Practice Questions

Q30.In the experimental set up of metre bridge shown in the figure, the null point is obtaine data distance of 40 cm from A. If a 10Ω resistor is connected in series with R1, the null point shifts by 10 cm. The resistance that should be connected in parallel with (R1 + 10)Ω such that the null point shifts back to its initial position is (1) 20Ω (2) 40Ω (3) 60Ω (4) 30Ω Q31. 25 mL of the given HCl solution requires 30 mL of 0.1M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2M aqueous NaOH solution? (1) 25 mL (2) 75 mL (3) 50 mL (4) 12.5 mL JEE Main 2019 (11 Jan Shift 2) JEE Main Previous Year Paper

Q36.A mixture of 100 m mol of Ca (OH)2 and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. What is the mass of calcium sulphate formed and the concentration of OH− in resulting solution, respectively? (Molar mass of Ca (OH)2, Na2 SO4 and CaSO4 are 74, 143 and 136 g mol−1, respectively; Ksp of Ca (OH)2 is 5 .5 ×10−6 ) (1) 1 .9 g, 0 .14 mol L−1 (2) 13 .6 g, 0 .28 mol L−1 (3) 1 .9 g, 0 .28 mol L−1 (4) 13 .6 g, 0 .14 mol L−1

Q37.Two solids dissociate as follows: A (s) ⇌B (g) + C (g); KP1 = x atm2 D (s) ⇌C (g) + E (g); KP2 = y atm2 The total pressure when both the solids dissociate simultaneously is: (1) √x + y atm (2) x2 + y2 atm (3) (x + y) atm (4) 2(√x + y) atm Q38. 50 mL of 0 .5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. What is the amount of NaOH in 50 mL of the given sodium hydroxide solution? (1) 2 g (2) 4 g (3) 1 g (4) 8 g

Q42.Which one of the following alkenes when treated with HCl yields majorly an anti Markovnikov product? (1) CH3O - CH = CH2 (2) Cl - CH = CH2 (3) H2N - CH = CH2 (4) F3C - CH = CH2

Q52.The correct order of the spin –only magnetic moment of metal ions in the following low-spin complexes, 4 - 4 - 3 + 2 + [V ( CN ) 6] , [Fe ( CN ) 6] , [Ru ( NH3 ) 6] and [Cr ( NH3 ) 6] , is: (1) V2 + > Cr2 + > Ru3 +(2)>Cr2Fe2++ > V2 + > Ru3 +(3)> Cr2Fe2++ > Ru3 + > Fe2 +(4)> V2V2++ > Ru3 + > Cr2 + > Fe2 + JEE Main 2019 (08 Apr Shift 1) JEE Main Previous Year Paper

Q53.The complex ion that will lose its crystal field stabilization energy upon oxidation of metal to +3 state is: (1) Niphen3 2 + (2) Fephen3 2 + (3) Cophen3 2 + (4) Znphen3 2 +

Q55.The major product of the following reaction is: JEE Main 2019 (10 Apr Shift 1) JEE Main Previous Year Paper (1) (2) (3) (4)

Q56. In the following compound, the favourable site/s for protonation is/are : JEE Main 2019 (11 Jan Shift 2) JEE Main Previous Year Paper (1) (a) and (e) (2) (b), (c) and (d) (3) (a) and (d) (4) (a)

Q58.With dehydrating agent present which dicarboxylic acid is least reactive towards forming an anhydride? JEE Main 2019 (10 Jan Shift 1) JEE Main Previous Year Paper (1) (2) (3) (4)

Q60.The increasing order of pKa of the following amino acids in aqueous solution is Glycine, Aspartate, Lysine, Arginine. (1) Arginine < Lysine < Glycine < Aspartate (2) Aspartate < Glycine < Arginine < Lysine (3) Glycine < Aspartate < Arginine < Lysine (4) Aspartate < Glycine < Lysine < Arginine

Q60.The correct structure of the product ‘P’ in the following reaction is NEt3 −Asn −Ser + (CH3 CO)2O → P (excess) JEE Main 2019 (10 Jan Shift 1) JEE Main Previous Year Paper (1) (2) (3) (4)

Q61.Consider the quadratic equation (c −5)x2 −2cx + (c −4) = 0, c ≠5. Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then the number of elements in S is (1) 11 (2) 12 (3) 18 (4) 10

Q61.If m is chosen in the quadratic equation (m2 + 1)x2 −3x + (m2 + 1)2 = 0 such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is: (1) 4√3 (2) 10√5 (3) 8√3 (4) 8√5

Q61.If λ be the ratio of the roots of the quadratic equation in x, 3m2x2 + m(m −4)x + 2 = 0, then the least value of m for which λ + λ1 = 1, is : JEE Main 2019 (12 Jan Shift 1) JEE Main Previous Year Paper (1) 2 −√3 (2) −2 + √2 (3) 4 −2√3 (4) 4 −3√2 α −

Q61.The number of real roots of the equation 5 + 2𝑥- 1 = 2𝑥2𝑥- 2 is : (1) 2 (2) 3 (3) 1 (4) 4 π

Q61.If α and β are the roots of the quadratic equation x2 + xsinθ −2sinθ = 0, θ ∈(0, 2π ) , then α12+β12 is equal to : (α−12+β−12).(α−β)24 (1) 26 (2) 212 (sinθ+8)12 (sinθ−4)12 (3) 212 (4) 212 (sinθ+8)12 (sinθ−8)6 , has magnitude , then −z is equal to:

Q62.If 𝑧 and 𝜔 are two complex numbers such that 𝑧𝜔= 1 and 𝑎𝑟𝑔𝑧- 𝑎𝑟𝑔( 𝜔) = 2, then: (1) 𝑧¯ω = 1 - 𝑖 (2) ¯𝑧𝜔= 𝑖 √2 (3) 𝑧¯ω = -1 + 𝑖 (4) ¯𝑧ω = - 𝑖 √2

Q64.Let Sn = 1 + q + q2 + … . +qn and Tn = 1 + ( q+12 ) ( q+12 ) ( q+12 ) and q ≠1. If 101C1 + 101C2 ⋅S1 + … . +101C101 ⋅S100 = αT100, then α is equal to : (1) 299 (2) 202 (3) 200 (4) 2100

Q65.If sin4α + 4cos4β + 2 = 4√2sinαcosβ, α, β ∈[0, π] , then cos(α + β) −cos(α −β) is equal to (1) −1 (2) −√2 (3) √2 (4) 0 JEE Main 2019 (12 Jan Shift 2) JEE Main Previous Year Paper

Q65.The sum of the following series 1 + 6 + 9(12+22+32)7 + 12(12+22+32+42)9 + 15(12+22+…+52)11 +. . . . is: (1) 7520 (2) 7510 (3) 7830 (4) 7820

Q65.Let 𝑎1, 𝑎2, … , 𝑎30 be an A.P., 𝑆= ∑𝑖=30 1 𝑎𝑖 and 𝑇= ∑𝑖=15 1 𝑎( 2𝑖- 1 ) . If 𝑎5 = 27 and 𝑆- 2𝑇= 75, then 𝑎10 is equal to: (1) 52 (2) 47 (3) 42 (4) 57

Q65.If 20C1 + (22) 20C2 + (32) 20C3+. . . . . +(202) 20C20 = A(2β), then the ordered pair (A, β) is equal to JEE Main 2019 (12 Apr Shift 2) JEE Main Previous Year Paper (1) (380, 19) (2) (420, 18) (3) (420, 19) (4) (380, 18) − 3 ) 6 is equal to x2

Q67.All the pairs (x, y), that satisfy the inequality 2√sin2x−2sinx+5 ⋅ 1 ≤1 also satisfy the equation: 4sin2y (1) 2 sin x = sin y (2) sin x = 2 sin y (3) |sin x| = |sin y| (4) 2|sin x| = 3 sin y

Q69.If a circle of radius R passes through the origin O and intersects the coordinate axes at A and B, then the locus of the foot of perpendicular from O on AB is : (1) (x2 + y2)(x + y) = R2xy (2) (x2 + y2)3 = 4R2x2y2 (3) (x2 + y2) 2 = 4R2x2y2 (4) (x2 + y2) 2 = 4Rx2y2

Q69.The locus of the centres of the circles, which touch the circle, 𝑥2 + 𝑦2 = 1 externally, also touch the 𝑦-axis and lie in the first quadrant, is: (1) 𝑦= √1 + 2𝑥, 𝑥≥0 (2) 𝑦= √1 + 4𝑥, 𝑥≥0 (3) 𝑥= √1 + 2𝑦, 𝑦≥0 (4) 𝑥= √1 + 4𝑦, 𝑦≥0