix the sign so that

\vec{E}\times\vec{B}

points along the propagation direction.

Step 1:Given:

\vec{B} = B_0 \sin(2\pi\nu t - 2\pi x/\lambda)\hat{j}

. The phase

(\omega t - kx)

with

k > 0

corresponds to propagation along

+\hat{i}

\hat{k}_{prop} = +\hat{i}

Step 2:By the E-B amplitude relation, the electric-field amplitude is

E_0 = cB_0

and

c = \nu\lambda

, hence

E_0 = \nu\lambda B_0

E_0 = cB_0 = \nu\lambda B_0

Step 3:Transverse condition:

\vec{E}

is perpendicular to both

\vec{B}\parallel\hat{j}

and

\hat{k}_{prop}\parallel\hat{i}

, so

\vec{E}\parallel \pm\hat{k}

\vec{E}\parallel \pm\hat{k}

Step 4:Choosing

\vec{E}\propto -\hat{k}

(-\hat{k})\times(\hat{j}) = -(\hat{k}\times\hat{j}) = -(-\hat{i}) = +\hat{i}

, which matches

\hat{k}_{prop}

. Hence the sign in front of the amplitude is negative.

(-\hat{k})\times(\hat{j}) = +\hat{i}

Step 5:Combine amplitude, direction, and the same phase as

\vec{B}

\vec{E} = -\nu\lambda B_0\sin\left(2\pi\nu t - \dfrac{2\pi x}{\lambda}\right)\hat{k}

Final answer:

\vec{E} = -\nu\lambda B_0 \sin\left(2\pi\nu t - \dfrac{2\pi x}{\lambda}\right)\hat{k}

Q41Single correctOptics

A convex lens is made from glass material having refractive index of 1.4 with same radius of curvature on both sides. The ratio of its focal length and radius of curvature is ___ .

(A)

0.5

(B)

2.5

(C)

0.8

(D)

1.25

Q42Single correctOptics

An unpolarized light of certain intensity passes through a combination of two polarizers whose transmission axes are at 30º and 90º, respectively, with respect to the horizontal axis. A third polarizer with its transmission axis at 60º with the horizontal axis is placed between the two existing polarizers. The ratio of the output intensities with and without the third polarizer is ____ .

(A)

3/4

(B)

4/3

(C)

9/4

(D)

4/9

SolutionAnswer: Option 39/4

Approach:

Compute the transmitted intensity in both configurations (two-polarizer and three-polarizer) using Malus's law sequentially, then take the ratio.

Step 1:Given: incident unpolarized intensity

I_0

. Polarizer axes (with respect to horizontal):

P_1

30^\circ

P_2

90^\circ

, optional middle

P_3

60^\circ

. Target:

I_{\text{with}}/I_{\text{without}}

\theta_{P1}=30^\circ,\ \theta_{P3}=60^\circ,\ \theta_{P2}=90^\circ

Step 2:Configuration without

P_3

: after

P_1

the intensity is

I_0/2

. The angle between

P_1

(

30^\circ

) and

P_2

(

90^\circ

) is

60^\circ

I_{without} = \dfrac{I_0}{2}\cos^2(60^\circ) = \dfrac{I_0}{2}\cdot\dfrac{1}{4} = \dfrac{I_0}{8}

Step 3:Configuration with

P_3

inserted: after

P_1

the intensity is again

I_0/2

. The angle between

P_1

(

30^\circ

) and

P_3

(

60^\circ

) is

30^\circ

I_{after\,P3} = \dfrac{I_0}{2}\cos^2(30^\circ) = \dfrac{I_0}{2}\cdot\dfrac{3}{4} = \dfrac{3I_0}{8}

Step 4:Light is now polarized along the

P_3

axis. The angle between

P_3

(

60^\circ

) and

P_2

(

90^\circ

) is

30^\circ

, so Malus's law gives the final intensity.

I_{\text{with}} = \dfrac{3I_0}{8}\cos^2(30^\circ) = \dfrac{3I_0}{8}\cdot\dfrac{3}{4} = \dfrac{9I_0}{32}

Step 5:Form the requested ratio.

\dfrac{I_{\text{with}}}{I_{\text{without}}} = \dfrac{9I_0/32}{I_0/8} = \dfrac{9}{32}\cdot 8 = \dfrac{9}{4}

Final answer: 9/4

Q43Single correctAtoms and Nuclei

In Rutherford's alpha-particle scattering experiment, only a few alpha particles rebound back because
A. The size of gold nucleus is very small as compared to the size of gold atom.
B. Alpha particle and gold nucleus have equal charge.
C. The impact parameter is minimum for a few alpha particles.
D. A few alpha particles have very high kinetic energy.
E. Only a few alpha particles undergo head-on collision with the nuclei.
Choose the correct answer from the options given below:

(A)

A, B Only

(B)

B, E Only

(C)

C, D Only

(D)

A, C, E Only

Q44Single correctDual Nature of Matter and Radiation

The de Broglie wavelength associated with an electron accelerated through a potential difference V is

\lambda_e

and the de Broglie wavelength associated with a proton accelerated through the same potential difference is

\lambda_p

. If their corresponding masses are

m_e

and

m_p

, respectively, then the ratio of their de Broglie wavelengths

\left(\dfrac{\lambda_e}{\lambda_p}\right)

is _____ .

(A)

\displaystyle \sqrt{\dfrac{m_p}{m_e}}

(B)

\displaystyle \sqrt{\dfrac{m_e}{m_p}}

(C)

\displaystyle \dfrac{m_p}{m_e}

(D)

\displaystyle \left(\dfrac{m_p}{m_e}\right)^2

Q45Single correctElectronic Devices

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: A diode under reverse-biased condition provides very small current which is nearly independent of voltage until a critical limit at which the current increases drastically.
Reason R: Below the critical voltage limit, only majority charge carriers flow which increases drastically above critical voltage.
choose the correct answer from the options given below

(A)

Both A and R are true and R is the correct explantion of A

(B)

Both A and R are true but R is NOT the correct explanation of A

(C)

A is true but R is false

(D)

A is false but R is true

Q46NumericalElectronic Devices

A diode has Zener voltage of 10 V and maximum power dissipation of 0.5 W, then the minimum resistance to be used in series with this diode for safety when it is connected to a 25 V power supply is ____

\Omega

Q47NumericalKinematics

A gun mounted on the ground fires bullets in all directions with same speed. The farthest distance the bullets could reach is 6.4 m. The speed of the bullets from the gun is ______ m/s. (take

g = 10\ \text{m/s}^2

)

Q48NumericalMagnetic Effects of Current and Magnetism

Two identical small bar magnets each of dipole moment

3\sqrt{5}\ \text{J/T}

are placed at a center to center separation of 10 cm , with their axes perpendicular to each other as shown in figure. The value of magnetic field at the point P midway between the magnets is

\alpha \times 10^{-3}

T. The value of

\alpha

is ___ . (

\mu_0 = 4\pi \times 10^{-7}\ \text{Tm/A}

)

Two bar magnets shown 10 cm apart center-to-center. The left magnet is horizontal with its S pole on the left and N pole on the right (axis along the line joining the magnets). The right magnet is oriented vertically with its S pole at the top and N pole at the bottom (axis perpendicular to the line joining the magnets). Point P is marked at the midpoint of the line joining the two magnet centers.

SolutionAnswer: 12

Approach:

Treat each bar magnet as an ideal dipole. P lies on the axial line of magnet 1 and on the equatorial line of magnet 2; compute the two field magnitudes, identify that the two contributions are mutually perpendicular, and combine vectorially.

Step 1:Given: dipole moment

M = 3\sqrt{5}\ \text{J/T}

, centre-to-centre separation

d = 10\ \text{cm}

, axes mutually perpendicular as per the figure. Midpoint P is at

r = d/2 = 5\ \text{cm} = 0.05\ \text{m}

from each magnet. Also

\mu_0/(4\pi) = 10^{-7}\ \text{T m/A}

. Target:

\alpha

such that the net field at P equals

\alpha\times 10^{-3}\ \text{T}

M = 3\sqrt{5}\ \text{J/T},\ r = 0.05\ \text{m},\ \dfrac{\mu_0}{4\pi} = 10^{-7}\ \text{T m/A}

Step 2:From the figure, magnet 1 has its axis along the line joining the centres, so P lies on its axial line. Magnet 2's axis is perpendicular to that line, so P lies on its equatorial line.

\text{P}:\ \text{axial to magnet 1, equatorial to magnet 2}

Step 3:Axial-field magnitude at P from magnet 1.

B_{ax} = \dfrac{10^{-7}\cdot 2\cdot 3\sqrt{5}}{(0.05)^3} = \dfrac{6\sqrt{5}\times 10^{-7}}{1.25\times 10^{-4}} = 4.8\sqrt{5}\times 10^{-3}\ \text{T}

Step 4:Equatorial-field magnitude at P from magnet 2.

B_{eq} = \dfrac{10^{-7}\cdot 3\sqrt{5}}{(0.05)^3} = \dfrac{3\sqrt{5}\times 10^{-7}}{1.25\times 10^{-4}} = 2.4\sqrt{5}\times 10^{-3}\ \text{T}

Step 5:

B_{ax}

is directed along the line of centres (parallel to magnet 1's axis). The equatorial field of an ideal dipole points anti-parallel to its moment, so

B_{eq}

is directed along magnet 2's axis, which is perpendicular to the line of centres. Hence the two contributions are mutually perpendicular.

B_{ax}\perp B_{eq}

Step 6:Combine by Pythagoras.

B = \sqrt{B_{ax}^2 + B_{eq}^2} = \sqrt{(4.8\sqrt{5})^2 + (2.4\sqrt{5})^2}\times 10^{-3} = \sqrt{115.2 + 28.8}\times 10^{-3} = \sqrt{144}\times 10^{-3}\ \text{T}

Step 7:Compare with the prescribed form

\alpha\times 10^{-3}\ \text{T}

\alpha\times 10^{-3} = 12\times 10^{-3}\;\Rightarrow\;\alpha = 12

Final answer: 12

Q49NumericalMagnetic Effects of Current and Magnetism

A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of

30^\circ

with the direction of the magnetic field. The torque acting on the coil is

\alpha \times 10^{-4}

N.m. The value of

\alpha

is ____ . (

\pi = 3.14

)

Q50NumericalOptics

In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index 1.56, the central fringe shifts to the position of

7^{th}

bright fringe, obtained with both slits uncovered. If the light source wavelength is 450 nm, the thickness of mica sheet is

\alpha \times 10^{-9}

m. The value of

\alpha

is ______.

Chemistry25 questions

Q51Single correctSome Basic Concepts in Chemistry

The correct order of total number of atoms present in (A) 2 moles of cyclohexane (B) 684 g of sucrose (C) 90.8 L of dihydrogen at STP is :

(A)

\displaystyle C > A > B

(B)

\displaystyle C > B > A

(C)

\displaystyle B > C > A

(D)

\displaystyle B > A > C

Q52Single correctAtomic Structure

The species having identical radii according to the Bohr's theory are: A. H (first orbit) B. He

^+

(first orbit) C. He

^+

(Second orbit) D. L

i^{2+}

(first orbit) E. B

e^{3+}

(Second orbit) Choose the correct answer from the options given below:

(A)

A and C Only

(B)

A and E Only

(C)

B and E Only

(D)

C and D Only

Q53Single correctChemical Bonding and Molecular Structure

Which of the following pictorial diagram most correctly represents the

\pi^{*}

(

\pi

antibonding) molecular orbital between two atoms if the internuclear axis is taken to be in the z-direction (

\xrightarrow{z\text{-axis}}

) ?

(A)

Sphere-shaped orbital (resembling a face) representing a sigma-bonding MO along the z-axis.

(B)

Two side-by-side lobes (one above and one below) on each atom of the same sign on each side, representing a pi-bonding MO perpendicular to z-axis (no node between atoms).

(C)

Two atoms with side-lobes of opposite phase between them (nodal plane perpendicular to z between the nuclei), representing the pi* antibonding MO.

(D)

Two p-orbitals along z-axis with a node between the nuclei (sigma* antibonding along z).

Q54Single correctSolutions

27^{\circ}\text{C}

0.1\,\text{M}, 1\,\text{L}\ K_{4}[Fe(CN)_{6}]

aqueous solution and

0.1\,\text{M}, 1\,\text{L}\ FeCl_{3}

aqueous solution are placed in a container separated by a semi permeable membrane AB. Assume complete dissociation of both the solutes. Which of the following statement is correct?

Rectangular container divided vertically by a semipermeable membrane labeled AB. Left compartment labeled 'side x' contains 0.1 M, 1 L K4[Fe(CN)6]; right compartment labeled 'side y' contains 0.1 M, 1 L FeCl3.

(A)

Blue color is formed on both sides.

(B)

Ionic solutes in aqueous solution can pass through semi-permeable membrane.

(C)

Solution on side '

\displaystyle y

' is hypotonic.

(D)

To cause the reverse flow of solvent during osmosis, external pressure (any value) should be applied to side '

\displaystyle x

SolutionAnswer: Option 3Solution on side '

y

' is hypotonic.

Approach:

Compute the van't Hoff factor and effective particle concentration on each side; assign hypertonic/hypotonic on that basis and then refute the remaining statements.

Step 1:Given: side x has 0.1 M

K_4[Fe(CN)_6]

, side y has 0.1 M

\text{FeCl}_3

at the same temperature 27

^{\circ}

C; assume complete dissociation. Target: identify the correct statement.

T=300\ \text{K},\ C_x=C_y=0.1\ \text{M}

Step 2:Dissociation on side x produces 5 ions per formula unit.

K_4[Fe(CN)_6] \rightarrow 4K^{+} + [Fe(CN)_6]^{4-};\ i_x = 5

Step 3:Dissociation on side y produces 4 ions per formula unit.

\text{FeCl}_3 \rightarrow Fe^{3+} + 3Cl^{-};\ i_y = 4

Step 4:Compare osmotic pressures via

\pi = i C R T

at common T. Since

C_{\text{eff},x} > C_{\text{eff},y}

\pi_x > \pi_y

. By definition, the side with lower osmotic pressure is hypotonic.

\pi_x = 0.5 R T,\ \pi_y = 0.4 R T \Rightarrow \pi_x > \pi_y

Step 5:Refute the other options. (1) Prussian-blue formation requires the

K_4[Fe(CN)_6]

and

Fe^{3+}

to mix; the semipermeable membrane blocks ions, so no colour forms on either side. (2) Ions are solute particles which a semipermeable membrane explicitly blocks. (4) Reverse osmosis requires an external pressure greater than the osmotic pressure difference applied on the hypertonic (concentrated) side x; an arbitrary 'any value' does not suffice, so the statement is wrong.

\Rightarrow \text{Only statement (3) is correct}

Final answer: Solution on side '

y

' is hypotonic.

Q55Single correctEquilibrium

20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X) ?

(\,pK_{a}

value of acetic acid is 4.75).

(A)

7.0

(B)

4.75

(C)

3.5

(D)

4.82

Q56Single correctSome Basic Principles of Organic Chemistry

Match the LIST-I with LIST-II

List-I (Reaction)	List-II (Mechanism)
A. Williamson Synthesis	I. Electrophilic addition
B. Friedel Craft Reaction	II. Free radical substitution
C. Bromination of vinyl benzene	III. Nucleophilic substitution
D. Chlorination of toluene in light	IV. Electrophilic substitution

(A)

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

(B)

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

(C)

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

(D)

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

Q57Single correctClassification of Elements and Periodicity in Properties

The

1^{st}

ionization enthalpy for Mg is

+737\ kJ/\text{mol}

. The most probable estimated value of the

2^{nd}

ionization enthalpy of Mg is ______ .

(A)

\displaystyle 906\ kJ/\text{mol}

(B)

\displaystyle 856\ kJ/\text{mol}

(C)

\displaystyle +1450\ kJ/\text{mol}

(D)

\displaystyle +590\ kJ/\text{mol}

Q58Single correctp-Block Elements

The electronegativity of a group 13 element ' E ' is same as that of Ge (on Pauling scale and upto one decimal point). The CORRECT statements about

E^{3+}

are A. It can act as a reducing agent. B. It can act as an oxidizing agent. C.

E^{3+}

is more stable than

E^{+}

. D. The standard electrode potential value for

E^{3+}/E

is positive. Choose the correct answer from the options given below:

(A)

A and C Only

(B)

B and C Only

(C)

B and D Only

(D)

A and D Only

Q59Single correctd- and f-Block Elements

Pairs of elements with the same number of electrons in their respective 4 f orbital are [Atomic number. Eu-63, Gd-64, Dy-66, Ho-67, Tm-69, Yb-70, Lu-71, Hf-72] A. (Eu and Gd) B. (Dy and Ho) C. (Yb and Hf) D. (Lu and Tm) Choose the correct answer from the options given below:

(A)

B and C Only

(B)

A and B Only

(C)

A and D Only

(D)

A and C Only

Q60Single correctCoordination Compounds

Consider the metal complexes

[Ni(en)_{3}]^{2+}\ (A),\ [\text{NiCl}_{4}]^{2-}\ (B)

and

[Ni(NH_{3})_{6}]^{2+}\ (C)

. Choose the CORRECT option by considering the number of unpaired electrons present in (A), (B) and (C) respectively and the order of frequency of absorption.

(A)

2, 2, 2 and

\displaystyle (A) > (C) > (B)

(B)

0, 2, 0 and

\displaystyle (A) > (C) > (B)

(C)

2, 2, 0 and

\displaystyle (B) > (C) > (A)

(D)

2, 2, 2 and

\displaystyle (C) > (A) > (B)

Q61Single correctChemical Bonding and Molecular Structure

Consider the following molecules/species: (x) cycloheptatrienone (tropone), (y) acetone

(CH_3)_2C=O

, (z) acetate ion

CH_3-C(=O)-O^{(-)}

. The correct order of carbon - oxygen double bond length is :

Three structures labelled (x), (y), (z). (x) is a seven-membered carbocyclic ring (cycloheptatrienone) with three conjugated C=C bonds and a C=O group. (y) is acetone with two CH3 groups attached to a central carbon doubly bonded to O. (z) is the acetate anion CH3-C(=O)-O(-) with one C-O double bond and one C-O single bond bearing a negative charge.

(A)

\displaystyle x > y > z

(B)

\displaystyle y > z > x

(C)

\displaystyle z > x > y

(D)

\displaystyle x > z > y

Q62Single correctd- and f-Block Elements

Consider

|x|

is the difference in oxidation states of Mn in highest manganese fluoride and highest manganese oxide. The ions with

|x|

number of unpaired electrons from the following are: A.

Sc^{3+}

Zn^{2+}

V^{2+}

Fe^{2+}

Co^{2+}

Choose the correct answer from the options given below:

(A)

A and B Only

(B)

C, D and E Only

(C)

C and E Only

(D)

B and E Only

Q63Single correctChemical Kinetics

Consider the given graph showing variation of reactant concentration with time. Three different reactions were started with identical initial concentration of reactants. Which of the following statement is correct?

Plot of [R] (mol L^-1) on the y-axis vs t/s on the x-axis starting from a common initial concentration [R]_0. Three curves labelled 1, 2, 3 decrease with time; curve 1 is a straight line (zero order), curve 2 is an exponential decay (first order), and curve 3 is a slower decay (second order or higher) below curves 1 and 2.

(A)

The order of all the three reactions is same.

(B)

The rate constant of reaction 3 is larger than the rate constant of reaction 2 if the order of reaction is same for both.

(C)

The SI unit of rate constant of reaction 1 is

\displaystyle s^{-1}

(D)

Thermal decomposition of HI on gold surface is an example of reaction 2.

SolutionAnswer: Option 2The rate constant of reaction 3 is larger than the rate constant of reaction 2 if the order of reaction is same for both.

Approach:

Read the relative position of the three [R]-vs-t curves to identify which reaction decays fastest, then test each statement against the integrated rate laws and the units of

k

Step 1:Given: three reactions with the same starting concentration

[R]_0

; curve 1 lies on top (slowest reactant loss), curve 2 in the middle, curve 3 at the bottom (fastest reactant loss) on the [R]-vs-t plot. Target: identify the only true statement.

[R]_1(t) > [R]_2(t) > [R]_3(t)\ \text{at any}\ t>0

Step 2:Evaluate statement (1). The three curves have visibly different shapes (line versus curved decays of different curvatures), so they cannot all have the same order. FALSE.

\text{Shape}(1)\neq\text{Shape}(2)\neq\text{Shape}(3)

Step 3:Evaluate statement (3). Curve 1, being a straight line, corresponds to zero order; its rate constant has units

\text{mol}\,L^{-1}\,s^{-1}

, not

s^{-1}

. FALSE.

[k_{\text{zero order}}] = \text{mol}\,L^{-1}\,s^{-1}

Step 4:Evaluate statement (4). Decomposition of HI on a hot gold surface is the textbook example of a zero-order surface reaction (saturated catalyst surface), corresponding to curve 1, not curve 2. FALSE.

2HI(g) \xrightarrow{\text{Au surface}} H_2 + I_2\ (\text{zero order})

Step 5:Evaluate statement (2). Suppose hypothetically reactions 2 and 3 had the same order n and the same initial concentration. From each integrated rate law,

[R]_2(t)

and

[R]_3(t)

depend monotonically on k; greater k produces a steeper drop, hence smaller [R] at every

t > 0

. Since curve 3 lies below curve 2, the conditional implication 'same order

\Rightarrow k_3 > k_2

' holds for every n. TRUE.

[R]_3(t) < [R]_2(t)\ \forall\,t>0,\ \text{same order} \Rightarrow k_3 > k_2

Step 6:Only statement (2) survives; it is the answer.

\Rightarrow \text{Option (2)}

Final answer: The rate constant of reaction 3 is larger than the rate constant of reaction 2 if the order of reaction is same for both.

Q64Single correctHydrocarbons

Compound (X) is subjected to the sequence of reactions as shown above. Molar mass of the major product (Y) formed is ______

\text{gmol}^{-1}

. (Given molar mass in

\text{gmol}^{-1}\; C:12, H:1, O:16

) Reagents: (i)

Br_2 / \text{CHCl}_3

, (ii)

\text{NaNH}_2

excess, (iii)

CH_3I

, (iv)

H_2,\;Na/NH_3(l)

Reaction scheme: starting compound (X) is styrene (a benzene ring attached to a CH=CH2 vinyl group). Arrow shows four reagents in sequence (i) Br2/CHCl3, (ii) NaNH2 excess, (iii) CH3I, (iv) H2, Na/NH3 (l) leading to the Major Product (Y).

(A)

(B)

118

(C)

160

(D)

125

SolutionAnswer: Option 2118

Approach:

Sequentially apply each reagent to styrene, identify the final hydrocarbon Y, and compute its molar mass from atomic contributions.

Step 1:Identify the starting material X = styrene (

\text{PhCH}=CH_2

) and the reagent sequence (i)

Br_2/\text{CHCl}_3

, (ii) excess

\text{NaNH}_2

, (iii)

CH_3I

, (iv)

Na/NH_3(l)

. Target: molar mass of the major product Y.

X: Ph-CH=CH_2; \; \text{atomic masses } C=12, H=1

Step 2:Electrophilic anti-addition of

Br_2

to the alkene gives the vicinal dibromide.

Ph-CH=CH_2 \xrightarrow{Br_2/CHCl_3} Ph-\text{CHBr}-CH_2Br

Step 3:Excess

\text{NaNH}_2

performs two successive

E2

eliminations on the dibromide, giving phenylacetylene; the excess base then deprotonates the terminal alkyne to give the sodium acetylide.

Ph-CHBr-CH_2Br \xrightarrow{NaNH_2 \text{ (excess)}} Ph-C\equiv C^{-}Na^{+}

Step 4:

S_N2

alkylation of the acetylide with

CH_3I

furnishes an internal alkyne.

Ph-C\equiv C^{-}Na^{+} + CH_3I \rightarrow Ph-C\equiv C-CH_3

Step 5:Dissolving-metal reduction with

Na/NH_3(l)

delivers two hydrogens trans across the internal alkyne, yielding the (E)-alkene.

Ph-C\equiv C-CH_3 \xrightarrow{Na/NH_3(l)} (E)\text{-}Ph-CH=CH-CH_3

Step 6:Compute the molar mass of

C_9H_{10}

M(Y) = 9(12) + 10(1) = 108 + 10 = 118 \; g\,\text{mol}^{-1}

Final answer: 118

Q65Single correctSome Basic Principles of Organic Chemistry

The following structures are

Two tetrahedral wedge-dash structures of a chiral carbon bearing four substituents Br, CH3, Me, Cl. Left structure: central C with Br (plain bond up), CH3 on dashed wedge (back, left), Me on plain bond (front, middle), Cl on solid wedge (forward, right). Right structure: central C with Br (plain bond up), Cl on solid wedge (forward, left), Me on plain bond (front, middle), CH3 on dashed wedge (back, right). The two structures are related by rotation and represent the same configuration at the stereocenter.

(A)

enantiomers.

(B)

identical molecules.

(C)

diastereomers.

(D)

meso compounds.

Q66Single correctOrganic Compounds Containing Oxygen

The descending order of acidity among the following compounds is: A. Phenol B. 4-nitrophenol C. 4-methoxyphenol D. 4-nitrobenzoic acid E. Benzoic acid. Choose the correct answer from the options given below:

Five aromatic compounds labelled A-E shown as benzene rings with substituents. A: phenol (benzene with -OH). B: 4-nitrophenol (benzene with -OH at C1 and -NO2 at C4, drawn as O2N- on the ring). C: 4-methoxyphenol (benzene with -OH at C1 and H3CO- at C4). D: 4-nitrobenzoic acid (benzene with -COOH at C1 and O2N- at C4). E: benzoic acid (benzene with -COOH).

(A)

\displaystyle B > D > E > A > C

(B)

\displaystyle D > B > E > A > C

(C)

\displaystyle C > A > B > D > E

(D)

\displaystyle D > E > B > A > C

Q67Single correctOrganic Compounds Containing Nitrogen

The strongest conjugate acid will result from:

(A)

Aniline: a benzene ring with -NH2 group attached.

(B)

4-methoxyaniline (p-anisidine): a benzene ring with -NH2 at C1 and -OCH3 at C4.

(C)

4-nitroaniline: a benzene ring with -NH2 at C1 and -NO2 at C4.

(D)

4-methylaniline (p-toluidine): a benzene ring with -NH2 at C1 and -CH3 at C4.

Q68Single correctBiomolecules

A D-aldotetrose on oxidation with concentrated

\text{HNO}_3

resulted in optically inactive dicarboxylic acid. The structure of the D-aldotetrose is:

(A)

Fischer projection of an aldotetrose: CHO at top, then C with HO on left and H on right, then C with HO on left and H on right, then CH2OH at bottom. (L-threose orientation: both OH on left.)

(B)

Fischer projection of an aldotetrose: CHO at top, then C with H on left and OH on right, then C with HO on left and H on right, then CH2OH at bottom. (D-threose: OH groups on opposite sides.)

(C)

Fischer projection of an aldotetrose: CHO at top, then C with H on left and OH on right, then C with H on left and OH on right, then CH2OH at bottom. (D-erythrose: both OH on right side.)

(D)

Fischer projection of an aldotetrose: CHO at top, then C with HO on left and H on right, then C with H on left and OH on right, then CH2OH at bottom. (L-erythrose orientation: OH groups on opposite sides.)

Q69Single correctPrinciples Related to Practical Chemistry

Among

Fe^{3+}, Pb^{2+}, Cu^{2+}

and

Mn^{2+}

, identify the one that gets precipitated out while passing

H_2S

in presence of

NH_4OH

as group reagent. The highest possible oxidation state of the corresponding metal is

(A)

(B)

(C)

(D)

Q70Single correctPrinciples Related to Practical Chemistry

Match the LIST-I with LIST-II

List-I Compound	List-II Test
A.	I. Hinsberg's reagent test
B.	II. Phthalein dye test
C.	III. Lucas test
D.	IV. Tollen's test

(A)

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

(B)

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

(C)

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

(D)

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

Q71NumericalChemical Thermodynamics

If 3.365 g of ethanol (l) is burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The

|\Delta H_f^\circ|

of ethanol at 298.15 K is _____

\times 10^2 \,\text{kJ mol}^{-1}

. (Nearest integer)

Given: Standard enthalpy for combustion of graphite

= -393.5 \,\text{kJ mol}^{-1}

Standard enthalpy of formation of water (l)

= -285.8 \,\text{kJ mol}^{-1}

Molar mass in

\text{g mol}^{-1}

of C, H, O are 12, 1 and 16 respectively

SolutionAnswer: 3

Approach:

Convert the calorimetric heat to

\Delta U_{comb}

per mole, add

\Delta n_g RT

to obtain

\Delta H_{comb}

, then invert Hess's law to extract

\Delta H_f^\circ

of ethanol.

Step 1:Given:

m = 3.365

q = 99.472

kJ (heat released),

T = 298.15

K. Molar mass of

C_2H_5OH

2(12)+6(1)+16 = 46

g/mol. Target:

|\Delta H_f^\circ(C_2H_5OH,l)|

in units of

10^2

kJ/mol.

M(C_2H_5OH) = 46\,\text{g mol}^{-1}

Step 2:Moles of ethanol burnt.

n = \dfrac{3.365}{46} = 7.3152\times 10^{-2}\,\text{mol}

Step 3:Molar

\Delta U

of combustion from the bomb-calorimeter (constant-volume) heat (sign: exothermic ==> negative).

\Delta U_{comb} = -\dfrac{99.472}{0.07315} = -1359.83\,\text{kJ mol}^{-1}

Step 4:Write the balanced combustion:

C_2H_5OH(l) + 3O_2(g) \to 2CO_2(g) + 3H_2O(l)

. Gaseous moles:

\Delta n_g = 2 - 3 = -1

\Delta n_g RT = (-1)(8.314\times 10^{-3})(298.15) = -2.479\,\text{kJ mol}^{-1}

Step 5:Convert to enthalpy of combustion.

\Delta H_{comb} = -1359.83 + (-2.479) = -1362.31\,\text{kJ mol}^{-1}

Step 6:Apply Hess's law with

\Delta H_f^\circ(O_2) = 0

-1362.31 = 2(-393.5) + 3(-285.8) - \Delta H_f^\circ(C_2H_5OH)

Step 7:Solve for

\Delta H_f^\circ

of ethanol.

\Delta H_f^\circ = -787.0 - 857.4 + 1362.31 = -282.09\,\text{kJ mol}^{-1}

Step 8:Express the magnitude in units of

10^2

kJ/mol and round to the nearest integer.

\dfrac{282.09}{10^2} = 2.8209 \Rightarrow 3

Final answer: 3

Q72NumericalEquilibrium

For the following reaction at

50^\circ\text{C}

and at 2 atm pressure,

2\,\text{N}_2\text{O}_5(g) \rightleftharpoons 2\,\text{N}_2\text{O}_4(g) + \text{O}_2(g)

\text{N}_2\text{O}_5

is 50% dissociated.
The magnitude of standard free energy change at this temperature is x.

x =

_____

\text{J mol}^{-1}

[Nearest integer].
Given:

R = 8.314 \,\text{J mol}^{-1}\text{K}^{-1}, \log 2 = 0.30, \log 3 = 0.48, \ln 10 = 2.303, \,^\circ\text{C} + 273 = \text{K}

SolutionAnswer: 2474

Approach:

Set up the ICE table with 50% dissociation, convert mole fractions to partial pressures at the stated total pressure, compute

K_p

, then apply

\Delta G^\circ = -RT\ln K_p = -2.303\,RT\log K_p

Step 1:Setup: start with

n_0

mol of

N_2O_5

; take

n_0 = 2

for convenient arithmetic. Degree of dissociation

\alpha = 0.5

means 1 mol of

N_2O_5

reacts.

T = 50 + 273 = 323\,K;\; P_{tot} = 2\,\text{atm};\; \alpha = 0.5

Step 2:Stoichiometry of

2\,N_2O_5 \to 2\,N_2O_4 + O_2

: reaction of 1 mol

N_2O_5

produces 1 mol

N_2O_4

and 0.5 mol

O_2

n_{N_2O_5}=1,\; n_{N_2O_4}=1,\; n_{O_2}=0.5,\; n_{tot}=2.5

Step 3:Mole fractions and partial pressures at

P_{tot}=2

atm.

x_{N_2O_5}=0.4,\; x_{N_2O_4}=0.4,\; x_{O_2}=0.2 \Rightarrow p_{N_2O_5}=0.8,\; p_{N_2O_4}=0.8,\; p_{O_2}=0.4\,\text{atm}

Step 4:Evaluate

K_p

(atm units).

K_p = \dfrac{(0.8)^2 (0.4)}{(0.8)^2} = 0.4

Step 5:Compute

\log K_p

using the supplied logarithm table.

\log 0.4 = \log 4 - \log 10 = 2\log 2 - 1 = 2(0.30) - 1 = -0.40

Step 6:Apply

\Delta G^\circ = -2.303\,RT\log K_p

\Delta G^\circ = -2.303 (8.314)(323)(-0.40) = +2.303 (8.314)(323)(0.40)

Step 7:Arithmetic.

2.303 \times 8.314 = 19.147;\; 19.147 \times 323 = 6184.5;\; 6184.5 \times 0.40 = 2473.8\,\text{J mol}^{-1}

Final answer: 2474

Q73NumericalRedox Reactions and Electrochemistry

An electrochemical cell, consist of the following two redox couples,

\text{M}^{x+}(aq)/\text{M}(s)\,[E_{red}^\ominus = +0.15\,\text{V}]

and

\text{Fe}^{3+}(aq)/\text{Fe}(s)\,[E_{red}^\ominus = -0.036\,\text{V}]

. The cell

\text{EMF}\,(E_{cell})

is recorded to be 0.2057 V. If the reaction quotient of the electrochemical reaction is found to be

10^{-2}

, then the value of x is _____. (Nearest integer)
[Given: M is a p-block metal and

\dfrac{2.303 RT}{F} = 0.059\,\text{V}

]

SolutionAnswer: 2

Approach:

Identify cathode and anode from standard potentials, compute

E^\circ_{cell}

, then apply the Nernst equation with the given reaction quotient to extract the number of electrons n; relate n to x from the balanced cell reaction.

Step 1:Given inputs:

E^\circ(M^{x+}/M)=+0.15\,V

E^\circ(Fe^{3+}/Fe)=-0.036\,V

E_{cell}=0.2057\,V

Q=10^{-2}

. Higher standard reduction potential means

M^{x+}

is reduced (cathode); Fe is oxidised to

Fe^{3+}

(anode).

\text{Cathode}: M^{x+}+xe^- \to M;\; \text{Anode}: Fe \to Fe^{3+}+3e^-

Step 2:Standard cell potential.

E^\circ_{cell} = 0.15 - (-0.036) = 0.186\,V

Step 3:Balance the overall reaction by taking LCM of

x

and 3 electrons: multiply the cathode half by 3 and the anode half by

x

. Total electrons transferred

n = 3x

3M^{x+} + \text{xFe} \to 3M + \text{xFe}^{3+};\; n=3x

Step 4:Substitute into the Nernst equation with

\log Q = \log(10^{-2}) = -2

0.2057 = 0.186 - \dfrac{0.059}{n}(-2) = 0.186 + \dfrac{0.118}{n}

Step 5:Solve for

n

\dfrac{0.118}{n} = 0.2057 - 0.186 = 0.0197 \Rightarrow n = \dfrac{0.118}{0.0197} = 5.99

Step 6:From

n = 3x

3x = 6 \Rightarrow x = 2

Step 7:Sanity check on identity:

x=2

with

E^\circ \approx +0.15\,V

matches the p-block couple

Sn^{2+}/Sn

(

E^\circ \approx +0.14\,V

), consistent with the hint that M is a p-block metal.

M \sim Sn,\; x=2

Final answer: 2

Q74NumericalChemical Kinetics

For a first order reaction

A \to B

t/\text{min}

[A]/M

|
| 0 | 0.6500 |
| x | 0.0650 |
| 20 | 0.00065 |

x =

_____ min. (Nearest integer)

Q75NumericalPrinciples Related to Practical Chemistry

In sulphur estimation,

2.0\times 10^{-3}

mol of an organic compound (X) (molar mass

76\,\text{g mol}^{-1}

) gave 0.4813 g of barium sulphate (molar mass

233\,\text{g mol}^{-1}

). The percentage of sulphur in the compound (X) is _____

\times 10^{-1}

% (Nearest integer)

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