sec: sr. iit iz jee-main date: 24-12-18 time: 09:00 am 12 ... · 12/24/2018 · narayana iit...

Sec: Sr. IIT_IZ Jee-Main Date: 24-12-18 Time: 09:00 AM –12:00 Noon GTM-1 Max.Marks:360

KEY SHEET

PHYSICS

1 2 2 3 3 2 4 3 5 3

6 3 7 2 8 2 9 4 10 4

11 2 12 2 13 2 14 4 15 2

16 1 17 2 18 1 19 1 20 4

21 3 22 2 23 4 24 4 25 2

26 4 27 2 28 4 29 4 30 2

CHEMISTRY

31 3 32 2 33 3 34 2 35 1

36 1 37 2 38 4 39 2 40 2

41 2 42 4 43 3 44 4 45 2

46 3 47 1 48 1 49 1 50 3

51 2 52 3 53 3 54 3 55 4

56 4 57 3 58 2 59 4 60 3

MATHS

61 2 62 3 63 1 64 3 65 2

66 2 67 2 68 3 69 4 70 3

71 2 72 3 73 2 74 1 75 2

76 3 77 2 78 1 79 3 80 3

81 1 82 3 83 3 84 3 85 2

86 1 87 4 88 1 89 3 90 3

Narayana IIT Academy 24-12-18_Sr.IIT_IZ_JEE-Main_GTM-1_Key & Sol’s

Sec: Sr. IIT_IZ

SOLUTIONS PHYSICS

1. Work function

hC hC hC

2 2= − =

2. As seen by B: F = mg + T

As seen by A F = mg + T - ma

3. fv 0.97v 0.97 440 427 Hz= = =

4.

The shown figure represents(C) By snell’s law:

2

1

sin i

sin r

=

For 1 4 1i ; r and 1= = =

12

4

sin

sin

=

5.

4Adh gh r

dt 8

=

6. 360 200 div→

0 18

15 200 div360

→

1nm

LC200

=

1

x 10 mm200

=

31

F 100 10 5mN20

−= =

7. B,gg cosˆ â g sin i j

= − −

2

P,gˆ â g sin i g cos j= − −

P,gg cos â j

= −

2

P,gˆ û cos i usin j= + 4 45 45

( )usin

Tg cos /

=

2 45

2

8.

2

0

4T

r 2

=

( )

2

22

0

4T 1 Q

r 2 4 r=


Sec: Sr. IIT_IZ

( )208TQ 4 r

r

=

( )2 02TQ 8 r

r

=

0Q 8 r 2T r=

9. f Fsin=

mg cos

F

= + =2 02

ll

mgcos

F

=2

mg Fcos= −

For maximum

du

d=

0

10. Conceptual

11. The direction of velocity is along the tangent

Hence

dy

2xdx

=

0 4 2

tan53 2x 2x x m3 3

= = =

2

2 2 4y x m

3 9

= = =

Hence ( )2 4ˆ ˆ ˆ ˆL r P M i j 15cos53i 15sin53i3 9

= = + +

( )2 4ˆ ˆ ˆ ˆ2 i j 9i 12j 83 9

= + + =

12. For perfectly reflecting mirror, the force exerted by the light of power P is

( )2 power

Fc

=

For equilibrium, ( )2 power

F mgc

= =

6mg c

Power 30 10 W2

= =

As only 30% of the power is given to the mirror

So, 100

P ' 30 100MW30

= =

13. Conceptual


Sec: Sr. IIT_IZ

14.

mag 2F N mg+ =

2 magN mg F= −

1 magN mg F= +

( )2 2 2 magF N F mg F= = −

( )1 1 1 magF N F mg F= = +

2 1F F

2Bi

−=

15.

2

2 2

1tan g

2 u cos

= −

( )

221

tan g 1 tan2 2g

= − +

2 4tan 4 tan 1 0

− + + =

If particle will not hit the target

( )2b 4ac 0−

416 4 1 0 4 3

− +

16. L mvr r=

( )A B Cmeanr r r

A B CL L L

17. Index error in u = +1cm

U=8cm

Index error in v= 1cm−

v= 17+1=18 cm

1 1 1

f 5.53cmf 18 18= + =

18. Conceptual

19.


Sec: Sr. IIT_IZ

mg sin f kx cos 0− − =

N kx sin mg sin 0− − =

f N=

20. Pressure at point 0 B 0

2T 2TB P P P

3R 3R= + − =

Pressure point 0 C 0

2T 2T 4T 32TC P P P

R3R R 3R

2

= + + + − =

Ratio of point C 0 B 0P P &P P 16− − =

21. Conceptual

22.

When suddenly current injected into the left loop flux through the IInd loop increase with respect to time. So,

according to Lenz law a induced current develop in the IInd loop.

M B clockwise =

23. Consider system of two disks to be short dipole.

3

2kdpdE

r=

dp dq dA= =

3 3

2k 2kE dp A

r r= =

22

3 3

0 0

2 RE R

4 r 2 r

= =

24. conceptual

25. C C S Sg 1m V m V− … (i)

( )C C C S Sg 2m V m m V+= …………(ii)

( )C S S C C S Sg2 2m m V m V m V'+ = − + …………(iii)

C Cg

S2S

2m VV '

m=

26. Internal loss 2i r=

ir R

=

+

Battery life time=battery cap

current


Sec: Sr. IIT_IZ

27. conceptual

28.

R

1

LC =

2 R from diagram

Circuit will behave as capacitance circuit

current leads voltage

29. E S E 1

GS G S R G 4R

G S

= + +

++

3RS

HR 3S

=−

R S

G 3S=

30. 2 20I

I cos 30 cos 302

=

09I

32=

CHEMISTRY

31. 3

O

CH3

OH

O

CH3

OH2

OCH3

O O OH

H +

⎯⎯→

H −tauto

32. 2

33. 3

34. 2

Higher the value of + on carbonyl carbon more prone the molecule will be for reactivity

35. 5.3 g and 4.2 g

Using PhH mmoles of Na2CO3 = 2.5 x 0.2 = 0.5

Using MeOH, mmoles of Na2CO3+mmoles of NaHCO3 = 2.5 x 0.4 = 1.0

Mmoles of NaHCO3= 0.5 and mmoles of Na2CO3= 0.5

For 3

WNaHCO 1000 0.5

84 =

W = 0.042g in 10 mL

W = 4.2 g in 1 litre

For 2 3

WNa CO 1000 0.5

106 =


Sec: Sr. IIT_IZ

W = 0.053 g in 10 mL

W = 5.3 g in 1 litre

36. 2NH−

Pyramidal

37. 2

+ 2 2CH Cl + 3AlCl

CH2 Cl + 3AlCl

2 2Ph CH

38. i) n 3 l 2 m 0 n 1/ 2= = = = + Allowed

ii) n 2 l 2 m 1 n 1/ 2= = = = + not allowed

(l as to be 0 .... n - 1)

iii) n 4 l 3m 1 n 1/ 2= = = − = −

iv) n 1 l 0 m 2 n 1/ 2= = = − = − not allowed

l(m l)=

v) n 3 l 2 m 3 n 1/ 2= = = = + not allowed

l(m l)=

39. 2CO

2 2NO,O ,O−

are paramagnetic according to

MOT, unparired e−

are in * *

px py,

40. 2

As carboxylic acid is preffered functional group in the given molecule so it is given lower locant

41. pH = 9.35 , pOH = 14-9.35=4.65

x

pOH 4.74 log 4.65100

= + =

x

log 0.09100

= −

x

0.813100

=

x 81.3mmoles=

millimoles of ( )4 42NH SO 40.65 132 5.365g= =

42. 4

43. It is the defination fo Nucleoside

44. 100

2(g) (g) (g)XY XY Y+

time t = 0600 0 0(Pressure in mmHg)

time t = teq(600-x) x x (Pressure in mmHg)

given: 2(g)XY XY(g) Y(g)P P P 800+ + =

( )600 x x 800 x 200 − + + = =

2

XY(g) Y(g)

P

XY (g)

P PK 100

P

= =


Sec: Sr. IIT_IZ

45. 120.6kcal mol−

|

(s) 2 2 2

CH3

5C 4H CH C CH CH+ → = − =

H ? =

( ) ( )R p

H Bondenergy Bondenergy = −

(s) (g) H H C C C H C C5 C 4 H 2 H 8H 2 H− − = − −= + − − −

5 171 4 104 2 147 8 98.8 2 83 + − − −

120.6Kcalmol−=

46. 3

Terylene and Dacron are same name of polymer obtained by polymerisation of ethylene glycol and

Terepthalic acid. It is an ester polymer

47. Sol: 0.12264nm

o150 150

1.5 1.22A 0.122nmV 100

= = = = =

48. 26 min

0

C2.303log kt

C= −

40% to completion means 60 % unreacted

( )42.303log0.6 3.3 10 t−= −

t 26min =

49. 1

NH

O

O

O

O

N

O

O

CH2 Br⎯⎯⎯→KOH − N K

Br CH2 Cl

50. 99.1%

9.8

9.8g Cu 2 0.308 mole63.5

−= =

1

1000 3 0.311mole96500

− =

If 0.311 mol electrons provided by the current 0.308 mol was used to deposit copper. The current

efficiency is 0.308

100 99.1%0.311

=

51. The aE of forward reaction is more then that of backward reaction

9K decreases with temperature. Hence H must be –ve as evident from the Eq.

9

2

d ln K H

dT RT

=

( ) ( )a aH E FR E BR = −

( )a aE FR E (BR)

52. 3


Sec: Sr. IIT_IZ

HPh

CH3

Ph Br

CH3

is

PhPh

3CH

3H C

H

Br Br

Ph

Ph

H

3CH

3CH

⎯⎯⎯→rotate ⎯⎯⎯→2

alc

KOHE Ph

CH3Ph

CH3

53. 40, 20

Let x and y be the molar mass of A and B respectively then

( )

f

f

T 8 1000 80m 1 .....(1)

K x 2y 100 x 2y

= = − =

+ +

And

( )

f

f

T 10 1000 100m 1 .....(2)

K 2x y 100 2x y

= = − =

+ +

Solving (1) and (2) x = 40 ; y = 20

54. 4 6Na XeO

In the compound 4NaHXeO , Xenon is in +6 oxidation state. If it disproportionate to give ‘Xe’ (zero)

then Xe (VIII) must be formed. In this reaction 4 6Na XeO . 28H O get precipitated.

55. 4

Ranitidine is antihistamine so. Ans is D. Rest all are artificial sweetners.

56. behaves as strong electrolyte Below CMC no micellization takes place.

Sodium oleate ionizes almost completely, aqueous solution

57. 2

58. 2

2 3Ph - CH - CH ⎯⎯⎯→2Br /hν

3Ph - CH - CH

*

Br , so Ans is B. C & D forms diastereomers

59. 2O −

ions are present at octahedral voids 2Na O has antiflourite structure in which 2O − ions are

present at ccp and Na+ ions are occupy all the tetrahedral voids which are located at body diagonals.

60. Extraction is done by electrolysis

MATHS

61. 2 3 4 2 2 3 2 2 21 + + = + + + = + + + +

2 2 2 2 3 & y 3 x= + + + + = + = +

( ) ( )3 2

y 3 y 3 1 − = − +

3 2y 10y 33y 37 0 − + − =

62. R1 , R2, R3

contain 1 person each

1 1 1C C C2 .2 .2 3! P=

8.6 48

316 16

= =

63. x y 1 0 + − = & ( ) ( )6 x 2 8 y 2 0+ + − + = represent some time

2 6− =+

2 2 8 2 10 0− = − −+ =

64. ( )2013 22 2 2k

1 2 1 2

k 0

2Z Z 2014 Z Z 2014 2

2014=

+ = + = =


Sec: Sr. IIT_IZ

67. 1D 0 tan 3

3

−

1b

tan 0 02a

−−

( ) ( )3

1af 0 0 2 co t 0 R− =

Minimum integral value is 2

68. ( )( )2 2 2

0 1 21 x x 1 x x ........ a a x a x ......+ + + + + = + + +

1 2a 2 1 1 a 1 1 1 3= = + = + + =

3 4a 1 1 1 3 a 1 1 1 3......= + + = = + + =

( )2010a 1 1 1 3.....2 2009 3 6029= + + = + =

69. k

1 1 12cos k.sin sin k sin k

2 2 2t

1 12sin 2sin

2 2

+ − −

= =

n

1 1sin n sin n

2 2S

12sin

2

+ − −

=

n

n

Slt 0

n→= as nS is a finite quantity for every n

70. ( ) ( )22 22g x x g 3 3= =

( )( )22 1f g 3 f (3)

3= =

2 1 1f (1) f (2) f (3) f (2)

3 3 3= + = − =

2 1f (0) f (1) f (2) f (0) 1

3 3= + = + =

71. ( ) ( ) ( ) ( ) ( )( )2g ' 0 a 5, g '' 0 a 3, g ' 0 g '' 0 a 2 a 1 0 a 1, 2+ − = + + = + − = = −

72. AC a c,BC b c,OA b,OB a,OC a b c= + = + = = = + − Use cosine rule and ABC & OAC of

ABC & OAC

73. conceptual

74. conceptual

75. ( )2 2cot cot 1 1 cot + = −

2 22

2

sin coscot cosec

sin

− =

cot cos2 = −

Now cos2 tan2−

2

2 tancot

1 tan

− −

−

2

1 2 tan

tan 1 tan

− −

−

( )

2 2

2

1 tan 2 tan

tan 1 tan

− + −

−


Sec: Sr. IIT_IZ

( )

( )22

22

1 tan1 tancot

1 tantan 1 tan

+ − = =

− −

sec2 cot= −

1

.cotcos 2

= −

1

.cot 1cot

= =−

79. f(g(x) = x

( )( ) ( )' ' 1f g x g x =

( )( ) ( )' 1 ' 1 1f g g =

( ) ( )' 0 ' 1 1f g =

( )1

' 13

g =

G.E = 1 + 3 + 5 = 9= ( )( )

2

1

' 1g

80. 2ac

0 ba c

= =+

Now , ( )

( )( ) ( )

43 3 2 23 33

3 3 3 3 3

a c a c a ac ca ct a c ......(I)

b 8a c 8a c

+ + − ++= = + =

Now, ( )2 2 2a c a 2ac c 0− = − +

( )2 2a ac c ac........ II − +

Also, ( ) ( )4 2 2a c 2 ac a c 16a c ........ III+ +

(II) (III)..

( ) ( )4 2 2 3 3a c a ac c 16a c + − +

( ) ( )

4 2 2

3 3

a c a ac c2

8a c

+ − +

t 2

81.

( ) 1

n

1 1 1 11 1 1 1 1 ....... 1

3 4 5 n 12 3 4 n2 .........

n 2 3 4 n

+ + + + + + + + + + +

( ) ( )1

nn nn

n Sn 1 S n n 1 n

n

+ + + −

82. ( ) ( )r x a b y 2b c= + + −

r.a 1 x 1= =

( )r xa b x 2y yc= + + −

2 2r 13y 14y 5= + +

min

4r

13=

87. x y2 2cos sina 2 b 2

+ + + +

+


Sec: Sr. IIT_IZ

= 2cos 1eq2

−−

→

x cos ysin p + = → 2eq

(1) = (2)

88. Centre of circle ( )a,0 and radius 2a

Equation of circle ( )2 2 2x a y 4a− + =

( )2 2 2 2x a x y 2ax 3a 0− + − − = and

2y 4ax= solving 2 2x 4ax 2ax 3a 0+ − − =

2 2x 2ax 3a 0+ − =

x 3a,a= − and y 2a=

Length of AB 4a=

89. Equation of tangent to the hyperbola : 2 2 2y mx m a a= −

Let ( )1 1,P x y be locus

2 2 2y mx m a a− = −

S.B.S

( )2 2 2 2 2

1 1 1 12 0m x a y x m y a− − + + =

2 2

1 1 11 2 1 22 2 2 2

1 1

2;

x y y am m m m

x a x a

++ = =

− −

0 1 2

1 2

tan 451

m m

m m

−=

+

( ) ( ) ( )2 2 2

1 2 1 2 1 2 1 21 4m m m m m m m m + = − = + −

2 2 2

1 1 1 1

2 2 2 2 2 2

1 1 1

21 4

y a x y y a

x a x a x a

+ + + = −

− − − 90. Let the two unknown items be x and y, then

Mean 1 2 6 x y

4 45

+ + + += =

x y 11+ = ……. (i)

and variance = 5.2

( )2 2 2 2 2

21 2 6 x ymean 5.2

5

+ + + +=

( )22 241 x y 5 5.2 4 + + = +

2 241 x y 106+ + =

2 2x y 65+ = ……. (ii)

Solving Equations (i) and (ii) for x and y, we get

x 4, y 7 or x 7, y 4= = = =

sec: sr. iit iz jee-main date: 24-12-18 time: 09:00 am 12 ... · 12/24/2018 · narayana iit...

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