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### Transcript of JEE MAINS-ENTRANCE EXAM-2019 (9th April-Evening) Solutions MAINS-ENTRANCE EX… · Vidyamandir...

Vidyamandir Classes

VMC | JEE Mains-2019 1 Solutions | Evening Session

SOLUTIONS Joint Entrance Exam | IITJEE-2019

09th APRIL 2019 | Evening Session

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VMC | JEE Mains-2019 2 Solutions | Evening Session

Joint Entrance Exam | JEE Mains 2019

PART-A PHYSICS

1.(1) Conservation of angular momentum

2.(3) We have confidence only till two places of decimal in actual measurement so the final answer accurate to the just 2 decimal place.

Note : It is actual measurement which using some device with least count.

3.(3)

4.(2) ,

, a = 0]

,

5.(4) Truth table can be formed as

Hence the equivalent is “OR” gate Method-2

OR (GATE)

6.(4) Limit of resolution

22 22

02 (0) 212 12 2ML ML Lm m

é ùé ù æ ö+ w + ´ wê úê ú ç ÷è øê ú ê úë û ë û

06

MM m

ww =

+

2mne

r =t

81.67 10 m-= ´ W

2 3( )x t at bt ct= + - 2( ) 2 3dx tv a bt ctdt

= = + -

2

2( ) 2 6d x ta b ct

dt= = -

3bgc

=2

2 33 3b bv a b cc c

æ ö æ ö= + -ç ÷ ç ÷è ø è ø

2

3bv ac

= +

Equivalent0 0 00 1 11 0 11 1 1

A B

1.22dl

=

9

21.22 600 10250 10

-

-´ ´

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7.(3)

When moving away from each other

8.(1) Electric field at

Applying binomial approximation

9.(3) T = MB

10.(1) ,

,

0c

c

v vf fv vé ù-

= ê ú+ë û

0340 202000340 20

f -é ù= ê ú+ë û

0 2250f Hz=

1 1 1 22 cos 2 cosp E E= q - q

2 2 2 2 1/2 2 2 2 2 1/22 2

( ) ( ) [(2 ) ] [(2 ) ]Kq D Kq D

d D d D d D d D= ´ - ´

+ + + +

2 2 3/2 2 2 3/22 ( ) (4 )KqD d D d D- -é ù= + - +ë û3/2 3/22 2

3 2 22 41 1KqD d dD D D

- -é ùæ ö æ öê ú= + - +ç ÷ ç ÷ç ÷ ç ÷ê úè ø è øë ûd D\ <<

2 2

3 2 22 3 3 41 1

2 2KqD d dD D D

é ùæ ö´= - - -ê úç ÷ç ÷ê úè øë û

2 2

3 2 22 12 3

2 2KqD d dD D D

é ù= -ê ú

ê úë û2

49KqdD

=

C NIABf =6

34 3

10 10175 10 10

B T-

-- -´p

= =´ ´

2( ) Krr

r =2

2Mv GMR R

¢=

22

02

( )4R

G r r dnMvR R

r p

=ò 2

22 4GK Rm RT Rp pæ ö =ç ÷

è ø

constantTR=

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11.(2)

12.(2)

13.(1) L = 2.0 m f = 240 Hz

,

v = 320,

14.(2)

15.(1) ,

16.(1)

Linear momentum conservation

, c

17.(2) , b

At t = 0 18.(4) Potential Energy of spring = Total Head Energy

( 2 )IP a rC

= +

2 34 4

IC

æ ö= ´ +ç ÷è ø

85 20 8 104IPC

-= = × ´

!2 215 (4 20 )r t i t j®= + -!

!2

2 30 ( 40)dt i jdt

®

= + -!

250 / seca m=

3 2402vfL

= =3 2404v=

0 802vf HzL

= =

2 13 ( ) ( )3

q -q q-q= =

1 2910 10q q

q = +

3R = W 2R lµ22 3 12R¢ = ´ = W

15 106

R Rp ¢= ´ = Wp

2 26

R Rp ¢= ´ = Wp

1 2

1 2

53

R RRR R

= = W+

2 2 0 22vm v mv m m¢

+ = ´ + ´ 2 2v v¢ =

2outer 0( )4tnKte R-af µ p [1 ]td Ce t

dt-a- f

e = = -a

induced[1 ]

(Resistance)

tCe ati-a- -

=

inducedi ve= -

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,

19.(3)

t = 2 sec

20.(4)

n = 2, z = 2

21.(3) Magnification is 2

If image is real,

If image is virtual,

22.(3)

When is filled between lens and mirror

23.(3)

21 1 2 2

12kA m s m s= Dq+ Dq

2

1 1 2 2

12kA

m s m sDq =

+

2

5

2400100 3.6 10

(4184 200)k-

æ ö´ç ÷è øDq = = ´+

0 tw =w +a

20tw =

21 12002Iw =

21 1.5 (20 ) 12002

t´ ´ =

2

213.6nzEn

= -

2

2213.6 13.62

E eV= - ´ = -

132fx =

2 2fx =

1

23:1x

x=

1

1 1 2 12 18 18f

= ´ =

1

2

1 ( 1)18f

µ --

-

11

2 2 2 2 2( 1)18 18 18

P - µ += - µ - =

1

182mF

æ ö= = -ç ÷-µè ø

12 6 3= - µ

13 4µ =

143

µ =

1 2P P P+ =

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24.(3)

25.(2) The physical size of antenna of receiver and transmitter both inversely proportional to carrier frequency.

26.(1) By conservation of L momentum

Conservation of M.E.

,

27.(4)

28.(4) Initially

After pouring of oil

29.(1) Q = Equivalent

x y

l l l- =

l l l

x y

y x

l ll =

l -l

( )g gI I S I G- ´ =

0.002 50 0.20.5 0.002

S ´æ ö= = Wç ÷-è ø

0 5mv mv¢=

0 05 5mv vvm

¢ = =

2 20

1 1 5 ( )2 2mv m v mgH¢= +

2025vHg

=

22 30 1

2 2 3/2 102( )q

i R rR x

-µf = ´p =

+2

20 12 2 3/22(r )pi r Rx

µf = ´p

+2 2 3/2

22 2 3/2 3

1

( )( ) 10

P P

Q

i R xi r x -

f + f= × =

f +

323 10

P-

f=

46.67 10P-f = ´

45V g mgrw =

02 2V Vmg g g= rw +r

0 42 2 5

rrw+ = rw

0 4 12 5 2r é ù= rw -ê úë û

0 8 52 10r -é ù= rwê úë û

035

r = rw

eqQ C V=

( 1)Q n CV= +

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After intersection dielectric equivalent

Conservation of charge

30.(4) For A

For B

PART-B CHEMISTRY 1.(2) Kieselghur is an amorphous form of slica.

2.(1) CO is diamagnetic according to molecular orbital theory.

3.(2) By passing 0.1 Faraday electricity, 0.1 gm-equivalents of will be discharged. Number of gm-equivalent = (n – factor) × number of moles 0.1 = 2 × number of moles

Number of moles

4.(2)

More the intensity of +ve charge on carbon atom of carbonyl group, more is the reactivity towards nucleophilic addition reaction. So propanal is more reactivity than acetone.

5.(2)

eqC

eqC KC nC¢ = +

( )eqC C n K¢ = +

1) 1( )n CV nV Vn n K C n K( + +æ ö= = ç ÷+ +è ø

7p vR C C= - =

4422 6.32 7vfRC f= = Þ = =

9p vR C C= - =

42212 9vfRC f= = Þ =

2Ni+

Þ

Þ0.1 0.052

= =

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6.(3)

7.(1) Noradrenaline is one of the neurotransmitters that plays a role in mood change

8.(1) Assertion is correct as Haematite ore is used for extraction of Fe. Haematite is an oxide ore, so reason is incorrect.

9.(1) Neptunium (Np) and plutonium (Pu) show maximum number of oxidation states starting from +3 to +7.

10.(2) 20% (mass/mass) means 100 gm of solution has 20 gm of KI 80 gm of solvent contains 20 gm of KI

Molality,

= 1.506

11.(3) According to the postulates of VBT we can’t predict the magnetic properties of complexes, it can’t distinguish ligands as weak and strong field thus is not able to explain colour of complexes.

12.(1) for inert gases

‘b’ depends on size of the gas atom the gas atom therefore steepest increase in plot of is maximum for Xenon.

13.(1) Boron trioxide is acidic

and amphoteric

and are basic

14.(2) Initially solution is of NaOH therefore pH is high then it starts decreasing with addition of HCl. After equivalence point pH will become constant and in acidic range due to presence of strong acid.

15.(1) Due to intermolecular hydrogen bonding HF has highest b.p. among hydrogen halides.

16.(1)

17.(3) As the distance from nucleus increases total energy increases and is minimum at distance

18.(1) Hinsberg’s reagent is benzenesulphonylchloride.

19.(2) Ceric ammonium nitrate test is for alcohols and carbylamines test is for amines. Both these functional groups are present in peptide formed by serine and lysine.

20.(4) More stable is the carbocation formed more is the reactivity for reactions.

Þ20 1000m166 80

= ´

1.51m»

Pbz 1RT

= +

z b,µ z

2 3(B O )

2 3Al O 2 3Ga O

2 3In O 2 3Tl O

U q WD = +

U ( 2) 10 8kJD = - + =

0a .

NS 1

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21.(3) i = 3

= 3 × 4 × 0.03 = 0.36 K

22.(4) According to given graph, activation enthalpy to from C is greater than that to from D.

23.(1) Bauxide , Malachite

Siderite , Calamine

24.(3) Stratosphere lies between 10 to 50 km from sea level.

25.(4) Number of particles

Total area = Area covered by one particle × number of particle 0.24 =

(a) edge length

26.(1) Common CN of transition elements = 6 Common CN of inner transition elements = 8 to 12.

27.(1) It is acid catalyzed intermolecular esterification reaction

28.(4) is a two step reaction in which first step is rate determining step and stable carbocation is formed. So, peak of first step is higher than second step.

29.(2) Electrophilic substitution is more in phenol ring.

30.(4) In vapour phase, exit as dimer

In solid state, has polymer chain structure

PART-C MATHEMATICS

22 4 4K SO 2K SO+ -= +

f fAl i K m= ´ ´

2 3Al O® 3 2CuCO .Cu(OH)®

3FeCO® 3ZnCO®

3

A310 10 N10

- ´=

323

310 10Area 6 1010

- ´´ ´ ´

20 2Area 4 10 cm-= ´102 10 cm 2pm-= ´ =

NS 1

2BeCl

2BeCl

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1.(4) ,

,

,

,

2.(4) ... (i)

... (ii)

Clearly circle (i) & circle (ii) touches internally because

Equation of common tangent Point (6, –2) passes through the tangent.

3.(1)

Put

Ans.

4.(3) Let number of people in city = 100

( )55 1

zwz

+=

-5 5 5w wz z- = +

5 55 3wzw-

=+

13/5

wzw

-=

+

1 13/5

wzw

-= <

+315

w w- < +

( ) 1Re5

w > ( )5Re 1w >

2 2 4x y+ =2 2 6 8 24 0x y x y+ + + - =

( ) ( )2 23 4 49x y+ + + =

1 2 1 2c c r r= -

1 2 0s s- =

1 2 6 8 20 0s s x y- = + - =

3 4 10x y+ =

( )1

1 2 4

0

cot 1I x x x dx-= × - +ò2x t=2xdx dt=

( )1

1 2

0

1 cot 12

I t t dt-= - +ò

( )

11

0

1 1tan2 1 1

dtt t

- æ ö= ç ÷ç ÷+ -è øò

( )( )

11

0

11 tan2 1 1

t tdt

t t- æ ö- -

= ç ÷ç ÷+ × -è øò

( ){ }1

1 1

0

1 tan tan 12

t t dt- -= + -ò

\ ( )1 1

1 1

0 0

tan tan 1t dt t dt- -= -ò ò1

1

0

tanI t dt-= ò111

200

tan1tt t dtt

-= × -+ò ( )

12

0

1 14 2

n tp= - +!

1 24 2

np= - !

17A BÇ =

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VMC | JEE Mains-2019 11 Solutions | Evening Session

30% of

40% of

50% of

Total = 5.1 + 4.8 + 4 = 13.9

5.(4) & ... (i)

... (ii)

Using (i) & (ii)

6.(1)

,

Equation (i)

Equation (ii)

7.(4)

Let We know that

,

,

8.(1)

( ) 5.1A BÇ =

12A BÇ =

( ) 4.8A BÇ =

8A BÇ =

( ) 4A BÇ =

2 4x ¹ 3 0x x- >2, 2x ¹ -

( )( )1 1 0x x x× - + >

( ) ( )1, 0 1,xÎ - È ¥

( ) ( ) ( )1, 0 1, 2 2,xÎ - È È ¥

2 3 11 2 02 1 1

D k-

= - =-

( )2 2 3 5 2 1 0k k- - ´ + + =

4 10k = 92

k =

2 3 0x zy y× + - =

2 1 0x zy y× - + =

12

xy

-=

2zy=

14

xz= -

12

x x z ky z y+ + + =

1ˆ cos3 2

a i p× = = =!

"

1ˆ cos4 2

a j mp× = = =!

! cosa k n× = = q"

2 2 2 1m n+ + =!21 1 1

4 2n+ + =

1 cos2

n = ± = q2,

3 3p p

q =

( ) ( )2199 2

2n n

n+

+ = -

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n = 19

No. Of balls used for the triangle = = 190.

9.(2) Ar(ABC)

.

10.(3)

11.(4)

.

12.(2) Let a point lying on the parabola is

Equation of tangent at to the parabola is

If this line also touches the ellipse then

(–ive sign rejected)

13.(2) upto 11th term

20192

´

12BC AN= ´ ´

!!!"

ˆˆ3 4i k= +!"

#

( ) ( )ˆ ˆˆ ˆ ˆ3 2 2 3 4

5

i j k i k- + ´ +=

ˆˆ ˆ8 6 6 1365 5

i j k- + += =

( ) 1 13652 5

Ar ABC = ´ ´ 34=

5 1 5 3a bp- + = - p +

( ) ( )5 1 5 3a b- p + = - p +

( )( )5 2a b- - p =

25

a b- =- p

5tan15d

° =

Þ52 3d

- =

Þ5

2 3d =

-( )5 2 3= +

2y x= ( )2,b b

( )2,b b 2y x=2

2xy +b

b =

Þ2 2xy b

= +b

2 21

1 1/ 2x y

+ = 2 2 2 2c a m b= +

2

21 11

4 24b

= × +b

Þ 4 21 2b = + b Þ 4 22 1 0b - b - =

( )22 1 2b - = Þ 2 1 2b - = ±

2 1 2b = ±

Þ 2 1 2b = + Þ 1 2a = +

1 2 3 3 5 4 7 ....+ ´ + ´ + ´ +(2 1)nT n n= -

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= 946

14.(1)

Sum of roots

is max. when m = 0

= 9 – 2 = 7

15.(2)

Given that

(Using L.H. Rule)

16.(4)

Differentiating both sides w.r.t. x

17.(3) Equation of required plane is

Equation of plane

Distance

18.(1) Mid point of AC lies on diameter

11 11 112

111 1 1

2nn n n

S T n n= = =

= = -å å å 11 12 23 11 1226 2

´ ´ ´= ´ -

( ) ( )22 2 21 3 1 0m x x m+ - + + =

223 , 11

mm

= = a +b ab = ++

a +b

Þ 3, 1a +b = ab = Þ ( )22 2 2a +b = a +b - ab

( )( )3 3 2 2a -b = a -b a +ab+b ( ) ( )2 2 24= a +b - ab a +ab+b

( )9 4 7 1 8 5= - + =

( )

26

2lim2

f x

x

t dtx® -ò

( )2 6f =

( )

( )6

2

2

lim2

f x

x

t dt

x®=

-

ò 0 form0æ öç ÷è ø

( ) ( )2

2 ' 0lim

1 0x

f x f x®

-=

-

( )2 6 ' 2f= ´

( )12 ' 2f=

( ) ( )( ) ( )sec 2 secsec tan sec tan secx xe x x f x x x x dx e f x c+ + = +ò

Þ ( )sec 2sec tan ( ) sec tan secxe x x f x x x x+ + ( ) ( )sec sec' sec tanx xe f x f x e x x= + ×

Þ ( ) 2' sec tan secf x x x x= + Þ ( ) sec tanf x x x K= + +

1 2 0P P+ l =

( )2 3 5 6 0x y z x y z+ + + + l + + - =

( ) ( ) ( ) ˆˆ ˆ2 3 1i j kl = + l + + l + + l!

! 0kl × ="

1l = -2 11 0x y= + + =

2 2

11

1 2=

+

115

=

151,2y+æ ö-ç ÷

è ø3 7y x= +

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Area of rectangle = 14 × 6 = 84

19.(1)

Solve and we get (8, 4) & (2, –2)

Ans. (1)

20.(3)

h = 2r

21.(3)

,

lines are and point of intersection of lines is

Distance from origin .

22.(3)

I.F.

,

,

,

Þ 153 1 72y+æ ö = - +ç ÷

è øÞ 1 1y = -

2 2 , 4y x x y£ £ +2 2y x= 4;x y= +

( )4 2

2

42yy dy

-

æ ö= + -ç ÷ç ÷

è øò

1 1tan2

-q =

1tan2

rh

q = = Þ

213

V r h= p

2 313 4 12h hV h p

= p × =

2312

dv dhhdt dt

p= ×

Þ 5 1004

dhdt

p= × ´ Þ

15

dhdt

=p

( )1 1x a y+ - =

22 1x a y+ = { }0, 1a RÎ -

1 2 1m m = -

21 2 11a a-

- × = --

3 22 a a- = -

3 2 2 0a a- + = Þ 1a = -

\ 2 1x y- = 2 1x y+ = \3 1,5 5æ ö-ç ÷è ø

9 1 10 225 25 25 5

= + = =

6tan ; 0,cos 2

dy xy x xdx x

pæ ö- = Îç ÷è ø

tan log cos cosdx xe e x

- += = =ò6cos coscosxy x xdxx

× = ×ò 2cos 3y x x c= +

, 03

y pæ öç ÷è ø

20 3

9C

æ öp= +ç ÷ç ÷

è ø2

3C p= -

22cos 3

3y x x p

= -

Vidyamandir Classes

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,

23.(1)

d = 4, – 4

24.(1)

... (i)

r = 2.

Average

25.(3) Slope of normal at P = slope of CP

Slope of normal

2 2 2 2 2 23cos 36 36 3 12 3 12 4

yæ öp p p p p p p

= - = - = - = -ç ÷ç ÷è ø

2 224 3 2 3

y p -p= - × =

, , Termsa d a a d- + ®

3 33 11a a= Þ =

( )( ) 1155a a d a d- + =

( )211 121 1155d- =

2121 105d- =2 16d = Þ

( )( )11 15 10 4 40 15 25T = + - = - + = -

1 1: : 2 :15 : 70n n nr r rC C C- + =

12: ,15

n nr rC C- =

1

152

nr

nr

CC -

=

1 152

n rr

- +=

2 2 2 15n r r- + =

2 17 2n r- = -

( )1 1 170 1415 1 3

nr

nr

n rCrC

+ - + += Þ =

+

141 3

n rr-

=+

Þ 3 3 14 14n r r- = +3 17 14 ...( )2 17 2 ...( )

16

n r iin r i

n

- =- = -

- + +=

48 17 14r- =17 34r =

Þ

16 16 16 171 2 3 3

17.16.151616 16 680 6961.2.3 2323 3 3 3 3

C C C C ++ + + += = = = = =

2 4y x=

2 ' 4yy =

2'yy

=

2 12-

= = -

211

b-- =

a -1 2-a + = b-

3b = -a

( ) ( )2 2 21 2a - + b- = b

2 2 22 4 5a +b - a - b+ = b

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... (i)

Area

26.(1) Verify with options

27.(4)

28.(2) 10, 22, 26, 29, 34, x, 42, 67, 70, y

Mean

... (i)

Median x = 36 y = 84

29.(1)

2 2 4 5 0a - a - b+ =2 2 4 5 0a - a - b+ =2 2 12 4 5 0a - a - + a + =2 2 7 0a + a - =

( )21 8a + =

1 2 2a + = ±

1 2 2, 1 2 2a = - - - +

\ ( )3 1 2 2b = - - +

4 2 2b = -

( )2= p b

( ) ( ) ( ) ( )2 24 2 2 4 2 2 4 4 2 4 2 8 3 2 2= p - = p - = p + - = p -

( )p q rÞ È

; ;p T q F r F= = =

( )T F FÞ È

T FÞ

sin10 sin30 sin50 sin70° ° ° °1cos80 cos40 cos202

æ ö° ° °ç ÷è ø

1 cos20 cos40 cos802

° ° °

( )3

3

sin 2 2012 2 sin 20

× °×

°

1 sin16016 sin 20

°°

( )sin 180 201 sin 20 116 sin 20 16sin 20 16

°- ° °= = =

° °

10 22 26 29 34 42 67 7010x y+ + + + + + + + +

=

3004210x y+ +

= Þ 420 300 x y= + +

120x y+ =

34 352

x += = \

84 736 3

yx= =

33TA A I=

0 2 2 0 2 1 1 0 02 2 1 3 0 1 01 1 1 2 1 0 0 1

x x yy y y R y

x y

é ù é ù é ùê ú ê ú ê ú- - =ê ú ê ú ê úê ú ê ú ê ú- - +ë û ë û ë û

Vidyamandir Classes

VMC | JEE Mains-2019 17 Solutions | Evening Session

;

Ans. 4

30.(4)

is continuous at x = 4.

2

2

8 0 0 3 0 00 6 0 0 3 00 0 3 0 0 3

x

y

é ù é ùê ú ê úê ú = ê úê ú ê úë ûê úë û28 3x =

38

x = ± 26 3y = Þ12

y = ±

( ) [ ]4xF x x é ù= - ê úë û

( ) ( ) [ ] ( )0 0 04

4lim lim 4 lim 4 lim 4 1 34h h hx

hF x F h h+ ® ® ®®

æ ö+é ù= + = + - = - =ç ÷ê úë ûè ø

( ) ( ) [ ] ( )0 0 04

4lim lim 4 lim 4 lim 3 0 34h h hx

hF x F h h- ® ® ®®

æ ö-é ù= - = - - = - =ç ÷ê úë ûè ø

( ) [ ] 44 4 4 1 34

F é ù= - = - =ê úë û

( )F x