JEE MAINS-ENTRANCE EXAM-2019 (9th April-Morning) Solutions · Vidyamandir Classes VMC | JEE...

Vidyamandir Classes

VMC | JEE Mains-2019 1 Solutions | Morning Session

SOLUTIONS Joint Entrance Exam | IITJEE-2019

9th APRIL 2019 | Morning Session

Vidyamandir Classes


Joint Entrance Exam | JEE Mains 2019

PART-A PHYSICS

1.(3) Given

2.(3)

3.(1)

4.(3) Clearly

Hence for 4th harmonic,

5.(4)

COM gives : ...(1)

For elastic collision ...(2)

From (1) and (2)

6.(3)

7.(3) We have

8.(3) Conservation of energy gives

5=m Þ 5 5-= Þ = -

v v uu

\1 1 1 32 cm5 40

+ = - Þ =-

uu u

2 2 6net 0 1B B B 10 904-= + = ´

\ 8 6net 3 10 10 904-= ´ = ´ ´ ´E C B

\ 4 8 610 3 10 10 904 0.9 N.- -= = ´ ´ ´ ´ =F QE

\ rns 0.6N2

= =FF

eq12 46 912 4´

= + = W+

R \ main72 8 amp.9

= =i

\ 104 8 2 amp.

12 4W = ´ =+

i

\ 6CV 10 10 (10 2) 200-= = ´ ´ ´ = µQ C

2 0.157p=

l\ 40ml =

4 80m2l

= ´ =l

002 2 mv

4= ´ +

vv

004

- =vv v

1.2 kg=m

system0

1 ( ) ( ) 2 ( )4 / 2 ( / 2)

é ù´ - ´ -= + +ê úpe + -ë û

Q q Q q qUD d D d d

2

20

1 24

é ù= - -ê úpe ë û

q qdd D

22 2

4æ ö

- =ç ÷è ø!

dD D

r.m.s.V Tµ

\200 273 127 V 100 5 m/sV 273 227

+= Þ =

+

2IAOR

1 I mgh2

w =

Vidyamandir Classes


For ring

For Cylinder,

For sphere

9.(3) Net repulsive force

10.(1) Theoretical

11.(3) For no drifting,

Hence direction with flow of rivers is

12.(1) Work-done will be equal to rise in potential energy of suspended section

13.(4) H-Cl is diatomic molecule hence degree of freedom is (translational) + 2 (rotational) + 1 (vibration) = 6

14.(3)

(where Clearly it is parabolic functions. Starting from ground momentum first decreases and then becomes zero at highest position. Therefore it

increases in –ve direction. Hence 3rd option is correct.

15.(4)

16.(2)

17.(4) At 0°C,

22

11 2 mgh2

æ ö´ ´ =ç ÷è øVmRR

Þ2

1 =vhg

2 2 2

2 21 R v 3 v3m mgh h2 2 R / 2 2 g

æ ö æ ö´ = Þ =ç ÷ ç ÷è ø è ø2 2

31 7 mgh2 5 4 / 4

æ ö æ ö× ´ =ç ÷ ç ÷è ø è øR Vm

R\

2

3710

=vhg

\ 1 2 33 7: : 1: : 10 :15 : 72 10

= =h h h

0 1 2 0 1 21 2 2 2 2

µ ´ µ= - = - ´

p pi i a i iF F F aa a

01 24

µ=

pi i

2 1sin4 2

q = = =m

uv

\ 30q = °

180 120a = - q = °

\ 22 2= ´ ´ =mg L L MgLWL n n n

3=f

\2

21 6 .2 2 6

= \ =BB

mvmv K T TK

2 2 2= -V u gh Þ 2 2 2 2 2mgh= -m v m u

Þ 2 = -p A Bh 2 2 , 2mg)= =A m u b

[ ]= +g gV i R R34 10 [50 5000] 20.2 volt-= ´ + =

20 volt.»

2 26

1 2

| | 3.75 10 J2 2

-= D = - = ´Q QW UC C

1000m/s3

w= =vK

\1000 / 3 273336 273

=+T

( )µ!V T \ 277 4 C» = °T K

Vidyamandir Classes


18.(4)

19.(3)

20.(1)

21.(3) If is contact angle, then for equilibrium of rises water in tube, we have Hence if is doubled, rises mass will also doubled.

22.(3)

23.(1)

24.(1) According to question shift = width of n fringe pattern

25.(2) ( Buoyancy )

26.(4) For a solenoid, self inductance is given as

27.(3) input resistance =

Voltage gain =

28.(3) Clearly

.

29.(1) Area under curve in process `A’ on P-V diagram is more. Hence

30.(4)

Also

PART-B CHEMISTRY

2 21 1 22 2w = q Þ ´ wI K I 2w q

´ = ´ q×d dKdt dt

\2w q

a = =d kdt I

&4 8

= = = = =AB BC CD DE ECR RR R R R R \

374 8 864

æ ö+ ´ç ÷è ø= =eq

R R R

R RR

2 22

2

2 2

1 12 3 488.88 488.9Å1 16602 4

-l= Þ l = »

-

qc cos 2 Mgq´ p =T rr

2 2 2.2

(3 ) (3 ) 3= + =pGM G M GMEa a a

sin ( ) sint = µ q = qB niA B 5 2.5 1100 3 1100 100 2

´= ´ ´ ´ ´

´0.265N 0.27N= »

Þ ( 1)µ - = ´ ×lDt na

\[ 1]

l=

µ -nDta

2 & ' 2( /16)

= p = p-

l lT Tg g g

!16

=mg

\1 1' 4

151116

= ´ = ´-

T T T

self 0æ ö= ´ ´µ ç ÷è øNL N AL

\ self1

µLL

1 610mVr15 10-

=´

3 20.67 10 6.0 10 0.67 K» ´ W = ´ W = W

Rc 3mA 1KB 300 Voltr 15 A 0.67K

W+ = + ´ = +

µ W

72 2

5 10-p p=

l ´\ 5000Ål =

\ = + fE eV Þ12375 2 0.475 V 0.48 volt5000

= - = »V

>A BW W \ &D > D D = DA B A BQ Q U U

4 33

10 10 kg/m(0.1)

+r = =

3r =Ml

\3 0.1 0.01 3

10 0.1Dr D D

= + = + ´r

M lM l

31 3 31 0.31kg/m100 10 100

= + = =

Vidyamandir Classes


1.(2) from given eq:

for option -1 6 gm reacts 28 gm i.e. both reactants utilized completely.

for option -2 6 gm reacts 28 gm

10 gm reacts

required, however amount. of given = 56 gm

i.e. present in excess.

So, utilizes completely & hence limiting reagent.

for option-3 6 gm reacts 28 gm

4 gm reacts

required. But given = 14 gm.

is L. R.

for option - 4 reacts 28 gm

6 gm reacts required but given = 35 gm.

2.(2)

major product should be

3.(3) q & w are path functions.

Option – 3 is correct

2 2 3N (g) 3H (g) 2NH (g)1mole 3mole1 28gm 3 2gm28gm 6gm

+ ¾¾®

´ ´= =

! 2H c 2N

! 2H c 2N

\ 2H 228 10gm N6´

246.67gm N= 2N

2N

2H

! 2H c 2N

\ 2H c 228 4 gmN6´

218.67gmN= 2N

2N\

6gm! 2H c 2N

\ 2H c 228 8 gm N6´

237.33gmN= 2N

\

Vidyamandir Classes


4.(4) Aerosol is a colloidal solution of solid in air. Eg. Smoke, dust etc. are considered as aerosol.


5.(1)


6.(2)

7.(4) from Raoult’s law: from Dalton’s law:

&

…… (i)

& …… (ii)

eq. (i) ÷ eq. (ii)

8.(1)

One molecule is present as water of crystallization.

9.(4) for EMR

From Planck’s formula,

air,3 2Mg MgO Mg ND¾¾¾® +

0M M MP x P= × M M TP Y P= ×

0N N NP x P= × N N TP Y P= ×

\ 0M M M Tx P Y P× = ×

0N N N Tx P Y P× = ×

0M M M T

0N TN N

x P Y PY Px P

× ×=

××

M M

N N

x Y450x 700 Y

´ =

M M

N N

x Y14x 9 Y

= ´

\ M M

N N

x Yx Y

>

4 2CuSO 5H O×

[ ]2 4 4 2Cu(H O) SO H O×

2H O

C = nl C\n =

l

Vidyamandir Classes


So,

10.(4)

11.(4)

2-chloro-1-methyl-4-nitrobenzene

12.(1) Increase in CO2 induces Global warming.

E h= nEh

\n =C E

h\ =l

( )( )

( )( )

max min max minLyman Lyman Lyman

Balmer max min max minBalmer Balmer

E E

E E

n - n -Dn= =

Dn n - n -

1 1 1 113.6 13.61 1 41 1 1 113.6 13.64 4 9

æ ö æ ö- - -ç ÷ ç ÷aè ø è ø=æ ö æ ö- - -ç ÷ ç ÷aè ø è ø

11 0 1 1 941 1 1 4 40 14 4 9

9

- - += = =

- - +

9 : 4\

Vidyamandir Classes


13.(4) Sucrose is formed due to bonding between.

Glucose from C1 position and fructose from C2 position.

14.(4)

15.(4)

16.(2)

17.(4)

I.P.1st <<< I.P. 2nd.

18.(3)

i.e.

19.(4) Cryolite =

-a -b

4LiAlH3 3 3 2

O||

CH CH CH C O CH CH CH CH CH OH- = - - - ¾¾¾¾® - = - -

DCl3 3

3 2

Cl|

CH C CH CH C CHD

DICl|

CH C CHD|I

- º ¾¾¾® - =

¯

- -

2 2 6 1K 1s 2s 2p 3s=

2 2 6K 1s 2s 2p

inert gas configuration

+ =

CN Cl Me OMeActivating nature

- - - -¾¾¾¾¾¾¾¾®

D A C B< < <

3 6Na [AlF ]

Vidyamandir Classes


20.(2)

21.(1)

No symmetry

22.(3) contains 12 Pentagons of 20 hexagons

23.(3)

Polyethylene

Ethanal

Iron oxide

24.(4)

25.(1)

For 1st order reaction

If

or

Similarly

2 2Cu Zn Zn Cu+ ++ ¾¾® +

n 2=

cellG nf E°D ° = - 2 96500 2= - ´ ´ 386 KJ= -

2 2( ( ) )M AA A B type

60C

2 5V O ——— 2 4H SO

4 3/ ( )TiCl Al Me ———

2PdCl ———

——— 3NH

R Pa xa x x

¾¾®

-

1( )dx k a xdt

+ = -

dx kdta x

=-ò òln( )a x kt c- - = +

0t =lnc a= - \ ln( ) lna x kt a- - = -

ln( ) lna x kt ay m x C- = - +

( )0( )¾¾®

AA

R P

Vidyamandir Classes


or

or

or

26.(3) For as complete dissociation takes place.

For

As

27.(3)

For Kr the value of is highest

Thus is also highest

28.

29.(3)

30.(1) For For (filling in BMO)

(filling in ABMO)

( )0[ ] [ ]

Ad R k Rdt

-=

( )d R kdt- =

[ ]d R kdt= -ò ò[ ]R kt c= - +

[ ] = - += +A kt cy mx c

xy2i =2xy CRTp = ´

2BaCli 3=

2BaCl 3 0.01 RT 0.03 RTp = ´ ´ =

2xy BaCl4p = ´p

2

xy

BaCl

2 C RT0.03RT

p ´ ´=

p

2

2

BaCl

BaCl

4 2 C0.03

´p ´=

p

24 0.03C 6 102 100

-´= = ´

´

C8aT27Rb

=

ab

æ öç ÷è ø

CT

2 1 4 3

2 2 2 3NO, N O, NO , N O+ + + +

2 2C , Z 12=

BO 2.5=

\ 2 2C , Z 13- =

–2C

BO 2.5=

–2O

BO 2.5=

Vidyamandir Classes


PART-C MATHEMATICS

1.(1) The points and

So, slope of line joining the two pts

Again, slope of tangent to is

2.(2) Since S.D. =

Now,

3.(3)

4.(1)

–NOBO 2.0=

–2F

BO 0.5=

(1, (1)) (1, 2)= -f ( 1, ( 1)) ( 1,0)- - = -f

2 12-

= = -

3 2 2= - -y x x x 23 2 2= - -dy x xdx

Þ 23 2 2 1- - = -x x Þ 23 2 1 0- - =x x Þ2 4 12 1 ,1

6 3± +

= = -x

22å åæ ö- ç ÷è ø

xi xix x

22 2 2 2( 1) 0 (1) ( 1) 0 (1)4 4

- + + + - + + +æ ö= - ç ÷è ø

K k

2 224 16+

= -K K 23 8 16 5 80Þ + = ´ =K

23 84+

=K 2 72 24

3Þ = =K

23 8 54+

=K 2 6Þ =K

32

0

sin ( )sin

=+òx dx I say

x cosxÞ

3/2

0

cossin cos

=+òxI dx

x xp

Þ 2 =I3 3/2

0

sin cossin

++òx x dxx cosx

pÞ

/2 2 20

2 (sin cos sin cos )= + -òI x x x x dxp

Þ/2

0

12 1 sin 22

æ ö= -ç ÷è øò

pI x dx Þ

/2

0

cos224

ù= + úû

xI xp

Þ1 1 12

2 4 4 2 2æ ö æ ö= - - = -ç ÷ ç ÷è ø è ø

p pI Þ14-

=pI

2 2cos 10 cos10 cos 50 cos 50° - ° ° + °

2 21 2cos 10 2cos10 cos 50 2 cos 502é ù= ° - ° ° + °ë û [ ]1 1 cos20 cos40 cos60 1 cos100

2= + ° - ° - ° + + °

( ) ( )1 3 cos 20 cos40 cos 90 102 2é ù= + ° - ° + ° + °ê úë û

1 3 32 sin30 sin 10 sin 102 2 4é ù= + × ° ° - ° =ê úë û

Vidyamandir Classes


5.(2)

6.(1) Point on the line

Now for p

7.(2)

8.(3)

9.(3)

For

10.(1)

11.(2) Let equation of the plane be

……(1)

22 cos 3 sin 0+ =q q Þ 22 2 sin 3sin 0- + =q q

Þ 22 sin 3 sin 2 0+ - =q q Þ 2 22 sin 4 sin sin 2 0- + - =q q q

(2sin 1) (sin 2) 0+ - =q q

Þ 1sin 2-=q Þ ,

7 11 5, ,6 6 6 6

- -=

p p p pq Þ7 11 5 12 2

6 6+ - -

= =p p p p p p

1 1 22 3 4- + -

= = = lx y zÞ (2 1, 3 1, 4 2)= + - +l l l

(2 1) 2(3 1) 3(4 2) 15+ + - + + =l l l Þ120 5 152

+ = Þ =l l

Þ1 1 1 92, , 4 4 16 202 4 4 2

æ ö= Þ = + + = + =ç ÷è ø

p OP

2

2 21( ) 1 \ [ 1, 0)

1 1= = - Þ = -

- -xf x A Rx x

Þ ( )( ) , 1 [0, )Î -¥ - È ¥f x

1 1 1 2 1 3 1 1 1 78.......

0 1 0 1 0 1 0 1 0 1-é ù é ù é ù é ù é ù

=ê ú ê ú ê ú ê ú ê úë û ë û ë û ë û ë û

n

Þ ( 1) 78S - =n ( 1) 782-

= =n n

Þ ( 1) 156 12 13- = = ´n n Þ 13=n

Þ11 13 1 13

0 1 0 1

- -é ù é ù=ê ú ê ú

ë û ë û

'( ) ( 1)( 1)= - +f x x x xl Þ 3( )¢ = -l lf x x x Þ4 2

( )4 2

= - +x xf x l l µ

( ) (0)=f x f

Þ4 2

04 2

- =l lx x

Þ2

2( 2) 04

- =lx x Þ 0, 2, 2= -x

22

(24 18) 7 3 6 3 24 824 18

+= Þ = -

-

M M MM

Þ 2 236 72 54= -M M Þ 2 290 725

= Þ ±M M

( 0) ( 1) ( 0) 0- + + + - =a x b y c z

Þ + + = -ax by cz b

Vidyamandir Classes


Now

……(2)

and

So, equation of plane

satisfy.

12.

Also, and

So, and

and

13.(3)

and

(0 0) (0 1) (1 0) 0- + + + - =a b c

Þ 0+ =b c

2 2 2

0 1 ( 1) 1cos4 22

´ + ´ + ´ -= =

+ +

pa b c

a b c

Þ 2 22- = +b c a b

2 22 2= +b a b

2 2 24 2= +b a b

Þ 2 22=a b Þ 2= ±a b

2 1± + - = -x y z

( 2, 4)+

!1 3= = +b la l l" " !i j Þ ! ! !2 3 (3 ) ( 3 ) .- + = - + + +! !i j k i x j kl µ l d

! !2 3= - +"

!i j kb µ µ µ

1 2= -b b b!" " "

Þ 3 2- =l µ 3= -d

3 1+ = -l µ

12=l 1

2-=µ

Þ !( )11 32

= +b"

!i j ! !2

1 3 32 2

= - + +b"

!i j k

! !

! !1 2

3 3 9 9 11 022 2 2 4 41 3 32 2

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

- -

b b

!

" "!

i j k

i j k ! !( )1 3 9 32

= - + +!i j k

1( 7)502 --

= + Þ = -n n nn nSn n A T S S

Þ [ ]50 ( 1) ( 7) ( 1) ( 8)2

= - + + - - - -nAT n n n n n n

[ ]50 7 9 82

= + - + -nAT n n

Þ 50 ( 4)= + -nT A n 5 4 (50 ) (50)= - = + - =d T T A A

Þ 50 50 46= +a A

Vidyamandir Classes


14.(4)

15.(2)

So, circle with radius 1.

16.(3)

17.(2)

So, g(x) is not differentiable at

18.() The other root is

The other root is

So,

19.(4)

50( , 9 ) ( , 50 46 )= +d A A

33

2

4 4 4

2 cos 1 1 2 cos 1lim lim lim 2 sin 2cot 1 2cosec 2® ® ®

- - 1æ ö= = = = ´ =ç ÷- è øx x x

x xk xx xp p p

| || | | | 1| |

+ += Þ = Þ =

- +a aa a

i iz z zi i

( )833log6 7

4 32 20 8æ ö= = ´ç ÷

è øxT C x

xÞ 8log 64=xx x

Þ( )8

3log7

320 8 20 9´ ´ = ´xx

xÞ 2

0 8(log8 ) log 64 log= +x x

Þ8log

64=xx

xÞ ( ) ( )8 8

1log 2 log 1 0 64,8

- + = Þ =x x x

( ) ( ( )) 15 | ( ) 10 |= = - -g x f f x f x

15 |15 | 10 | 10 |= - - - -x

Þ ( ) 15 | 5 | 10 |= - - -g x x Þ ( ) 15 |15 | ; 10= - - ³g x x x

15 | 5 | 10- - <x x Þ ( ) 10 ; 5= + <g x x x

20 ; 5 10- £ <x x

; 10 15£ <x x

30 ; 15- £x x

5,10,15=x

, Îp q R 2 3+

. Î Þp q R 2 3+

4 1= - =p q

2 4 16 4 12- = - =p q

2 4 12 0- - =p q

22/3 4/3

4/5 2/5 4/3secsec cos

sin cos (tan )= = =ò ò ò

dx xdxI x ecx xdxx x x

Vidyamandir Classes


20.(4)

Now

lies on the curve

So, curve is

21.(4) Mid-point

Now line as: is layout

So, locus:

22.(3) Path

23.(2)

and

Point of intersection

Þ

413

4/3(tan ) (tan )

4tan 13

-

= = +-

òd x xI C 1/33(tan )-= - +x C

3= + -y x ax b 23Þ = +dy x adx

(1, 5)1

-

= -dydx

Þ 3 1+ = -a Þ 4= -a

(1, 5)-

Þ 5 1 4- = - - b

5 3 2= - =b

3 4 2= - -y x x

Þ , ,2 2

æ ö= = ç ÷è ø

h kA B P

Þ 2 , 2= =h x k y

1+ =x yh x

Þ

2 2

0 0 11

1 1

+ -=

+

h k

h k

Þ 2 21 11

+ =h x

2 21 14 4

1+ =

x yÞ 2 2 2 2 2 2 2 24 4 0+ = Þ + - =x y x y x y x y

1 1 11 1 1 1 12 3 4 31æ öæ öæ öæ ö= - - - - -ç ÷ç ÷ç ÷ç ÷

è øè øè øè ø

1 2 3 7 7 251 12 3 4 8 32 32

= - ´ ´ ´ = - =

{ }2( , 4); 2= £ £ +lA x y x

2³y x 2 0- - £y x

Þ 2 22 2 0= + Þ - - =x x x x

1, 2= -x

Vidyamandir Classes


24.(2)

Also,

25.(2) Roots of are

26.(3)

I.F. =

So,

So, set

( )( )22 2 3

2

1 1

8 12 2 2 4 22 3 3 2 3

- -

é ù 1æ ö æ ö= + - = + - = + - - - +ê ú ç ÷ ç ÷è ø è øê úë û

òx xA x x dx x

18 32

= - -1 952 2

= - =

( ) ( ) ( ) ( )+ = × Þ = xf x y f x f y f x a

(1) 2 ( ) 2= Þ = xf f x

( )1 10102 10

1

2 1( ) 2 2 2 ... 2

2 1

+

=

-é ù+ = + + + =ë û -å

aa

k

zf a k

( ) ( )1 10 102 2 1 16 2 1+ - = -a

Þ 1 4 3+ = Þ =a a

2 1 0+ + =x x 2,w w

Þ 2, 1 0= = Þ + + =a w b w a b

1 11 1 0 1

1 1 0 1

++ = + = + -

+ + - + -

y yy y y y y

y y y y

a b a b a ba b b a bb a a a a b

1 1 2 3( )® + +C C C C 2 2 1( )® -R R R 3 3 1( )® -R R R

1

0 ( 3) 1

0 1 ( 3)

= - -

- +

a b

b

a

y y i

y i

2

2 3

{( 3) (1 )}

{ 3 (2 1)}

= + - - - +

= + - + =

a b aby y

y y y

2 22 æ ö+ = Þ + =ç ÷è ø

dy dyx y x y xdx dx x

22=ò dx

xe x

42 3 2

4= Þ = +ò

xyx x dx yx c

1 3(1) 1 14 4

= Þ = + Þ =y C C

2

23

4 4= +xy

x

Vidyamandir Classes


27.(2) if point is (1, 4)

Again, for focal chord

So, length of focal chord =

28.(4) Let the equation of line be

For,

29.(3)

So, ]

30.(3)

2 16=y x 112

Þ =t 21 1( , 2 )= at t

1 2 1= -t t 2 2Þ = -t

2

1 21( ) 4 2 252

æ ö- = + =ç ÷è ø

a t t

2 cos , 3 sin= + = +x r y rq q

24=r

(2 4 cos ) (3 4 sin ) 7+ + + =q q

Þ 4(cos sin ) 2+ =q q Þ 2 24(cos sin 2sin cos ) 1+ + =q q q q

Þ 2 24 tan 8 tan 4 sec+ + =q q q Þ 3tan 2 8tan 3 0+ + =q q

Þ8 64 36tan

6- ± -

=q

Þ8 2 7 2 7 8tan6 6

- ± -= Þ =q qtan

( )27 12 7 8 7 1tan6 (7 1) 7 1

- -- - -= = =

- +q

1 7tan1 7-

=+

q

( ) ( ~ ) ( )Ù º Ù º!pV p q pV p pVq pVq

~ ( ) (~ ) ( )º Ù !pVq p q

8 5 8 5 8 56 5 7 4 8 3= ´ + ´ + ´ =M C C C C C C n Þ 78= =M n

JEE MAINS-ENTRANCE EXAM-2019 (9th April-Morning) Solutions · Vidyamandir Classes VMC | JEE...

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Transcript of JEE MAINS-ENTRANCE EXAM-2019 (9th April-Morning) Solutions · Vidyamandir Classes VMC | JEE...