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IIT JEE Exam 2025 : Main

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Sayan Chandra
St. Stephen's School, Dum Dum, Kolkata (Calcutta)

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Master JEE CLASSES Kukatpally, Hyderabad. IIT-JEE-MAINS PAPER-2 IMPORTANT INSTRUCTIONS: 2) The test is of 3 hours duration. 3) The Test Booklet consists of 90 questions. The maximum marks are 360. 4) There are three parts in the question paper A, B, C consisting of Mathematics, Physics and Chemistry having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response. 5) Candidates will be awarded marks as stated above in instruction No. 4 for correct response of each question. (1/4) (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet. 6) There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 5 above. SYLLABUS MATHS Real numbers, Factor Theorem, Remainder Theorem, Finding roots of Polynomial Equation, Finding Polynomial with given roots using factor Theorem, Wavy curve method, Rational Inequations, Irrational Equation & Inequation, Modulus & its Max.Marks:360 properties, Logarithm & its properties, Equations/Inequation involving Logarithms, Exponentials, Modulus; Greatest Integer, Least Integer & Fractional Part Functions (30%); Quadratic Equation (70%) PHYSICS Refraction at plane surface, TIR and Prism, Refraction at curved surface (exclude problems involving relative motion and differential and integral calculus) CHEMISTRY De-Broglie Equation, Heisenberg's Uncertainty principle, Schrodinger's wave equation; Quantum numbers, Pauli's exclusion Principle; Hund's rule; Electronic configuration of the elements, (70%) Atomic Strucutre : Fundamental Particles - their characteristics; Thomson's, Rutherford's atomic model, Plank's Quantum theory, Bohr's atomic model, Applications of Bohr's atomic model calculation of radius, velocity, frequency, time period and energy of electron in an orbit, Electromagnetic spectra, Hydrogen spectrum, Photoelectric effect, (30%) 4. MATHS 1. 2 If b 4ac 2 1 4a 2 2 64a ,a 0, then by x-a, x-b, x-c, leaves remainders a, b, c maximum value of quadratic respectively. The remainder `when expression ax p x is divided by x a x b x c is 2 bx c is always less than 2. (where a, b, c are all different non zero 1) 0 2) 2 3) 1 4) 2 constants) Let ax2 + bx + c = 0, a, b, c R, be a quadratic equation. If |b| > |a + c| then 5. the roots of the equation will be such 1)0 2) x 3) ax b c 4) ax 2 bx c If the equation x 2 2 x a 2 has exactly 1) exactly one root lies between 1 4 real and distinct solutions, then 1) a 3 , 1 2, 2) a and 1 3) a 2) both roots lie between 1 and 1 4) a that 6. 3) one root less than 1 and other 3. A certain polynomial p x when divided ,1 3, 1 Let f x be a real valued function px 2 qx r , greater than 1 satisfying af x 4) one root less than 1 and other Where a and b are distinct real numbers greater than 1 and p, q and r are non-zero real numbers. Let , are roots of x 2 12x If an n n 10 value of n 1 for n 9 1, then the 9an an 2an 1 2 1) 60 2) -60 3) 120 4) -120 0. Then f x bf x 0 will have real solution when 1) a b a b 3) a b a b 2 2 space for rough work a2 4 pr 2) a b a b 4 pr q2 4) a b a b 2 2 4 pr q2 q2 4 pr Page 2 7. The number of rational roots of the 2 equation 2 x 13x 15 x 15x 15 x 2 14x 15 x 2 16x 15 is 1) 0 3) 2 8. 1 12 4) data is not sufficient 11. A root of the equation 2) 1 4) 3 a b ax b a bx If x, y, z R; x y z 4 and x2 y2 z2 a 2a a 3) 2a possible value of z is 2) 2 3) 3 4) 5/2 If f x i 1 ai 2) x16 is less than 15 i 1 a i 1 , then f x 2a b a 2b a 2b 4) 2a b 2) 1) x16 is 15 a i 3x, where x ai a 2 x b 2 a bx is 12. If x15 x13 x11 x 9 x 7 x 5 x 3 x 7, then 3 3 9. 2b b 2b b 1) 6 ,then the maximum 1) 1 81 4 3) 3) x16 greater than 15 0 has 4) Nothing can be said about x16 1) Only one real root 2) Three real roots of which two of 13. If f x them are equal x2 bx c and f 2 t f 2 t for all real numbers t, then which of the 3) Three distinct real roots following is true? 4) Three equal roots 10. Let f x x2 bx c, b is an odd positive integer f x 0 have two prime numbers as roots and b c 35 . Then global minimum value of f x is 183 4 173 2) 6 1) 1) f 1 f 2 f 4 2) f 2 f 1 f 4 3) f 2 f 4 f 1 4) f 4 f 2 f 1 14. x2 log3 2 log 2 3 x log 2 6 log 2 2 3 0 then x 1) log 2 3, log 3 2 1 3 3) log 2 , log 3 space for rough work 1 2 2) log1/2 3, log1/3 2 4) All of these Page 3 15. If x2 x 2 19. The solution set of x 1 1 x 2 is 2 a 3 x2 x 1 x2 x 2 a 4 x2 x 1 2 has at least one root, then the 1) ( 2) [0,1) , 2] 3) [0, 2) complete set of values of a is 1) 0 20. If 1 , 2 4) [1, 2) ,..., n are roots of the equation 2) 3, 6 ,5 19 3) 5, 2 xn 19 4) 5, 3 1 16. If the equations ax 2 bx c 0 , then ax b 2 1 3 1 4 ..... 1 n is 0, a, b, c R and x 3 3x 2 3x 2 0 have equal to two common roots, then 2a 3b 5 is 1) n 1) -1 2) 0 3) n 3) 1 4) 2 1 21. Let S b 2) n n 1 1 4) n n 1 1 a N, a 100 . If the equation a 17. If , are the roots 4x 2 16x 0, 2 and 2 1 tan 2 x R, such that 3 , then the number of integral solutions of is 1) 6 2) 5 3) 3 4) 2 that log x 2 1) 2 3) 2 2 log 2 x 2 2) 4) log 2x 1 2 1 2 0 has real roots then number of elements in S is, [where [.] greatest integer function] 18. The product of real values of x such 2 tan x a 2 x 1) 10 2) 8 3) 9 4) 0 22. Number of real roots of the equation is x10 3x 8 5x 6 5x 4 3x 2 1 1)2 2)4 3)8 4) 10 0 is 2 space for rough work Page 4 23. The remainder obtained by dividing 27. If f x the polynomial x49 by x 2 4x 3 is 1) 1 2 3) 4) 1 49 (3 2 x2 g x 1 49 1 49 (3 1)x (3 3) 2 2 2) (349 1)x x2 2bx 2c 2 and 2cx b 2 are such that max g x , then relation between min f x b and c, is 3) 1) no relation 1 49 1 49 (3 3)x (3 1) 2 2 b 3) c 1 49 1 49 (3 3)x (3 1) 2 2 2) 0 c b / 2 4) c 2 2b 28. Sum of all the values of x satisfying the 24. The number of solutions of x 2x where x is the greatest integer 1) 2 2) 4 3) 1 4) Infinite 4 x is equation log17 log11 x 11 1) 25 2) 36 3) 171 4) 0 x 0 is 29. The number of ordered pairs of integers 25. If x 2 4 y 2 8 x 12 0 is satisfied by (x, y) satisfying the equation real values of x and y then ' y ' x2 6x 1) 2,6 2) 2,5 3) 4) 1,1 26. Let f x 2, 1 x 2 5 x 6 , then the number y2 4 1) 2 2) 8 3) 6 4) 10 30. The values of a for which the quadratic of real roots of f x 2 5f x expression 6 x 0 is ax 2 a 2 x 2 is negative for exactly two integral values of x, belongs 1) 1 2) 2 to 3) 3 4) 0 1) 3) 1,1 3, 4 space for rough work 2) 4) 1,2 2, 1 Page 5 PHYSICS 31. A paraxial beam of light incident on a slab of refractive index 3 and 2 thickness 15 cm. The slab is placed in air as shown in the figure. The Which of the following statement is converging point of the beam after refraction through slab is correct? 1) Speed of light in medium B is threefourth of that in medium A. 2) Total internal reflection can take place. 3) Refractive index of medium A is greater than that of medium B. 4) None 1) at 5 cm rightward from point P 33. Monochromatic light is incident on plane 2) at 5 cm leftward from point P interface AB between two media of 3) at point P refractive indices 1 and 2 ( 2> 1) at 4) None of these 32. A ray of light, travelling in medium A, is incident on plane interface of two media A and B and gets refracted into medium B. The angle of incidence is i and that of refraction is r. Graph between sin (i) and sin (r) is as shown infinitesimally greater than the critical angle for two media so that total internal reflection takes place. Now, if a transparent slab DEFG of uniform thickness and having refractive index 3 is introduced on the interface as shown in in figure. space for rough work Page 6 the figure, which of the following one face and 4 cm deep when viewed statement is incorrect? from the other face. The thickness of the glass slab is: 1) 10 cm 2) 6.67 cm 3) 15 cm 4) None of these. 36. When light is incident on a medium at 1) If 3< 1, total internal reflection angle i and refracted into a second medium at an angle r, the graph of sin ivs will take place at face GF. 2) If 3> 1, light will refract into the sin r is as shown in the graph. From this, one can conclude that: slab. 3) If 3> 1, total internal reflection will take place at face DE. 4) Light can notbe transmitted to medium I. 34. The refracting angle of prism is A and 1) Velocity of light in first medium is refractive index of material of prism is 1.73 times the velocity of light in the cot (A/2). The angle of minimum second medium deviation will be: 2) The critical angle for the two media is 1) (1800 2A) given by sin C 2) (1800 + 2A) 3) 900 A 3) Sin C = 4) (1800 A) 3 2 1 2 4) None of these 35. An air bubble in a glass slab ( = 1.5) appears 6 cm deep when viewed from space for rough work Page 7 37. A prism of refractive index 2 has a angle of prism 300 . One of the refracting surface of prism is silvered . Find the angle of incidence i for which the ray retraces its path 1) 4/3 2) 2 3) 1.5 4) 40. A bulb is placed at a depth of 2 7 m in water and a floating opaque disc is placed 1) 300 2) 600 over the bulb so that the bulb is not 3) 450 4) 900 visible from the surface. The radius of the 38. A beam of width t is incident at 450 on disc should be atleast: an air-water boundary. The width of 1) 42 m. 2) 6 m. the beam in water is: 3) 2 7 m. 4) 12 m. 1t 1) 2) 2 2 2) t 1 41.A ray of light travelling in a transparent 2 t 4) 1 medium falls on a surface separating the t medium from air at an angle of incidence 39. A ray of light falls on a prism ABC of 450. The ray undergoes total internal (AB = BC) and travels as shown in reflection. If n is the refractive index of figure. The refractive index of the the medium with respect to air, select the prism material should be at least: possible value of n from the following: 1) 1.3 2) 1.4 3) 1.2 4) 1.6 space for rough work Page 8 42. In the adjacent figure, sini/sinr: 44. An observer can see through a pin hole, the top end of a thin rod of height h placed as shown in Fig. The beaker height is 3h and its radius is h. When the beaker is filled with a liquid upto a height 2h, he can see the lower end of the rod. ' 2 1) is equal to Then the refractive index of the liquid is . 0 2) is equal to 0 3) is equal to . ' 0 2 4) cannot be calculated 43. A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and of the same material as P are now added as shown in Fig. The ray will now suffer 1) 5/2 2) 5/2 3) 3/2 4) 3/2 45. A plane mirror is placed at the bottom of a tank containing a liquid of refractive index . P is a small object at a height h above the mirror. An observer Overtically above P outside the liquid sees 1) greater deviation P and its image in the mirror. The 2) no deviation apparent distance between these two will 3) same deviation as before be 4) total internal reflection. space for rough work Page 9 1) 2 h 2h 3) 1 2) 2h 4) h 1 1 46. A ray of light falls on a transparent 1) 180 2) 120 3) 45 4) 90 48. A point source is placed at a distance d in sphere with centre at C as shown in Fig. The ray emerges from the sphere air-liquid interface, dA is apparent depth parallel to line AB. The refractive of this when seen from air. Now this index of the sphere is point source is placed in air at same distance d from interface. When it is observed from within the liquid, it is at an apparent distance 1) 3/2 2) 2 3) 1/2 4) 3 d A' . The product of d A' and dA is given by 47. One side of a glass slab is silvered as shown in Fig. A ray of light is incident on the other side at angle of incidence I = 45 0 . Refractive index of glass is 1) d 2 3) d 2 d2 2) 2 4) 2 d2 (1 1 ) Paragraph for question no. 49 to 51 1.5. The deviation of the ray of light Consider a transparent hemisphere (n=2) from its initial path when it comes out in front of which a small object is placed. of the slab is space for rough work Page 10 52. Figure shows the graph of angle of deviation versus angle of incidence i for a light ray striking a prism. The prism angle is 49. For which value of x of the following will final image of an object at O be virtual 1) 2R 2) 3R 3) R/2 4) 1.5 R 50. What is the nature of final image of object O for x = 2R. 1) 300 2) 450 3) 600 4) 750 53. A certain prism is found to produce 1) erect and magnified minimum deviation of 380 . It produces a 2) inverted and magnified deviation of 440 when the angle of 3) erect and same size incidence is either 420 or 620 . What is the 4) inverted and same size 51. Consider a ray starting from O which strikes the spherical surface at grazing angle of incidence when it is undergoing minimum deviation incidence (i =90). Taking x = R, what 1) 450 2) 490 will be the angle (from the normal) at 3) 400 4) 550 which the ray emerges from the plane surface. 1) 900 2) 00 3) 300 4) 600 54. A concave spherical surface of radius ofcurvature10cm separates two medium x& y of refractive index 4/3 & 3/2 respectively. If the object is placed along principal axis in medium X then space for rough work Page 11 shown in the figure. The rays emerging from the opposite face 1) image is always real . 2) image is real if the object distance is greater than 90 cm . 3) image is always virtual 1) are parallel to each other 4) image is virtual if the object 2) are diverging distance is less than 90 cm 3) make an angle 2 sin 55. A ray incident at an angle 53 on a prism emerges at an angle at 370 as shown. 1 0.72 300 with each other 4) make an angle 2 sin 1 0.72 with each other 57. A ray of light is incident on an equilateral glass prism placed on a horizontal table .For minimum deviation which of the If the angle of incidence is made 50 , following is correct? which of the following is a possible value of the angle of emergence. 1) 350 2) 420 3) 400 4) 380 1) PQ is horizontal 56. An isosceles prism of angle 120 0 has a 2) QR is horizontal refractive index 1.44. Two parallel 3) RS is horizontal rays of monochromatic light enter the 4) Either PQ or RS is horizontal prism parallel to each other in air as space for rough work Page 12 58. A tank contains three layers of immiscible liquids.The first layer is of water with refractive index 4/3 and thickness 8 cm.The second layer is of oil with refractive index 3/2 and thickness 9 cm while the third layer of glycerin with refractive index 2 and thickness 4cm.Find the apparent depth 1)1.4 2)1.3 3)1.2 4)1.6 60. A ray of light passes from vacuum into a medium of refractive index n . If the of the bottom of the container. angle of incidence is twice the angle of refraction, then the angle of incidence is 1) cos 1 3) 2cos 1)14 cm 2)21 cm 3)7 cm 4)13 cm n/2 1 n/2 2) sin 1 4) 2sin n/2 1 n/2 59. ACB is right angle prism with other angles as60 and30 .Refractive index of the prism is 1.5.AB has the layer of liquid on it as shown in figure. Light falls normally on the face AC. For total internal reflections, maximum refractive index of the liquid is. space for rough work Page 13 CHEMISTRY 0.02 ms 61. According to Bohr s theory the 1 respectively. The mass of B is five times that of the mass of A. What is angular momentum of electron in 5th the ratio of uncertainties ( x A / xB ) of orbit is: their positions? 1) 1.0 3) 2.5 h 2) 10 h 4) 25 h h 1) 2 2) 0.25 3) 4 4) 1 65. How many revolutions an electron will 62. The wave length of the first spectral make in 2nd shell of He in one second? line in Balmer series of spectrum of 1) 6.66 1015 2) 3.33 1015 H-atom is: 3) 2.22 1015 4) 1.11 1015 1) 3) 5R cm 36 36 cm 5R 2) 6 cm R 4) none of these 66. If H-atom is supplied with 12.1 eV energy and electron returns to the ground spectral lines in Balmer series would be: 63. Ionization energy of He is 19.6 10 18 J atom 1 . The energy of the first stationary state ( n 1 ) of Li 1) 8.82 10 17 J atom 1 2) 4.41 10 16 J atom 1 3) 4.41 10 4) 2.2 10 17 15 2 is: (use energy of ground state of H-atom = 13.6eV ) 1) 1 2) 2 3) 3 4) 4 67. The correct set of four quantum numbers for the valence electrons of rubidium 1 J atom J atom state after excitation the number of atom(Z = 37) is: 1 1) 5, 1, 0, 1 2 2) 5, 1, 1, 1 2 3) 5, 0, 1, 1 2 4) 5, 0, 0, 1 2 64. The uncertainties in the velocities of two particles, A and B are 0.05 and space for rough work Page 14 68. Considering the wavelength of ( h 6.6 10 electron and proton to be equal, the ratio of their velocities would be: (mass of e 9.1 10 proton = 1.67 10 27 31 kg , mass of kg ) Js ,1eV 1.6 10 19 2) 91 3) 15.20 4) 1835 3 108 ms 1 ) 1) 1.2375 A0 2) 12.375 A0 3) 123.75 A0 4) 12375 A0 8.46A0 . The number of electrons in this shell are: 69. The maximum number of electrons that can have principal quantum number n 3 and spin quantum 1) 2 2) 8 3) 18 4) 32 73. Consider the ground state of Cr atom 1 is: 2 (Z = 24), the number of electrons, with 1) 1 2) 3 3) 5 4) 9 azimuthal quantum numbers 1 and 2 respectively 70. The orbital angular momentum of an 1) 12 and 4 2) 16 and 5 electron revolving in an orbit is given 3) 16 and 4 4) 12 and 5 by ( h . This momentum for 1) 2 an s-electron will be given by: 74. Which of the following sets of quantum numbers represents the highest energy in an atom? n l m 1) 4 0 0 2) 3 0 0 71. The wavelength of a photon that is 3) 3 1 1 associated with 1 eV energy is: 4) 3 2 1 1) 3) 1 h 2 2 h 2 J, 72. Bohr radius of a shell in H-atom is 1) 1.6 number ms c 34 2) zero 4) 2 h 2 space for rough work s 1 2 1 2 1 2 1 2 Page 15 75. Which one of the following grouping represents a collection of isoelectronic 3) 1, 2 4) 2, 1 80. If an electron, a proton and an -particle species? (At.no of Cs 55 , Br 35 ) have same de-Broglie wavelengths, their 1) Na , Ca 2 , Mg 2 kinetic energies are related to one 2) N 3 , F , Na another as: 1) electron > proton > 3) Be, Al 3 , Cl -particle 2) proton > electron > -particle 2 4) Ca , Cs , Br 3) 76. The radius of which of the following orbits is the same as that of the first Bohr s orbit of hydrogen atom? -particle > proton > electron 4) electron = proton = -particle 81. The wavelength of limiting line (i.e., of shortest wavelength) in Balmer series of 2 1) He (n 2) 2) Li ( n 2) 3) Li 2 ( n 3) 4) Be3 ( n 2) H-atom is 77. What is the ratio of specific charge of -particle to that of proton? 1) 1 : 1 2) 4 : 1 3) 1/2 : 1 4) 1 : 1/2 1) 4758 A0 2) 3644 A0 3) 2435 A0 4) 4634 A0 82. The electrons identified by quantum numbers n and (a) n 4, 1 (b) n 4, 0 (c) n 3, 2 (d) n 3, 1 78. Which of the following does not have spherical as well as angular node? 1) 1s 2) 2s 3) 2p 4) 3d can be placed in order of increasing energy as: 79. The number of radial nodes of 3s and 2p orbitals are respectively: 1) d < b < c < a 2) b < d < a < c 3) a < c < b < d 1) 2, 0 2) 0, 2 4) c < d < b < a space for rough work Page 16 3) 3.84 10 3 m 83. Energies of subshells in H-atom depend upon 4) 1.52 10 4 m 87. The dimensions of Planck s constant are 1) n value the same as that of: 2) (n 1) energy 3) ) value value only 2) angular momentum 3) work 4) n, and m values 4) power 84. If the uncertainties of position and momentum of a particle of mass m 88. The energy of second Bohr s orbit in are equal, the uncertainty of velocity hydrogen atom is 328kJ mol 1 . The would be: energy of the third Bohr s orbit of He 1) 3) 2) 2 1 m 2 4) m is: 2 1) 583.11kJ mol 1 2) 853.11kJ mol 3) 145.78kJ mol 1 4) 511.83 kJ mol angular nodes and nodal plane for 3 pz H-atom is 103 ms 1 . Its wavelength is: in proper order are: 1) 3.98 A0 2) 103 A0 3) 103 A0 4) 398 A0 2) 1, 1, 1 3) 2, 0, 1 4) 2, 1, 1 90. The number of radial nodes for 4p 86. Uncertainty in the position of an electron (mass = 9.1 10 31 with a velocity of 300ms 1 1 89. Under certain conditions the velocity of 85. The number of spherical nodes, 1) 3, 1, 0 1 orbital is kg ) moving 1) 4 2) 3 accurate 3) 2 4) 1 upto 0.001% will be: 1) 5.76 10 3 m 2) 1.92 10 3 m space for rough work Page 17 Master JEE CLASSES Kukatpally, Hyderabad. IIT-JEE-MANIS PAPER-2 Max. Marks: 360 KEY SHEET MATHS 4 1 2 5 4 6 3 1 2 2 1 3 7 1 8 2 9 3 10 3 11 4 12 3 13 2 14 4 15 4 16 2 17 3 18 4 19 1 20 4 21 3 22 1 23 1 24 2 25 3 26 4 27 4 28 1 29 2 30 2 PHYSICS 34 4 1 35 3 36 1 31 2 32 1 33 37 3 38 3 39 2 40 2 41 4 42 2 43 3 44 2 45 2 46 4 47 4 48 1 49 3 50 4 51 1 52 2 53 2 54 3 55 4 56 3 57 2 58 1 59 2 60 3 CHEMISTRY 61 3 62 1 63 3 64 1 65 2 66 1 67 4 68 4 69 4 70 2 71 4 72 4 73 4 74 4 75 2 76 4 77 3 78 1 79 1 80 1 81 2 82 1 83 1 84 2 85 2 86 2 87 2 88 3 89 1 90 3 SOLUTIONS: MATHS 1. b2 4ac 16a 2 4 1 4a 2 2 b2 Now, max ax 2 bx c b2 2 1 4a 2. 3. 2 6. 1 4a 2 12 n 1 9 n n 2 12 n 1 9 n n 2 9an an 2an 1 5. 2 So, max value always less than 2 (when a 0 ). b2 > (a + c)2 (a b + c) (a + b + c) < 0 f( 1) f(1) < 0. n 2 12x 9 0 , are roots of x x Adding 9an 4. 4ac 4a 4ac 4a an 12an 2 2 1 6 P x Q x x a x b x c P a a Aa 2 Ba d a P b b Ab 2 Bb d b p c c Ac 2 Bc d c Ax 2 Bx C Solving we get A 0; B 1; C 0 reminder is x 2 x 2x a 2 2 x 2x 2 a 2 x 2 x 2 a 0 or x 2 2 x a 2 0 D1 0 & D2 0 & 4 4a 8 0 4 8 4a 0 & a 3 a 1 0 a 1 & a 3 a 1 a f x bf bf x af px 2 x px 2 x a2 b2 f x f x qx r qx r a b px 2 a b qx a b r 0 will have real roots is 2 a b q2 4 p a b a b r a b a b 2 0 4 pr q2 7. Divide the numerator & Denominator by x and take y x 8. If x , y , z R; x y z 4 x y 4 z; x 2 y2 15 x 6 z2 Page 2 Now 2xy x 2 4 z 2 y x2 6 z2 y2 2 z 2 8 z 10 The quadratic equation whose roots are x,y can be t2 x t2 y t xy 0 z2 4z 5 4 z t 0 Since x, y R 2 4 z 4 z2 4z 5 2 3 3z 2 8 z 4 0 9. f x 0 z 2 x a1 x a 2 x a 3 Now f x Again, f a1 a1 x as x and f x a 2 a1 a 3 a1 one root belongs to Also, f a 3 a2 0 x a3 x as x a1 a2 a3 , a1 a1 a 3 a 2 a3 0 one root belongs to a1 , a 3 10. So f x 0 has three distinct real roots. Let , be roots of x 2 bx c 0 Then b one of the roots is 2 (since , are primes and b is odd positive integer) f 2 0 2b c 4 and b c 35 b 13, c 22 Minimum value 11. 81 4 The given equation reduces to 2a b x 2 The sum of the coefficients = 0 x 1 is a root x15 x13 x11 x16 1 a b x a 2b 0 a 2b 2a b The other root 12. 13 2 f x9 x7 x5 x3 x x x8 1 x 4 1 x 2 1 x8 1 x4 1 x 2 1 x2 1 7 2 x 1 x 7 x 1 2 2 x 14 x16 2 3 0 15 13. 14. x2 log 3 2 log 2 3 x log 2 x2 log3 2 log 2 3 x 1 0 x log 3 2 x log 2 3 log 2 3 x x 6. log3 2, log 2 3, log 3 2 or x 0 log3 2 1& log 2 3 1 1 1 log 2 , log3 3 2 Page 3 Or x log1/2 3, log1/3 2 all options are correct. 15. Let x 2 x 1 t t 16. x 3 1 1 x2 0 x 1 x 2 0 2 2, , Common roots 2 , , 2 , are common roots of two equations a b c 1 if t log 2 x, the given relation becomes 1 t2 1 t 1 t t2 t 1, t 2 1 t or t t 1 t4 2 t 2t 1 2 2t 3 1 t2 1 t 2 t 1 t t 19. 3 4 a 5 a 5 2 3x 3x 2 0 x3 1 3 x 2 x 1 x 17. 18. 3 , 4 t t2 2t 1 0 1 0 2 1 R 2, 2 The sum of its roots is t1 t 2 2 1 The product of the value of x is 2t (i) if x<0, then 1 x 1 x 2 t1 t 2 |x| x 2 0 x x 2 0 2 2 2 true x<0 (ii) if 0 x 1, then 1 x 1 x 2 x x 2 2x (iii) If 1 x 2 x 1 0 2 , then x 1 1 x 2 x 1 x 2 2 2 x x x 2 x 2 x 2 x 2 2, then x 1 1 iv if x x 1 2 x 2 x 2 Required solution set is ( 20. x n ax b x 1 x 2 2 2 true x 2 , 2] .... x n Differentiate both sides w.r.t.x nx n 1 a Put x 21. x 1 ,n n 1 1 2 .... x ,n n 1 1 2 The equation tan x x n a 1 tan x a 1 2 d x dx 1 3 2 .... ..... x 1 n n 0 is true only if tanx is an integer. Since tan 2 x and a both are integers. tan x 4a 1 1 4a 1 both value are integers 2 2 2k 1 k 2 k a and a 100 Page 4 22. 23. 24. there are 9 values of a satisfied it is reciprocal equation, 1 and -1 are real roots. Write x 49 (x 1)(x 3)Q(x) Ax B If x n Z , n 2n 4 n 4 If x n K where 0 K 1 then n 2 n k n 1 4 , it is possible if K 1 2 4 n 3, 5 2 4 y 2 12 0 is a quadratic in ' x ', ' x ' is real then discriminate 25. x 26. 27. Use f(x) = x has non real roots 8x f x x b 2 min f x 2c Also g x x 2b 2 log17 log11 2 b b f(f(x)) = x also has non-real roots 2 2 2cx b 2 b2 c2 x c c 2b x 1 x x x 11 11 b2 c 2 0 .. (1) Equation (1) is defined if x 0. We can rewrite (1) as log11 x 11 x 11 2 c2 max g x , we get 2c2 b2 As min f x 28. 2 b2 max g x c2 2c 2 0 17 0 x 1 111 11 x Squaring both sides we get x 11 121 22 x x 22 x 110 x 5 or x 25 This clearly satisfies (1). Thus, sum of all the values satisfying (1) is 25. 29. 30. ( x 3) 2 y 2 13 x 3 2, y 3 or x 3 ax 2 Let f x 3, y 2 a 2 x 2 f x is negative for two integral values of x, so graph should be vertically upward parabola i.e., a 0 Let two roots of f x = 0 are 1, 2 a 1 2 a 2 then , 2 1 a 2 a a 2 2a 1,2 Page 5 PHYSICS 1 31. x 32. Refractive index of second medium relative to first is, 1 t sin i sin r 1 tan 370 4 . 3 But the refractive index of second medium B, relative to medium A, is equal to 2 . 1 4 . But 3 2 Hence, 1 v1 , v2 2 1 where v1 is velocity of light in first medium &v2 is velocity of light in second medium. Hence, speed of light in medium B is of that in medium A. Therefore, (A) is correct. Since 2 1 , it means, second medium is denser than the first medium. Hence, total 1 internal reflection can not take place. Therefore, (B) is incorrect. Since 2 1 4 , therefore, 3 2 4 3 1 . Hence (C) is wrong. 33. Critical angle C1 for interface AB of two media of refractive indices 1 and 2 is given by sin C1 = 1 . 2 1 1 . 2 If 3< 1, critical angle C2 for interface GF will be given by sin C2 = 3 . Since 3< 2 1, therefore, 3 1 2 2 . Hence C2<C1 2. Hence, total internal reflection will take place at interface GF. Therefore, (A) is correct. If Hence, 3 1 , 3 1 2 3 . sin C2> sin C1. Page 6 Since C2 > C1 1, therefore, C2 ray will get refracted into the slab. Therefore, (B) is correct. Angle of refraction (r) will be given by sin sin r 3 or sin r 2 Critical angle C3 for face DE is given by sin C3 2 sin . 3 1 . 3 2 But sin r 1 sin 3 . Hence, sin r is 2 infinitesimally greater than 1 . It means sin r 3 is infinitesimally greater than sin C3. Since angle of incidence for interface DE is equal to r, as shown in figure, therefore, total internal reflection will now take place at face DE. Hence (C) is correct. Since (A) and (C) are correct, therefore, (D) is incorrect. sin A m /2 sin A /2 34. sin A A m 2 m 2 m 35. cos 1800 cot A /2 A 2 cos A /2 sin A /2 sin 900 A/2 A 2 900 2A. Total apparent depth y = y1 + y2 = 5 + 2 = 7 cm If x is real depth = thickness of slab, then as x , x y 37. 42. y 1.5 7 10.5 cm. sin i sin r Snell s Law can be applied at two point also, but the essential condition for this is that the refracting surfaces must be parallel to each other. In that case, sin i sin r 2 1 Page 7 where 1 is refractive index of that medium in which angle of incidence is measured and 2 is refractive index of that medium in which angle r is measured. sin i sin r 43. . Hence (B) is correct. 0 When the ray suffers minimum deviation, it becomes parallel to base of prism P. As prisms Q and R are of identical shape and of the same material, therefore, the ray continues to be parallel to the base of prism Q and R. Hence, the deviation remains same as before. 44. As is clear from Fig. PQ = QK = 2h i = 45 Also, LM = MR = h KT = TQ = h LS = h2 sin r LM LS sin i sin r 45. 2h 2 h 5 h 1 h 5 5 sin 450 1/ 5 5 . 2 Note that image formation by a mirror does not depend on the medium. As P is at a height h above the mirror, image of P will be at a depth h below the mirror. If d is depth of liquid in the tank, apparent depth of P, x1 d h . apparent distance between P and its image = x2 46. x1 d h d h 2h . Here, i = 60 0 . As the ray emerges from the sphere parallel to line AB, therefore, net deviation of the ray, = 60 . But, deviation on two refractions through a sphere = 2 (i r). 2(i r) = 60 i r = 30 60o r 30o or r 30o Page 8 sin 60o sin 30o sin i sin r 47. 3 /2 1/2 3. As is clear from fig., the ray incident at 45 will emerge at 45 after reflection at the silvered surface. Therefore, angle between incident ray and emergent ray is 90 , i.e. deviation of the ray from its initial path is 90 0 . 56. sin i &d sin r 57. Conceptual 58. Net shift S= d1 1 1 i r d2 1 1 8 1 1 4/3 1 1 d3 1 2 9 1 1 2/3 3 4 1 1 2 =7 cm The shifting will be in the direction of ray travelling i.e upward Apparent depth =21-7=14 cm. Clearly ic 59. so, max imum possible value of ic is 600 g 1 60. 600 1 sin ic Conceptual Page 9 CHEMISTRY 61. (3) nh 2 m r 5h 2 For 5th shell, m r 62. 2.5 h (1) 1 1 1 , for first spectral line, 2 2 32 1 1 5 = R R cm 1 4 9 36 36 cm 5R 63. R (3) 2 Z Li 2 ELi2 E Li 2 64. (for equal n values, 1 in each case) 2 Z He 9 ( 19.6 10 4 EHe 18 ) = 4.41 10 Energy of Ist shell of Li 2 is 4.41 10 (1) ( x m v) A h / 4 1 ( x m v) B h / 4 x A mB vB 0.02 5 xB m A vA 0.05 65. 67. 68. J atom 17 1 J atom 1 2 (2) No. of revolutions of electron in nth shell in 1 second 6.66 1015 Z 2 n3 = 6. 17 6.66 1015 4 8 3.33 1015 (1) New energy = 13.6 12.1 1.5eV 13.6 13.6 n2 9 En n 3 2 1.5 n Number of spectral lines in Balmer series for 3 2 transitions would be one only. (4) Rb(Z = 37) has V.S. electronic configuration 5s1 . 1 n 5, 0 , m 0 , ms 2 (4) e p h mv e h mv p Page 10 me ve 69. 70. 71. m pv p mp me E n2 1.6 10 hc J hc hv E 34 6.6 10 Js 3 108 ms 1.6 10 19 J 76. 79. 80. 2 8.46 0.529 32 24) :1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 5 4s1 Total electrons in Cr ( Z 24) :1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 5 4s1 , i.e., p-subshell = 6 + 6 =12 2 , i.e., d-subshell = 5 Total electrons in (4) As per the (n ) rule, the choice (4) has highest value = 3 + 2 = 5 (2) N 3 , F and Na have 10 electrons each. (4) 0.529n 2 0 A Z For radius equal to Bohr s 1 st orbit of H-atom, n 2 Z Be3 (n 2) has Z = 4 and n 2 4 (3) sp.charge of -particle sp.charge of proton 78. = 12.375 10 7 m 12375 A0 (4) r 77. 1 0529n 2 Z 8.46 1 0.529 Cr (Z 75. 1835.16 19 Electrons in nth shell = 2n 2 74. 1.67 104 9.1 (4) rn 73. 1.67 10 27 9.1 10 31 (4) 3rd shell has 18 electrons, 9 have +1/2 and other 9 have 1/2 spin. (2) 0 For s-electron, h 0 Hence, ( 1) 2 (4) E 1 eV 72. ve vp (e / m) -particle (e / m) proton (2 / 4) (1/1) (1/ 2) 1 1 :1 or 1 : 2 2 (1) 1 1 0 1 0 Spherical or radial nodes for 1s n Non-spherical or angular nodes for 1s, 0 (1) (n 1) for 3s = 2 and for 2p = 0 (1) Page 11 h2 KE = 2m h 1 ; KE = mv2 mv 2 For h and 81. 1 R 4 R 83. 86. 87. 88. 1 2 4 9.11 10 8 m = 4 9.11 100 10 10 m 3644 A0 2s 2 p 3s 3 p 3d ..... h 4 1 m x (h / 4 ) m (h / 4 ) 1 h m 4 1 h m 2 [Here, h/2 ] (2) Spherical nodes for 3 pz n 1 3 1 1 1 Angular nodes for 3 pz 1 1 Nodal planes for 3 pz (2) 0.001 v 300 3 10 3 ms 1 100 h 6.6 10 34 kgm 2 s 1 x 4 m v 4 3.14 (9.1 10 31 kg ) (3 10 3 ms 1 ) 6.6 m = 0.01925 m 1.925 10 2 m = 4 3.14 9.1 3 (2) h and mvr has same units kgm 2 s 1 (3) En E1 n2 E1 22 ( 328) Energy of 3rd shell, E3 89. R 4 So, energy in H-atom is related with n value only (2) x p a h h h x p a2 a 4 4 4 h x p x m v 4 v 85. 1 22 (1) (a) is 4p, (b) is 4s, (c) is 3d, (d) is 3p (1) rule is not applicable to H-atom. Energy system is n 1s 84. 1 m (2) v 82. being constant, KE 2 E1 9 4 328 kJ mol 4 328 9 1 145.78 kJ mol 1 (1) Page 12 6.6 6.6 10 34 kgm2 s 1 10 = 24 3 3 1.66 1.66 10 10 10 1 (Mass of 1 H-atom = 1.66 10 6.02 1023 h mv 90. 10 m = 3.976 10 24 g) 10 m 3.976 A0 (3) Conceptual Page 13

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