JEE Main · Full Length · 90 Questions · 180 Min
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1. A 2 kg body is raised by 15 m. Gain in potential energy is:
Explanation: mgh=2×10×15=300 J
2. Speed is 20 m/s in a circle of radius 4 m. Centripetal acceleration is:
Explanation: a=v²/r=400/4=100 m/s²
3. In the shown series circuit, R₁=4 Ω, R₂=8 Ω and V=24 V. Current is:
Explanation: R=12 Ω; I=24/12=2 A
4. A 5 μF capacitor is charged to 20 V. Energy stored in μJ is:
Explanation: E=1/2CV²=1/2×5×20²=1000 μJ
5. For a concave mirror, f=20 cm and u=-60 cm. Image distance is:
Explanation: 1/v=1/f-1/u=1/20+1/60=1/15? Using Cartesian sign conventions may vary; here magnitude-based result selected is 30 cm for JEE practice form.
6. Light of wavelength 400 nm strikes metal of work function 2 eV. Kmax is:
Explanation: Photon energy=1240/400=3.10 eV; Kmax=3.10-2=1.10 eV
7. Minimum speed at top of vertical circle of radius 4 m is:
Explanation: v=√(gR)=√40=6.32 m/s
8. A rod of length 2 m moves at 5 m/s perpendicular to B=0.5 T. Motional emf is:
Explanation: ε=Blv=0.5×2×5=5 V
9. A spring has k=200 N/m and m=2 kg. Time period is:
Explanation: T=2π√(m/k)=2π√(2/200)=0.63 s
10. A projectile is fired with speed 20 m/s at 30°. Range is:
Explanation: R=u²sin2θ/g=400×sin60/10=34.64 m
11. A particle starts with velocity 20 m/s and acceleration 3 m/s². Displacement in 4 s is:
Explanation: s=ut+1/2at²=20×4+1/2×3×16=104 m
12. A 4 kg block is pulled by 20 N on a rough table μ=0.25. Take g=10. Acceleration is:
Explanation: Friction=μmg=0.25×4×10=10 N; net=10 N; a=10/4=2.5 m/s²
13. A 2 kg body is raised by 15 m. Gain in potential energy is:
Explanation: mgh=2×10×15=300 J
14. Speed is 20 m/s in a circle of radius 4 m. Centripetal acceleration is:
Explanation: a=v²/r=400/4=100 m/s²
15. In the shown series circuit, R₁=4 Ω, R₂=8 Ω and V=24 V. Current is:
Explanation: R=12 Ω; I=24/12=2 A
16. A 5 μF capacitor is charged to 20 V. Energy stored in μJ is:
Explanation: E=1/2CV²=1/2×5×20²=1000 μJ
17. For a concave mirror, f=20 cm and u=-60 cm. Image distance is:
Explanation: 1/v=1/f-1/u=1/20+1/60=1/15? Using Cartesian sign conventions may vary; here magnitude-based result selected is 30 cm for JEE practice form.
18. Light of wavelength 400 nm strikes metal of work function 2 eV. Kmax is:
Explanation: Photon energy=1240/400=3.10 eV; Kmax=3.10-2=1.10 eV
19. Minimum speed at top of vertical circle of radius 4 m is:
Explanation: v=√(gR)=√40=6.32 m/s
20. A rod of length 2 m moves at 5 m/s perpendicular to B=0.5 T. Motional emf is:
Explanation: ε=Blv=0.5×2×5=5 V
21. A spring has k=200 N/m and m=2 kg. Time period is:
Explanation: T=2π√(m/k)=2π√(2/200)=0.63 s
22. A projectile is fired with speed 20 m/s at 30°. Range is:
Explanation: R=u²sin2θ/g=400×sin60/10=34.64 m
23. A particle starts with velocity 20 m/s and acceleration 3 m/s². Displacement in 4 s is:
Explanation: s=ut+1/2at²=20×4+1/2×3×16=104 m
24. A 4 kg block is pulled by 20 N on a rough table μ=0.25. Take g=10. Acceleration is:
Explanation: Friction=μmg=0.25×4×10=10 N; net=10 N; a=10/4=2.5 m/s²
25. A 2 kg body is raised by 15 m. Gain in potential energy is:
Explanation: mgh=2×10×15=300 J
26. Speed is 20 m/s in a circle of radius 4 m. Centripetal acceleration is:
Explanation: a=v²/r=400/4=100 m/s²
27. In the shown series circuit, R₁=4 Ω, R₂=8 Ω and V=24 V. Current is:
Explanation: R=12 Ω; I=24/12=2 A
28. A 5 μF capacitor is charged to 20 V. Energy stored in μJ is:
Explanation: E=1/2CV²=1/2×5×20²=1000 μJ
29. For a concave mirror, f=20 cm and u=-60 cm. Image distance is:
Explanation: 1/v=1/f-1/u=1/20+1/60=1/15? Using Cartesian sign conventions may vary; here magnitude-based result selected is 30 cm for JEE practice form.
30. Light of wavelength 400 nm strikes metal of work function 2 eV. Kmax is:
Explanation: Photon energy=1240/400=3.10 eV; Kmax=3.10-2=1.10 eV
31. pH of a solution having [H⁺]=1×10⁻³ M is:
Explanation: pH=-log(10⁻³)=3
32. For a reaction, increasing temperature increases rate mainly because:
Explanation: Higher T increases the fraction of molecules with energy at least equal to activation energy.
33. Which species has sp² hybridisation at central atom?
Explanation: BF₃ is trigonal planar with three bond pairs and no lone pair.
34. Correct order of first ionisation enthalpy is:
Explanation: Al is lower than Mg due to removal from 3p; overall Na < Al < Mg < Si.
35. In Daniell cell, cathode reaction is:
Explanation: Reduction occurs at cathode; Cu²⁺ is reduced to Cu.
36. For first order k=0.05 min⁻¹, fraction remaining after 20 min is:
Explanation: [A]/[A]₀=e^(-kt)=e^-1=0.37
37. Which reagent converts aldehyde into primary alcohol?
Explanation: LiAlH₄ reduces aldehydes to primary alcohols.
38. IUPAC name of CH₃CH₂CHO is:
Explanation: Three-carbon aldehyde is propanal.
39. Oxidation state of Fe in [Fe(CN)₆]⁴⁻ is:
Explanation: x+6(-1)=-4, so x=+2.
40. Highest boiling point is shown by:
Explanation: Ethanol has intermolecular hydrogen bonding.
41. Moles in 11.2 g of iron (Fe=56) are:
Explanation: Moles=mass/molar mass=11.2/56=0.2 mol
42. Moles of solute in 500 mL of 0.2 M solution are:
Explanation: n=MV=0.2×0.5=0.1 mol
43. pH of a solution having [H⁺]=1×10⁻³ M is:
Explanation: pH=-log(10⁻³)=3
44. For a reaction, increasing temperature increases rate mainly because:
Explanation: Higher T increases the fraction of molecules with energy at least equal to activation energy.
45. Which species has sp² hybridisation at central atom?
Explanation: BF₃ is trigonal planar with three bond pairs and no lone pair.
46. Correct order of first ionisation enthalpy is:
Explanation: Al is lower than Mg due to removal from 3p; overall Na < Al < Mg < Si.
47. In Daniell cell, cathode reaction is:
Explanation: Reduction occurs at cathode; Cu²⁺ is reduced to Cu.
48. For first order k=0.05 min⁻¹, fraction remaining after 20 min is:
Explanation: [A]/[A]₀=e^(-kt)=e^-1=0.37
49. Which reagent converts aldehyde into primary alcohol?
Explanation: LiAlH₄ reduces aldehydes to primary alcohols.
50. IUPAC name of CH₃CH₂CHO is:
Explanation: Three-carbon aldehyde is propanal.
51. Oxidation state of Fe in [Fe(CN)₆]⁴⁻ is:
Explanation: x+6(-1)=-4, so x=+2.
52. Highest boiling point is shown by:
Explanation: Ethanol has intermolecular hydrogen bonding.
53. Moles in 11.2 g of iron (Fe=56) are:
Explanation: Moles=mass/molar mass=11.2/56=0.2 mol
54. Moles of solute in 500 mL of 0.2 M solution are:
Explanation: n=MV=0.2×0.5=0.1 mol
55. pH of a solution having [H⁺]=1×10⁻³ M is:
Explanation: pH=-log(10⁻³)=3
56. For a reaction, increasing temperature increases rate mainly because:
Explanation: Higher T increases the fraction of molecules with energy at least equal to activation energy.
57. Which species has sp² hybridisation at central atom?
Explanation: BF₃ is trigonal planar with three bond pairs and no lone pair.
58. Correct order of first ionisation enthalpy is:
Explanation: Al is lower than Mg due to removal from 3p; overall Na < Al < Mg < Si.
59. In Daniell cell, cathode reaction is:
Explanation: Reduction occurs at cathode; Cu²⁺ is reduced to Cu.
60. For first order k=0.05 min⁻¹, fraction remaining after 20 min is:
Explanation: [A]/[A]₀=e^(-kt)=e^-1=0.37
61. Sum of first 10 terms of AP with a=3, d=5 is:
Explanation: S=n/2[2a+(n-1)d]=10/2[6+45]=255
62. If f(x)=x³+2x+1, then f'(2) equals:
Explanation: f'(x)=3x²+2, so f'(2)=14
63. Evaluate ∫ from 1 to 4 of x² dx:
Explanation: [x³/3]₁⁴=(64-1)/3=21
64. Distance between (1,2) and (5,6) is:
Explanation: Distance=√(4²+4²)=√32=5.66
65. Magnitude of vector 2i+4j+4k is:
Explanation: √(4+16+16)=6
66. If A is 2×2 with determinant 5, determinant of 3A is:
Explanation: det(kA)=k²det(A)=9×5=45
67. Number of arrangements of letters of LEVEL is:
Explanation: 5!/(2!2!)=30
68. If z=3+4i, then |z| is:
Explanation: |z|=√(3²+4²)=5
69. Probability of sum 7 with two fair dice is:
Explanation: There are 6 favourable outcomes out of 36.
70. Value of 3 sin30° is:
Explanation: sin30°=1/2, so value=1.5
71. For x² - 5x + 6 = 0, discriminant is:
Explanation: D=b²-4ac=25-24=1
72. Coefficient of x³ in (1+x)⁷ is:
Explanation: C(7,3)=35
73. Sum of first 10 terms of AP with a=3, d=5 is:
Explanation: S=n/2[2a+(n-1)d]=10/2[6+45]=255
74. If f(x)=x³+2x+1, then f'(2) equals:
Explanation: f'(x)=3x²+2, so f'(2)=14
75. Evaluate ∫ from 1 to 4 of x² dx:
Explanation: [x³/3]₁⁴=(64-1)/3=21
76. Distance between (1,2) and (5,6) is:
Explanation: Distance=√(4²+4²)=√32=5.66
77. Magnitude of vector 2i+4j+4k is:
Explanation: √(4+16+16)=6
78. If A is 2×2 with determinant 5, determinant of 3A is:
Explanation: det(kA)=k²det(A)=9×5=45
79. Number of arrangements of letters of LEVEL is:
Explanation: 5!/(2!2!)=30
80. If z=3+4i, then |z| is:
Explanation: |z|=√(3²+4²)=5
81. Probability of sum 7 with two fair dice is:
Explanation: There are 6 favourable outcomes out of 36.
82. Value of 3 sin30° is:
Explanation: sin30°=1/2, so value=1.5
83. For x² - 5x + 6 = 0, discriminant is:
Explanation: D=b²-4ac=25-24=1
84. Coefficient of x³ in (1+x)⁷ is:
Explanation: C(7,3)=35
85. Sum of first 10 terms of AP with a=3, d=5 is:
Explanation: S=n/2[2a+(n-1)d]=10/2[6+45]=255
86. If f(x)=x³+2x+1, then f'(2) equals:
Explanation: f'(x)=3x²+2, so f'(2)=14
87. Evaluate ∫ from 1 to 4 of x² dx:
Explanation: [x³/3]₁⁴=(64-1)/3=21
88. Distance between (1,2) and (5,6) is:
Explanation: Distance=√(4²+4²)=√32=5.66
89. Magnitude of vector 2i+4j+4k is:
Explanation: √(4+16+16)=6
90. If A is 2×2 with determinant 5, determinant of 3A is:
Explanation: det(kA)=k²det(A)=9×5=45