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JEE Main 26 August 2021 Shift 2 Question Paper with Answers

26 August 2021 · August session · 87 questions

87 of the 90 questions from the JEE Main 26 August 2021 Shift 2 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

3 questions are held back from this paper while we re-check the transcription or the answer key. We would rather show you nothing than show you an answer we are not confident is right.

Physics
29
Chemistry
28
Mathematics
30

Physics — JEE Main 26 August 2021 Shift 2

Q1·PhysicsSingle correct
The temperature of equal masses of three different liquids x,y and z are 10°C, 20°C and 30°C respectively. The temperature of mixture when x is mixed with y is 16°C and that when y is mixed with z is 26°C. The temperature of mixture when x and z are mixed will be :
  1. (A)28.32° C
  2. (B)25.62° C
  3. (C)23.84°C
  4. (D)20.28°C

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correct
The de-Broglie wavelength of a particle having kinetic energy E is λ\lambdaλ. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value ?
  1. (A)19E\frac{1}{9}\mathrm{E}91​E
  2. (B)79E\frac{7}{9}\mathrm{E}97​E
  3. (C)E
  4. (D)169E\frac{16}{9}\mathrm{E}916​E

Correct answer: (B)

Step-by-step solution →
Q3·PhysicsSingle correct
A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r such that r=L2r = \frac{L}{\sqrt{2}}r=2​L​ .The speed of particle will be :
  1. (A)rg\sqrt{rg}rg​
  2. (B)2rg\sqrt{2rg}2rg​
  3. (C)2rg2\sqrt{rg}2rg​
  4. (D)rg2\sqrt{\frac{rg}{2}}2rg​​

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correct
A cylindrical container of volume 4.0×10−34.0 \times 10^{-3}4.0×10−3 m3^{3}3 contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is 400 K. The pressure of the mixture of gases is : [Take gas constant as 8.3 J mol−1^{-1}−1 K−1^{-1}−1]
  1. (A)249×101249 \times 10^{1}249×101 Pa
  2. (B)24.9×10324.9 \times 10^{3}24.9×103 Pa
  3. (C)24.9×10524.9 \times 10^{5}24.9×105 Pa
  4. (D)24.9 Pa

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
The angle between vector (A⃗)\left(\vec{A}\right)(A) and (A⃗−B⃗)\left(\vec{A} - \vec{B}\right)(A−B) is :
  1. (A)tan⁡−1(−B2A−B32)\tan^{-1}\left(\frac{-\frac{B}{2}}{A - B\frac{\sqrt{3}}{2}}\right)tan−1(A−B23​​−2B​​)
  2. (B)tan⁡−1(A0.7 B)\tan^{-1}\left(\frac{A}{0.7\,B}\right)tan−1(0.7BA​)
  3. (C)tan⁡−1(3B2A−B)\tan^{-1}\left(\frac{\sqrt{3}B}{2A - B}\right)tan−1(2A−B3​B​)
  4. (D)tan⁡−1(Bcos⁡θA−Bsin⁡θ)\tan^{-1}\left(\frac{B\cos\theta}{A - B\sin\theta}\right)tan−1(A−BsinθBcosθ​)

Correct answer: (C)

Step-by-step solution →
Q6·PhysicsSingle correct
A light beam is described by E=800 sin⁡ω(t−xc)E = 800\ \sin\omega\left(t - \frac{x}{c}\right)E=800 sinω(t−cx​) .An electron is allowed to move normal to the propagation of light beam with a speed of 3×1073 \times 10^{7}3×107 ms−1^{-1}−1. What is the maximum magnetic force exerted on the electron ?
  1. (A)1.28×10−181.28 \times 10^{-18}1.28×10−18 N
  2. (B)1.28×10−211.28 \times 10^{-21}1.28×10−21 N
  3. (C)12.8×10−1712.8 \times 10^{-17}12.8×10−17 N
  4. (D)12.8×10−1812.8 \times 10^{-18}12.8×10−18 N

Correct answer: (D)

Step-by-step solution →
Q7·PhysicsSingle correct
The two thin coaxial rings, each of radius 'a' and having charges +Q and –Q respectively are separated by a distance of 's'. The potential difference between the centres of the two rings is :
  1. (A)Q2πε0[1a+1s2+a2]\frac{Q}{2\pi\varepsilon_{0}}\left[\frac{1}{a} + \frac{1}{\sqrt{s^{2} + a^{2}}}\right]2πε0​Q​[a1​+s2+a2​1​]
  2. (B)Q4πε0[1a+1s2+a2]\frac{Q}{4\pi\varepsilon_{0}}\left[\frac{1}{a} + \frac{1}{\sqrt{s^{2} + a^{2}}}\right]4πε0​Q​[a1​+s2+a2​1​]
  3. (C)Q4πε0[1a−1s2+a2]\frac{Q}{4\pi\varepsilon_{0}}\left[\frac{1}{a} - \frac{1}{\sqrt{s^{2} + a^{2}}}\right]4πε0​Q​[a1​−s2+a2​1​]
  4. (D)Q2πε0[1a−1s2+a2]\frac{Q}{2\pi\varepsilon_{0}}\left[\frac{1}{a} - \frac{1}{\sqrt{s^{2} + a^{2}}}\right]2πε0​Q​[a1​−s2+a2​1​]

Correct answer: (D)

Step-by-step solution →
Q8·PhysicsSingle correct
If you are provided a set of resistances 2Ω, 4Ω, 6Ω and 8Ω. Connect these resistances so as to obtain an equivalent resistance of 463Ω\frac{46}{3}\Omega346​Ω.
  1. (A)4Ω and 6Ω are in parallel with 2Ω and 8 Ω in series
  2. (B)6Ω and 8Ω are in parallel with 2Ω and 4Ω in series
  3. (C)2Ω and 6Ω are in parallel with 4Ω and 8 Ω in series
  4. (D)2Ω and 4Ω are in parallel with 6Ω and 8Ω in series

Correct answer: (D)

Step-by-step solution →
Q9·PhysicsSingle correct
The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CD parallel to AB as shown in figure is 2.7 kg m2^{2}2.
  1. (A)14.9 kg / m3^{3}3
  2. (B)7.5×1017.5 \times 10^{1}7.5×101 kg / m3^{3}3
  3. (C)7.5×1027.5 \times 10^{2}7.5×102 kg/m3^{3}3
  4. (D)1.49×1021.49 \times 10^{2}1.49×102 kg / m3^{3}3

Correct answer: (D)

Step-by-step solution →
Q10·PhysicsSingle correct
A parallel - plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K1K_{1}K1​ and K2K_{2}K2​ of same area A/2 and thickness d/2 are inserted in the space between the plates. The capacitance of the capacitor will be given by :
  1. (A)ε0Ad(12+K1K2K1+K2)\frac{\varepsilon_{0}A}{d}\left(\frac{1}{2} + \frac{K_{1}K_{2}}{K_{1} + K_{2}}\right)dε0​A​(21​+K1​+K2​K1​K2​​)
  2. (B)ε0Ad(12+K1K22(K1+K2))\frac{\varepsilon_{0}A}{d}\left(\frac{1}{2} + \frac{K_{1}K_{2}}{2\left(K_{1} + K_{2}\right)}\right)dε0​A​(21​+2(K1​+K2​)K1​K2​​)
  3. (C)ε0Ad(12+K1+K2K1K2)\frac{\varepsilon_{0}A}{d}\left(\frac{1}{2} + \frac{K_{1} + K_{2}}{K_{1}K_{2}}\right)dε0​A​(21​+K1​K2​K1​+K2​​)
  4. (D)ε0Ad(12+2(K1+K2)K1K2)\frac{\varepsilon_{0}A}{d}\left(\frac{1}{2} + \frac{2\left(K_{1} + K_{2}\right)}{K_{1}K_{2}}\right)dε0​A​(21​+K1​K2​2(K1​+K2​)​)

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correct
A bomb is dropped by fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a :
  1. (A)hyperbola
  2. (B)parabola in the direction of motion of plane
  3. (C)straight line vertically down the plane
  4. (D)parabola in a direction opposite to the motion of plane

Correct answer: (C)

Step-by-step solution →
Q12·PhysicsSingle correct
At time t = 0, a material is composed of two radioactive atoms A and B, where NA(0)=2NB(0)N_{A}(0) = 2N_{B}(0)NA​(0)=2NB​(0). The decay constant of both kind of radioactive atoms is λ\lambdaλ. However, A disintegrates to B and B disintegrates to C. Which of the following figures represents the evolution of NB(t)N_{B}(t)NB​(t) / NB(0)N_{B}(0)NB​(0) with respect to time t ? [NA(0)=No. of A atoms at t=0NB(0)=No. of B atoms at t=0]\left[\begin{array}{l} N_{A}\left(0\right) = \text{No. of A atoms at } t = 0 \\\\ N_{B}\left(0\right) = \text{No. of B atoms at } t = 0 \end{array}\right]​NA​(0)=No. of A atoms at t=0NB​(0)=No. of B atoms at t=0​​
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q13·PhysicsSingle correct
A transmitting antenna at top of a tower has a height of 50 m and the height of receiving antenna is 80 m. What is range of communication for Line of Sight (LoS) mode ? [use radius of earth = 6400 km]
  1. (A)45.5 km
  2. (B)80.2 km
  3. (C)144.1 km
  4. (D)57.28 km

Correct answer: (D)

Step-by-step solution →
Q14·PhysicsSingle correct
A refrigerator consumes an average 35 W power to operate between temperature –10°C to 25°C. If there is no loss of energy then how much average heat per second does it transfer ?
  1. (A)263 J/s
  2. (B)298 J/s
  3. (C)350 J/s
  4. (D)35 J/s

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correct
An electric bulb of 500 watt at 100 volt is used in a circuit having a 200 V supply. Calculate the resistance R to be connected in series with the bulb so that the power delivered by the bulb is 500 W.
  1. (A)20 Ω
  2. (B)30 Ω
  3. (C)5 Ω
  4. (D)10 Ω

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsSingle correct
Four NOR gates are connected as shown in figure. The truth table for the given figure is :
  1. (A)A B | Y 0 0 | 1 0 1 | 0 1 0 | 1 1 1 | 0
  2. (B)A B | Y 0 0 | 0 0 1 | 1 1 0 | 1 1 1 | 0
  3. (C)A B | Y 0 0 | 0 0 1 | 1 1 0 | 0 1 1 | 1
  4. (D)A B | Y 0 0 | 1 0 1 | 0 1 0 | 0 1 1 | 1

Correct answer: (D)

Step-by-step solution →
Q17·PhysicsSingle correct
Match List–I with List–II. Choose the most appropriate answer from the options given below :
List-IList-II
a.Magnetic Inductioni.ML2T−2A−1ML^{2}T^{-2}A^{-1}ML2T−2A−1
b.Magnetic Fluxii.M0L−1AM^{0}L^{-1}AM0L−1A
c.Magnetic Permeabilityiii.MT−2A−1MT^{-2}A^{-1}MT−2A−1
d.Magnetizationiv.MLT−2A−2MLT^{-2}A^{-2}MLT−2A−2
  1. (A)(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
  2. (B)(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
  3. (C)(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
  4. (D)(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correct
If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is:
  1. (A)86.4 s
  2. (B)4.32 s
  3. (C)43.2 s
  4. (D)8.64 s

Correct answer: (C)

Step-by-step solution →
Q19·PhysicsSingle correct
Two blocks of masses 3 kg and 5 kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is 24π×102\frac{24}{\pi} \times 10^{2}π24​×102 Nm−2^{-2}−2. What is the minimum radius of the wire? (Take g = 10 ms−2^{-2}−2)
  1. (A)125 cm
  2. (B)1250 cm
  3. (C)12.5 cm
  4. (D)1.25 cm

Correct answer: (C)

Step-by-step solution →
Q20·PhysicsNumerical
Two waves are simultaneously passing through a string and their equations are : y1=A1sin⁡k(x−vt)y_1 = A_1 \sin k(x-vt)y1​=A1​sink(x−vt), y2=A2sin⁡k(x−vt+x0)y_2 = A_2 \sin k(x-vt + x_0)y2​=A2​sink(x−vt+x0​). Given amplitudes A1=12A_1 = 12A1​=12 mm and A2=5A_2 = 5A2​=5 mm, x0=3.5x_0 = 3.5x0​=3.5 cm and wave number k=6.28k = 6.28k=6.28 cm−1^{-1}−1. The amplitude of resulting wave will be .......... mm.

Correct answer: 7

Step-by-step solution →
Q21·PhysicsNumerical
A source of light is placed in front of a screen. Intensity of light on the screen is I. Two Polaroids P1P_1P1​ and P2P_2P2​ are so placed in between the source of light and screen that the intensity of light on screen is I/2. P2P_2P2​ should be rotated by an angle of ........... (degrees) so that the intensity of light on the screen becomes 3I8\frac{3I}{8}83I​.

Correct answer: 30

Step-by-step solution →
Q22·Physics·Magnetic Field of CurrentNumerical
If the maximum value of accelerating potential provided by a ratio frequency oscillator is 12 kV. The number of revolution made by a proton in a cyclotron to achieve one sixth of the speed of light is ............ [mp=1.67×10−27[m_p = 1.67 \times 10^{-27}[mp​=1.67×10−27 kg, e=1.6×10−19e = 1.6 \times 10^{-19}e=1.6×10−19 C, Speed of light =3×108= 3 \times 10^8=3×108 m/s]]]

Correct answer: 543

Step-by-step solution →
Q23·PhysicsNumerical
The acceleration due to gravity is found upto an accuracy of 4% on a planet. The energy supplied to a simple pendulum to known mass 'm' to undertake oscillations of time period T is being estimated. If time period is measured to an accuracy of 3%, the accuracy to which E is known as ..........%

Correct answer: 14

Step-by-step solution →
Q24·PhysicsNumerical
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s−1^{-1}−1 in a uniform horizontal magnetic field of 3.0×10−23.0 \times 10^{-2}3.0×10−2 T. The maximum emf induced the coil will be .......... ×10−2\times 10^{-2}×10−2 volt (rounded off to the nearest integer)

Correct answer: 60

Step-by-step solution →
Q25·PhysicsNumerical
Two simple harmonic motions are represented by the equations x1=5sin⁡(2πt+π4)x_1 = 5 \sin\left(2\pi t + \frac{\pi}{4}\right)x1​=5sin(2πt+4π​) and x2=52 (sin⁡2πt+cos⁡2πt)x_2 = 5\sqrt{2}\,(\sin 2\pi t + \cos 2\pi t)x2​=52​(sin2πt+cos2πt). The amplitude of second motion is ............. times the amplitude in first motion.

Correct answer: 2

Step-by-step solution →
Q26·Physics·Magnetic Field of CurrentNumerical
A coil in the shape of an equilateral triangle of side 10 cm lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field 20 mT. The torque acting on the coil when a current of 0.2 A is passed through it and its plane becomes parallel to the magnetic field will be x×10−5\sqrt{x} \times 10^{-5}x​×10−5 Nm. The value of x is...........

Correct answer: 3

Step-by-step solution →
Q27·PhysicsNumerical
For the given circuit, the power across zener diode is ............ mW.

Correct answer: 120

Step-by-step solution →
Q28·PhysicsNumerical
An object is placed at a distance of 12 cm from a convex lens. A convex mirror of focal length 15 cm is placed on other side of lens at 8 cm as shown in the figure. Image of object coincides with the object. When the convex mirror is removed, a real and inverted image is formed at a position. The distance of the image from the object will be ........(cm)

Correct answer: 50

Step-by-step solution →
Q29·PhysicsNumerical
The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is .......N. (take g=10g = 10g=10 ms−2^{-2}−2)

Correct answer: 15

Step-by-step solution →

Chemistry — JEE Main 26 August 2021 Shift 2

Q30·ChemistrySingle correct
Which one of the following phenols does not give colour when condensed with phthalic anhydride in presence of conc. H2SO4\mathrm{H_2SO_4}H2​SO4​ ?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q31·ChemistrySingle correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Photochemical smog causes cracking of rubber. Reason (R) : Presence of ozone, nitric oxide, acrolein, formaldehyde and peroxyacetyl nitrate in photochemical smog makes it oxidizing. Choose the most appropriate answer from the options given below :
  1. (A)Both (A) and (R) are true but (R) is not the true explanation of (A)
  2. (B)(A) is false but (R) is true.
  3. (C)(A) is true but (R) is false
  4. (D)Both (A) and (R) are true and (R) is the true explanation of (A)

Correct answer: (D)

Step-by-step solution →
Q32·ChemistrySingle correct
The interaction energy of London forces between two particles is proportional to rx\mathrm{r^x}rx, where r is the distance between the particles. The value of x is :
  1. (A)3
  2. (B)−3-3−3
  3. (C)−6-6−6
  4. (D)6

Correct answer: (C)

Step-by-step solution →
Q33·ChemistrySingle correct
The number of non-ionisable hydrogen atoms present in the final product obtained from the hydrolysis of PCl5\mathrm{PCl_5}PCl5​ is :
  1. (A)0
  2. (B)2
  3. (C)1
  4. (D)3

Correct answer: (A)

Step-by-step solution →
Q34·ChemistrySingle correct
The bond order and magnetic behaviour of O2−\mathrm{O_2^-}O2−​ ion are, respectively :
  1. (A)1.5 and paramagnetic
  2. (B)1.5 and diamagnetic
  3. (C)2 and diamagnetic
  4. (D)1 and paramagnetic

Correct answer: (A)

Step-by-step solution →
Q35·ChemistrySingle correct
Given below are two statements : one is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A) : Sucrose is a disaccharide and a non-reducing sugar. Reason (R) : Sucrose involves glycosidic linkage between C1\mathrm{C_1}C1​ of β\betaβ-glucose and C2\mathrm{C_2}C2​ of α\alphaα-fructose. Choose the most appropriate answer from the options given below :
  1. (A)Both (A) and (R) are true but (R) is not the true explanation of (A)
  2. (B)(A) is false but (R) is true.
  3. (C)(A) is true but (R) is false
  4. (D)Both (A) and (R) are true and (R) is the true explanation of (A)

Correct answer: (C)

Step-by-step solution →
Q36·ChemistrySingle correct
Match List-I with List-II : Choose the most appropriate match :
List-I (Chemical Reaction)List-II (Reagent used)
a.CH3COOCH2CH3→CH3CH2OH\mathrm{CH_3COOCH_2CH_3} \rightarrow \mathrm{CH_3CH_2OH}CH3​COOCH2​CH3​→CH3​CH2​OHi.CH3MgBr\mathrm{CH_3MgBr}CH3​MgBr / H3O+\mathrm{H_3O^+}H3​O+ (1.equivalent)
b.CH3COOCH3→CH3CHO\mathrm{CH_3COOCH_3} \rightarrow \mathrm{CH_3CHO}CH3​COOCH3​→CH3​CHOii.H2SO4\mathrm{H_2SO_4}H2​SO4​ / H2O\mathrm{H_2O}H2​O
c.CH3C≡N→CH3CHO\mathrm{CH_3C \equiv N} \rightarrow \mathrm{CH_3CHO}CH3​C≡N→CH3​CHOiii.DIBAL-H/H2O\mathrm{H_2O}H2​O
d.CH3C≡N→\mathrm{CH_3C \equiv N} \rightarrowCH3​C≡N→iv.SnCl2\mathrm{SnCl_2}SnCl2​, HCl/H2O\mathrm{HCl/H_2O}HCl/H2​O
  1. (A)a-ii, b-iv, c-iii, d-i
  2. (B)a-iv, b-ii, c-iii, d-i
  3. (C)a-ii, b-iii, c-iv, d-i
  4. (D)a-iii, b-ii, c-i, d-iv

Correct answer: (C)

Step-by-step solution →
Q37·ChemistrySingle correct
The major product in the above reaction is :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correct
Indicate the complex/complex ion which did not show any geometrical isomerism :
  1. (A)[CoCl2(en)2]\mathrm{[CoCl_2(en)_2]}[CoCl2​(en)2​]
  2. (B)[Co(CN)5(NC)]3−\mathrm{[Co(CN)_5(NC)]^{3-}}[Co(CN)5​(NC)]3−
  3. (C)[Co(NH3)3(NO2)3]\mathrm{[Co(NH_3)_3(NO_2)_3]}[Co(NH3​)3​(NO2​)3​]
  4. (D)[Co(NH3)4Cl2]+\mathrm{[Co(NH_3)_4Cl_2]^+}[Co(NH3​)4​Cl2​]+

Correct answer: (B)

Step-by-step solution →
Q39·ChemistrySingle correct
The sol given below with negatively charged colloidal particles is :
  1. (A)FeCl3\mathrm{FeCl_3}FeCl3​ added to hot water
  2. (B)KI added to AgNO3\mathrm{AgNO_3}AgNO3​ solution
  3. (C)AgNO3\mathrm{AgNO_3}AgNO3​ added to KI solution
  4. (D)Al2O3.xH2O\mathrm{Al_2O_3.xH_2O}Al2​O3​.xH2​O in water

Correct answer: (C)

Step-by-step solution →
Q40·ChemistrySingle correct
Given below are two statements : Statement I : Sphalerite is a sulphide ore of zinc and copper glance is a sulphide ore of copper. Statement II : It is possible to separate two sulphide ores by adjusting proportion of oil to water or by using 'depressants' in a froth flotation method. Choose the most appropriate answer from the options given below :
  1. (A)Statement I is true but Statement II is false.
  2. (B)Both Statement I and Statement II are true.
  3. (C)Statement I is false but Statement II is true.
  4. (D)Both Statement I and Statement II are false.

Correct answer: (B)

Step-by-step solution →
Q41·ChemistrySingle correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Heavy water is used for the study of reaction mechanism. Reason (R) : The rate of reaction for the cleavage of O - H bond is slower than that of O-D bond. Choose the most appropriate answer from the options given below :
  1. (A)Both (A) and (R) are true but (R) is not the true explanation of (A).
  2. (B)Both (A) and (R) are true and (R) is the true explanation of (A).
  3. (C)(A) is false but (R) is true.
  4. (D)(A) is true but (R) is false.

Correct answer: (D)

Step-by-step solution →
Q42·ChemistrySingle correct
Arrange the following Cobalt complexes in the order of increasing Crystal Field Stabilization Energy (CFSE) value. Complexes : [CoF6]3−\mathrm{[CoF_6]^{3-}}[CoF6​]3− (A), [Co(H2O)6]2+\mathrm{[Co(H_2O)_6]^{2+}}[Co(H2​O)6​]2+ (B), [Co(NH3)6]3+\mathrm{[Co(NH_3)_6]^{3+}}[Co(NH3​)6​]3+ (C) and [Co(en)3]3+\mathrm{[Co(en)_3]^{3+}}[Co(en)3​]3+ Choose the correct option :
  1. (A)A < B < C < D
  2. (B)B < A < C < D
  3. (C)B < C < D < A
  4. (D)C < D < B < A

Correct answer: (B)

Step-by-step solution →
Q43·ChemistrySingle correct
Chlordiazepoxide The class of drug to which chlordiazepoxide with above structure belongs is :
  1. (A)Antacid
  2. (B)Analgesic
  3. (C)Tranquilizer
  4. (D)Antibiotic

Correct answer: (C)

Step-by-step solution →
Q44·ChemistrySingle correct
Chalcogen group elements are :
  1. (A)Se, Tb and Pu.
  2. (B)Se, Te and Po.
  3. (C)S, Te and Pm.
  4. (D)O, Ti and Po.

Correct answer: (B)

Step-by-step solution →
Q45·Chemistry·IsomerismSingle correct
The number of stereoisomers possible for 1,2-dimethyl cyclopropane is :
  1. (A)One
  2. (B)Four
  3. (C)Two
  4. (D)Three

Correct answer: (D)

Step-by-step solution →
Q46·ChemistrySingle correct
Consider the given reaction, Identify 'X' and 'Y' :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q47·ChemistrySingle correct
Consider the given reaction, the product A is :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q48·ChemistryNumerical
In the sulphur estimation, 0.471 g of an organic compound gave 1.44 g of barium sulphate. The percentage of sulphur in the compound is ______%. (Nearest integer) (Atomic Mass of Ba = 137 u)

Correct answer: 42

Step-by-step solution →
Q49·ChemistryNumerical
The equilibrium constant KcK_cKc​ at 298 K for the reaction A+B⇌C+DA + B \rightleftharpoons C + DA+B⇌C+D is 100. Starting with an equimolar solution with concentrations of A, B, C and D all equal to 1M, the equilibrium concentration of D is ______ ×10−2\times 10^{-2}×10−2 M. (Nearest integer)

Correct answer: 182

Step-by-step solution →
Q50·ChemistryNumerical
For water ΔvapH=41\Delta_{vap} H = 41Δvap​H=41 kJ mol−1^{-1}−1 at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ______ kJ mol−1^{-1}−1 [Use : R = 8.3 J mol−1^{-1}−1 K−1^{-1}−1]

Correct answer: 38

Step-by-step solution →
Q51·ChemistryNumerical
A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is 4.3×10144.3 \times 10^{14}4.3×1014 Hz. The velocity of ejected electron is ______ ×105\times 10^{5}×105 ms−1^{-1}−1 (Nearest integer) [Use : h =6.63×10−34= 6.63 \times 10^{-34}=6.63×10−34 Js, me=9.0×10−31m_e = 9.0 \times 10^{-31}me​=9.0×10−31 kg]

Correct answer: 5

Step-by-step solution →
Q52·ChemistryNumerical
For the galvanic cell, Zn(s)+Cu2+(0.02 M)→Zn2+(0.04 M)+Cu(s)\mathrm{Zn(s)} + \mathrm{Cu}^{2+} (0.02\ \mathrm{M}) \rightarrow \mathrm{Zn}^{2+} (0.04\ \mathrm{M}) + \mathrm{Cu(s)}Zn(s)+Cu2+(0.02 M)→Zn2+(0.04 M)+Cu(s), Ecell=E_{cell} =Ecell​= ______ ×10−2\times 10^{-2}×10−2 V. (Nearest integer) [Use : ECu/Cu2+0=−0.34E^{0}_{\mathrm{Cu/Cu}^{2+}} = -0.34ECu/Cu2+0​=−0.34 V, EZn/Zn2+0=+0.76E^{0}_{\mathrm{Zn/Zn}^{2+}} = +0.76EZn/Zn2+0​=+0.76 V, 2.303 RTF=0.059\frac{2.303\ RT}{F} = 0.059F2.303 RT​=0.059 V]

Correct answer: 109

Step-by-step solution →
Q53·ChemistryNumerical
100 mL of Na3PO4\mathrm{Na_3PO_4}Na3​PO4​ solution contains 3.45 g of sodium. The molarity of the solution is ______×10−2\times 10^{-2}×10−2 mol L−1^{-1}−1 . (Nearest integer) [Atomic Masses - Na : 23.0 u, O : 16.0 u, P : 31.0 u]

Correct answer: 50

Step-by-step solution →
Q54·ChemistryNumerical
The overall stability constant of the complex ion [Cu(NH3)4]2+[\mathrm{Cu(NH_3)_4}]^{2+}[Cu(NH3​)4​]2+ is 2.1×10132.1 \times 10^{13}2.1×1013. The overall dissociations constant is y×10−14y \times 10^{-14}y×10−14. Then y is ______.(Nearest integer)

Correct answer: 5

Step-by-step solution →
Q55·ChemistryNumerical
83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is ________K. (Nearest integer) [Use : Molal Freezing point depression constant of water = 1.86 K kg mol−1^{-1}−1] Freezing Point of water = 273 K Atomic masses : C : 12.0 u, O : 16.0 u, H : 1.0 u]

Correct answer: 269

Step-by-step solution →
Q56·ChemistryNumerical
The reaction rate for the reaction [PtCl4]2−+H2O⇌[Pt(H2O)Cl3]−+Cl−[\mathrm{PtCl_4}]^{2-} + \mathrm{H_2O} \rightleftharpoons [\mathrm{Pt(H_2O)Cl_3}]^{-} + \mathrm{Cl}^{-}[PtCl4​]2−+H2​O⇌[Pt(H2​O)Cl3​]−+Cl− was measured as a function of concentrations of different species. It was observed that −d[[PtCl4]2−]dt=4.8×10−5[[PtCl4]2−]−2.4×10−3[[Pt(H2O)Cl3]−][Cl−]\frac{-d\left[[\mathrm{PtCl_4}]^{2-}\right]}{dt} = 4.8 \times 10^{-5} \left[[\mathrm{PtCl_4}]^{2-}\right] - 2.4 \times 10^{-3} \left[[\mathrm{Pt(H_2O)Cl_3}]^{-}\right]\left[\mathrm{Cl}^{-}\right]dt−d[[PtCl4​]2−]​=4.8×10−5[[PtCl4​]2−]−2.4×10−3[[Pt(H2​O)Cl3​]−][Cl−]. where square brackets are used to denote molar concentrations. The equilibrium constant Kc=K_c =Kc​= ______. (Nearest integer)

Correct answer: 0.02

Step-by-step solution →
Q57·ChemistryNumerical
A chloro compound "A". (i) forms aldehydes on ozonolysis followed by the hydrolysis. (ii) when vaporized completely 1.53 g of A, gives 448 mL of vapour at STP. The number of carbon atoms in a molecule of compound A is ________.

Correct answer: 3

Step-by-step solution →

Mathematics — JEE Main 26 August 2021 Shift 2

Q58·MathematicsSingle correct
Let [t] denote the greatest integer less than or equal to t. Let f(x)=x−[x]f(x) = x - [x]f(x)=x−[x], g(x)=1−x+[x]g(x) = 1 - x + [x]g(x)=1−x+[x], and h(x)=min⁡{f(x),g(x)}h(x) = \min\{f(x), g(x)\}h(x)=min{f(x),g(x)}, x∈[−2,2]x \in [-2, 2]x∈[−2,2]. Then h is :
  1. (A)continuous in [−2,2][-2, 2][−2,2] but not differentiable at more than four points in (−2,2)(-2, 2)(−2,2)
  2. (B)not continuous at exactly three points in [−2,2][-2, 2][−2,2]
  3. (C)continuous in [−2,2][-2, 2][−2,2] but not differentiable at exactly three points in (−2,2)(-2, 2)(−2,2)
  4. (D)not continuous at exactly four points in [−2,2][-2, 2][−2,2]

Correct answer: (A)

Step-by-step solution →
Q59·MathematicsSingle correct
Let A=(100011100)A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix}A=​101​010​010​​. Then A2025−A2020A^{2025} - A^{2020}A2025−A2020 is equal to :
  1. (A)A6−AA^{6} - AA6−A
  2. (B)A5A^{5}A5
  3. (C)A5−AA^{5} - AA5−A
  4. (D)A6A^{6}A6

Correct answer: (A)

Step-by-step solution →
Q60·MathematicsSingle correct
The local maximum value of the function f(x)=(2x)x2f(x) = \left(\frac{2}{x}\right)^{x^{2}}f(x)=(x2​)x2, x>0x > 0x>0, is
  1. (A)(2e)1e\left(2\sqrt{e}\right)^{\frac{1}{e}}(2e​)e1​
  2. (B)(4e)e4\left(\frac{4}{\sqrt{e}}\right)^{\frac{e}{4}}(e​4​)4e​
  3. (C)(e)2e\left(e\right)^{\frac{2}{e}}(e)e2​
  4. (D)111

Correct answer: (C)

Step-by-step solution →
Q61·MathematicsSingle correct
If the value of the integral ∫05x+[x]ex−[x]dx=αe−1+β\int_{0}^{5} \frac{x + [x]}{e^{x - [x]}} dx = \alpha e^{-1} + \beta∫05​ex−[x]x+[x]​dx=αe−1+β, where α,β∈R\alpha, \beta \in \mathbf{R}α,β∈R, 5α+6β=05\alpha + 6\beta = 05α+6β=0, and [x][x][x] denotes the greatest integer less than or equal to x; then the value of (α+β)2(\alpha + \beta)^{2}(α+β)2 is equal to :
  1. (A)100100100
  2. (B)252525
  3. (C)161616
  4. (D)363636

Correct answer: (B)

Step-by-step solution →
Q62·MathematicsSingle correct
The point P(−26,3)P\left(-2\sqrt{6}, \sqrt{3}\right)P(−26​,3​) lies on the hyperbola x2a2−y2b2=1\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1a2x2​−b2y2​=1 having eccentricity 52\frac{\sqrt{5}}{2}25​​. If the tangent and normal at P to the hyperbola intersect its conjugate axis at the point Q and R respectively, then QR is equal to :
  1. (A)434\sqrt{3}43​
  2. (B)666
  3. (C)636\sqrt{3}63​
  4. (D)363\sqrt{6}36​

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
Let y(x)y(x)y(x) be the solution of the differential equation 2x2dy+(ey−2x)dx=02x^{2}dy + (e^{y} - 2x)dx = 02x2dy+(ey−2x)dx=0, x>0x > 0x>0. If y(e)=1y(e) = 1y(e)=1, then y(1)y(1)y(1) is equal to :
  1. (A)000
  2. (B)222
  3. (C)log⁡e2\log_{e} 2loge​2
  4. (D)log⁡e(2e)\log_{e}(2e)loge​(2e)

Correct answer: (C)

Step-by-step solution →
Q64·MathematicsSingle correct
Consider the two statements : (S1):(p→q)∨(∼q→p)(S1) : (p \rightarrow q) \vee (\sim q \rightarrow p)(S1):(p→q)∨(∼q→p) is a tautology. (S2):(p∧∼q)∧(∼p∨q)(S2) : (p \wedge \sim q) \wedge (\sim p \vee q)(S2):(p∧∼q)∧(∼p∨q) is a fallacy. Then :
  1. (A)only (S1) is true.
  2. (B)both (S1) and (S2) are false.
  3. (C)both (S1) and (S2) are true.
  4. (D)only (S2) is true.

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
The domain of the function cosec−1(1+xx)\mathrm{cosec}^{-1}\left(\frac{1 + x}{x}\right)cosec−1(x1+x​) is :
  1. (A)(−1,−12]∪(0,∞)\left(-1, -\frac{1}{2}\right] \cup (0, \infty)(−1,−21​]∪(0,∞)
  2. (B)[−12,0)∪[1,∞)\left[-\frac{1}{2}, 0\right) \cup [1, \infty)[−21​,0)∪[1,∞)
  3. (C)(−12,∞)−{0}\left(-\frac{1}{2}, \infty\right) - \{0\}(−21​,∞)−{0}
  4. (D)[−12,∞)−{0}\left[-\frac{1}{2}, \infty\right) - \{0\}[−21​,∞)−{0}

Correct answer: (D)

Step-by-step solution →
Q66·MathematicsSingle correct
A fair die is tossed until six is obtained on it. Let X be the number of required tosses, then the conditional probability P(X≥5∣X>2)P\left(X \geq 5 \mid X > 2\right)P(X≥5∣X>2) is :
  1. (A)125216\frac{125}{216}216125​
  2. (B)1136\frac{11}{36}3611​
  3. (C)56\frac{5}{6}65​
  4. (D)2536\frac{25}{36}3625​

Correct answer: (D)

Step-by-step solution →
Q67·MathematicsSingle correct
If ∑r=150tan⁡−112r2=p\sum_{r=1}^{50} \tan^{-1} \frac{1}{2r^{2}} = p∑r=150​tan−12r21​=p, then the value of tan⁡p\tan ptanp is :
  1. (A)101102\frac{101}{102}102101​
  2. (B)5051\frac{50}{51}5150​
  3. (C)100100100
  4. (D)5150\frac{51}{50}5051​

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correct
Two fair dice are thrown. The numbers on them are taken as λ\lambdaλ and μ\muμ, and a system of linear equations x+y+z=5x + y + z = 5x+y+z=5 x+2y+3z=μx + 2y + 3z = \mux+2y+3z=μ x+3y+λz=1x + 3y + \lambda z = 1x+3y+λz=1 is constructed. If p is the probability that the system has a unique solution and q is the probability that the system has no solution, then :
  1. (A)p=16p = \frac{1}{6}p=61​ and q=136q = \frac{1}{36}q=361​
  2. (B)p=56p = \frac{5}{6}p=65​ and q=536q = \frac{5}{36}q=365​
  3. (C)p=56p = \frac{5}{6}p=65​ and q=136q = \frac{1}{36}q=361​
  4. (D)p=16p = \frac{1}{6}p=61​ and q=536q = \frac{5}{36}q=365​

Correct answer: (B)

Step-by-step solution →
Q69·MathematicsSingle correct
The locus of the mid points of the chords of the hyperbola x2−y2=4x^{2} - y^{2} = 4x2−y2=4, which touch the parabola y2=8xy^{2} = 8xy2=8x, is :
  1. (A)y3(x−2)=x2y^{3}(x - 2) = x^{2}y3(x−2)=x2
  2. (B)x3(x−2)=y2x^{3}(x - 2) = y^{2}x3(x−2)=y2
  3. (C)y2(x−2)=x3y^{2}(x - 2) = x^{3}y2(x−2)=x3
  4. (D)x2(x−2)=y3x^{2}(x - 2) = y^{3}x2(x−2)=y3

Correct answer: (C)

Step-by-step solution →
Q70·MathematicsSingle correct
The value of 2sin⁡(π8)sin⁡(2π8)sin⁡(3π8)sin⁡(5π8)sin⁡(6π8)sin⁡(7π8)2\sin\left(\frac{\pi}{8}\right)\sin\left(\frac{2\pi}{8}\right)\sin\left(\frac{3\pi}{8}\right)\sin\left(\frac{5\pi}{8}\right)\sin\left(\frac{6\pi}{8}\right)\sin\left(\frac{7\pi}{8}\right)2sin(8π​)sin(82π​)sin(83π​)sin(85π​)sin(86π​)sin(87π​) is :
  1. (A)142\frac{1}{4\sqrt{2}}42​1​
  2. (B)14\frac{1}{4}41​
  3. (C)18\frac{1}{8}81​
  4. (D)182\frac{1}{8\sqrt{2}}82​1​

Correct answer: (C)

Step-by-step solution →
Q71·MathematicsSingle correct
If (3+i)100=299(p+iq)\left(\sqrt{3} + i\right)^{100} = 2^{99}\left(p + iq\right)(3​+i)100=299(p+iq), then p and q are roots of the equation :
  1. (A)x2−(3−1)x−3=0x^{2} - \left(\sqrt{3} - 1\right)x - \sqrt{3} = 0x2−(3​−1)x−3​=0
  2. (B)x2+(3+1)x+3=0x^{2} + \left(\sqrt{3} + 1\right)x + \sqrt{3} = 0x2+(3​+1)x+3​=0
  3. (C)x2+(3−1)x−3=0x^{2} + \left(\sqrt{3} - 1\right)x - \sqrt{3} = 0x2+(3​−1)x−3​=0
  4. (D)x2−(3+1)x+3=0x^{2} - \left(\sqrt{3} + 1\right)x + \sqrt{3} = 0x2−(3​+1)x+3​=0

Correct answer: (A)

Step-by-step solution →
Q72·MathematicsSingle correct
A hall has a square floor of dimension 10m ×\times× 10m (see the figure) and vertical walls. If the angle GPH between the diagonals AG and BH is cos⁡−115\cos^{-1}\frac{1}{5}cos−151​, then the height of the hall (in meters) is :
  1. (A)555
  2. (B)2102\sqrt{10}210​
  3. (C)535\sqrt{3}53​
  4. (D)525\sqrt{2}52​

Correct answer: (D)

Step-by-step solution →
Q73·MathematicsSingle correct
Let P be the plane passing through the point (1,2,3) and the line of intersection of the planes r⃗⋅(i^+j^+4k^)=16\vec{r} \cdot \left(\hat{i} + \hat{j} + 4\hat{k}\right) = 16r⋅(i^+j^​+4k^)=16 and r⃗⋅(−i^+j^+k^)=6\vec{r} \cdot \left(-\hat{i} + \hat{j} + \hat{k}\right) = 6r⋅(−i^+j^​+k^)=6. Then which of the following points does NOT lie on P ?
  1. (A)(3,3,2)(3, 3, 2)(3,3,2)
  2. (B)(6,−6,2)(6, -6, 2)(6,−6,2)
  3. (C)(4,2,2)(4, 2, 2)(4,2,2)
  4. (D)(−8,8,6)(-8, 8, 6)(−8,8,6)

Correct answer: (C)

Step-by-step solution →
Q74·MathematicsSingle correct
A 10 inches long pencil AB with mid point C and a small eraser P are placed on the horizontal top of a table such that PC=5PC = \sqrt{5}PC=5​ inches and ∠PCB=tan⁡−1(2)\angle PCB = \tan^{-1}(2)∠PCB=tan−1(2). The acute angle through which the pencil must be rotated about C so that the perpendicular distance between eraser and pencil becomes exactly 1 inch is :
  1. (A)tan⁡−1(34)\tan^{-1}\left(\frac{3}{4}\right)tan−1(43​)
  2. (B)tan⁡−1(1)\tan^{-1}(1)tan−1(1)
  3. (C)tan⁡−1(43)\tan^{-1}\left(\frac{4}{3}\right)tan−1(34​)
  4. (D)tan⁡−1(12)\tan^{-1}\left(\frac{1}{2}\right)tan−1(21​)

Correct answer: (A)

Step-by-step solution →
Q75·MathematicsSingle correct
The value of ∫−π2π2(1+sin⁡2x1+πsin⁡x)dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{1 + \sin^{2} x}{1 + \pi^{\sin x}}\right) dx∫−2π​2π​​(1+πsinx1+sin2x​)dx is
  1. (A)π2\frac{\pi}{2}2π​
  2. (B)5π4\frac{5\pi}{4}45π​
  3. (C)3π4\frac{3\pi}{4}43π​
  4. (D)3π2\frac{3\pi}{2}23π​

Correct answer: (C)

Step-by-step solution →
Q76·MathematicsSingle correct
A circle C touches the line x=2yx = 2yx=2y at the point (2,1)(2,1)(2,1) and intersects the circle C1:x2+y2+2y−5=0C_{1} : x^{2} + y^{2} + 2y - 5 = 0C1​:x2+y2+2y−5=0 at two points P and Q such that PQ is a diameter of C1C_{1}C1​. Then the diameter of C is :
  1. (A)757\sqrt{5}75​
  2. (B)151515
  3. (C)285\sqrt{285}285​
  4. (D)4154\sqrt{15}415​

Correct answer: (A)

Step-by-step solution →
Q77·MathematicsSingle correct
lim⁡x→2(∑n=19xn(n+1)x2+2(2n+1)x+4)\lim_{x \to 2}\left(\sum_{n=1}^{9} \frac{x}{n(n + 1)x^{2} + 2(2n + 1)x + 4}\right)limx→2​(∑n=19​n(n+1)x2+2(2n+1)x+4x​) is equal to :
  1. (A)944\frac{9}{44}449​
  2. (B)524\frac{5}{24}245​
  3. (C)15\frac{1}{5}51​
  4. (D)736\frac{7}{36}367​

Correct answer: (A)

Step-by-step solution →
Q78·MathematicsNumerical
The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit "1" and they all are multiple of 11, is ________.

Correct answer: 7744

Step-by-step solution →
Q79·MathematicsNumerical
Let a and b respectively be the points of local maximum and local minimum of the function f(x)=2x3−3x2−12xf(x) = 2x^{3} - 3x^{2} - 12xf(x)=2x3−3x2−12x. If A is the total area of the region bounded by y=f(x)y = f(x)y=f(x), the x-axis and the lines x=ax = ax=a and x=bx = bx=b, then 4A is equal to ________.

Correct answer: 114

Step-by-step solution →
Q80·MathematicsNumerical
If the projection of the vector i^+2j^+k^\hat{i} + 2\hat{j} + \hat{k}i^+2j^​+k^ on the sum of the two vectors 2i^+4j^−5k^2\hat{i} + 4\hat{j} - 5\hat{k}2i^+4j^​−5k^ and −λi^+2j^+3k^-\lambda\hat{i} + 2\hat{j} + 3\hat{k}−λi^+2j^​+3k^ is 1, then λ\lambdaλ is equal to ________.

Correct answer: 5

Step-by-step solution →
Q81·MathematicsNumerical
Let a1,a2,.....,a10a_{1}, a_{2}, ....., a_{10}a1​,a2​,.....,a10​ be an AP with common difference −3-3−3 and b1,b2,.....,b10b_{1}, b_{2}, ....., b_{10}b1​,b2​,.....,b10​ be a GP with common ratio 2. Let ck=ak+bkc_{k} = a_{k} + b_{k}ck​=ak​+bk​, k=1,2,...,10k = 1, 2, ..., 10k=1,2,...,10. If c2=12c_{2} = 12c2​=12 and c3=13c_{3} = 13c3​=13, then ∑k=110ck\sum_{k=1}^{10} c_{k}∑k=110​ck​ is equal to ________.

Correct answer: 2021

Step-by-step solution →
Q82·MathematicsNumerical
Let Q be the foot of the perpendicular from the point P(7,−2,13)P(7, -2, 13)P(7,−2,13) on the plane containing the lines x+16=y−17=z−38\frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}6x+1​=7y−1​=8z−3​ and x−13=y−25=z−37\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}3x−1​=5y−2​=7z−3​. Then (PQ)2(PQ)^{2}(PQ)2, is equal to ________.

Correct answer: 96

Step-by-step solution →
Q83·MathematicsNumerical
Let (nk)\begin{pmatrix} n \\ k \end{pmatrix}(nk​) denotes nCk^{n}C_{k}nCk​ and [nk]={(nk),if 0≤k≤n0,otherwise\begin{bmatrix} n \\ k \end{bmatrix} = \begin{cases} \begin{pmatrix} n \\ k \end{pmatrix}, & \text{if } 0 \le k \le n \\ 0, & \text{otherwise} \end{cases}[nk​]=⎩⎨⎧​(nk​),0,​if 0≤k≤notherwise​ If Ak=∑i=09(9i)[1212−k+i]+∑i=08(8i)[1313−k+i]A_{k} = \sum_{i=0}^{9} \begin{pmatrix} 9 \\ i \end{pmatrix} \begin{bmatrix} 12 \\ 12-k+i \end{bmatrix} + \sum_{i=0}^{8} \begin{pmatrix} 8 \\ i \end{pmatrix} \begin{bmatrix} 13 \\ 13-k+i \end{bmatrix}Ak​=∑i=09​(9i​)[1212−k+i​]+∑i=08​(8i​)[1313−k+i​] and A4−A3=190 pA_{4} - A_{3} = 190\,pA4​−A3​=190p, then p is equal to :

Correct answer: 49

Step-by-step solution →
Q84·MathematicsNumerical
Let λ≠0\lambda \ne 0λ=0 be in R\mathbf{R}R. If α\alphaα and β\betaβ are the roots of the equation x2−x+2λ=0x^{2} - x + 2\lambda = 0x2−x+2λ=0, and α\alphaα and γ\gammaγ are the roots of equation 3x2−10x+27λ=03x^{2} - 10x + 27\lambda = 03x2−10x+27λ=0, then βγλ\frac{\beta\gamma}{\lambda}λβγ​ is equal to ________.

Correct answer: 18

Step-by-step solution →
Q85·MathematicsNumerical
Let the mean and variance of four numbers 3, 7, x and y(x>y)y (x > y)y(x>y) be 5 and 10 respectively. Then the mean of four numbers 3+2x3 + 2x3+2x, 7+2y7 + 2y7+2y, x+yx + yx+y and x−yx - yx−y is ________.

Correct answer: 12

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Q86·MathematicsNumerical
Let A be a 3×33 \times 33×3 real matrix. If det⁡(2Adj(2 Adj(Adj(2A))))=241\det(2\mathrm{Adj}(2\ \mathrm{Adj}(\mathrm{Adj}(2A)))) = 2^{41}det(2Adj(2 Adj(Adj(2A))))=241, then the value of det⁡(A2)\det(A^{2})det(A2) equal ________.

Correct answer: 4

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Q87·MathematicsNumerical
The least positive integer n such that (2i)n(1−i)n−2,i=−1\frac{(2i)^{n}}{(1-i)^{n-2}}, i = \sqrt{-1}(1−i)n−2(2i)n​,i=−1​ is a positive integer, is ________.

Correct answer: 6

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Chapters tested in this paper

Open any chapter to practise every past-year question on that topic across all papers, and see how often it has actually appeared.

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Coordination Compounds 176/186
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Kinematics 156/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Equilibrium 163/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Kinetic Theory of Gases 135/186
  • Alcohols and Ethers 106/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Oscillations 117/186
  • Nuclei 116/186
  • Straight Lines 114/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Statistics 118/186
  • Isolation of Metals 106/186
  • Differentiability 91/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Environmental Chemistry 83/186
  • Hydrogen 81/186
  • Hyperbola 77/186
  • Electric Potential 63/186
  • Chemistry in Everyday Life 60/186
  • States of Matter: Gases and Liquids 52/186
  • Isomerism 51/186
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