narayana iit academy · narayana iit academy 27-01-18_sr.iit_iz_ph-i_jee-main_gtm-4_key&sol’s...

Narayana IIT Academy INDIA

Sec: Sr. IIT_IZ JEE-MAIN Date: 27-01-18 Time: 07:30 AM to 10:30 AM GTM-4 Max.Marks: 360

KEY SHEET

PHYSICS

1 1 2 2 3 3 4 2 5 4 6 3

7 4 8 4 9 3 10 3 11 2 12 3

13 4 14 2 15 2 16 2 17 2 18 4

19 1 20 3 21 2 22 2 23 2 24 3

25 1 26 2 27 3 28 4 29 1 30 4

CHEMISTRY

31 2 32 4 33 2 34 3 35 3 36 1

37 3 38 4 39 2 40 1 41 3 42 1

43 4 44 4 45 4 46 4 47 2 48 3

49 3 50 3 51 4 52 1 53 1 54 4

55 3 56 4 57 1 58 2 59 2 60 3

MATHS

61 4 62 3 63 1 64 1 65 2 66 1

67 3 68 3 69 4 70 2 71 2 72 1

73 2 74 2 75 3 76 4 77 4 78 2

79 1 80 2 81 1 82 4 83 3 84 2

85 3 86 2 87 2 88 4 89 3 90 1

Narayana IIT Academy 27-01-18_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol’s

Sec: Sr.IIT_IZ

SOLUTIONS PHYSICS

1. At any instant let be the angle made by v with u and let they meet after line T

0

cos T

v dt uT and 0

cos T

v u dt l 0 0

cos T T

v dt u dt l 0

cos T

vT u dt l

uTvT u lv

2 2

2 2

v u lvT l Tv v u

2. 10 VSD=9MSD=9mm.

Least count =1MSD-1VSD=0.1mm.

Zero error= 0.1 10 8 0.2 mm.

Thickness= Main scale reading +L.CV.C+error.

i.e, Thickness=12 0.1 3 0.2 12.5mm

3.

4. The initial extension of spring is 0mgxk

. Just after collision of B and A the speed of

combined mass is 2v

( |2mv m v ). For spring to just attain natural length the combined mass must rise up

by 0mgxk

(see figure) and come to rest. Now applying conservation of energy

2

20 0

1 12 22 2 2

vm k x m gx

and 0mgxk

.


Sec: Sr.IIT_IZ

So get 6mv gk

.

5.

1) As mutual repulsive force between the particles is internal for the system and as there

is no other external force on the system ,linear momentum of the system is conserved in

any direction

2) As the forces on the particles due to one on the other are equal in magnitude,

opposite in direction and act along the line joining them always, net torque on the

system due to these forces about any point in space is zero. therefore angular

momentum of the system remains constant about any point in space

3) As center of mass of the system lies on the line joining the particles always and force

on any of them is passing through C.M always, torque due to this force on any particle

about C.M is zero. Hence angular momentum of any particle about C.M is conserved

individually.

4) About any other point except C.M, torque on any individual particle is not zero.

Hence angular momenta of individual particles change but total angular momentum of

the system remains constant.

6.

7. 0cos2L v and sin

2L v 0 tanv v


Sec: Sr.IIT_IZ

8. From conservation of angular momentum of the mass about the centre of the black

hole, 0mv R mva and from conservation of energy 2 20

1 12 2

GMmmv mva

.

9.

sinF mg tanmg ( is small) . i.e, F 2

40dy xmg mgdx

2xa 1

2

10.

2g

g5g 2g

P0

5effg g . Therefore pressure at centre= 0 5P Rg 11. If is temperature of the body at any instant and 0 temperature of surroundings, from

Newton’s law of Cooling 30 04eAd

dt ms

. i ,e At any temperature 1ddt m as

, , , se A are same for both.

As area is same for both mass of cube is less than that of sphere. Slope of the curve

at any for cube is greater than that of sphere

12. At triple point substance exits is three states in equilibrium


Sec: Sr.IIT_IZ

Keeping pressure constant if temperature is increased substance will be in vapour state

and if temperature is decreased substance will be solid state. Keeping temperature

constant if pressure increased substance will be in liquid state and pressure decreased

substance will be vapour state .

13. Internal energy 52

U nRT and PV nRT

52

U PV U P 1

2

1

2

i

i

U PU P

( V is

same for both) . As final pressures are equal 1 2f fU U

14. Let the wave equation is siny A t kx .

At the given instant particle at 0.09x m is moving towards mean position from

positive extreme.

i.e , 1 33 4sin sin 0.5894

t kx t kx

rad. its phase is 0.589 rad .

At the given instant particle at 0x is at mean position and is moving towards positive

extreme.

its phase is 0 or 2 rad or 4 rad etc. As wave is moving along positive x-direction

its phase is more than that at 0.09x m and as these are within one wave length range

phase of particle 0x should be 2 rad.

Phase difference between the particles 2 0.589 0.589 3.729 rad

3.7290.09

k x k . Wave velocity 12 6.28 0.04 3.793.729 0.09

Tv msk k

15. dVEdx

16. Potential at any point inside the shell=potential at any point on the surface

potential at A=potential at C due to ‘q’ and induced charges =


Sec: Sr.IIT_IZ

0 0 0 0

1 1 1 14 4 4 4

q q q qr R R r

17. When the switches open charges on differtent plates are as shown

When all the switches are closed the two capacitors will be in parallel and net charge

on the isolated plates B and C together (-100+40=-60 c )is shared by them in the the

ratio of their capacities

final charges on different plates are as shown

80 C flows through s1 fromplate A to ground, 80 C charge flows through s2 from plate

C to B and 80 C charge flows through s3 from ground to plate D.

18. 0E DV V (as they are earthed )

2 , 10 2 12 , 12 6 6B C AV V V V V V

Current through 2 22

A BV V A

Current through 3 43

C DV V A

From Kirchhoff’s junction rule we can show that current through wire DE=0

19. Electric field is induced and is the form of concentric circles about centre of cylindrical

region.


Sec: Sr.IIT_IZ

2 22 2 2

d dB r dB rE r r r Edt dt dt

. Consider a small path length dl along

path AB at distance r from the centre

tan 30 ABL

. cos2r Lde E dl Edl dl

r

2

2 2 2 3 2 3ABL L L L Le de dl AB

20. When the right wire starts moving, due to motional emf developed in it anti clock wise

current flows in the frame, due to this there is mutual attraction between the wires and

due to this the wires attain common velocity after some times. When they attain

common velocity there will be no net emf hence no current flow, hence wires continue

to move with the common velocity

00 2

2vmv mv v = common velocity. Loss in

22 200 0

1 1 122 2 2 4

vkE mv m mv

21. 0 0 0NiB niL

22. For permanent magnet we prefer a material with high retentivity (so as to make a

stronger magnet) and high coercivity (so that magnetization any not be wiped out

easily). For electromagnet we prefer high saturated magnetism low coercivity and least

possible area of hysteresis loop so that electromagnet develops high magnetization, is

easily demagnetized and energy loss in a magnetization cycle is least. Therefore, P is

suitable for making permanent magnets and Q for making electromagnets.

23. 0 sincJ E t kx and 0 0E

d

d dEi Adt dt

0 0 cosAE t kx

dJ 0 0 cosE t kx 0

c

d

JJ


Sec: Sr.IIT_IZ

24. Replacing it with string block system

Let at initial position 2F force is a applied then

W.E.T. from A to B

0S FW W

3FXK

Net elongation in spring max3 3F Q CEK

25. 0

0

1e

v Dmu f

0 0

0 0

25 2525 15 6

v vu u

For eye lens 1 1 1 1 1 1 1 6 25 25 5 25 6e

e e e e e

u cmv u f u u

059146ev u and 0

06 5925 25vu cm .

For objective 00 0 0 0

1 1 1 6 25 1 59 59 59 31

f cmv u f f

.

26.

Due to total internal reflection light comes out through a cone of semi vertex angle

which is equal to critical angle. Total light energy is distributed over 4 solid angle.


Sec: Sr.IIT_IZ

Solid angle made by cone at source 2 1 cos . 5 sin 14

053 .

02 1 cos53 1 2 14 2 5 5T

PP

27. For 1st minimum sin sin 30b b ……(1)

For 1st secondary maximum 3sin2

b ……..(2)

1 sin 30 sin 3 432 sin

2

1sin 3 4

28. Charge is conserved. In order to fully convert an electron into energy, a positron (the

electron’s antiparticle must be involved). That is, electron+positron energy , Not

electron positron+energy

29. For excitation with least KE of colliding particle the collision should be perfectly in

elastic and all the loss in KE is to be used for excitation.

0 ( ) mumu M M m v vM m

where v=common velocity after collision .

maximum loss in KE is

2 22 2

2

1 1 1( )2 2 2

m u M MK mu M m mu K KM m M mM m

4. 10.2 10.2 4 63.75

4 1i e K K eV

30. .y A B B CHEMISTRY

31. NCERT-XI, VOL-II, PAGE NO : 403, 404

32. NCERT-XII, VOL-II, PAGE NO : 452, 453

33. NCERT-XII, VOL-I, PAGE NO : 229

34. Cis - 3 24Co NH Cl - violet and trans- 3 24

Co NH Cl - green

35.

Ionization energy (in Kj/mol)

H 1312

2H 1488

N 1402


Sec: Sr.IIT_IZ

2N 1503

O 1314

2O 1175

F 1681

2F 1515

If energy of electrons in HOMO of molecule is higher than energy of last election

in its atom then ionization energy of molecule will be less than its atom


37. NCERT-XII, VOL-I, PAGE NO : 250,251,252,254

In a group pairing energy of 3d elements greater than that of 4d and 5d series elements

because 3d-oribital is more compact ( small size) than 4d-and 5d


39. NCERT-XI, VOL-II, PAGE NO : 310

40. From Si to P to S to Cl, P d bond strength with oxygen increases due to decrease in

size(or energy) difference of 3d of central atom and 2p of oxygen. As a result

polymerization tendency decreases.


42. Weak base and strong nuclophiles like I and RS give best yield for substitution than

elimination

43. -I,+M groups 3 2 2 3 2 6 5, , , , , , ,OH OR OCOCH NH NR NHCOCH CH CH C H activates

– O and –p positions due to +M effect but deactivates m-position due to –I effect, But

+I –groups like alkyl groups activates all –O, m and –p positions.

44.


Sec: Sr.IIT_IZ

45. Sucrose is non-reducing sugar, polysaccharides are non-reducing non-sugar

46.

2 4

2

,Conc H SO

E

eH H

Less stable alkenes rearrange to more stable alkene by acid calatyst but not by

base .

47. Molecule which is non superinposable on its mirror image is chiral molecule.

Structures I, II and IV are nonsuperinposable on their mirror images. But compound

one is optically inactive because its enantiomers are inter convertable ( conformers)

compound two is optically inactive due to pyramidal inversion .

48. NCERT-XII, VOL-II, PAGE NO : 129,130

49. NCERT-XII, PRACTICAL MANUAL , PAGE NO : 90

Compounds Phthalein dye test

OH3CH

Red

OH

3CH

Bluish purple

OH

3CH

No colour

OH

OH

Blue but takes longer time to

appear

OH

OH

Green fluorescent colour of

fluorescein


Sec: Sr.IIT_IZ

Take 0.1g of organic compound and 0.1g of phthalic anhydride in a clean dry test tube

and add 1-2 drops of 2 4.Canc H SO heat the test tube for about 1 min in oil bath. Cool and

pour the reaction mixture carefully into a beaker containing 15 ml of dil. NaOH .

Appearance of pink, blue, green, red etc. Colour indicates the presence of phenol –OH

group in the compound

50. 0E

A B AG VdP , area of ABCA and CDEC are equal

51. In case of acetones at any concentration of it slow step is nucleophilic attack of

carbanion at carbonyl carbon because ketones are less reactive in nucleophilic addition

52. Number of milli moles of 3 50 0.1 5AgNO

Number of milli moles of 50 0.2 10NaCl

3 3

5 10 0 05 5 0 10 5 5 5 5

AgNO NaCl AgCl NaNO

25 /100 5 10Cl M

sAgCl Ag Cl

10 210 5 10Ag

92 10Ag M

53. NCERT-XI, VOL-I, PAGE NO : 48

54. NCERT-XI, VOL-I, PAGE NO : 150

55. I, II, III, IV

56. 22 4

0 0 01 , 1 , 2 2Na K C O

X X Y Y Z Z

2 4

0 02 42 , 2 / 2NaKC O X Y Z NaKC O X Y Z

57.

Number of atoms per unit cell= 1 1 2 14 4 1 1 1 26 12 3 3

X


Sec: Sr.IIT_IZ

58. For inilial solution 0.53Tb

10000.53 0.5360 1060w

w

60ureaw g

For final solution left 2.12Tb

60 10002.12 0.5360 x

250x g

Weight of water vaporized =1000-250=750g


60. NCERT-XII, VOL-I, PAGE NO : 100, 102, 104

MATHS 61. (4) Since , are the roots of the equation

2x px q 0

p and a q.

Now, 1/4 1/4 4 1/4 1/4 2 2( ) [( ) ]

1/2 1/2 1/4 2[ 2( ) ]

2

1/42 2( )

2

1/4p 2 q 2(q)

1/4p 6 q 4q p 2 q

1/4

1/4 1/4 1/4p 6 q 4q p 2 q

62. (3)

The total numbers of factors of xnyz…. is equal to the number of ways of selecting one

or more out of n identical quantities of one type and remaining m distinct quantities.

Hence, the required numbers of factors = (n + 1)2m

63. (1) F(x) is defined for

2[x] [x] 6 0 ([x] 3)([x] 2) 0

[x] 2 or [x] 3


Sec: Sr.IIT_IZ

But [x] 2 [x] 3, 4, 5,......

x 2

Domain of f ( , 2) [4, )

64. (1) Let the roots of the given equation be 1 + ip and 1 – ip, where p R

product of roots

2(1 ip)(1 ip) 1 p 1, p R

(1, )

65. (2) Given condition 2 2A B (A B)(A B)

2 2 2 2A B A AB BA B

AB BA .

66. (1) Since 1 1 1 2 2 2a b c , a b c and 3 3 3a b c are divisible by k, therefore

1 1 1 1

2 2 2 2

3 3 3 3

100a 10b c n k100a 10b c n k100a 10b c n k

where n1, n2, n3 are integers.

Now 1 1 1

2 2 2

3 3 3

a b ca b ca b c

1 1 1 1 1

2 2 2 2 2

3 3 3 3 3

a b 100a 10b ca b 100a 10b ca b 100a 10b c

(Applying 3 3 2 1C C 10C 100C )

1 1 1

2 2 2

3 3 3

a b n ka b n ka b n k

= k 1 1 1

2 2 2 1

3 3 3

a b na b n ka b n

is divisible by k

(Since elements of 1 are integers, 1 is an integer)

67. (3) The four digits 3, 3, 5, 5 can be arranged at four even places in 4! 62!2!

ways and the

remaining digits viz., 2, 2, 8, 8, 8 can be arranged at the five odd places in 5! 102!3!

ways. Thus, the number of possible arrangements is (6) (10) = 60.


Sec: Sr.IIT_IZ

68. (3) term independent of x will be 10 5 55C (sin ) (cos )

10

555

C sin (2 )2

So maximum value will be 10

55

C2

69. (4) Given: 4 4 4

1 1 1....1 2 3

4 4 4 4i 1

1 1 1 1 .....(2i 1) 1 3 5

4 4 4 4 4 4

1 1 1 1 1 1... ...1 2 3 4 2 4

4 4 4 4

1 1 1 1 ...2 1 2 3

1 1516 16

70. (2) A.M. > G.M.

1/3

x y z x y z/ 3 . .y z x y z x

x y z 3y z x

71. (2)

x y is an integer

x x 0 is an integer A is Reflexive

x y is an integer y x is an integer A is symmetric x y, y z are integers

As sum of two integers is an integer.

(x y) (y z) x z is an integer

A is transitive. Hence statement 1 is true.

Also x 1x is a rational number B is reflexive

xx is rational y

x need not be rational

i.e., 01

is rational 10

is not rational


Sec: Sr.IIT_IZ

Hence B is not symmetric

B is not an equivalence relation.

72. (1) Let Ei denote the event that Ai dies in a year, then iP(E ) p and 'iP(E ) 1 p for I = 1,

2, …..n P (none of A1, A2, …. A3 dies in a year)

' ' ' ' ' ' n1 2 n 1 2 nP(E E E ) P(E )P(E )....P(E ) (1 p)

Because E1, E2,……En are independent.

Let E denote the event that at least one of 1 2 nA , A ,.......A dies in a year, then ' ' ' n1 2 nP(E) 1 P(E E .... E ) 1 (1 p)

Let F denote the event that A1 is the first to die, then 1P(F / E) .Also,n

P(F) = P(E).P(F/E)

= n1 1 (1 p)n

73 (2) Since, 1P(A B)3

1 1P(A B) 1 P(A B)3 3

2 2P(A B) P(A) P(B) P(A B)3 3

1 2p 2p2 3

74 (2)

2 23

4x 0 3 2 3

x z (z x)lim8xz 4x 8xz

3 2

4x 0 3 33 3

x 2zx xlimx 8z 4x 8z x

4/3 3 3

4 4x 0 4/3 3 3 3

x 2z x 2zlimx 8z 4x 8z 2 8z

23/3

12 .z

75. (3) In the neighbourhood of x = 0, f(x) = log 2 – sin x

g(x) = f(f(x)) = log2 – sin (f(x))

log 2 – sin(log 2 – sin x)

Since g(x) is differentiable at x = 0,


Sec: Sr.IIT_IZ

g ' (x) = – cos(log2 – sinx) (–cosx)

g '(0) cos(log 2)

76. (4) 3 2f (x) x x f '(1) xf "(2) f "'(3)

2f '(x) 3x 2xf '(1) f "(2)

f "(x) 6x 2f '(1)

f "'(x) 6

Now, f '(1) 3 2f '(1) f "(2)

f "(2) f '(1) 3 0 (1)

Again f "(2) 12 2f '(1)

f "(2) 2f '(1) 12 0 (2)

Again f "'(3) 6 (3)

From (1) and (2)

2f "(2) f "(2) 6 0 f "(2) 2 (4)

(1) gives f '(1) 2 3 0

f '(1) 5

3 2f (x) x 5x 2x 6 (5)

f (0) 6

f (1) 1 5 2 6 4

f (2) 8 20 4 6 2

f (3) 27 45 6 6 6

f (0) f (2) 6 2 f (1)

(a) is false f (0) f (3) 6 6 0

f (1) f (3) (1 5 2 6) 6

2 f (2)

f (1) f (3) 2 f (0) [ f (0) 6]

(4) is false

77. (4) Let, 3f (x) x 3x a

2f '(x) 3x 3 3(x 1)(x 1)

Now, f (1) a 2,f ( 1) a 2


Sec: Sr.IIT_IZ

The roots would be real and distinct if, f (1)f ( 1) 0

(a 2) (a 2) 0 or

2 a 2

Thus the given equation would have real and distinct roots if a ( 2, 2)

78. (2) Let n 1 n n 1 2

0 1 2 n 1 nx x x xf (x) a a a ... a a xn 1 n n 1 2

Then f(x) is continuous and differentiable in [0, 1], as it is a polynomial function of x.

Also, f(0) = 0

and 0 1 2 n 1n

a a a af (1) ... a 0n 1 n n 1 2

. (Given)

Hence, by Rolle’s theorem, there exists atleast one real number c (0,1) such that

f '(c) 0 i.e., c is a root of the equation n n 10 1 n 1 na x a x ... a x a 0

.

79. (1) 2

2 1

11 dxxI

1 1(x ) 1 tan xx x

Put 1x tx

80. (2) Let 1100

r 1 0

I f (r 1 x)dx

1 1 1 1

0 0 0 0

I f (x)dx f (x 1)dx f (2 x)dx ... f (99 x)dx

1 2 3 100

0 1 2 99

I f (x)dx f (x)dx f (x)dx ... f (x)dx

100

0

I f (x)dx a(given)

81. (1) We have, sin x x for x 0

sin(cos x) cos x for 0 < x < / 2

/2 /2

0 0

sin(cos x) dx cos x dx

3 2I I (1)


Sec: Sr.IIT_IZ

Now, x cos x cos and x, 0,2

x sin x

cos x cos(sin x)

/2 /2

0 0

cos x dx cos(sin x)dx

3 1I I (2)

from (1) and (2) 1 3 2I I I

82. (4) 2 1 1 2 2 2a and ah . .2 2 33 3 3

abc (2)(1)(1)(3) 3R4 3(4) 2 2 2

83. (3) 2 42

11 cos cos .... 4 41 cos

2cos ec 4

2 21sin sin4 6

n6

85. (3) Given equation of lines 2 23y 4xy 3x 0

2 23y 3xy xy 3x 0

( 3y x)(y 3x) 0

xy , y 3x3

oAPO 75

In AMP, o AMsin 753

oAM 3sin 75

Now, length of chord of contact AB = 2AM

o o2(3sin 75 ) 6sin 75


Sec: Sr.IIT_IZ

3 1 ( 3 1)6 32 2 2

86. (2) 2 2y 4b(x (a 2b)) or y 4bX where x (2a b) X

2 2x 4a(y (a 2b)) or x 4aY where y (a 2b) Y

For 2y 4bX, extremities of latus rectum (b, 2b) and (b, –2b) w.r.t. X-Y axis.

i.e., (2a, 2b) and (2a, – 2b)w.r.t. x-y axis.

For 2x 4aY, extremities of latus rectum (2a, – a)

and (–2a, –a) w.r.t. X-Y axis.

i.e., (2a, 2b) and (–2a, 2b)

hence the common end of latus rectum (2a, 2b)

Now for 1st parabola

dy2y 4bdx

1

dy 2b 1dx y

at (2a, 2b)

Also for 2nd parabola

dy dy x2x 4a ordx dx 2a

Hence parabolas intersect orthogonally at (2a, 2b)

87. (2) 2 2 29(x 3) 9(y 4) y

2 29(x 3) 8y 72y 144 0

2 29(x 3) 8(y 9y) 144 0

2

2 9 819(x 3) 8 y 144 02 4

2

2 99(x 3) 8 y 162 144 182


Sec: Sr.IIT_IZ

2

98 y9(x 3) 2 1

18 18

2

9y(x 3) 2 192

4

2 2.4 1e 19 9

1e3

88. (4) 2 2y x 11 1

16 9

Locus will be the auxiliary circle

2 2 1x y16

89. (3) Range of 1 2f (x) tan (3x bx c) is 0,2

if and only if range of 2g(x) 3x bx c is

[0, ) . This is possible only when discriminate of the equation 23x bx c 0 is equal to

zero.

i.e. b2 = 12c

90. (1) Vector ˆ ˆ ˆ ˆ ˆ ˆ((3i 2j 4k) (4i 3j 4k)) is perpendicular to ˆ ˆ ˆ2i j mk .

3 2 14 3 4 02 1 m

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