All Stream Narayana Jee-main Gtm-8 Final Key & Sol

8/11/2019 All Stream Narayana Jee-main Gtm-8 Final Key & Sol

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NARAYANA IIT ACADEMYINDIA

Sec:Sr.IIT- ALL STREAMS PHASE 1 Date: 16-02-14GRAND TEST - 8

Max Marks:360

KEY SHEETPHYSICS

1)4 2)2 3)2 4)1 5)2 6)3 7)38)3 9)3 10)3 11)2 12)3 13)3 14)315)3 16)3 17)1 18)4 19)1 20)4 21)222)2 23)4 24)2 25)4 26)1 27)1 28)229)2 30)3

MATHEMATICS31)2 32) 1 33) 2 34)1 35)1 36)4 37) 438) 1 39)1 40)3 41)2 42)1 43)3 44)1

45)1 46)2 47)2 48)3 49)1 50)4 51)452)4 53)1 54)1 55) 4 56)2 57)1 58)159)4 60)3

CHEMISTRY61)4 62)4 63)2 64)4 65)3 66)4 67)168)4 69)1 70)3 71)3 72)4 73) 3 74)375)2 76)4 77)2 78)4 79)3 80)1 81)282)1 83)1 84)4 85)1 86)4 87)4 88) 289)3 90)2


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Narayana IIT Academy 16-02-14_Sr.IIT-ALL STREAMS_PHASE-I_JEE MAIN_GTM-8

Sr.IIT-ALL STREAMS_JEE-MAIN_Solutions Page 2

HINTS &SOLUTIONS

PHYSICS

1 20 0 00

1 2

2

II E C or E

C

0 12 82 48.8 10 3 10E

155.5NC 2 Conceptual3 from input signals , we have

A B OUTPUT nandGATE

0 0 11 0 10 0 1

1 1 00 0 1

The output signal is shown at (b)4

A B Y1Y 2Y Y

0 0 1 1 1 00 1 1 1 0 11 0 1 0 1 11 1 0 1 1 0

56

1 1.54

t and drift t km

6

20

50cos37 v

R 2

40 ; 20 /10

vv m s

7 conceptual


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8

9 30 20 5di di

l i t vedt dt

30 30 2sec; 1di

L at t i ampdt

10 Basic concept of phasor11 This error occur in all the instruments which utilizes screw or nut-bolt system

12 L.C of vernier =M.S.D V.S.D 45 4.9 10 0.01m mm

L.C. of screw gauge =0.01mm

13

Let slope of line AC is 1 &m that of line DB is 2m , then

2 21 2

1 1

ln 2 lnm m

nRT nRT m m

1

2

2T

T

14 The temperature of these objects being identical the cooling rate will depend upon surface area. Thecircular pate has the largest area and hence will cool faster than cube and sphere, Sphere having lesssurface area than a cube will cool at the lower rate

152

4 4 0.052 ; 10

2 10

SR cm P Pa

R


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16

0 3tan378 4

r

6r m

20F P h g r

510 10 8000 10 36 5 71.8 36 10 2 10

17 since it is a conductor

0V = potential at surface=0

0 0

0

4 4 2

q Q

a a

2

Qq

18A at surface B C O

V V V V V

Charge is on the outer surface hence insideV remains constant

19 The potential at centre of sphere in which q charge is uniformly distributed throughout the volume is

0

1 34 2

CqV

R

By symmetry the potential at centre due to half sphere will be half of the complete sphere

0

1 3 / 2

4 2C

qV

R

0

1 3

4 2 2

Q qQ

R

20 1 2 12

1 2

;2

r

v v vA A v

v v

energy reflected

1

9 energy incident 21


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36 /1.2

30

m sl m

0.3m 22 Conservation of angular momentum

2 2 2

0 '2 2 3

i f

mR mR mlL L

20

2 2

3'

3 2

R

R l

23 1 2 2 1 2(1 cos )cos 2 cos 2 cos 2 cos 2E a f t f t a f t a f t f t

2 1 2 1 21 1

cos 2 cos 2 cos 22 2

E a f t a f f t a f f t

This is a complex vibration consists of harmonic vibrations of frequencies 2 1 2 1 2,f f f and f f

The highest is 1 2 max.f f So hv T

max 1 2T h f f

3415 15

19

6.6 103.6 10 1.2 10 2.35

1.6 10

17.45ev 24 Apply conservation of momentum & energy

25 Let total sum of number of isotopes at the formation of the planet was 0N and currently is N

0 1 0.22

t

X

NN e N

.(i)

20 0.82

t

y

NN e N

..(ii)

Dividing equation (ii) by (i) 1 2

4t

e

1 2 ln 4t

9

ln 2 1 12ln2

10 2 4t

98 10t years26

At P, 8 3

, d d

xD

For 2ndmaxim,224

2 2d

xD


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27 0 0 02 2 2

I I IB i j k

R R R

0 32

IB

R

28 In steady state the capacitor is fully charges and is treated as open circuit, so no current flowsthrough branch containing capacitor in steady state, So, the circuit can be redrawn as

Potential difference across capacitor in steady state6 6V V V

(-ve sign signifies that left hand plate is of negative polarity )

Charge 1 6 6CV C 29 In steady state current through capacitor is zero. For zero deflection A BV V

So, 11

qIR

C

And 1 222 2 1

q R CIR

C R C

30 4

3 2

02

4

T xU x x dx x Since there is no loss energy , so

42 2

1

13

4 2

xx E mv

4 24 12 0x x

2 4 16 48

2x

4 86, 2

2

2x


7/13


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35 30 4 27C

36 conceptual

37 We have

5 0 0 1 0 0

0 1 0 0 1 00 10 2 5 0 0 1

x

ABx x

1/ 5x

38

1

1 0

1

a a

b b

c c

Applying 2 2 1 3 3 1C C C and C C C

And expand to obtain

1 1 1 1 1 1 0a b c b a c a b

Divide by 1 1 1a b c 39 Two tallest boy can be different groups in 2 2 11 1n nC C ways

40 , 22

413

204 13 | sin

3xt at t x

x

2 12 39 3sin /x x a x

2 12 39 3sina

x xx

2

6 3 sin 1

a

x x

sin 1, 66

ax

42 From 2sin sin 1,x x we get 2sin cosx x . Now, the given expression is equal to

6 6 4 2cos cos 3cos 3cos 1 1x x x x

36 2cos cos 1 1x x

33sin sin 1 1x x

32sin sin 1 1 1 0x x

43 We have 1 1 sin 2S x 2 1 2 cos2S x x 3 1 2 3 cosS x x x 4 1 2 3 4 sinS x x x x

So that4

1 1 1 3

1 2 4

tan tan1

i

i

S Sx

S S


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1 sin 2 costan1 cos 2 sin

1 cos 2 sin 1tan

sin 2sin 1

1 1tan cot tan tan / 2

/ 2 44 33 87y m x

1 87 33

33 872

mA m

m

2

20 0

dA d A

dm dm

45 Perpendicular distance from centre to the line1

2

common difference

11

2 1 2 222 2 4

or or

46 Common tangent nearer P is x=1directrix of ellipse is x=1e=1/2

1,1

2s

47 Perpendicular bisector of the base must meet the parabola at the vertex of the triangle.48 Let P( a,b,c) be the foot of the perpendicular from the origin to the plane, then direction rations of

OP are

0, 0, 0a b c i.e, a,b,cSo the equation of the plane passing through P(a,b,c) the direction ratios of the normal to which area,b,c is

0a x a b y b c z c 2 2 2

ax by cz a b c

49 Since ' 2 cos 0 ,f x x for all x R so f is one-to-one. Moreover,

f x as x and f x as x , hence the range of f is R. Therefore, f is onto as

well50 Conceptual

51

0'9 limh

f x h f x

f x h

0 0

0lim lim

0

h

h h

f x f h f x e g h

h h

0

lim 'h hh

e g h e g h

' 0 0 4g g

Hence 4 10 10 0f x x c but f g

So 4f x x


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52 Conceptual

53

1/ 1/

1 21n n

n n

f x xf o f x

xf x f x

Similarly

1/1 3

nn

xf o f o f x

x

and likewise

1/

.....1

nn

xf o f o f n times x

nx

1

1/ 02

1/. 1

1

nn

n n

nn

xx g x dx dx Put nx t

nx

We get the above integral as

111 1

1

nnt dt t K

n t n n

111

11

n nnx Kn n

54 h x h I

Where sin4x

I t dt

Put ,t so that

4

0sin

x

I d

4

0sin

x

d h x h x h h x

55 conceptual

56 2 2 2 ' ' 2 0d

f x g x f x f x g x g x f x g x f xdx

Hence 2 2f x g x is constant . Thus

22 2 2 2 216 16 2 2 2 ' 2 16 16 32f g f g f f

57 Let 1/

1 x

y x

2 3 41 1

log log 1 ...2 3 4

x x xy x x

x x

2 3

1 ..2 3 4

x x x

2 2

1 .. ..2 3 2 3

x x x xa

y e ee

22 21

1 ..... ..... ....2 3 2! 2 3

x x x xe

2

21 1 1 10 0 .....2 3 2 2

y e ex ex x x

(0,(x) in terms containing x)


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20

12 1 1 112lim

3 8 24x

y ex

e ex

58 Conceptual59 Conceptual60 Vertex is (-2,1), length of the latus return =4a where a=1/2 latus rectum is

parallel to the tangent x+2=0 at the vertex, at a distance from it. So itsequation is x=-3/2,1+1) and (-3/2, 1-1) i.e (-3/2,2) and (-3/2,0). The point not

lying on 2 3 0 3/ 2, 2x y is


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CHEMISTRY

61

4 2

1

f

f

T KCl i KCl

T X i x i x

20.5

4

ix

for association of 3 molecules 1

1 1i xn

11 1 0.5

3

0.75

62 The boyles temperature of both 2H and He is very low.

63 Along the body diagonal two B ions and one A are removed64 At high concentration of sulphide ions the cations of both II and IV groups are precipitated.65 Follow CIP rules.

66 2 r n 2 12 nr n

67 Depends on stability of carbocations

68 (i) Molecules move faster for whichT

Mgreater. obviously 2H molecule move faster.

69 Cannizzarrows reaction70 Rate of reaction depends on leaving ability of the group71 Nucleophilic substitution72 In the partial hydrolisys of 6Xe F different products formed or 4 2 2XeO F and Xe O F

73 Tollens reagent oxidizes aldehyde group only

74 3B PH 2 2A NaH PO 75 Conductivity 6 6 64 2 10 1 10 1 10BaSO

2 24 4

0 0 0BaSO Ba SO

Molar conductivities =1 2 160 2 20 100ohm cm mol

In the case of sparingly soluble salt4

60 1000 1000 10 100

BaSO V

K

C C

3510 10 /

100C moles litre

76 1C epimers are called anomers

77 a) basic strength in vapour state depends on +I effectc) +R group increase basic strength at para and decrease the basic strength at meta due to I.

d) Lesser the surface volume more the basic strength, hence NH3is more basic. ROH > H2O due to+I effect of R group

78 Conceptual79 Conceptual80 3B NH 2C N 3 2D Mg N

81 Conceptual


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82

83 Iodoform reaction84 Teflon, polystyrene and PVC are formed by addition polymerization85 Conceptual86 Ratio of concentration of reactants and products is same in all three.87 0 tKt C C

88 Follow the structure of dichromate ion.

89

2CH

2H N

20

/

0.5.........NaNO HCl

C

N N

2N

H

or

H

90 Sod. Benzoate is metabolized to hippuric acid 6 5 2 .C H CO NH CH COOH

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