JEE Main 2026 6th April Morning Shift Question Paper Question Paper with Solutions | JEE Main Mock Test | JeeMate

Q 1PhysicsSection A • MCQunits and measurements

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The density $\rho$ of a uniform cylinder is determined by measuring its mass $m$ , length $l$ and diameter $d$ . The measured values of $m$ , $l$ and $d$ are 97.42 $\pm$ 0.02 g, 8.35 $\pm$ 0.05 mm and 20.20 $\pm$ 0.02 mm, respectively. Calculated percentage fractional error in $\rho$ is _______.

A

0.63%

B

0.82%

C

0.72%

D

0.25%

Q 2PhysicsSection A • MCQunits and measurements

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The potential energy of a particle changes with distance $x$ from a fixed origin as $V = \frac{A\sqrt{x}}{x + B}$ , where $A$ and $B$ are constants with appropriate dimensions. The dimensions of $AB$ are _______.

A

$[M^1 L^{5/2} T^{-2}]$

B

$[M^{3/2} L^{5/2} T^{-2}]$

C

$[M^1 L^2 T^{-2}]$

D

$[M^1 L^{7/2} T^{-2}]$

Q 3PhysicsSection A • MCQwork power and energy

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The rain drop of mass 1 g, starts with zero velocity from a height of 1 km. It hits the ground with a speed of 5 m/s. The work done by the unknown resistive force is _______ J.
(take g = 10 m/s $^2$ )

A

$-8.75$

B

$-8.35$

C

$-9.55$

D

$-9.98$

Q 4PhysicsSection A • MCQlaws of motion

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Two blocks (P and Q) with respectively masses 2 kg and 1.5 kg are joined by a massless thread. These blocks are mounted on a frictionless pully which is fixed on the edge of a cube (S), as shown in the figure below. Block P is positioned on the top surface which has no friction and block Q is in contact with side-surface, having coefficient friction $\mu$ . The cube (S) moves towards the right with acceleration of $\frac{g}{2}$ , where g is gravitational acceleration. During this movement the block P and Q remain stationary. The value of $\mu$ is _______.
(take g = 10 m/s $^2$ )

A

0.33

B

0.67

C

1

D

0.5

Q 5PhysicsSection A • MCQproperties of matter

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A lift of mass 1600 kg is supported by thick iron wire. If the maximum stress which the wire can withstand is $4 \times 10^8$ N/m $^2$ and its radius is 4 mm, then maximum acceleration the lift can take is _______ m/s $^2$ .
(take g = 10 m/s $^2$ and $\pi$ = 3.14)

A

2.56

B

3.89

C

4.32

D

5.16

Q 6PhysicsSection A • MCQrotational motion

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A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of 1200 rpm. It is brought to rest in 10 s by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are _______ and _______ respectively.

A

$0.128\pi$ Nm, 100

B

$0.0128\pi$ Nm, 50

C

$0.128\pi$ Nm, 50

D

$0.0128\pi$ Nm, 100

Official Correct Answer:Option (D)

Step-by-Step Solution:

Initial angular velocity, $\omega_0 = 1200\text{ rpm}$

Final angular velocity, $\omega = 0\text{ rad/s}$ (since it is brought to rest)

Time taken to stop, $t = 10\text{ s}$

$\omega_0 = 1200 \times \frac{2\pi}{60}$

$\omega_0 = 20 \times 2\pi = 40\pi\text{ rad/s}$

Using the first equation of rotational kinematics:

$\omega = \omega_0 + \alpha t$

$0 = 40\pi + \alpha (10)$

$10\alpha = -40\pi$

$\alpha = -4\pi\text{ rad/s}^2$

The magnitude of angular deceleration is $4\pi\text{ rad/s}^2$ .

The formula for the moment of inertia of a solid sphere about an axis passing through its center is:

$I = \frac{2}{5}MR^2$

$I = \frac{2}{5} \times 5 \times (0.04)^2$

$I = 2 \times 0.0016$

$I = 0.0032\text{ kg}\cdot\text{m}^2 = 3.2 \times 10^{-3}\text{ kg}\cdot\text{m}^2$

Using Newton's second law for rotation (magnitude of torque):

$\tau = I |\alpha|$

$\tau = (0.0032) \times (4\pi)$

$\tau = 0.0128\pi\text{ N}\cdot\text{m}$

$\tau = 1.28\pi \times 10^{-2}\text{ N}\cdot\text{m}$

First, find the total angular displacement ( $\theta$ ):

$\theta = \left(\frac{\omega_0 + \omega}{2}\right) t$

$\theta = \left(\frac{40\pi + 0}{2}\right) \times 10$

$\theta = 20\pi \times 10 = 200\pi\text{ rad}$

Now, convert radians to the number of rotations:

$N = \frac{\theta}{2\pi}$

$N = \frac{200\pi}{2\pi}$

$N = 100$

Q 7PhysicsSection A • MCQcircular motion

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A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body is released from height $h$ as shown. If the body exerts a force of three times its weight on the plane at the highest point of circle then the height $h = \alpha R$ . The value of $\alpha$ is _______.

A

2

B

4

C

3

D

6

Q 8PhysicsSection A • MCQcenter of mass

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The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid point (p) of normal bisector, as shown in the figure is _______.

A

$\left(\frac{\sqrt{3}}{4}, 1.25\right)$

B

$\left(\frac{\sqrt{3}}{4}, 1.0\right)$

C

$(0, 0)$

D

$(1.25, 0)$

Q 9PhysicsSection A • MCQproperties of matter

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The two wires $A$ and $B$ of equal cross-section but of different materials are joined together. The ratio of Young's modulus of wire $A$ and wire $B$ is 20/11. When the joined wire is kept under certain tension the elongations in the wires $A$ and $B$ are equal. If the length of wire $A$ is 2.2 m, then the length of wire $B$ is _______ m.

A

1.1

B

2.22

C

1.21

D

4.44

Q 10PhysicsSection A • MCQheat and thermodynamics

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Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure 90 kPa and temperature 400 K. Keeping the temperature of one vessel constant at 400 K the second vessel temperature is raised to 500 K. The final pressure in the vessels is _______ kPa.

A

100

B

120

C

90

D

105

Q 11PhysicsSection A • MCQwave optics

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In interference experiment the path difference between two interfering waves at a point $A$ on the screen is $\lambda/3$ , where $\lambda$ is the wavelength of these waves, and at another point $B$ the path difference is $\lambda/6$ . The ratio of intensities at points $A$ and $B$ is _______.

A

3

B

4

C

1/3

D

1/4

Q 12PhysicsSection A • MCQsimple harmonic motion

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A particle is executing simple harmonic motion. Its amplitude is $A$ and time period is 5 sec. The time required by it to move from $x = A$ to $x = \frac{A}{\sqrt{2}}$ is _______ sec.

A

1/4

B

5/4

C

5/8

D

3/8

Q 13PhysicsSection A • MCQelectrostatics

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A thin half ring of radius 35 cm is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at centre of the half ring is 100 V/m, then the value of $Q$ is _______ nC.
$(\varepsilon_0 = 8.85 \times 10^{-12}$ C $^2$ /Nm $^2$ and $\pi = 3.14)$

A

2.14

B

2.44

C

3.25

D

0.7

Q 14PhysicsSection A • MCQsemiconductor devices and logic gates

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The maximum rated power of the LED is 2 mW and it is used in the circuit with input voltage of 5 V as shown in the figure below. The current through resistance $R_S$ is 0.5 mA.
The minimum value of the resistance of $R_S$ , to ensure that the LED is not damaged is _______ k $\Omega$ .

A

6

B

2

C

4

D

5

Q 15PhysicsSection A • MCQelectromagnetic waves

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A point light source emits E.M. waves in free space. A detector, placed at a distance of $L$ m, measures the intensity as $I_0$ . The detector is now shifted to another location on the same spherical surface ensuring the angle between original location and new location as 45 $^\circ$ . The measured intensity at new location will be _______.

A

$\frac{I_0}{4}$

B

$I_0$

C

$\frac{I_0}{\sqrt{2}}$

D

$\frac{I_0}{2}$

Q 16PhysicsSection A • MCQray optics

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A spherical interface lens of radius $R$ separates two media of refractive indices 1 and 1.4 respectively as shown in the figure below. A point source is placed at a distance of $4R$ in front of spherical interface. The magnitude of the magnification of point source image is _______.

A

1.66

B

2.33

C

2.66

D

1.33

Q 17PhysicsSection A • MCQmoving charges and magnetism

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A small cube of side 1 mm is placed at the centre of a circular loop of radius 10 cm carrying a current of 2 A. The magnetic energy stored inside the cube is $\alpha \times 10^{-14}$ J. The value of $\alpha$ is _______.
$(\mu_0 = 4\pi \times 10^{-7}$ Tm/A, $\pi = 3.14)$

A

6.28

B

$6.28 \times 10^{-6}$

C

628

D

$6.28 \times 10^{-4}$

Q 18PhysicsSection A • MCQelectromagnetic induction

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An inductor of inductance 10 mH having resistance of 100 $\Omega$ is connected to battery of E.M.F. 1.0 V through a switch as shown in the figure below. After switch is closed, the ratio of instantaneous voltages across the inductor when the current passing through it is 2 mA and 4 mA is _______.

A

4/3

B

3/4

C

5/3

D

3/5

Q 19PhysicsSection A • MCQatoms and nuclei

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The ratio of momentum of the photons of the 1 $^{st}$ and 2 $^{nd}$ line of Balmer series of Hydrogen atoms is $\alpha/\beta$ . The possible values of $\alpha$ and $\beta$ are:

A

27 and 20

B

3 and 16

C

5 and 36

D

20 and 27

Q 20PhysicsSection A • MCQalternating current

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A LCR series circuit driven with $E_{rms} = 90$ V at frequency $f_d = 30$ Hz has resistance $R = 80$ $\Omega$ , an inductance with inductive reactance $X_L = 20.0$ $\Omega$ and capacitance with capacitive reactance $X_C = 80.0$ $\Omega$ . The power factor of the circuit is _______.

A

0.8

B

0.64

C

0.9

D

0.5

Q 21PhysicsSection B • Numericalcurrent electricity

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Refer to the circuit diagram given below. The heat generated across the 6 $\Omega$ resistance in 100 second is $\frac{\alpha}{100}$ J. The value of $\alpha$ is _______. (Nearest integer)

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{6 \times 4}{6 + 4} = \frac{24}{10} = 2.4\ \Omega$

$E_{\text{eq}} = \frac{\frac{E_1}{R_1} + \frac{E_2}{R_2}}{\frac{1}{R_1} + \frac{1}{R_2}} = \frac{\frac{3}{6} + \frac{0}{4}}{\frac{1}{6} + \frac{1}{4}} = \frac{0.5}{\frac{5}{12}} = \frac{0.5 \times 12}{5} = 1.2\text{ V}$

$R_{\text{total}} = 3 + R_{\text{eq}} = 3 + 2.4 = 5.4\ \Omega$

$E_{\text{total}} = 2\text{ V} - 1.2\text{ V} = 0.8\text{ V}$

$I_{\text{total}} = \frac{E_{\text{total}}}{R_{\text{total}}} = \frac{0.8}{5.4} = \frac{4}{27}\text{ A}$

$V_{\text{parallel}} = E_{\text{eq}} + I_{\text{total}}R_{\text{eq}}$

$V_{\text{parallel}} = 1.2 + \left(\frac{4}{27} \times 2.4\right) = \frac{6}{5} + \left(\frac{4}{27} \times \frac{12}{5}\right) = \frac{6}{5} + \frac{16}{45} = \frac{54 + 16}{45} = \frac{70}{45} = \frac{14}{9}\text{ V}$

$I = \frac{V_{\text{parallel}} - 3}{6} = \frac{\frac{14}{9} - 3}{6} = \frac{-\frac{13}{9}}{6} = -\frac{13}{54}\text{ A}$

$H = I^2 R t = \left(\frac{13}{54}\right)^2 \times 6 \times 100$

$H = \frac{169}{2916} \times 600 = \frac{101400}{2916} \approx 34.77\text{ J}$

$\frac{\alpha}{100} = 34.77 \implies \alpha \approx 3477$

Q 22PhysicsSection B • Numericalwave optics

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An unpolarized light of intensity $I_0$ passes through polarizer and then through a certain optically active solution and finally it goes to analyser. If the angle between analyser and polariser is 0 $^\circ$ and intensity of light emerged from analyser is $\frac{3}{8}I_0$ , the angle of rotation of the light by the solution with respect to analyser is _______ degrees.

Answer Format:Enter numerical value rounded to the nearest integer

Q 23PhysicsSection B • Numericalatoms and nuclei

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The energy released when $\frac{7}{17.13}$ kg of $^7_3$ Li is converted into $^4_2$ He by proton bombardment is $\alpha \times 10^{32}$ eV. The value of $\alpha$ is _______. (Nearest integer)
(Mass of $^7_3$ Li = 7.0183 u, mass of $^4_2$ He = 4.004 u, mass of proton = 1.008 u and 1 u = 931 MeV/c $^2$ and Avogadro number = $6.0 \times 10^{23}$ )

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The nuclear reaction for the proton bombardment of Lithium-7 is:

$^{7}_{3}\text{Li} + ^{1}_{1}\text{H} \rightarrow 2 ^{4}_{2}\text{He}$

$\Delta m = m(^{7}_{3}\text{Li}) + m(\text{proton}) - 2 \times m(^{4}_{2}\text{He})$

$\Delta m = 7.0183 \text{ u} + 1.008 \text{ u} - 2(4.004 \text{ u})$

$\Delta m = 8.0263 \text{ u} - 8.0080 \text{ u}$

$\Delta m = 0.0183 \text{ u}$

The energy released per reaction ( $Q$ ) is calculated using the given conversion factor $1 \text{ u} = 931 \text{ MeV/c}^2$ :

$Q = 0.0183 \times 931 \text{ MeV}$

$Q = 17.0373 \text{ MeV}$

$Q = 17.0373 \times 10^6 \text{ eV}$

Next, calculate the number of Lithium atoms ( $N$ ) in the given sample mass.

The given mass is $m = \frac{7}{17.13} \text{ kg} = \frac{7000}{17.13} \text{ g}$

The molar mass of $^{7}_{3}\text{Li}$ is approximately $7 \text{ g/mol}$ .

Calculate the number of moles ( $n$ ):

$n = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{7000 / 17.13}{7}$

$n = \frac{1000}{17.13} \text{ mol}$

Calculate the total number of atoms ( $N$ ) using Avogadro's number ( $N_A = 6.0 \times 10^{23}$ ):

$N = n \times N_A$

$N = \left( \frac{1000}{17.13} \right) \times 6.0 \times 10^{23}$

$N = \frac{6.0 \times 10^{26}}{17.13} \text{ atoms}$

The total energy released ( $E$ ) is the total number of atoms multiplied by the energy released per reaction:

$E = N \times Q$

$E = \left( \frac{6.0 \times 10^{26}}{17.13} \right) \times (17.0373 \times 10^6 \text{ eV})$

Rearranging the terms for simpler calculation:

$E = 6.0 \times \left( \frac{17.0373}{17.13} \right) \times 10^{32} \text{ eV}$

Notice that the fraction is very close to 1 ( $\frac{17.0373}{17.13} \approx 0.9946$ ):

$E \approx 6.0 \times 0.9946 \times 10^{32} \text{ eV}$

$E \approx 5.967 \times 10^{32} \text{ eV}$

Rounding to the nearest integer, we get:

$E = 6 \times 10^{32} \text{ eV}$

Comparing this with the given format $\alpha \times 10^{32} \text{ eV}$ , we find the value of $\alpha$ :

$\alpha = 6$

Q 24PhysicsSection B • Numericalelectrostatics

+4 / -1

A three coulomb charge moves from the point (0, -2, -5) to the point (5, 1, 2) in an electric field expressed as $\vec{E} = 2x\hat{i} + 3y^2\hat{j} + 4\hat{k}$ N/C. The work done in moving the charge is _______ J.

Answer Format:Enter numerical value rounded to the nearest integer

Q 25PhysicsSection B • Numericalheat and thermodynamics

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A certain gas is isothermally compressed to $\left(\frac{1}{3}\right)^{rd}$ of its initial volume ( $V_0 = 3$ litre) by applying required pressure. If the bulk modulus of the gas is $3 \times 10^5$ N/m $^2$ , the magnitude of work done on the gas is _______ J.

Answer Format:Enter numerical value rounded to the nearest integer

Q 26ChemistrySection A • MCQsome basic concepts of chemistry

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An oxide of iron contains 69.9% iron, its empirical formula, is:
(Given: Molar mass of Fe and O are 56 and 16 g mol $^{-1}$ respectively.)

A

FeO

B

Fe $_2$ O $_3$

C

Fe $_3$ O $_4$

D

FeO $_3$

Q 27ChemistrySection A • MCQatomic structure

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If shortest wavelength of hydrogen atom in Lyman series is $x$ , then longest wavelength in Balmer series of He $^-$ is:

A

$\frac{9x}{5}$

B

$\frac{36x}{5}$

C

$\frac{x}{4}$

D

$\frac{5x}{9}$

Q 28ChemistrySection A • MCQatomic structure

+4 / -1

Match the LIST-I with LIST-II

List-I Orbital		List-II Radial nodes and nodal plane
A.	2s	I.	1 Radial node + two nodal planes
B.	3s	II.	1 Radial node + one nodal plane
C.	3p	III.	2 Radial nodes + No nodal plane
D.	4d	IV.	1 Radial node + No nodal plane

Choose the correct answer from the options given below:

A

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

B

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

C

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

D

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

Q 29ChemistrySection A • MCQchemical bonding and molecular structure

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The pairs among
$A = [SO_3^{2-}, CO_3^{2-}]$ , $B = [O_2^{2-}, F_2]$ , $C = [CN^-, CO]$ , $D = [NH_3, H_3O^+]$ and $E = [MnO_4^{2-}, CrO_4^{2-}]$ that do not have similar Lewis dot structure are:

A

A, B and E

B

A and E

C

B, C and D

D

C and D

Official Correct Answer:Option (B)

Step-by-Step Solution:

Species having the same number of atoms and the same total number of valence electrons are isoelectronic and generally possess similar Lewis dot structures.

For the pair ([SO_3^{2-},,CO_3^{2-}]),

$SO_3^{2-}:6+3(6)+2=26\ \text{valence electrons},$

$CO_3^{2-}:4+3(6)+2=24\ \text{valence electrons}.$

Since the total number of valence electrons is different, their Lewis dot structures are not similar.

For the pair ([O_2^{2-},,F_2]),

$O_2^{2-}:2(6)+2=14\ \text{valence electrons},$

$F_2:2(7)=14\ \text{valence electrons}.$

Both are isoelectronic and therefore have similar Lewis structures.

For the pair ([CN^-,,CO]),

$CN^-:4+5+1=10\ \text{valence electrons},$

$CO:4+6=10\ \text{valence electrons}.$

Both are isoelectronic and possess similar Lewis dot structures.

For the pair ([NH_3,,H_3O^+]),

$NH_3:5+3(1)=8\ \text{valence electrons},$

$H_3O^+:6+3(1)-1=8\ \text{valence electrons}.$

Both have the same electron count and similar trigonal pyramidal Lewis structures.

For the pair ([MnO_4^{2-},,CrO_4^{2-}]), although both contain tetrahedral oxyanions, the central atoms have different electronic configurations. In (MnO_4^{2-}), manganese is in the (+6) oxidation state with a (d^1) configuration, whereas in (CrO_4^{2-}), chromium is in the (+6) oxidation state with a (d^0) configuration. Hence, their Lewis dot structures are not considered similar.

Therefore, the pairs that do not have similar Lewis dot structures are A and E.

Hence, the correct answer is A and E.

Q 30ChemistrySection A • MCQthermodynamics

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Arrange the following isothermal processes in order of the magnitude of the work (p - V) involved between states 1 and 2.

A. Expansion in single stage $w_A$
B. Expansion in multi stages $w_B$
C. Compression in single stage $w_C$
D. Compression in multi stages $w_D$

Choose the correct option.

A

$|w_B| > |w_A| > |w_C| > |w_D|$

B

$|w_C| > |w_D| > |w_A| > |w_B|$

C

$|w_C| > |w_D| > |w_B| > |w_A|$

D

$|w_B| > |w_A| > |w_D| > |w_C|$

Q 31ChemistrySection A • MCQsolutions

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When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vapor pressure of solution was $x$ % of vapor pressure of pure solvent. What is $x$ %?

A

50%

B

60%

C

70%

D

80%

Q 32ChemistrySection A • MCQchemical equilibrium

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One mole each of He and A(g) are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium.
A(g) $\rightleftharpoons$ B(g)
K $_c$ for this reaction at 400 K is 4.0. The partial pressures (in atm) of He and B(g) are respectively (at equilibrium)
(Assume He, A(g) and B(g) behave as ideal gases)
(Given: R = 0.082 L atm K $^{-1}$ mol $^{-1}$ )

A

3.28, 2.624

B

2.624, 3.28

C

3.28, 0.656

D

0.656, 6.56

Q 33ChemistrySection A • MCQelectrochemistry

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Consider the following data.

Electrolyte	$\Lambda^\circ_m$ (S cm $^2$ mol $^{-1}$ )
BaCl $_2$	$x_1$
H $_2$ SO $_4$	$x_2$
HCl	$x_3$

BaSO $_4$ is sparingly soluble in water. If the conductivity of the saturated BaSO $_4$ solution is $x$ S cm $^{-1}$ then the solubility product of BaSO $_4$ can be given as
(Here $\Lambda_m = \Lambda^\circ_m$ )

A

$\frac{10^6 x^2}{a^2(x_1 + x_2 - 2x_3)^2}$

B

$\frac{x^2}{(x_1 + x_2 - 2x_3)^2}$

C

$\frac{a^2(x_1 + x_2 - 2x_3)^2}{10^6 x^2}$

D

$\frac{x^2}{(x_1 + x_2 + 2x_3)^2}$

Official Correct Answer:Option (A)

Step-by-Step Solution:

According to Kohlrausch's Law, the limiting molar conductivity of an electrolyte is equal to the sum of the limiting ionic conductivities of its constituent ions.

Given,

$\Lambda_m^0(BaCl_2)=x_1=\lambda_{Ba^{2+}}^0+2\lambda_{Cl^-}^0,$

$\Lambda_m^0(H_2SO_4)=x_2=2\lambda_{H^+}^0+\lambda_{SO_4^{2-}}^0,$

$\Lambda_m^0(HCl)=x_3=\lambda_{H^+}^0+\lambda_{Cl^-}^0.$

Hence,

$\Lambda_m^0(BaSO_4)=\lambda_{Ba^{2+}}^0+\lambda_{SO_4^{2-}}^0,$

$\Lambda_m^0(BaSO_4)=\Lambda_m^0(BaCl_2)+\Lambda_m^0(H_2SO_4)-2\Lambda_m^0(HCl),$

$\Lambda_m^0(BaSO_4)=x_1+x_2-2x_3.$

For a sparingly soluble salt,

$\Lambda_m^0=\frac{1000,\kappa}{S},$

where $\kappa=x\ \text{S cm}^{-1}$ and (S) is the solubility in $\text{mol L}^{-1}$ .

Therefore,

$S=\frac{1000x}{\Lambda_m^0(BaSO_4)},$

$S=\frac{1000x}{x_1+x_2-2x_3}.$

Since

$BaSO_4(s)\rightleftharpoons Ba^{2+}(aq)+SO_4^{2-}(aq),$

the solubility product is

$K_{sp}=[Ba^{2+}][SO_4^{2-}]=S^2.$

Substituting the value of (S),

$K_{sp}=\left(\frac{1000x}{x_1+x_2-2x_3}\right)^2,$

$K_{sp}=\frac{10^6x^2}{(x_1+x_2-2x_3)^2}.$

Hence, the correct expression for the solubility product is

$K_{sp}=\frac{10^6x^2}{(x_1+x_2-2x_3)^2}.$

Q 34ChemistrySection A • MCQp block elements

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Given below are two statements:

Statement I: Aluminium is more electropositive than thallium as the standard electrode potential value of E $^\circ_{Al^{3+}/Al}$ is negative and E $^\circ_{Tl^{3+}/Tl}$ is positive.

Statement II: The sum of first three ionization enthalpies of boron is very high when compared to that of aluminium. Due to this reason boron forms covalent compounds only and aluminium forms Al $^{3+}$ ion.

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

A

Both Statement I and Statement II are true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

D

Statement I is false but Statement II is true

Q 35ChemistrySection A • MCQd and f block elements

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The correct statements among the following are.

A. Basic vanadium oxide is used in the manufacture of H $_2$ SO $_4$ .
B. The spin-only magnetic moment value of the transition metal halide employed in Ziegler-Natta polymerization is 2.84 BM.
C. The p-block metal compound employed in Ziegler-Natta polymerization has the metal in +3 oxidation state.
D. The number of electrons present in the outer most 'd' orbital of metal halide employed in Wacker process is 8.

Choose the correct answer from the options given below:

A

A and B Only

B

A, C and D Only

C

C and D Only

D

B, C and D Only

Q 36ChemistrySection A • MCQcoordination compounds

+4 / -1

Match the LIST-I with LIST-II

List-I Electronic configuration of tetrahedral metal ion		List-II Crystal Field Stabilization Energy ( $\Delta_t$ )
A.	d $^2$	I.	-0.6
B.	d $^4$	II.	-0.8
C.	d $^6$	III.	-1.2
D.	d $^8$	IV.	-0.4

Choose the correct answer from the options given below:

A

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

B

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

C

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

D

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

Q 37ChemistrySection A • MCQcoordination compounds

+4 / -1

Which of the following are true about the energy of the given d-orbitals of a tetrahedral complex?

A. $d_{xy} = d_{xz} > d_{x^2-y^2}$
B. $d_{xy} = d_{yz} > d_{z^2}$
C. $d_{x^2-y^2} > d_{z^2} > d_{xz}$
D. $d_{x^2-y^2} = d_{z^2} < d_{xz}$

Choose the correct answer from the given below:

A

A, B and D only

B

A and B only

C

B and D only

D

B, C and D only

Q 38ChemistrySection A • MCQbasics of organic chemistry

+4 / -1

R $_f$ value for 2-methylpropene in a solvent system (Ethyl acetate + ether) is 0.42. 2-methylpropene is treated with dilute H $_2$ SO $_4$ to give major organic product (X). R $_f$ value for (X) in the same solvent system under identical condition will be:

A

0.42

B

0.82

C

0.62

D

0.12

Q 39ChemistrySection A • MCQbasics of organic chemistry

+4 / -1

Given below are two statements:

Statement I: 2,6-diethylcyclohexanone and 6-methyl-2-n-propylcyclohexanone are metamers.

Statement II: 2,2,6,6-tetramethylcyclohexanone exhibits keto-enol tautomerism.

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

A

Both Statement I and Statement II are true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

D

Statement I is false but Statement II is true

Q 40ChemistrySection A • MCQhydrocarbons

+4 / -1

Given below are two statements:

Statement I: Methane can be prepared by decarboxylation of sodium ethanoate, Kolbe's electrolysis of sodium acetate and reaction of CH $_3$ MgBr with water.
Statement II: Methane cannot be prepared from unsaturated hydrocarbons and by Wurtz reaction.

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

A

Both Statement I and Statement II are true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

D

Statement I is false but Statement II is true

Q 41ChemistrySection A • MCQhydrocarbons

+4 / -1

Given below are two statements:

Statement I: 3-phenylpropene reacts with HBr and gives secondary alkyl bromide having a chiral carbon atom as the major product.

Statement II: Aryl chlorides and aryl cyanides can be prepared by Sandmeyer reaction as well as Gattermann reaction.

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

A

Both Statement I and Statement II are true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

D

Statement I is false but Statement II is true

Q 42ChemistrySection A • MCQalcohols phenols and ethers

+4 / -1

Consider the following sequence of reactions

The major product P is:

A

qImage----Jeemate

B

qImage----Jeemate

C

qImage----Jeemate

D

qImage----Jeemate

Q 43ChemistrySection A • MCQalcohols phenols and ethers

+4 / -1

Arrange the following compounds according to increasing order of boiling points.
n-C $_4$ H $_9$ OH (A), n-C $_4$ H $_9$ NH $_2$ (B), n-C $_4$ H $_{10}$ (C) and C $_2$ H $_5$ NHC $_2$ H $_5$ (D).

A

C < B < A < D

B

D < C < B < A

C

C < D < B < A

D

D < B < A < C

Q 44ChemistrySection A • MCQbiomolecules

+4 / -1

Match the LIST-I with LIST-II

List-I Deficiency Disease		List-II Vitamin
A.	Scurvy	I.	Pyridoxine
B.	Convulsions	II.	Vitamin A
C.	Cheilosis	III.	Ascorbic Acid
D.	Xerophthalmia	IV.	Riboflavin

Choose the correct answer from the options given below:

A

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

B

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

C

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

D

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

Q 45ChemistrySection A • MCQbiomolecules

+4 / -1

Match the LIST-I with LIST-II

List-I Amino acid		List-II Positive reaction/Test for functional group present in side chain of amino acid
A.	Glutamine	I.	Hinsberg's test
B.	Lysine	II.	Neutral FeCl $_3$ test
C.	Tyrosine	III.	Ceric ammonium nitrate test
D.	Serine	IV.	Hoffman bromamide degradation

Choose the correct answer from the options given below:

A

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

B

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

C

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

D

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

Q 46ChemistrySection B • Numericalperiodic classification of elements

+4 / -1

First and second ionization enthalpies of lithium are 520 kJ mol $^{-1}$ and 7297 kJ mol $^{-1}$ respectively. Energy required to convert 3.5 mg lithium (g) into Li $^{2+}$ (g) [Li(g) $\to$ Li $^{2+}$ (g)] is _______ kJ mol $^{-1}$ . (nearest integer)
[Molar mass of Li = 7 g mol $^{-1}$ ]

Answer Format:Enter numerical value rounded to the nearest integer

Q 47ChemistrySection B • Numericalcompounds containing nitrogen

+4 / -1

Consider the following sequence of reactions.

The percentage of nitrogen in the yellow product (X) formed is _______ %. (Nearest Integer)
(Given Molar mass in g mol $^{-1}$ H:1, C:12, N:14)

Answer Format:Enter numerical value rounded to the nearest integer

Q 48ChemistrySection B • Numericalalcohols phenols and ethers

+4 / -1

4.7 g of phenol is heated with Zn to give product X. If this reaction goes to 60% completion then the number of moles of compound X formed will be _______ $\times 10^{-2}$ . (Nearest Integer)
(Given molar mass in g mol $^{-1}$ : H:1, C:12, O:16)

Answer Format:Enter numerical value rounded to the nearest integer

Q 49ChemistrySection B • Numericalchemical kinetics

+4 / -1

Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with $t_{1/2} = 3$ hour. The percentage of sucrose remaining after 6 hours is _______. (Nearest integer)
(Given: log 2 = 0.3010 and log 3 = 0.4771)

Answer Format:Enter numerical value rounded to the nearest integer

Q 50ChemistrySection B • Numericalthermodynamics

+4 / -1

Consider the reaction X $\rightleftharpoons$ Y at 300 K. If $\Delta H^\theta$ and K are 28.40 kJ mol $^{-1}$ and $1.8 \times 10^{-7}$ at the same temperature, then the magnitude of $\Delta S^\theta$ for the reaction in J K $^{-1}$ mol $^{-1}$ is _______. (Nearest integer)
(Given: R = 8.3 J K $^{-1}$ mol $^{-1}$ , ln 10 = 2.3, log 3 = 0.48, log 2 = 0.30)

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

Step 1: Convert Given Units to SI System

Temperature $T=300K$
Enthalpy $\triangle H^{\circ\ }=28.40kJmol^{-1}=28400\ Jmol^{-1}\$
Equilibrium Constant $K=1.8\ \times\ 10^{-7}$

Step 2: Calculate the Standard Gibbs Free Energy Change $\triangle\ G^{\circ\ }$

The relation between Gibbs free energy and the equilibrium constant is given by:

$\triangle\ G^{\circ\ }$ $=RTlnK$

Using the relation $\ln K=\ln10\times\ \log_{10}K$ :

$\log_{10}K=\log_{10}\left(1.8\times\ 10^{-7}\right)=\log_{10}\left(1.8\right)-7$

We can express $\log_{10}\left(1.8\right)$ using given values:

$\log_{10}\left(1.8\right)=\log_{10}\left(\frac{18}{10}\right)=\log_{10}\left(2\times\ 3^2\right)-\log_{10}\left(10\right)$

$\log_{10}\left(1.8\right)=\log_{10}2+2\log_{10}3-1$

$\log_{10}\left(1.8\right)=0.30+2\left(0.48\right)-1=0.30+0.96-1=0.26$

Now, substitute this back to find $\log_{10}K:$

$\log_{10}K=0.26-7=-6.74$

Next, convert to natural logarithm:

$\ln K=2.3\times\ \left(-6.74\right)=-15.502$

Now, solve for $\triangle\ G^{\circ\ }:$

$\triangle\ G^{\circ}=-\left(8.3\ J\ K^{-1\ }mol^{-1}\right)\times\ \left(-15.502\right)=+38600\ J\ mol^{-1}$

Step 3: Calculate the Standard Entropy Change $\triangle\ S^{\circ\ }$

Using thermodynamic relation:

$\triangle\ G^{\circ\ }=\triangle\ H^{\circ\ }-T\triangle\ S^{\circ\ }$

$T\triangle\ S^{\circ\ }=\triangle\ H^{\circ\ }-\triangle\ G^{\circ\ }$

$300\times\ \triangle\ S^{\circ\ }=28400-38600\$

$300\times\ \triangle\ S^{\circ\ }=-10200\$

$\triangle\ S^{\circ\ }=-\frac{10200}{300}=-34\ J\ K^{-1\ }mol^{-1}$

Therefore, the magnitude of $\left|\triangle\ S^{\circ\ }\right|=34$

Therefore, the correct is 34.

Q 51MathematicsSection A • MCQfunctions

+4 / -1

Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f(x) = \sin^{-1}\left(\frac{x + [x]}{3}\right)$ is $[\alpha, \beta)$ , then $\alpha^2 + \beta^2$ is equal to:

A

2

B

5

C

10

D

13

Q 52MathematicsSection A • MCQquadratic equations

+4 / -1

Let one root of the quadratic equation in x:
$(k^2 - 15k + 27)x^2 + 9(k - 1)x + 18 = 0$
be twice the other. Then the length of the latus rectum of the parabola $y^2 = 6kx$ is equal to:

A

4

B

6

C

8

D

12

Q 53MathematicsSection A • MCQhyperbola

+4 / -1

Let $e_1$ and $e_2$ be two distinct roots of the equation $x^2 - ax + 2 = 0$ . Let the sets
$\{a \in \mathbb{R} : e_1, e_2 \text{ are the eccentricities of hyperbolas}\} = (\alpha, \beta)$ , and
$\{a \in \mathbb{R} : e_1, e_2 \text{ are the eccentricities of an ellipse and a hyperbola, respectively}\} = (\gamma, \infty)$ .
Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:

A

18

B

22

C

26

D

34

Q 54MathematicsSection A • MCQcomplex numbers

+4 / -1

Let the set of all values of $k \in \mathbb{R}$ such that the equation $z(\bar{z} + 2 + i) + k(2 + 3i) = 0$ , $z \in \mathbb{C}$ , has at least one solution, be the interval $[\alpha, \beta]$ . Then $9(\alpha + \beta)$ is equal to:

A

$-10$

B

$-8$

C

$10\sqrt{13}$

D

$8\sqrt{13}$

Q 55MathematicsSection A • MCQsequences and series

+4 / -1

The value of $1^3 - 2^3 + 3^3 - ... + 15^3$ is:

A

1706

B

1856

C

1982

D

2403

Q 56MathematicsSection A • MCQsequences and series

+4 / -1

The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a G.P. is 8. If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to common difference of the A.P., then the sum of all possible values of the first term of the G.P. is:

A

$\frac{34}{9}$

B

$\frac{34}{13}$

C

$\frac{32}{9}$

D

$\frac{32}{13}$

Q 57MathematicsSection A • MCQpermutations and combinations

+4 / -1

The number of 4-letter words, with or without meaning, each consisting of two vowels and two consonants that can be formed from the letters of the word INCONSEQUENTIAL, without repeating any letter, is:

A

2670

B

2840

C

2920

D

3600

Q 58MathematicsSection A • MCQbinomial theorem

+4 / -1

If the coefficients of the middle terms in the binomial expansions of $(1 + \alpha x)^{26}$ and $(1 - \alpha x)^{28}$ , $\alpha \neq 0$ , are equal, then the value of $\alpha$ is:

A

1

B

$\frac{14}{13}$

C

$\frac{27}{7}$

D

$\frac{7}{27}$

Q 59MathematicsSection A • MCQstatistics

+4 / -1

A data consists of 20 observations $x_1, x_2, ..., x_{20}$ . If $\sum_{i=1}^{20}(x_i + 5)^2 = 2500$ and $\sum_{i=1}^{20}(x_i - 5)^2 = 100$ , then the ratio of mean to standard deviation of this data is:

A

2:1

B

3:1

C

3:2

D

4:1

Q 60MathematicsSection A • MCQprobability

+4 / -1

A bag contains (N + 1) coins - N fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$ , then N is equal to:

A

5

B

7

C

8

D

9

Q 61MathematicsSection A • MCQhyperbola

+4 / -1

If the eccentricity $e$ of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ , passing through $(6, 4\sqrt{3})$ , satisfies $15(e^2 + 1) = 34e$ , then the length of the latus rectum of the hyperbola $\frac{x^2}{b^2} - \frac{y^2}{2(a^2 + 1)} = 1$ is:

A

10

B

20

C

25

D

30

Q 62MathematicsSection A • MCQparabola

+4 / -1

Let chord PQ of length $3\sqrt{13}$ of the parabola $y^2 = 12x$ be such that the ordinates of points P and Q are in the ratio 1:2. If the chord PQ subtends an angle $\alpha$ at the focus of the parabola, then $\sin \alpha$ is equal to:

A

$\frac{3}{5}$

B

$\frac{4}{5}$

C

$\frac{5}{13}$

D

$\frac{12}{13}$

Q 63MathematicsSection A • MCQinverse trigonometric functions

+4 / -1

Let $0 < \alpha < 1$ , $\beta = \frac{1}{3\alpha}$ and $\tan^{-1}(1 - \alpha) + \tan^{-1}(1 - \beta) = \frac{\pi}{4}$ . Then $6(\alpha + \beta)$ is equal to:

A

6

B

7

C

8

D

9

Q 64MathematicsSection A • MCQtrigonometric equations

+4 / -1

Let $S = \left\{\theta \in (-2\pi, 2\pi) : \cos\theta + 1 = \sqrt{3}\sin\theta\right\}$ . Then the $\sum_{\theta \in S} \theta$ is equal to:

A

$\frac{2\pi}{3}$

B

$\frac{4\pi}{3}$

C

$-\frac{2\pi}{3}$

D

$-\frac{4\pi}{3}$

Q 65MathematicsSection A • MCQ3d geometry

+4 / -1

Let the image of the point P(1, 6, a) in the line L: $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - a + 1}{b}$ , $b > 0$ , be $\left(\frac{a}{3}, 0, a + c\right)$ . If S( $\alpha, \beta, \gamma$ ), $\alpha > 0$ , is the point on L such that the distance of S from the foot of perpendicular from the point P on L is $2\sqrt{14}$ , then $\alpha + \beta + \gamma$ is equal to:

A

19

B

20

C

21

D

22

Official Correct Answer:Option (C)

Step-by-Step Solution:

$L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}$

The foot of the perpendicular $M$ is the midpoint of the segment joining point $P(1, 6, a)$ and its image $Q\left(\frac{a}{3}, 0, a+c\right)$ :

$M = \left(\frac{1 + \frac{a}{3}}{2}, \frac{6 + 0}{2}, \frac{a + a + c}{2}\right) = \left(\frac{3+a}{6}, 3, \frac{2a+c}{2}\right)$

Since $M$ lies on line $L$ , substituting $y = 3$ into the line equation gives: $\frac{x}{1} = \frac{3-1}{2} = 1 \implies x = 1$

$\frac{3+a}{6} = 1 \implies a = 3$

$P = (1, 6, 3)$ and $M = \left(1, 3, \frac{6+c}{2}\right)$

$\frac{z - a + 1}{b} = 1 \implies \frac{\left(\frac{6+c}{2}\right) - 3 + 1}{b} = 1 \implies \frac{c}{2} = b \implies c = 2b$

$\vec{PM} = M - P = (1-1, 3-6, 1+b-3) = (0, -3, b-2)$

The direction vector of line $L$ is $\vec{d} = (1, 2, b)$ . Since vector $\vec{PM}$ is perpendicular to line $L$ : $\vec{PM} \cdot \vec{d} = 0 \implies (0)(1) + (-3)(2) + (b-2)(b) = 0$

$-6 + b^2 - 2b = 0 \implies b^2 - 2b - 6 = 0 \quad \text{--- (1)}$

Let any general point $S(\alpha, \beta, \gamma)$ on line $L$ be expressed in terms of parameter $t$ : $S = (t, 2t+1, bt + 2)$

$SM^2 = (2\sqrt{14})^2 = 56$

$(t-1)^2 + (2t+1-3)^2 + (bt+2-(1+b))^2 = 56$

$5(t-1)^2 + (b(t-1) + 1)^2 = 56$

Let $k = t-1$ : $5k^2 + b^2k^2 + 2bk + 1 = 56 \implies k^2(5 + b^2) + 2bk - 55 = 0$

$k^2(2b + 11) + 2bk - 55 = 0$

$2bk(k+1) + 11k^2 - 55 = 0$

$b = 3$ and $k = 2$

$t = 3$

$\alpha = t = 3$

$\beta = 2t + 1 = 7$

$\gamma = bt + 2 = 3(3) + 2 = 11$

$\alpha + \beta + \gamma = 3 + 7 + 11 = 21$

Q 66MathematicsSection A • MCQ3d geometry

+4 / -1

Let a line L be perpendicular to both the lines
$L_1: \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ and $L_2: \frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 6}{7}$ .
If $\theta$ is the acute angle between the lines L and
$L_3: \frac{x - \frac{8}{7}}{2} = \frac{y - \frac{4}{7}}{1} = \frac{z}{2}$ , then $\tan\theta$ is equal to:

A

$\frac{3}{2}\sqrt{2}$

B

$\frac{5}{2}\sqrt{2}$

C

$\frac{5}{3}\sqrt{2}$

D

$\frac{4}{3}\sqrt{2}$

Q 67MathematicsSection A • MCQlimits continuity and differentiability

+4 / -1

The value of $\lim_{x \to 0}\left(\frac{x^2 \sin^2 x}{x^2 - \sin^2 x}\right)$ is:

A

2

B

3

C

4

D

6

Q 68MathematicsSection A • MCQdefinite integration

+4 / -1

The value of the integral $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{32\cos^4 x}{1 + e^{\sin x}}\right)dx$ is:

A

$4\pi + 2$

B

$3\pi + 8$

C

$3\pi + 4$

D

$4\pi + 3$

Official Correct Answer:Option (B)

Step-by-Step Solution:

$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x}{1 + e^{\sin x}} \, dx \quad \text{--- (Equation 1)}$

$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4(-x)}{1 + e^{\sin(-x)}} \, dx$

$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x \cdot e^{\sin x}}{e^{\sin x} + 1} \, dx \quad \text{--- (Equation 2)}$

$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x \left(1 + e^{\sin x}\right)}{1 + e^{\sin x}} \, dx$

$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 32 \cos^4 x \, dx$

$I = 16 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^4 x \, dx$

$I = 32 \int_{0}^{\frac{\pi}{4}} \cos^4 x \, dx$

$\cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1 + 2\cos 2x + \cos^2 2x}{4}$

$\cos^4 x = \frac{1}{4} \left( 1 + 2\cos 2x + \frac{1 + \cos 4x}{2} \right) = \frac{1}{8} (3 + 4\cos 2x + \cos 4x)$

$I = 32 \int_{0}^{\frac{\pi}{4}} \frac{1}{8} (3 + 4\cos 2x + \cos 4x) \, dx$

$I = 4 \left[ 3x + 2\sin 2x + \frac{\sin 4x}{4} \right]_{0}^{\frac{\pi}{4}}$

$I = 4 \left( \frac{3\pi}{4} + 2(1) + 0 \right)$

$I = 3\pi + 8$

Q 69MathematicsSection A • MCQarea under the curves

+4 / -1

The area of the region $\{(x, y) : 0 \leq y \leq 6 - x, y^2 \geq 4x - 3, x \geq 0\}$ is:

A

8

B

9

C

12

D

15

Q 70MathematicsSection A • MCQfunctions

+4 / -1

Let $e$ be the base of natural logarithm and let $f : \{1, 2, 3, 4\} \to \{1, e, e^2, e^3\}$ and $g : \{1, e, e^2, e^3\} \to \left\{1,\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right\}$ be two bijective functions such that $f$ is strictly decreasing and $g$ is strictly increasing. If $\phi(x) = \left[f^{-1}\left\{g^{-1}\left(\frac{1}{2}\right)\right\}\right]^x$ , then the area of the region R = {(x, y): $x^2 \leq y \leq \phi(x)$ , $0 \leq x \leq 1$ } is:

A

$\frac{3 - \log_e(2)}{3\log_e(2)}$

B

$\frac{1}{3\log_e(2)}$

C

$3 + \log_e(2)$

D

$\frac{3 + \log_e(2)}{2 + \log_e(3)}$

Q 71MathematicsSection B • Numericalmatrices and determinants

+4 / -1

Let $A=\begin{bmatrix} -1 & 1 & -1\\ 1 & 0 & 1\\ 0 & 0 & 1\end{bmatrix}$ satisfy $A^2+\alpha\bigl(\operatorname{adj}(\operatorname{adj}(A))\bigr) + \beta\bigl(\operatorname{adj}(A)\operatorname{adj}(\operatorname{adj}(A))\bigr) = \begin{bmatrix} 2 & -2 & 2\\ -2 & 0 & -1\\ 0 & 0 & -1 \end{bmatrix}$ for some $\alpha,\beta\in\mathbb{R}$ . Then $(\alpha-\beta)^2$ is equal to _______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

$A = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}$

$|A| = 1 \cdot \begin{vmatrix} -1 & 1 \\ 1 & 0 \end{vmatrix} = 1 \cdot (0 - 1) = -1$

$\operatorname{adj}(\operatorname{adj}(A)) = |A|^{3-2}A = |A|A$

$\operatorname{adj}(A)\operatorname{adj}(\operatorname{adj}(A)) = \operatorname{adj}(A) \cdot (|A|A) = |A| \cdot \big(\operatorname{adj}(A)A\big) = |A| \cdot (|A|I) = |A|^2 I$

$\operatorname{adj}(\operatorname{adj}(A)) = -A$

$\operatorname{adj}(A)\operatorname{adj}(\operatorname{adj}(A)) = (-1)^2 I = I$

$A^2 + \alpha(-A) + \beta(I) = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$

$A^2 - \alpha A + \beta I = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix} \quad \text{--- (Equation 1)}$

$A^2 = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$(A^2)_{12} - \alpha(A)_{12} + \beta(I)_{12} = -2$

$-1 - \alpha(1) + 0 = -2 \implies -\alpha = -1 \implies \alpha = 1$

$(A^2)_{33} - \alpha(A)_{33} + \beta(I)_{33} = -1$

$1 - (1)(1) + \beta(1) = -1 \implies \beta = -1$

$(\alpha - \beta)^2 = (1 - (-1))^2 = (2)^2 = 4$

Q 72MathematicsSection B • Numericalcircle

+4 / -1

Let the centre of the circle $x^2 + y^2 + 2gx + 2fy + 25 = 0$ be in the first quadrant and lie on the line $2x - y = 4$ . Let the area of an equilateral triangle inscribed in the circle be $27\sqrt{3}$ . Then the square of the length of the chord of the circle on the line $x = 1$ is _______.

Answer Format:Enter numerical value rounded to the nearest integer

Q 73MathematicsSection B • Numericalvector algebra

+4 / -1

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = \hat{j} - \hat{k}$ and $\vec{c}$ be three vectors such that $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$ , then $\vec{c} \cdot (\vec{a} - 2\vec{b})$ is equal to _______.

Answer Format:Enter numerical value rounded to the nearest integer

Q 74MathematicsSection B • Numericaltrigonometric ratios and identities

+4 / -1

For the functions $f(\theta) = \alpha \tan^2\theta + \beta \cot^2\theta$ , and $g(\theta) = \alpha \sin^2\theta + \beta \cos^2\theta$ , $\alpha > \beta > 0$ , let $\min_{0 < \theta < \frac{\pi}{2}} f(\theta) = \max_{0 < \theta < \pi} g(\theta)$ . If the first term of a G.P. is $\left(\frac{\alpha}{2\beta}\right)$ , its common ratio is $\left(\frac{2\beta}{\alpha}\right)$ and the sum of its first 10 terms is $\frac{m}{n}$ , $\gcd(m, n) = 1$ , then $m + n$ is equal to _______.

Answer Format:Enter numerical value rounded to the nearest integer

Q 75MathematicsSection B • Numericaldifferential equations

+4 / -1

Let $y = y(x)$ be the solution of the differential equation $(x^2 - x\sqrt{x^2 - 1})dy + (y(x - \sqrt{x^2 - 1}) - x)dx = 0$ , $x \geq 1$ . If $y(1) = 1$ , then the greatest integer less than $y(\sqrt{5})$ is _______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

$\left(x^2 - x\sqrt{x^2 - 1}\right)dy + \left(y\left(x - \sqrt{x^2 - 1}\right) - x\right)dx = 0$

$\left(x^2 - x\sqrt{x^2 - 1}\right)\frac{dy}{dx} + \left(x - \sqrt{x^2 - 1}\right)y = x$

$\frac{dy}{dx} + \frac{x - \sqrt{x^2 - 1}}{x\left(x - \sqrt{x^2 - 1}\right)}y = \frac{x}{x\left(x - \sqrt{x^2 - 1}\right)}$

$\frac{dy}{dx} + \frac{1}{x}y = \frac{1}{x - \sqrt{x^2 - 1}}$

$\frac{dy}{dx} + \frac{1}{x}y = \frac{x + \sqrt{x^2 - 1}}{x^2 - (x^2 - 1)} = x + \sqrt{x^2 - 1}$

$\text{I.F.} = e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x$

$y \cdot x = \int x\left(x + \sqrt{x^2 - 1}\right) \, dx$

$xy = \int x^2 \, dx + \int x\sqrt{x^2 - 1} \, dx$

For $\int x\sqrt{x^2 - 1} \, dx$ , substitute $u = x^2 - 1 \implies du = 2x\,dx$ :

$\int \frac{1}{2}\sqrt{u} \, du = \frac{1}{2} \cdot \frac{2}{3}u^{3/2} = \frac{1}{3}(x^2 - 1)^{3/2}$

$xy = \frac{x^3}{3} + \frac{1}{3}(x^2 - 1)^{3/2} + C$

$1 = \frac{1}{3} + 0 + C \implies C = \frac{2}{3}$

$xy = \frac{x^3}{3} + \frac{1}{3}(x^2 - 1)^{3/2} + \frac{2}{3}$

$\sqrt{5} \cdot y = \frac{(\sqrt{5})^3}{3} + \frac{1}{3}((\sqrt{5})^2 - 1)^{3/2} + \frac{2}{3}$

$y = \frac{5 + \frac{10}{\sqrt{5}}}{3} = \frac{5 + 2\sqrt{5}}{3}$

$y \approx \frac{5 + 2(2.236)}{3} = \frac{5 + 4.472}{3} = \frac{9.472}{3} \approx 3.157$

$\lfloor y(\sqrt{5}) \rfloor = 3$

JEE Main 2026 6th April Morning Shift Question Paper with Solutions

1. Core Formula

2. Case 1: Shortest Wavelength of Hydrogen in Lyman Series

3. Case 2: Longest Wavelength of He⁺ in Balmer Series

4. Substitute R = 1 / x

Correct Answer: