sri chaitanya iit academy., india

Sri Chaitanya IIT Academy., India A.P, TELANGANA, KARNATAKA, TAMILNADU, MAHARASHTRA, DELHI, RANCHI

A right Choice for the Real Aspirant ICON CENTRAL OFFICE, MADHAPUR - HYD

Sec: Sr. ICON ALL Jee-Main Date: 30-12-18 Time: 09:00 AM to12:00 Noon CODE-C Max.Marks:360

KEY SHEET PHYSICS

1 1 2 1 3 2 4 2 5 3 6 4 7 1 8 2 9 2 10 3 11 2 12 1 13 1 14 3 15 2 16 4 17 3 18 3 19 3 20 1 21 4 22 4 23 2 24 3 25 2

26 1 27 2 28 4 29 3 30 2

MATHS 31 2 32 1 33 2 34 1 35 1 36 1 37 2 38 3 39 4 40 3 41 2 42 3 43 4 44 2 45 1 46 3 47 4 48 4 49 2 50 2 51 3 52 4 53 3 54 2 55 3 56 3 57 2 58 1 59 3 60 2

CHEMISTRY

61 3 62 3 63 3 64 4 65 3 66 1 67 3 68 2 69 2 70 3 71 4 72 4 73 3 74 3 75 3 76 3 77 3 78 2 79 2 80 3 81 1 82 3 83 1 84 1 85 3 86 2 87 3 88 4 89 2 90 1

Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s

SOLUTIONS PHYSICS

1. Density,

Mass Md

Volume v

Here m = 2.42g, m 0.01g 3v 4.7cm V 0.1cc

Maximum error in density, d 100d

m V 0.01 0.1100 100 100 100 0.413% 2.127% 2.54%m V 2.42 4.7

2. A0v 0

1

2

l 3l 2

A1

A 2

V 3 3V 3 2 5

A1vv 3x 0.6v5

3. Conceptual 4. The acceleration of the horizontal movable pulley will be 6m/s2. So for the block B. 23 6, 2 /B Ba a m s 5. Free body diagram of the two bodies are as follows

Maximum friction between the two blocks is maxf mg , where m = 2kg maxf (0.5)(2)(10) 10N Let acceleration of both the blocks be a towards left. Then

f 2 20 fa

2 4

2f 4 20 f f 8N Now since, maxf f i.e., static region of 2kg on 4kg, hence friction force between the

two blocks is 8 N. 6. Conceptual 7. From work energy theorem W K

212

Pt mv

Or 2 ...(1)Ptvm

4kg2F 20 N

2kg1F 2N

f

f


Sec: Sr. ICON ALL

1/22ds Ptdt m

Or 1/2

0 0

2s tPds t dtm

Or 3/22 2 .... 23

Ps tm

From Eqs.(1) and (2) 23

s tv

8. From parallel axis theorem.

X CM

2

1 12MLI

22 1I I Mx

22 3

12 2 2ML LM

211

24ML

9. About point of contact I

23 0 8mgr mg mr

8gr

mg

10. 2 2 0 2cos 45 gO g Rw wR

11. If an bubble is formed, its radius is equal to that of capillary


Sec: Sr. ICON ALL

Required pressure = 02sghr

12. PP TT

,

oo

o

PT

3 32 2

oB o

o

PT

13. 3PT Constant

VPTConstant

2VT Constant 1/2TV Constant

11 1.52

14. Apparent Frequency of direct waves received by observer is

1s

Vv vV V

Apparent frequency of reflected waves received by the observer is 2s

Vv vV V

No.of beats per second 1 2v v v . 15. The wave pulse is travelling along positive x –a xis. Hence at and bx should have

opposite signs. Further wave speed

int

Coefficient of tvCoeffic of x

41

Coefficient of t

Coefficient of 14t s 16. S-Force of interaction

3 3

0

2 1 14 1

pqd d l

17. Conceptual 18. 0/ 2 ',E r where is the linear charged density of the inner cylinder.

And 0

ln2

b

a

bV Eda

Now, . .l J dA E dA


0

22

r drr

19.

The current through ammeter 4 3 2 1I I I I I

6 4 14 66 0 8 62 4 4 8

1 13 1 3 1 22 2

A

When only 4 resistance is shunted / 5gi i 20. / 5 4 4 / 5 16G i or G

21. For a circular wire

22. Conceptual 23. The emf induced in a conductor does not depend on its shape but only on its end

points. We replace the actual conductor by an imaginary straight conductor joining its two ends.

24. 00 0 0

2

3

hc hc hceV


Sec: Sr. ICON ALL

0 0

1 2.4

Q hceR

25. Conceptual 26. If screen is perpendicular to the line joining the sources the fringes will be circular and

central fringe will be bright if 1 2S S n 27. When light passes through a medium of refractive index , the optical path it travels is

t Path difference t b b 1 For a small element ‘dx’, path difference, x 1 ax 1 dx axdx

For the whole length,

2

0

atx axdx2

For a minima to be at’O’,

x 2n 12

2at 2n 1

2 2

For minimum ‘a’, n = 0

2

2at a2 2

28. When a real object is placed 25 cm from a lens, a real image is formed. Mark the is

INCORRECT statement from the following. 1) The lens is a converging lens 2) The image may be magnified or diminished 3) The focal length of the lens is less than 25 cm 4) The focal length of the lens may be greater than 25 29. CONCEPTUAL

30. 8

8 19 10 1.5 106

v msk

Refractive index 8

8

3 10 21.5 10

cnv

Also r rn For a non – magnetic medium 1.r Therefore 4r rn


Sec: Sr. ICON ALL

MATHS

31. 2 2

2 15 5cosx y

2

221

1 bae = 1

225cos 1 cos ;

5ly encentricity of the elipse

2 2

2 125cos 25

x y is

22 22

25cos1 sin ;25e put 1 3e 2 2

2 1 23e e e

2 2 2 11 cos 3sin 2 4sin sin ]

2

32. conceptual

33. 2A=XY sin ; 4

2 2 2 2 24 sin ; ( ) ; '( ) 0 ]2 1 3

xA x y f x f x xx

34. Use sin x expansion, sin x 3 5

.......3! 5!x xx

35. 3 4 3 4 1 02 3 2 3 0 1

BC BC I

2( )

2 2r r rA At A t t

+……….

2

1 1( ) ( ) ......2 2r r rt A t A t A

( ) 2 ( ) 2(2 1) 6

1 (1 / 2)r

rt A t A

36. Conceptual 37. 2 3 2 2

0 1 2 1 2(1 2 5 10 )[ ......] 1 ......X X X C C X C X a x a x

1 2a n and 2( 1) 2 5

2n na n

Put 221

2aa

2( 2) ( 1) 4 10n n n n

2 24 4 5 10n n n n

6n 38. CONCEPTUAL 39. Lines are 1 0;x y 4 3 4 0x y and 0x y where 2 2 2

1 1 14 3 41

=0

Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s 1 3 – 4 – 1 4 – 4 1 4 – 3 3 – 4 – 4 4 4 – 3

– 1 0 1 1 ] 40. Conceptual 41. Conceptual

42. f (x) = 23 ;

4t x x

x

f ' (x) =

2

2

( 4)(3 2 ) ( 3 )( 4)

x x t x xx

for maximum or minimum, f ' (x) = 0 – 2x2 + 11x – 12 – t – 3x + x2 = 0 – x2 + 8x – (12 + t) = 0 43. Shifting the origin at A equation is X2 = – 8Y Now (x – 2)2 = – 8(y – 2)

44. q2 - 4 p r = 0, p > 0, p > 0 f(x) = log (px3 + (p + q) x2 + (q + r) x + r) Let g(x) = px3 + (p + q) x2 + (q + r) x + r g(x) = (x + 1) (px2 + qx + r) Discreminant of px2 + qx + r = q2 – 4pr = 0 Domain (x + 1) (px2 + qx + r) > 0

p(x + 1) > 2 x – and x > – 1

x R – [(– , –1]

45. f is not differentiable at x = 12

g is not continuous in [0, 1] at x = 0 h is not continuous in [0, 1] at x = 1 & 0 k (x) = = (x + 3)p where 2 < p < 3 ] 46. x = cos , [0, ]

cos–1(cos ) + cos–1 =3

2

p2qx

p2q

5n2)3x( l

sin3

sincos3

cos


cos + cos–1cos 3

= 3 can hold only if 0

3 –

– 3 – – 2

3 x 1 ,1

2

0 03 ]

47. No of integral solutions 1 2 2 4 2 * 5 * 7 *11x x x x CASE-I ALL 0xi No of solutions = 4(4!) CASE-II All 0xi No of solutions = 4(4!)

CASE –III Two positive two negative 42C = 6 COMBINATIONS

No of solutions = 6* 4(4!)

Total no of integral solutions = 8* 4(4!) 48. Hint : find ‘c’ assuming the line is the tangent to the circle. And plot the inequalities

on x-y plane. 49. 22 + (x1 – 1)2 = x12

4 + x12 + 1 – 2x1 = x12 5 = 2x1 or x1 = 5/2 Equation of (1) from (2, 5/2) to the given base y – 5/2 = 2 (x – 2)

2y – 5 = 4 (x – 2) at y = 1 –3/4 = x – 2 or x = 5/4 ]

50. f '(x) =

Q f (0) = 0 Þ f ' (x) = f ' (x) = f ' (0) + | x | = | x | ] 51. Conceptual

52. [Hint :A = = =

f(a) = now f '(a) = = 0 Þ a3 = 1 Þ a = 1 53. Conceptual 54. Conceptual

hxhh|x|)h(Lim

h)x()hx(Lim

2

0h0h

fff

xh|x|h

)0()h(Lim0h

ff

dxx1

6xa2

a2

a2

a

2

x1

12x

a1

12a

a21

3a 22

a21

4a2

2a21

2a

Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s 55. Conceptual 56. Conceptual

57. u = put x = tan q

= = = = ln 2 – u

u = ln 2 4u = ln 2 ....(1)

again v = (put 2x = t) ; v = = v = – ln 2 ....(2) (1) + (2) Þ 4u + v = 0 ] 58. Conceptual 59. x2y = c3

x2 + 2xy = 0 = = equation of tangent at (x,y)

Y – y =

Y = 0, gives , X = = a

and X = 0 , gives , Y = 3y = b Now a2b = = ] 60. Conceptual

CHEMISTRY 61. Meq of salt = Meq of 2 3Na SO 50 0.1 n 25 0.1 2 n 1 changein O.N.

3 2M e M 62. Polarity in a molecule gives rise to an increase in forces of attraction among molecules

and thus, the boiling point increases.

63. Let w g of each be taken, then initial mole of w wP ;molof Q10 20

Final mole of 1wP5 10

Final mole of 14wQ20 5

For 0 1N t

N

PP e

P For 0 2N t

N

QQ e

Q

For 1 20

2

w 5 10P e10 w

For 2 20

1

w 20 5Q e20 w 4

By Eqs. (i) and (ii) 1 2 204 e 1 2 e20 log 4

1

02 dx

1x)1x(nl

4

0

d)tan1(nl

4

0

dtan1tan11nl

4

0

dtan12nl

4

8

2

2

0

dx)x2(sinnl

0

dt)t(sinn21 l

2

0

dx)t(sinnl

dxdy

dxdy

xy2

)xX(xy2

2x3

y3.4x9 2

32 c4

27yx4

27


Sec: Sr. ICON ALL

or e1/ 2

0.693 0.69320 log 410 t

t

64. Higher vapour pressure of 2H O in atmosphere will derive 2H O vapours to the solute particles

65. phenomenon of conversion of freshly precipitated mass into colloidal state by the action of solute or solvent is called peptization

66. 2 5 2 2d N O d NO 2d O1dt 2 dt dt

2

1 2 5 2 5 3 2 5kk N O N O 2k N O2

67.

22 32 5

c2 4

2 10 / 2NOK 1 10

0.2N O2

68. For NaX, 21 h h h

X H O HX OH

14

wh5

a

KK 10hC K C 10 0.1

8 410 10 4 2%H 10 100 10 0.01 69. T 300K, V 10 1 9L H E pV E 2 RT 0 2 8.314 300 4.98kJ E=0 for isothermal process 70. This gives rise to higher e nuclear charge in Na and the size of Na becomes smaller

due to more effective pull of valence shells towards nucleus. 71. Allotropes of an element have the same chemical properties but have different

arrangement of atoms and physical properties 72. N a is soluble in NaOH whereas 3cFe OH is insoluble. 73. 4 2 4 2FeSO 7H O FeSO 7H O; 4 2 3 2 32FeSO Fe O SO SO 74. 2 5 2 3HO SO OH PCl Cl SO Cl POCl 2HCl 75. The solubility of nobel gases increases with increase in molecular weight due to

increase in van der Waals’ forces. However, these are sparingly soluble. 76. According to Werner’s theory, only those ions are precipitated which are attached to

the metal atoms with ionic bonds and are present outside the coordination sphere. 77. PbS is black and 2S reacts with 2 2 7K Cr O to give 2 4 3Cr SO solution which is green.

78. cellE for the given change = 2

2

H I2

H II 1

H p0.059 log2 p H

Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s Thus, for positive

H 2II2 Icell P HE P 79.

80. In D-(+)-tartaric acid, the (+) is due to positive optical rotation and is derived from

D glyceraldehydes.

81. It undergoes dehydration easily as the product obtained is conjugated, and is more

stable.

82. 2 2 2

2H O H O H

2 2 2 3 3 2NiHg catalystCaC C H CH CHO CH CH OH

83.

2 2

2H O Br

3 3 3 3 3NaOHH .HgCH C CH CH COCH CHBr CH COONa

Since, B and D are different, thus B is 3 2CH CH CHO and so A is 3 2 2CH CH CHCl .

84. Must be a tertiary alcohol as it gives alkene on treatment with Cu. Thus, 4 8C H O is a

ketone.

85. Follow applications of inductive effect. The negative charge on carboxylate ion is

dispersed more due to IE of F atom.

The carboxylate ion thus becomes more stable and the acid becomes more reactive.

86.

Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s 87.

88.

89. DNA has deoxyribose sugar; RNA has ribose sugar with three bases common as

adenine, guanine and cytosine, DNA has fourth base thymine; RNA has uracil

90. Bakelite is a step-growth polymer i.e., the condensation involving the reaction of

functional group e.g. terylene, Bakelite etc.

sri chaitanya iit academy., india

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