Sol.Paper-1

WORKSHOP FOR NTSE STAGE-I _TEST PAPER-1 Ques.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

Ans.

A

D

D

D

A

D

A

A

D

B

C

C

D

A

D

A

D

D

D

C

Ques. 21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

C

D

D

C

D

D

B

C

D

C

B

C

B

B

B

C

B

B

D

B

Ques. 41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

B

A

B

C

B

B

A

C

A

C

B

D

A

B

D

A

B

C

C

B

Ques. 61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

D

A

D

C

A

B

A

B

C

C

A

B

A

C

A

C

D

C

B

B

Ques. 81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99 100

D

B

A

D

C

B

C

C

B

D

D

A

A

B

C

A

A

D

C

Ans.

Ans.

Ans.

Ans.

6.

Let the velocity after 2 second is v, then v = u – gt v = 20 – 10 × 2 = 0 Kinetic energy after 2s,

Change in linear momentum = 5%

 K=

P2 2m

K P 2 = 2 × 5 = 10% K P 7.

Momentum of the gun = mv = 0.050 × 400 = 20 kg × m/s Change in momentum of the gun in 1s due to n bullet = Force

K.E. = 14.

30 1     n  60 2  

P12 P22 P1  2m  2m  P  1 2 2

 n =

m1 < m2 P1 < P2 9.

11.

Acceleration of the body, a = 5 cm/s = 0.05 m/ s Mass of the body, m = 20 g = 0.02 kg  F = ma F = 0.05 × 0.02 = 10–3 N

v=

1 mv2 2

2k = m

expansions/second

Given : t1 = t 2 = t s1 = v1 t , s2 = v2 t

s1  s2 v t  v 2 t v1  v 2 = 1 = 2t 2t 2

20.

The displacement between X and Y = r + r = 2r

21.

Consider the atomic number of x = z then number of electrons present in x = z so 2z + 5z = 21 7z = 21 z=3 mass number (A) = 3 + 4 = 7

24.

pH = – log [H+]

S.I. unit of energy Joule 10 7 erg   C.G.S. unit of energy erg erg

Mass of the body, m = 2 kg Initial velocity of the body , u = 20 m/s Time taken, t = 2s

4 2.4

4 × 60 = 100 expansions/min. 2.4

vav =

2 = 1.414 m/s

= 107 13.

16.

Mass, m = 1 kg Kinetic energy, K = 1 J K =

12.

n=

W t

2.4n 1

4=

m1 m2

1 1 mv2 = × 2 × (0)2 = zero 2 2

Work done during each expansion = 2.4 J Suppose lungs expands n times per second, then work done in 1 second, W = 2.4 nJ Time, t = 1s, Power, P = 4 W P=

1 F =  20  10N 2 8.

C

1 pH  H    B > A > C, decreasing acidic strength. WORK SHOP_NTSE_ STAGE-I _PAGE # 1

30.

 2NaHCO3  Na2CO3 + H2O + CO2

33.

Number of electrons in CO32 = 6 + 24+2= 32

83.

(A) Series is abc/ aabbcc/ aaa

84.

The series is aab / aab / aab / aab. So, pattern aab is repeated.

Number of electrons in NO3 = 7 + 24 +1 = 32 38.

Ca(OH)2 + CO2  Ca(HCO3)2 (Excess)

71.

(4 × 2) – 1, (7 × 2) – 1, (13 × 2) – 1, (25 × 2) – 1 = 49

72.

73.

76.

The pattern of the series is : – 3, – 4, – 5, – 6, – 7, ..... = 31,24. 2 is substracted in each no.

I : 5, 9, 13, 17 [4 is added to each no.] II : 7, 9, 11,13, ....[4 is added to each no.] The first letter of the first, second, third, ...... terms are respectively moved two, four, six, ..... steps forward, second letter of each term is moved one step forward and the third letter of the first, second, third, ..... terms are moved one, three, five, ...... steps forward to obtain the corresponding letters of the suceessive terms.

79.

(B) 4 + 72 = 53, 6 + 92 = 87  1 + 72 = 50

80.

(B) (4 × 3) – (5 × 1) = 7, (4 × 3) – (2 × 3) = 6 Similarly , (2 × 1) - (3 × 0) = 2

81.

In the first row N moves one step forward into O which again moves one step forward into P. In the second row S moves one step backward into R which again moves one step backward into Q. In the third row T moves one step forward into U which again moves one step forward into V. So, missing letter in second column is Q.

82.

85.

The series is abcd / dabc / cdab / bcda.

86.

(B) a a b c/b b c a/c c a b/a a b c/b b c a

87. 0, 0, 1, 3, 10, 15, 21 5 Series should be +0 +1 +2 +3 +4 +5 +6

3 + 3 6 5 is a wrong term. 88.

(C) Given series is :

89.

It is a combination of two series : I : 1, 3, 5, 7 [+2, +2,.....] 4, ×2

32,

64

×2

×2

4 must be replaced by 8.  4 is wrong term. 90.

(D)

Hence, the wrong number is 132 and should be replaced by 133. 92.

In every term first second and third letter is in alphabetical order to its next term respectively. Fourth term is not following the same rule. The wrong term is HTY. So, the answer would be HSY.

93.

(A) In every term, first second and third letter is moved two steps forward to its next term respectively. Fourth term is not following the same rule. The wrong term is VVT. So, the answer would be VUT.

94.

(B) First letter of every term is moved three

In 1st figure, (16 – 6)2 + (5 – 2)2 = (10)2 + (3)2 = 109. In 2nd figure, (22 – 15)2 + (21 – 19)2 = (7)2 + (4)2 = 53. So, in 3rd figure, the digits of the missing number are : (17 – 13)2 + (51 – 48)2 = (4)2 + (3)2 = 25.

16,

II :

steps forward in each next term. Second number of every term of the pattern  × 2 + 1, × 2 + 2, × 2 + 3,............and third letter of every term is moved two steps backward. The wrong term is J10R. So, the answer would be J9R. WORK SHOP_NTSE_ STAGE-I _PAGE # 2

95.

Given expression = 0.2 + 0.2 – 1 × 0.04 = 0.4 – 0.04 = 0.36

96.

(A) By making the interchanges given in (A), we get the equation as 2 – 5 + 3 = 0 or 0 = 0 which is true. By making the interchanges given in (B), we get the equation as 3 – 2 + 5 = 0 or 6 = 0, which is false. By making the interchanges given in (C), we get the equation as 5 – 3 + 2= 4 or 4 = 0 which is not true.

WORK SHOP_NTSE_ STAGE-I _PAGE # 3