KVPY SA Stream Solution 2009

HINTS & SOLUTIONS (YEAR-2009) ANSWER KEY  Ques.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

D

A

A

A

C

C

B

D

D

B

D

B

D

B

C

Ques.

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

Ans.

C

B

C

B

A

A

D

A

D

C

D

C

B

D

B

Ques.

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

Ans.

D

C

A

D

B

C

A

C

D

B

C

B

A

B

C

Ques.

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

Ans.

D

C

C

B

A

B

A

A

D

B

B

A

A

C

C

Ques.

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

Ans.

 A

B

B

C

D

C

A

D

D

A

A

C

D

B

C

Ques.

76

77

78

79

80

Ans.

C

D

A

A

B

PART-I (1 Mark) MATHEMATICS x  5  0 x 1 – x x + 5 > 1 + x 2 – 2x x2 – 3x – 4 < 0 (x – 4) (x +1) < 0 x  (–  (–1, 4) ....(ii) Using (i) & (ii) x  (–1, 1) 3.

x = x2 + y2 & y = 2 xy  when y = 0 ⇒ x = x2 ⇒ x = 0, x = 1  solutions are (0,0) (1,0) 1  when x = 2 ⇒

 4.

1 1  –  = y 2 2 4 solutions

 – 1  9 2

1  = y2 4

⇒

y=±

1 2

 1 1   1 1   ,   ,–   2 2   2 2 

x3 – 3 |x| + 2 = 0 x3 – 3x + 2 = 0 (x – 1) (x2 + x – 2) = 0 x = 1, x =

⇒

[Let x > 0]

 = 1

Now, let x < 0 x3 + 3x + 2 = 0 no solution x = 1 only one solution

CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892

Page # 104

5.

(1 + 2x) 20 = put x = 1 x = –1 subtract add

a0 + a1x + a2x2 + ------- + a20x20 320 = a0 + a1 + a2 + ------- + a20 1 = a0 – a1 + a2 – a3 ------- + a20 320 – 1 = 2(a1 + a3 + -------- + a19] 320 + 1 = 2 (a 0 + a2 + -------- +a20] 2 (a1 + a3 + ------- + a19) + 3 [a0 + a2 + --------- + a20] = (320 –1) +

5.320 = 2 6.

3 20 (3 + 1) 2

1

Points which are equidistant from O and P1 lies on perpendicular bisector of OP 1

P5 P1

P4 O

P2

P3

Similarly for others  In figure points on pentagon are equdistant from points P 1, P2, ...... and O  interior region of pentagon is closed to ‘O 7.

Let (x, y) be any point Using given condition 2 ( x  2)2

 y 2  < 3 ( x  0)2  ( y  3)2

5x2 + 5y2 – 54y + 16x + 65 > 0 x2 + y2 – 10.8y + 3.2x + 13 > 0 It is a circle, radius =

18.72

Centre (– 1.6, 5.4) Hence region is exterior of (x + 1.6)2 + (y – 5.4) 2 = 18.72. 8.

9.

(r + 3)2 + (r + 10)2 = (13)2 ⇒ r 2 + 9 + 6r + r 2 + 100 + 20r = 169 2r 2 + 26r – 60 = 0 ⇒ ⇒ r 2 + 13r – 30 = 0 ⇒ (r + 15) (r –2) = 0 ⇒ r=2

C

3 D

3

10

r  B

r 

10

 A

To find possible intergal value of 4th sier minimum possible value of 4th side greater than 0 is 1. For maximum possible value.

D 20 C  A

B

let angle ,   are slightly smaller than 180º if    ,  = 180º  AD = 35  maximum value of 4th side is 34  1, 2, 3, -------- ----- 34 are possible value of 4th side  34 values are possible.

CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892

Page # 105

10.

4 R’3 V’ 3  =  = 4 100 V R 3 3 172 .8

R’  = 1.2 R Now, ratio of surface area =

=

S’ S

S’ S

 =

4 R’2 4 R 3

=

 = 1.44

Hence surface area increased by 44% 11.

x = 0.d25 d25d25 -----x = 0. d25 1000 x = d25. d25 999x = d25 x=

d25 999

x=

d25 37.27

take d = 9 then x =

25 27

d = 9 n = 25  d + n = 34 12.

 Let at x minute past 10 oclock they become symmetric We know k now the speed of hour hand and minute hand is 1 : 12



When minute m inute hand moves x mimute distance then hour hand moves x



x  minute distance 12

x1 12

x  = 10 – x 12

11

10 – x

2 10



13x  = 10 12



x=



x = 9 minute 13.8 second therefore required time is 10h 9m 13.8 sec.

120 13

CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892

Page # 106

13.

14.

15.

P = LLLLLL W P=

9 8 7 6 5 4 1  ×  ×  × 10 9 8 7 6 5 4

P=

1 10

let x student like all three games.  If x minimum then a + b + c will be maximum and p + q + r should be min minimum. mum.  p+q+r=0 a + b + x = 74 ..... (i) a + c + x = 76 ......(ii) b + c + x = 82 ......(iii) and a + b + c + x = 100 a + b + c = 100 – x  Add (i), (ii) & (iii) 2 (a + b + c) + 3x = 232 2 (100 – x) + 3x = 232 200 – 2x + 3x = 232 x = 32.

Cricket

p b

a x

q

Football

c

r  Tennis

Number of integers between 2 n and 2n + 1 is 2n + 1 – 2n – 1 and  term = 2n + 1 last term = 2n+1 – 1

2 

n 1



Sn =

2

2 

n 1

=

 2n  1

 2n  1 2

[2n  + 1 + 2 n + 1 – 1]

 (2n)(1 + 2)

 2  .3 n

n

= (2  – 1)

2

Sn = 9 ;      3 × 2n – 1 × (2n – 1) = 9 2n – 1 × (2n  – 1) = 3  2n(2n – 1) = 6 It is possble when n is even.

PHYSICS 17.

given mass of planet A = m A and mass of planet B = m B m B = 8 m A g A =

Gm A GmB  and g = B r  r B2  A 2

g A or  gB 



m A  r B     mB   r   A  

2

.......(i)

 mB = 8 m A

4 3  4   r B d  8  r  A3 d  3  3  

g A gB

1 8

 4 

r B

 r  = 2  A

1 2

gB = 2g A

CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892

Page # 107

25.

 A =  r 2

 A  A

 A  A

 A  A

2

 A r 

  A       100   r   

%  2

%  2  0.15 = 0.30%

27.

207 Pb  724 He  4 1e0 82 so, n = 4 (  –1 particles)

29 .

Bulbs are connected in series and resistance of 100W is greater then that of 200 watt.  In series, P = I2 R (Here I is constant) So, PR So, power of 100W will be greater in the com bination.

30.

235 92 U

 As masses are equal for cude and sphere So, m s = v s × density m e = ve × density  vs × density  ve × density as m s = m e  vs = ve

4 3 r   s3 3 surface area comparison 6s 2 > 4r 2



CHEMISTRY 31.

Silicon is tetravalent, so it forms SiCl 4 .

32.

 NH4Cl undergoes sublimation while NaCl does not.

Oxidation 34.

36.

SO2 + 2H2S 3S + 2H2O Reduction

 As per Boyles law PV = constant and P 

38.

39.

40.

1 V

CaC2 + 2H2O   Ca(OH)2 + C2H2 Calcium Water Calcium Ethyne carbide hydroxide (acetylene)  As we move downwards downwards in a group, atomic radii increases. So the order will be Li < Na < K < Cs. CH3 – C  CH (Pr (Prop opy yne) ne) and and

(cycl (cyclo o pro prope pene ne)) are are two two poss possib ible le struc structu tural ral isome isomers rs of of C 3H4 .

CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892

Page # 108

Propionic acid and methyl acetate both have same molecular formula ( C 3H6O2) but different functional groups, so they are isomers.

41.

N2

42

Initial no. of  moles  After  reaction

+

3H2

2NH 2NH3

0.5

0

1 1 – 0.167

0.5 – 0.5

0.334 mole

0.334 moles of NH 3 = 2.0 × 10 23 molecules 43.

CH2OC OCOR OR CHOCOR

CH2 – OH + 3NaOH

 Alkaline

CH2OC OCOR OR

CH – OH + 3RCOONa Soap CH2 – OH OH

Triester 

Glycerol

Hydrolysis

Copper cannot displace lead from its solution as it is less reactive than lead..

44.

PART-II (2 Mark) MATHEMATICS 61.

 62.

loga b = 4 & logcd = 2 b = a4 d =c 2 4 2 a  – c = 7 (a2 – c) (a2 + c) = 7 a2 – c = 7 & a 2 + c =1 not possible or a 2 – c = 1 & a2 + c = 7 2a2 = 8 a = ±2 a =2 c =3 c –a =1 P(x) = 1 + x + x 2 + x 3 + x 4 + x5 =

63.

1 x6

1 x It has 5 roots let  1, 2, 3, 4, 5 they are 6 th roots of unity except unity Now, P(x12) =1 + x12 + x24 + x36 + x48 + x60 = P(x)  Q(x) + R(x) Here R(x) is a polynomial polynomial of maximum degree 4 Put x =   1,   2, .......5 we get R(1) = 6, R(2) = 6, R( 3) = 6, R(4) = 6, R(5) = 6 i.e. R(x) – 6 = 0 has 6 roots Which contradict that R(x) is maximum of degree 4 So, it is an identity R(x) = 6  BD  b and CE  c  A c sin A  b ...(i) and b sin A  c ...(ii) c D E (i) + (ii) (c + b) sin A  b + c k h sin A  1 B a  A = 90º  Now from (i) and (ii) c  b and b  c c=b Hence angles are 45º, 45º, 90º 

CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892

b C

Page # 109

64.

cos   =

(37) 2

 (7)2  – (33)2 2.37.7

 =

(37)2

 A

 x 2  – (33)2 2.37.x

37

[(37)2 + (7)2 – (33)2] x = 7(31)2 + 7x2 – 7(33)2 x = 40

65.

B

7 D

37

33 x

C

Case - I

a

r 

 A1 = Case-II

r  a/2

a/2

r – a/4

Case-III

r 

3      0      º      

a/2

a/4 a/2

a/2 r + a/4

   

 A3 =   a  r 

a   r    = 2 ra 4 4 

a

Hence A1 = A2 = A3

CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892

Page # 110

PHYSICS C

u=0

h=?

67. d

Time taken by the photon, t = d/c

1 2 gt 2

h=0×t+

1 d2 h= g 2 c2 gd2

h=

69.

2c 2

For lens L1, u = –0.40 m f = 0.20 m

1 v

 

1



   1     0.20  0.40 



2 1 0.40

v

1 v

1 1 f  u 1

 v = 0.40 m

 for lens L2 , u = 0.1 m f = – 0.1 m

1 v



v= 70.

1 0.1

1 0.1





 After removing charge from P, net force on central charge will be be : F=

Kq1q2 r 2



9  109  10 5  5  10 5 12

F = 4.5 N 

m = 0.5 kg

so, acceleration, a=

F M



4.5  = 9 m/s2 upwards 0.5

CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892

Page # 111

CHEMISTRY 71.

  Na2SO4 + 2HCl(g) 2NaCl + H2SO4    (X) 2HCl + CaCO3      CaCl2 + H2O + CO2 (X)

(Y)

Ca(OH) 2

+ CO2     CaCO3 + H2O

lime water

72.

(Y)

Milky Suspension

Initially concentration of salt in solution

=

222  10 –3 111  10  10 – 3

= 0.2M

On dilution the final concentration of CaCl 2 will be M1V1 = M2V2 0.2 × 10 = M 2 × 100 M2 = 0.02 M CaCl2     Ca2+  + 2Cl – 0.02 0.02 M

0.02 0.02 M

2 × 0.02 0.02

 –

[Cl ] = 0.04M = 0.04 mole/L 73.

     3Mn + 2Al 2O3 4Al + 3MnO2    To reduce 3 m oles of MnO 2 required moles of Al = 4 So, for one mole of MnO2 required moles of Al will be = 4/3

74.

       CH3COOH CH3 – CH CH2 – OH   

 Alkaline

Ethanol

KMnO 4

Acetic acid (X) 

H CH3COOH + CH3OH        CH3COOCH3 + H2O

 Acetic acid

75.

Methanol

Methyl acetate (Y)

On dilution, mili equivalent of the solute r emains constant. Initially pH of HCl = 4 so normality of HCl = 10 –4 N after dilution pH of HCl = 5 so normality of HCl will be = 10 –5 N N1V1 = N2V2 10 –4 × 10 = 10 –5 × V V = 100 mL So, 90 mL of water should be added for this pH change

CAREER POINT : 128, Shakti Nagar, Kota-324009 (Raj.), Ph: 0744-2503892

Page # 112