JEE-Main-2023 Past Paper (24th Jan-Second Shift)-PCM-1 PDF
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2023
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This is a past paper for the JEE-Main 2023 exam (24th January - Second Shift). The paper covers Physics, Chemistry, and Mathematics and includes multiple choice and numerical questions. Students preparing for the exam can use this paper to practice and gain familiarity with the format and nature of JEE Main questions.
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JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-1 FIITJEE Solutions to JEE(Main) -2023 Test Date: 24th January 2023 (Second Shift) PHYSICS, CHEMISTRY & MATHEMATICS Paper-1 Time Allotted...
JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-1 FIITJEE Solutions to JEE(Main) -2023 Test Date: 24th January 2023 (Second Shift) PHYSICS, CHEMISTRY & MATHEMATICS Paper-1 Time Allotted: 3 Hours Maximum Marks: 300 ▪ Please read the instructions carefully. You are allotte d 5 minutes specifically for this purpose. Important Instructions: 1. The test is of 3 hours duration. 2. This test paper consists of 90 questions. Each subject (PCM) has 30 questions. The maximum marks are 300. 3. This question paper contains Three Parts. Part-A is Physics, Part-B is Chemistry and Part-C is Mathematics. Each part has only two sections: Section-A and Section-B. 4. Section – A : Attempt all questions. 5. Section – B : Do any 5 questions out of 10 Questions. 6. Section-A (01 – 20) contains 20 multiple choice questions which have only one correct answer. Each question carries +4 marks for correct answer and –1 mark for wrong answer. 7. Section-B (1 – 10) contains 10 Numerical based questions. The answer to each question is rounded off to the nearest integer value. Each question carries +4 marks for correct answer and –1 mark for wrong answer. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-2 PART – A (PHYSICS) SECTION - A (One Options Correct Type) This section contains 20 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE option is correct. Q1. If the distance of the earth from sun is 1.5 × 10 6 km. Then the distance of an imaginary planet from Sun, if its period of revolution is 2.83 years is : (A) 6 106 km (B) 3 106 km (C) 6 10 km 7 (D) 3 107 km Q2. Let 1 be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and 2 be the similar ratio of diatomic gas. 1 Considering the diatomic gas molecule as a rigid rotator, the ratio, is : 2 27 35 (A) (B) 35 27 25 21 (C) (D) 21 25 Q3. Given below are two statements : one is labelled as Assertion A and other is labelled as Reason R Assertion A : Steel is used in the construction of buildings and bridges. Reason R : Steel is more elastic and its elastic limit is high. In the light of above statements, choose the most appropriate answer from the given below (A) Both A and R are correct and R is the correct explanation of A (B) Both A and R are correct but R is NOT the correct explanation of A (C) A is not correct but R is correct (D) A is correct but R is not correct Q4. When a beam of white light is allowed to pass through convex lens parallel to principal axis, the different colours of light converge at different point on the principle axis after refraction. This is called : (A) Scattering (B) Spherical aberration (C) Polarisation (D) Chromatic aberration FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-3 Q5. In an Isothermal change, the change in pressure and volume of a gas can be represented for three different temperature : T3 T2 T1 as : (A) (B) T T2T3 P T3T2 1 P T1 V V (C) (D) T2 T1 P P T1 T2 T3 T3 V V Q6. An metallic rod of length ‘L’ is rotated with an angular xxxxxxxxxxxxx x x x x speed of ‘ ’ normal to a uniform magnetic field ‘B’ about x xxxxxxxxxxx x x x x x xx xxxx x x x x x x x x an axis passing through one end of rod as shown in xx x x x x x xx x x x xx x figure. The induced emf will be : x x x x x x x x x x x x xx x x 1 x x x x x xxxL x x x x xxx x (A) BL2 x x x x x xx x x x x x x xx x 2 x x xx x x x x x x x xxxx x 1 x x x x x x x x x x x x x xx x (B) BL2 x x x x x x x x x x x x x x xx 4 x x x x x x x x x x x x xx 1 (C) B2L x x x x x x x x x x x x 4 1 (D) B2L2 2 Q7. If two vectors P = ˆi + 2mjˆ + mkˆ and Q = 4iˆ − 2ˆj + mkˆ are perpendicular to each other. Then, the value of m will be : (A) -1 (B) 2 (C) 1 (D) 3 Q8. Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : A pendulum clock when taken to Mount Everest becomes fast. Reason R : The value of g (acceleration due to gravity) is less at Mount Everest than its value on the surface of earth. In the light of the above statements. choose the most appropriate answer from the options given below (A) A is correct but R is not correct (B) A is not correct but R is correct (C) Both A and R are correct and R is the correct explanation of A (D) Both A and R are correct but R is NOT the correct explanation of A FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-4 Q9. An -particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength : (A) p e (B) p e (C) = p = e (D) p e +5V Q10. The logic gate equivalent to the given circuit diagram is : (A) OR (B) NOR (C) NAND A1 o (D) AND 1 (Glow) B1 o Y Q11. The electric potential at the centre of two concentric half + + + + ++ ++ rings of radii R1 and R2, having same linear charge +++ ++ density is : + + + +++++++ +++ + + R2 ++ + ++ + 2 + ++ + ++ (A) (B) + + + R1 + + 4 0 0 + + + + + + + + + O (C) (D) 0 2 0 Q12. A cell of emf 90V is connected across series combination of two resistors each of 100 resistance. A voltmeter of resistance 400 is used to measure the potential difference across each resistor. The reading of the voltmeter will be : (A) 40 V (B) 80 V (C) 90 V (D) 45 V Q13. A body of mass 200g is tied to a spring constant 12.5 N/m, while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad/s. Then the ratio of extension in the spring to its natural length will be : (A) 2 : 5 (B) 1 : 1 (C) 1 : 2 (D) 2 : 3 Q14. The frequency ( ) of an oscillating liquid drop may depend upon radius (r) of the drop, density ( ) of liquid and the surface tension (s) of the liquid as : v = r a b sc. The value of a, b, c respectively are 3 1 1 3 1 1 (A) , , − (B) − , − , 2 2 2 2 2 2 3 1 1 3 1 1 (C) − , , (D) , − , 2 2 2 2 2 2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-5 Q15. Given below are two statements : Statement I : Acceleration due to earth’s gravity decreases as you go ‘up’ or ‘down’ from earth’s surface. Statement II : Acceleration due to earth’s gravity is same at a height ‘h’ and depth ‘d’ from earth’s surface, if h = d. In the light of above statements, choose the most appropriate answer from the options given below (A) Statement I is incorrect but statement II is correct (B) Both Statement I and Statement II are incorrect (C) Statement I is correct but statement II is incorrect (D) Both Statement I and II are correct Q16. The electric field and magnetic field components of an electromagnetic wave going through vacuum is described by E x = Eo sin ( kz − t ) B y = Bo sin ( kz − t ) Then the correct relation Eo and Bo is given by (A) EoBo = k (B) Eo = kBo (C) E0 = kBo (D) kEo = Bo Q17. A photon is emitted in transition from n = 4 to n = 1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h = 4 10−15 eVs ) : (A) 94.1 nm (B) 974 nm (C) 99.3 nm (D) 941 nm Q18. For a certain radioactive process the graph between n (R ) and t(sec) is obtained as shown in the figure. Then the value n (R ) + of half life for the unknown radioactive material is approximately : R=decay rate (A) 2.62 sec (B) 9.15 sec 8 (C) 4.62 sec (D) 6.93 sec 6 4 2 10 20 30 40 50 60 time, t ( sec) Q19. A long solenoid is formed by winding 70 turns cm -1. If 2.0A current flows, then magnetic field produced inside the solenoid is______ ( 0 = 4 10−7 TmA −1 ) (A) 176 10−4 T (B) 88 10−4 T (C) 352 10−4 T (D) 1232 10−4 T FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-6 Q20. The velocity time graph of body moving in a straight 8 6 line is shown in figure. 4 The ratio of displacement to distance travelled by the V(ms ) 2 -1 0 body in time 0 to 10s is : -2 2 4 6 8 10 (A) 1 : 1 -4 -6 (B) 1 : 4 time (s) (C) 1 : 3 (D) 1 : 2 SECTION - B (Numerical Answer Type) This section contains 10 Numerical based questions. The answer to each question is rounded off to the nearest integer value. Q1. Three identical resistors with resistance R = 12 and two L R identical inductors with self inductance L = 5 mH are R connected to an ideal battery with emf of 12 V shown in figure. The current through the battery long after the R L switch has been closed will be _________ A. () 12 V Q2. A parallel plate capacitor with air between the plate has a capacitance of 15pF. The separation between the plate become twice and the space between them is filled with a x medium of dielectric constant 3.5. Then the capacitance becomes pF. The value of x is 4 ________. Q3. A Spherical ball of radius 1mm and density 10.5g/cc is dropped in glycerine of coefficient of viscosity 9.8 poise and density 1.5 g/cc. Viscous force on the ball when it attains constant 22 velocity is 3696 × 10-x N. The value of x is (Given, g = 9.8 m/s 2 and = ) 7 Q4. A uniform solid cylinder with radius R and length L has moment inertia I 1, about the axis of R L the cylinder. A concentric solid cylinder of radius R ' = and length L ' = is carved out of 2 2 the original cylinder. If I 2 is the moment of inertia of the carved out portion of the cylinder I then 1 = __________. I2 (Both I1 and I2 are about the axis of the cylinder) Q5. A convex lens of refractive index 1.5 and focal length 18cm in air is immersed in water. The change of focal length of the lens will be __________cm. 4 (Given refractive index of water = ) 3 Q6. A body of mass 1kg begins to move under the action of a time dependent force ( ) F = tiˆ + 3t 2 ˆj N, where î and ĵ are the unit vectors along x and y axis. The power developed by above force, at the time t = 2s, will be ________W. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-7 Q7. If a copper wire is stretched to increase its length by 20%. The percentage increase in resistance of the wire is ________%. Q8. A single turn current loop in the shape of a right angle triangle with sides 5cm, 12cm, 13cm is carrying a current of 2A. The loop is in a uniform magnetic field of magnitude 0.75 T whose direction is parallel to the current in the 13cm side of the loop. The magnitude of the x magnetic force on the 5cm side will be N. The value of x is ________. 130 Q9. The energy released per fission of nucleus of 240 X is 200 MeV. The energy released if all the atoms in 120g of pure 240 X undergo fission is ________ 1025 MeV. Q10. A mass m attached to free end of a spring executes SHM with a period of 1s. If the mass is increased by 3kg the period of oscillation increases by one second, the value of mass m is _________kg. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-8 PART – B (CHEMISTRY) SECTION - A (One Options Correct Type) This section contains 20 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE option is correct. Q1. The number of s-electrons present in an ion with 55 protons in its unipositive state is (A) 8 (B) 9 (C) 12 (D) 10 Q2. K2Cr2O7 paper acidified with dilute H2SO4 turns green when exposed to (A) Hydrogen sulphide (B) Sulphur trioxide (C) Carbon dioxide (D) Sulphur dioxide Q3. Find out the major products from the following reactions. Hg(OAc)2, H2O BH3, THF B NaBH4 H2O2 / OH OH (A) A = OH , B = (B) A = OH , B = OH OH OH OH OH (C) A = , B= (D) A = , B= Q4. A solution is 0.1 M in Cl− and 0.001 M in CrO42 −. Solid AgNO3 is gradually added to it. Assuming that the addition does not change in volume and K sp ( AgCl ) = 1.7 10 −10 M2 and K sp ( Ag2CrO4 ) = 1.9 10−12 M3. Select correct statement from the following : (A) Ag2CrO4 precipitates first because the amount of Ag+ needed is low. (B) Ag2CrO4 precipitates first as its K sp is low. (C) AgCl will precipitate first as the amount of Ag+ needed to precipitate is low. (D) AgCl precipitates first because its K sp is high. Q5. What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species: N2; N2+; O2; O2+? (A) 0,1,2,1 (B) 2,1,2,1 (C) 2,1,0,1 (D) 0,1,0,1 Q6. Which one amongst the following are good oxidizing agents? (A) Sm2+ (B) Ce2+ (C) Ce4+ FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-9 (D) Tb4+ Choose the most appropriate answer from the options given below: (A) C only (B) D only (C) C & D only (D) A & B only Q7. Which of the following cannot be explained by crystal field theory? (A) The order of spectrochemical series (B) Colour of metal complexes (C) Stability of metal complexes (D) Magnetic properties of transition metal complexes Q8. Given below are two statements: Statement I: Pure Aniline and other arylamines are usually colourless. Statement II: Arylamines get coloured on storage due to atmospheric reduction In the light of the above statements, choose the most appropriate answer from the options given below: (A) Both Statement I and Statement II are correct (B) Statement I is correct but Statement II is incorrect (C) Statement I is incorrect but Statement II is correct (D) Both statement I and Statement II are incorrect Q9. In the following sequence of reactions, H+ /H2 O K O C3H6 ⎯⎯⎯⎯ → A ⎯⎯⎯ →B + C dil KOH The compounds B and C respectively are: (A) CI3COOK, HCOOH (B) CI3COOK, CH3I (C) CH3I, HCOOK (D) CHI3, CH3COOK Q10. Given below are two statements: O H2N Statement I: under Clemmensen reduction conditions will give HOOC O O Statement II: under Wolff-Kishner reduction condition will give Cl Cl In the light of the above statements, choose the correct answer from the options given below: (A) Both Statement I and Statement II are false (B) Statement I is false but Statement II is true (C) Statement I is true but Statement II is false (D) Both Statement I and Statement II are true FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-10 Q11. Given below are two statements: Statement I : The nucleophilic addition of sodium hydrogen sulphite to an aldehyde or a ketone involves proton transfer to form a stable ion. Statement II : The nucleophilic addition of hydrogen cyanide to an aldehyde or a ketone yields amine as final product. In the light of the above statements, choose the most appropriate answer from the options given below: (A) Statement I is true but Statement II is false. (B) Both Statement I and Statement II are true. (C) Both Statement I and Statement II are false. (D) Statement I is false but Statement II is true. Q12. In which of the following reactions the hydrogen peroxide acts as a reducing agent? (A) Mn2+ + H2O2 → Mn4+ + 2OH− (B) PbS + 4H2O2 →PbSO4 + 4H2O (C) 2Fe2+ H2O2→ 2Fe3+ + 2OH− (D) HOCl + H2O2 → H3O+ + Cl−+O2 Q13. Which will undergo deprotonation most readily in basic medium? O O O O O O MeO OMe OMe a. b c. (A) b only (B) a only (C) Both a and c (D) c only Q14. Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Benzene is more stable than hypothetical cyclohexatriene Reason R: The delocalized electron cloud is attracted more strongly by nuclei of carbon atoms. In the light of the above statements, choose the correct answer from the options given below: (A) A is true but R is false (B) Both A and R are correct but R is NOT the correct explanation of A (C) Both A and R are correct and R is the correct explanation of A (D) A is false but R is true Q15. A student has studied the decomposition of a gas AB3 at 25o C. He obtained the following data. P(mm Hg) 50 100 200 400 Relative t1/2(s) 4 2 1 0.5 The order of the reaction is (A) 0.5 (B) 0 (zero) (C) 1 (D) 2 Q16. The hybridization and magnetic behaviour of cobalt ion in [Co(NH3)6]3+ complex, respectively is (A) d2sp3 and diamagnetic (B) sp3d2 and paramagnetic (C) sp3d2 and diamagnetic (D) d2sp3 and paramagnetic FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-11 Q17. Choose the correct representation of conductometric titration of benzoic acid vs sodium hydroxide. (A) (B) Conductance Conductance VNaOH VNaOH (C) (D) Conductance VNaOH Conductance VNaOH Q18. The potassium ferrocyanide solution gives a Prussian blue colour, when added to: (A) CoCl3 (B) FeCl3 (C) FeCl2 (D) CoCl2 Q19. Identify the element for which electronic configuration in +3 oxidation state is [Ar]3d 5: (A) Ru (B) Mn (C) Fe (D) Co Q20. Identify A in the following reaction. NH2 K2Cr2O7 A O NO2 (A) H (B) O NH2 NO2 (C) (D) KO SECTION - B FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-12 (Numerical Answer Type) This section contains 10 Numerical based questions. The answer to each question is rounded off to the nearest integer value. Q1. When 0.15 g of an organic compound was analyzed using carius method for estimation of bromine, 0.2397 g of AgBr was obtained. The percentage of bromine in the organic compound is ________..(Nearest Integer) (Atomic mass: Ag = 108, Br = 80) Q2. Maximum number of isomeric monochloro derivatives which can be obtained from 2,2,5,5- tetrametylhexane by chlorination is___________ Q3. The number of units, which are used to express concentration of solutions from the following is____________. Mass percent, Mole, Mole fraction, Molarity, ppm, Molality Q4. An aqueous solution of NiCl2 was heated with excess sodium cyanide in presence of 2− strong oxidizing agent to form Ni ( CN)6 . The total change in number of unpaired electrons on metal centre is ________. Q5. PCl 5 ( g) → PCl3 ( g) + Cl2 ( g) In the above first order reaction, the concentration of PCl5 reduces from initial concentration 50mol L−1 to 10 mol L−1 in 120 minutes at 300K. The rate constant for the reaction at 300K is x 10−2 min−1. the value of x is ________. Given log5 = 0.6989 Q6. The total pressure observed by mixing two liquids A and B is 350 mm Hg when their mole fraction are 0.7 and 0.3 respectively. The total pressure becomes 410 mm Hg if the mole fractions are changed to 0.2 and 0.8 respectively for A and B. The vapour pressure of pure A is _______mm Hg. (Nearest integer). Consider the liquids and solutions behave ideally. Q7. Following figure shows spectrum of an ideal black body T1 at four different temperatures. The number of correct T2 statement/s from the following is______. Energy distribution A. T4 > T3 > T2 > T1 T3 B. The black body consists of particles performing simple harmonic motion. T4 C. The peak of the spectrum shifts to shorter wavelength as temperature increases. Wavelength T T T D. 1 = 2 = 3 constant 1 2 3 E. The given spectrum could be explained using quantisation of energy Q8. If the pKa of lactic acid is 5, then the pH of 0.005 M calcium lactate solution at 25o C is _______10−1(Nearest integer) H Lactic acid CH3 COOH OH FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-13 Q9. Total number of tripeptides possible by mixing of valine and proline is____________ Q10. One mole of an ideal monoatomic gas is 1 2 1.0 subjected to changes as shown in the grap. The magnitude of the work done (by the Pressure (bar) system or on the system) is_____ J (nearest integer) 0.5 3 Given: log 2 = 0.3 ln 10 = 2.3 20 40 Volume (L) FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-14 PART – C (MATHEMATICS) SECTION - A (One Options Correct Type) This section contains 20 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE option is correct. Q1. Let y = y (x) be the solution of the differential equation (x 2 ) − 3y 2 dx + 3xy dy = 0, y (1) = 1. Then 6y 2 ( e ) is equal to 3 2 (A) e (B) 3e2 2 (C) e2 (D) 2e2 1 1 Q2. The number of real solutions of the equation 3 x 2 + 2 − 2 x + + 5 = 0 , is x x (A) 4 (B) 0 (C) 2 (D) 3 Q3. Let f ( x ) be a function such that f ( x + y ) = f ( x ) f ( y ) for all x,y N. If f (1) = 3 and n f (k ) = 3279 , then the value of n is k =1 (A) 8 (B) 7 (C) 9 (D) 6 Q4. If the system of equations x + 2y + 3z = 3 4x + 3y − 4z = 4 8x + 4y − z = 9 + has infinitely many solutions, then the ordered pair ( , ) is equal to : 72 21 72 21 (A) − − (B) , − 5 5 5 5 72 21 72 21 (C) , (D) − , 5 5 5 5 Q5. The number of integers, greater than 7000 that can be formed, using the digits 3,5,6,7,8 without repetition, is (A) 48 (B) 168 (C) 120 (D) 220 The locus of the mid points of the chords of the circle C1 : ( x − 4 ) + ( y − 5 ) = 4 which 2 2 Q6. subtend an angle 1 at the centre of the circle C1 , is a circle of radius r1. If 2 1 = , 3 = and r12 = r22 + r32 , then 2 is equal to 3 3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-15 3 (A) (B) 4 6 (C) (D) 4 2 Q7. If f ( x ) = x 3 − x 2 f ' (1) + xf " ( 2 ) − f "' ( 3 ), x R , then (A) 2f ( 0 ) − f (1) + f ( 3 ) = f ( 2 ) (B) f ( 3 ) − f ( 2 ) = f (1) (C) 3f (1) + f ( 2 ) = f ( 3 ) (D) f (1) + f ( 2 ) + f ( 3 ) = f ( 0 ) 60! ( ) ( ) ( ) ( ) 2 2 2 2 Q8. If 30 C1 +2 30 C2 +3 30 C3 +... + 30 30 C30 = then is equal to : ( 30!) 2 (A) 15 (B) 10 (C) 60 (D) 30 Q9. Let in a right angled triangle, the smallest angle be . If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then sin is equal to : 2 −1 5 +1 (A) (B) 2 4 5 −1 5 −1 (C) (D) 2 4 Q10. Let the six numbers a1,a2 ,a3 ,a4 ,a5 ,a6 , be in A.P. and a1 + a3 = 10. If the mean of these 19 six numbers is and their variance is 2 , then 82 is equal to : 2 (A) 105 (B) 220 (C) 210 (D) 200 Q11. Let A, B and C be three events such that the probability that exactly one of A and B occurs is (1 – k), the probability that exactly one of B and C occurs is (1 – 2k), the probability that exactly one of C and A occurs is (1 – k) and the probability of all A, B and C occur simultaneously is k2, where 0 < k < 1. Then the probability that at least one of A, B and C occur is : 1 1 1 (A) greater than (B) greater than but less than 2 4 2 1 1 (C) exactly equal to (D) greater than but less than 2 8 4 Q12. The set of all values of a for which lim x − 5 − 2x + 2 = 0 , where denotes the x →a ( ) greatest integer less than or equal to is equal to (A) ( −7.5, − 6.5 (B) −7.5, − 6.5 (C) ( −7.5, − 6.5 ) (D) −7.5, − 6.5 ) 22x 1 2 2022 Q13. If f ( x ) = , x R , then f + f 2023 +.... + f 2023 is equal to 2 +2 2x 2023 (A) 1010 (B) 1011 (C) 2011 (D) 2010 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-16 Q14. The equations of the sides AB and AC of a triangle ABC are ( + 1) x + y = 4 and x + (1 − ) y + = 0 respectively. Its vertex A is on the y-axis and its orthocentre is (1,2 ). The length of the tangent from the point C to the part of the parabola y 2 = 6x in the first quadrant is : (A) 2 2 (B) 2 (C) 6 (D) 4 3 2 2 1 + sin 9 + icos 9 Q15. The value of is 1 + sin 2 − icos 2 9 9 (A) 1 2 ( 1− i 3 ) (B) − 1 2 1− i 3 ( ) (C) 1 2 ( 3 +i ) (D) − 1 2 3 −i ( ) r r r r r Q16. Let = 4iˆ + 3ˆj + 5kˆ and = ˆi + 2ˆj − 4kˆ. Let 1 be parallel to and 2 be perpendicular to ( ) r r r r r . If = 1 + 2 , then the value of 52 ˆi + ˆj + kˆ is (A) 7 (B) 9 (C) 6 (D) 11 3 3 48 Q17. 4 3 2 9 − 4x 2 dx is equal to 4 (A) (B) 2 3 (C) (D) 2 6 Q18. ( ) Let A be a 3 x 3 matrix such that adj adj ( adj A ) = 124. Then A −1 adjA is equal to (A) 2 3 (B) 1 (C) 6 (D) 12 sec 2 x 4 f(x)dx Q19. Let f: R → R be a continuous function. The lim 2 is equal to: x→ 2 4 x2 − 16 (A) 2f(2) (B) f(2) (C) 4f(2) (D) 2f( 2 ) Q20. The number of square matrices of order 5 with entries from the set 0,1 , such that the sum of all the elements in each row is 1 and the sum of all the elements in each column is also 1, is (A) 120 (B) 125 (C) 225 (D) 150 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-17 SECTION - B (Numerical Answer Type) This section contains 10 Numerical based questions. The answer to each question is rounded off to the nearest integer value. 3 Q1. If the point on the curve y 2 = 6x, nearest to the point 3, is ( , ) , then 2 ( + ) is 2 equal to……… Q2. The minimum number of elements that must be added to the relation R= ( ), (b,c ), (b,d) on the set a,b,c,d so that it is an equivalence relation, is……. a,b Q3. The equations of the sides AB, BC and CA of a triangle ABC are : 2x + y = 0,x + py = 21a, ( a 0 ) and x − y = 3 respectively. Let P ( 2,a ) be the centroid of ABC. Then (BC) is equal to 2 Q4. Let a curve y = y ( x ) be given by the solution of the differential equation 1 2 ( ) cos cos−1 e− x dx = e2x − 1 dy If it intersects y-axis at y = −1, and the intersection point of the curve with x-axis is ( , 0 ) , then e is equal to…….. Q5. Let f be a differentiable function defined on 0, 2 such that f ( x ) 0 and 2 ( f ( x ) + f ( t ) 1 − loge f ( t ) ) x 2 dt = e, x 0, . Then 6loge f is equal to…… 0 2 6 Q6. If the shortest between the lines x+ 6 y− 6 z− 6 x− y−2 6 z+2 6 = = and = = is 6, then the square of sum of 2 3 4 3 4 5 all possible values of is Q7. If the area of the region bounded by the curves y 2 − 2y = − x, x + y = 0 is A, then 8A is equal to Q8. Three urns A, B and C contain 4 red, 6 black; 5 red, 5 black; and red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn C is 0.4 then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola y 2 = x with one vertex at the vertex of the parabola, is Q9. Let the sum of the coefficients of the first three terms in the expansion of n 3 x − x 2 , x 0, n N , be 376. Then the coefficient of x is…… 4 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-18 r r r r r Q10. Let a = ˆi + 2ˆj + k, ˆ ar cr = 7,2b cr + 43 = 0, ar cr = b cr. Then ar b is equal to ˆ b = 3iˆ − 5ˆj − k, FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-19 FIITJEE KEYS to JEE (Main)-2023 PART – A (PHYSICS) SECTION - A 1. B 2. C 3. A 4. D 5. D 6. A 7. B 8. B 9. A 10. C 11. D 12. A 13. D 14. B 15. C 16. D 17. A 18. C 19. A 20. C SECTION – B 1. 3 2. 105 3. 7 4. 32 5. 54 6. 100 7. 44 8. 9 9. 6 10. 1 PART – B (CHEMISTRY) SECTION - A 1. D 2. D 3. A 4. C 5. A 6. C 7. A 8. B 9. D 10. C 11. A 12. D 13. B 14. C 15. D 16. A 17. C 18 B 19. C 20. B SECTION - B 1. 68 2. 3 3. 5 4. 2 5. 1 6. 314 7. 2 8. 85 9. 8 10. 620 PART – C (MATHEMATICS) SECTION - A 1. D 2. B 3. B 4. B 5. B 6. D 7. A 8. A 9. C 10. C 11. A 12. C 13. B 14. A 15. D 16. A 17. B 18 A 19. A 20. A SECTION - B 1. 9 2. 13 3. 29 4. 2 5. 27 6. 384 7. 36 8. 432 9. 405 10. 8 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-20 FIITJEE Solutions to JEE (Main)-2023 PART – A (PHYSICS) SECTION – A Sol1. T a 2 3 T a3/2 ( ) 3/2 1 1.5 106 2.83 ( a ) 3/2 3/2 1 1.5 106 = 2.83 a 1 1.5 106 a = 1.5 ( 2.83 ) 106 2/3 = ( 2.83 ) 2/3 a = 3 106 km 5 7 Sol2. 1 = , 2 = 3 5 25 1 = 2 21 Sol3. Steel is more elastic so it is used in buildings and bridges. Sol4. It is called chromatic aberration. Sol5. For constant volume: P3 > P2 > P1 and graph will be hyperbolic in nature. 1 Sol6. Emf across its end = BL2 2 Sol7. P.Q = 0 4 − 4m + m2 = 0 (m − 2) = 0 2 m=2 Sol8. At Mount Everest g will decrease so time period will increase and clock will slow. h h Sol9. = = mv 2km h m FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-21 p e Sol10. When bath switch is open or any one of the switch is 5V closed bulb will glow. If bath switch is closed; It will not A B glow. 0 0 1 1 0 1 0 1 1 1 1 0 K ( R1 ) ( R2 ) Sol11. V = + R1 R2 = 2K = 20 Sol12. Requ = 180 − V 1 i= A 2 v = 40v Sol13. Kx = mw 2 ( + x ) 1 12.5x = 25 ( + x ) 5 x 5 2 = = + x 15.5 5 5x = 2 + 2x x 3x = 2 = 2:3 Sol14. V = r a sb sc ( )( ) (MT ) b C T −1 = La ML−3 −2 b + c = 0 b = −c a − 3b = 0 1 −1 = 2c c = 2 GM Sol15. g = for outside r2 gMr g = 3 for inside R Sol16. E = Bv FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-22 w E=B k −13.6 1 15 Sol17. E = E = 13.6 1 − 16 = 13.6 16 eV n2 hc 4 10−15 3 108 12 10−7 = = = = 94.1nm E 15 15 13.6 13.6 16 16 dN Sol18. R = − = N = N0 e −t n (R ) = n ( N0 ) − t , so dt 6 3 −1 = tan = = s Slope of graph 40 20 0.693 0.693 20 t1 = = = 4.62s 2 3 Sol19. B = 0ni = 4 10−7 7000 2 22 = 56 10−4 7 = 176 16−4 Sol20. Displacement = 8 2 − 2 4 + 4 4 − 2 4 = 32 − 16 = 16 Distance = 8 2 + 2 4 + 4 4 + 2 4 = 48 SECTION - B Sol1. At steady state: V 3V i= = = 3A R R 3 A Sol2. Co = = 15pf d K A K A 105 C= 0 = 0 = pf 2d 2 d 4 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-23 Sol3. fr = mg − B B = vg − vg = ( − ) vg 9 10−3 4 3 1 = 3 9.8 10 −6 3 10 Fr 4 22 1 = 9 103 9.8 3 7 109 = 369.6 10−6 v = 3696 10−7 Mg MR2 1 Sol4. I1 = = ( AL ) R 2 2 2 2 1 A L R I2 = 2 4 2 2 I1 1 = = 32 R L I2 1 L R 2 32 2 I1 3 1 2 1 1 Sol5. = − 1 − 18 1 R1 R2 3 1 2 1 1 = − 1 − f 4 R1 R 2 3 1 f = 2 =4 18 1 8 f = 72 f = 54cm Sol6. ( ) f = tiˆ + 3t 2 ˆj N a = tiˆ + 3t j 2ˆ t 2 ˆ 3t 3 ˆ t 2 ˆ 3 ˆ v= i+ j= i+t j 2 3 2 3 t 8 p = f.v = + 3t 5 = + 3 ( 2 ) = 4 + 96 = 100 5 3 2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-24 Sol7. R= 0 A0 A0 0 A0 = 1.2 0A A= 1.2 1.2 0 R' = = 1.44 0 A0 A0 1.2 R = 44% R Sol8. f = (Bcos ) i B 3 12 1 = 2 13cm 4 13 20 2A 9 5cm = 130 12cm 120 1 Sol9. n= = 240 2 1 = NA 2 1 = 6.2 1023 200MeV 2 = 6.2 1025 MeV m Sol10. T = 2 k m 1 = 2 k m+3 2 = 2 k 1 m 1 m = = m + 3 = 4m m = 1 2 m+3 4 m+3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-25 PART – B (CHEMISTRY) SECTION – A Sol1. Z = 55. The element is Cs. Cs ( 55 ) = Xe 6s1 Cs+ = Xe 6so The number of s-orbitals is 1s, 2s, 3s, 4s and 5s Total electron = 5 2 = 10 Sol2. K 2Cr2O7 + SO2 + H2SO4 ⎯⎯→ Cr2 (SO4 )3 + K 2SO4 + H2O (green) +3 Due to formation of Cr - salt green colour is obtained. Sol3. (1) BH3 , THF (Anti Markovnikov's product) (2) H2O2 / OH (A) OH OH (1) Hg(OAc)2 , H2O ( Markovnikov's product) (2) NaBH4 Sol4. Cl − = 10−1 M CrO4 −− = 10− 3 M 1.7 10 −10 for AgCl ppt 1, Ag+ req = M 10 −1 Ag + = 1.7 10− 9 M req 1.9 10−12 For Ag2CrO4 ppt2, Ag+ req = M 10−3 Ag+ = 4.3 10− 5 M req AgNO3 (Solid) Solution of Solution of Cl − & CrO42− Being lower concentration of [Ag+] in case of AgCl, it will precipitate first. Sol5. HOMO is highest occupied molecular orbital N2 = 1s 2 1s 2s *2 *2 2 2s 2px 2py 2pz → HOMO orbital 2 2 2 No. of u.e = 0 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-26 N2+ = 1s 2 1s *2 2 2s 2px 2py zpz → HOMO 2s *2 2 2 1 No. of u.e = 1 O2 = 1s 2 1s 2s 2s *2 2 2pz 2px *2 2 2 2py 2 2px *1 2py *1 → HOMO orbital No. of u.e = 2 O2+ = 1s 2 1s 2s 2s *2 2 2pz 2px *2 2 2 2py 2 2px *1 2py *0 → HOMO No. of u.e = 1 Sol6. The most stable O.S of Lanthanoid = (+3) Hence Ce+4 and Tb+4 tends to get (+3) O.S thus behaves like O.A Ce+4 + e− ⎯⎯→ Ce+3 Tb+4 + e− ⎯⎯→ Tb+3 Sol7. C.F.T does not explain the order of Spectrochemical series because it is an experimentally determined series. C.F.T introduces spectrochemical series on the basis of experiment value of . Sol8. Amines & anilines are almost colourless It becomes cloured compound, when exposed in air for a long time due to oxidation. O2 Aniline ⎯⎯⎯⎯⎯ Long time → dark brown Sol9. Here, C3H6 is propene H+ / H2O CH2 CH CH3 H3C CH CH3 OH A KIO H3C CH CH3 CHI3 + CH3 COOK dil. KOH B C OH A Second reaction is iodoform reaction. Sol10. O Zn - Hg H2N C HO C Conc. HCl O O H It reduces C O group to C H But acid sensitive groups are not present in compound O H2N-NH2 / KOH glycol Cl → It reduces >C=O group to >CH2 but further elimination takes place due to OH− group. Sol11. Nucleophilic addition of sodium hydrogen sulphite to aldhyde or ketone is as; NaHSO3 Na+ + HSO3− FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-27 O C O + S C O HO O S OH O O O Proton transfer OH C C SO3H SO3 Less stable more stable or less acidic or more acidic So; nucleophilic addition of sodium hydrogen sulphite to an aldehyde or a ketone involves proton transfer to form a stable ion. Addition of hydrogen cyanide; H+ OH C O + HCN C CN Final product is cyanohydrin. Sol12. Oxidation +1 (-1) + (0) HOCl + H2O2 H3O + Cl + O2 Reduction → H2O2 undergoes oxidation thus behaves like reducing agent. Sol13. The compound will be easily deprotonated in which conjugate base is readily stabilized. O O O O OH Maximum stable conjugate base. In other compound (+M) effect of RO − group decreases the stability of conjugate base. Sol14. The HHydrogenation of cyclohexatriene 36K.Cal/mole more than benzene therefore. Benzene is more stable than cyclohexatriene. 1−n Sol15. t1/2 (P0 ) 1−n 50 4 = 100 2 1−n 1 = 21 2 n−1 ( 2) = 21 n −1 = 1 n=2 Order of reaction is 2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-28 +3 Sol16. Co (NH3 )6 O.S.of Co = ( +3 ) Co+3 = 4so 3d6 +3 Co = 3d 4s 4p In presence of strong ligand NH3 pairing takes place. +3 Co = 3d 4s 4p Hybridization = d2 sp3 u.e = o (Diamagnetic) Sol17. C6H5 COOH + NaOH ⎯⎯→ C6H5 COONa + H2O A to B → Free H ions are replaced by slow moving Na ion thus conductance (D) decreases B to C → Undissociated benzoic acid forms (A) salt with NaOH which increases ions thus Conductance conductance increases. C to D → After equivalence point NaOH further (C) increases fast moving OH− thus (B) conductance increases. Sol18. FeCl3 gives prussion blue colour on reacting with potassium ferrocynaide solution as 4FeCl3 + 3K 4 Fe ( CN)6 ⎯⎯ → Fe4 Fe ( CN)6 + 12KCl 3 Potassium Ferric ferrocyanide ferrocyanide (Pr ussionblue) Sol19. Electronic configuration of Fe is [Ar] 4s 23d6 and in +3 oxidation state it has [Ar] 4s 03d5 configuration. Sol20. O NH2 K2Cr2O7 oxidation O (Benzoquinone) FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-29 SECTION – B Sol1. 0.15 gm (organic compounds) ⎯⎯⎯⎯ Ag NO3 → AgBr (s) 0.2397 gm 80 Weight of Br in AgBr = 0.2397 = 0.102 gm 188 0.102 100 % Br in compound = = 68% 0.15 Sol2. CH3 CH3 H3C C CH2 CH2 C CH3 2,2,5,5 -Tetramethyl hexane. CH3 CH3 Cl2 / h CH3 CH3 CH3 CH3 CH3 C CH CH2 C CH3 CH2 C CH2 CH2 C CH3 CH3 Cl CH3 Cl CH3 CH3 2 Product (One product) Total isomeric product = 3 Sol3. Concentration of solution expressed is (1) Mass percent (2) Mole fraction (3) Molarity (4) ppm (5) Molality 2− Sol4. NiCl2 + NaCN ⎯⎯⎯ strong O.A. → Ni ( CN)6 4+ Complex has Ni and strong ligand, hence following are the metal ion electronic configuration. 4+ 2+ 8 0 Ni (strong ligand) Ni 3d 4s 6 0 3d 4s Change of unpaired electron = 2 Sol5. PCl5 (g) ⎯⎯ → PCl3 (g) + Cl2 (g) (first order reaction) 2.303 50 K= log 120 10 2.303 log5 2.303 0.6989 K = = = 0.0134min−1 120min 120 Rate constant at 300K = 1.34 10−2 min−1 [Nearest integer = 1.0] FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-30 Sol6. V.P of pure A = PAo V.P of pure B = PBo Case-I PT = PAo X A + PBo XB 350 = 0.7 PAo + 0.3 PBo ----------------------------------- (I) Case-II 410 = 0.2 PAo + 0.8 PBo ----------------------------------- (II) (II)- (i) 60 = −0.5 PAo + 0.5 PBo 60 600 PBo − PAo = = = 120 0.5 5 ( PBo = 120 + PAo ) ( 350 = 0.7 PAo + 0.3 120 + PAo ) 350 = PAo + 36 PAo = 350 − 36 = 314 mm Hg PBo = 120 + 314 = 434 mm Hg Sol7. T1 > T2 > T3 > T4 T1 When temperature increases peak of spectrum shifted to T2 shorter wavelength or higher frequency Energy distribution → Spectrum of black body radiation is explained by using T3 quantization of energy. T4 Wavelength Sol8. Ca (Lac ) = 0.05M = 5 10−3 (M) 2 Lac − = 2 5 10−3 = 10−2 (M) Calcium lactate is salt of weak acid and strong base. 1 pH = 7 + (Pka + logC ) 2 1 = 7 + 5 + log 10−2 2 ( ) 1 3 = 7 + (5 − 2) = 7 + 2 2 −1 = 8.5 = 85 10 Sol9. Total number of tripeptide by mixing valine and proline = 2 3 = 8 Val− Val − Val Pro − Pro −Pro Val−Pro−Pro Pro − Val − Pro Val−Val−Pro Val − Pro −Val Pro−Pro−Val Pro − Val − Val Sol10. 1→2 ⎯⎯→ Isobaric process FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-31 2→3 ⎯⎯→ Isochoric process 3→1 ⎯⎯→ Isothermal process Total work = W1→2 +W2→3 +W3→1 V = −P ( V2 − V1 ) + 0 + −P1V1 n 2 V1 20 = − ( 40 − 20 ) − 1 20 n 40 40 = −20 + 20 n 20 = −20 + 20 n2 = −20 + 20 0.3 2.3 = −6.2bar lit Magnitude of work done = 6.2 bar lit = 620J FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-32 PART – C (MATHEMATICS) SECTION – A Sol1. (x 3 ) − 3y 2 dx + 3xy dy = 0, y (1) = 1 dy 3y 2 − x 2 = dx 3xy Put y = vx dv 3v 2 − 1 v+x = dx 3v dv 3v 2 − 1 x = −v dx 3v dx 3vdv = − x 3 2 3 v = − loge x + 2 2 2 3y 3 = − loge x + c = − loge x + 2 x 2 3 y (1) = I, c = 2 2 3y 3 = − loge x + 2 x 2 x=e e2 y2 ( e) = 3 e2 6y 2 ( e ) = 6 = 2e 2 3 1 1 Sol2. 3 x2 + 2 − 2 x + + 5 = 0 x x 1 2 1 3 x + − 2 − 2 x + + 5 = 0 x x 1 t = x+ x t ( −, − 2 2, ) 3t 2 − 2t − 1 = 0 1 t = 1, t = − not lies on range. 3 Sol3. f ( x + y ) = f ( x ) f ( y ) x,y N f (1) = 3 ( f (1 + 1) = f ( 2 ) = f (1) f (1) = f (1) ) 2 = 32 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-33 f ( 3 ) = 33 f (1) + f ( 2 )....... + f (n ) = 3279 3 + 32........ + 3n = 3279 3n − 1 3 = 3279 3 −1 3n+1 = 6561 3n+1 = 38 n +1= 8 n=7 Sol4. x + 2y + 3z = 3 4x + 3y − 4z = 4 8x + 4y − z = 9 + D=0 1 2 3 4 3 −4 = 0 8 4 − 72 = 5 D3 = 0 1 2 3 4 3 =0 8 8 9+4 −21 = 5 Sol5. Formed digit greater then 7000. (i) using all digit (5 digit number > 4 digit number) = 5 = 120 (ii) using 4 digit = ………….= 2C1 4C3 3 = 48 Total digit = 168 (h − 4 ) + (k − 5 ) ( 4,5 ) 2 2 Sol6. = ri2 ri r 2 i cos i = , ri = 2cos i 2 2 2 2 ( h,k ) i (h − 4 ) + ( k − 5 ) = 4cos2 2 2 2 ri2 = 4cos2 i 2cos i ri = 2 i = 600 r1 = 3 r3 = 1 r12 = r22 + r32 3 = r22 + 1 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-34 4cos2 2 2= 2 cos2 2 1 = 2 2 2 = 450 2 2 = 900 Sol7. f ( x ) = x 3 − x 2 f ' (1) + xf " ( 2 ) − f "' ( 3 ) x R Let f ' (1) = 9, f " ( 2 ) = b, f "' ( 3 ) = c f ( x ) = x 3 − ax 2 + bx − c f ' ( x ) = 3x 2 − 2ax + b f " ( x ) = 6x − 29 f "' ( 3 ) = 6 = c f ' (1) = a 3a − b = 3 f " ( 2) = b 6 2 − 2a = b 2a + b = 12 a = 3, b = 6 ( ) ( ) ( ) ( ) 2 2 2 2 Sol8. S = 0 30 C0 + 1 30 C1 +2 30 C2......30 30 C30 S = 30 ( ) ( ) ( ) 2 2 2 30 C30 + 29 30 C29...... + 0 30 C0 2S = 30 C30 2 + C229..... + C30 2 S = 15 C02 + C12..... + C30 2− 24 C02 + C12..... + C2x = xx 60 S = 15 30 30 = 15 Sol9. Let smallest angle is C = θ A Therefore angle A = 900 – θ And angle B = 900 i.e. b > a > c c b Here , a =2Rcosθ , b = 2R , c = 2Rsinθ b2 = a2 + c2 C according to question , B a 1 1 1 2 = 2 + 2 c a b a 2b 2. 2 = a2 + b2 c FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-35 cos2 = cos2 + 1 sin2 ( 1 − sin2 = sin2 2 − sin2 ) sin4 − 3sin2 + 1 = 0 3− 5 sin2 = 2 5 −1 sin = − 2 a1 + a2 + a3....a6 19 Sol10. = 3 2 a1 + a2..... + a6 = 57 a1 = a, a2 = a + d, a3 = a + 2d 6a + 15d = 19 a+d=5 a = 2, d = 3 Number are 2, 5, 8, 11, 14, 17 Variance 2 = ( ) 105 4 8 105 82 = = 210 4 Sol11. P(A B) + P(A B) = 1 − k , P(A C) + P(A C) = 1 − 2k , P(B C) + P(B C) = 1 − k and P(A B C) = k 2 P(A) + P(B) − 2P(A B) = 1 − k ………..(i) P(B) + P(C) − 2P(B C) = 1 − k ………..(ii) P(A) + P(C) − 2P(A C) = 1 − 2k ……….(iii) −4k + 3 Adding (i),(ii) and (iii) we get P(A B C) = + k2 2 2k 2 − 4k + 2 + 1 2(k − 1)2 + 1 1 P(A B C) = = P(A B C) 2 2 2 ( Sol12. lim x − 5 − 2x + 2 = 0 x →a ) Put −7 5 −7 5 − 5 − −15 + 2 −12 5 − −13 −13 + 13 = 0 Put a = 6 5 −6 5 − 5 − −13 + 2 −11 5 − −11 Put a = −64 −6 4 − 5 − −12 8 + 2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493,Fax 26513942 website: www.fiitjee.com. JEE-MAIN-2023 (24th Jan-Second Shift)-PCM-36 −11 4 − −10 8 22x 4x Sol13. f ( x ) = = 22x + 2 4 x + 2 41− x f (1 − x ) = 1− x 4 +2 f ( x ) + f (1 − x ) = 1 1 2 2022 f + f 2023 ...... + f 2023 2023 1 2022 2 f +f +f +...... = 1011 2023 2023 2023 Sol14. H (1,2 ) A ( 0, ) AB :: ( + 1) x + y = 4 D AC x + (1 − ) y + = 0 H = 4 …………(i) (A lies on AB & AC) B C = ( − 1) =2 A ( 0,2 ) −1 2 ,1 P y 2 = 6x wrt equation (i) 2x − 3y + 4 = 0 C 1 Solving AC & CD − , 1 2 3 P ,3 2 3 ty = x + at 2 , a = 2 −1 passing ,1 2 3t − 2t − 1 = 0 2 1 t = 1, − 3 3 P ,3 2 PC = 4 + 4 = 2 3 2 2 sin 9 + icos 9 Sol15. 1 + sin 2 − i cos 2 9 9 1+ z , zz = 1 1
