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IIT JEE Exam 2025 : Main : exam lakshya

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Agaaz Aggarwal
Yadavindra Public School (YPS), SAS Nagar, Mohali
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DATE : 02/09/2020 Time : 3 hrs. Answers & Solutions for SEPT - JEE - 2020 Max. Marks : 300 JEE (2-SEPT)-2020 (MORNING) PART : PHYSICS SECTION 1 : (Maximum Marks : 80) Straight Objective Type This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its answer, out of which Only One is correct. 1. A particle of mass m with an initial velocity u i collides perfectly elastically with a mass 3 m at rest. It moves with a velocity v j after collision, then, v is given by (1) v 2 u 3 (2) v Ans. (4) Sol. From momentum conservation u 3 (3) v 1 6 u (D) v u 2 mu i 0 mv j 3m v ' v' u v i j 3 3 From kinetic energy conservation Solving v 2. u 2 v 2 1 1 1 mu 2 mv 2 (3m) 3 2 2 2 3 u 2 Two identical strings X and Z made of same material have tension Tx and TZ in then If their fundamental frequencies are 450 Hz and 300 Hz, respectively, then the ratio Tx/TZ is : (1) 1.25 (2) 0.44 Ans. (4) Sol. fx 1 2 Tx fy 1 2 Ty fx 450 = fy 300 (3) 1.4 Tx Ty Tx/Ty = 9/4 = 2.25 1 of 30 (4) 2.25 JEE (2-SEPT)-2020 (MORNING) 3. A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is 5 cm and the angular speed of rotation is rad s 1. The difference in the height, h (in cm) of liquid at the centre of vessel and at the will be : 5 2 2g (1) Ans. Sol. (2) 2 2 25 g (3) 25 2 2g (4) 2 2 5g (3) dr 2r gdh R h 2 rdr g dh 0 2 0 2 R gh 2 h= 4. Ans. Sol. 25 2 2R 2 = 2g 2g The least count of the main scale of a vernier callipers is 1 mm. Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the 7th division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of cylinder the zero of the vernier scale between 3.1 cm and 3.2 cm and 4th VSD coincides with a main scale division. The length of the cylinder is (VSD is vernier scale division) (1) 2.99 cm (2) 3.07 cm (3) 3.21 cm (4) 3.2 cm (2) Zero Error = 0 + 7 0.1 = 0.070 Vernier reading = (3.1 + 4 0.01) 0.07 = 3.07 2 of 30 JEE (2-SEPT)-2020 (MORNING) Ans. Sol. 6. Ans. Sol. to be ideal and the oxygen bond to be rigid, the total internal energy (in units of RT) of the mixture is : (1) 15 (2) 13 (3) 20 (4) 11 (1) f1n1RT1 f2n 2RT2 5 5 3 RT 3RT 15 2 2 2 2 Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source ( = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to : (1) 2.87 (2) 2 nm (3) 1.27 m (4) 2.05 m (3) P = dsin = d dy 10 3 1.270mm = = 1.27 m D 1m = 7. An amplitude modulated waves is represented by expression vm = 5 (1 + 0.6 cos 6280t) sin(211 104t) volts. The minimum and maximum amplitudes of the amplitude modulated wave are, respectively: 5 3 (1) 5V, 8V (2) V , 8V (3) 3V, 5V (4) V , 5V 2 2 (2) From Given Equation = 0.6 Am = Ac A max . A min A c 5 ......(1) 2 A max . A min 3 ......(2) 2 From Equation (1) + (2) Amax = 8 From Equation (1) (2) Amin = 2 Ans. Sol. Ans. K over a large distance 'r' from its center. In that region, r a small star is in a circular orbit of radius R. Then the period of revolution, T depends on R as : 1 (1) T 2 3 (2) T2 R (3) T R (4) T2 R3 R (2) Sol. M dV 8. The mass density of a spherical galaxy varies as r R0 M 0 k 4 r 2 dr r 3 of 30 JEE (2-SEPT)-2020 (MORNING) R0 M 4 k rdr 0 M 4 kR 02 = 2 kR 2 2 M R FG m GMm m 20R R 20 G T 4 kR 2 2 02R 2 R 2 2 R 0 2 KG 0 2 KG R 2 R KG T2 R 9. Ans. Sol. A beam of protons with speed 4 105 ms 1 enters a uniform magnetic field of 0.3T at an angle of 60 to the magnetic field. the pitch of the resulting helical path of protons is close to : (Mass of the proton = 1.67 10 27 kg, charge of the proton = 1.69 10 19 C) (1) 12 cm (2) 2 cm (3) 4 cm (4) 5 cm (3) Pitch = (Vcos )T 2 m = (Vcos ) eB = (4 105 cos60 ) 2 1.67 10 27 0.3 10 1.69 1019 = 4 cm 10. Ans. A uniform cylinder of mass M and radius R is to be pulled over a step of height a (a < R) by applying a force F at its centre 'O' perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of F required is : R a (1) Mg 1 R (1) 2 (2) Mg 1 a2 R 2 R (3) Mg 1 R a 4 of 30 (4) Mg a R JEE (2-SEPT)-2020 (MORNING) F R N R Sol. mg a f FR > mg cos R F > mgcos F > mg 11. 2 A bead of mass m stays at point P (a, b) on a wire bent in the shape of a parabola y = Cx2 and rotating with angular speed (see figure). The value of is (neglect friction) (1) Ans. Sol. R 2 (R a) 2 R a Mg 1 R R (2) 2 gC 2gC ab (3) (4) y m 2acos N a mg mgsin m 2a x mgcos m 2acos = mgsin g tan a y = 4cx2 dy tan = = 8xC dx (tan )a,b = 8aC = g 8ac = 2 2gc a 5 of 30 2g C (4) 2 2gC JEE (2-SEPT)-2020 (MORNING) 12. Ans. Sol. If speed V, area A and force F are chosen as fundamental units, then the dimension of Young's modulus will be : (1) FA2V 3 (2) FA2V 1 (3) FA2V 2 (4) FA 1V0 (4) F Y FaVbAc Y= A MLT 2 (M1L1T 2 )a (L1T 1 )b (L2 )c L2 M1L 1T 2 Ma La + b + 2c T 2a b a + b + 2c = 1 2a + b = 2 a = 1, b = 0, c = 1 Y = F1v0A 1 13. A charged particle (mass m and charge q) moves along X axis with velocity V0. When it passes through the origin it enters a region having uniform electric field E E j which extends upto x = d. Equation of path of electron in the region x > d is : (1) y Ans. Sol. qEd mV02 x (2) y qEd ( x d) 2 (3) y mV0 (3) y x t=0 x Vx t (x1 y) Vy Straight line x > d path is straight line 1 y at 2 at 2 = V0 x d 1 2 at x d 2 at V0 y y 1 d x d at 2 V0 V0 V0 6 of 30 qEd d x mV02 2 (4) y qEd2 mV02 x JEE (2-SEPT)-2020 (MORNING) myV0 x d qEd V0 2 V0 y qEd x d mV0 V0 2 V0 y 14. Ans. Sol. In a reactor, 2 kg of 92U235 fuel is fully used up in 30 days. The energy released per fission is 200 Mev. given that the Avogadro number, N = 6.023 1026 per kilo mole and 1 eV = 1.6 10 19 J. The power output of the reactor is close to : (1) 54 MW (2) 60 MW (3) 125 MW (4) 35 MW (2) E P t 15. qEd d x mV02 2 2 6.023 10 26 200 1.6 10 19 = 60W 235 30 24 60 60 A plane electromagnetic wave, has frequency of 2.0 1010 Hz and its energy density is 1.02 10 8 J/m3 in vacuum. The amplitude of the magnetic field of the wave is close to ( Ans. of light = 3 108 ms 1) : (1) 160 nT (2) 180 nT (1) Sol. Energy Density = (3) 190 nT 1 Nm 2 9 10 9 2 and speed 4 0 C (4) 150 nT 1 B2 2 0 B 2 0 Energy density B 2 4 10 7 1.02 10 8 160 10 9 = 160 nT 16. Ans. Sol. Train A and train B are running on parallel tracks in the opposite directions with speed of 36 km/hour and 72 km/hour, respectively. A person is walking in train A in the direction opposite to its motion with a speed of 1.8 km/hour. Speed (in ms 1) of this person as observed from train B will be close to : (take the distance between the tracks as negligible) (1) 30.5 ms 1 (2) 29.5 ms 1 (3) 31.5 ms 1 (4) 28.5 ms 1 (2) VA = 36 km/hr = 10 m/s VB = 72 km/hr = 20 m/s 7 of 30 JEE (2-SEPT)-2020 (MORNING) VMA = 1.8 km/hr = 0.5 m/s Vman, B = Vman, A + VA, B = Vman, A + VA VB = 0.5 + 10 ( 20) = 0.5 + 30 = 29.5 m/s 17. Ans. Sol. 18. Magnetic materials used for making permanent magnets (P) and magnets in a transformer (T) have different properties of the following, which property best matches for the type of magnet required ? (1) T : Large retentivity, large coercivity (2) P : Large retentivity, large coercivity (3) T : Large retentivity, small coercivity (4) T : Small retentivity, large coercivity (2) Based on theory Shown in the figure is rigid and uniform one meter long rod AB held in horizontal position by two strings tied to its ends and attached to the ceiling. The rod is of mass 'm' and has another weight of mass 2m hung at a distance of 75 cm from A. The tension in the string at A is : 100 0 25 50 75 A B 2m Ans. Sol. (1) 0.75 mg (2) 1 mg (2) net about B is zero at equilibrium (3) 0.5 mg (4) 2mg TB TA 50 25 B A mg 2mg TA 100 mg 50 2 mg 25 = 0 TA 100 = 100 mg TA = 1 mg 19. Ans. Consider four conducting materials copper, tungsten, mercury and aluminum with resistivity C, T, M and A respectively. Then : (1) C > A > T (2) A > T > C (3) A > M > C (4) M > A > C (4) 8 of 30 JEE (2-SEPT)-2020 (MORNING) Sol. m = 98 10 8 A = 2.65 10 8 C = 1.724 10 8 T = 5.65 10 8 20. Ans. Sol. A spherical mirror is obtained as shown in the figure from a hollow glass sphere. if an object is positioned infront of the mirror, what will be the nature and magnification of the image of the object? (Figure drawn as schematic and not to scale) (1) Inverted, real and magnified (3) Inverted, real and unmagnified (3) hi O C h2 F Image (2) Erect, virtual and unmagnified (4) Erect, virtual and magnified P h2 < hi SECTION 2 : (Maximum Marks : 20) This section contains FIVE (05) questions. The answer to each question is NUMERICAL VALUE with two digit integer and decimal upto one digit. If the numerical value has more than two decimal places truncate/round-off the value upto TWO decimal places. Full Marks : +4 If ONLY the correct option is chosen. Zero Marks : 0 In all other cases 21. A 5 F capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 F capacitor. If the energy change during the charge X redistribution is J then value of X to the nearest integer is : 100 Ans. 4 Sol. C1 = 5 F V1 = 220 Volt C2 = 2.5 F V2 = 0 Heat loss; H = Ui Uf = 1 c 1c 2 ( v 1 v 2 )2 2 c1 c 2 = 1 5 2 .5 (220 0 )2 J 2 ( 5 2 .5 ) = 5 22 22 100 10 6J 2 3 9 of 30 JEE (2-SEPT)-2020 (MORNING) 5 11 22 55 22 10 4 J = 10 4 J 3 3 1210 1210 = 10 4 J = 10 3J = 4 10 2 3 3 According to questions = x = 4 10 2 100 So, x = 4 22. A small block starts slipping down from a point B on an inclined plane AB, which is making an angle with the horizontal section BC is smooth and the remaining section CA is rough with a coefficient of friction . It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If BC = 2AC, the coefficient of friction is given by = ktan . The value of k is . Ans. Sol. 3 Let AC = BC = 2 AB = 3 B Smooth Rough C 3 sin ) 23. Ans. Sol. A Apply work Energy theorem Wf + Wmg = KE mg (3 )sin mgcos ( ) = 0 + 0 mgcos = 3mg sin = 3tan = ktan k=3 An engine takes in 5 moles of air at 20 C and 1 atm, and compresses it adiabatically to 1/10th of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be XkJ. The value of X to the nearest integer is : 46 T1V1 1 = T2V2 1 V T2 = T1 1 V2 7 1 1 = T1 (10 ) 5 T2 = T1(10)2/5 5 5 V nR ; 5 3[10 2 / 5 1](293 ) 2 2 625 1.5 293 461440 6 46ks 10 of 30 JEE (2-SEPT)-2020 (MORNING) 24. Ans. Sol. When radiation of wavelength A is used to illuminate a metallic surface, the stopping potential is V. When V the same surface is illuminated with radiation of wavelength 3A, the stopping potential is . If the 4 threshold wavelength for the metallic surface is n then value of n will be : 9 hc eV (1) hc eV 3 4 .(2) from (1) & (2) hc 1 3 1 eV 3 4 hc 2 3 eV 3 4 eV 8 hc 9 hc 8 hc 9 = hc hc 9 th th = 9 k=9 25. A circular coil of radius 10 cm is placed in a uniform magnetic field of 3.0 10 5 T with its plane perpendicular to the field initially. It is rotated at constant angular speed about an axis along the diameter of coil and perpendicular to magnetic field so that it undergoes half of rotation in 0.2 s. The maximum value of EMF induced (in V) in the coil will be close to the integer Ans. 15 Sol. Flux as a function of time B.A AB cos( t ) Emf induced, e d AB sin( t ) dt Max. value of Emf = AB = R2B = 3.14 0.1 0.1 3 10 5 = 15 0.2 11 of 30 JEE (2-SEPT)-2020 (MORNING) PART : CHEMISTRY SECTION 1 : (Maximum Marks : 80) Straight Objective Type This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its answer, out of which Only One is correct. 1. The figure that is not a direct manifestation of the quantum nature of atom is : (1) (2) (3) (4) Ans. Sol. (4) 1, 2 and 3 are according to quantum theory but (4) is statement of kinetic theory of gases. 2. Which one of the following graphs is not correct for ideal gas ? (1) Ans. Soln. (3) (2) (4) d = Density, P = Pressure, T = Temperature (4) For ideal gas PM = dRT PM 1 d= R T So graph between d Vs T is not straight line. 3. Ans. For the following Assertion and Reason, the correct option is Assertion (A) : When Cu (II) and sulphide ions are mixed, they react together extremely quickly to give a solid. Reason (R) : The equilibrium constant of Cu2+ (aq) + S2 (aq) CuS(s) is high because the solubility product is low. (1) (A) is false and (R) is true. (2) Both (A) and (R) are false. (3) Both (A) and (R) are true but (R) is not the explanation for (A). (4) Both (A) and (R) are true and (R) is the explanation for (A). (3) 12 of 30 JEE (2-SEPT)-2020 (MORNING) Sol. Rate of chemical reaction has nothing to do with value of equilibrium constant. 4. The major product in the following reaction is : H3C CH=CH2 H O 3 Heat OH CH3 H3C CH3 CH3 (1) CH3 H3C (2) CH3 (3) (4) CH3 Ans. Sol. (3) CH3 CH3 CH=CH2 CH+ CH3 H2O E.A.R Mark. rule acc. to 5. Ans. Sol. CH3 + Rearrangment CH3 H+ While titration dilute HCl solution with aqueous NaOH, which of the following will not be required ? (1) Clamp and phenolphthalein (2) Burette and porcelain tile (3) Bunsen burner and measuring cyclinder (4) Pipette and distilled water (3) In this acid base Titrating there is no use of Bunsen burner and measuring cylinder other laboratory equipments will be required for getting the end point of titration. Ans. On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with H2 in the presence of a catalyst gives another gas (C) which is basic in nature. (A) should not be : (1) (NH4)2Cr2O7 (2) NH4NO2 (3) Pb(NO3)2 (4) NaN3 (3) Sol. (1) Pb(NO3)2 PbO + 2NO2 6. (2) NH4NO2 N2 + 2H2O (3) (NH4)2Cr2O7 N2 + Cr2O3 + H2O (4) NaN3 3N2 + 2Na 300 C 7. Which of the following compounds will show retention in configuration on nucleophile substitution by OH ion ? Br (1) CH3 CH CH2Br C2H5 Ans. (2) CH3 CH Br (3) CH3 C H C6H13 CH3 (1) 13 of 30 (4) CH3 CH Br C6H5 JEE (2-SEPT)-2020 (MORNING) OH CH3 C CH3 C2H5 Sol. CH3 CH CH2 Br SN1 C2H5 8. C2H5 C2H5 The major aromatic product C in the following reaction sequence will be : O HBr O3 ( excess ) ( i ) KOH ( Alc .) A B C Zn / H3 O ( ii) H OH Br OH (2) (1) (3) CO2H CHO Ans. + CH3 C CH3 Rearrangement CH3 CH CH3 Br (4) CHO CO2H (1) OH O HBr Sol. Br Excess Br alc. KOH/ OH CHO + HCHO + OH O3 Zn CHO CHO (B) (C) 9. The increasing order of the following compounds towards HCN addition is : CHO H3CO CHO (ii) (i) NO2 CHO (iii) Ans. Sol. 10. Ans. O2N CHO (iv) OCH3 (1) (iii) < (i) < (iv) < (ii) (2) (i) < (iii) < (iv) < (ii) (3) (iii) < (iv) < (ii) < (i) (4) (iii) < (iv) < (i) < (ii) (1) I, M effect of NO2 increase reactivity towards nucleophilic addition reaction with HCN. In general, the property (magnitudes only) that shows an opposite trend in comparison to other properties across a period is : (1) Electronegativity (2) Electron gain enthalpy (3) Ionization enthalpy (4) Atomic radius (4) 14 of 30 JEE (2-SEPT)-2020 (MORNING) Sol. On moving Left to Right along a period. Atomic Radius decreases. Electronegativity Increases. Electron gainenthalpy Increases. Ionisation Enthalpy Increases. 11. Ans. Sol. 12. For octahedral Mn(ll) and tetrahedral Ni(ll) complexes, consider the following statements : (l) both the complexes can be high spin. (ll) Ni(ll) complex can very rarely be of low spin. (lll) with strong field ligands, Mn(ll) complexes can be low spin. (lV) aqueous solution of Mn(ll) ions is yellow in color. The correct statements are : (1) (l) and (ll) only (2) (l), (ll) and (lll) only (3) (l), (lll) and (lV) only (4) (ll), (lll) and (lV) only (2) With weak field ligands Mn(II) will be of high spin and with strong field ligands it will be of low spin. Ni(II) tetrahedral complexes will be genrally of high spin due to sp3 hybridisation. Mn(II) is of light pink color in aqueous solution. In Carius method of estimation of halogen, 0.172 g of an organic compound showed presence of 0.08 g of bromine. Which of these is the correct structure of the compound ? (2) H3C CH2 Br (1) H3C Br Ans. (3) Sol. Mole of Bromine = M 13. Ans. Sol. (4) 0.08 10 3 mole 80 Molar mass of compound = Molar mass of (3) 0.172 10 3 0.172 10 3 M 172 gm = 80 + 72 + 6 + 14 = 172 gm An open beaker of water in quilibrium with water vapour is in a sealed container. When a few grams of glucose are added to the beaker of water, the rate at which water molecules : (1) leaves the vapour decreases (2) leaves the solution decreases (3) leaves the vapour increases (4) leaves the solution increases (3) The vapour pressure of solution will be less than vapour pressure of pure solvent, so some vapour molecules will get condensed to maintain new equilibrium. 15 of 30 JEE (2-SEPT)-2020 (MORNING) 14. Ans. Sol. The statement that is not true about ozone is : (1) in the stratosphere, CFCs release chlorine free radicals (Cl) which reacts with O 3 to give chlorine dioxide radicals. (2) in the stratosphere, it forms a protective shield against UV radiation. (3) it is a toxic gas and its reaction with NO gives NO2 (4) in the atmosphere, it is depleted by CFCs. (1) In presence of sunlight CFC s molecule divides & release chlorine free radical, which react with ozone give chlorine monoxide radical ( ClO ) and oxygen. UV CF2Cl2(g) C l(g) + C F2Cl(g) Cl (g) + O3 (g) ClO (g) + O2(g) ClO (g) + O(g) Cl (g) + O2(g) 15. Ans. Soln 16. Ans. Sol. 17. Ans. Sol. 18. Ans. The metal mainly used in devising photoelectric cells is : (1) Li (2) Rb (3) Cs (4) Na (3) Cesium has lowest ionisation enthalpy and hence it can show photoelectric effect to the maximum extent hence it is used in photo electric cell. Which of the following is used for the preparation of colloids ? (1) Ostwald process (2) Van Arkel Method (3) Mond Process (4) Bredig s Arc Method (4) Bredig s Arc Method is used for preparation of colloidal sol s of less reactive metal like Au, Ag, Pt. Consider that d6 metal ion (M2+) forms a complex with aqua ligands, and the spin only magnetic moment of the complex is 4.90 BM. The geometry and the crystal field stabilization energy of the complex is : (1) tetrahedral and 1.6 t + 1P (2) octahedral and 2.4 0 + 2P (3) tetrahedral and 0.6 t (4) octahedral and 1.6 0 (3) Since spin only magnetic moment is 4.90 BM so number of unpaired electrons must be 4. so If the complex is octahedral, then it has to be high spin complex with configuration t2g2,1,1eg1,1, in that case CFSE = 4 X (-0.4 0) + 2 X 0.6 0 = -0.4 0 If the complex is tetrahedral then its electronic configuration will be = eg2,1t2g1,1,1 and CFSE will be = 3 X (-0.6 t) + 3 X (0.4 t) = -0.6 t If AB4 molecule is a polar molecule, a possible geometry of AB4 is : (1) Square pyramidal (2) Rectangular planar (3) Square planar (4) Tetrahedral (1) 16 of 30 JEE (2-SEPT)-2020 (MORNING) Soln. For AB4 compound possible geometry are S. No. 1 2 3 Bond pair Lone pair 4 0 4 1 4 2 Total 4 5 6 Hybridisation Geometry SP3 Tetrahedral SP3d Sea-saw sp3d2 Square Planar Polarity non polar Polar non polar Square pyramidal can be polar due to lone pair moment as the bond pair moments will get cancelled out. 19. Ans. The IUPAC name for the following compound is : (1) 2, 5-dimethyl-6-carboxy-hex-3-enal (3) 6-formyl-2-methyl-hex-3-enoic (4) 6 CHO (2) 2, 5-dimethyl-5-carboxy-hex-3-enal (4) 2, 5-dimethyl-6-oxo-hex-3-enoic acid 3 Sol. 20. 5 2 (2, 5-dimethyl-6-oxo-hex-3-enoic acid COOH 1 Consider the following reactions : 4 x eq. of dry HCl (i) Glucose + ROH Acetal acetyl derivative ( CH3 CO )2 O y eq. of Ni / H2 (ii) Glucose A acetyl derivative ( CH3CO )2 O z eq. of (iii) Glucose acetyl derivative ( CH3CO )2 O Ans. Sol. x , y and z in these reactions are respectively. (1) 5, 6 & 6 (2) 4, 5 & 5 (3) 4, 6 & 5 (3) 17 of 30 (4) 5, 4 & 5 JEE (2-SEPT)-2020 (MORNING) COOH COOH HNO3 Oxidation (CHOH)4 (Ac2)O COOH 4-mole (Ac2)O require COOH CHO (CHOH)4 (CH OAc)4 CHO (Ac2)O (CHOAc)4 5-mole (Ac2)O require CH2OH CH2OAc CH2OAc CH2OH H2/Ni (CHOH)4 (Ac2)O CH2OH (CH OAc)4 CH2OAc 6-mole (Ac2)O require SECTION 2 : (Maximum Marks : 20) This section contains FIVE (05) questions. The answer to each question is NUMERICAL VALUE with two digit integer and decimal upto one digit. If the numerical value has more than two decimal places truncate/round-off the value upto TWO decimal places. Full Marks : +4 If ONLY the correct option is chosen. Zero Marks : 0 In all other cases 21. The mass of gas adsorbed, x, per unit mass of adsorbate, m, was measured at various pressures, p. A Ans. x and log p gives a straight line with slope equal to 2 and the intercept equal to m x at a pressure of 4 atm is : 0.4771. The value of m (Given log 3 = 0.4771) (48) graph between log 18 of 30 JEE (2-SEPT)-2020 (MORNING) Soln. 1 x = k(P) n m x log = logk + logP n m Slope = = 2 So n = . 2 n Intercept logk = 0.477 So k = Antilog (0.477) = 3 1 x So k(P) n m = 3[4]2 = 48 x log m Slope=2 logk=0.477 logP 22. The oxidation states of iron atoms in compounds (A), (B) and (C), respectively, are x, y and z. The sum of x, y and z is Na4 [Fe(CN)5 (NOS)] Na4 [FeO4 ] [Fe 2 (CO)9 ] Ans. Sol. (6) The oxidation states of iron in these compounds will be A = +2 B = +4 C=0 The sum of oxidation states will be = 6. 23. The internal energy change (in J) when 90 g of water undergoes complete evaporation at 100 C is (Given : Hvap for water at 373 K = 41 kJ/mol, R = 8.314 JK 1 mol 1) (189494) H = U + ngRT 41000 X 5 = U + 5 X 8.314 X 373 205000 = U + 15505.61 U = 189494.39 J = 189494 J (A) Ans. Sol. (B ) (C) 24. The Gibbs energy change (in J) for the given reaction at [Cu2+] = [Sn2+] = 1 M and 298 K is : Cu(s) + Sn2+(aq.) Cu2+ (aq.) + Sn(s); 0 0 = 0.16 V, ECu = 0.34 V, Take F = 96500 C mol 1) ( ESn 2 2 |Cu |Sn Soln 96500 0 E 0cell E Sn E 0Cu2 / Cu 2 / Sn = 0.16 0.34 = 0.50V G0 = nF E 0cell = 2 96500 ( 0.5) = 96500J = 96.5 KJ = 96500 J 19 of 30 JEE (2-SEPT)-2020 (MORNING) 25. The number of chiral carbons present in the molecule given below is Ans. (5) * * Sol. * * * 20 of 30 JEE (2-SEPT)-2020 (MORNING) PART : MATHEMATICS SECTION 1 : (Maximum Marks : 80) Straight Objective Type This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its answer, out of which Only One is correct. 1. The sum of the first three terms of a G.P. is S and their product is 27. Then all such S lie in (1) ( , 3] [9, ) (2) ( , 9] [3, ) (4) ( , 9] Ans. (3) [ 3, ) (1) Sol. Let terms are a , a, ar r a=3 then a3 = 27 3 Now + 3 + 3r = S r 1 3 r 3 S r r+ 1 2 r 1 3 r 3 ( , 3] [9, ) r 2. Box I contains 30 cards numbered I to 30 and Box II contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box I is : (1) 2 3 (2) 2 5 Ans. (3) Sol. P(B1) = 3. Area (in sq. units) of the region outside Ans. (1) 3( 2) (3) (3) 8 17 (4) 4 17 1 = P(B2) 2 P(Non-prime) = P(B1). P(N.P/B1) + P(B2). P(N.P/B2) 1 20 1 15 = . . 2 30 2 20 1 20 . 8 2 30 P(B1/N.P) = 1 20 1 15 17 . . 2 30 2 20 (2) 3(4 ) x2 y2 |x| |y| 1 is : 1 and inside the ellipse 4 9 2 3 (3) 6( 2) 21 of 30 (4) 6(4 ) JEE (2-SEPT)-2020 (MORNING) (0, 3) Sol. ( 2, 0) (2, 0) (0, 3) Area of Ellipse = ab = 6 |x| |y| Area enclose by 1 is 2 3 1 1 (d1 d2) = (4)(6) = 12 2 2 so required Area is = 6 12 = 4. | x | 5 The domain of the function f(x) = sin 1 2 is ( , a] [a, ), Then a is equal to : x 1 (1) Ans. Sol. 1 17 2 (2) 17 1 2 (3) 17 2 (4) 17 1 2 (1) | x | 5 x2 1 1 |x| + 5 x2 + 1 x2 |x| 4 0 Let |x| = t | x | 17 1 2 t2 t 4 0 | x | 17 1 0 2 17 1 17 1 , x , 2 2 a= 5. If |x| < 1, |y| < 1 and x y, then the sum to infinity of the following series (x + y) + (x 2 + xy + y2) + (x3 + x2y + xy2 + y3) + .. is : (1) Ans. Sol. 1 17 2 x y xy (1 x )(1 y ) (2) x y xy (1 x )(1 y ) (3) x y xy (1 x )(1 y ) (2) (x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3 ) + .. = x y xy 1 x 2 y 2 1 x 2 x 2 y y 2 xy 2 (1 x)(1 y) x y 1 x 1 y x y (1 x)(1 y) 22 of 30 (4) x y xy (1 x )(1 y ) JEE (2-SEPT)-2020 (MORNING) 3 6. 2 2 1 sin i cos 9 9 is : The value of 2 2 1 sin i cos 9 9 (1) 1 ( 3 i) 2 (2) 1 ( 3 i) 2 (3) 1 (1 i 3 ) 2 Ans. (1) Sol. 5 5 5 5 2 5 i sin i2 sin . cos 2 cos 1 cos 18 18 36 36 36 = 5 5 5 5 2 5 i sin i 2 sin . cos 1 cos 2 cos 18 18 36 36 36 3 1 (4) (1 i 3 ) 2 3 3 5 5 6 i sin cos 36 36 cos 5 i sin 5 = 5 5 36 36 i sin cos 36 36 5 5 5 5 = cos 6 + i sin i sin 6 = cos 36 36 6 6 7. Let y = y(x) be the solution of the differential equation, and 3 1 i 2 2 2 sinx dy . = cosx, y > 0, y(0) = 1. If y( ) = a y 1 dx dy at x = is b, then the ordered pair (a, b) is equal to : dx 3 (1) 2, 2 Ans. (4) Sol. dy cos x dx 1 y 2 sinx (2) (1, 1) (3) (2, 1) (4) (1, 1) n(1 + y) = n(2 + sinx) + nc (1 + y)(2 + sinx) = c 4.1=c c=4 4 4 1+y= y= 1 2 sinx 2 sinx y( ) = 2 1 = 1 = a 4 dy . cos x = 1 at x = b = 1 = dx (2 s inx ) 2 (a, b) = (1, 1) 8. Ans. Sol. The plane passing through the points (1, 2, 1), (2, 1, 2) and parallel to the line, 2x = 3y, z = 1 also passes through the point : (1) (2, 0, 1) (2) (0, 6, 2) (3) (0, 6, 2) (4) ( 2, 0, 1) (4) Plane passes through (2, 1, 2) is a(x 2) + b(y 1) + (z 2) = 0 23 of 30 JEE (2-SEPT)-2020 (MORNING) it also passes through (1, 2, 1) a + b c = 0 a b+c=0 .(1) given line x y z 1 is parallel to (1) 3 2 0 3a + 2b + c0 = 0 a b c 0 2 3 0 2 3 a b c 2 3 2 3 a b c 2 3 5 9. plane is 2x 4 3y + 3 5z + 10 = 0 2x 3y 5z + 9 = 0 Let > 0, > 0 be such that 3 + 2 = 4. If the maximum value of the term independent of x in the binomial expansion of x1/ 9 x 1/ 6 Ans. Sol. 10 (1) 176 (2) 336 (2) Tr + 1 = 10Cr ( x1/9)10 r ( x 1/6)r is 10k, then k is equal to : (3) 352 (4) 84 10 r r 9 6 Tr + 1 = 10Cr 10 r r x Term independent of x 10 r r =0 9 6 T5 = 10C4 6 4 Now Let 3, 2 A G 10. Ans. r=4 are 2 numbers. 3 2 ( 3 2)1/2 2 3 2 4 6 4 16 T5 10 C 4 6 T5 16 . 10C4 T5 max = 16 10C4 = 10 K K = 336 If R = {(x, y) : x, y Z, x2 + 3y2 8} is a relation on the set of integers Z, then the domain R 1 is ; (1) { 1, 0, 1} (2) { 2, 1, 1, 2} (3) {0, 1} (4) { 2, 1, 0, 1, 2} (1) 24 of 30 JEE (2-SEPT)-2020 (MORNING) Sol. x2 y2 1 8 8/3 0, 8 3 ( 2 2, 0) (2 2, 0) 0, 8 3 Domain of R 1 Range of R Value of y { 1, 0, 1} 11. Ans. Sol. If p(x) be a polynomial of degree three that has a local maximum value 8 at x = 1 and a local minimum value 4 at x = 2; then p(0) is equal to (1) 24 (2) 12 (3) 6 (4) 12 (2) Clearly P'(x) = (x 1) (x 2) where > 0 x3 3x 2 P(x) = 2x C 2 3 1 3 given P(1) = 8 2 + C = 8 3 2 5 C 8 6 (i) 8 also P(2) = 4 6 4 C 4 3 2 C = 4 3 (ii) By (i) and (ii) C = 12 P(0) = 12 12. A line parallel to the straight line 2x y = 0 is tangent to the hyperbola x2 y2 = 1 at the point (x1, y1). 4 2 Then x12 5y12 is equal to : Ans. Sol. (1) 10 (2) 5 (4) Tangent at (x1, y1) xx1 2yy1 4 = 0 This is parallel to 2x y = 0 x1 =2 2 y1 x1 = 4y1 Point (x1, y1) lie on hyperbola. x 12 x2 1 1 = 0 4 2 (3) 8 .(1) .(2) 25 of 30 (4) 6 JEE (2-SEPT)-2020 (MORNING) On solving eq. (1) and (2) We get x12 5y12 = 6 13. Ans. Sol. 14. Ans. The contrapositive of the statement If I reach the station in time, then I will catch the train is: (1) If will catch the train, then I reach the station in time. (2) If do not reach the station in time, then will not catch the train. (3) If I do not reach the station in time, then I will catch the train. (4) If I will not catch the train, then I do not reach the station in time. (4) Contrapositive of p q is ~q ~p Let P(h, k) be a point on the curve y = x2 + 7x + 2, nearest to the line, y = 3x 3. Then the equation of the normal to the curve at P is : (1) x + 3y 62 = 0 (2) x + 3y + 26 = 0 (3) x 3y 11 = 0 (4) x 3y + 22 = 0 (2) Common normal y = 3x 3 Sol. P(h,k) Tangent at p(h, k) will be parallel to given line dy = 2h + 7 = 3 dx (h,k ) h = 2 Point P lies on curve K = ( 2)2 7 2 + 2 = 8 Normal at P( 2, 8), normal slop = 1 3 x + 3y + 26 = 0 15. Ans. Sol. Let A be a 2 2 real matrix with entries from {0, 1} and |A| 0. Consider the following two statements ; (P) If A I2, then |A| = 1 (Q) If |A| 1, then tr(A) = 2 where I2 denotes 2 2 identity matrix and tr(A) denotes the sum of the diagonal entries of A. Then : (1) (P) is true and (Q) are false (2) Both (P) and (Q) are true (3) Both (P) and (Q) are false (4) (P) is false and (Q) is true (2) a b Let A = c d |A| = ad bc 0 ad = 1, bc = 0 (P) If A I2 (Q) If A = I a, b, c, d {0, 1} or ad = 0, bc = 1 ad 1 ad = 0, bc = 1 ad = 1 ad = 1, bc = 0 26 of 30 |A| = 1 tr(A) = 2 (P) is true. (Q) is true. JEE (2-SEPT)-2020 (MORNING) 16. Ans. Sol. Let S be the set of all R for which the system of linear equations 2x y + 2z = 2 x 2y + z = 4 x + y + z = 4 has no solution. then the set S (1) is a singleton (2) contains exactly two elements (3) contains more than two elements (4) is an empty set (2) For no. solution = 0 and at least one of 1, 2, 3 is non-zero. 2 1 2 = 1 2 = ( 1) (2 + 1) 1 1 2 1 2 1 = 4 2 = 2( 2 + 6 4) 4 1 =0 = 1, Hence, S = {1, 17. Ans. Sol. 1 2 1 } 2 3 1 If the tangent to the curve y = x + siny at a point (a, b) is parallel to the line joining 0, and , 2 , 2 2 then (1) |b a| = 1 (2) b = +a (3) |a + b| = 1 (4) b = a 2 (1) y = x + siny 1 0 1 dy = = 2 =1 3 1 cos y dx 2 2 cos y = 0 y = (2n + 1) 2 Point lie on curve b = a + sin y b a = sin y |b a| = 1 18. Ans. If a function f(x) defined by aex be x , 1 x 1 f ( x) cx 2 , 1 x 3 ax 2 2cx , 3 x 4 be continuous for some a, b, c R and f'(0) + f'(2) = e, then the value of a is : 1 e e e (1) 2 (2) 2 (3) 2 (4) 2 e 3e 13 e 3e 13 e 3e 13 e 3e 13 (2) 27 of 30 JEE (2-SEPT)-2020 (MORNING) Sol. Continuous at x = 1, 3 f(1) = f(1+) + f(3) = f(3 ) From (1) and (2) b = ae(3 e) ae + be 1 = c 9c = 9a + 6c ..(1) c = 3a ..(2) ..(3) ae x be x 1 x 1 2cx 1 x 3 f (x) = 2ax 2c 3 x 4 f (0) = a b, f (2) = 4c Given f'(0) + f'(2) = e a b + 4c = e by using eq. (1), (2), (3) & (4) e a= 2 e 3e 13 19. Ans. Sol. ..(4) Let and be the roots of the equation, 5x2 + 6x 2 = 0. If Sn = n + n, n = 1, 2, 3, .., then : (1) 6S6 + 5S5 = 2S4 (2) 5S6 + 6S5 = 2S4 (3) 5S6 + 6S5 + 2S4 = 0 (4) 6S6 + 5S5 + 2S4 = 0 (2) 5 2 + 6 = 2 5S6 + 6S5 = 4(5 2 + 6 ) + 4(5 2 + 6 ) = 2( 4 + 4) = 2S4 Ans. Let X = {x N : 1 1 17} and Y = {ax + b : x X and a, b R, a > 0}. If mean and variance of elements of Y are 17 and 216 respectively then a + b is equal to : (1) 7 (2) 9 (3) 7 (4) 27 (3) Sol. B x ax b 20. a 1 2 3 .... 17 17 a.(17.18) b 17 17.2 9a + b = 17 b 17 .(i) 2 17.18.35 17.18 6.17 2.17 = 105 81 = 24 2B = a2 A2 = a2.24 = 216 216 a2 = 9 24 a = 3 b = 17 27 b = 17 27 b = 10 a+b= 7 28 of 30 JEE (2-SEPT)-2020 (MORNING) SECTION 2 : (Maximum Marks : 20) This section contains FIVE (05) questions. The answer to each question is NUMERICAL VALUE with two digit integer and decimal upto one digit. If the numerical value has more than two decimal places truncate/round-off the value upto TWO decimal places. Full Marks : +4 If ONLY the correct option is chosen. Zero Marks : 0 In all other cases 21. Ans. Sol. 2 2 2 2 Let a, b and c be three unit vectors such that a c a c 8 . Then a 2b a 2c is equal to (02.00) a | b | c 1 2 a b | a c |2 8 a . b a . c 2 2 2 a 2b a 2c = 2| a |2 + 4| b |2 + 4| c |2 + 4( a . b a . c ) = 2 Now x x 2 x 3 ... x n n = 820, (n N) then the value of n is equal to . x 1 22. If lim Ans. (40.00) Sol. lim x 1 x x 2 x3 .... xn n 0 820 x 1 x 1 0 lim 1 2x x 1 3x 2 ..... nxn 1 820 1 1 + 2 + 3 + + n = 820 n(n 1) 820 2 n2 + n 1640 n2 + n 1640 = 0 n = 40 n N 29 of 30 JEE (2-SEPT)-2020 (MORNING) 23. Ans. Sol. The number of integral values of k for which the line, 3x + 4y= k intersects the circle, x2 + y2 2x 4y + 4 = 0 at two distinct points is . (09.00) If line cuts circle then p < r Centre of circle (1, 2), r = 1 3 8 k <1 5 k (6, 16) k = 7, 8, 9, 10, 11, 12, 13, 14, 15, . 24. Ans. Sol. If the letters of the word 'MOTHER' be permuted and all the words so formed (with or without meaning) be listed as in a dictionary, then the position of the word 'MOTHER' is (309.00) MOTHER 3 4 6 215 2 5! + 2 4! + 3 3! + 2! + 1 = 240 + 48 + 18 + 2 + 1 = 309 2 25. The integral || x 1| x | dx is equal to ; 0 Ans. Sol. (01.50) 2 x 1 x dx = 0 = 1 1 x x dx 0 2 | x 1 x dx 1 1 2x dx 2x 1 dx dx 1/ 2 1 0 1/ 2 = x x2 x 1 2 0 2 x 2 1 1 1 2 2 x 1 = 1 1 1 1 1 1 3 (1 1) 2 1 = 1 = 2 4 4 4 2 4 2 30 of 30

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