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SOLUTIONS Joint Entrance Exam | IITJEE-2019

8th April 2019 | Evening Session

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Joint Entrance Exam | JEE Mains 2019

PART-A PHYSICS

1.(4)

Net magnetic field,

since Hence force = 0

2.(4)

Pot. Diff

3.(4) For uniformly accelerated motion

4.(4) isobaric isothermal adiabatic isochoric

5.(4)

For maximum power

( )0

32 ˆ

42

pxMB id

µ=

p

( )

( )0

32 ˆ

42

pyM

B jd

µ=

p

( )( )0

32 ˆ ˆ

42

MB i jd

µ= +

p

!

||v B!!

Þ

1 11

Q C VC F

µ= = =

µ

a®b®c®d ®

EiR r

=+

( )

22

2E RP i RR r

= =+

0dP R rdR

= Þ =

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6.(2) Conserving momentum

7.(4)

The block will experience angular acceleration for a very small duration of time

Time taken to reach ground

angle rotated,

rad

8.(1) Limit of resolution

rad.

9.(3)

Maximum ht =

Similarly

10.(2)

1 1 2 2 1 3 2 4m v m v m v m v+ = +! ! ! !

1 2 1 11 1 1 42 2 2

m v v mm v m v+ = +

Þ 1 4 21 4 22 2

v v v v v v-= Þ = -

2 32 3 2L mL gmg

L= aÞ a =

32gT TL

w = aD Þw =

2 1sec.hg

= =

32g TL

q =

Þ3 10 10.01 0.52 0.3 2´

q = ´ = =´

1.22Dl

=

99

21.22 500 10 305 10200 10

--

-´ ´

= = ´´

22 2 2 2

21 1 1 1 22 2 2 2 5sph

vKE mv I mv mRR

æ öæ ö= + w = + ç ÷ç ÷è øè ø

2 21 2 7 112 5 5 2mv mvæ ö æ ö= + =ç ÷ ç ÷

è ø è ø

sphh

227 1 7

5 2 5 2sph sphvmgh mv hg

æ öæ ö= Þ = ç ÷ç ÷è ø è ø

232 2cylvhg

æ ö= ç ÷

è ø

Þ7 1453 152

sph

cyl

hh

= =

( )( )6 7 152

ˆ ˆ1.6 10 cos 2 10 6 10 2 WbB z t i jm

-= ´ ´ + ´ +!

0 0E cB=8 6 23 10 1.6 10 4.8 10-= ´ ´ ´ = ´

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From the equation of we can conclude that the direction of wave is

Direction of direction of

Hence

11.(3)

Linear momentum,

Comparing we get Solving we get

12.(3)

13.(1)

14.(1) Radius

B!

k̂-

E =!

B v´! !

( ) ( )ˆˆ ˆ ˆ ˆ2 2i j k i j+ ´ - = - +

( )( )2 7 15 ˆ ˆ4.8 10 cos 2 10 6 10 2 /E z t i j V m= ´ ´ + ´ - +!"

[ ] 2S MT -é ù= ë û

[ ] 2I MLé ù= ë û

[ ] 2 1h ML T -é ù= ë û

[ ] 1P MLT -é ù= ë û

[ ] [ ] [ ] [ ]a b cP S I h=

2 2 2a b c b c b cM L T+ + + - -é ù= ë û

1a b c+ + =2 2 1b c+ =2 1b c- - = -

1 1, , 02 2a b c= = =

64 4E m E MV V R R= Þ =

4.3 4

M M EE E M

E E

g R gg G R gg R

p= r Þ = Þ =

2eE v E gRµ Þ µ

1 14 4

M M M M

E E E

E g R EE g R E

= Þ = ´

Þ16MEE =

2 2 21 2 1 2 1 2

1 2 1 2

2 cos

25 9 25 2 15cos3cos10

3 9. cos 3 510 2

A A A A A A

A A A A

+ = + + q

= + + ´ q-

Þ q =

- -æ öÞ = q = ´ ´ =ç ÷è ø

! ! ! ! ! !

! ! ! !

( ) ( ) 2 21 2 1 2 1 2 1 22 3 . 3 2 6 3 5 .A A A A A A A A+ - = - +! ! ! ! ! ! ! !

954 150 5 118.52-æ ö= - + = -ç ÷

è ø

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Volume, mass

hence mass densities i.e. will be equal

15.(3)

16.(1) Since current leads emf, hence circuit will have capacitor and resistor.

17.(1)

Equating the two we get

18.(2)

19.(1)

20.(2)

21.(4) Amplitude

1/3R AµÞ V Aµ

AµMV

FllAY

D =

1 22

1 1 2 2

2 1.54 47

L LAY A Y R

= Þ =´´

Þ64 1.721

R mm= !

31 10 10100cX R C FC

= Þ = Þ = µ

3 3b b Arms

k T k N TVm M

= =

2eV gR=

410T K=

354 4 4

3 12

a a am m mbx

m

æ ö æ ö æ ö+ +ç ÷ ç ÷ ç ÷è ø è ø è ø= =

354 4 4

3 12

b b bm m mby

m

æ ö æ ö æ ö+ +ç ÷ ç ÷ ç ÷è ø è ø è ø= =

e

3 0.5R 6q

Vi A= = =

( )( )0.5 0.5 0.25AJ AJV iR VD = = =

2

242 l lT g

g Tp

= p Þ =

100 100 2 100g l Tg l TD D D

´ = ´ + ´

1 1100 2 100550 30

= ´ + ´ ´

6.8%»

/20

bt mA A e-=

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Solving we get ,

22.(1) Using lens formula

Since the image is formed at the same point, it must be at center of curvature of mirror. Max. distance for virtual image = 10cm

23.(4)

24.(4)

25.(4)

26.(4)

( )20 202

bmA A e

-=

0 201000

b tmA A e-

=

20 .t s=

1 1 1 1 1 1 1 130 20 60v u f v v

- = Þ - = Þ =-

60v cm=

R 20 10mirror mirrorcm f cm= Þ =

Þ

e160R3q =

e

15 9R 160 / 3 32q

Vi = = =

1Er

µ

1Fr

µ

1ar

µ

1vdv drr

µ

2

0ln rvr

æ öµ ç ÷

è ø

0ln rvr

æ öµ ç ÷

è ø

E Ax B= +dV E dx= -

( )2

1

5

1

20 10V

V

dV x dx-

= - +ò ò

2 1 180V V- = -

1 2 180V V V- =

2 2T Rd Rh Rh= +

3 6 650 10 2 6.4 10 70 2 6.4 10 Rh´ = ´ ´ ´ + ´ ´ ´

Þ 32Rh m=

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27.(1)

Putting the values we get

28.(2)

29.(3)

As is small

where,

30.(1)

As

CCC

C

VIR

D =

C

B

II

Db =

D

.C CC

BC

I VIR

DD = =

b b

40BI AD = µ

( ) ( )0 0ˆ ˆ2 2I IB k kd d

µ µ= + -

p p

!

0B =!

sinPEt = q. sinI q dEa = q

sinqdEI

a = q

qqdEI

a = q

,qdEI

w =2

2.4dI m=

. .A A B B C Cm u m u m u= -

,B Cm m= .2B

B B BA

h um u m= -l

1 1.2 2B B

A B

h hm u= =l l

2A

Bl

l =

C Al = l

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PART-B CHEMISTRY

1.(2)

2.(2)

Ione pair on “N” marked as “C” is most nucleophilic and form

with

3.(4)

4.(1)

Bond order

5.(1)

3CH I

4CH

cn 1 mole=

Hn 4mole=

C

C H

n 1%C 100 100 20%n n 1 4

= ´ = ´ =+ +

2 2 * 2 2 * 2 2 2 22C 1s , 1s , 2s , 2s , 2Px 2Py 2Pz- é ù¾¾®s s s s p = p së û

( )10 4 3 diamagnetic2-

= =

2 2 2 2 * 2 2 2 2 1 12 x y z x yN 1s , *1s , 2s , 2s , 2P 2P 2P *2P *2P- é ù é ù®s s s s p = p s p = pë û ë û

( )10 6B.O 2 Paramagnetic2-

= =

2 2 2 2 2 2 2 2 2 22 z x y x yO 1s , *1s , 2s , *2s , 2P 2P 2P *2P *2P- é ù é ù®s s s s s p = p p = pë û ë û

( )10 8B.O 1 Diamagnetic2-

= =

2 2 2 2 2 2 2 1 12 z x y x yO 1s , *1s , 2s , *2s , 2P 2P 2P *2P *2Pé ù é ù®s s s s s p = p p = pë û ë û

( )10 6B.O 2 Paramagnetic2-

= =

1B.OBL

µ

v5 28 100U nC T 14KJ1000

´ ´D = D = =

( ) 5 8 100PV nR T 4KJ1000´ ´

D = D = =

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6.(3) Interstitial compound are almost inert.

7.(3)

8.(4)

9.(1) is very strong-I group hence it will favour anti-Markownikaff’s product.

(anti-Markownikoff’s product)

10.(4) According to Fajan’s rule when size of cation decrease, then covalent character increase. So is covalent in nature.

11.(2)

12.(3) ...(1)

…(2)

...(3)

13.(1) Seliwanoff’s test: Seliwanoff’s reagent is (0.5%) resorcinol in 3N HCl. It gives red solution with fructose and sucrose but no change in colour with glucose.

14.(3) 119 Ununenium (Uue)

15.(3) Mass of fatty acid = 0.027g Density = 0.9 g/cc

Volume of fatty acid

Area of plate

Volume of fatty acid = area of plate × height of fatty acid layer

( ) ( ) ( )50 C4Ni s 4CO g Ni CO

Impure

° é ù+ ¾¾¾®ë û

( ) ( ) ( )200 250 C4Ni CO Ni s 4CO g

pure

°- °é ù¾¾¾¾¾® +ë û

[ ] [ ] [ ]1 2d B

K A K Bdt

= -

[ ]d B0

dt=

Þ [ ] [ ]1

2

KB AK

=

3CFHX

3 2 3 2 2 3 2 2CF CH CH CF CH CH CF CH CH X+- = ¾¾¾® - - ¾¾® - -

2BeX

[ ][ ] [ ]

23

eq 22 2

SOK

SO O=

[ ][ ]

5221

2

SOK 10

O= =

[ ][ ]

21293

2 32

SOK 10

O= =

[ ][ ]

22 10421 2

2

SOK 10

O= =

( )

129252

eq 2 1041

K 10K 1010K

= = =

®

30.027 0.03cm0.9

= =

( )22 2r 3 10 300cm= p = ´ =

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16.(1)

17.(3) It is example of intramolecular aldol condensation.

18.(3)

19.(1) Nylon-6 is derived from

20.(2) In Friedel craft alkylation, the alkylated product obtained is more activated than reactant, hence undergoes poly substitution.

21.(1)

22.(1)

30.03cm 300 h= ´ Þ 40.03h cm 10 cm300

-= = 610 m-=

32ICl sp d- ®

32BrF sp d- ®

3 36IF sp d- ®

3 3eff A eff B

A B3

4 4z r z r3 3P.F.

a

æ ö æ ö´ p + ´ pç ÷ ç ÷è ø è ø=

( )A B2 r r 3a+ =

Þ ( )A A2 r 2r 3a+ =

Þ A2 3r a=

Þ( )3 3

A A

3A

4 4 41 r 8r 93 3 3p.f .8 3 3r 8 3 3 2 3

´ p + p ´ p p= = =

´ ´

P.F 100 90%2 3p

= ´ =

Volume strengthM11.2

=

M 1=W 1000 1M.W. V

´ =

W MW.100V 10´ =

W 34% 3.4%V 10= =

gas H gasP K .X=

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For higher , slow will be more negative

23.(2)

No. of unpaired

.

No. of unpaired

24.(1)

25.(4) It is factual, taken from NCERT.

26.(1)

( )2gas H H OP K . 1 X= -

2gas H H H OP K K .X= -

y c mx= +

HK

( ) 4

6Fe CN-

é ùë û2 6Fe 3d+ ®

CN S.F.L.- ®222 00t2g eg

e 0- =B.M. 0=

( ) 22 6Fe H O

+é ùë û

2 6Fe 3d+ ®

2H O W.F.L®2,1,1 1,1

2gt eg

e 4- =

4 6 24 B.M. 4.9BMµ = ´ = =

maxE W KE= +2hc PW2m

= +l

( )W 0®2

2 12

1 2

PP

l=

l

( )

22

2P1.5P

l=

l

2 49

l=

l

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27.(4)

28.(4)

29.(1)

Due to intramolecular H-bonding enol content is maximum

30.(3)

For given cell reaction.

2 01Fe Fe G 2Fy+ ® D = -

3 02Fe Fe G 3Fz+ ® D = -

\ 2 3Fe Fe+ +® 0 03G 1FED = -

0 0 03 1 2G G GD = D -D

FE 2Fy 3Fz E 2y 3z- ° = - + Þ ° = -

( ) ( ) ( ) ( )2 3Fe aq Ag aq Fe aq Ag s+ + ++ ® +

0cellE x 2y 3z= + -

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PART-C MATHEMATICS

1.(3)

2.(3)

Slope of Slope of

and

3.(2) Total ways

4.(4)

2 201 2 20............2 2 2

S = + + +

2 3 211 2 20................

2 2 2 2S= + + +

2 3 20 211 1 1 1 20...............

2 2 2 2 2 2S= + + + + -

20

21

1121 2012 212

æ öæ ö-ç ÷ç ÷ç ÷è øè ø= --

20 1922 112 22 2

S = - = -

1 ,3

OP = 3PQ = -

( )1 3 3 3 3y x x- = - - = - +

Þ 3 4x y+ = 4 ,03

Q æ öç ÷è ø

23

OPQD =

4 8 72 216 94 216 310= + + + = + =

( ) ( )( ) ( )

1

0

1 3 3lim

1 2 2

x

x

f x ff x f®

æ ö+ + -ç ÷ç ÷+ - -è ø

Þ( ) ( )( ) ( )0

1 3 3 11lim 1 2 2x

f x ff x f xe ®

æ ö+ + --ç ÷ç ÷+ - -è ø=

( ) ( )' 3 ' 2 0 1f fe e+= = =

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5.(2)

Put in both lines

6.(1)

Passes through

7.(4)

Required point in quadrant first is (1, 2) Required equation is and now check option]

8.(4)

Hence infinitely many values

9.(No answer)

Where

No solution.

10.(3) …(i) …(ii) …(iii) Add (i) & (iii)

( ) ( )11: 2 1 2 5 02

L y x x y- = - - = + - =

( ) ( )2 : 3 2 4 2 5 0L y x x y- = - = - - =

,h k

( ) ( ) 1, 3,13

kh kh

= Þ =

22 2ln lndy y y C

dx xx= Þ = - +

( )1,1

20 ln ,ln 21

C C-= + =

2ln 2yx

= - +

( )ln 2 1x y x= -

2 4 5 0x x+ - =

( )( )5 1 0x x+ - =

1 0x y- + =

( ) ( )( )2 24 1 3 4 1 1 8D m m m= + - + +

( )( )2 2 34 1 9 6 1 8 8m m m m m= + + - + + +

( )2 34 8 2 8m m m= - -

( )28 2 1 0m m= - - <

10

32 16 4

3 1 log 200xC xx +

1æ ö =ç ÷è ø

( )101 3 1 log4 2 10

xx

- +=

( )1 3 1 14 2

t tæ ö- + =ç ÷è ø

10logt x=26 5 4 0t t+ + =0,D <

2 1x y kz- + =

2 2x y z+ + =

3 3x y kz- - =

4 3 4 0x y- - =

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11.(4)

or

12.(4)

13.(2) Let hyperbola be

also (from )

and Equation of tangent at is

( )2 3cosh = q

3sinr = q2v r h= p29sin .6cos= p q q

54 sin2 cosV = p q q

0dvd

=q

Þ 2 32sin cos sin 0q q- q =

Þ ( )2 32 1 0s s s- - =

Þ 3 32 2 0s s s- - =

Þ 32 3 0s s- =

Þ 0s = 22 3 0s- =

23

s = ±

\ 2 1cos 13 3

q = - =

163

h æ ö= ç ÷è ø

2 3h =

( )1 2

x xa af x-+

=

Þ ( ) ( )1 1f x y f x y+ + -

( ) ( )1 12 f x f x=

2 2

2 2 1x ya b

- =

Þ 2 216 36 1a b

- =

2 216 36 1a b

- = 2e =

Þ 2 4a = 2 12b =

( )4,6

2 2 0.x y- - =

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14.(3) are in

Put

15.(3)

Distance from origin

16.(2)

17.(4)

roots of are equal i.e.

Divide by

are in A.P.

, ,a b c .A P2C AÐ = Ð

Þ sin sin 2C A=sin 2cossinC AA=

Þ( )2 2 2

22

b c aca bc

+ -=

,a b d= - ,c b d= + 0d >

Þ5bd =

4 6,5 5

ba b C= =

: : 4 :5 : 6a b c =

8 2 41

kk+

=+

Þ 12

k =

Þ 2y = - 6z =

\ ( )4, 2,6R - 2 14.=

( )2/33 61

dx

x xÞ

+ò 2/3 6

76

1111

dx tx

xx

æ öÞ + =ç ÷è øæ ö+ç ÷

è ø

ò

7 2/36 1

6dtdx dt

x t- = Þ- ò

( )( )( )

1/31 61/3 3 1/366 6

11 1 1 11 116 2 23

xt C C xf x xx x

æ öé ù +ç ÷ æ öê ú- = - + + = - + = +ç ÷ ç ÷ê úè øç ÷ ë û

è ø

( ) 312

f xx

= -

2b ac=

2 2 0ax bx c+ + = ba

-

2

2 0b bd e fa a

æ ö æ ö- + - + =ç ÷ ç ÷è ø è ø2 22 0db bea fa- + =

2 0dc eb fa- + =

ac

2 22 2 20 0 0dc eb fa d eb fa d e f

ac ac a ac a b cb b- + = Þ - + = Þ - + =

, ,d e fa b c

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18.(1)

Hence not a tautology

19.(1)

is even is odd

Also

Now

20.(3)

21.(4)

Put

22.(3) Height of two tower are 20 m 80 m ……………….

p q p qÚ ~ q ~p qÚ ( ) ( )( )~p q p qÚ ® Ú

T T T F T TT F T T T TF T T F F FF F F T T T

( ) ( )0

x

f x g t dt= ò( )g x ( ).fn f x\ .fn

( ) ( )'f x g x=

( ) ( )5f x g x+ =

( ) ( ) ( ) ( )5 5f x g x g x f x- = - = = +

( )0

x

f t dtò5t z= + dt dz=

( ) ( )5 5

5 5

5x x

f z dz g z dz- -

- -

+ =ò ò

( ) ( ) ( )5

5

' 5 5x

f z dz f x f-

-

= = - - -ò( ) ( )5 5f f x= - -

( ) ( )5 5f f x= - +

( ) ( )5 5

5 5

'x x

f t dt g t dt+ +

= =ò ò

ˆˆ ˆ

3 21 1 1

i j ka b x´ =

-

!!

22 2 38a b x x r´ = - + =!!

Þ752

r ³

Þ352

r ³

( )( )( ) ( )( )2y f f f x f x= +

( )( )( ) ( )( ) ( ) ( ) ( )' . ' . ' 2 'dy f f f x f f x f x f x f xdx

= +

( )( )( ) ( )( ) ( ) ( ) ( )1 ' 1 . ' 1 . ' 1 2 1 . ' 1 27 6 33x f f f f f f f f= Þ + = + =

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23.(2) minimum value of is 4

24.(3)

25.(3)

26.(3)

20 80tan , .20 80

h h h hx hxy x yy y x x y x= b Þ = = Þ = = -

-

120 80 20 80hx hx h hx+ = Þ + =

5 80 16h h= Þ =

1 9 11 2 10102 2

nn n- > Þ Þ > \ n

3 cos sin2 2 6 6

iz ip p= + = +

Þ 5 32 2

iz -= +

8 1 32 2iz = - -

( )95 81 iz z izÞ + + +

91 3 cos3 sin32 2i i

æ ö= + = p + pç ÷è ø1.= -

41 45 54 57 43 48 486

x x+ + + + += Þ =

( )2 2 2 2 2 2 2 2148 41 45 54 57 43 486

s + = + + + + +

2 14024 7012 7012 6912 1002304 23046 3 3 3

-s = - = - = =

103

s =

2 25 3 75be b e= Þ =2 2 75b a- =

( )( ) 75b a b a- + =

15b a+ =10, 5b a= =

22 2 25 510

aLRb

´= = =

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27.(1)

28.(3)

are in A.P.

29.(4)

Plane is

30.(1)

Discontinuous at

2 4y x=

( )3/2

3/2

00

4 82 23 / 2 3xS I xdx

ll ù= = = lú

ûò

( )( )

3/2

3/22 2

4 5 54SSl l

= Þ =

2/3 1/32 44 45 25æ ö æ öl = =ç ÷ ç ÷è ø è ø

( ) ( )( ) ( )

( ) ( )( ) ( )2 2

2 2 2 2

1 1 1 1 0 0 2 22 2 2 2

4 44 4 4 4

b cb c b c

b cb c b c

- -= - - =

- -- -

( )( ) ( )1 1

2 22 2

b cb c

= - -+ +

( )( )( )2 2A b c c b= - - -

2, ,b c2,2 ,2 2d d+ +

( ) [ ] [ ] [ ] [ ]3 32 2 2,16 1,8 1,2 2 2,4A d d d d d d d= = Î Þ Î Þ Î Þ Î

[ ]2 2 4,6d+ Î

( ) ( )1 2 3 4 5 0x y z x y z+ + - + l + + - =

( ) ( ) ( )1 2 1 3 1 4 1 5 0x y z+ l + + l + + l - - l =

Þ 13l = -

2 0x z- + =

( )ˆˆ. 2 0r i k- + =!

( )

( )1 , 1 0, 0 1

2 , 1 22 , 2 33 , 3

x xx x

f x x xx xx x

- + - £ <ìï

£ <ïï £ <íï + £ <ï

+ =ïî\

0,1,3x =