JEE Mains 2023 Past Paper (Mathematics, Physics, Chemistry) PDF

Answers & Solutions Time : 3 hrs. for M.M. : 300 JEE (Main)-2023 (Online) Phase-2 (Mathematics, Physics and Chemistry) 15/04/2023...

Answers & Solutions Time : 3 hrs. for M.M. : 300 JEE (Main)-2023 (Online) Phase-2 (Mathematics, Physics and Chemistry) 15/04/2023 Morning IMPORTANT INSTRUCTIONS: (1) The test is of 3 hours duration. (2) The Test Booklet consists of 90 questions. The maximum marks are 300. (3) There are three parts in the question paper consisting of Mathematics, Physics and Chemistry having 30 questions in each part of equal weightage. Each part (subject) has two sections. (i) Section-A: This section 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. (ii) Section-B: This section contains 10 questions. In Section-B, attempt any five questions out of 10. The answer to each of the questions is a numerical value. Each question carries 4 marks for correct answer and –1 mark for wrong answer. For Section-B, the answer should be rounded off to the nearest integer. -1-  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning MATHEMATICS Sol. |A| = m – n, where 4m + n = 22 …(i) SECTION - A and 17m + 4n = 93 …(ii) Multiple Choice Questions: This section contains 20 Solving (i) and (ii) multiple choice questions. Each question has 4 choices m = 5, n = 2 (1), (2), (3) and (4), out of which ONLY ONE is correct. Order = 5 Choose the correct answer: |A| = 3 1. Let S be the set of all (, ) for which the vectors  det (n adj (adj(mA))) iˆ − jˆ + kˆ, iˆ + 2 jˆ + kˆ and 3iˆ − 4 ˆj + 5kˆ, where = |2adj (adj (5A))|  –  = 5, are coplanar, then  ( 80 2 + 2 ) is = 25 |5A|16 = 25 580 |A|16 = 25· 316· 580 ( , )S = 311 580 65 equal to So, a + b + c = 96 (1) 2210 (2) 2130 3. If (, ) is the orthocenter of the triangle ABC with (3) 2290 (4) 2370 vertices A(3, –7), B(–1, 2) and C(4, 5), then Answer (3) 9 – 6 + 60 is equal to  −1 1 (1) 25 (2) 35 Sol.  1 2  =0 (3) 30 (4) 40 3 −4 5 Answer (1)  10 + 4] + 1 [5 – 3 + 1 [–10] = 0  4 + 10 – 3 = 5 …(i) And  –  = 5 So, by (1) Sol. 4(5 + ) + 10(5 + ) – 3 = 5  42 + 27 + 45 = 0  42 + 15 + 12 + 45 = 0  (4 + 15) – ( + 3) = 0 −15   = −3, 4 −5 5 AD : y + 7 = ( x − 3)  = 2, 3 4  3y + 21 = –5x + 15  ( )  225 25   80 2 + 2 = 80 ( 9 + 4 ) + 16 + 16  5x + 3y + 6 = 0 …(i) s −1 BE : y − 2 = ( x + 1)  125  12 = 80 13 + = 10  229  8   12y – 24 = –x – 1  x = 23 – 12y = 2290 by (ii) 115 – 60y + 3y + 6 = 0 2. Let the determinant of a square matrix A of order m 57y = 121 be m – n, where m and n satisfy 4m + n = 22 and 17m + 4n = 93. If det(n adj(adj (mA))) = 3a5b6c, then 121 121 y= , x = 23 − 12  a + b + c is equal to 57 57 (1) 84 (2) 96 121 121  9 – 6 + 60 = 9 × 23 – 108  − 6 + 60 (3) 101 (4) 109 57 57 Answer (2) = 207 – 242 + 60 = 25 -2-  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning 4. Let ABCD be a quadrilateral. If E and F are the mid Sol. points of the diagonals AC and BD respectively and B(9, 3, 4) ( AB – BC ) + ( AD – DC ) = k FE , then k is equal to A(–1, –2, –3) (1) 4 (2) –2 (3) 2 (4) –4 C(9, –2, 1) Answer (4) Sol. AC = 10i + 4k D C AB = 10i + 5 j + 7k i j k AC  AB = 10 0 4 10 5 7 = –20i – 30j + 50k A B Equation of plane 2x + 3y – 5z = d Let position vector of A, B, C and D are a, b, c and d Put (–1, –2, –3) respectively –2 – 6 + 15 = d d=7 OC + OA c + a  2x + 3y – 5z = 7  Position vector of E = = 2 2 Foot of ⊥r x −3 y +2 z+9  38  b+d = = = −  Position vector of F = 2 3 −5  38  2 x = 1, y = –5, z = –4 ( ) Now, AB − BC + AD − DC Q(1, –5, –4) Distance from origin = 1 + 25 + 16 (  b−a− c −b +d −a− c −d ) ( ) = 42 6. The number of common tangents, to the circles  2b − 2a − 2c + 2d x2 + y2 – 18x –15y + 131 = 0 and x2 + y2 – 6x – 6y – 7 = 0, is (  2 b+d −2 a+c ) ( ) (1) 3 (2) 1 (3) 4 (4) 2 b + d a + c  Answer (1)  4 −  = 4 OF − OE   2 2  Sol. x 2 + y 2 − 18 x − 15 y + 131 = 0  15  225 5  4 EF = − 4 FE C1  9,  , r1 = 81 + − 131 =  2  4 2  k = –4 x 2 + y 2 − 6 x − 6y − 7 = 0 5. Let the foot of perpendicular of the point C2 (3, 3), r2 = 9 + 9 + 7 = 5 P(3, –2, –9) on the plane passing through the points 2  15  81 15 (–1, –2, –3), (9, 3, 4), (9, –2, 1) be Q(, , ). Then d = C1C2 = (9 − 3)2 +  − 3  = 36 + =  2  4 2 the distance Q from the origin is 5 15 r1 + r2 = +5 = (1) 42 (2) 38 2 2  C1C2 = r1 + r2 (3) 35 (4) 29  Circles touch each other externally, 3 common Answer (1) tangents. -3-  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning 7. The total number of three-digit numbers, divisible Put x = 3 by 3, which can be formed using the digits 1, 3, 5, 8, if repetition of digits is allowed, is = ( ) 8 9 + y 2 − 16y 2 64 + 64 y 2 (1) 21 (2) 20 (3) 22 (4) 18 f (y ) = ( 1 9−y 2 ) Answer (3) 8 1+ y 2 Sol. Sum of digits 3 : (1, 1, 1) Range of f(y) = (– 0.125, 1.125] Sum of digits 9 : (1, 3, 5) or (3, 3, 3)  = – 0.125 Sum of digits 12 : (1, 3, 8)  = 1.125 Sum of digits 15 : (5, 5, 5)  –  = 1.25 Sum of digits 18 : (5, 5, 8) 24( – ) = 30 Sum of digits 21 : (5, 8, 8) ( ) 10 20 Sum of digits 24 : (8, 8, 8) 9. Let a + bx + cx 2 =  pi x i , a, b, c . If i =10 Possible numbers are p1 = 20 and p2 = 210, then 2(a + b + c) is equal to 3! 3! = 1 + 3!+ 1 + 3!+ 1 + + + 1 (1) 6 (2) 15 2! 2! (3) 12 (4) 8 = 22 Answer (3) z z zz ( ) 8. If the set Re :z , Re z 3 is 10! r3 2 3z 5z Sol. General term : ( a )r1 ( bx )r2 cx 2 r1 ! r2 ! r3 ! equal to the interval (, ], then 24( – ) is equal For coeff. of x : r2 + 2r3 = 1 to r1 r2 r3 (1) 36 (2) 27 9 1 0 (3) 30 (4) 42 10! 9 1 Answer (3)  coff of x = a b = 20 9!  z − z + zz   a9  b = 2 …(i) Sol. Re   2 − 3z + 5z    10! 8 2 10! Coff of x2 : a b +  a9  c = 210 8!2!0! 9!0!1!  x + iy − ( x − iy ) + x 2 + y 2  Re    45a8  b2 + 10  a9  c = 210  2 − 3 ( x + iy ) + 5 ( x − iy )     9a8b2 + 2a9  c = 42  x 2 + y 2 + i ( 2y )  as a, b, c N Re    2 + 2x − 8iy   a = 1, b = 2, c = 3   2(a + b + c) = 2(3 + 2 + 1) = 12 Re  ( )  x 2 + y 2 + 2yi ( 2 (1 + x ) + 8iy )   10. The number of real roots of the equation    ( 2 (1 + x ) ) + ( 8y ) 2 2   xx 5x 2 6 0, is ( ) (1) 5 (2) 4 2 x 2 + y 2 (1 + x ) − 16 y 2 = (3) 6 (4) 3 4 (1 + x ) + ( 8 y ) 2 2 Answer (4) -4-  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning Sol. x|x| – 5|x + 2| + 6 = 0 1  b 4 Case-I : G1 = a   a x < –2 2 –x2 + 5(x + 2) + 6 = 0  b 4 G2 = a    x2 – 5x – 16 = 0 a 5 25 64 3  x 2  b 4 G3 = a   a 5 89  x is accepted 2 (G1 )4 + (G2 )4 + (G3 )4 + (G1 )2  (G3 )2 Case-II : b b2 b3 b2 –2  x < 0  a4  + a4  + a4  + a4  a a2 a3 a2 –x2 – 5(x + 2) + 6 = 0 = ba3 + b2a2 + b3a + a2b2  x2 + 5x + 4 = 0 = ab(a2 + b2 + 2ab) = ab(a + b)2  (x + 1) (x + 4) = 0 x = –1 is accepted ( A1 + A3 )2  G1G3 = ( a + b )2  ab Case-III :  Option (1) is correct. x0 12. Let [x] denote the greatest integer function and x2 – 5(x + 2) + 6 = 0 f ( x ) = max 1 + x +  x , 2 + x, x + 2  x  , 0  x  2.  x2 – 5x – 4 = 0 Let m be the number of points in [0, 2], where f is 5 25 16 not continuous and n be the number of points in x 2 (0, 2), where f is not differentiable. Then 5 41 (m + n)2 + 2 is equal to 2 (1) 2 5 41 (2) 11 x is accepted 2 (3) 6  3 real roots are possible. (4) 3 Option (4) is correct. Answer (4) 11. Let A1 and A2 be two arithmetic means and G1, G2 max  x + 1, x + 2, x 0  x 1 and G3 be three geometric means of two distinct  Sol. f ( x ) = max  x + 2, x + 2, x + 2 1  x  2 positive numbers. Then G14 G24 G34 G12G32 is  max 5, 4, 6 x=2  equal to x + 2 0  x 1 (1) (A1 + A2)2 G1G3   f ( x ) = x + 2 1 x  2 (2) 2(A1 + A2) G1G3 6  x=2 (3) ( A1 + A2 ) G12G32 x + 2 0x2  f (x) =  (4) 2 ( A1 + A2 ) G12G32 6 x=2 Answer (1) f is not continuous at x = 2 Sol. Let the two numbers are a, b. f is differentiable in (0, 2) b a 2a b A1 a  m = 1, n = 0 3 3 (m + n)2 + 2 = 1 + 2 = 3. b a a 2b A2 a 2 Option (4) is correct 3 3 -5-  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning 13. Let S be the set of all values of , for which the 15. Let the system of linear equations shortest distance between the lines – x + 2y – 9z = 7 x y 3 z 6 x y z 6 and – x + 3y + 7z = 9 0 4 1 3 4 0 – 2x + y + 5z = 8 is 13. Then 8 is equal to – 3x + y + 13z =  S has a unique solution x = , y = , z = . Then the (1) 306 (2) 304 distance of the point (, , ) from the plane (3) 308 (4) 302 Answer (1) 2x – 2y + z =  is x – y –3 z+6 (1) 11 (2) 7 Sol. = = 0 4 1 (3) 9 (4) 13 x+ y z–6 Answer (2) = = 3 –4 0 Sol. –x + 2y – 9z = 7...(i) d= ( a2 – a1 )  ( n1  n2 ) = 13 –x + 3y + 7z = 9...(ii) n1  n2 –2x + y + 5z = 8...(iii) iˆ ˆj kˆ –3x + y + 13z = ...(iv) n1  n2 = 0 4 1 From (i), (ii), (iii) 3 –4 0 x = –3, y = 2, z = 0 = 4iˆ + 3 ˆj – 12kˆ Substitute in (iv) 3×3+2= ( 2iˆ + 3 jˆ – 12kˆ )  ( 4iˆ + 3 ˆj – 12kˆ ) =8 16 + 9 + 144  = 11 8 + 9 + 144 = 104 Point: (–3, 2, 0) Plane: 2x – 2y + z = 11 8 + 153 = 104 8 = ±104 – 153 –6 – 4 – 11 21 d= = =7 2 2 –49 –257 2 + 2 +1 3 = , 8 8 16. A bag contains 6 white and 4 black balls. A die is 8   = 8  49 + 257  = 306 rolled once and the number of balls equal to the  s 8 8  number obtained on the die are drawn from the bag 14. Negation of p  ( q   ( p  q ) ) is at random. The probability that all the balls drawn are white is (1) (  ( p  q ))  p (2) p  q 1 11 (3)  (pq) (4) (  ( p  q ))  q (1) 4 (2) 50 Answer (1) 1 9 (3) (4) Sol.  p  ( q  ) ( p  q ) ) 5 50  p  ( q  ( p  q ) )  Answer (3) Sol. Bag have 6 white and 4 black balls  p  ( ( q  p )  (q  q ) )  Probability all drawn balls are white  p  ( q  p )  1  6C1 6 C2 6 C3 6 C4 6 C5 6 C6  =  10 + 10 + 10 + 10 + 10 + 10  p (q  p) 6  C1 C2 C3 C4 C5 C6  p  ( q  p) 504 1 = = p (pq )  p 2520 5 -6-  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning 1 1 1   + 1 Sol. 4 x 2 + 11x + 6  0  ( 4 x + 3 )( x + 2 )  0 0 dx = loge  , (5 + 2x − 2x2 )(1+ e(2−4x ) ) 17. If      −3   x  ( −, − 2 )   ,   …(i) ,  > 0, then 4 – 4 is equal to  4  (1) 19 (2) –21 −1  4 x + 3  1  −4  4 x  −2 (3) 0 (4) 21 Answer (4) 1 −1  x  − …(ii) 1 2 1 1 Sol. I =  dx 20  −  1  − 10 x + 6 5 (  ) −1   1  − 3  10 x + 6  3 4  2 x    2 − x − x  1 + e 2  3      −9  10 x  −3 1 1 1 =  dx −9 −3 20 2  −4 x −    1 x  11   1   …(iii)    −  x −   1+ e  2   10 10  4   2        −3 −1 (i), (ii), (iii)  x 1 4 2 Let x − = t , dx = dt 2 −3 −1 1 = , = 2 4 2 1 1 =  2 −1   2  dt 5  2   11  − t  1+ e   2  2  −4 t ( ) 36  +  = 36  4 = 45   19. Let x = x(y) be the solution of the differential 1 1 2 1 1 equation =  2 0  2  +  2  2 ( y + 2 ) loge ( y + 2 ) dx + ( x + 4 − 2loge ( y + 2))dy = 0,   (   11  − t 2  1 + e −4t   2  )   11  − t 2  1 + e 4t   2  ( )       ( ) ( y > –1 with x e 4 − 2 = 1. Then x e9 − 2 is equal ) 1 1 11 2 to 2 +t  1 1 1 1 2 0  11 2 = dt =  ln 2  (1) 3 2  11  11    − t2 2   −t  4  2  2  (2)  2  0 9  ( ) 2  1  11 + 1  1  11 + 1  32 = ln  = (3)  ln   2 11  11 − 1  2 11  10 9    10 (4) 1  11 + 1  3 = ln   11  10  Answer (3)   = 11,  = 10   −  = 21 4 4 Sol. Let x + 4 = u, y + 2 = v 18. If the domain of the function dx = du, dy = dv ( 2 ) f ( x ) = loge 4 x + 11x + 6 + sin (4 x + 3) + −1 (2v lnv)du = –(u – 2 lnv)dv  10 x + 6  cos−1   is (, ], then 36  +  is equal to 2v ln v du + u = 2ln v  3  dv (1) 54 (2) 72 du 1 1 (3) 63 (4) 45 + u = dv 2v ln v v Answer (4) -7-  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning 1 1 1 SECTION - B  ln( ln v ) 1 IF = e 2 v ln v = e 2 = ( ln v ) 2 Numerical Value Type Questions: This section 1 1 contains 10 questions. In Section B, attempt any five 1 u  ( ln v ) 2 =   ( ln v ) 2 dv questions out of 10. The answer to each question is a v NUMERICAL VALUE. For each question, enter the 1 3 2 u  ( ln v ) 2 = (ln v ) 2 + c …(i) correct numerical value (in decimal notation, 3 truncated/rounded-off to the second decimal place; y = e4 – 2  x = 1 e.g., 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the mouse and the on-screen virtual numeric keypad in  v = e4  u = 5 the place designated to enter the answer.  1 3 2 5   42  =  ( 4)2 + c 21. Let f ( x ) =  dx ,x  2. If f(0) = 0   3   (3 + 4x ) 2 4 – 3x 2 3 16 1  10 = +c and f (1) = tan–1   , ,  > 0, then 2 + 2 is 3   14 equal to _______. c= Answer (28) 3 dx y = e9 – 2  v = y + 2 = e9 Sol. f ( x ) =  (3 + 4 x 2 ) 4 − 3 x 2 2 14 14 1 1 (i)  u  3 =  27 + = 18 + Put x = , dx = − 2 dt 3 3 3 t t 14 −dt x +4 =u = 6+ f (x) =  9  4 3 t2 3 + 2  4 − 2  t  t 14 32 x = 2+ = − t dt 9 9 = (3t + 4) 4t 2 − 3 2 20. The mean and standard deviation of 10 4t 2 − 3 =  2  8t dt = 2 d  observations are 20 and 8 respectively. Later on, it  d was observed that one observation was recorded f (x) = −  as 50 instead of 40. Then the correct variance is   2 + 3   4 ·  3    + 4 ·   (1) 11 (2) 13   4   d d 1 d =− = − 3 (3) 12 (4) 14 =− Answer (2) 3 + 9 + 16 2 3 + 25 2 2 + 25 3 x1 + x2 +... + x9 + 50  3 Sol. = 20 1 3 =−  tan–1  +c 10 3 5  5    x1 + x2 + …+x9 = 150 3  3 (4 − 3 x 2 )  x12 + x22 +... + x92 + 2500 f (x) = − tan−1   +c 64 = − 400 15  5x    10 3 x12 + x22 +... + x92 = 2140 f (0) = 0  c = + 30 150 + 40 − 3  3 3  New mean = = 19 f (1) = tan−1   +  15  10  5  15 2 2140 + 1600 3  3 3  5  1  5  − (19 ) cot –1  tan–1  tan–1  2 = = New  =  5  15 =  10 15    3  5 3  3  = 13  2 + 2 = 28 -8-  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning 22. Let an ellipse with center (1, 0) and latus rectum of  = 11, (a, b)/(c, d) → (4, 7)/(5, 6) → 8 numbers 1  = 12, (a, b)/(c, d) → (5, 7) Not possible length have its major axis along x-axis. If its 2  = 13, (a, b)/(c, d) → (6, 7) Not possible minor axis subtends an angle 60º at the foci, then the square of the sum of the lengths of its minor and  = 14, (a, b)/(c, d) → (7, 7) Not possible major axes is equal to _______.  Total numbers = 72 Answer (9) 24. The number of elements in the set {n   : 10  n  Sol. 100 and 3n – 3 is a multiple of 7} is _______. Answer (15.00) Sol. 3n − 3 = 7k 3  3(mod 7) 32  2(mod 7) 33  –1 (mod 7) 36  1 (mod 7) 2b2 1 b 37  3 (mod 7) = , tan30 = a 2 ae a 1 b2 a2 b2 = , =  a2 − b2 = 3b2  b2 = 313  3 (mod 7) 4 3 a2 − b2 4 n can be 13, 19, 25, … 97 1 1  a = 1, b2 = b= Total 15 such n exist 4 2 25. Let A = {1, 2, 3, 4} and R be a relation on the set  (2a + 2b)2 = 9 A × A defined by R = {((a, b), (c, d)) : 2a + 3b = 4c 23. A person forgets his 4-digit ATM pin code. But he + 5d}. Then the number of elements in R is remembers that in the code all the digits are _______. different, the greatest digit is 7 and the sum of the Answer (06.00) first two digits is equal to the sum of the last two digits. Then the maximum number of trials Sol. 2a + 3b = 4c + 5d necessary to obtain the correct code is________. Maximum value of 2a + 3b = 20 at (4, 4) Answer (72) Minimum value of 4c + 5d = 9 at (1, 1) Sol. abcd a+b=c+d So, 4c + 5d can be equal to 9, 13, 14, 17, 18, 19 0, 1, 2, 3, 4, 5, 6, 7 2a + 3b can be 9  (a, b) = (3, 1) Let a + b = c + d =  (c, d) = (1, 1) There must be 7 2a + 3b can be 13  (a, b) = (2, 3)  a+b=c+d=7 (c, d) = (2, 1)  = 7, (a, b)/(c, d) → (0, 7)/(1, 6) 2a + 3b can be 14  (a, b) = (4, 2) OR (1, 4) or (2, 5)(3, 4) → 24 numbers (c, d) = (1, 2)  = 8, (a, b)/(c, d) → (1, 7)/(2, 6) or (3, 5) → 16 numbers 2a + 3b can be 17  (a, b) = (4, 3)  = 9, (a, b)/(c, d) → (2, 7)/(3, 6) (c, d) = (3, 1) or (4, 5) → 16 numbers 2a + 3b can be 18  (a, b) = (3, 4)  = 10, (a, b)/(c, d) → (3, 7)/(4, 6) → 8 numbers (c, d) = (2, 2) -9-  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning 26. Consider the triangles with vertices A(2, 1), B(0, 0) Answer (42.00) and C(t, 4), t  [0,4]. If the maximum and the minimum perimeters of such triangles are obtained Sol. at t =  and t =  respectively, then 6 + 21 is equal to ___________. Answer (48.00) Sol. To minimize CA + CB, take image of B in y=4 B = (0, 8)  2y 2   dy −  ( 2 )  3/2 2 1 A=  (3 − y ) − 0  3  8 AB 36 − 9 − 6  21  = − = − 8 4 8 4 28. If the line x = y = z intersects the line xsinA + ysinB + zsinC – 18 = 0 = xsin2A + ysin2B + zsin2C – 9, where A, B, C are the angles of a  A B C triangle ABC, then 80  sin sin sin  is equal to  2 2 2 −7 y–8= ( x − 0) __________. 2 Answer (5) when y = 4 Sol. x = y = z = k (let) −7 −4 = (x) 2  k (sin A + sin B + sin C ) = 18 8 8  A B C x= =  k  4cos  cos  cos  = 18 …(i) 7 7  2 2 2 Maximization will be possible if  = 0 or 4 k (sin 2 A + sin 2B + sin 2C ) = 9 When compared  = 4  k (4 sin A  sin B  sin C ) = 9 …(ii) 6 + 21 = 48 (ii)/(i) A B C 9 27. If the area bounded by the curve 2y2 = 3x, lines 8 sin  sin  sin = 2 2 2 18 x + y = 3, y = 0 and outside the circle (x – 3)2 + y2 = 2 A B C is A, then 4( + 4A) is equal to __________.  80 sin  sin  sin = 5 2 2 2 - 10 -  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning 29. If the sum of the series Sol. Plane containing the line 2x + y – 3 – 3 = 0 = 5x –  1 1  1 1 1   1 1 1 1  3y + 4z + 9 is 2x + y – z – 3 + (5x – 3y + 4z + 9) =  – + 2 – + + – + – + 2 3 2 2  3 32   23 22  3 2  32 33  0  1 1 1 1 1   4 – 3 + – +  +....  x(2 + 5) + y(1 – 3) + z(4 – 1) + 9 – 3 = 0 2 2 3 2 2 3 2 3 23 34   This plane is parallel to the line is , where  and  are co-prime, then  + 3 is  x +2 3−y z−7 = = equal to _________. 2 −4 5 Answer (7)  (2 + 5)(2) + (1 – 3)(4) + (4 – 1)5 = 0 1 1 Sol. Let a = , b = 2 3  4 + 10 + 4 – 12 + 20 – 5 = 0 Given: (a – b) + (a2 – ab + b2) + (a3 – a2b + ab2 – b3) −1  18 = –3   = +… 6  1 a+b (( ) ( ) ( ) ) a2 − b2 + a3 + b3 + a 4 − b 4 +...  Plane P : 7 6 3 5 9 x+ y − z− =0 2 3 2  1  2    a + b  ( 3 4 2 ( 3 4  a + a + a.... − b − b + b.... ))  7x + 9y – 10z – 27 = 0  1  a b2  2 =    −   a + b   1 − a 1 + b   1 1   4 1 9  =  −  1 1 1 1 +  1− 1 +  2 3 2 3 x − 8 y + 1 z + 19 61 1  AB : = = =k = − −3 4 12 5  2 12  B = (–3k + 8, 4k – 1, 12k – 19) 6 5  1  =  = = 5  12  2  B lies on plane P.  + 3 = 1 + 6 = 7 7(–3k + 8) + 9(4k – 1) –10(12k – 19) = 27 30. Let the plane P contain the line 2x + y – z – 3 = 0  –21k + 56 + 36k – 9 –120k + 190 = 27 = 5x – 3y + 4z + 9 and be parallel to the line x+2 3–y z–7  –105k = –210 = =. Then the distance of the 2 –4 5  k=2 point A(8, – 1, – 19) from the plane P measured x y –5 2–z  B = (2, 7, 5) parallel to the line = = is equal to –3 4 –12 AB = 36 + 64 + 576 ___________. Answer (26) = 676 = 26 - 11 -  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning PHYSICS SECTION - A Gmm Sol. mω2a = 2 4a Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices Gm ω= (1), (2), (3) and (4), out of which ONLY ONE is correct. 4a3 Choose the correct answer: 34. The position vector of a particle related to time t is 31. A flask contains Hydrogen and Argon in the ratio given by  2 : 1 by mass. The temperature of the mixture is r = (10t iˆ + 15t 2 jˆ + 7kˆ ) m 30°C. The ratio of average kinetic energy per The direction of net force experienced by the molecule of the two gases (K argon/K hydrogen) is: particle is : (Given : Atomic Weight of Ar = 39.9) (1) Positive x-axis (2) In x-y plane (1) 2 (2) 1 (3) Positive y-axis (4) Positive z-axis 39.9 (3) 39.9 (4) Answer (3) 2  Answer (2) Sol. F = 30 jˆ 1 2 4RT 35. The half-life of a radioactive nucleus is 5 years. The = Sol. KE = mv avg 2 π fraction of the original sample that would decay in 15 years is : KEH ⇒ 2 =1 1 1 KE Ar (1) (2) 8 4 32. A 12 V battery connected to a coil of resistance 6 Ω 3 7 through a switch, drives a constant current in the (3) (4) 8 4 circuit. The switch is opened in 1 ms. The emf induced across the coil is 20 V. The inductance of Answer (3) − t /T1/2 the coil is : Sol. N = N0 2 (1) 10 mH (2) 8 mH N0 (3) 5 mH (4) 12 mH ⇒ N= 8 Answer (1) Sol. V = 12 volt 7N0 ⇒ Decayed amount = R=6Ω 8 t = 1 ms 36. Match List-I with List II of Electromagnetic waves dφ dI with corresponding wavelength range: =L dt dt List I List II 2 20= L × (A) Microwave (I) 400 nm to 1 nm 10−3 L = 10 mH (B) Ultraviolet (II) 1 nm to 10–3 nm 33. Two identical particles each of mass ‘m’ go round a (C) X-Ray (III) 1 mm to 700 nm circle of radius a under the action of their mutual (D) Infra-red (IV) 0.1 m to 1 mm gravitational attraction. The angular speed of each Choose the correct answer from the options given particle will be : below: Gm Gm (1) (A)-(IV), (B)-(I), (C)-(II), (D)-(III) (1) (2) a 3 8a3 (2) (A)-(IV), (B)-(I), (C)-(III), (D)-(II) (3) (A)-(IV), (B)-(II), (C)-(I), (D)-(III) Gm Gm (3) (4) (4) (A)-(I), (B)-(IV), (C)-(II), (D)-(III) 3 4a 2a3 Answer (1) Answer (3) Sol. λX < λUV < λIR < λMW - 12 -  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning 37. A wire of length 'L' and radius 'r' is clamped rigidly Sol. 50 = (RG + R)10–3 at one end. When the other end of the wire is pulled 5000 = 54 + R by a force f, its length increases by 'l'. Another wire R ≈ 50 kΩ of same material of length '2L' and radius '2r' is pulled by a force '2f'. Then the increase in its length 10–3 × 54 = r × 9 × 10–3 will be: r=6Ω (1) 4l (2) l/2 40. The height of transmitting antenna is 180 m and the (3) 2l (4) l height of the receiving antenna is 245 m. The Answer (4) maximum distance between them for satisfactory communication in line of sight will be: l F Sol. σ= = (given R = 6400 km) L πr 2 y (1) 96 km (2) 56 km l′ 2F ′ σ= = ′ l ⇒ l= (3) 48 km (4) 104 km 2L 4πr 2 y Answer (4) 38. The de Broglie wavelength of an electron having kinetic energy E is λ. If the kinetic energy of Sol. Rmax = 2 × 180 × 6400 × 103 E electron becomes , then its de-Broglie + 2 × 245 × 6400 × 103 4 wavelength will be: = 6 × 80 × 102 + 7 × 80 × 102 λ = 104 km (1) 2λ (2) 2 41. A thermodynamic system is taken through cyclic λ process. The total work done in the process is : (3) (4) 2λ 2 Answer (4) h Sol. λ = 2mKE 39. For designing a voltmeter of range 50 V and an ammeter of range 10 mA using a galvanometer which has a coil of resistance 54 Ω showing a full scale deflection for 1 mA as in figure. (1) 200 J (2) 300 J (3) 100 J (4) Zero Answer (2) 2 × 300 Sol. W = J 2 = 300 J 42. A single slit of width a is illuminated by a (A) for voltmeter R ≈ 50 kΩ monochromatic light of wavelength 600 nm. The (B) for ammeter r ≈ 0.2 Ω value of ‘a’ for which first minimum appears at (C) for ammeter r ≈ 6 Ω θ = 30° on the screen will be : (D) for voltmeter R ≈ 5 kΩ (1) 1.2 µm (2) 3 µm (E) for voltmeter R ≈ 500 Ω (3) 1.8 µm (4) 0.6 µm Choose the correct answer from the options given Answer (1) below: (1) (A) and (C) (2) (C) and (E) Sol. dsinθ = λ ⇒ d = 2λ (3) (C) and (D) (4) (A) and (B) Answer (1) = 1.2 µm - 13 -  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning 43. A vector in x – y plane makes an angle of 30° with Choose the correct answer from the options given y-axis. The magnitude of y-component of vector is below: 2 3. The magnitude of x-component of the vector (1) (C) and (D) only will be : (2) (A), (C) and (D) only 1 (3) (A), (B) and (C) only (1) (2) 6 3 (4) (A), (B) and (D) only (3) 2 (4) 3 Answer (3) Answer (3) Sol. F = –kx Sol. ay = 2 3 a = –ω2x ∴ ax = ay tan30° Velocity is maximum at mean position. 1 47. The speed of a wave produced in water is given by = 2 3× 3 ν = λagbρc. Where λ, g and ρ are wavelength of wave, acceleration due to gravity and density of water respectively. The values of a, b and c 44. The position of a particle related to time is given by respectively, are x = (5t2 – 4t + 5) m. The magnitude of velocity of the particle at t = 2 s will be : (1) 1, –1, 0 (1) 06 ms–1 (2) 14 ms–1 1 1 (2) , 0, (3) 10 ms–1 (4) 16 ms–1 2 2 Answer (4) (3) 1, 1, 0 Sol. x = 5t2 – 4t + 5 1 1 (4) , ,0 ⇒ v = 10t – 4 2 2 ⇒ v = 16 m/s Answer (4) 45. The electric field due to a short electric dipole at a Sol. [ν] = [λagbρc] large distance (r) from center of dipole on the ⇒ [LT–1] = [L]a[LT–2]b[ML–3]c equatorial plane varies with distance as : 48. A body is released from a height equal to the radius (1) r (R) of the earth. The velocity of the body when it 1 strikes the surface of the earth will be: (2) r2 (Given g = acceleration due to gravity on the earth.) 1 (3) (1) 2gR r3 (2) gR 1 (4) r (3) 4gR Answer (3) gR 2kp (4) Sol. E = 2 r3 Answer (2) 46. In a linear Simple Harmonic Motion (SHM) (A) Restoring force is directly proportional to the GMm GMm 1 Sol. − = − + mv 2 displacement. 2R R 2 (B) The acceleration and displacement are 1 GMm ⇒ mv 2 = opposite in direction. 2 2R (C) The velocity is maximum at mean position. GM (D) The acceleration is minimum at extreme points. =v = gR R - 14 -  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning 49. Given below are two statements: 51. The fundamental frequency of vibration of a string Statement I : The equivalent resistance of resistors between two rigid support is 50 Hz. The mass of the in a series combination is smaller than least string is 18 g and its linear mass density is 20 g/m. resistance used in the combination. The speed of the transverse waves so produced in Statement II: The resistivity of the material is the string is _______ ms–1. independent of temperature. Answer (90) In the light of the above statements, choose the correct answer from the options given below: v Sol. = 50 (1) Both Statement I and Statement II are false 2l (2) Both Statement I and Statement II are true (3) Statement I is true but Statement II is false 100 × 18 v= 100 × = l = 90 m/s (4) Statement I is false but Statement II is true 20 Answer (1) 52. A block of mass 10 kg is moving along x-axis under Sol. Rseries > R1 or R2 the action of force F = 5x N. The work done by the as Rseries = R1 + R2 force in moving the block from x = 2 m to 4 m will ρ = ρ0 (1 + α∆T) be _______ J. 50. In the given circuit, the current (I) through the battery will be Answer (30) Sol. F = 5x 5 5 ( ) W = xf2 − xi2 = × 12 = 30 J 2 2 53. A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations respectively (ksph : kcyl) is 2 : x. The (1) 2.5 A (2) 1 A (3) 2 A (4) 1.5 A value of x is _______. Answer (4) Answer (5) Sol. Considering rotational axis as the diametrical axis for sphere and axis of cylinder. Then Sol. 2 2 1 K12 = R and K 22 = R 2 5 2 K1 2/5 4 ∴= = 20 K2 1/ 2 5 ⇒ Req = Ω 3 K1 2 30 ⇒ = =I = A 1.5 A K2 5 20 SECTION - B ∴ x=5 Numerical Value Type Questions: This section 54. An electron in a hydrogen atom revolves around its contains 10 questions. In Section B, attempt any five nucleus with a speed of 6.76 × 106 ms–1 in an orbit questions out of 10. The answer to each question is a of radius 0.52 Å. The magnetic field produced at the NUMERICAL VALUE. For each question, enter the nucleus of the hydrogen atom is _______ T. correct numerical value (in decimal notation, Answer (40) truncated/rounded-off to the second decimal place; µ0I µ0 e × ω e.g., 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using Sol. = B = 2r 2π × 2 × r the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. 10−7 × 1.6 × 6.76 × 106 × 10 −19 = = 40 0.52 × 0.52 × 10-20 - 15 -  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning 55. There is an air bubble of radius 1.0 mm in a liquid 58. As per given figure A, B and C are the first, second of surface tension 0.075 Nm–1 and density 1000 kg and third excited energy levels of hydrogen atom m–3 at a depth of 10 cm below the free surface. The respectively. If the ratio of the two wavelengths amount by which the pressure inside the bubble is  λ1  7 greater than the atmospheric pressure is _______  i.e.  is , then the value of n will be  λ2  4n Pa (g = 10 ms–2) Answer (1150) 2T 2 × 0.075 Sol. ∆P =ρgh += 1000 + r 10−3 = 1000 + 150 = 1150 56. In the given figure the total charge stored in the combination of capacitors is 100 µC. The value of Answer (5) ‘x’ is ______________. 1 1  7 λ1  9 − 16  16 = Sol. = λ2  1 1 5  −  4 9 4 λ1 7 ⇒ = λ2 20 59. The refractive index of a transparent liquid filled in an equilateral hollow prism is 2. The angle of minimum deviation for the liquid will be _______°. Answer (30)  δ sin  30 +  Sol. 2=  2 Answer (5) 1 Sol. 10(2 + x + 3) = 100 2 ⇒ x=5 δ 57. A 20 cm long metallic rod is rotated with 210 rpm ⇒ 45°= 30° + 2 about an axis normal to the rod passing through its δ = 30° one end. The other end of the rod is in contact with 60. A network of four resistances is connected to 9 V a circular metallic ring. A constant and uniform battery, as shown in figure. The magnitude of magnetic field 0.2 T parallel to the axis exists voltage difference between the points A and B is everywhere. The emf developed between the __________ V. centre and the ring is _________mV. 22 Take π = 7 Answer (88) B ωl 2 Sol. ε = 2 210 × 2π (0.2)2 0.2 × = × 60 2 0.2 × 210 × 22 × 0.04 = Answer (3) 7 × 60 Sol. IA = IB = 1.5 A = 88 mV VA – VB = 4 × 1.5 – 2 × 1.5 = 3 - 16 -  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning CHEMISTRY SECTION - A Multiple Choice Questions: This section contains 20 Sol. multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct. Choose the correct answer: 61. The product formed in the following multistep 63. Which of the following statement(s) is/are correct? reaction is: (A) The pH of 1 × 10–8 M HCl solution is 8. (i) B2H6 (B) The conjugate base of H2PO 4− is HPO24−. (ii) H2O2 , NaOH CH3 − CH = CH2 ⎯⎯⎯⎯⎯⎯⎯→ (C) Kw increases with increase in temperature. (iii) PCC (iv) CH3MgBr (D) When a solution of a weak monoprotic acid is titrated against a strong base at half O || 1 (1) CH3 − CH2 − C− OCH3 neutralisation point, pH = pK a. 2 OH Choose the correct answer from the options given | below: (2) CH3 − C − CH3 | (1) (B), (C) (2) (A), (D) CH3 (3) (A), (B), (C) (4) (B), (C), (D) OH Answer (1) | (3) CH3 − CH2 − CH− CH3 Sol. H2PO −4 HPo −2 4 + H+ acid Conjugate base (4) CH3 − CH2 − CH2 − CH2 − OH Kw increase with increase in temperature as Answer (3) dissociation of water increases with increase in Sol. temperature. When weak monoprotic acid is titrated against strong base, at half neutralization point, the solution becomes acidic buffer and its pH is given by [Salt] pH = pK a + log [Acid]  pH = pKa + 0 as [salt] = [acid] 64. Which is not true for arginine? 62. The major product formed in the Friedel-Craft (1) It has a fairly high melting point. acylation of chlorobenzene is (2) It is associated with more than one pKa values. Cl Cl (3) It has high solubility in benzene. COCH 3 (4) It is a crystalline solid. (1) (2) CH3 Answer (3) COCH 3 O Sol. O CH 3 Cl Cl (3) (4) CH 3 O O CH 3 Answer (4) Since arginine is polar, it is not soluble in benzene. - 17 -  JEE (Main)-2023 : Phase-2 (15-04-2023)-Morning 65. The number of P – O – P bonds in H4P2O7, Sol. Complex CFSE (HPO3)3 and P4O10 are respectively [Ti(OH2)6]3+ –0.4 0 (1) 0, 3, 6 (2) 0, 3, 4 [Cr(H2O)6]3+ –1.2 0 (3) 1, 2, 4 (4) 1, 3, 6 [Mn(H2O)6]3+ –0.6 0 Answer (4) [Fe(H2O)6]3+ 0 Sol. H4P2O7: 68. During