Practice Questions
3,296 questions across 23 years of JEE Main — find and practise any topic!
Found 3,296 results
Q31.Following figure shows spectrum of an ideal black body at four different temperatures. The number of correct statement/s from the following is _______. JEE Main 2023 (24 Jan Shift 2) JEE Main Previous Year Paper A. T4 > T3 > T2 > T1 B. The black body consists of particles performing simple harmonic motion. C. The peak of the spectrum shifts to shorter wavelength as temperature increases. D. T1 = T2 = T3 ν1 ν2 ν3 E. The given spectrum could be explained using quantisation of energy
Q31.When a hydrocarbon A undergoes complete combustion it requires 11 equivalents of oxygen and produces 4 equivalents of water. What is the molecular formula of A? (1) C11H8 (2) C11H4 (3) C5H8 (4) C9H8
Q31.The shortest wavelength of hydrogen atom in Lyman series is λ. The longest wavelength in Balmer series of He+ is (1) 5 (2) 9λ 9λ 5 (3) 36λ (4) 5λ 5 9
Q31.The radius of the 2nd orbit of Li2+ is x . The expected radius of the 3rd orbit of Be3+ is (1) 9 4 x (2) 94 x (3) 27 16 x (4) 2716 x JEE Main 2023 (25 Jan Shift 1) JEE Main Previous Year Paper
Q31.Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: 3 . 1500 𝑔 of hydrated oxalic acid dissolved in water to make 250 . 0 𝑚𝐿 solution will result in 0 . 1 𝑀 oxalic acid solution. Reason R: Molar mass of hydrated oxalic acid is 126 𝑔 𝑚𝑜𝑙– 1 . In the light of the above statements, choose the correct answer from the options given below: (1) Both A and R are true but R is NOT the correct (2) A is true but R is false explanation of A (3) Both A and R are true and R is the correct (4) A is false but R is true explanation of A
Q31.The wave function (Ψ) of 2 s is given by 1/2 − Ψ2s = a0 1 ( a01 ) (2 r )e−r/2a0 2√2π At r = r0 , radial node is formed. Thus, r0 in terms of a0 (1) r0 = a0 (2) r0 = 4a0 (3) r0 = a02 (4) r0 = 2a0
Q31.Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from n = 4 to n = 2 of He+spectrum (1) n = 2 to n = 1 (2) n = 1 to n = 3 (3) n = 1 to n = 2 (4) n = 3 to n = 4
Q31.Number of hydrogen atoms per molecule of a hydrocarbon A having 85. 8% carbon is (Given : Molar mass of A = 84 g mol−1 )
Q31.Given below are two statements Statement I : According to Bohr’s model of hydrogen atom, the angular momentum of an electron in a given stationary state is quantised. Statement II : The concept of electron in Bohr’s orbit, violates the Heisenberg uncertainty principle. In the light of the above statements, choose the most appropriate answer from the options given below (1) Statement I is incorrect but Statement II is (2) Both Statement I and Statement II are correct correct (3) Both Statement I and Statement II are incorrect (4) Statement I is correct but Statement II is incorrect
Q31.A metal chloride contains 55. 0% of chlorine by weight. 100 mL vapours of the metal chloride at STP weigh 0. 57 g . The molecular formula of the metal chloride is (Given: Atomic mass of chlorine is 35. 5 u ) (1) MCl4 (2) MCl3 (3) MCl2 (4) MCl
Q31.The molality of a 10%(v/V) solution of di-bromine solution in CCl4 (carbon tetrachloride) is x '. x = _____ × 10−2M. (Nearest integer) [Given : molar mass of Br2 = 160 g mol−1 atomic mass of C = 12 g mol−1 atomic mass of Cl = 35. 5 g mol−1 density of dibromine = 3. 2 g cm−3 density of CCl4 = 1. 6 g cm−3]
Q31.If the radius of the first orbit of hydrogen atom is a0 , then de Broglie’s wavelength of electron in 3rd orbit is (1) πa0 (2) πa0 6 3 (3) 6πa0 (4) 3πa0
Q31.An organic compound gives 0. 220 g of CO2 and 0. 126 g of H2O on complete combustion. If the % of carbon is 24 then the % of hydrogen is _________ × 10–1. (Nearest integer)
Q31.For a concentrated solution of a weak electrolyte (Keq = equilibrium constant) A2B3 of concentration ‘C’, the degree of dissociation ‘α’ is (1) Keq 15 (2) Keq 15 5c4 108c4 (3) Keq 15 (4) Keq 15 25c2 6c5 JEE Main 2023 (06 Apr Shift 1) JEE Main Previous Year Paper
Q32.For OF2 molecule consider the following: (A) Number of lone pairs on oxygen is 2. (B) FOF angle is less than 104 . 5°. (C) Oxidation state of O is -2. (D) Molecule is bent 'V' shaped. (E) Molecular geometry is linear. Correct options are: (1) C, D, E only (2) B, E, A only (3) A, C, D only (4) A, B, D only
Q32.What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : N2, N +2 , O2, O+2 ? (1) 0, 1, 2, 1 (2) 2, 1, 2, 1 (3) 0, 1, 0, 1 (4) 2, 1, 0, 1
Q32.Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A: In the photoelectric effect, the electrons are ejected from the metal surface as soon as the beam of light of frequency greater than threshold frequency strikes the surface. Reason R : When the photon of any energy strikes an electron in the atom, transfer of energy from the photon to the electron takes place. In the light of the above statements, choose the most appropriate answer from the options given below : (1) Both A and R are correct and R is the correct (2) A is correct but R is not correct explanation of A (3) Both A and R are correct but R is NOT the (4) A is not correct but R is correct correct explanation of A JEE Main 2023 (11 Apr Shift 1) JEE Main Previous Year Paper
Q32.The correct increasing order of the ionic radii is (1) Cl- < Ca2 + < K+ < S2 - (2) K+ < S2 - < Ca2 + < Cl- (3) S2 - < Cl- < Ca2 + < K+ (4) Ca2 + < K+ < Cl- < S2 -
Q32.Given below are two statements: Statement I: The decrease in first ionization enthalpy from B to Al is much larger than that from Al to Ga. Statement II: The d orbitals in Ga are completely filled. In the light of the above statements, choose the most appropriate answer from the options given below JEE Main 2023 (29 Jan Shift 2) JEE Main Previous Year Paper (1) Statement I is incorrect but statement II is (2) Both the statements I and II are correct correct. (3) Statement I is correct but statement II is incorrect (4) Both the statements I and II are incorrect
Q32.Consider the following statement (A) NF3 molecules has a trigonal planar structure. (B) Bond Length of N2 is shorter than O2 . (C) Isoelectronic molecules or ions have identical bond order. (D) Dipole moment of H2S is higher than that of water molecule. Choose the correct answer from the options given below: (1) (A) and (B) are correct (2) (A) and (D) are correct (3) (C) and (D) are correct (4) (B) and (C) are correct
Q32.The volume of hydrogen liberated at STP by treating 2. 4 g of magnesium with excess of hydrochloric acid is ...... ×10−2 L. Given Molar volume of gas is 22. 4 L at STP. Molar mass of magnesium is 24 g mol−1
Q32.Assume carbon burns according to following equation : 2C(s) + O2( g) →2 CO (s) when 12 g carbon is burnt in 48 g of oxygen, the volume of carbon monoxide produced is ______ × 10−1 L at STP [nearest integer] [Given : Assume CO as ideal gas, Mass of C is 12 g mol−1 , mass of O is 16 g mol−1 and molar volume of an idal gas at STP is 22. 7 L mol−1 ]
Q32.Given below are two statement : one is labelled as Assertion A and the other is labelled as Reason R Assertion A : 5f electron can participate in bonding to a far greater extent than 4f electrons Reason R : 5f orbitals are not as buried as 4f orbitals In the light of the above statements, choose the correct answer from the options given below (1) A is false but R is true (2) Both A and R are true and R is the correct explanation of A (3) A is true but R is false (4) Both A and R are true but R is NOT the correct explanation of A
Q32.Match Lis-I with List-II. List-1 List-II Hexamethylenediamine A. Weak intermolecular forces of attraction I. + adipic acid B. Hydrogen bonding II. AlEt3 + TiCl4 C. Heavily branched polymer III. 2 - chloro - 1, 3 - butadiene D. High density polymer IV. Phenol + formaldehyde Choose the correct answer from the options given below (1) A-IV, B-II, C-III, D-I (2) A-IV, B-I, C-III, D-II (3) A-II, B-IV, C-I, D-III (4) A-III, B-I, C-IV, D-II
Q32.Decreasing order of the hydrogen bonding in following forms of water is correctly represented by A. Liquid water B. Ice C. Impure water (1) A = B > C (2) B > A > C (3) C > B > A (4) A > B > C