Jee - Main_Final_ 24 - Feb - 202 1 _Shift - 01 PHYSICS Max Marks: 100 (SINGLE CORRECT ANSW ER TYPE) This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and – 1 in all other cases. 1. In the given figure, the energy levels of hydrogen atom have been shown along with some transitions marked A, B, C, D and E. The transitions A, B and C respectively represent : 1) The first member of the Lyman series, third member of Balmer series and second member of Paschen series. 2) The series limit of Lyman series, second member of Balmer series and second member of Paschen series. 3) The ionization potential of hydrogen, second member of Balmer series and third member of Paschen series 4) The series limit of Lyman series, third member of Balmer series and second member of Paschen series Key: 4 Sol: A  S eries limit of laymen B  3 rd line of Balmer C  2 nd line of paschan Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 2 | P a g e 2. A cube of side ‘a’ has point charges +Q located at each of its vertices except at the origin where the charge is – Q. The field at the cen tre of cube is : 1)   2 0 ˆ ˆ ˆ 3 3 Q x y z a     2)   2 0 2 ˆ ˆ ˆ 3 3 Q x y z a     3)   2 0 2 ˆ ˆ ˆ 3 3 Q x y z a    4)   2 0 ˆ ˆ ˆ 3 3 Q x y z a    Key: 2 Sol: If we consider two point charges +q and – q at position of – q charge, then after interchanging – q charge with +q charge, net electric field at centre of cube is zero due to symmetry. Now remaining charges are – 2q so net electric field at centre is 2 8 3 kq a        3. A current through a wire depends on time as 2 0 i t t     where 0 20 /  A s  and 2 8   As  . Find the charge crossed through a section of the wire in 15 s. 1) 2100 C 2) 2250 C 3) 260 C 4) 11250 C Key: 4 Sol:   2 20 8   dq t t dt   15 2 0 20 8     dq t t dt 15 2 3 0 8 20 2 3          t t q     2 3 20 15 8 15 2 3     11250   q C 4. Consider two satellites 1 S and 2 S with periods of revolution 1 hr. and 8 hr respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite 1 S to be 1) 2 : 1 2) 1 : 8 3) 1 : 4 4) 8 : 1 Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 3 | P a g e Key: 1 Sol: r atio of time period 1 2 T 1 T 8  1 2 2 1 2 8      1 2 8    5. Match List I with List II. List I List II a) Isothermal i) Pressure constant b) Isochoric ii) Temperature constant c) Adiabatic iii) Volume constant d) Isobaric iv) Heat content is constant Choose the correct answer from the options given below : 1)                 a ii , b iv , c iii , d i     2)                 a iii , b ii , c i , d iv     3)                 a i , b iii , c ii , d iv     4)                 a ii , b iii , c iv , d i     Key: 4 Sol: P - iv ; Q - iii ; R - ii ; S - i 6. Given below are two statements : Statement I : Two photons having equal linear momenta have equal wavelengths. Statement II : If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease. In the light of the above statements, choose the correct answer from the options given below 1) Both Statement I and Statement II are false 2) St atement I is false but Statement II is true 3) Both Statement I and Statement II are true 4) Statement I is true but Statement II is false Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 4 | P a g e Key: 4 Sol:  h p  7. If the velocity time graph has the shape AMB, what would be the shape of the corresponding acceleration - time graph? 1) 2) 3) 4) Key: 1 Sol:    v mt C   dv m dt    v mt C dv m dt 8. In a Young’s double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit width. Find the ratio of the maximum to the minimum intensity in the interference pattern 1) 2 : 1 2) 1 : 4 3) 3 : 1 4) 4 : 1 Key: 4 Sol: 1 2 1 3  A A 1 2 , 3   A x A x Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 5 | P a g e         2 2 1 2 max 2 2 min 1 2 4 16 4 :1 4 2             A A x I I x A A 9. Each side of a box made of metal sheet in cubic shape is ‘a’ room temperature ‘T’, the coefficient of linear expansion of the metal sheet is ‘a’. The metal sheet is heate d uniformly, by a small temperature T  , so that its temperature is T T   . Calculate the increase in the volume of the metal box. 1) 3 4 3  a T   2) 3 3  a T  3) 3 4  a T   4) 3 4  a T  Key: 2 Sol: V T V    3 T   3 V 3a T    10. In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new am plitude of oscillation will be : 1) M m A M  2) M A M m  3) M A M m  4) M m A M  Key: 2 Sol: Velocity at mean position is A   Conserving momentum   1 0 MA M m V      1 0 MA K V A' M m M m        K MA M m M M A ' A M m K M m       Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 6 | P a g e 11. The work done by a gas molecule in an isolated system is given by, 2 2 x W e KT    , where x is the displacement k is the Boltzmann constant and T is the temperature,  and  are constants. Then the dimensions of  will be : 1) 0 0 M LT     2) 2 2 M LT     3) 2 MLT      4) 2 2 ML T      Key: 3 Sol: 2 x KT   dimensions 2 L KT   2 2 1 2 2 1 2 2 L L M L T M T v M        work = 2   (dimensionless) 1 1 2 1 1 2 2 M L T .L M T     3KT v M  2 v .M KT 3  1 2 M LT    12. The focal length f is related to the radius of curvature r of the spherical convex mirror by : 1) 1 f r 2   2) 1 f r 2   3) f r  4) f r   Key: 2 Sol: R f 2  13. n mole of a perfect gas and undergoes a cyclic process ABCA (see figure) consisting of the following processes. A B  : Isothermal expansion at temperature T so that the volume is doubled from 1 V to 2 1 V 2V  and pressure changes from 1 P to 2 P B C  : Isobaric compression at pressure 2 P to initial volume 1 V Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 7 | P a g e C A  : Isochoric change leading to change of pressure from 2 P to 1 P Total work done in the complete cycle ABCA is : 1) 1 nRT ln 2 2        2) 0 3) nRT ln 2 4) 1 nRT ln 2 2        Key: 4 Sol: AB 1 1 W 2P V n2   BC 1 1 W P V   CA W 0    ABCA 1 1 1 1 W 2P V n2 P V      nRT 2 n2 1    14. If Y, K and  are the values of Young’s modulus, bulk and modulus of rigidly of any material respectively. Choose the correct relation for those parameters. 1) 2 9K Y N / m 2 3K     2) 2 3YK N / m 9K Y    3) 2 Y K N / m 9 3Y     4) 2 9K Y N / m 3K     Key: 3 Sol: 15. Moment of inertia (M. I ) of four bodies, having same mass and radius, are reported as; 1 I  M.I. of thin circular ring about its diameter, 2 I  M.I. of circular disc about an axis perpendicular to disc and going through the centre, 3 I  M.I of solid cylinder about its axis and 4 I  M.I. of solid sphere about its diameter. Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 8 | P a g e 1) 1 3 2 4 I I I I    2) 1 2 3 4 5 I I I I 2    3) 1 2 3 4 I I I I    4) 1 2 3 4 I I I I    Key: 4 Sol: 2 1 MR I 2  2 2 MR I 2  2 3 MR I 2  2 4 2 I MR 5  16. Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be: 1) 2 : 1 2) 1 : 2 3) 4 : 1 4) 1 : 4 Key: 4 Sol: 1 2 C 1 C 4  17. Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be : 1)   G 1 2 2  2)   1 2 2 G 2  3)   G 2 2 1 2  4)   G 1 2 2 2  Key: 2 Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 9 | P a g e Sol: Net force on one particle net 1 2 F F 2F cos 45º centripetal force        2 2 2 2 2 GM 2GM MV cos 45º R 2R 2R                  1 GM 1 V 1 2 2 V G 1 2 2 2 R 2     18. A cell 1 E of emf 6V and internal resistance 2  is connected with another cell 2 E of emf 4 V and internal resistance 8  (as shown in the figure). The potential difference across points X and Y is : 1) 10.0 V 2) 3.6 V 3) 5.6 V 4) 2.0 V Key: 3 Sol: Current 6 4 1 I A 10 5    x y 1 v 4 8 v 5     x y v v 5.6v    Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 10 | P a g e 19. If an emitter current is changed by 4 mA, the collector current changes by 3.5 mA. The value of  will be : 1) 0.875 2) 3.5 3) 7 4) 0.5 Key: 3 Sol: E I 4   C I 3.5   C E I 3.5 7 I 4 8                   7 8 7 7 1 1 8         20. Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is: 1) 3 1 d 2 3Gm  2) 3 3Gm 2 d  3) 3 1 3Gm 2 d  4) 3 d 2 3Gm  Key: 4 Sol:     2 2 G m 2m 2d m d 3    2 2 2Gm 2d d 3    2 3 3Gm d   3 3 3Gm d ;T 2 d 3Gm     (NUMERICAL VALUE TYP E) This section contains 10 questions. Each question is numerical value type. For each question, enter the correct numerical val ue (in decimal notation, truncated/rounded - off to second decimal place. (e.g. 6.25, 7.00, 0.33, 30, 30.27, 127.30). Attempt any fi ve questions out of 10. Marking scheme: +4 for correct answer, 0 if not attempted and 0 in all other cases. 21. A ball with a speed of 9 m/s collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of 30 º with the original direction. The ratio of velocities of the balls after collision is x : y, where x is ______ Key: 1 Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 11 | P a g e Sol: Using linear momentum conservation in y - direction i P 0  f 1 2 1 1 P m v m v 2 2     1 2 v v  22. A common transistor radio set requires 12 V (D.C) for its operation. The D.C. source is constructed by using a transformer and a rectifier circuit, which are operated at 220V (A.C) on standard domestic A.C. supply. The numbe r of turns of secondary coil are 24, then the number of turns of primary are _______ Key: 440 Sol: P P S S N V N V  P P N 220 ; N 440 24 12   23. The coefficient of static friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall will be _______ N. 2 g 10ms       Key: 25 Sol: F = N, f = 0.2× N 0.2N 5  N 25  24. An unpolarized light beam is incident on the polarizer of a polarization experiment and the intensity of light beam emerging from the analyzer is measured as 100 Lumens. Now, if the analyzer is rotated around the horizontal axis (direction of light) by 30º in clockwise direction the intensity of emerging light will be ______Lumens. Key: 75 Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 12 | P a g e Sol: 2 2 0 o o I I cos I cos 30    o 3 I 4        =75 25. A hydraulic press can lift 100 kg when a mass ‘m’ is placed on the smaller piston. It can lift ________ kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass ‘m’ on the smaller piston. Key: 25600 Sol: Initially   1 2 100g mg .......... i A A  Initially   2 1 Mg mg .......... ii A 16A 16        100 16 1 M 25600kg M 16     26. A resonance circuit having inductance and resistance 4 2 10 H   and 6.28  respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is ______.   3.14   Key: 200 Our key is : 2000 Sol: L x L 2 fL Q R R R      6 4 2 10 10 2 10 Q 2000 6.28         Q 2000  27. An electro magnetic wave of frequency 5 GHz is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are 2. Its velocity in this medium is________ 7 10 m / s  Key: 15 Sol: r r n 2     8 7 C 3 10 v 15 10 m / s n 2      x 15  Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 13 | P a g e 28. In connection with the circuit drawn below , the value of current flowing, through 2k  resistor is _______ 4 10 A   Key: 2 5 Sol: Zener diode breakdown 3 3 5 i 2.5 10 2 10        4 3 x 10 2.5 10      x 2.5mA  29. An inclined plane is bent in such a way that the vertical cross – section is given by 2 x y 4  where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction 0.5   , the maximum height in cm at which a stationary block will not slip downward is _______cm. Key: 25 Sol: dy dx x tan dx 4 2       x 0.5 2  x 1  4y 1  2 y 1  1 y 4  30. An audio signal   m v 20sin 2 1500t   amplitude modulates a carrier   c v 80sin 2 100,000t   The value of percent modulation is __________ Key: 25 Sol: m c A 20 m% 100 100 25% A 80      Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 14 | P a g e CHEMISTRY Max Marks: 100 (SINGLE CORRECT ANSW ER TYPE) This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and – 1 in all other cases. 3 1. Which of the following ore is concentrated using group 1 cyanide salt ? 1) Calamine 2) Sideri te 3) Sphalerite 4) Malachite Key: 3 Sol: NaCN is used as depressant in froth flotation process for the separation of galena & Sphalerite 3 2. 2 3 Al O was leached with alkali to get X. The solution of X on passing of gas Y, forms Z. X, Y and Z respectively are : 1)   2 2 3 2 3 , , X Al OH Y SO Z Al O xH O    2)   2 2 3 2 4 , , X Na Al OH Y CO Z Al O xH O        3)   2 2 3 2 4 , , X Na Al OH Y SO Z Al O xH O        4)   2 2 3 3 , , X Al OH Y CO Z Al O    Key: 2 Sol:     2 2 3 2 3 2 4 4 CO Al NaOH Na Al OH Na CO Al O xH O        3 3. Out of the following, which type of interaction for the stabilisation of   helix structure of protons ? 1) Covalent bonding 2) Hydrogen bonding 3) Ionic bonding 4) vander Waals forces Key: 2 Sol: Hydrogen bond is responsible for the stacking of   helix structure of protein. 3 4. Match List – I with List – II. List - I List – II (Monomer Unit) (Polymer) (a) Caprolactum (i) Natural rubber (b) 2 - Chloro - 1, 3 - buta diene (ii) Buna – N (c) Isoprene (iii) Nylon 6 (d) Acrylonitrile (iv) Neoprene Choose the correct answer from the options given below : 1) , , , a i b ii c iii d iv     2) , , , a iv b iii c ii d i     3) , , , a ii b i c iv d iii     4) , , , a iii b iv c i d ii     Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 15 | P a g e Key: 4 Sol: , , , a iii b iv c i d ii     3 5. 2 2 3 2 ( ) A HOCl H O H O Cl O       2 2 2 2 ( ) 2 2 2 B I H O OH I HO O        Choose the correct option 1) H 2 O 2 acts as reducing and oxidizing agent respectively in equations (A) and (B) 2) H 2 O 2 acts as reducing agent in equations (A) and (B) 3) H 2 O 2 acts as oxidixing agent in equations (A) and (B) 4) H 2 O 2 acts as oxidixing and reducing agent respectively in equations (A) and (B) Key: 2 Sol: In 1 & 2 H 2 O 2 acts as reducing agent 3 6. What is the final product (major) ‘A’ in the given reaction ? 3 CH C H O H 3 CH HCl  ' A ' ( m a j o r p r o d u c t ) 1) 3 CH C H C l 3 CH 2) 3 CH C H 2 CH 3) 3 CH C H 3 CH 4) 3 CH C l 3 CH 2 CH Key: Sol: Cl   Cl      Cl       OH   H    2 OH ( + ) 2 H O       H Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 16 | P a g e 3 7. What is the product formed by HI on reaction with 3 CH 3 CH C H 2 ? CH C 3 CH 1) 3 2 3 CH CH CH CH CH     3 CH I 2) 3 3 CH C CH CH    3 CH I 3 CH 3) 3 3 CH C CH CH    3 CH I 3 CH 4) 3 2 CH C CH CH I    3 CH H 3 CH Key: 2 Sol: + H I 3 CH 3 CH C H 2 ? CH C 3 CH + + I I - 3 8. Which of the following compound gives pink colour on reaction with phthalic anhydride in conc. H 2 SO 4 followed by treatment with NaOH ? 1) 3 H C HO 3 CH OH 2) HO 3 CH 3) 3 CH OH 4) HO 3 CH HO 3 CH Key: 3 Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 17 | P a g e Sol: 3 9. Which of the following are isostructural pairs ? A ) 2 2 4 4 & SO CrO   B ) 4 4 & SiCl TiCl C ) 3 3 & NH NO  D ) 3 3 & BCl BrCl 1) A and B only 2) B and C only 3) A and C only D) C and D only Key: 1 Sol: 2 2 4 4 4 4 & , & SO CrO SiCl TiCl   The two pairs are isostructural, all are tetrahedral 4 0. In Freundlich adsorption isotherm, slope of AB line is : log / x m log P A B 1) 1 1 0 1 with to n n        2) log n with (n > 1) 3) n with (n, 0.1 to 0.5) 4)   1 log 1 with n n  Key: 1 Sol: 1/ n x kp m  1 log log log x k p m n     1 0 1 Slope to n  Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 18 | P a g e 4 1. The gas released during anaerobic degradation of vegetation may lead to : 1) Corrosion metals 2) Ozone hole 3) Global warming and cancer 4) Acid rain Key: 3 Sol: The gas CH 4 evolved due to anaerobic degradation o f vegetation which causes global warming and cancer 4 2. The product formed in the first step of the reaction of   3 2 2 3 2 2 5 / : CH CH CH CH CH CH withexcess Mg Et O Et C H is       B r B r 1) 3 2 2 3 CH CH CH CH CH CH      3 2 2 3 CH CH CH CH CH CH      2) 3 2 2 3 CH CH CH CH CH CH      3 2 2 3 CH CH CH CH CH CH      3) 3 CH CH  2 CH 3 CH CH  4) 3 2 2 3 CH CH CH CH CH CH     M g B r M g B r Key: 4 Sol: B r B r Mg EtOH   3 2 2 3 CH CH CH CH CH CH     M g B r M g B r Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 19 | P a g e 4 3. Which of the following regents is used for the following reaction ? ? 3 2 3 3 2 CH CH CH CH CH CHO   1) Potassium permanganate 2) Molybdenum oxide 3) Manganese acetate 4) Copper at high temperature and pressure Key: 2 Sol: Control o xidizi ng reagent is molybdenum oxide 4 4. Consider the elements Mg, Al, S, P ad Si, the correct increasing order of their first ionization enthalpy is : 1) Mg < Al < Si < S < P 2) Mg < Al < Si < P < S 3) Al < Mg < S < Si < P 4) Al < Mg < Si < S < P Key: 4 Sol: Al < Mg < Si < S < P 4 5. Identify products A and B. 3 CH 3 4 273 CrO dil KMnO K A B     1) 3 CH A : B : O H O H 3 CH O H O 2) 3 CH A : B : O H O H 3 CH O 3) 2 2 2 3 2 2 2 3 : : A OHC CH CH CH C CH B HOOC CH CH CH C CH       || || O O 4) 3 CH A : B : O H 3 CH O Key: 1 Sri Chaitanya IIT Academy, India. Final_Jee - Main_ 24 - Feb - 202 1 _Shift - 01_Q.Paper 20 | P a g e Sol: 4 6. ‘A’ and ‘B’ in the following reactions are : 2 NH     2 3 2 / / / SnCl HCl H O NaNO HCl KCN A B      1) A : B : C N C H O 2) A : B : C N C l 3) A : B : C H O 2 N Cl   4) A : B : C l 2 N Cl   Key: 1

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