KVPY SA Stream Solution 2008

HINTS & SOLUTIONS (YEAR-2008) ANSWER KEY  Ques.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

 A

B

D

A

C

C

C

A

C

D

B

B

D

B

A

Ques.

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

Ans.

D

D

A

A

B

C

A

D

A

B

C

B

D

C

A

Ques.

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

Ans.

B

A

A

C

D

A

C

A

D

C

A

B

B

C

D

Ques.

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

Ans.

B

B

B

B

A

C

A&C

C

A

D

C

A

D

B

D

Ques.

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

Ans.

B

A

D

A

C

D

A

B

A

A

C

B

D

B

A

Ques.

76

77

78

79

80

Ans.

B

C

C

A

D

PART-I (1 Mark) MATHEMATICS 1.

Obvious (A) is greatest

2.

S= S

1 10



2 10

1

 =

10 2 Subtracting, 10

9S 10

1

 =

10



2

 +

 +

3 10

3

2 10 3 1

10

2

 +

 ...  +

n 10 n

 .... 

.........  

1 10 3

+

.........  

1 9S 10  = 1 10 1 – 10 9S 10 S= 3.

 =

1 9

10 81

(1024)1024 = (16)16n (210)1024 = (24)16n 10 × 1024 = 4 × 16n n=

10  1024

4  16 n = 160 4.

x2 + 6x + 8 x    R x2 – 2x – 8  0 x2 – 2x – 8 =

x2 + 2x – 4x – 8 x(x + 2) – 4(x + 2)   0

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2

x + 6x 6x + 8  –4

 –3

 –2 0

2

4

x    [–2, 4] clearly min value of expression is 0 at x = – 2 5.

6. 7.

Check by option P12 = {24, 36, 60, 84, ....} P20 = {40, 60, 100, ......} P12  P20 has common element  All even values of a i.e. 50 and 1, 9, 25, 49, 81, total 55 If any statement is true then remaining 2 are false. C D

8.

F  A

4 2

P

4 2

4 2

B

E



 Angle bisector  bisector 

Incircle is formed whose radius = 4 2

PE = r = 4 2



PF = r = 4 2 also PF = AE  APE, (AP)2 = (AE)2 + (PE) 2 = ( 4 2 )2 + ( 4 2 )2 = 64

 9.

 AP = 8

1 dd 2 1 2 Let one diagonal = x

D

 Area of rhombus =

1 = ×(x)(2x) = x2 2  A = x2 Let side of rhombus = y & height = h

BFC side BF =

y2

In  AFC, (y +

y2

 h 2 )2 + h2 = (AC)2 = 4x2

DEB (y –

 h2 )2 + h2 = (BD)2 = x2

y2

C

y

y

h

y

 A 2

y –h

2

E

B

2

y –h

2

F

 h2

 Adding  Adding 4y2 = 5x 2 y=

5x 2 4

 =

5 A 2

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 A(2a) (a) E P 10.

B

D

C (3b) (2b) 



Let B is origin and the position vector of A and C are 2a  and 3b 



Then P.V. of E = a  and P.V. of D = 2b Now, let P divides AD in   : 1 ratio and P divides EC in  : 1 







3b  a 2b  2a  =  1  1

 















2b  + 2b  + 2a  + 2a  = 3b  + a  + 3b  + a 



a (2 + 2 –  – 1) = b (3 + 3 – 2 – 2) 



But a  and b  are not collinear.. 2 –  + 1 = 0 and  + 3 – 2 = 0 We get  = 1 



a  3b

Now, P.V. of P is =

2 



ar PED  = ar  ABC

Now,





        1    a  a  3b    2b  a  3b    2   2     2    1  2a  3b 2

   1  a  3b  b  a 4 1   =  = 6ab 12



N

D

11.

O

 A

 

y



C

M P L

B

Let and then and

AB = a, BC = b PL = h1 PN = b – h1 OP = h2, then PM = a – h 2



ar(PAB + PCD) = =

1 a(h1 + b – h 1) 2

ab

and area (PBC + PAD) =

2 1

× b × (h2 + a – h2) =

2 From this only option B is correct

ab 2

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12.

Let x, x + 1, x + 2, x + 3, x + 4, x + 5, x + 6 & x + 7, x + 8, x + 9, x + 10, x + 11 7x + 21 = 5x + 45 2x = 24 x = 12 largest = x + 11 = 23

13.

Let x minite will be taken. I n one minute A can fill the 1 40

1 60

 part of tanker and in one minute B can fill the

 part.

Both can fill in t t 60 t=

t

 +

40

 = 1

60  40 100

t = 24 both can fill in one minute

1=

1=

x=

14.

1  part of tanker.. 24

x   1    x  1     +  2  24   2  40 x 80

 +

x 48

80  48 128

 = 30 30

Given a < b < c   6 three digit number are possible with distinct a, b, c. a b c

    

a c

b

b a c b c c

a

a b

c b a

1 5 5 4

= 100[2(a + b + c)] + 10[2(a + b + c)] + [2(a + b + c)] = 1554 111[2(a + b + c)] = 1554 a+b+c=7  Given a < b < c   1, 2, 4 only satisly above two condition. Hence, c = 4. 15.

Since factor of 128 are = 1, 2, 4, 8 , 16, 32, 64, 128 Hence it will be incr eased by 8.

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PHYSICS 16.

speed will not decrease, so answer is (D)

17.

For electron, t1 =

2s ap

For protion, t 2 =

t2 or  t 1

ae ap



eE me



mp eE

mp

= 18.



2s ae

me

Focal length, f = 6 cm u = 1.5m = 150 cm v =?

1 1 1   f  v u 1 6



1 1  v 150

1 v



1 1  6 150



25  1 150

150 75 25    6.25 24 12 4 change in distance = 6.25 – 6 = 0.25 cm = 0.25cm = 2.5 mm decreased

v=

19.

Initial momentum, P1 = mvcos30 and final momentum, P2 = mvcos30 change in momentum P = – 2mv cos30

P = – 3 mv Force on wall-1

F1 =

2mv t

Force on wall-2

3mv , so F 1 > F 2 t

F2 = 22.

 A1u1 = A2u2

u1 u2 23.



 A 2  A1



1 16

resultant force at centre is zero. On removing the charge from the position 6, the resultant force at centre will be

kq r 2

 downward.

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25.

v V



dw dL

1 2



dw 840

R.D. = 26.

420 10 3

 0.42

RS = nR for maximum resistance RP = R/n for minimum resistance or 

27.

 dw = 420 kg/m 3

RS  = n2 RP

For block 2kg T – 2g = 2a ...... . .(i) For 6 kg 6g – T = 6a ...... . .(ii) From (i) and (ii) T = 30 N

T

T a

a 2g 6g

v e  

28.



29.

R.H. =

2gR with height g will change, so answer is (D)

partial pressure saturated vapour  pressure

partial pressure 90  = 100 0.0169  10 5 partial pressue = 0.0152 × 10 5 Pa

CHEMISTRY 31.

NaOH N1V1 = 0.5 × V = V = 40 mL

35.

Ethanol (C2H5OH) and dimethyl ether (CH3 –O – CH3) have same molecular formula but different functional groups, so they are isomers.

36.

For the elements belonging to one period, increase in atomic number results in decrease in atomic radius. So Li has the largest atomic radius.

37.

  2H2O + O2 2H2O2   

39.

S

+

HCl N2V2 2 × 10

O2

   

1 mole

1 mole

1 mole 2

1  mole 2

SO2 1 mole

1  mole 2

3.01 × 1023 0.5 mole ? 23  3.01 × 10 molecules of SO 2 will be formed.

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40.

Zn and Pb are placed above hydrogen in the metal activity series, so they will produce hydrogen gas with dilute acids.

41. 42.

45.

Milk of magnesia is basic, water is neutral and lemon juice is acidic in nature.  As pressure is increased, solubility of gas in liquid increases.

CH3 – CH CH –CH –CH2 – CH CH2 – CH CH3

Conc.. H2SO4 Conc  –H 2O

OH

CH3 – CH CH = CH – CH CH2 – CH CH3 + CH CH2 = CH CH – CH2 – CH CH 2 – CH CH 3 2-Pentene (Major) 1-Pentene (minor)

2-Pentanol

PART-II (2 Mark) MATHEMATICS 61.

x

1

 2 x1  1  +   x 2  4 x 2  4   +   x 3  6 x 3  9   +   x 4  8 x 4  16   +   x 5  8 x 5  25   = 0





   2

x 1  1  1  +

x2

x1  1  – 1 = 0,

Now,

 4  2  +   x 3  9  3  +   x 4  16  4  +   x 5  25  5 2

x 2   4  – 2 = 0,

2

x 3   9  – 3 = 0,

2

x 4   16  – 4,

2

x 5   25  – 5 = 0

x1 = 2, x 2 - 8, x 3 = 18, x4 = 32, x5 = 50

x1  x 2



 x3  x4  x5 2

 = 55

62.

(1 + 2x + 3x 2 + .... 21x20) (21x (21x20 + 20x19 + 19x18 + ........ 2x + 1) coeff. of x 30 is 11.21 + 12.20 + ..................... + 21.11 = 2[11.21 + 12.20 + 13.19 + 14.18 + 15.17] + 16.16 = 2[231 + 240 + 247 + 252 + 255] + 256 = 2[1225] + 256 = 2450 + 256 = 2706

63.

If we follow the pattern according accordin g the rule that angle made by incident ray with normal is equal to the angle made by reflected ray with normal then we find that ball will not go in any hole.

64.

C can be collin ear with A & B.

65.

Let initial prize is P

   

after X% increment  P 

Px  



100 

after decrement y%

  Px     Px   y   P  100  –  P  100  100   = P          1+

x 100

x – y 100

 =

 –

y 100

 –

xy (100)2

 = 1

xy (100 )2

1 1 1  –  = y 100 x

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PHYSICS 66.

mass of water, m 1 = 0.4 kg temperature of water, water, 1 0ºC mass of ice, m 2 = 0.1 kg tempeature of ice,  2 = – 15ºC mass of steam = m kg final temperature of mixture  = 40ºC specific heat heat of ice, s ice = 2.2 × 10 3 J/kg×k latent heat of fusion, L f  = 333 × 103  J/kg latent heat of vaporisation, L V = 2260 × 103 J /kg Heat given = Heat taken By steam of 100ºC to water of 100 ºC + By water from100ºC to 40ºC = By water from 0ºC to 40ºC + By ice from –15ºC to ice of 0ºC + By ice of 0ºC to water to 0ºC + By water of 0ºC to 40ºC mL v + mSw (1100 – 40) = m 1 × Sw (40 – 0) + m2sice(15) + m 2 × Lf  + m2Sw (40–0) m (2260 × 103 + 4200 × 60) = 0.4 ×4200 × 40 +0.1 ×2.2 × 10 3 ×15 + 0.1 × 333 × 10 3 + 0.1 × 4200 × 40 m (2512 × 103) = 67200 + 3300 + 33300 + 16800 m=

120600 251200



603 kg = 48 g 12560

100

67.

u

v

 As for question u + v = 100 cm .....(i) after displacing lens by 40 cm u and v will be u + 40, v - 40 for Ist condition (i)

1 1  f  v



1 ( u)



1 v



1 u

1 u  v 100   f  uv uv (ii) For second condition 1 1  f  v  40 =



1

 (u  40)

u  40  v  40 uv  = ( v  40)(u  40) ( v  40)(u  40)

From (i) and (ii)

100 uv



100 ( v  40)(u  40 )

v– u = 40 .....(ii) v + u = 100 from equation (i) and (ii) 2v = 140 v =70, u = 30

1 u v 100   = f  uv 2100

  f = 21 cm

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68.

we know that  A1V1 = A2V2 = Q (volume flowing per second) 25 × 0.6 = 12 × v 2 v2 = 15 m/s ...... . .(i)  – 2 2 Q = A1v1= (5 ×10 )  × 0.6 = 4.71g force, F = rate of change of momentum F = mv 2 – mv1 = 4.71 × 15 – 4.71 × 0.6 0.6 (m = 4.71, 4.71, mass flow flow per per unit time) F = 67.9 N

CHEMISTRY 71.

Consider the volume of the solution = x cm 3 Then the mass of the solution will be = 1.13x (mass = density × volume) The solution contains 18% of NaCl by weight

18 × 1.13x = 36 100 3600 x=  = 177 cm 3 18  1.13



+

72.

CH3 – CH2 – COOH + CH2 – OH Propanoic acid

73.

H

CH3 – CH CH2 – COOCH2 CH3 + H2 O Ethyl propanoate

CH3 Ethanol

Consider that the salt contains x molecules of water . Molecular weight of anhydrous salt = 160 g so molecular weight of hydrated salt will be = 160 + 18x g Then, no. of m oles of water present in 10x gm of hydrated salt = and weight of water present in 10 gm of hydrated salt =

10 ×x 160  18 x

10 x 160

 18 x

× 18

     Anhydrous salt + Water 

Hydrated salt 10g

6.4 g

3.6g

180 x

 = 3.6 160  18 x 180x = 576 + 64.8 x x= 5

74.

   6Fe3+ + 2Cr 3+ + 7H2O Cr 2 O72 – + 6Fe2+ + 14H+    Change in oxidation number of Cr is = 6 – 3 = 3 Change in oxidation number of Fe is = 3 – 2 = 1

75.

 Ca(HCO3)2 CaCO 3 + H2O + CO 2    

Ca(HCO3)2    CaCO3 + H2O + CO2

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