NARAYANA IIT/NEET ACADEMY (1)

NARAYANA I I T / N E E T A C A D E M Y

C o m m o n P r a c t i c e T e s t – 2 2 XII STD BATCHES [CF] Date: 06.11.17

AANNSSWWEERR PHYSICS CHEMISTRY MATHEMATICS

1. (4) 2. (4) 3. (2) 4. (3) 5. (3) 6. (1) 7. (2) 8. (1) 9. (2) 10. (2) 11. (2) 12. (3) 13. (2) 14. (4) 15. (1)

16. (2) 17. (2) 18. (1) 19. (3) 20. (4) 21. (3) 22. (4) 23. (1) 24. (1) 25. (4) 26. (2) 27. (4) 28. (2) 29. (4) 30. (1)

31. (2) 32. (4) 33. (2) 34. (1) 35. (2) 36. (1) 37. (1) 38. (3) 39. (4) 40. (2) 41. (3) 42. (2) 43. (1) 44. (3) 45. (4)

46. (1) 47. (3) 48. (1) 49. (3) 50. (2) 51. (1) 52. (2) 53. (2) 54. (2) 55. (1) 56. (4) 57. (1) 58. (3) 59. (2) 60. (1)

61. (2) 62. (2) 63. (2) 64. (2) 65. (4) 66. (4) 67. (2) 68. (2) 69. (2) 70. (3) 71. (2) 72. (2) 73. (1) 74. (3) 75. (4)

76. (1) 77. (3) 78. (2) 79. (1) 80. (4) 81. (1) 82. (2) 83. (4) 84. (4) 85. (4) 86. (1) 87. (4) 88. (4) 89. (2) 90. (2)

(Hint & Solution)

PART A : PHYSICS 1. (4) 1/3

0r r A 1/3 1/3

1 1

2 2

64 4125 5

r Ar A

2. (4) For a wide range of values of load resistance, the current in the zener diode may change

but the voltage across it remains unaffected. Thus the output voltage across the zener diode is a regulated voltage.

3. (2) The diode in lower branch is forward biased and diode in upper branch is reverse biased

5 520 30 50

i A

4. (3) Because N-side is more positive as compared to P-side. 5. (3) In intrinsic semiconductors, the creation or liberation of one free electron by the thermal

energy has created one hole. Thus intrinsic semiconductors ne = nh. 6. (1) When NPN transistor is used as an amplifier, majority charge carrier electrons of N-type

emitter move from emitter to base and then base to collector. 7. (2) it is used to convert ac into dc (rectifier)

NARAYANA IIT/NEET ACADEMY (2)

8. (1) The arrow head in the transistor symbol always shows the direction of hole flow in the

emitter region. 10. (2) ‘OR’ gate X = A + B

i.e. 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 1 11. (2) Depletion layer is more in less doped side. 12. (3)

Y A B According to De morgan’s theorem

. .Y A B A B A B This is the output equation of ‘AND’ gate

13. (2) 7N14 + 2He4 8O17 + 1H1

15. (1) Penetration power of is 100 times of , while that of is 100 times of . 16. (2) Due to the large concentration of electrons in N-side and holes in P-side, they diffuses

from their own side to other side. Hence depletion region produces. 19. (3) During -decay, a neutron is transformed into a proton and an electron. 20. (4) 6C11 5B11 + + + v because + = 1e0 21. (3) Negative -decay is expressed by the equation

n = p+ + e + v 23. (1)

2 2

1 213.6 13.6 47.2

4 9Z ZE E E

Z = 5 25. (4) 234N gm

10

10 3.7 10 2343.7 10 ,xgm xN

26. (2) 5

0

1 12 2

nNN

29. (4) Resistance of conductors (Cu) decrease with decrease in temperature while that of semi-

conductors (Ge) increase with decrease in temperature. 30. (1) For the stability of the nucleus it should have high binding energy per nucleon.

NARAYANA IIT/NEET ACADEMY (3)

PART B : CHEMISTRY 31. (2) 3 2 3 33 3PCl H O H PO HCl 32. (4) 4 2 2 72HClO H O Cl O Thus, Cl2O7 is anhydride of HClO4 33. (2) I Cl Chlorine is more electronegative than iodine 34. (1) Out of HCl, HBr and HI, HCl is thermally most stable due to high bond dissociation energy. HOCl is a highly unstable compound. 22 2HOCl HCl O 36. (1)

In basic medium, the following reaction takes place 4 2 4 22KMnO 2KOH 2K MnO H O [O]

Due to the presence of nascent oxygen [O], KMnO4 (in basic medium) behaves like a strong oxidizing agent.

37. (1) Given, X = [Ar}d4 The complete configuration of the ion, x3+ = 1s2, 2s2, 2p6, 3s2, 3p6, 3d4 x = 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s2 the atomic number of the element is 25 and the element is Mn. 38. (3) The number of unidentate ligand in the complex ion is called coordination number. 39. (4) Transition metals can form ionic or covalent compounds and their melting and boiling points

are high. Their compounds are generally coloured and they usually exhibit variable valency. 40. (2) Maximum oxidation state of transition metals = number of electrons in (n – 1)d orbital +

number of electrons in ns orbital. The electronic configuration of Os = [Xe]4f14, 5d6, 6s2 Maximum oxidation state = 6 + 2 = 8 The highest oxidation state exhibited by transition metal is +8 e.g., OsO4. 41. (3) CO is a neutral ligand, so the oxidation state of metal in metal carbonyls is always zero. [Ni(CO)4] x + (0 × 4) = 0 x = 0 43. (1) In blood, He is much less soluble than nitrogen, hence He – O2 mixture is used by deep sea

divers in preference to N2 – O2 mixture. 44. (3) Option (3) has noble gas configuration as it has 8 electrons in valence shell. 49. (3) Py is a neutral ligand and Cl– an anionic ligand carries one unit negative charge.

NARAYANA IIT/NEET ACADEMY (4)

50. (2) [Ni(CN)4]2– is a square planar complex and trans [Pt(NH3)2Cl2] both have zero dipole

moment. 51. (1) Based on facts 58. (3) Fluorine reacts with cold and dilute sodium hydroxide solution to give oxygen difluoride

(OF2). However with hot and concentrated sodium hydroxide, it gives oxygen. 59. (2) In 1868, Lockyer studied the spectrum of the sum during the total solar eclipse. From his

investigations he concluded that a new element was present in the atmosphere of the sun. He called this new element as Helium.

60. (1) On liquefying Helium changes property so much that it is called Helium - II.

PART C : MATHEMATICS

61. Given 00

xp y x yyp z y z

xp y yp z

Operating C1 – pC2 – C3

2

00 0

2 7

x yy z

xp py xp y yp z

(xz – y2)(xp2 + 2py + z) = 0 xz – y2 = 0 [also for this D = 0 for other quad. Factor in p which shows second factor is a perfect square] y2 = xz x, y, z are in G.P.

62. Given that 1 32 2

i

2 1 32 2

i

Also 1 + + 2 = 0 and 3 = 1 Now given det. is

2 2 2

2 4 2

1 1 1 1 1 11 1 11 1

Operating C1 C1 + C2 + C3

2

2

3 1 100

(as 1 + + 2 = 0)

Expanding along C1, we get 3(2 - 4) = 3(2 - ) = 3( - 1)

NARAYANA IIT/NEET ACADEMY (5)

63. For infinitely many solutions the two equations becomes identical

1 8 4 13 3 1

k k k

k k k

64. Since the system has no solution, = 0 and any one among x, y, z is non zero.

2 1 21 2 1 0 11 1

Also, 2 1 21 2 4 6 01 1 4

z

65. x + y + z = 1 x + y + z = 1 x + y + z = 1

1 1

1 11 1

= (2 – 1) – 1 ( - 1) + 1(1 - ) = ( - 1)( + 1) – 1 ( - 1) – 1 ( - 1) For infinite solutions, = 0 ( - 1)[2 + 1 – 1] = 0 ( - 1)[2 + a – 2] = = - 2, 1; For infinitely may solution for no solution = 1 = – 2 66. The given equation are x + cy + bz = 0 cx – y + az = 0 bx + ay – z = 0 x, y, z are not all zero The above system should not have unique (zero) solution

1

0 1 01

c bc cb a

(infinitely many sol)

1(1 – a2) – c(c – ab) + b(ac + b) = 0 1 + a2 + b2 + c2 + 2abc = 0 a2 + b2 + c2 + 2abc = 1

67.

2 1

1 1 1 1 11 1 1 1 1 01 1 1 1 1n n n

a a a a b cb b b b b cc c c a b c

2

1

1 1 11 11 1 1 1 1 01 1 1 1 1

n

n

n

a a aa a ab b b b b b

c c c c c c

21 0

1 1 2 1 2 1 01 1

nab b b

c

NARAYANA IIT/NEET ACADEMY (6)

21 1 2 1 2 1 0n a c b b

4b (a + c)[1 + (1)n+2] = 0 1 + (-1)n+2 = 0 as b(a + c) 0 n should be an odd integer.

68. Consider 0p b ca q ca b r

Operating R1 R1 – R2 and R2 R2 – R3 we get

0 0p a q b c

q b c ra b r

Taking (p – q), (q – p) and (r – c) common from C1, C2 and C3 resp, we get

1 1 00 1 1 0p a q b r ca b r

p a q b r c

Expanding along R1

1 0r b ap a q b r cr c q b p a

As given that , ,p a q b r c

0r b ar c q b p a

2p q rp a q b r c

69. As given equation is an identity in , it must be true for all value of . For = 0 also. Putting = 0 we get

0 1 31 0 4 03 4 0

t

70. .b b c c

b a b a a a b bc c a a

2 0,

b b ca b a a b

c c a [by C2 C2 + C3].

71. 2 1 1y z x y

z x z x x y z z x z xx y y z x y y z

[by R1 R1 + R2 + R3]

1 1 1

;x y z x z xx y z

[by C1 C1 – C2]

= (x + y + z).{(z2 – xy) – (xz – x2) + (xy – xz)} = (x + y + z)(x – z)2 k = 1.

NARAYANA IIT/NEET ACADEMY (7)

72. 1 1 3 22 cosC C C C dx given

2 21 2 cos0 cos cos0 sin sin

a a dx a apx p d xpx p d x

21 2 cos sin ,a a dx dx (which is independent of p). 73. The determinant can be expanded as

cos cos sin sin cos cos sin sin cos cos sin sincos cos sin sin cos cos sin sin cos cos sin sincos cos sin sin cos cos sin sin cos cos sin sin

A P A P A Q A Q A R A RB P B P B Q B Q B R B RC P C P C Q C Q C R C R

This determinant can be written as 8 determinants and the value of each of these 8 determinants is zero

cos sin 0 cos cos coscos sin 0 sin sin sin 0 0 0cos sin 0 0 0 0

A A P Q RB B P Q RC C

74. Here, 1 cos cos

2cos sin 1 sin cos 01 cos sin

x xx x x x

x x

or 1 cos cos

2cos sin 0 sin cos 0 00 0 sin cos

x xx x x x

x x

or 22cos sin sin cos 0x x x x

tan 2,1.x But tan 2x in ,4 4

tan 1.x So, .4

x

75. We have 1 2 3 na ,a ,a ,....., a in G.P.

Then, 2 n 1 n 2

1 n n 1

a a ar i.e., r ......a a a

Hence, n 1 n n 2 n 1log r log a log a log a log a ...

Now, n n 1 n 2

n 3 n 4 n 5

n 6 n 7 n 8

log a log a logaloga loga log alog a loga loga

operate 2 2 1 3 3 2C C C and C C C

n n 1 n n 2 n 1 n

n 3 n 4 n 3 n 5 n 4 n 3

n 6 n 7 n 6 n 8 n 7 n 6

log a loga log a loga loga log a log r log rloga loga loga loga log a log a log r log r 0log a log a log a loga log a log a log r log r

76. From Venn diagram, we can see that

3 1 1 1P B C P B P A B C P A B C

4 3 3 12

NARAYANA IIT/NEET ACADEMY (8)

77. The number of ways of choosing the fist square is 64 and that for the second square is 63. Therefore the number of square of chosing the first and second square is 64 × 63 = 4032. Now we proceed to find the number of favourable ways. If the first happens to be any of the four square in the corner, the second square can be chosen in two ways. If the first square happens to be any of the 24 square on either side of the chess borad the second square can be chosen in 3 ways. If the first square happens to be any of the 36 remaining squares, Therefoe the number of favourable ways is

(4)(2) + (24)(3) + (36)(4) = 244

Hence the required probability 224 14032 18

78. Since events are mutually exclusive and exhaustive

1 3 1 3 1 12 3 6

p p p

Also 1 30 12

p and 1 40 1

3p

and 10 1.6

p

1 3 1 1 1 4 1 10 1 , , 0 1 , ,2 3 3 3 4 2

p pp p

10 1 1, 56

p p

Hence the set of value satisfying all the above inequalities are 1 1,4 3

79. Probability of getting head 12

and probability of throwing 5 to 6 with a dice 2 1.6 3

He

starts with a coin and alternately tosses the coin and throws the dice and he will win if he get a head before he get 5 or 6.

Probability 1 1 2 1 1 2 1. . . ...2 2 3 2 2 3 2

21 1 1 1 1 31 ...... .2 3 3 2 1 1/3 4

80. Consider two events: Ai getting a number i on first die. Bi getting a number more than i on second die. The required probability 1 1 2 2 3 3 4 4 5 5P A B P A B P A B P A B P A B

5 5

1 1i i i i

i iP A B P A P B

1 2 51 ....6

P B P B P B 1 5 4 3 2 1 5 .6 6 6 6 6 6 12

81. Required probability 100 101330 133

82. The total number of ways in which n person can sit at a round table = (n – 1)! Favourable number of cases = 2! (n 2)!

NARAYANA IIT/NEET ACADEMY (9)

Thus the required probability

2! 2 ! 21 ! 1

nn n

Hence the total against are 1 : 3 : 2p p or n 83. The minimum of two numbers will be less than 4 or at least one of the number is less than 4. P (at least one numbers < 4) = 1 – P(both the number 4) 3 21

6 5 61

30

1 415 5

84. Let x and y cases are in favour of B and A B.

According to question 4.10 10 10y x

5y = 2x x can be 5 or 10

85. Required probability =

5

3

5256365

6

86. Suppose the coin is tossed n times 1 0 0.8P X

1 10.2 2 52 5

nn

This least value of n is 3 87. E1 = event that white ball is drawn in first draw E2 = event that a black ball is drawn in first draw E = event that a white ball is drawn in second draw

1 21 2

E EP E P E P P E PE E

m m k n m mm n m n k m n m n k m n

88. Required probability 6

6

6 66

51

324

89. Required probability 5

2 25

C

10 2 1120 6