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Vidyamandir Classes VMC/JEE Mains-2018 1 Solutions SOLUTIONS Joint Entrance Exam | IITJEE-2018 Paper Code - B 8th April 2017 | 9.30 AM – 12.30 PM

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### Transcript of SOLUTIONS Main 2018... · Vidyamandir Classes VMC/JEE Mains-2018 2 Solutions Joint Entrance Exam |...

Vidyamandir Classes

VMC/JEE Mains-2018 1 Solutions

SOLUTIONS Joint Entrance Exam | IITJEE-2018

Paper Code - B 8th April 2017 | 9.30 AM – 12.30 PM

Vidyamandir Classes

VMC/JEE Mains-2018 2 Solutions

Joint Entrance Exam | JEE Mains 2018

PART-A PHYSICS

1.(3) For collision with deuterium:

1 22mv o mv mv (Conservation of momentum ) ......... (1)

2 1v v v ( 1e ) ......... (2)

By (1) and (2) 1 3vv

2 2

1

2

1 182 2 0.89

1 92

d

mv mvP

mv

For collision with carbon Nucleus

1 20 12mv mv mv (Conservation of momentum ) ......... (1)

2 1v v v ( 1e ) ......... (2)

By (1) and (2)

11113

v v

22

2

1 1 11482 2 13 0.281 169

2

mv m vPc

mv

2.(3) Change in momentum of a single molecule.

0 0 22

uP m

Total change in momentum per second

0 0. 2P n P n m u

Pressure 0 2nm uFA A

Substituting values: 3 22.35 10 / .P N m

3.(1) PB VV

V PV B

Also, 3V rV r

(As 24V r r )

3

r Pr B

3mgKa

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4.(4)

Equivalent circuit is

Where,

1 2 21 2 3eqr

1 2

1 2

1 2

12.331 1eq

V Vr r

v

r r

1010AB eq

eqV V

r

= 11.55 volts

5.(1) 22KUr

3dU KF r rdr r

2

3K mv

rr (As Force towards center

2mvr

)

K. E. = 22

12 2

KmVr

Total energy = KE + PE 2 22 2

K Kr r

= Zero

6.(4) For 1m to be at rest

5T g

For 2&m m to be at rest

5f T g

( )f N

20.15( )f m m g

23.33m kg

Amongst the options minimum mass that can be kept for no motion is 27.3 kg 7.(2) For Series limit of Lyman : 1 1n and 2n

2 1 11P RcZ

For Series limit of Pfund: 1 5n and 2n

2 1 125 25

LP RcZ

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8.(1) When an unpolarized light of intensity I passes through a polarizer for the 1st time, intensity of output is 2I

(irrespective of orientation of polarizer) So,

i.e., polarizers A and B have axes parallel to each other. Now let the axis of C make an angle with A, and with B.

4cos

2 8I I

Solving, 45

9 (3). 22 2

1 1 11n

RZn

1

2 21 11n

RZ n

Since n is very large, using binomial

2 21 11n

RZ n

2 2 21 1 1

nRZ RZ n

2nn

BA

As 2 2

2 22 12

4n

r n h nn nmZe

10 (1).

Voltage across Si diode in forward bias is 0.7 volts. Hence voltage across 200 resister is 3 – 0.7 = 2.3V

2.3200

I = 11.5 mA

11.(4) 2mv mKrqB qB

2 e

em K

reB

2 4

2pm K

reB

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2 p

pm K

reB

Comparing (1), (2) and (3) e pr r r

12.(3) iq CV

fq KCV

( 1)induced f iq q q K CV 125 1 90 10 20 1.23

nC

13.(3) Quality factor 02

Q

0LR

14.(1) Overall bandwidth use for transmission 10% of C

Number of telephonic channel Total bandwidth

Channel bandwidth

95

3

10 10 10100 2 10

5 10

15.(3) 12

c Yfl

10

31 9.27 10

2 0.6 2.7 10

71 9.27 10

1.2 2.7

34.88 10 Hz 5kHz

16.(2) 2 2

20 6 (2 )

2 2

MR MRI M R 2 21 553 242 2

MR MR

O is the centre of mass of the system. Applying parallel axis theorem between O & P.

2 2 20

557 (3 ) 632PI I M R MR MR 2181

2MR

17.(4) CA BB

kQkQ kQVb b c

2 2 2( )(4 ) ( )(4 ) ( )(4 )a b ck

b b c

2 2

0

1 44

a b cb b

2 2

0

a b cb

18.(4) Let potential difference per unit length of potentiometer wire be x. In case-I (52)( )x …(i)

In case-II

5

ir

(40)( )ir x

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405

r xr

5 405

xr

…(ii)

From (i) & (ii)

552 405

rx x

52 13 151 5 1 1.540 5 10 10

r r

19.(1) From wave equations :

In air: 22 , kc

In medium: , 2 kc k k

,

k kc c 2

2

cc

c c

2 2 1 10 0 0 0

1 1 12

r r r r

Medium and air are non-magnetic

1 21 ; 1r r

1

2 1 2

1 1 14 4

r

r r r

20.(3) Angular width of central maxima 2a

; (where a is slit width and is wavelength)

23a

… (i)

In YDSE, fringe width

D

d [where d is slit separation and D is distance of screen from slits)

6

6

21 3.14 102 6 10

d

6100 104

25 m

21.(4) 12 1102

fT

122 10

2 km

22 3 1223

108 10 2 10 7 16 023 10

k m ..

22.(3) 2

2 29 232 2 3

RMM R RI M

2 2 2

29 42 18 9

MR MR MRI MR

23.(4)

It is given that final total kinetic energy has increased, so some internal energy of the system must have been

converted into kinetic energy.

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2 2 21 2 0

1 1 1mv mv 1.5 mv2 2 2

2 2 21 2 0v v 1.5v ...... (i)

Since, there is no external force, momentum can be conserved 1 2 0mv mv mv

1 2 0v v v ...... (ii)

From (i) & (ii)

2 21 0 1 0v v v 1.5v

2 21 0 1 02v 2v v 0.5v 0

Relative velocity 2 1v v Difference of roots 0D 2 va

24 (1). 201 1 1

1;

2

IB m I r

r

20

2 2 22

;2

IB m I r

r

22221 1

rm

m r

22

12 r

r

2

12r

r

1 2

2 12B r

B r

25 (1). 3M MV L

3M LM L

Maximum % error in density 1.5% 3 1% 4.5%

26.(1) Let the resistances in left and right slot be r and 1000 r respectively Initial: (100 ) (1000 )( )r x r x ..................(1)

After interchanging: (1000 )[100 ( 10)] ( 10)r x r x

(1000 )(110 ) ( 10)r x r x ..................(2)

From (1): 100 1000r rx x rx 10r x

From (2): (1000 ) 110 1010 10r rr r

2(1000 )(1100 ) 100r r r r

2 21000 1100 2100 100r r r r 1000 1100 5502000

r

27.(4) cosrms rmsP V I ; 4

100 20 12 2 2

P 1000

2

Wattless current, 20 1sin 102 2rmsI I

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28.(4) The (1), (2) and (3) graphs can represent the motion of a ball that is thrown in vertically upward direction. Initially speed decreases, becomes zero and then on the return trip, speed increases. Slope of graph in option (4) does not explain it.

29.(1) For mono atomic gas 53

Using 1 constantTV

2 23 3(300) 2V T V

2

3

300 1892

T K

3 32 8.314 189 300 27682 2

U n R T 2.7 kJ

30.(1) 1nF

R

2

nk mVF

RR 2

1nkV

mR

(1 )2n

V R

Now 1

2(1 )

2

2n

nR RT R

VR

Vidyamandir Classes

VMC/JEE Mains-2018 9 Solutions

PART-B MATHEMATICS

31.(2) 7 62

yx 2 7 12x y 2 5x y 2 5 0x y

Also, centre of the circle is 8 6 , and the radius is 64 36 c

16 6 5 1005

c

5 100 95c c

32.(4) 2 2 3 2 0x y z

1 3 2 2 1 3 0x y z

221 3 2

2 1 3 2 3 3

2

So the equation of plane is 7 7 8 3 0x y z

Now, distance from origin equal to 2 2 2

3 13 27 7 8

33.(1) 2 1 0x x 21 32

x , (where and 2 are non-real cube roots of unity)

and 2

107101 2 101 214 2 1

34.(3) Equation of PQ, 4 0 3 36x y

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VMC/JEE Mains-2018 10 Solutions

12y

Area of 1 15 6 5 45 52

TPQ

35.(2) 2 6yy'

1

6 32

y'y y

1 118 2 0x by y'

1 1 1 12 2

1 1 1 1

18 9 27 2712

x x x xy' bby by by y

21 1

962

y x b

36.(4) 1 33 2 02 4 3

kk

72

k

3 0x ky z …..(i)

3 2 0x ky z ….(ii)

2 4 3 0x y z …..(iii)

On solving (i) and (ii) 2 5 0x z …..(iv) On solving (iii) and (iv) 4 2y z

2 2

52 10

4

z zxzy z

37.(1) 2 3 6 6 0x x x

Case-I: 3x

2 3 6 6 0x x x 4 0x x 0 16x ,

As 9 16x x

Case-II: 3 2 6 6 6 0x x x x 8 12 0x x

6 2 0x x 36 4x ,

As, 3 4x x

There are exactly two elements in the given set.

38.(4) 2 2 18cos · cos sin 16 2

x x

23 18cos 1 cos 14 2

x x

28cos 4cos 1 2 14

x x

3cos3 4cos 3cosx x x

Vidyamandir Classes

VMC/JEE Mains-2018 11 Solutions

2 cos3 1x

1cos32

x

3 0, 3x

3 , 2 , 23 3 3

x Sum = 13

9 .

39.(4)

Total probability 4 1 6 1 2. .

10 2 10 3 5

40.(2) Let 1g x x tx

211 0g' xx

20 0t R ; t ,

2

2 22

1 2 2 2f x x x t ,xx

f xh x

g x

2 2 2f x t tg x t t

Let 2h t tt

221h' tt

Local minimum value occurs at 2t

Local minimum value 22 2 2 22

h

41.(4) Since Set A is, | a 5 | 1 4 < a < 6

Vidyamandir Classes

VMC/JEE Mains-2018 12 Solutions

and | b 5 | 1 4 < b < 6

Now B is 2 2(a 6) (b 5) 1

9 4

It can be seen that all vertices of rectangle lie inside the ellipse, therefore A B

42.(3) ( ) ( )p q p q

p q ( )p q p q ~ p

T F F F F T F F F F F T F T T F F T F T

43.(4) The equation of tangent at P

1y 16 (x 16)2

A ( 16, 0)

The normal is y 16 2(x 16)

B (24, 0) Since APB

2

AB is the diameter. Center of the circle C (4, 0)

Slope of 1PB 2 m

Slope of 24CP m3

2 1

2 1

m mtan 2

1 m m

44.(1) 2x 4 2x 2x2x x 4 2x (A Bx)(x A)2x 2x x 4

Put x = 0

34 0 0

0 4 0 A0 0 4

A 4

Put x = 1

23 2 2

2 3 2 (A B)(1 A)2 2 3

3(9 4) 2( 6 4) 2(4 6)

Vidyamandir Classes

VMC/JEE Mains-2018 13 Solutions

15 20 20 ( 4 B)25

1 ( 4 B)

B = 5

45.(2) Let 3 1x y

5 5x y x y

5 5 5 4 5 5 5 5 4 50 1 5 0 1 55 5 C x C x y ....... C y C x C x y ........ C y

5 5 3 2 5 4 5 3 2 5 40 2 4 0 2 42 5 2 5 C x C x y C xy C x C x y C xy

25 3 3 3 5 6 3 6 32 10 1 5 1 2 10 10 5 1 2 x x x x x x x x x x x

5 6 3 7 42 10 10 5 5 10 2 1 10 5 5 2 x x x x x x

46.(1) 1 5 9 49 416a a a ........a 24 32a d ……(i)

9 43 66 25 33a a a d ….(ii)

From (i) and (ii) 1d and 8a

Now, 2 2 21 2 17 140a a ......a m

217

18 1 140

rr m

17

2

17 140 4760 140 34

rr m m m

47.(1) Let, ( , )R h k

(0, )P k

( ,0)Q h

Equation of line would be,

1x yh k . . . (i)

2 3 1h k

2 3k h hk

Locus of (h, k) is 2 3y x xy

48.(2) Given/ 2 2

/ 2

sin1 2x

x dx

2 22sin 2 (sin )( ) ( ) sin

1 2 1 2

x

x xx xf x f x x

/ 22

0

sin x dx

/ 2

2

0

sin4

x dx

49.(3) 2( ) cosg x x

( )f x x

( ( ) ) cosg f x x

Given, 2 218 9 0x x (6 ) (3 ) 0x x

,6 3

x

Area = / 3

/ 6

3 1cos2

x dx

Vidyamandir Classes

VMC/JEE Mains-2018 14 Solutions

50.(1) 0

1 2 15lim ...........x

xx x x

0

1 1 2 2 15 15lim .......x

xx x x x x x

= 0 0

1 2 15lim (1 2 3 ........ 15) lim ........x x

xx x x

Now 0 1x x R 120

51.(1) Variance = 245 (1)9 5 1 4

Variance 2

52.(4) 2 2

5 3 2 3 2 5 2sin x cos x dx

(sin x cos x sin x sin x cos x cos x)

2 6

25 2 3

tan x sec xdx

tan x tan x tan x 1

Put tan x t 2 dtsec x

dx

2 2 2

3 2 2 2t (1 t ) dt

(t 1) (t 1)

3t 1 y

2 dy3tdt

21 dy 1 C3 3(y)y

31 C

3(tan x 1)

53.(3) Doubtful points for differentiability are 0 and

At x = 0

| || | ( 1) sin | | 0(0 ) limh

h o

h e hfh

( ) ( 1) sinlimh

h o

h e hh

0

sinhlim 1h h

and lim 1 0h

h oe

(0 ) 0 1 0f

| || | ( 1) sin | | 0(0 ) lim

h

h o

h e hfh

0

( ) ( 1) sinlimh

h

h e hh

0

sinhlim 1h h

and lim 1 0h

h oe

(0 ) 0 1 0f

Vidyamandir Classes

VMC/JEE Mains-2018 15 Solutions

(0 ) (0 ) 0f f

Similarly ( ) ( ) 0f f

Hence ( )f x is differentiable x R

54.(1) 4 4dy y cot x x cosec x d y sin x xdxdx

Integrating both sides we get: 22y sin x x c

Also, 02

y

2

2c

2

222

y sin x x

286 9

y

55.(3) ( ) 0;u a b 0 u a and 24.u b

Let ˆ ˆ ˆ ˆ( ) ( )b b a a b u u

2 2 2ˆ ˆ| | ( ) ( )b b a b u

2

2 22ˆ( )ˆ| | ( )

ˆ| |b ub b a

u

2

22 (24)27 ˆ| |u

2| | 336u

56.(2) 5 1 41 1 1

x y z

5 1 4 P , ,

P is foot of perpendicular from A to plane 3 8 7

13

14 4 113 3 3

P , ,

4 1 31 1 1

x y z

4 1 3Q , ,

Q is foot of perpendicular from B to plane 3 6 7

13

13 2 103 3 3

Q , ,

1 4 1 6 23 3 3

PQ

57.(3) 13

hx

3x h

2 2200 3h h

Vidyamandir Classes

VMC/JEE Mains-2018 16 Solutions

2 24 (200)h

24 40000h 100h

58.(3) 6 34 1 1 4!C C

6 5 3 24 45 24 10802

59.(4) 2 2 2 2 2 21 2 2 3 2 4 2 20A . . ........ A .

2 2 2 2 2 2 2 21 2 3 4 20 2 4 20......... .......

20 21 41 10 11 2146 6

2870 1540 4410 2870 1540 4410

40 41 81 4 20 21 416 6

B 540 41 41 280 41 820 33620

33620 8820 100 100 24800 248 60.(1)

2 3 3

3a

2 12 6a a

2 5 33

b 2 4 2b b

2 2(6 3) 3AC

Diameter 81 9AC 90

Radius 3 10 3 102 2 2

53

2

PART-C CHEMISTRY

61.(1) 3I is - 3sp d hybridised

- linear shape

62.(4) CH3COOK is a salt of a weak acid and a strong base

Most basic

63.(1)

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64.(1) Amidines, are stronger organic bases.

65.(1) Methyl orange is used for titration of strong acid and weak base. 66.(1) 67.(2) x y zC H O has z oxygen atom

x y 2 2 2y y

C H x O x CO H O4 2

O atoms required for combustiony

2 x4

1 y

z 2 x2 4

y

z x4

68.(1) During reduction 2 2 2H O H O

During oxidation 2 2 2H O O

69.(2)

Option (2) is correct [NCERT Class XII Part-II, Page No.-340]

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70.(1) 2 6 2 2 3 2B H 3O B O 3H O

2 627.66

nB H 127.66

2On required = 3

2 2 22H O 2 H O

n-factor for O2 = 4 Number of equivalent = 3 4 12F 12 96500C

i t 12 96500

12 96500

t s100

12 96500

h 3.2 hr100 3600

71.(3) G H T S RT nk H T S

H SnkRT R

Slope is HR

Since H is ve

Slope is positive.

72.(3) nr k A

n1 k 363 0.95 ….(i)

n0.5 k 363 0.67 ….(ii)

From (i) and (ii) n 2 73.(3)

Option (3) is correct [NCERT Class XII Part-II, Page No 405] 74.(4) Case - I

Case - II

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VMC/JEE Mains-2018 19 Solutions

Two isomers (fac and mer) are produced if reactant complex ion is a cis isomer. Only one isomer (fac) is formed if reactant complex ion is a trans isomer.

75 (4).

76 (3).

OH OH

+ CO2NaOH H O3 +

COOH

(X)

OH OCOCH3

+ (CO CO) O3 2

H SO2 4

COOH

Aspirin

COOH

Cat

(X)

+ CH COOH3

[NCERT class XII part II/Page No. 330]

77.(1) 2 2 104 4 spBaSO (s) Ba (aq) SO (aq) K 10

22 4 4Na SO 2Na SO

Conc. of 24SO in final solution 50 1 0.1M

500

For final solution

2 2 104Ba SO 10 2 9Ba 10 M

i i f fM V M V

9 9C 450 10 500 C 1.1 10 M

78.(4) Kjeldahl method is not applicable to compounds containing nitrogen in nitro (NO2) and azo (N = N –) groups and nitrogen present in the ring (pyridine) as nitrogen of these compounds does not change to ammonium sulphate.

[NCERT Class XI part II/Page No. 358]

79.(1) NaOH23(M) white gel ppt

(X)

Al NaOH Al OH NaAlO

2 33Al OH Al O

2 3Al O is used in chromatography as an absorbent. (Refer NCERT Class XIth/Part-II, Page-352)

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80.(4) HCl H Cl 0.2M 0.2 M

2H S H HS 71K 10

2HS H S 132K 1.2 10

22H S 2H S 1 2K K K

201.2 10

2 2

2

H SK

H S

H 0.2M , 2H S 0.1

2 2

200.2 S

1.2 100.1

2 20S 3 10 M

81.(1) (Refer NCERT Class XIth Part-II, Page-407)

The F ions make the enamel on teeth much harder by converting hydroxyapatite, 4 2 23Ca PO Ca OH ,

into much harder fluorapatite i.e. 4 223Ca PO CaF

82.(2) 4 2 2 2NH NO N 2H O

4 4 3 2 42NH SO NH H SO

3 22Ba N Ba 3N

4 2 7 2 2 2 32NH Cr O N 4H O Cr O

83.(3) NCERT Class XII/Part-II, Page No. 443

84.(1) 2 36Cr H O Cl

x 0 3 0 x 3

6 6 2Cr C H

x 0 0 x 0

2 2 32 2K Cr CN O O NH

2 x 2 4 2 0 0 x 6

85.(1) Pressure of cation in interstitial sites is ‘Frenkel’ defect.

86.(2) 6 6 2 2 215C H ( ) O (g) 6CO (g) 3H O( )2

(g)3n2

(g)H U n RT

1.5 8.314 2983263.91000

3267.6 kJ / mol

87.(2) 3BCl and 3AlCl are e deficient and thus act as Lewis acid

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88.(1) KCl exist as K and Cl

89.(2) Depression in freezing pt

f fT i K m

Less the value of i,

Higher the value of freezing pt.

For (2) i = 1 (min)

90.(2) 22H does not exist as Bond order is zero

Electronic configuration of 2 2 22 1s 1sH : *

2 2B.O 02