Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Main PYQs
  3. /2021
  4. /26 Feb Shift 2

JEE Main 26 February 2021 Shift 2 Question Paper with Answers

26 February 2021 · February session · 90 questions

The complete JEE Main 26 February 2021 Shift 2 paper — all 90 questions with the correct answer for each, tagged to the chapter it tests. Free to read, no account needed.

Physics
30
Chemistry
30
Mathematics
30

Physics — JEE Main 26 February 2021 Shift 2

Q1·PhysicsSingle correct
A tuning fork A of unknown frequency produces 5beats/s with a fork of known frequency 340 HZ. When fork A filed, the beat frequency decreases to 2beats/s. What is the frequency of fork A?
  1. (A)342 Hz
  2. (B)335 Hz
  3. (C)338 Hz
  4. (D)345 Hz

Correct answer: (B)

Step-by-step solution →
Q2·PhysicsSingle correct
The trajectory a projectile in a vertical plane is y = αx − βx2^{2}2, where α and β are constants and x & y are respectively the horizontal and vertical distance of the projectile from the point of projection. The angle of projection θ and the maximum height attained H are respectively given by:
  1. (A)tan⁡−1α, α24β\tan^{-1}\alpha,\ \frac{\alpha^{2}}{4\beta}tan−1α, 4βα2​
  2. (B)tan⁡−1β, α22β\tan^{-1}\beta,\ \frac{\alpha^{2}}{2\beta}tan−1β, 2βα2​
  3. (C)tan⁡−1(βα), α2β\tan^{-1}\left(\frac{\beta}{\alpha}\right),\ \frac{\alpha^{2}}{\beta}tan−1(αβ​), βα2​
  4. (D)tan⁡−1α, 4α2β\tan^{-1}\alpha,\ \frac{4\alpha^{2}}{\beta}tan−1α, β4α2​

Correct answer: (A)

Step-by-step solution →
Q3·PhysicsSingle correct
A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance ‘h’, the square of angular velocity of wheel will be:
  1. (A)2ghI+mr2\frac{2gh}{I+mr^{2}}I+mr22gh​
  2. (B)2gh
  3. (C)2mghI+2mr2\frac{2mgh}{I+2mr^{2}}I+2mr22mgh​
  4. (D)2mghI+mr2\frac{2mgh}{I+mr^{2}}I+mr22mgh​

Correct answer: (D)

Step-by-step solution →
Q4·PhysicsSingle correct
Find the peak current and resonant frequency of the following circuit (as shown in figure)
  1. (A)0.2 A and 100 Hz
  2. (B)2 A and 50 Hz
  3. (C)2 A and 100 Hz
  4. (D)0.2 A and 50 Hz

Correct answer: (D)

Step-by-step solution →
Q5·PhysicsSingle correct
The incident ray, reflected ray and the outward drawn normal are denoted by the unit vectors a⃗,b⃗\vec{a}, \vec{b}a,b and c⃗\vec{c}c respectively. Then choose the correct relation for these vectors.
  1. (A)b⃗=2a⃗+c⃗\vec{b} = 2\vec{a} + \vec{c}b=2a+c
  2. (B)b⃗=a⃗−c⃗\vec{b} = \vec{a} - \vec{c}b=a−c
  3. (C)b⃗=a⃗+2c⃗\vec{b} = \vec{a} + 2\vec{c}b=a+2c
  4. (D)b⃗=a⃗−2(a⃗⋅c⃗)c⃗\vec{b} = \vec{a} - 2\left(\vec{a}\cdot\vec{c}\right)\vec{c}b=a−2(a⋅c)c

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsSingle correct
A radioactive sample is undergoing α decay. At any time t1_{1}1​, its activity is A and another time t2_{2}2​, the activity is A5\frac{A}{5}5A​. What is the average life time for the sample?
  1. (A)t2−t1ln⁡5\frac{t_{2}-t_{1}}{\ln 5}ln5t2​−t1​​
  2. (B)ln⁡(t2+t1)2\frac{\ln\left(t_{2}+t_{1}\right)}{2}2ln(t2​+t1​)​
  3. (C)t1−t2ln⁡5\frac{t_{1}-t_{2}}{\ln 5}ln5t1​−t2​​
  4. (D)ln⁡5t2−t1\frac{\ln 5}{t_{2}-t_{1}}t2​−t1​ln5​

Correct answer: (A)

Step-by-step solution →
Q7·PhysicsSingle correct
A particle executes S.H.M., the graph of velocity as a function of displacement is:
  1. (A)a circle
  2. (B)a parabola
  3. (C)an ellipse
  4. (D)a helix

Correct answer: (C)

Step-by-step solution →
Q8·PhysicsSingle correct
A scooter accelerates from rest for time t1_{1}1​ at constant rate a1_{1}1​ and then retards at constant rate a2_{2}2​ for time t2_{2}2​ and comes to rest. The correct value of t1t2\frac{t_{1}}{t_{2}}t2​t1​​ will be:
  1. (A)a1+a2a2\frac{a_{1}+a_{2}}{a_{2}}a2​a1​+a2​​
  2. (B)a2a1\frac{a_{2}}{a_{1}}a1​a2​​
  3. (C)a1+a2a1\frac{a_{1}+a_{2}}{a_{1}}a1​a1​+a2​​
  4. (D)a1a2\frac{a_{1}}{a_{2}}a2​a1​​

Correct answer: (B)

Step-by-step solution →
Q9·PhysicsSingle correct
Draw the output Y in the given combination of gates.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correct
An inclined plane making an angle of 30° with horizontal is placed in a uniform horizontal electric field 200NC200\frac{N}{C}200CN​ as shown in the figure. A body of mass 1 kg and charge 5mC is allowed to slide down from rest at a height of 1m. If the coefficient of frication is 0.2, find the time taken by the body to reach the bottom. [g=9.8 m/s2, sin⁡30∘=12; cos⁡30∘=32]\left[g = 9.8\,m/s^{2},\ \sin 30^{\circ} = \frac{1}{2};\ \cos 30^{\circ} = \frac{\sqrt{3}}{2}\right][g=9.8m/s2, sin30∘=21​; cos30∘=23​​]
  1. (A)2.3 s
  2. (B)0.46 s
  3. (C)1.3 s
  4. (D)0.92 s

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correct
If ‘C’ and ‘V’ represent capacity and voltage respectively then what are the dimensions of λ where C/V = λ?
  1. (A)[M−2L−4I3T7]\left[M^{-2}L^{-4}I^{3}T^{7}\right][M−2L−4I3T7]
  2. (B)[M−2L−3I2T6]\left[M^{-2}L^{-3}I^{2}T^{6}\right][M−2L−3I2T6]
  3. (C)[M−1L−3I−2T−7]\left[M^{-1}L^{-3}I^{-2}T^{-7}\right][M−1L−3I−2T−7]
  4. (D)[M−3L−4I3T7]\left[M^{-3}L^{-4}I^{3}T^{7}\right][M−3L−4I3T7]

Correct answer: (A)

Step-by-step solution →
Q12·PhysicsSingle correct
Given below are two statements: One is labeled as Assertion A and the other is labeled as Reason R. Assertion A : For a simple microscope, the angular size of the object equals the angular size of the image. Reason R : Magnification is achieved as the small object can be kept much closer to the eye than 25 cm and hence it subtends a large angle. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both A and R are true but R is NOT the correct explanationof A
  2. (B)Both A and R are true and R is the correct explanation of A
  3. (C)A is true but R is false
  4. (D)A is false but R is true

Correct answer: (B)

Step-by-step solution →
Q13·PhysicsSingle correct
The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be:
  1. (A)4.17 m/s
  2. (B)4.34 m/s
  3. (C)219 m/s
  4. (D)3.25 m/s

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsSingle correct
Two masses A and B, each of mass M are fixed together by a massless springs. A force acts on the mass B as shown in figure. If the mass A starts moving away from mass B with acceleration ‘a’, than the acceleration of mass B will be:
  1. (A)F+MaM\frac{F+Ma}{M}MF+Ma​
  2. (B)F−MaM\frac{F-Ma}{M}MF−Ma​
  3. (C)Ma−FM\frac{Ma-F}{M}MMa−F​
  4. (D)MFF+Ma\frac{MF}{F+Ma}F+MaMF​

Correct answer: (B)

Step-by-step solution →
Q15·PhysicsSingle correct
A wire of 1Ω has a length of 1 m. It is stretched till its length increases by 25%. The percentage change in a resistance to the nearest integer is:
  1. (A)25%
  2. (B)12.5%
  3. (C)76%
  4. (D)56%

Correct answer: (D)

Step-by-step solution →
Q16·PhysicsSingle correct
Given below are two statements : Statement (1) :- A second's pendulum has a time period of 1 second. Statement (2) :- It takes precisely one second to move between the two extreme positions. In the light of the above statements, choose the correct answer from the options give below.
  1. (A)Both Statement I and Statement II are false
  2. (B)Statement I is true but Statement II is false
  3. (C)Statement I is false but Statement II is true
  4. (D)Both Statement I and Statement II is true

Correct answer: (C)

Step-by-step solution →
Q17·PhysicsSingle correct
An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5×10−42.5 \times 10^{-4}2.5×10−4 Wb/m2^22 and the angle of dip is 60°. The emf induced between the tips of the plane wings will be __________.
  1. (A)88.37 mV
  2. (B)62.50 mV
  3. (C)54.125 mV
  4. (D)108.25 mV

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correct
The length of metallic wire is l1l_1l1​ when tension in it is T1T_1T1​. It is l2l_2l2​ when the tension is T2T_2T2​. The original length of the wire will be :
  1. (A)l1+l22\frac{l_1 + l_2}{2}2l1​+l2​​
  2. (B)T1l1−T2l2T2−T1\frac{T_1 l_1 - T_2 l_2}{T_2 - T_1}T2​−T1​T1​l1​−T2​l2​​
  3. (C)T2l1+T1l2T1+T2\frac{T_2 l_1 + T_1 l_2}{T_1 + T_2}T1​+T2​T2​l1​+T1​l2​​
  4. (D)T2l1−T1l2T2−T1\frac{T_2 l_1 - T_1 l_2}{T_2 - T_1}T2​−T1​T2​l1​−T1​l2​​

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsSingle correct
The internal energy (U), pressure (P) and volume (V) of an ideal gas are related as U = 3PV + 4. The gas is :
  1. (A)polyatomic only
  2. (B)monoatomic only
  3. (C)either monoatomic or diatomic
  4. (D)diatomic only.

Correct answer: (A)

Step-by-step solution →
Q20·PhysicsSingle correct
Given below are two statements : Statement – I : An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere. Statement – II : If R is the radius of a solid metallic sphere and Q be the total charge on it. The electric field at any point on the spherical surface of radius r ( < R) is zero but theelectric flux passing through this closed spherical surface of radius r is not zero. In the light of the above statements. Choose the correct answerfrom the option given below :
  1. (A)Statement I is true but Statement II is false
  2. (B)Statement I is false but Statement II is true
  3. (C)Both Statement I and Statement II are true
  4. (D)Both Statement I and Statement II are false

Correct answer: (A)

Step-by-step solution →
Q21·PhysicsNumerical
If thehighest frequency modulating a carrier is 5 kHz, then the number of AM broadcast stations accommodated in a 90 kHz bandwidth are ____________________.

Correct answer: 9

Step-by-step solution →
Q22·PhysicsNumerical
1 mole of rigid diatomic gas performs a work of Q5\frac{Q}{5}5Q​ when heat Q is supplied to it. The molar heat capacity of the gas during this transformation is xR8\frac{xR}{8}8xR​. The value of x is __________.

Correct answer: 25

Step-by-step solution →
Q23·PhysicsNumerical
A particle excutes S.H.M with amplitude 'a' and time period T. The displacement of the particle when its speed is half of maximum speed is x a2\frac{\sqrt{x}\,a}{2}2x​a​. The value of x is ______________

Correct answer: 3

Step-by-step solution →
Q24·PhysicsNumerical
Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x : y. The value of x is ________.

Correct answer: 1

Step-by-step solution →
Q25·PhysicsNumerical
A point source of light S, placed at a distance 60 cm infront of the centre of plane mirror of width 50 cm, hangs vertically on a wall. A man walks infront of the mirror along a line parallel to the mirror at a distance 1.2 m from it (see in the figure). The distance between the extreme points where he can see the image of the light source in the mirror is _______________cm

Correct answer: 150

Step-by-step solution →
Q26·PhysicsNumerical
The zener diode has a VZV_ZVZ​= 30 V. The current passing through the diode for the following ciruit is ________________mA.

Correct answer: 9

Step-by-step solution →
Q27·PhysicsNumerical
In the reported figure of earth, the value of acceleration due to gravity is same at point A and C but it is smaller than that of its value at point B (surface of the earth). The value of OA : AB will be x : y. The value of x is ________________.

Correct answer: 4

Step-by-step solution →
Q28·PhysicsNumerical
27 similar drops of mercury are maintained at 10 V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ___________________ times that of a smaller drop.

Correct answer: 243

Step-by-step solution →
Q29·PhysicsNumerical
The volume V of a given mass of monatomic gas changes with temperature T according to the relation V=KT23V = KT^{\frac{2}{3}}V=KT32​. The work done when temperature changes by 90 K will be xR. The value of x is ____________________. [R =universal gas constant]

Correct answer: 60

Step-by-step solution →
Q30·PhysicsNumerical
Time period of a simple pendulum is T. The time taken to complete 58\frac{5}{8}85​ oscillations starting from mean position is αβ\frac{\alpha}{\beta}βα​T. The value of α\alphaα is _____________.

Correct answer: 7

Step-by-step solution →

Chemistry — JEE Main 26 February 2021 Shift 2

Q31·ChemistrySingle correct
2,4-DNP test can be used to identify:
  1. (A)aldehyde
  2. (B)halogens
  3. (C)ether
  4. (D)amine

Correct answer: (A)

Step-by-step solution →
Q32·ChemistrySingle correct
Identify A in the following chemical reaction.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q33·ChemistrySingle correct
The nature of charge on resulting colloidal particles when FeCl3_33​ is added to excess of hot water is:
  1. (A)positive
  2. (B)neutral
  3. (C)sometimes positive and sometimes negative
  4. (D)negative

Correct answer: (A)

Step-by-step solution →
Q34·ChemistrySingle correct
Match List-I with List-II List-I (c) 2CH3_33​CH2_22​Cl + 2Na →Ether\xrightarrow{\text{Ether}}Ether​ C2_22​H5_55​– C2_22​H5_55​ + 2NaCl (d) 2C2_22​H5_55​Cl +2Na →Ether\xrightarrow{\text{Ether}}Ether​ C6_66​H5_55​– C6_66​H5_55​ + 2NaCl List-II (i) Wurtz reaction (ii) Sandmeyer reaction (iii) Fitting reaction (iv) Gatterman reaction Choose the correct answer from the option given below:
  1. (A)(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
  2. (B)(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
  3. (C)(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
  4. (D)(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

Correct answer: (C)

Step-by-step solution →
Q35·ChemistrySingle correct
In CH2_22​ = C = CH–CH3_33​ molecule, the hybridization of carbon 1, 2, 3 and 4 respectively are:
  1. (A)sp2^22, sp, sp2^22, sp3^33
  2. (B)sp2^22, sp2^22, sp2^22, sp3^33
  3. (C)sp2^22, sp3^33, sp2^22, sp3^33
  4. (D)sp3^33, sp, sp3^33, sp3^33

Correct answer: (A)

Step-by-step solution →
Q36·ChemistrySingle correct
Match List-I with List-II. List-I (a) Sucrose (b) Lactose (c) Maltose List-II (i) β-D-Galactose and β-D-Glucose (ii) α-D-Glucose and β-D-Fructose (iii) α-D- Glucose and α-D-Glucose Choose the correct answer from the options given below:
  1. (A)(a)-(iii), (b)-(ii), (c)-(i)
  2. (B)(a)-(iii), (b)-(i), (c)-(ii)
  3. (C)(a)-(i), (b)-(iii), (c)-(ii)
  4. (D)(a)-(ii), (b)-(i), (c)-(iii)

Correct answer: (D)

Step-by-step solution →
Q37·ChemistrySingle correct
Which pair of oxides is acidic in nature?
  1. (A)N2_22​O, BaO
  2. (B)CaO, SiO2_22​
  3. (C)B2_22​O3_33​, CaO
  4. (D)B2_22​O3_33​, SiO2_22​

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correct
Calgon is used for water treatment. Which of the following statement is NOT true about calgon?
  1. (A)Calgon contains the 2nd^{nd}nd most abundant element by weight in the earth's crust.
  2. (B)It is also known as Graham's salt.
  3. (C)It is polymeric compound and is water soluble.
  4. (D)It doesnot remove Ca2+^{2+}2+ ion by precipitation.

Correct answer: (A)

Step-by-step solution →
Q39·ChemistrySingle correct
Ceric ammonium nitrate and CHCl3_33​/alc. KOH are used for the identification of functional groups present in __________and_________respectively.
  1. (A)alcohol, amine
  2. (B)amine, alcohol
  3. (C)alcohol, phenol
  4. (D)amine, phenol

Correct answer: (A)

Step-by-step solution →
Q40·ChemistrySingle correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: In TlI3_33​, isomorphous to CsI3_33​, the metal is present in +1 oxidation state. Reason R: Tl metals has fourteen f electrons in its electronic configuration. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both A and R are correct and R is the correct explanation of A
  2. (B)A is not correct but R is correct
  3. (C)Both A and R are correct R is NOT the correct explanation of A
  4. (D)A is correct but R is not correct

Correct answer: (C)

Step-by-step solution →
Q41·ChemistrySingle correct
The correct order of electron gain enthalpy is:
  1. (A)S > Se > Te > O
  2. (B)O > S > Se > Te
  3. (C)S > O > Se > Te
  4. (D)Te > Se > S > O

Correct answer: (A)

Step-by-step solution →
Q42·ChemistrySingle correct
Identify A in the given chemical reaction.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q43·ChemistrySingle correct
Match List-I with List-II List-I (a) Siderite (b) Calamine (c) Malachite (d) Cryolite List-II (i) Cu (ii) Ca (iii) Fe (iv) Al (v) Zn Choose the correct answer from the options given below:
  1. (A)(a)-(i), (b)-(ii), (c)-(v), (d)-(iii)
  2. (B)(a)-(iii), (b)-(v), (c)-(i), (d)-(iv)
  3. (C)(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
  4. (D)(a)-(iii), (b)-(i), (c)-(v), (d)-(ii)

Correct answer: (B)

Step-by-step solution →
Q44·ChemistrySingle correct
Identify A in the given reaction
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q45·ChemistrySingle correct
Match List-I with List-II. List-I (a) Sodium Carbonate (b) Titanium (c) Chlorine (d) Sodium hydroxide List-II (i) Deacon (ii) Caster-Kellner (iii) Van-Arkel (iv) Solvay Choose the correct answer from the option given below:
  1. (A)(a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
  2. (B)(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
  3. (C)(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
  4. (D)(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)

Correct answer: (B)

Step-by-step solution →
Q46·ChemistrySingle correct
Match List-I with List-II. List-I (Molecule) List-II (Bond order) (a) Ne2Ne_2Ne2​ (i) 1 (b) N2N_2N2​ (ii) 2 (c) F2F_2F2​ (iii) 0 (d) O2O_2O2​ (iv) 3 Choose the correct answer from the options given below:
  1. (A)(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
  2. (B)(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
  3. (C)(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
  4. (D)(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

Correct answer: (A)

Step-by-step solution →
Q47·ChemistrySingle correct
Which of the following forms of hydrogen emits low energy β−\beta^-β− particles?
  1. (A)Proton H+H^+H+
  2. (B)Deuterium 12H_1^2H12​H
  3. (C)Protium 11H_1^1H11​H
  4. (D)Tritium 13H_1^3H13​H

Correct answer: (D)

Step-by-step solution →
Q48·ChemistrySingle correct
A. Phenyl methanamine B. N, N-Dimethylaniline C. N-Methyl aniline D. Benzenamine Choose the correct order of basic nature of the above amines.
  1. (A)D > C > B > A
  2. (B)D > B > C > A
  3. (C)A > C > B > D
  4. (D)A > B > C > D

Correct answer: (D)

Step-by-step solution →
Q49·ChemistrySingle correct
Considering the above reaction, the major product among the following is:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q50·ChemistrySingle correct
Seliwanoff test and Xanthoproteic test are used for the identification of ___________ and ____________ respectively
  1. (A)ketoses, proteins
  2. (B)proteins, ketoses
  3. (C)aldoses, ketoses
  4. (D)ketoses, aldoses

Correct answer: (A)

Step-by-step solution →
Q51·ChemistryNumerical
The NaNO3NaNO_3NaNO3​ weighed out to make 50 mL of an aqueous solution containing 70.0 mg Na+Na^+Na+ per mL is_________g. (Rounded off to the nearest integer) [Given: Atomic weight in g mol−1mol^{-1}mol−1. Na: 23; N: 14; O : 16]

Correct answer: 13

Step-by-step solution →
Q52·ChemistryNumerical
The number of stereoisomers possible for [Co(ox)2(Br)(NH3)]2−[Co(ox)_2(Br)(NH_3)]^{2-}[Co(ox)2​(Br)(NH3​)]2− is _____________[ox = oxalate]

Correct answer: 3

Step-by-step solution →
Q53·ChemistryNumerical
The average S−F bond energy in kJ mol−1mol^{-1}mol−1 of SF6SF_6SF6​ is ____________. (Rounded off to the nearest integer) [Given : The values of standard enthalpy of formation of SF6SF_6SF6​(g), S(g) and F(g) are - 1100, 275 and 80 kJ mol−1mol^{-1}mol−1 respectively.]

Correct answer: 309

Step-by-step solution →
Q54·ChemistryNumerical
Emf of the following cell at 298 K in V is x ×10−2\times10^{-2}×10−2. Zn∣Zn2+Zn|Zn^{2+}Zn∣Zn2+ (0.1 M)∣∣Ag+||Ag^+∣∣Ag+(0.01 M)∣|∣ Ag The value of x is ___________. (Rounded off to the nearest integer) [Given: EZn2+/Zn0E^0_{Zn^{2+}/Zn}EZn2+/Zn0​ = −0.76V; EAg+/Ag0E^0_{Ag^+/Ag}EAg+/Ag0​ = +0.80V; 2.303RTF\frac{2.303RT}{F}F2.303RT​ = 0.059]

Correct answer: 147

Step-by-step solution →
Q55·ChemistryNumerical
A ball weighing 10g is moving with a velocity of 90ms−1^{-1}−1. If the uncertainty in its velocity is 5%, then the uncertainty in its position is _________×10−33\times10^{-33}×10−33m. (Rounded off to the nearest integer) [Given : h = 6.63×10−346.63\times10^{-34}6.63×10−34 Js]

Correct answer: 1

Step-by-step solution →
Q56·ChemistryNumerical
In mildly alkaline medium, thiosulphate ion is oxidized by MnO4−MnO_4^-MnO4−​ to "A". The oxidation state of sulphur in "A" is_________.

Correct answer: 6

Step-by-step solution →
Q57·ChemistryNumerical
When 12.2 g of benzoic acid is dissolved in 100g of water, the freezing point of solution was found to be −0.93°C (KfK_fKf​ (H2OH_2OH2​O) = 1.86 K kg mol−1mol^{-1}mol−1). The number (n) of benzoic acid molecules associated (assuming 100% association ) is_______________.

Correct answer: 2

Step-by-step solution →
Q58·ChemistryNumerical
If the activation energy of a reaction is 80.9 kJ mol−1mol^{-1}mol−1, the fraction of molecules at 700K, having enough energy to react to form products is e−xe^{-x}e−x. The value of x is _______. (Rounded off to the nearest integer) [Use R = 8.31 JK−1JK^{-1}JK−1 mol−1mol^{-1}mol−1]

Correct answer: 14

Step-by-step solution →
Q59·ChemistryNumerical
The pH of ammonium phosphate solution, if pkapk_apka​ of phosphoric acid and pkbpk_bpkb​ of ammonium hydroxide are 5.23 and 4.75 respectively, is_______________.

Correct answer: 7

Step-by-step solution →
Q60·ChemistryNumerical
The number of octahedral voids per lattice site in a lattice is ___________. (Rounded off to the nearest integer)

Correct answer: 1

Step-by-step solution →

Mathematics — JEE Main 26 February 2021 Shift 2

Q61·MathematicsSingle correct
Let L be a line obtained from the intersection of two planes x + 2y + z = 6 and y + 2z = 4. If point P(α, β, γ) is the foot of perpendicular from (3, 2, 1) on L, then the value of 21(α + β + γ) equals :
  1. (A)142
  2. (B)68
  3. (C)136
  4. (D)102

Correct answer: (D)

Step-by-step solution →
Q62·MathematicsSingle correct
The sum of the series ∑n=1∞n2+6n+10(2n+1)!\sum_{n=1}^{\infty} \frac{n^{2} + 6n + 10}{(2n+1)!}∑n=1∞​(2n+1)!n2+6n+10​ is equal to :
  1. (A)418e+198e−1−10\frac{41}{8} e + \frac{19}{8} e^{-1} - 10841​e+819​e−1−10
  2. (B)−418e+198e−1−10-\frac{41}{8} e + \frac{19}{8} e^{-1} - 10−841​e+819​e−1−10
  3. (C)418e−198e−1−10\frac{41}{8} e - \frac{19}{8} e^{-1} - 10841​e−819​e−1−10
  4. (D)418e+198e−1+10\frac{41}{8} e + \frac{19}{8} e^{-1} + 10841​e+819​e−1+10

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
Let f(x) be a differentiable function at x = a with f′(a)=2f'(a) = 2f′(a)=2 and f(a) = 4. Then lim⁡x→axf(a)−af(x)x−a\lim_{x \to a} \frac{xf(a) - af(x)}{x - a}limx→a​x−axf(a)−af(x)​ equals :
  1. (A)2a + 4
  2. (B)2a − 4
  3. (C)4 − 2a
  4. (D)a + 4

Correct answer: (C)

Step-by-step solution →
Q64·MathematicsSingle correct
Let A (1, 4) and B(1, −5) be two points. Let P be a point on the circle (x−1)2+(y−1)2=1(x - 1)^{2} + (y - 1)^{2} = 1(x−1)2+(y−1)2=1 such that (PA)2+(PB)2(PA)^{2} + (PB)^{2}(PA)2+(PB)2 have maximum value, then the points, P, A and B lie on :
  1. (A)a parabola
  2. (B)a straight line
  3. (C)a hyperbola
  4. (D)an ellipse

Correct answer: (B)

Step-by-step solution →
Q65·MathematicsSingle correct
If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x2+y2=1x^{2} + y^{2} = 1x2+y2=1 is a circle of the radius r, then r is equal to :
  1. (A)14\frac{1}{4}41​
  2. (B)12\frac{1}{2}21​
  3. (C)1
  4. (D)13\frac{1}{3}31​

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsSingle correct
Let slope of the tangent line to a curve at any point P(x, y) be given by xy2+yx\frac{xy^{2} + y}{x}xxy2+y​. If the curve intersects the line x + 2y = 4 at x = − 2, then the value of y, for which the point (3, y) lies on the curve, is :
  1. (A)−1811-\frac{18}{11}−1118​
  2. (B)−1819-\frac{18}{19}−1918​
  3. (C)−43-\frac{4}{3}−34​
  4. (D)1835\frac{18}{35}3518​

Correct answer: (B)

Step-by-step solution →
Q67·MathematicsSingle correct
Let A1A_{1}A1​ be the area of the region bounded by the curves y = sinx, y = cos x and y-axis in the first quadrant. Also, let A2A_{2}A2​ be the area of the region bounded by the curves y = sin x, y = cos x, x-axis and x= π2\frac{\pi}{2}2π​ in the first quadrant. Then,
  1. (A)A1=A2A_{1} = A_{2}A1​=A2​ and A1+A2=2A_{1} + A_{2} = \sqrt{2}A1​+A2​=2​
  2. (B)A1:A2=1:2A_{1} : A_{2} = 1 : 2A1​:A2​=1:2 and A1+A2=1A_{1} + A_{2} = 1A1​+A2​=1
  3. (C)2A1=A22A_{1} = A_{2}2A1​=A2​ and A1+A2=1+2A_{1} + A_{2} = 1 + \sqrt{2}A1​+A2​=1+2​
  4. (D)A1:A2=1:2A_{1} : A_{2} = 1 : \sqrt{2}A1​:A2​=1:2​ and A1+A2=1A_{1} + A_{2} = 1A1​+A2​=1

Correct answer: (D)

Step-by-step solution →
Q68·MathematicsSingle correct
If 0 < a, b < 1, and tan⁡−1a+tan⁡−1b=π4\tan^{-1} a + \tan^{-1} b = \frac{\pi}{4}tan−1a+tan−1b=4π​, then the value of (a+b)−(a2+b22)+(a3+b33)−(a4+b44)+…(a + b) - \left(\frac{a^{2} + b^{2}}{2}\right) + \left(\frac{a^{3} + b^{3}}{3}\right) - \left(\frac{a^{4} + b^{4}}{4}\right) + \ldots(a+b)−(2a2+b2​)+(3a3+b3​)−(4a4+b4​)+… is :
  1. (A)log⁡e2\log_{e} 2loge​2
  2. (B)log⁡e(e2)\log_{e}\left(\frac{e}{2}\right)loge​(2e​)
  3. (C)e
  4. (D)e2−1e^{2} - 1e2−1

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsSingle correct
Let F1F_{1}F1​(A, B, C) = (A ∧ ~B) ∨ [~C ∧ (A ∨ B)] ∨ ~A and F2F_{2}F2​(A, B) = (A ∨ B) ∨ (B → ~A) be two logical expressions. Then :
  1. (A)F1F_{1}F1​ is not a tautology but F2F_{2}F2​ is a tautology
  2. (B)F1F_{1}F1​ is a tautology but F2F_{2}F2​ is not a tautology
  3. (C)F1F_{1}F1​ and F2F_{2}F2​ both area tautologies
  4. (D)Both F1F_{1}F1​ and F2F_{2}F2​ are not tautologies

Correct answer: (A)

Step-by-step solution →
Q70·MathematicsSingle correct
Consider the following system of equations : x + 2y − 3z = a 2x + 6y − 11 z = b x − 2y + 7z = c, Where a, b and c are real constants. Then the system of equations :
  1. (A)has a unique solution when 5a = 2b + c
  2. (B)has infinite number of solutions when 5a = 2b +c
  3. (C)has no solution for all a, b and c
  4. (D)has a unique solution for all a, b and c

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correct
A seven digit number is formed using digit 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is :
  1. (A)67\frac{6}{7}76​
  2. (B)47\frac{4}{7}74​
  3. (C)37\frac{3}{7}73​
  4. (D)17\frac{1}{7}71​

Correct answer: (C)

Step-by-step solution →
Q72·MathematicsSingle correct
If vectors a⃗1=xi^−j^+k^\vec{a}_{1} = x\hat{i} - \hat{j} + \hat{k}a1​=xi^−j^​+k^ and a⃗2=i^+yj^+zk^\vec{a}_{2} = \hat{i} + y\hat{j} + z\hat{k}a2​=i^+yj^​+zk^ are collinear, then a possible unit vector parallel to the vector xi^+yj^+zk^x\hat{i} + y\hat{j} + z\hat{k}xi^+yj^​+zk^ is :
  1. (A)12(−j^+k^)\frac{1}{\sqrt{2}} (-\hat{j} + \hat{k})2​1​(−j^​+k^)
  2. (B)12(i^−j^)\frac{1}{\sqrt{2}} (\hat{i} - \hat{j})2​1​(i^−j^​)
  3. (C)13(i^−j^+k^)\frac{1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k})3​1​(i^−j^​+k^)
  4. (D)13(i^+j^−k^)\frac{1}{\sqrt{3}} (\hat{i} + \hat{j} - \hat{k})3​1​(i^+j^​−k^)

Correct answer: (C)

Step-by-step solution →
Q73·MathematicsSingle correct
For x>0, if f(x)=∫1xlog⁡et(1+t)dtf(x) = \int_{1}^{x} \frac{\log_{e} t}{(1+t)} dtf(x)=∫1x​(1+t)loge​t​dt, then f(e)+f(1e)f(e) + f\left(\frac{1}{e}\right)f(e)+f(e1​) is equal to :
  1. (A)12\frac{1}{2}21​
  2. (B)−1
  3. (C)1
  4. (D)0

Correct answer: (A)

Step-by-step solution →
Q74·MathematicsSingle correct
Let f : R → R be defined as f(x)={2sin⁡(−πx2),if x<−1∣ax2+x+b∣,if −1≤x≤1sin⁡(πx)if x>1f(x) = \begin{cases} 2\sin\left(-\frac{\pi x}{2}\right), & \text{if } x < -1 \\ |ax^{2} + x + b|, & \text{if } -1 \le x \le 1 \\ \sin(\pi x) & \text{if } x > 1 \end{cases}f(x)=⎩⎨⎧​2sin(−2πx​),∣ax2+x+b∣,sin(πx)​if x<−1if −1≤x≤1if x>1​ If f(x) is continuous on R, then a + b equals :
  1. (A)3
  2. (B)−1
  3. (C)−3
  4. (D)1

Correct answer: (B)

Step-by-step solution →
Q75·MathematicsSingle correct
Let A = {1,2,3……,10} and f: A→ A be defined as f(k)={k+1if k is oddkif k is evenf(k) = \begin{cases} k + 1 & \text{if k is odd} \\ k & \text{if k is even} \end{cases}f(k)={k+1k​if k is oddif k is even​ Then the number of possible functions g : A→A such that gof = f is :
  1. (A)10510^{5}105
  2. (B)10C5^{10}C_{5}10C5​
  3. (C)555^{5}55
  4. (D)5!

Correct answer: (A)

Step-by-step solution →
Q76·MathematicsSingle correct
A natural number has prime factorization given by n=2x3y5zn = 2^{x} 3^{y} 5^{z}n=2x3y5z, where y and z are such that y+z=5y + z = 5y+z=5 and y−1+z−1=56y^{-1} + z^{-1} = \frac{5}{6}y−1+z−1=65​, y>zy > zy>z. Then the number of odd divisors of n, including 1, is :
  1. (A)11
  2. (B)6x
  3. (C)12
  4. (D)6

Correct answer: (C)

Step-by-step solution →
Q77·MathematicsSingle correct
Let f(x)=sin⁡−1xf(x) = \sin^{-1} xf(x)=sin−1x and g(x)=x2−x−22x2−x−6g(x) = \frac{x^{2}-x-2}{2x^{2}-x-6}g(x)=2x2−x−6x2−x−2​. If g(2)=lim⁡x→2g(x)g(2) = \lim_{x \to 2} g(x)g(2)=limx→2​g(x), then the domain of the function fog is :
  1. (A)(−∞,−2]∪[−43,∞)(-\infty, -2] \cup \left[-\frac{4}{3}, \infty\right)(−∞,−2]∪[−34​,∞)
  2. (B)(−∞,−1]∪[2,∞)(-\infty, -1] \cup [2, \infty)(−∞,−1]∪[2,∞)
  3. (C)(−∞,−2]∪[−1,∞)(-\infty, -2] \cup [-1, \infty)(−∞,−2]∪[−1,∞)
  4. (D)(−∞,−2]∪[−32,∞)(-\infty, -2] \cup \left[-\frac{3}{2}, \infty\right)(−∞,−2]∪[−23​,∞)

Correct answer: (A)

Step-by-step solution →
Q78·MathematicsSingle correct
If the mirror image of the point (1,3,5) with respect to the plane 4x−5y+2z=84x-5y+2z = 84x−5y+2z=8 is (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ), then 5(α+β+γ)5(\alpha + \beta + \gamma)5(α+β+γ) equals:
  1. (A)47
  2. (B)39
  3. (C)43
  4. (D)41

Correct answer: (A)

Step-by-step solution →
Q79·MathematicsSingle correct
Let f(x)=∫0xetf(t)dt+exf(x) = \int_{0}^{x} e^{t} f(t)dt + e^{x}f(x)=∫0x​etf(t)dt+ex be a differentiable function for all x∈Rx \in Rx∈R. Then f(x) equals.
  1. (A)2e(ex−1)−12e^{(e^{x}-1)} - 12e(ex−1)−1
  2. (B)e(ex−1)e^{(e^{x}-1)}e(ex−1)
  3. (C)2eex−12e^{e^{x}} - 12eex−1
  4. (D)eex−1e^{e^{x}} - 1eex−1

Correct answer: (A)

Step-by-step solution →
Q80·MathematicsSingle correct
The triangle of maximum area that can be inscribed in a given circle of radius 'r' is:
  1. (A)A right angle triangle having two of its sides of length 2r and r.
  2. (B)An equilateral triangle of height 2r3\frac{2r}{3}32r​.
  3. (C)An isosceles triangle with base equal to 2r.
  4. (D)An equilateral triangle having each of its side of length 3\sqrt{3}3​ r.

Correct answer: (D)

Step-by-step solution →
Q81·MathematicsNumerical
The total number of 4-digit numbers whose greatest common divisor with 18 is 3, is

Correct answer: 1000

Step-by-step solution →
Q82·MathematicsNumerical
Let α\alphaα and β\betaβ be two real numbers such that α+β=1\alpha + \beta = 1α+β=1 and αβ=−1\alpha\beta = -1αβ=−1. Let Pn=(α)n+(β)nP_{n} = (\alpha)^{n} + (\beta)^{n}Pn​=(α)n+(β)n, Pn−1=11P_{n-1} = 11Pn−1​=11 and Pn+1=29P_{n+1} = 29Pn+1​=29 for some integer n≥1n \geq 1n≥1. Then, the value of Pn2P_{n}^{2}Pn2​ is ________________.

Correct answer: 324

Step-by-step solution →
Q83·MathematicsNumerical
Let X1X_{1}X1​, X2X_{2}X2​,……….. X18X_{18}X18​ be eighteen observation such that ∑i=118(Xi−α)=36\sum_{i=1}^{18}\left(X_{i}-\alpha\right) = 36∑i=118​(Xi​−α)=36 and ∑i=118(Xi−β)2=90\sum_{i=1}^{18}\left(X_{i}-\beta\right)^{2} = 90∑i=118​(Xi​−β)2=90, where α\alphaα and β\betaβ are distinct real numbers. If the standard deviation of these observations is 1, then the value of ∣α−β∣|\alpha - \beta|∣α−β∣ is ________________.

Correct answer: 4

Step-by-step solution →
Q84·MathematicsNumerical
In Im,n=∫01xm−1(1−x)n−1dxI_{m,n} = \int_{0}^{1} x^{m-1}\left(1-x\right)^{n-1} dxIm,n​=∫01​xm−1(1−x)n−1dx, for m, n ≥\geq≥ 1 and ∫01xm−1+xn−1(1+x)m+ndx=αIm,n\int_{0}^{1}\frac{x^{m-1}+x^{n-1}}{\left(1+x\right)^{m+n}}dx = \alpha I_{m,n}∫01​(1+x)m+nxm−1+xn−1​dx=αIm,n​, α∈R\alpha \in Rα∈R, then α\alphaα equals______________ .

Correct answer: 1

Step-by-step solution →
Q85·MathematicsNumerical
Let L be a common tangent line to the curves 4x2+9y2=364x^{2} + 9y^{2} = 364x2+9y2=36 and (2x)2+(2y)2=31(2x)^{2} + (2y)^{2} = 31(2x)2+(2y)2=31. Then the square of the slope of the line L is ________________.

Correct answer: 3

Step-by-step solution →
Q86·MathematicsNumerical
If the matrix A=[10002030−1]A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{bmatrix}A=​103​020​00−1​​ satisfies the equation A20+αA19+βA=[100040001]A^{20} + \alpha A^{19} + \beta A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix}A20+αA19+βA=​100​040​001​​ for some real numbers α\alphaα and β\betaβ, then β−α\beta - \alphaβ−α is equal to________________.

Correct answer: 4

Step-by-step solution →
Q87·MathematicsNumerical
If the arithmetic mean and geometric mean of the pthp^{th}pth and qthq^{th}qth terms of the sequence −16-16−16, 8, −4-4−4, 2, ………… satisfy the equation 4x2−9x+5=04x^{2} - 9x + 5 = 04x2−9x+5=0, then p+q is equal to ________________.

Correct answer: 10

Step-by-step solution →
Q88·MathematicsNumerical
Let the normals at all the points on a given curve pass through a fixed point (a, b). If the curve passes through (3, -3) and (4,−22)\left(4, -2\sqrt{2}\right)(4,−22​), and given that a−22 b=3a - 2\sqrt{2}\ b = 3a−22​ b=3, then (a2+b2+ab)(a^{2}+b^{2}+ab)(a2+b2+ab) is equal to________________.

Correct answer: 9

Step-by-step solution →
Q89·MathematicsNumerical
Let z be those complex number which satisfy ∣z+5∣≤4|z+5| \leq 4∣z+5∣≤4 and z(1+i)+zˉ(1−i)≥−10,i=−1z(1+i) + \bar{z}\left(1-i\right) \geq -10, i = \sqrt{-1}z(1+i)+zˉ(1−i)≥−10,i=−1​. If the maximum value of ∣z+1∣2|z+1|^{2}∣z+1∣2 is α+β2\alpha + \beta\sqrt{2}α+β2​, then the value of (α+β)\left(\alpha + \beta\right)(α+β) is ________________.

Correct answer: 48

Step-by-step solution →
Q90·MathematicsNumerical
Let a be an integer such that all the real roots of the polynomial 2x5+5x4+10x3+10x2+10x+102x^{5}+5x^{4}+10x^{3}+10x^{2}+10x+102x5+5x4+10x3+10x2+10x+10 lie in the interval (a, a + 1). Then, ∣a∣|a|∣a∣ is equal to________________.

Correct answer: 2

Step-by-step solution →

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
  • 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
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Electric Field and Coulomb's Law 133/186
  • 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
  • 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
  • Alternating Currents 108/186
  • Nuclei 116/186
  • Organic Compounds Containing Halogens 109/186
  • Waves 109/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Statistics 118/186
  • Ellipse 103/186
  • Isolation of Metals 106/186
  • s-Block Elements 88/186
  • Surface Chemistry 98/186
  • Inverse Trigonometric Functions 93/186
  • Hydrogen 81/186
  • Electric Potential 63/186
  • Solid State 63/186
← 26 Feb Shift 1 2021All papers16 Mar Shift 1 2021 →

Attempt this paper under exam timing.

Take the 26 February 2021 Shift 2 paper as a timed mock and Jarvis marks it, then tells you which errors were conceptual gaps, which were silly mistakes, and which pattern you have now repeated. Step-by-step solutions for every question included.

Attempt this paper freeSee pricing

Free plan, no time limit · No credit card needed

Jarvis OS

AI-powered JEE preparation.

Question papers

JEE Main PYQsJEE Advanced PYQsPhysics PYQsChemistry PYQsMaths PYQs

Product

TourPricingSign inCreate account

Legal

Privacy PolicyTerms of ServiceRefund & CancellationShipping & DeliveryContact Us

Operated by

Venyou Craft Private Limited

info@jarvisos.net

© 2026 Jarvis OS