NSEJS answer key

NATIONAL STANDARD EXAMINATION IN JUNIOR SCIENCE

NSEJS_STAGE-I (2013-14) CODE : 514

HINTS & SOLUTIONS ANSWER KEY Ques.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

Ans

a

d

b

b

d

b

d

c

c

a

b

d

b

b

b

a

b

a

c

a

Ques.

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

Ans

c

a

c

d

c

b

c

b

d

b

c

a

a

d

c

d

b

b

c

b

Ques.

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

Ans

d

d

a

c

b

d

d

a

c

**

c

b

c

c

b

a

b

d

c

c

Ques.

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

Ans

d

*

d

c

*

c

c

a

c

*

a

a

c

b

a

c

b

d

a

d

coordination number = no. of ligands that form coordinate bond with central metal atom.  coordination number = 4 4.

(b) Pea is a kharif crop and it cannot be grown in summur.

5.

(d) Food was distrubted and fed equally.

6.

(b) Grasshopper has not stored food for future use.

1.

 = 180 – 2 × 45 = 90 41

= 6.40 ... 5 41 = 5  6.40

= 32.0

3.

Pt Cl2 (NH3)2 OS  x–2=0 x = +2

7.

H 20 m

2.

P.E. = K.E. mgh = 100 mg (20sin) = 100 (1) (10) (20sin) = 100 sin = 1/2  = 30º

NSEJS_STAGE-I _ SOL.-2013-14_PAGE # 1

4x2 – 4x – x – 1 = 5

8.

4x2 – 5x – 6 = 0 Area of triangle =

3

3 2 a x– 2, x = - 3/4 this is 4

not possible

3

3  142 so x = 2, y =3 4

=

= 84.868 It is nearer to 85 2 way

3

3

3

12.

0.025 KOH solution [OH–] = 25 × 10–3 M pOH = –log [25 × 10–3] pOH = – log(5)2 + 3log10 pOH = (–2 × 6990) + 3  log 10 = 1 pOH = –1.39 + 3 pOH = 1.602 pH + pOH = 14 pH + 1.602 = 14 pH = 14 – 1.602 pH = 12.398 = 12.40

14.

14 ny2 = a3, yz2 = b3

3

.. (i)

zn2 = c3 multiply (i) , (ii) and (iii) n3y3z3 = a3b3c3 nyz = abc

..(iv)

can (i) divided by (ii) 2 way

2 way

Total 6 ways

xy 2 yz 2

9.

11.

b3

O > N > S > Br [Decreasing order of EN] xy

10.

a3

=

F = P(A) F =  gh(A) 3xy + 3y – 1 4x2 + y – 5

=

z2

a3 b3

a3

xy =

b3

z2

..(v)

substitute in (iii)

4x + 2y 4x + 2y = 3x + 3y y=x+1

z3

a3 b3

= abc

4x2 + y – 5 = 4x + 2y 4x2 – 4x – y = 5

z3 =

zz3 =

abc  b 3 a3 b4c a2 NSEJS_STAGE-I _ SOL.-2013-14_PAGE # 2

15.

nC12H22O11 =

3.4 342

k 1 1 > if a,b real positive k k 1

= 0.0099  1 mole C12H22O111 = 22 × NA H-atoms  0.0099 mole C12O22O11 = 22 × 6.023 × 1023× 0.0099 = 1.3 × 1023 16.

17.

(a) Mountain streems has most fresh water and diatoms grow well in fresh water & it sewage canal has more pollution so the diatoms will grow less, Lake water has less diversity because of because the of little pollution.

k 1

21.

Standard enthalpy of formation is 0 kJ/mole for Na(s).

22.

Initially acc. is +ve and constant so s = ut +

(b) Because of Water pollution different varities of diatom will not grow

18.

(a) Different species have different structure.

19.

On Surface of earth gs =

At height R, gh =

1 2 a t and then particle decelerates so s = 2 0

ut -

1 2 at 2 0

GM R2

GM ( h = R) (R  R)2 

23.

No. of integer sided triangle with perimeter n is given by

gh =

g =

GM 4R 2

GM GM – 4R 2 R 2

1  3  1 = – g 4 s 4 

closest integer to

n2 , n is even 48

closest integer to

(n  3 ) 2 , n is odd 48

g = gs 

m=

g 3 =– × 100 = – 75 decreased by75%. gs 4 n = 20.

x=k+

x=k+

x=1+

=

 1  1  k  1 –   k 1  k

324 (15  3)2 = , closest integer = 7 48 48 256 16 2 = , closest integer = 5 48 48

m-n=7–5 =2 24.

Molality(m) =

25.

 R50 > R100

1 1 – –k+ ? k k 1 1 1 – k k 1

k 1 1 – k k 1

No. of moles of solute mass of solvent in kg

() So in Parallel P =

V 2  P  1 so R R

100W will glow more () in series P = I2 R  P  R so 50 W will glow more. m=

100 624  .7979

m = 0.201 mol kg–1.

NSEJS_STAGE-I _ SOL.-2013-14_PAGE # 3

26.

abc 3 11

30.

At 4 o’ clock angle between min and hour hand = 120º

(b + 3) (a + c) max value 12

max value

let at 4 hours x min angle is 120º

18

min value 3

min value

11 xº – 30º  4 = 120º 2

so angle is

1

No. of ways a  c b  3 1 1  1

18

7

2 1  2

17

6

3 1  3

16

5

4 1  4

15

4

5

14

3

7

12

12

8

11

11

8

10

10

9

9

9

8

8

8

7

7

7

6

6

6

5

5

5

4

4

4

3

3

3

1

1

12

x =

7 min = 43 min 38 sec] 11 Ans (B) = 43

31.

=

atm

255 L 1000

T = 288 K R = 0.0821 L atm K–1 mol–1 m = 0.5755 g M=

M=

PV mRT 50.01 10 3  255 5

1.01325  10  1000  0.5755  0.0821  288

M = 108 g  The molecular formula of compound is SF4.

10 –3  10 M3 = 100

(b) Full course of Antibiotics should be taken to cure the diseases.

1.01325  10 5

V=

M2 = 10–3 M2V2 = M3V3 10–3 × 10 = M3 × 100

28.

50.01 10 3

V = 255 mL

0.1 1 100

M3 = 10–4 pH = – log [10–4] pH = 4

mRT M

P = 50.01 KPa = 50.01 × 103 Pa

M1V1 = M2V2 0.1 × 1 = M2 × 100 M2 =

Following PV = nRT PV =

81 27.

480 11

32.

(a) zees.

33.

PA = PB

Humans didnot evolve from chimpan-

1.6 × g × 26. 6 =  B × g × 50

B

=

1.6  26.6 = 0.8512 50

NSEJS_STAGE-I _ SOL.-2013-14_PAGE # 4

34.

a,b are two positive real numbers

40.

(b) Both the explanation for the squirt of horned lizard are correct.

a2  b2 =6 ab

41.

Q = mS  96 = m × 0.8 × 6



a b  b a

Let x =

m=

a b

42.

96 = 20 g 0 .8  6

x3 – 3ax2 + 3ax – a = 0 a3 – 3a3 + 3a2 – a = 0

1 x+ =6 x  x2 – 6x + 1 = 0

– 2a3 + 3a2 – a = 0 a(2a2 – 3a + 1) = 0 a20, 2a2 – 2a – a + 1 = 0

6  32 x= 2 1

2a (a + 1) – 1 (a – 1) = 0

64 2 2

a20, 1, 1/2 Molecular formula = C4H8

x=

43.

x = 3 2 2

Isomers

x =3+2 2 x=3-2 2 x = 3 + 2  1.41 x = 3 – 2  1.414 x = 3 + 2.82 x = 3 - 2.828 x = 5.82 x = 0.172 So not in option

Total 5 isomers

x=

35.

(a – 1) (2a – 1) = 0

a lies between 5 and 6 b

38.

45.

EK = hf –  On comparing this with y = mx + c Slope m = h & intercept c = – 

46.

(d) Biodiversity means variaties species in a particular area and different species has different gens and different ecosystem

ATQ x2 + y2 +z2 = 1000 (x + y + z)2 = 2500 x2 + y2+ z2 + 2(xy + yz + zx) = 1500 2(xy + yz + zx) = 1500 xy + yz + zx = 750

Ans (B) 39.

(c) Iodine is important for the actvation of thyroxin hormone.

and

 Position of Cl is different  both are position isomers. 36.

44.

Ag is less reactive than hydrogen Ag + dil. HCl  No reaction

A outer A inner

(r1 ) 2 2 r12 25 = (r2 ) 2 = 2 = r2 24 2

r1 5  r = 2 6 2 perimeter outer 2r1 r1 5 = = = peremeterinner 2r2 r2 2 6

47.

Boiling can remove only temporary hardness.

48.

(a) Companion cell and sieve tube originate same mother cell so they are associated with each other.

NSEJS_STAGE-I _ SOL.-2013-14_PAGE # 5

next, x = 50, y = 10 51.

4PCl3 t=0 1 t = equi. 1–x KC =

KC =

So maximum of x + y = 60, is means

P4 + 6Cl2 0 x

60 max. items can be purchased and minimum

0 6x

of x + y = 38  (x + y)max = 60 = m

[P4 ][Cl2 ]6

(x + y)min = 38 = n

[PCl3 ]4

 m + n = 98

( 6 x )6 x (1 – x )4

59.

It is a biochemistry question and according to concept of chemistry when ph changed it

52. 53.

54.

(b) Placenta is layer that provides food, respiration and elemination of the waste of the foetes. altitude = 350 × 6 = 2100 m. = 2.1 km.

r    A = P 1   100 

caused loss of tertiary structure. 60.

(c) ties.

Nucleus controls all the cellular activi-

2

r    96800 = P 1  100  

2

...(1)

62.

r1 r2

P lies an y =

r    compound half yearly, 97240 = P 1  200  

4

... (2) After solving we got

P (3K,K) Now OAP is isoscles if (i) OP = PA (ii) PA = OA (iii) OA = OP Coee (i) OP = PA  OP2 = PA2 (3K – 0)2 + (K – 0)2 = (3K – 5)2 + (K – 0)2 9k2 + k2 = 9k2 – 30k + 25 + k2  30K = 25

r = 10% K= 55.

Polyalkanes are not inflammable.

56.

(a) Receptor receive the message that is taken by sensory neuron to the CNS and the reply of the message is given by motor neuron to the effector. Body B & C should have same kind of charge while body A may either have opposite charge or be neutral.

57.

58.

Equation formed is : 13x + 35y = 1000 x  y both should be integer So, x = 15, y = 23 1st integer value which satisfy the equation

x h = 50 K =  h = 3K 3 3

25 5 P 5 , 5    K= 30 6 2 3

Case (ii) PA = OA PA2 = OA2 (3K – 5)2 + (K – O)2 = 52  9K2 – 30 K + 25 + K2 = 25  10K2 – 30K = 0  10K ( K – 3) = 0 K=0/K=3 P (0,0) P (a,3) Total 4 points are there Case (iii) OA2 = OP2 52 = (3k – O)2 + (K – O)2 25 = 9k2 + k2  10k2 = 25 NSEJS_STAGE-I _ SOL.-2013-14_PAGE # 6

5 2

K2 = K=±

–

(a2 – b)x = ab + a + 1

5 2

x=

 5 5   P 3 2 , 2     5 – 5   P – 3 2, 2   

67.

63.

–[A] = 2[C]

64.

(c) Ovum has X-chromosome and sperms has X-chromosome so the child will be girl.

2kg

a2  b

Cu2+(aq) + M(s)  Cu(s) + M2+(aq) EºCell = Eºcathode – Eºanode

EºM2 = 0.34 – 0.75 EºM2 = – 0.410 V

F = 20N

68.

2kg v=0

(a) Deodar and pinus belong to gymnosperms.

x = 10m

f = 6N

Initially, u = 0

Ff 20  6 a= = = 7m/s2 2 2

m = 2kg

69.

m = 3kg

u1 = 4m/s

m1 + m2 ter collision

V=

m1 u1 + m2 u2 = (m1 + m2) v 2 × 4 – 3× 1 = 5v v = 1 m/s

56 m/s 1 1 mv2 = × 2 × 56 = 56J 2 2

v

Initial K.E. K1 =

x3 = a + 1 + x+

b =a x

x2  b =a x

(1) – (2)  x x3 – a – 1 = 0 x3 – ax2 – bx = 0 –

+

_________________ ax2 – bx – a – 1 = 0 _________________ ax – bx – a – 1 = 0 2

1 5 (2 +3) (1)2 = J 2 2

Loss in energy = 17.5 – 2.5 = 15 J 70.

given (ab)2 = (bc)4 = (ca)x = abc = k (Let) ab = k1/2 bc = k1/4 ca = k1/x abc = k now. a2b2c2 = k (abc)2 = k

1 1 1   2 4 x

1 1 1   2 4 x

(4) – (2)  a 2

1 1 2(4)2 + 3(1)2 = 16 2 2

3 = 17.5 J 2

Final K.E. K2 =

x2 – ax + b = 0

af-

u2 = 1m/s

V2 = u2 + 2aS V2 = 02 + 2 × 7 × 4 V2 = 56

K.E. =

66.

ab  a  1

0.75 = 0.34 – EºM2

N = 20N

65.

+

__________________

k

2

= k

1 1 1   2 4 x

2

ax – a x + ab = 0 NSEJS_STAGE-I _ SOL.-2013-14_PAGE # 7

2

=

x=

4 5

1 1 1   2 4 x

75.

Answer is not in the options

Due to presence of lp shape is triangular pyramidal like NH3.

71.

Alpha (due to high mass)

CO32–, NO3–, SO3 all are sp2 hybridised.

72.

(a) Cell  Nucleus Chromosomes  DNA Protein

73.

Current due to electrons I1 16

=

2  10  1.6  10 2

76.

(c) To conserved the environment person should first minimize the use of plastics

78.

Let radius, height and slant height be

19

r, h and l so, slant area of original cane = rl

Current due to protons I2 =

2  1016  1.6  10 19 2

ATQ 2r

So total current I = I1 + I2 =

 4h2 + 16r2 = 9h2 + 9r2  7r2 = 5h2

mA

7r 2 = h2 5

Let only 5 terms of the series 1–

=

h2  r 2

 4(h2 + 4r2) = 9(h2 + r2)

2  1016  1.6  10 19 = 6.4 × 10–3 = 6.4

74.

h 2  4r 2 = 3r

Now 4r

1 1 1 1 + – + 2 3 4 5

(i)

h 2  16r 2 = K r

h2  r 2

16 (h2 + 16r2) = K2 (h2 + r2)

60  30  20  15  12 47 = 60 60

7r 2 5

put h2 =

according this our option become (a)

=

=

 7r 2  2  16   5  16r  = K2  

30  20  15  12 60

67 60

(b)

(c)

1 1 1 1 + + + 2 3 4 5

1 1 1 + + 3 4 5

87r 2 12r 2 = K2 5 5

 16  =

20  15  12 47 = 60 60



1 1 1 643 13 + + = = 2 3 4 12 12



 7r 2   r2    5 

16  87 = K2 12

116

integer closest to K is 11

1 1 43 7 (d) + = = 3 4 12 12

79.

Mole =

0.2 =

Given mass Gram molecular mass

W 32

W = 6.4 g 80.

(d) Wuchereria is an organisms that belong to nematoda phylum. NSEJS_STAGE-I _ SOL.-2013-14_PAGE # 8