jee mains-entrance exam-2019 (10th april-evening) solutions mains... · 2019-12-31 · am 0 012...

Vidyamandir Classes

VMC | JEE Mains-2019 1 Solutions | Evening Session

SOLUTIONS Joint Entrance Exam | IITJEE-2019

10th APRIL 2019 | Evening Session

Vidyamandir Classes


Joint Entrance Exam | JEE Mains 2019

PART-A PHYSICS

1.(4) Breaking stress

2.(1) square loop

circular loop

(radius of loop)

3.()

No option match

4.(3) Beat frequency =

. Between two maximum.

5.(4)

(at constant pressure) =

6.(2)

7.(1)

8.(4)

9.(3)

(Radius of new sphere = )

24´

= =p

F FA d

Þ 6 32

400 4 379 10 1.16 10-´= ´ Þ = ´

p´d

d1.16mm=

2=m Ia2' = pm I r

24 2= p Þ =paa r r Þ

22' 4æ ö= p =ç ÷p pè ø

a mm I

212

= +i fU V mv

9 12 9 129 2

3 39 10 10 9 10 10 1 4 10

210 9 10

- --

- -´ ´ ´ ´

= + ´ ´ ´´

v Þ 46.32 10 /= ´v m s

1 2 2- =f f Hz

1 2

1 0.5D = =-

t sf f

52

æ ö= Dç ÷è ø

RQ n T

'Q 72

D = ´ DPRnC T n T Þ

7'5

=QQ

( )= ´! ! !L m r v

2ˆ ˆ ˆ ˆ2 3 2 6= - Þ = -!r ti t j v i tj ˆ ˆ4 12= -i j ˆ ˆ2 12= -i j Þ ˆ48= -

!L k

2

32.5 1020 10

-

-- ´

=´

a 2 2 2= +v u as2

2 23

2 2.5 101 20 1020 10

--

-´ ´

= - ´ ´´

0.5=

0.7m/s!v

1 2: 4 :1=I I

( )( )

22

1 2max2

min 1 2

3 91

+ æ ö= = =ç ÷è ø-

I II II II I

( )2 21

27 78 2 4

= ´ =RMI MR

2

225 8 2

æ ö æ ö= ´ ´ç ÷ ç ÷è ø è ø

M RI2R 21

80= MR

1 2: 140=I I

Vidyamandir Classes


10.(3)

11.(2)

12.(4)

13.(3)

14.(3)

15.(4) gets excited to n = 4

16.(4)

17.(1) …(i)

…(ii)

21 1

44r r é ù= = = -ê úpp ë ûò ò

b

a

drR dRa br

6=LV V 33

6 1.5 104 10

-= = ´´Li A

8=BV V 33

8 6 2 101 10

--= = ´

´Rii

3 3 32 10 1.5 10 0.5 10- - -= ´ - ´ = ´zeneri A

16=BV V 33

16 6 10 101 10

--= = ´

´iiR

3 310 10 1.5 10- -= ´ - ´zeneri 38.5 10-= ´

2 20 1 0 2

1 12 2

+ r + r = + rP gh v P v

Þ 2 22 1 2gh= +V V

2 2m/s=V

1 1 2 2=AV A V

Þ4

5 2210 5 10 m2

--= = ´A

50

- l= tAN N e 0

-l= tBN N e

5

2 21 1 4 2

- l

-l= Þ = Þ l =t

At

B

N e tN e e e

12

=l

t

( ) 62 1 2 1 3.03 10- = - r = ´P P d d g

6

2 1 33.03 10 303m 300m10 10

´- = =

´!d d

-e

22 2

13.6 13.631 4

é ù- =ê ú lë û

v ncÞ 10.5nml =

3 9

34 82 10 500 106.6 10 3 10

- -

-l ´ ´ ´

= = =æ ö ´ ´ ´ç ÷lè ø

P Pnhc hc

161.5 10= ´

3 30.3 wr = ´ ´rba g a g3 3

wr + = ´rba g mg a g

Þ 3 30.7 10 0.0875w= r ´ ´ = ´m a 87.5= kg

Vidyamandir Classes


18.(3)

19.(4)

20.(2)

21.(3)

22.(3) max. acceleration of 1 kg block

When maximum F is applied on 3 kg.

23.(2) Radiation pressure

Momentum transferred

24.(4) T = Time of flight

Range Along the plane

25.(1)

Slope

26.(1)

225 2 105

w -w ´ pa = = = p

Df i radt s

22 2 55 1.96 10

4 4-æ ö

t = a = + a = = ´ç ÷ç ÷è ø

MRI MR MR 52.0 10 Nm-´!

[ ]03 sin 60 sin 604µ

= ´ ° + °ppiBd

12 3

=d Þ 618 10 18-= ´ = µpB T T

00 3

224 60 60

p= Þw = = =

´ ´vGM GM nv

r r rÞ 11=n

25=x yz

[ ] [ ] 1 2 4 2

2 2 4 225

- -

- -= Þ = =é ùë û

xx M L T Ay yz M T Az

3 2 8 4- -=M L T A

2max 1 2 /1 1

µ´ ´= = =f g m s

0.2 4 10 8= ´ ´ =kf

8 (3 1) 2- = + ´F

16=F N

=IPC

= ´DF t = ´ ´Dp A t

= ´ ´DI A tC

4 4

825 10 25 10 40 60

3 10

-

-´ ´ ´ ´ ´

=´

35 10-= ´ Ns

24sin15cos30

°=

°g

214cos15 9sin302

= °´ - °´R T T 20cm!

1 1 1 1- = Þ = = -v vm

v u f u f

1-= =

cf b

=bfc

/ /0 0 0(1 ) 0.8 (1 )- -= - Þ = -Rt L Rt Li i e i i e

Vidyamandir Classes


= 0.016 s

27.()

No option Match

28.(4)

29.(3)

(after crossing)

=

30.(1)

/ 0.2 5- = Þ = !Rt L Rte nL

310 10 1.65(0.9 0.1)

-´ ´= =

+!

Lt nR

31 2 0.2 10-D = + = ´!! !

F FY A Y A

3

1 20.2 10-

æ ö+ = ´ç ÷ç ÷

è ø

! !FA Y Y

39

3 0.2 10120 10

-æ ö = ´ç ÷´è ø

FA

628 10-= ´

F NA m

00 1 2

= Þ =PV V P

00 2

728

= Þ =PV V P

2 2 1 12 1

-D = - =

PV PVT T TnR

0 00 0

7 28 2´ -

=

P PV V

nR0 054

=PVR

350 1000350 50æ ö= =ç ÷-è ø

apf f

'apf

350 1000 3000 350350 50 350 400

´æ ö = ´ç ÷+è øf 3000 750Hz

4= =

22 æ ö= + ç ÷

è øeff

qEg gm

2= p!

effT

g

Vidyamandir Classes


PART-B CHEMISTRY

1.(2)

2.(2) Glucose Open chain Closed

3.(4) ‘a’ is correct

(M is Alkali/Alkaline earth metal) ‘b’ is correct

‘c’ are called molecular hydrias

4.(4)

5.(3)

Given

From

pH ?= 9Kb 10-= Þ kbP 5= Þ 10g2 0.301=

40.02MNH Cl

4 2 4AcidicsaltNH Cl H O NH OH HCl+ +

( )kb1pH 7 p logC2

= - + ( )217 5 log2 102

-= - + ´ ( )17 3.3012

= - 7 1.6505= -

pH 5.3495=

2 2 2 2MH 2H O M(OH) 2H+ ¾¾® +

2 2MH H O MH H+ ¾¾® +

3 4 6 34BF 3LiAlH 2BrH 3LiF 3AF+ ¾¾® + +

4HF,CH

T 373

(hydratedSolid) (Monohydrate) Anhydrouswhite PowderX Y Z>¾¾® ¾¾¾®

2 3 2 2 3 2 2 2 3WashingSoda Soda Ash

Na CO 10H O Na CO .H O 9H O Na CO¾¾® + ¾¾®

4 3 1 11 1K 2.5 10 dm mol s T 327 C 600K- - -= ´ = ° Þ

3 1 12 2K 1.0dm mol s T 527 C 800K- -= = ° Þ

AE in kJ / moleÞ

R 8.314J / k-mole=

2

1 1 2

K Ea 1 1logK 2.303R T T

æ ö= -ç ÷

è ø

4

1 2

1 Ea 1 1log 102.5 2.303R T T

- æ ö´ = -ç ÷

è ø

41 Ea 1 1log

2.303 8.314 600 8002.5 10-æ ö= -ç ÷´´ è ø

E 165.54kJ=

Vidyamandir Classes


6.(2) CFSE Of

weak field Ligand i.e. do not pair up the unpaired electron.

.

CFSE

weak field Ligand do not pair up

7.(1) Heptane

Since temperature T is given assume it at room temperature.

8.(1) Colloidal particles in lyophobic sols are charged particles which can move towards anode/cathode on application of electric field hence their precipitation can occure.

9.(1) Highest possible oxidation state of

(Refer NCERT)

10.(2) Since in acylic compound angle strain is not present so option (2) is correct.

11.(2) Molar conductivity versus

is less hydrated than so for is more than of

12.(1) 12 five membered rings in and 4 number trigons (triangles) in (white) phosphorous.

13.() (Radon do not occur in nature)

It is obtained as a decay product of

Technically no option is matching.

( )2 26OctahedralComplex

Fe H O Clé ùë û

2H O

( ) 22 6Fe H O

+é ùë û

2 2 2 2 2 6 6Fe 1s 2s 2p 3s 3p 3d+

0 04 0.4 2 0.6Þ- ´ D + ´ D

0 01.6 1.2= - D + D

00.4= - D

[ ]2 4K NiCl

[ ]32

24

spNi

NiCl TetrahedralComplex+

-

Cl- Þ2Ni+ 2 2 6 2 6 81s 2s 2p 3s 3p 3d

t tCFSE 0.6 4 0.4 4= - ´ D + ´ D t t2.4 1.6= - D + D

tCFSE 0.8= - D

( )7 16 2(g) 2(g) 2 ( )C H 11 O 7CO 8H O+ ¾¾® + !!

(g)H U n RTD = D + D

H U 4RTD - D = -

(g)n 7 11 4D = - = -

22Uranium UO 6+® Þ +

Plutonium 7® +

c

K + Na+ mL K +mL Na .+

60C 4P

nR

aR226 222 4

88 86 2(Radium) (Radon)R R Hea n¾¾® +

Vidyamandir Classes


14.(3) Given

(1) From as T increases, K decreases (2) Equilibrium constant is not a affected by change in volume. (3) Although is large but it doesn’t mean reaction go for completion. (4) Equilibrium will shift in forward direction as pressure increases. Hence (3) is incorrect.

15.(2) The correct order of first ionization enhalpies.

Hence option (2) is correct.

16.(2) , where R universal gas constant Temperature Molar mass

Greater the ratio, greater will be the speed and higher the speed, the graph will shift towards right

17.(1) Aniline is a froth stabilizer.

18.(1) The gemstone, ruby has ions having repeating unit . The formula of beryl is

.

19.(3) Greater the stability of negative charge lesser will be the nucleophilicity. Neutral molecules are weaker nucleophiles as compared to the negatively charge species having same nucleophilic centre. Therefore

order of nucleophilicity will be

20.(1) Refer NCERT

21.(1)

No. of moles of required for one gram of

No. of moles of required for one gram of Mg



So least amount is needs for the first reaction and is equal to

22.(4) The shortest wavelength for Lyman series

2 2 32SO ( ) O ( ) 2SO ( )g g g+ à àÜá à àH = 57.2 kJ/molD -

161.7 10cK = ´H

RTK A eD

-¯ =

cK

HiD661 716 908 (kJ/mol)736

Ti < Mn < Ni < Zn¯ ¯ ¯¯

mp2RTVM

= Þ Þ Þ

TM

3Cr +2 3Al O Cr

3 2 6 18Be Al Si O

2 3 3 3 2H CH SO CH CO OH-- -< < <

2 2 34Fe(s) 3O (g) 2Fe O (g)+ ¾¾®

2O3Fe moles224

=

22Mg(s) O (g) 2MgO(s)+ ¾¾®

2O1 moles48

=

4(s) 2 4 10(s)P 5O P O+ ¾¾®

2O 45P moles124

=

3 8 2 2 2C H (g) 5O (g) 3CO (g) 4H O( )+ ¾¾® + !

2O 3 85C H moles44

=

3 32 0.4285g224

´ =

Vidyamandir Classes


The shortest wavelength for paschen series

23.(1)

24.(4) 25.(3)

H 2 2Lyman

1 1 1R1é ù= -ê úl ¥ë û

LymanH

1R

l =

H 2 2Paschen

1 1 1R3é ù= -ê úl ¥ë û

PaschenH

9R

l =

Paschen H

Lyman H

9 / R 91/ R

l= =

l

Vidyamandir Classes


26.(2) Higher value means higher volatile means low adsorption.

27.(2) Since the molality for both ‘A’ and ‘B’ is same

28.(3) It is intermolecular friedel craft reaction and will go via formation of carbocation

29.(2) Air pollution that occurs in sunlight is oxidizing smog

30.(2)

PART-C MATHEMATICS

1.(1) For given numbers mean = 16 Standard deviation ( ) = 4 Mean of

Standard deviation of (as standard deviation is not affected by shifting)

2.(3) solution will exist only for as for R.H.S. is negative. For

Let

only one solution. 3.(4) …(1)

…(2)

Distance between (1) and (2)

fR

b bT K mD = ´

b b

b b

T (A) K (A) 1K 1:5T (B) K (B) 5K

D= = Þ

D

( )µs

( 4) 16 4 12ix - = - =

( 4) 16ix - =2

2 2( 4)(12) (16)

50ix - - =å

2( 4)256 144 400.

50ix - = + =å

5 | 2 1| 2 (2 2)x x x+ - = - 0x > 0x <0x >

2x t=2 3 4 0t t- - =( 4)( 1) 0t t- + =

4, 1t t= ¹ -

2 4 2x x= Þ =

2 2 3 0x y z- + + =

4 2 4 0 2 2 02l

- + + l = Þ - + + =x y z x y z

Vidyamandir Classes


…(3)

Distance between (1) and (3)

4.(1) Equation of line parallel to this line

Distance of this line from origin.

Equation of lines will be

…(1) and …(2) Option (1) is satisfying first line.

5.(4)

Squaring on both sides

6.(2)

tangent is parallel to line

312 3 1

3 3 2

l-

l= Þ - =

8 or 4l = l =2 2 0x y z- + + µ =

| 3 | 23 3-µ

= Þ 3 2 or 3 2-µ = -µ = -

1 or 5µ = µ =

max( ) 13.l + µ =

4 3 2 0x y- + =

4 3 0x y- + l =

| |5l 3

5=

3l = ±

4 3 3 0x y- + =

4 3 3 0x y- - =

1 1cos cos2yx- - æ ö- = aç ÷

è ø

21 2cos 1 . 12 4xy yx-é ùê ú- - - = aê úë û

221 . 1 cos

2 4xy yx- - - = a

22cos 1 . 1

2 4xy yx- a = - -

2 2 22 2cos cos (1 ) 1

4 4x y yxy x

æ ö+ a - a = - -ç ÷ç ÷

è ø2 2

4x y 2 2 2

2 2cos cos 14 4y x yxy x+ a - a = - - +

22 2cos sin

4yx xy+ - a = a

2 2 24 4 cos 4 sin .x xy y- a + = a2

2 2( 3) (2 )( 3)

dy x x xdx x

- -=

-

!

Vidyamandir Classes


Put

for

for

which satisfies option (2).

7.(2) is a tangent to

…(1)

and …(2) equation of common tangent to (1) and (2)

is also tangent to (1)

So

Common tangent will be

On comparing

8.(1)

9.(1)

2

2 23 1

3( 3)xx- -

= --

\2

2 23 1

3( 3)xx

+=

-2x t=

23( 3) 9 6t t t+ = + - Þ 23 9 9 6t t t+ = + - Þ 2 9 0t t- =

0, 9t =

\ 2 0, 9x = Þ x = ±3

3 13;6 2

x y= = =

3 13;6 2

x y - -= - = =

ax y c+ =2 2 1x y+ =2 4 2y x=

2y mxm

= +

2

2

11

m

m=

+Þ 2

22 (1 )mm

= + Þ 4 2 2 0m m+ - =

2 2( 2)( 1) 0m m+ - =

1m = ±

2 and 2y x y x= + = - -

and y ax c y ax c= - + = - +

1, 2 and 1 and 2a c a c= - = = = -

| | 2.c =

2.tan 2 tandy y x x x xdx

+ = +

tan log | sec |I.F. = secx dx xe e x= =ò

2(sec ) (2 tan )secy x x x x x dx c= + +ò 22 sec tan secx x dx x x x dx= +ò ò 2 secx x c= +

2 .cos (0) 1 1y x c x y c= + Þ = Þ = Þ 2 cosy x x= + Þ 2 siny x x¢ = -

Þ1

4 2 2y p pæ ö¢ = -ç ÷è ø

Þ1

4 2 2y p -pæ ö¢ - = +ç ÷è ø

| | 1zw =

arg2

zw

pæ ö =ç ÷è ø

Vidyamandir Classes


10.(3)

11.(3)

At given function is maximum.

12.(2)

13.(3)

arg ( )2

zw p= -

as | | 1zw i zw= - =

350 cm /mindvdt=

34 (10 )3

v x= p +

43

dvdt= .3p 2(10 ) .dxx

dt+

250 50

4 (15)dxdt= =

p

102

42.15p3.153

1 cm/min.18

=p

1 4 5f a a a=

6 2a =

5 2a d+ =( 3 )( 4 ) (2 5 )(2 5 3 )(2 5 4 )a a d a d d d d d d= + + = - - + - +

2(2 5 )(2 2 )(2 ) 2(2 5 )(2 2 )d d d d d d d= - - - = - - - +2 2 3 22(2 5 )( 3 2) 2(2 6 4 5 15 10 )d d d d d d d d= - - + = - + - + -

3 2 3 22( 5 17 16 4) 2[5 17 16 4]d d d d d d= - + - + = - - + -22[15 34 16]f d d¢ = - - +

215 34 16 0d d- + =215 24 10 16 0d d d- - + =

3 (5 8)(3 2) 0d d d- - =

8 2,5 3

d =

2[30 34] 2 30f d¢¢ = - - = -6 8

5´ 34 28

é ù- = -ê ú

ê úë û85

d =

2

15 15

1 1

( 1)12

( 1) 22

n n

n n

s nn n= =

+æ öç ÷è ø= -

+å å

15 15

1 1

( 1) 12 2n n

n n n= =

+= -å å

152

1

1 12 2 n

n n=

= + å15

1

12 n

n=

- å15

1n=å 1 15 16 31 620.

2 6´ ´

= ´ =

6 12 3 3 03 2 2

xx xx x

- -- - =

- +

Vidyamandir Classes


All three real roots

14.(3) Let

Let

15.(3) Area of triangle

angle , angle are in A.P.

, A = 30°

, C = 90°

area =

16.(3)

Tangent :

Normal :

17.(2)

6 1 6 12 3 3 0 2 3 3 05 5 5 3 1

x xx x x xx x

- - - -- - = Þ - - =

- -

( 3x x- 2 3x x- + ) 6(2 3) 1(2 3 ) 0x x x+ + - - - =3 6 6 0x x x- + - + =3 7 6 0x x- + - = 3 7 6 0 1, 2, 3x x xÞ - + = Þ = -

0.s =

2 22 2( ) ( )x xI x e x dx g x e c- -= = +ò2 , 2x t xdx dt= = 2 21 1 2

2 2t t tt e dt e t t e dt- - -é ù= = - +ë ûò ò 21 . 2 2

2t t te t t e e dt- - -é ù= - - +ë ûò

21 [ . 2 2 ]2

t t te t te e- - -= - - -22 4 22 2 2 2

2 2t xt t x xe e- -é ù é ù- - - - - -

= =ê ú ê úê ú ê úë û ë û

4 22 2( )2

x xg x - - -= Þ

1 2 2 5( 1) .2 2

g - - -- = = -

1 sin B2ac=

B = 60°Ð

sin A sin B sin A = sin Baa b b

= Þ ´

sin C sin Ac a

=

1 4 22

a = ´ =

1 2 4 sin 60 2 3.2´ ´ ´ ° =

2 23 5 32x y+ =

3 5 16x y+ = \16 , 03

Qæ öç ÷è ø

52 ( 2)3

y x- = - Þ 5 3 4 0x y- - = \4 , 05

Ræ öç ÷è ø

2 2 11 16 680 1 .2 3 154 0 15

Ar PQRD = =

2 2 51 3

1( )r

n n r n n rr r rT C x C x

x- -

+æ ö= =ç ÷è ø

\ 2 5 1n r- =

Vidyamandir Classes


18.(4) Eqn. of line :

DR’s of

Given line

B(–4, 0, 0) A(–1, 2, 6) AB = 7.

19.(3)

…(1)

20.(2) For infinitely many solution

which satisfies

21.(3)

Least n = 7.

22.(1) Negation of

23.(2) No. of beams = no. of diagonals of a 20 sided polygon

24.(3)

2 15nr -

=

2 1 23 585n n-

= Þ =

2 1 23 385nn n-æ ö- = Þ =ç ÷

è ø\ 38.n =

2 3 46 3 4x y z- - +

= = = l-

AB 6 3, 3 1, 4 10l + l + - l -

AB^

\ 6(6 3) 3(3 1) 4( 4 10) 0l + + l + - - l - =

Þ 1l = -\

2 21

9 16x y

- =

2 2 2( 1)b a e= -

2 516 9( 1)3

e e= - Þ =

9 95 5

a ae e

-- = Þ =

\ ( , 0) ( 5, 0)ae- = -

1 2 3 0D = D = D = D =

1 1 10 4 0

3 2 4D = Þ l -l =

-3Þ l = 2 6 0.l - l - =

01 991 ( 0) 12 100

nnP x C æ ö- = = - >ç ÷

è ø

\1 1100 2n

>

2 100n > \

~ (~ )s r sÚ Ù ~ (~ (~ ))s s r s= Ú Ù

( ~ )s r s s r= Ù Ú = Ù

202 20 170.C= - =

( ) log (sin )ef x x=1( ) sin ( )xg x e- -=

1( ( )) log (sin sin )xf g x e x- -= = -

Vidyamandir Classes


satisfies option (3).

25.(3)

…(1)

…(2) From (1) and (2)

26.(2)

Put

When when

27.(4) Let centre of circle

Locus of centre of circle

28.(2) Let

Any point P on the given line is given by

PQ is perpendicular to the plane

Equation of PQ :

Q lies on both planes as well as …(1)

Also …(2) Solving (1) and (2), we get Point

29.(2)

Let

is

( ) 1fog a¢ = - =

( )( )fog ba = -a =

1 and a b= - = -a

2

1lim 5

1x

x ax bx®

- +=

-1 .1 0a b- + =

1

2lim 51x

x a®

-=

2 5 3a a- = Þ = -

4b = -\ 7.a b+ = -

!

2/3 /3

2/3 4/3 4/3/6 /6

seccos .sin tan

dx x dxIx x x

p p

p p= =ò ò

2tan secx t x dx dt= Þ =

1, ;6 3

x tp= = , 3

3x tp= = \

7 53 31/3 6 6

4/31/ 31

3

3 3 3 .dtI tt

- ù= = - = -úûò

( , )h kº

! 0, 0h k³ ³

Þ 2 21h h k+ = +

Þ 2 2 22 1h h h k+ + = +

Þ 2 2 1k h= +

Þ 2 1k h= +

\ 2 1, 0.y x x= + ³

1 12 1 1x y z- +

= = = l-

(2 1, 1, )P l + - l - l

! 3.x y z+ + =

\(2 1) ( 1) (say)1 1 1

x y z k- l + + l + - l= = =

\ Q ( 2 1, 1, ).k k kº + l + - l - + l

! 3x y z+ + = 3.x y z- + =

Þ ( 2 1) ( 1) ( ) 3 3 2 3k k k k+ l + + - l - + + l = Þ + l =

( 2 1) ( 1) ( ) 3 3 1k k k k+ l + - - l - + + l = Þ + l =

0, 1kl = = \ Q (2, 0,1).º

, , G.P.a b c®2and b ar c ar= =

\ 3 , 7 ,19a b c 23 , 7 ,15 A.P.a ar ar ®22.7 3 15ar a ar= +

Vidyamandir Classes


30.(1)

Þ3 1,5 3

r =

!10,2

r æ ùÎç úè ûÞ

13

r =

1 27 . 33 3

ad a a -= - =

423 3.3aa a a-æ ö= + =ç ÷

è ø

1

0

3 1( 1) 22 ln 2

xA x dx= + - = -ò

jee mains-entrance exam-2019 (10th april-evening) solutions mains... · 2019-12-31 · am 0 012...

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