JEE Main 2024 January Session Question Papers PDF

JEE Main 2024 January Question Paper with Answer 27th, 29th, 30th, 31st January & 1st February (Shift 1 & Shift 2) JEE Main 2024 – 27th Jan Shift 1 Question Paper Page No. 2 to 23 JEE Main 2024 – 27th Jan Shift 2 Question Paper Page No. 24 to 48 JEE Main 2024 –...

JEE Main 2024 January Question Paper with Answer 27th, 29th, 30th, 31st January & 1st February (Shift 1 & Shift 2) JEE Main 2024 – 27th Jan Shift 1 Question Paper Page No. 2 to 23 JEE Main 2024 – 27th Jan Shift 2 Question Paper Page No. 24 to 48 JEE Main 2024 – 29th Jan Shift 1 Question Paper Page No. 49 to 71 JEE Main 2024 – 29th Jan Shift 2 Question Paper Page No. 72 to 95 JEE Main 2024 – 30th Jan Shift 1 Question Paper Page No. 96 to 120 JEE Main 2024 – 30th Jan Shift 2 Question Paper Page No. 121 to 144 JEE Main 2024 – 31st Jan Shift 1 Question Paper Page No. 145 to 170 JEE Main 2024 – 31st Jan Shift 2 Question Paper Page No. 171 to 197 JEE Main 2024 – 1st Feb Shift 1 Question Paper Page No. 198 to 224 JEE Main 2024 – 1st Feb Shift 2 Question Paper Page No. 225 to 250 Download more JEE Main Previous Year Question Papers: Click Here 1  FINAL JEE–MAIN EXAMINATION – JANUARY, 2024 (Held On Saturday 27th January, 2024) TIME : 9 : 00 AM to 12 : 00 NOON MATHEMATICS TEST PAPER WITH SOLUTION SECTION-A Sol. B = (2 +7, 3 – 2, 6 + 11) 1. n 1 Cr   k 2  8 n Cr 1 if and only if : (1) 2 2  k  3 (2) 2 3  k  3 2 (3) 2 3  k  3 3 (4) 2 2  k  2 3 Ans. (1) Sol. n-1 Cr = (k2 – 8) nCr+1 r  1  0, r  0 x  6 y  4 z 8 r 0 Point B lies on   n 1 1 0 3 Cr n  k2  8 2  7  6 3  2  4 6  11  8 Cr 1   1 0 3 r 1  k2  8 3 – 6 = 0 n  k2  8  0  = –2  k  2 2  k  2 2   0 B  (3, 4, –1) k   ,  2 2    2 2,    7  3   4  2   11  1 2 2 2 …(I) AB  r 1  16  36  144  n  r  1, 1 n  196  14  k2 – 8 1 3. Let x = x(t) and y = y(t) be solutions of the k2 – 9 0 –3 k  3 ….(II) dx differential equations  ax  0 and From equation (I) and (II) we get dt    k   3,  2 2  2 2, 3 dy dt  by  0 respectively, a, b  R. Given that 2. The distance, of the point (7, –2, 11) from the line x(0) = 2; y(0) = 1 and 3y(1) = 2x(1), the value of t, x  6 y  4 z 8   along the line for which x(t) = y(t), is : 1 0 3 x  5 y 1 z  5 (1) log 2 2 (2) log 4 3   , is : 3 2 3 6 (1) 12 (2) 14 (3) log 3 4 (4) log 4 2 3 (3) 18 (4) 21 Ans. (2) Ans. (4) 2  dx Sol. Equation of CE Sol.  ax  0 dt y – 1 = (x – 3) dx x+y=4  adt x dx  x  a  dt ln | x | at  c at t = 0, x = 2 ln 2  0  c orthocentre lies on the line x + y = 4 ln x  at  ln 2 so, a + b = 4 x  eat b 2 I1   x sin  x(4  x)  dx …(i) x  2e  at ….(i) a dy Using king rule  by  0 b dt I1    4  x  sin  x(4  x)  dx …(ii) dy  bdt a y (i) + (ii) ln | y | bt   b t = 0, y = 1 2I1   4sin  x(4  x)  dx 0=0+ a y = e–bt ….(ii) 2I1 = 4I2 According to question I1 = 2I2 3y(1) = 2x(1) I1 3e–b = 2(2 e–a) 2 I2 4 ea  b  36I1 3  72 For x(t) = y(t) I2  2e–at = e–bt 5. If A denotes the sum of all the coefficients in the 2 = e(a – b)t expansion of (1 – 3x + 10x2)n and B denotes the t 4 sum of all the coefficients in the expansion of 2  3 (1 + x2)n, then : log 4 2  t (1) A = B3 (2) 3A = B 3 3 (3) B = A (4) A = 3B 4. If (a, b) be the orthocentre of the triangle whose Ans. (1) vertices are (1, 2), (2, 3) and (3, 1), and b b Sol. Sum of coefficients in the expansion of I1   x sin  4x  x 2  dx , I 2   sin  4x  x 2  dx (1 – 3x + 10x2)n = A a a then A = (1 – 3 + 10)n  8n (put x = 1) I1 and sum of coefficients in the expansion of , then 36 is equal to : I2 (1 + x2)n = B (1) 72 (2) 88 then B = (1 + 1)n = 2n (3) 80 (4) 66 A = B3 Ans. (1) 3 6. The number of common terms in the progressions x  4 y 1 z Sol.   3 th 4, 9, 14, 19,...... , up to 25 term and 3, 6, 9, 12, 1 2......., up to 37th term is : x   y 1 z  2   (1) 9 (2) 5 2 4 5 (3) 7 (4) 8 the shortest distance between the lines a  b  d  d  Ans. (3) Sol. 4, 9, 14, 19, …., up to 25th term  1 2 T25 = 4 + (25 – 1) 5 = 4 + 120 = 124 d1  d 2 3, 6, 9, 12, …, up to 37th term T37 = 3 + (37 – 1)3 = 3 + 108 = 111 4 0 2 st Common difference of I series d1 = 5 1 2 3 nd Common difference of II series d2 = 3 2 4 5  First common term = 9, and i j kˆ their common difference = 15 (LCM of d1 and d2) 1 2 3 then common terms are 2 4 5 9, 24, 39, 54, 69, 84, 99 If the shortest distance of the parabola y2 = 4x from 7.     4  10  12   0  2  4  4  the centre of the circle x2 + y2 – 4x – 16y + 64 = 0 2i  1j  0kˆ is d, then d2 is equal to : (1) 16 (2) 24 6 2    4 (3) 20 (4) 36  5 5 Ans. (3) 3 = | – 4| Sol. Equation of normal to parabola y = mx – 2m – m3  – 4 = ±3 this normal passing through center of circle (2, 8)  = 7, 1 8 = 2m – 2m – m 3 Sum of all possible values of  is = 8 m = –2 1 1 So point P on parabola  (am2, –2am) = (4, 4) 9. If  0 3  x  1 x dx  a  b 2  c 3 , where And C = (2, 8) a, b, c are rational numbers, then 2a + 3b – 4c is PC = 4  16  20 equal to : 2 d = 20 (1) 4 (2) 10 8. If the shortest distance between the lines (3) 7 (4) 8 x  4 y 1 z x   y 1 z  2 Ans. (4)   and   is 1 2 3 2 4 5 3  x  1 x 1 1 1 6 , then the sum of all possible values of  is : Sol.  0 3  x  1 x dx   0  3  x   1  x  dx 5 1    1 1 2  0 (1) 5 (2) 8  3  x dx   1  x dx  (3) 7 (4) 10 0  Ans. (2) 4   3 3  1 11. If S = {z  C : |z – i| = |z + i| = |z–1|}, then, n(S) is: 1   3  x  2 2 1  x  2  2  (1) 1 (2) 0 2 3 3    0 (3) 3 (4) 2 1 2 2  32     Ans. (1)  8  3 3   2  1  2  3 3   Sol. |z – i| = |z + i| = |z – 1| 1 8  3 3  2 2  1 3 2  3 3  2  a b 2 c 3 3 2 ABC is a triangle. Hence its circum-centre will be a  3, b   , c  1 3 the only point whose distance from A, B, C will be 2a + 3b – 4c = 6 – 2 + 4 = 8 same. 10. Let S = {l, 2, 3,... , 10}. Suppose M is the set of all So n(S) = 1 the subsets of S, then the relation R = {(A, B): A  B  ; A, B  M} is : 12. Four distinct points (2k, 3k), (1, 0), (0, 1) and (1) symmetric and reflexive only (0, 0) lie on a circle for k equal to : (2) reflexive only 2 3 (1) (2) (3) symmetric and transitive only 13 13 (4) symmetric only 5 1 Ans. (4) (3) (4) 13 13 Sol. Let S = {1, 2, 3, …, 10} Ans. (3) R = {(A, B): A  B  ; A, B  M} Sol. (2k, 3k) will lie on circle whose diameter is AB. For Reflexive, M is subset of ‘S’ So  M for    =   but relation is A  B   So it is not reflexive. (x – 1) (x) + (y – 1) (y) = 0 For symmetric, x2 + y 2 – x – y = 0 …(i) ARB A  B  , Satisfy (2k, 3k) in (i)  BRA  B  A  , (2k)2 + (3k)2 – 2k – 3k = 0 So it is symmetric. 13k2 – 5k = 0 For transitive, 5 If A = {(1, 2), (2, 3)} k = 0, k  B = {(2, 3), (3, 4)} 13 C = {(3, 4), (5, 6)} 5 hence k  ARB & BRC but A does not relate to C 13 So it not transitive 5 13. Consider the function. 10  a  7x  12  x 2  Sol. a k 1 k  50  , x3  b x  7x  12 2 a1 + a2 + … + a10 = 50 ….(i)   f (x)   sin  x 3 2 x [x ] , x 3 a a k  j k j  1100....(ii)  , x 3  b If a1 + a2 + … + a10 = 50.  (a1 + a2 + … + a10)2 = 2500   10  a 2 i  2 a k a j  2500 Where [x] denotes the greatest integer less than or i 1 k j equal to x. If S denotes the set of all ordered pairs 10 (a, b) such that f(x) is continuous at x = 3, then the  a i 1 2 i  2500  2 1100  number of elements in S is : 10 (1) 2 (2) Infinitely many a i 1 2 i  300 , Standard deviation ‘’ (3) 4 (4) 1 2 Ans. (4)    a   a i  300  50 2 2 a  7x  12  x  2 i Sol. f 3    (for f(x) to be cont.)       b x 2  7x  12 10  10  10  10      a  x  3 x  4  a  f(3–) = ;x  3   30  25  5 b  x  3 x  4  b x 2 y2 a   1, Hence f 3    b 15. The length of the chord of the ellipse 25 16  sin  x 3   2 lim   whose mid point is  1,  , is equal to :  Then f 3  2  x3  x 3   2 and  5 f(3) = b. 1691 2009 (1) (2) Hence f(3) = f(3+) = f(3–) 5 5 a 1741 1541  b=2=  (3) (4) b 5 5 b = 2, a = –4 Ans. (1) Hence only 1 ordered pair (–4, 2). Sol. Equation of chord with given middle point. 14. Let a1, a2, ….. a10 be 10 observations such that T = S1 10 x y 1 1 a k 1 k  50 and a k  j k  a j  1100. Then the    25 40 25 100 standard deviation of a1, a2,.., a10 is equal to : 8x  5y 8  2  200 200 (1) 5 (2) 5 10  8x (3) 10 (4) 115 y …(i) 5 Ans. (2) 6  x 2 10  8x  2  1 (put in original equation) 25 400 16x 2  100  64x 2  160x 1 400 4x2 – 8x – 15 = 0 8  304 x m1  m2 8 tan   1  m1m 2 8  304 8  304 x1  ; x2  8 8 6 30 10  18  304 2  304 tan   5  Similarly, y   16 41 5 5 1 25 2  304 2  304 30 y1  ; y2  tan   5 5 41 Distance =  x1  x 2    y1  y2  2 2 17.   Let a  i  2j  k , b  3 i  j  k. Let c be the 4  304 4  304 1691 vector such that a  c  b and a  c  3. Then    64 25 5 16. The portion of the line 4x + 5y = 20 in the first    a  c  b  b  c is equal to : quadrant is trisected by the lines L1 and L2 passing (1) 32 (2) 24 through the origin. The tangent of an angle (3) 20 (4) 36 between the lines L1 and L2 is : Ans. (2)   8 25 (1) (2) Sol. a   c  b  b  c 5 41   (3) 2 (4) 30   a  c b  a b  a c ….(i) 5 41 Ans. (4) given a  c  b 5 8 a  c  b  b  b  b 2 Sol. Co-ordinates of A   ,    27  3 3 Co-ordinates of B    10 4  ,         a  c  b  a c b   a  c  b  27 …(ii)  3 3 Now a  b  3  6  3  0 …(iii) 8 Slope of OA = m1  a c  3 …(iv) (given) 5 2 By (i), (ii), (iii) & (iv) Slope of OB = m2  5 27 – 0 – 3 = 24 7  1 1 x4  2  cos x sin x 0  18. If a  lim x 0 x4 and Sol. f ( x)    sin x cos x 0   0 0 1  sin 2 x b  lim , then the value of ab3 is : x 0 2  1  cos x 1 0 0  (1) 36 (2) 32 (3) 25 (4) 30 f (x)  f ( x)  0 1 0   I Ans. (2) 0 0 1  1 1 x4  2 Sol. a  lim Hence statement- I is correct x 0 x4 Now, checking statement II 1  x4 1  lim cos y  sin y 0  x 4  1  1  x 4  2  x 0   f (y)   sin y cos y 0   0 0 1  x4  lim x 4  1  1  x 4  2  x 0    1 x4 1  cos(x  y)  sin(x  y) 0  f (x)  f (y)   sin(x  y) cos(x  y) 0  1 Applying limit a   0 0 1  4 2 sin 2 x  f (x)  f (y)  f (x  y) b  lim x 0 2  1  cos x Hence statement-II is also correct. 1  cos x   2 2  1  cos x  20. The function f : N – {1} N; defined by f(n) =  lim x 0 2  1  cos x  the highest prime factor of n, is : b  lim 1  cos x  x 0  2  1  cos x  (1) both one-one and onto (2) one-one only Applying limits b  2  2 2 4 2 (3) onto only 1   3 Now, ab3   4 2  32 (4) neither one-one nor onto 4 2 Ans. (4) cos x  sin x 0    Sol. f : N – {1} N 19. Consider the matrix f (x)  sin x cos x 0.    0 f(n) = The highest prime factor of n. 0 1  Given below are two statements : f(2) = 2 Statement I: f(–x) is the inverse of the matrix f(x). f(4) = 2 Statement II: f(x) f(y) = f(x + y).  many one In the light of the above statements, choose the correct answer from the options given below 4 is not image of any element (1) Statement I is false but Statement II is true  into (2) Both Statement I and Statement II are false Hence many one and into (3) Statement I is true but Statement II is false (4) Both Statement I and Statement II are true Neither one-one nor onto. Ans. (4) 8  SECTION-B 23. If the solution of the differential equation 21. The least positive integral value of , for which the (2x + 3y – 2) dx + (4x + 6y – 7) dy = 0, y(0) = 3, is x + y + 3 loge |2x + 3y – | = 6, then  + 2 + 3 angle between the vectors i  2j  2k and is equal to _____. i  2 j  2k is acute, is _____. Ans. (29) Ans. (5) Sol. 2x + 3y –2 = t 4x + 6y – 4 = 2t Sol. cos    ˆi  2ˆj  2kˆ    ˆi  2ˆj  2kˆ  23 dy dt  4x + 6y – 7 = 2t – 3 dx dx  44 2   4  4 2 2 dy   2x  3y  2    4  4 2  cos   dx 4x  6y  7 2  8 5 2  4 dt 3t  4t  6 t  6   2 – 4– 4 > 0 dx 2t  3 2t  3  2 – 4+ 4 > 8  – 2)2 > 8 2t  3    2  2 2 or   2  2 2  t 6 dt   dx   2  2 2 or   2  2 2  2t  12 9    t 6  t  6   dt  x – , – 0.82)  (4.82, ) 2t + 9 ln (t – 6) = x + c Least positive integral value of 5 2(2x + 3y – 2) + 9ln(2x + 3y – 8 ) = x + c 22. Let for a differentiable function f : (0, ) R , x = 0, y = 3 x c = 14 f (x)  f (y)  log e    x  y ,  x, y  (0, ). 4x + 6y – 4 + 9ln (2x + 3y – 8) = x+14  y x + 2y + 3 ln (2x + 3y – 8) = 6 20  1  Then  f '  2  is equal to _____. 1, 2,  = 8 n 1 n   3 = 1 + 4 + 24 = 29 Ans. (2890) 24. Let the area of the region {(x, y) : x – 2y + 4  0, Sol. f(x) – f(y) ln x – lny + x – y m x + 2y2  0, x + 4y2  8, y  0} be , where m f x  f y ln x  ln y n  1 xy xy and n are coprime numbers. Then m + n is equal to Let x > y _____.   1 lim f ' x    1 y x x …..(1) Ans. (119) Let x < y y x   1 lim f ' x    1 ….. (2) x (  , 0) –2 –1 1 – 1 + f (x ) = f (x ) 1 f1  x   1 Sol. x 1  1 f ' 2    x 1 2    A    8  4y2  2y2  dy   x  0 3/2 x   8  4y    2y  4  dy 20 20 1   x  20 2 2 2 x 1 x 1 1 1 3/ 2 20  21  41  2y 3   4y 3  107 m   20  8y    12y  y  2    6  3 0  3 1 12 n = 2890 m + n = 119 9 25. If 27. Let the set of all a  R such that the equation 1 1 1 cos 2x  a sin x  2a  7 has a solution be [p, q] 8  3 3  p   2 3  2p   3 3  3p  .... , 4 4 4 1 then the value of p is _____. and r  tan 9  tan 27   tan81 , then cot 63 Ans. (9) pqr is equal to _____. 1 p 3 4 Ans. (48) Sol. 8  1  1 2 1 4  1  4  Sol. cos2x + a·sinx = 2a – 7 a dr a  sin x  2   2  sin x  2  sin x  2  (sum of infinite terms of A.G.P =  ) 1  r 1  r 2 sin x  2, a  2  sin x  2  4p  4 p9  a  [2, 6] 9 26. A fair die is tossed repeatedly until a six is p=2 q=6 obtained. Let X denote the number of tosses r = tan 9° + cot 9° – tan27 – cot27 required and let a = P(X = 3), b = P(X  3) and c = 1 1 r  bc sin 9  cos9 sin 27  cos27 P(X  6 |X > 3). Then is equal to _____. a  4 4   2   Ans. (12)  5 1 5  1 5 5 1 25 r=4 Sol. a  P  X  3     6 6 6 216 p. q. r = 2 × 6 × 4 = 48 3 4 5 5 1 5 1 5 1 b  P  X  3            ...... 28. Let f (x)  x 3  x 2f '(1)  xf ''(2)  f '''(3) , x  R. 6 6 6 6 6 6 6 25 Then f '(10) is equal to _____. 25 6 25  216    5 216 1 36 Ans. (202) 1 6 5 6 Sol. f(x) = x3 + x2 · f (1) + x · f (2) + f(3) 5 1 5 1 P  X  6          ....... 6 6 6 6 f '(x)  3x 2  2xf '(1)  f ''(2) 5 5 1 f ''(x)  6x  2f '(1) 6 6 5 5     1 5 6 f '''(x)  6 6 5 f(1) = –5, f (2) = 2, f(3) = 6 5 6 f(x) = x3 + x2 · (–5) + x · (2) + 6 c   3  25 5 36 f(x) = 3x2 – 10x + 2 6   f(10) = 300 – 100 + 2 = 202 2 2 5 5  b  c  6   6   2 = 12 a 5 1 6 6   10  2 0 1 x1 = 1, y1 = – 1, z1 = –1 29.   Let A  1 1 0 , B = [B1, B2, B3], where B1, 2 0 1   x 2  2    AB2  1 1 0   y 2    3  1 0 1  1 0 1   z 2  0  1  x2 = 2, y2 = 1, z2 = –2   B2, B3 are column matrices, and AB1  0 ,   2 0 1   x 3   3 0  AB3  1 1 0   y 3   2  1 0 1   z3  1  2  3 AB2   3 , AB3   2   x3 =2, y3 = 0, z3 = –1 0  1  1 2 2 B   1 1 0  If  = |B| and  is the sum of all the diagonal  1 2 1 elements of B, then 3 + 3 is equal to _____. |B| = 3 Ans. (28)  1 2 0 1   3 + 3 = 27 + 1 =28 Sol. A  1 1 0  B = [B1, B2, B3] 30. If  satisfies the equation x2 + x + 1 = 0 and 1 0 1  (1 + )7 = A + B + C2, A, B, C  0, then  x1  x2  x3  5(3A – 2B – C) is equal to _____. B1   y1  , B2   y 2  , B3   y 3   z1   z 2   z3  Ans. (5) 2 0 1   x1  1  Sol. x2 + x + 1 = 0  x = , 2 =  AB1  1 1 0   y1   0  Let  =  1 0 1   z1  0  Now (1 + )7 =  = 2 = 1 +  A = 1, B = 1, C = 0 5(3A – 2B – C) 5(3 – 2 – 0) = 5 11  PHYSICS TEST PAPER WITH SOLUTION SECTION-A 33. If the refractive index of the material of a prism is 31. Position of an ant (S in metres) moving in Y-Z A cot   , where A is the angle of prism then the plane is given by S  2t 2 ˆj  5kˆ (where t is in 2 angle of minimum deviation will be second). The magnitude and direction of velocity  of the ant at t = 1 s will be : (1)   2A (2)  2A 2 (1) 16 m/s in y-direction  (3)   A (4)  A (2) 4 m/s in x-direction 2 Ans. (1) (3) 9 m/s in z-direction  A  min  (4) 4 m/s in y-direction sin   Sol. A cot   2  Ans. (4) 2 A sin 2 ds Sol. v  4t j A  A  min  dt  cos  sin   2  2  At t = 1 sec v  4 j A  min  A   32. Given below are two statements : 2 2 2 min    2A Statement (I) :Viscosity of gases is greater than 34. A proton moving with a constant velocity passes that of liquids. through a region of space without any change in its Statement (II) : Surface tension of a liquid velocity. If E and B represent the electric and decreases due to the presence of insoluble magnetic fields respectively, then the region of space may have : impurities. (A) E  0, B  0 (B) E  0, B  0 In the light of the above statements, choose the (C ) E  0, B  0 (D) E  0, B  0 most appropriate answer from the options given Choose the most appropriate answer from the below : options given below : (1)(A), (B) and (C) only (1) Statement I is correct but statement II is (2) (A), (C) and (D) only incorrect (3) (A), (B) and (D) only (2) Statement I is incorrect but Statement II is (4) (B), (C) and (D) only correct Ans. (3) Sol. Net force on particle must be zero i.e. (3) Both Statement I and Statement II are incorrect qE  qV  B  0 (4) Both Statement I and Statement II are correct Possible cases are Ans. (2) (i) E & B  0 Sol. Gases have less viscosity. (ii) V  B  0, E  0 Due to insoluble impurities like detergent surface (iii) qE  qV  B tension decreases E  0&B 0 12 35. The acceleration due to gravity on the surface of 38. Identify the physical quantity that cannot be earth is g. If the diameter of earth reduces to half of measured using spherometer : its original value and mass remains constant, then (1) Radius of curvature of concave surface acceleration due to gravity on the surface of earth (2) Specific rotation of liquids would be : (3) Thickness of thin plates (1) g/4 (2) 2g (4) Radius of curvature of convex surface (3) g/2 (4) 4g Ans. (2) Ans. (4) Sol. Spherometer can be used to measure curvature of GM 1 Sol. g g 2 surface. R2 R 39. Two bodies of mass 4 g and 25 g are moving with g 2 R12  equal kinetic energies. The ratio of magnitude of g1 R 22 their linear momentum is :  R1  (1) 3 : 5 (2) 5 : 4 g2 = 4g1  R 2   2  (3) 2 : 5 (4) 4 : 5 36. A train is moving with a speed of 12 m/s on rails Ans. (3) which are 1.5 m apart. To negotiate a curve radius P12 P2 Sol.  2 400 m, the height by which the outer rail should be 2m1 2m 2 raised with respect to the inner rail is (Given, g = 10 m/s2) : P1 m1 2   (1) 6.0 cm (2) 5.4 cm P2 m2 5 (3) 4.8 cm (4) 4.2 cm 40. 0.08 kg air is heated at constant volume through Ans. (2) 5°C. The specific heat of air at constant volume is v2 12  12 Sol. tan    0.17 kcal/kg°C and J = 4.18 joule/cal. The change Rg 10  400 in its internal energy is approximately. (1) 318 J (2) 298 J (3) 284 J (4) 142 J Ans. (3) h tan   Sol. Q = U as work done is zero [constant volume] 1.5 U = ms  T h 144   = 0.08 × (170 × 4.18) × 5 1.5 4000 h = 5.4 cm 284 J 37. Which of the following circuits is reverse - biased ? 41. The radius of third stationary orbit of electron for Bohr's atom is R. The radius of fourth stationary orbit will be: 4 16 (1) R (2) R 3 9 (1) (2) 3 9 (3) R (4) R 4 16 Ans. (2) n2 Sol. r Z (3) (4) r4 4 2  Ans. (4) r3 32 Sol. P end should be at higher potential for forward 16 biasing. r4  R 9 13 42. A rectangular loop of length 2.5 m and width 2 m 45. A body of mass 1000 kg is moving horizontally is placed at 60° to a magnetic field of 4 T. The with a velocity 6 m/s. If 200 kg extra mass is loop is removed from the field in 10 sec. The added, the final velocity (in m/s) is: average emf induced in the loop during this time is (1) 6 (2) 2 (1) – 2V (2) + 2V (3) 3 (4) 5 (3) + 1V (4) – 1V Ans. (4) Ans. (3) Sol. Momentum will remain conserve Change in flux  Sol. Average emf= =– 1000 × 6 = 1200 × v Time t v = 5 m/s 0   4   2.5  2  cos60  46. A plane electromagnetic wave propagating in = – 10 x-direction is described by = +1V Ey = (200 Vm–1) sin[1.5 × 107t – 0.05 x] ; 43. An electric charge 10–6C is placed at origin (0, 0) The intensity of the wave is : m of X –Y co-ordinate system. Two points P and (Use 0 = 8.85 × 10–12 C2N–1m–2) Q are situated at ( 3, 3)m and ( 6,0)m (1) 35.4 Wm–2 (2) 53.1 Wm–2 respectively. The potential difference between the (3) 26.6 Wm–2 (4) 106.2 Wm–2 points P and Q will be : Ans. (2) (1) 3V 1 (2) 6V Sol. I  0 E02  c 2 (3) 0 V 1 (4) 3 V I   8.85 1012  4 104  3 108 2 Ans. (3) I = 53.1 W/m2 KQ KQ Sol. Potential difference =  47. Given below are two statements : r1 r2 Statement (I) : Planck's constant and angular  3   3 2 2 r1  momentum have same dimensions. Statement (II) : Linear momentum and moment of  6 2 r2  0 force have same dimensions. As r1  r2  6m In the light of the above statements, choose the So potential difference = 0 correct answer from the options given below : 44. A convex lens of focal length 40 cm forms an (1) Statement I is true but Statement II is false image of an extended source of light on a photo- (2) Both Statement I and Statement II are false electric cell. A current I is produced. The lens is (3) Both Statement I and Statement II are true replaced by another convex lens having the same (4) Statement I is false but Statement II is true diameter but focal length 20 cm. The photoelectric Ans. (1) current now is : Sol. [h] = ML2T–1 I (1) (2) 4 I [L] = ML2T–1 2 (3) 2 I (4) I [P] = MLT–1 Ans. (4) [] = ML2T–2 Sol. As amount of energy incident on cell is same so (Here h is Planck's constant, L is angular current will remain same. momentum, P is linear momentum and  is moment of force) 14 48. A wire of length 10 cm and radius 7  10–4 m SECTION-B connected across the right gap of a meter bridge. 51. A particle starts from origin at t = 0 with a velocity When a resistance of 4.5  is connected on the left gap 5iˆ m / s and moves in x-y plane under action of a by using a resistance box, the balance length is found force which produces a constant acceleration of to be at 60 cm from the left end. If the resistivity of the (3iˆ  2 ˆj)m / s2. If the x-coordinate of the particle wire is R × 10–7m, then value of R is : (1) 63 (2) 70 at that instant is 84 m, then the speed of the particle (3) 66 (4) 35 at this time is  m / s. The value of  is ______. Ans. (3) Ans. (673) Sol. For null point, Sol ux = 5 m/s ax = 3 m/s2 x = 84 m 4.5 R  v  u  2ax 2 2 60 40 x x   v2x  25  2  3 84  Also, R   2 A r Vx = 23 m/s 0.1 vx  u x  a x t 4.5  40    60  7  108 23  5 t  6s 7 3   66  10  m vy  0  a y t = 0 + 2 × (6) = 12 m/s 49. A wire of resistance R and length L is cut into 5 equal parts. If these parts are joined parallely, v2  v2x  v2y  232  122  673 then resultant resistance will be : v  673 m/s 1 1 52. A thin metallic wire having cross sectional area of (1) R (2) R 25 5 10–4 m2 is used to make a ring of radius 30 cm. A (3) 25 R (4) 5 R positive charge of 2 C is uniformly distributed Ans. (1) over the ring, while another positive charge of 30 R pC is kept at the centre of the ring. The tension in Sol. Resistance of each part = 5 the ring is ______ N ; provided that the ring does 1 R R not get deformed (neglect the influence of gravity). Total resistance =   5 5 25 1 50. The average kinetic energy of a monatomic (given,  9  109 SI units) 40 molecule is 0.414 eV at temperature : (Use KB = 1.38 × 10–23 J/mol-K) Ans. (3) (1) 3000 K Fe (2) 3200 K (3) 1600 K T T d (4) 1500 K Ans. (2) Sol. q0 Sol. For monoatomic molecule degree of freedom = 3. 3  Kavg  K BT 2 0.414  1.6  10 19  2 T d kq 0 3  1.38  10 23 2Tsin  2 Rd 2 R = 3200 K  Q    2R    15  Kq 0 Q 55. In a nuclear fission process, a high mass nuclide  T  R   2 2 (A  236) with binding energy 7.6 MeV/Nucleon dissociated into middle mass nuclides (A  118),   9 10  2 30 10  9 12 having binding energy of 8.6 MeV/Nucleon. The  0.30  2  2 energy released in the process would be ____ MeV. Ans. (236) 9  10 3  30   3N Sol. Q = BEProduct – BERectant 9  10 2 = 2(118) (8.6) – 236(7.6) 53. Two coils have mutual inductance 0.002 H. The = 236 × 1 = 236 MeV current changes in the first coil according to the 56. Four particles each of mass 1 kg are placed at four corners of a square of side 2 m. Moment of inertia relation i = i0 sin t, where i0 = 5A and  = 50 of system about an axis perpendicular to its plane rad/s. The maximum value of emf in the second and passing through one of its vertex is _____  kgm2. coil is V. The value of  is ____.  Ans. (2) Sol.  = Mi = Mi0sint di Ans. (16) EMF = – M  0.002  i0  cos t  dt a m m EMFmax = i0  (0.002) =  5 50  0.002  Sol. a a  EMFmax  V 2 m m a 8 54. Two immiscible liquids of refractive indices 5  2a  2 I  ma 2  ma 2  m 3 and respectively are put in a beaker as shown in 2 = 4ma2 the figure. The height of each column is 6 cm. A = 4 × 1 × (2)2 = 16 coin is placed at the bottom of the beaker. For near 57. A particle executes simple harmonic motion with normal vision, the apparent depth of the coin is an amplitude of 4 cm. At the mean position,  velocity of the particle is 10 cm/s. The distance of cm. The value of is______. the particle from the mean position when its speed 4 becomes 5 cm/s is  cm, where  = ______. Ans. (12) Sol. Vat mean position  A  10  4 5  2 v   A2  x 2 Ans. (31) 5 2 5 4  x 2  x2  16  4 h1 h 2 6 6 2 15 31 Sol. h app      4   cm x  12 cm 1  2 3 / 2 8 / 5 4 4 16 58. Two long, straight wires carry equal currents in 1A opposite directions as shown in figure. The 10/3A Sol. separation between the wires is 5.0 cm. The magnitude of the magnetic field at a point P midway between the wires is ____ T (Given : 0= 4× 10–7 TmA–1) VA  10 1 – 6 1  VB 3 10 8 VA  VB  6   volt 3 3 Q  C  VA  VB  8  150   400C 3 Ans. (160) 60. If average depth of an ocean is 4000 m and the 7 i 4 10  10 Sol. B   0 2  bulk modulus of water is 2 × 109 Nm–2, then  2 a  5     102  2  V fractional compression of water at the bottom V  16 105  160T of ocean is  × 10–2. The value of  is _____ 59. The charge accumulated on the capacitor (Given, g = 10 ms–2,  = 1000 kg m–3) connected in the following circuit is ____ C Ans. (2) (Given C = 150 F) P Sol. B  V   V     V  gh 1000  10  4000   B   V  2  10 9 = 2 × 10–2 [–ve sign represent compression] Ans. (400) 17  CHEMISTRY TEST PAPER WITH SOLUTION SECTION-A 64. Which of the following is strongest Bronsted base? 61. Two nucleotides are joined together by a linkage known as : (1) Phosphodiester linkage (2) Glycosidic linkage (1) (3) Disulphide linkage (4) Peptide linkage (2) Ans. (1) Sol. Phosphodiester linkage (3) (4) Ans. (4) 62. Highest enol content will be shown by : Sol. : (1) (2) N localised lone pair& sp3 H 65. Which of the following electronic configuration (3) (4) would be associated with the highest magnetic moment ? (1) [Ar] 3d7 (2) [Ar] 3d8 Ans. (2) (3) [Ar] 3d3 (4) [Ar] 3d6 O OH Ans. (4) Sol. HO OH 3d7 3d8 3d3 3d6 Sol. O O No.of. 3 2 3 4 unpaired e– Aromatic 63. Element not showing variable oxidation state is : (1)Bromine (2)Iodine Spin only 15 8 15 24 (3)Chlorine (4)Fluorine Magnetic BM BM BM BM Ans. (4) moment Sol. Fluorine does not show variable oxidation state. 18 66. Which of the following has highly acidic hydrogen? 68. Consider the following complex ions P = [FeF6]3– (1) Q = [V(H2O) 6]2+ R = [Fe(H2O)6] 2+ The correct order of the complex ions, according to (2) their spin only magnetic moment values (in B.M.) is : (1) R < Q < P (2) R < P< Q (3) (3) Q < R < P (4) Q < P < R Ans. (3) Sol. [FeF6]3– : Fe+3 : [Ar] 3d5 (4) F : Weak field Ligand No. of unpaired electron's = 5 Ans. (4) = 5(5  2) O O C C = 35 BM H3C CH2 CH2 CH3 [V(H2O)6]+2 : V+2 : 3d3 Sol. Active methylene gro up O O No. of unpaired electron's = 3 = 3(3  2) C C H3C CH CH2 CH3 = 15 BM Θ [Fe(H2O)6]+2 : Fe+2 : 3d6 Conjugate base is more stable due to more resonance of negative charge. H2O : Weak field Ligand No. of unpaired electron's = 4 67. A solution of two miscible liquids showing 4(4  2) = negative deviation from Raoult's law will have : (1) increased vapour pressure, increased boiling = 24 BM point (2) increased vapour pressure, decreased boiling 69. Choose the polar molecule from the following : point (1) CCl4 (2) CO2 (3) decreased vapour pressure, decreased boiling (3) CH2 = CH2 (4) CHC13 point (4) decreased vapour pressure, increased boiling Ans. (4) point H Ans. (4) Sol. Solution with negative deviation has Sol. C Cl Cl PT< PA0XA + PB0XB Cl PA< PA0XA  0 PB< PB0XB CHCl3 is polar molecule and rest all molecules are If vapour pressure decreases so boiling point increases. non-polar. 19 70. Given below are two statements : 72. The ascending order of acidity of –OH group in the Statement (I) : The 4f and 5f - series of elements following compounds is : (A) Bu – OH are placed separately in the Periodic table to preserve the principle of classification. (B) Statement (II) :S-block elements can be found in pure form in nature. In the light of the above (C) statements, choose the most appropriate answer from the options given below : (D) (1) Statement I is false but Statement II is true (2) Both Statement I and Statement II are true (E) (3) Statement I is true but Statement II is false (4) Both Statement I and Statement II are false Choose the correct answer from the options given Ans. (3) below : Sol. s-block elements are highly reactive and found in (1) (A) < (D) < (C) < (B) < (E) (2) (C) < (A) < (D) < (B) < (E) combined state. (3) (C) < (D) < (B) < (A) < (E) (4) (A) < (C) < (D) < (B) < (E) 71. Given below are two statements : Ans. (4) Statement (I) : p-nitrophenol is more acidic than m-nitrophenol and o-nitrophenol. Sol. Statement (II) : Ethanol will give immediate turbidity with Lucas reagent. In the light of the above statements, choose the correct answer from the options given below : 73. Given below are two statements : one is labelled as (1) Statement I is true but Statement II is false Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Melting point of Boron (2453 K) (2) Both Statement I and Statement II are true is unusually high in group 13 elements. (3) Both Statement I and Statement II are false Reason (R) : Solid Boron has very strong (4) Statement I is false but Statement II is true crystalline lattice. Ans. (1) In the light of the above statements, choose the Sol. Acidic strength most appropriate answer from the options given below ; (1) Both (A) and (R) are correct but (R) Is not the correct explanation of (A) (2) Both (A) and (R) are correct and (R) is the correct explanation of (A) (3) (A) is true but (R) is false Ethanol give lucas test after long time (4) (A) is false but (R) is true Ans. (2) Statement (I)correct Sol. Solid Boron has very strong crystalline lattice so Statement (II) incorrect its melting point unusually high in group 13 elements 20  77. IUPAC name of following compound (P) is : 74. Cyclohexene is _________ type of an organic compound. (1) Benzenoid aromatic (2) Benzenoid non-aromatic (3) Acyclic (4) Alicyclic (1) l-Ethyl-5, 5-dimethylcyclohexane Ans. (4) (2) 3-Ethyl-1,1-dimethylcyclohexane (3) l-Ethyl-3, 3-dimethylcyclohexane Sol. is Alicyclic (4) l,l-Dimethyl-3-ethylcyclohexane Ans. (2) 75. Yellow compound of lead chromate gets dissolved on treatment with hot NaOH solution. The product of lead formed is a : Sol. (1) Tetraanionic complex with coordination number six (2) Neutral complex with coordination number 3-ethy 1, 1 -dimethylcyclohexane four (3) Dianionic complex with coordination number 78. NaCl reacts with conc. H2SO4 and K2Cr2O7 to give six reddish fumes (B), which react with NaOH to give (4) Dianionic complex with coordination number yellow solution (C). (B) and (C) respectively are ; four (1) CrO2Cl2, Na2CrO4 (2) Na2CrO4, CrO2Cl2 Ans. (4) (3) CrO2Cl2, KHSO4 (4) CrO2Cl2, Na2Cr2O7 Sol. PbCrO4 + NaOH (hot excess) [Pb(OH)4]-2 + Ans. (1) Na2CrO4 Sol. NaCl + conc. H2SO4 + K2Cr2O7 Dianionic complex with coordination number four CrO2Cl2 + KHSO4 + NaHSO4 +H2O 76. Given below are two statements : (B) Statement (I) : Aqueous solution of ammonium Reddish brown carbonate is basic. CrO2Cl2 + NaOH Na2CrO4 + NaCl +H2O Statement (II) : Acidic/basic nature of salt solution (C) of a salt of weak acid and weak base depends on Ka Yellow colour and Kb value of acid and the base forming it. 79. The correct statement regarding nucleophilic In the light of the above statements, choose the most substitution reaction in a chiral alkyl halide is ; appropriate answer from the options given below : (1) Retention occurs in SNl reaction and inversion (1) Both Statement I and Statement II are correct occurs in SN2 reaction. (2) Statement I is correct but Statement II is (2) Racemisation occurs in SNl reaction and incorrect retention occurs in SN2 reaction. (3) Both Statement I and Statement II are incorrect (3) Racemisation occurs in both SN1 and SN2 (4) Statement I is incorrect but Statement II is reactions. correct (4) Racemisation occurs in SN1 reaction and Ans. (1) inversion occurs in SN2 reaction. Sol. Aqueous solution of (NH4)2CO3is Basic Ans. (4) pH of salt of weak acid and weak base depends on Sol. SN1 – Racemisation Ka and Kb value of acid and the base forming it SN2 – Inversion 21 80. The electronic configuration for Neodymium is: 83. Mass of methane required to produce 22 g of CO2 [Atomic Number for Neodymium 60] after complete combustion is ______g. (1)[Xe] 4f4 6s2 (2) [Xe] 5f