2019-SOLUTIONS-JEE ENTRANCE...

Vidyamandir Classes

VMC | Paper-2 1 Joint Entrance Exam-2019 | Advanced

SOLUTIONS 2019-Joint Entrance Examination - Advanced | Paper-2

PART-I PHYSICS

SECTION-1 1.(1,4)

Increment in velocity after 1st collision = 2 V

If piston moves by dL time taken to move dL

tV

Time for one collision 2

o

L

V

no. of collision in one second 2

oV

L

N = no. of collision in t sec 2 2

o oV VdL dL

L V V L

Increment in velocity in after N collision 22

oo

V dL dLV V

V L L

If piston moves by dL increment in velocity o odL

dV VL

o odL

dV VL

[as L is decreasing]

/ 2

f o

o o

V Lo

oV L

dv dL

V L

1

ln ln ln 22

Vf

oV

2f oV V

K.E. becomes 4 times

Correct options are (1,4)

2.(1,2,4) 1 1 2 2min

1 2

5 3 1 5 15 5 20 10

5 1 6 6 3

n f n ff

n n

2 2 6 10 6

1 1 1 1.610 10 103

mixmixf

1

0 0 4

mixmix o

fV

T V T

1

1.6 1 0.600 0 0

0(4) (4)

4

mix

fV

T T T TV

Vidyamandir Classes


0.6

0.600 0

6 (4 1)| | 10 (4 1) 10 (2.3 1)

1 0.6

nR T R TW RT RT

013RT

Avg. kinetic energy 2mix

mix ff

n R T 0 0 0

10103 6 2.3 6 2.3 23

2 6R T R T RT

Final pressure 1.600 0 0

0(4) 9.2

4

mix

fV

P P P PV

3.(1,3) (I) 130

201.5

H cmH cm

n

(II) 1 1.5 1 1.5 1 1.5 0.5

V u R V H R

1 0.5 1.5 10 14.5 300

20.68300 30 10 300 14.5

V cmV

2 20.68H cm

(III) 1 15 1 1.5

V u R

1 15.5

19.35300

V cmV

3 19.35H cm


4.(3,4) As it is an equipotential surface, E will be perpendicular to the surface, i.e. it is radially outwards. 0 0V volts

0BV volts [due to dipole = 0, due to Eo = 0]

0A BV V

020A

KpV E R

R

1

30 2

o

Kp KpE R R

ER

00 0 0 03 2

2 2 1 23A

Kp Kp E RE E E E E

R RR R 0

3 ˆ ˆ( )2

AE E i j

0 0 030B

KpE E E E

R


5.(1) 2 ...(1)Mg T Ma

32 3 ...(2)Mg T Ma

12 2 ...(3)T Kx Ma

From (1) and (2)

Vidyamandir Classes


2 33

( ) 22 2

T TM a a Mg T Mg Mg

11

4

3

3 8 32 2 2 ...(4)

2 3 3

T Ma MaMa Mg T

18 8

(3) (4) 23 3

Mg Mkx M a

114 8

3 3

M Mga K x

K

13 8

14 3

K Mga x

M K

03 8, 2

14 3 2

K Mg xA A

M K [Maximum elongation = 2A]

max8 3

3 14

Mg KV A

K M

1 4ox

a at x 3 4 2

14 3 7

K Mg g

M K

Correct Answer is (1) 6.(1,3,4)

Loss in GPE = gain in KE

2

21

2 4 2 3

mmg

2

21 3

4 2 3 2

m gmg

23

2 2 3p Pm

I mg

3 3

4

g

2 3 3

2 2 2 4Rg g

a

2

, cos60 cos302 2CM yMg N MA M

3 3 3 3

8 8 2

gMg N M g

3 9 15

8 16 16

MgMg

16

MgN

Correct options are (1),(3),(4)

Vidyamandir Classes


7.(1) P.d. at O d

.yd

P d at P dD

0.36 0.36

deg ree180

rad

0.360.002

180rad

3. 0.3 10 0.002P d at O 76 10 600m nm

constructive interference at O

As d n where n = 1

600nm = (1) (600 nm)

P.d at 3 311 10 0.3 10

600 600 3300 39001

p nm nm nm nm

At P: (2 1) 78

3900 600 2 1 132 6

nn

Destructive interference at P

Correct option is (1)

8.(3,4) 1 4

1 1513.6 1 13.6

16 16 a

hcE eV

24

1 113.6

16m e

hcE

m

2

1 113.6

16 115 513.616

a

e

m

2 2

5 5 15 5 202

16 16 64m

m m

5a e

e a

p

p

12420 12420 16

4873 13.6 313.6

16

e nm

Å Å


SECTION-2 1.(135) (226.005 222.000 4.00)931Q MeV

(0.005)931 4.655MeV MeV

TE Total Energy available 4

AE

A

226

4.44 4.52222

MeV

4.655 4.52 135r TE Q E MeV KeV

2.(1.5) When 60

2 Cr

0 1sin sinn r

11

0

sinsinr

n

12 1

0

sinsin Cr A r A

n

Vidyamandir Classes


1 3 / 275 sin 75 30 45

3C

1 2

0 0

1 3sin 45 1.5

22

n nn

n n

3.(1.38) L.C. of scale of optical bench 1

0.254

cmcm

2 1u x x

1 2 0.5u x x cm

0.5v cm

30 60 1800

2030 60 90

uvf cm

u v

1 1 1

f v u (lens equation)

2 2 2

df dv du

f v u

2 2 2 2 2

100 100f v u f v u

fff v u v u

0.5 0.5

100 2060 60 30 30

1 4100 20 0.5

3600

2000 0.5 5

1.383600

4.(4) 0 11 0 2

2 sincos

u RR u

g

0 0 12 2

2 sincos

u u RR

g

13 4

RR

2

1 11 2 3 2

2

......1 11

TotalR a R

R R R R

12 sinou

Tg

12

2 sinou TT

g

13 2

TT

1 11 2 3 .....

1 11Total

T TT T T T

11

1

RV

T and Total

avg entire motionTotal

RV

T

21

1 121

10.8

11

RV V

T

0.8 0.8

0.2 0.8 4

5.(0.63) After closing the switch, moving rod behaves as a source of EMF

1 1Rt Rt

L Lo

Ei i e e

R

33

12 10

101 1 10 0.1( 1 ) 1

1i t ms e

Vidyamandir Classes


3 3110 1 0.63 10 A

e

0.63x

6.(1) If N photons strike the mirror, then momentum acquired by mirror 2

oh

N p mv

Loss in K.E of mirror = gain Elastic potential energy of spring

2 21 1

2 2o o o om

mv kx x vK

1

op

xm

2 1

oNh

xm

2

ox mN

h

6 6 248 10 1 10 10

2 4

1210

1x

SECTION-3

1.(1) 1 01 0

0 0

1

2 2

C Tf f

L L

2

2 2 22 0

02 0

33 3

32 22

C C Cf fo f

LL L

2 1 1 2 02

0 0

1

2 2 2 4

C C C T TC

L L

0

2 2

TT

3 3 3 33 3

03 0 0

5 5 2 25

52 24

C C C Cf fo fo

LL L L

313 3

1 3

4 3 16 16o

oT TC

C T T

4 4 4 14 4 0

04 0 0

14 414 14

72 224

C C C Cf fo f

LL L L

1 4 0 04 4

1

8 4 64 16

C T T TC T


2.(2) To keep fundamental frequency higher

01

1

2 o

Tfo fo

L

;

0

20

1

3 2 2

T fofo

L

03

0

1

2 3 3

T fofo

L

; 0 0

40

1

2 4 24

T fo ffo

L


3.(4) 1 2 Isothermal process

2 3 Isochoric process

2 0 01 2

1 0

2ln 1 ln

3

V T VW nRT R

V V 0 ln 23

RT

01 2 3 2 3 0

3 3 2(1)

2 2 3

TU U nR T R RT

Vidyamandir Classes


1 2 3 1 2 2 3Q W U 0 (3 ln 2)3

RT

01 2 1 2 ln 2

3

RTQ W


4.(2) 1 2 Isobaric process 2 3 Isochoric process

01 2 3 1 2 0 0 0 0 0(2 )

3

RTW W P V V P V

1 2 3 1 2 2 3U U U

2 2 1 1 3 3 2 23 3

2 2P V PV P V P V 33 3

2 2 (2 )2 2 2

oo o o o o o

PP V V V P V

00 0 0 0 0 0 0

3 33 3

2 2 3

RTP V P V P V RT

1 2 3 1 2 2 3Q Q Q

1 2 1 2 2 3 0 0 0 0 0 03 3

2 2W u u P V P V P V 0

0 04

43

RTP V

1 2 0 0 05 5

2 6Q P V RT


PART-II CHEMISTRY

SECTION-1 1.(2, 3, 4) Aqua regia is 3 part HCl and 1 Part 3HNO

3 2 23HCl HNO NOCl Cl 2H O (Yellow colouration is due to NOCl and 2Cl )

aqua4regia

Au HAuCl (+3 oxidation state of Au)

2.(1, 2, 4) 2 4 4 2 2

(G) (R) (X)Zn H SO ZnSO SO H O

2 2 2(T) (Q)

Zn 2NaOH Na ZnO H

4 2 4 (s) 2 4 2 4(X) (Y)(Z)

ZnSO H S NH OH ZnS H O (NH ) SO

3.(2, 4) (1)

(2)

(3)

(4)

Vidyamandir Classes


4.(2, 4)

5.(1, 3, 4)

2

NaCN Zn2 2 4O (Q) (T)

(R) (Z)Au Na[Au(CN) ] Au Na [Zn(CN) ]

6.(2, 4) nE for 2

2

ZHe 13.6

n

2

413.6 3.4

n

2n 16 n 4

2 ; m 0

It’s 4d orbital.

Radial nodes = n 1 4 2 1 1

Angular nodes 2

He is a single e species ; hence nuclear charge experienced is 2e

7.(1, 2)

(1)

(2)

(3) Cellulose is a straight chain polysaccharide composed of -D-glucose units held together by

1 4C C glycosidic linkage.

(4) Natural rubber is polyisoprene containing cis stereochemistry alkene units

Vidyamandir Classes


8.(1, 4)

(1)

(2)

(3)

(4)

SECTION-2

1.(6.00)

2.(10.00)

Vidyamandir Classes


3.(288) 8 3 2 4 2 2S conc.HNO H SO NO H O

On Balancing :

8 3 2 4 2 2S 48HNO 8H SO 48NO 16H O

1 mole of 8S produces 16 mole of 2H O

Mass of 2H O produced 16 18g 288g

4.(6.00)

5.(2.30) 2 5(g) 2 4(g) 2(g)2N O 2N O O

At time t = 0 1 atm 0 0

At time 3t y 10 s (1 x) atm x x/2

Pressure of container at time t is 1.45 atm 1 x x x / 2 1.45

1 x / 2 1.45

x = 0.9

2.303 1

t log2k 1 x

4

2.303 1log

0.12 5 10

3 31

4.606 10 y 102

4.60

y 2.302

(truncated value)

or 4.61

y 2.312

(Round off)

6.(2.98) Mole fraction of urea = 0.05

Number of moles of 2900

H O 5018

urea

urea

n0.05

n 50

urean 2.63

uream 2.63 60g 157.8 g

Total mass of solution = 157.8 + 900 = 1057.8 g

Volume of solution 1057.8

mL1.2

= 881.5 mL

Molarity no. of moles

vol. in Litres

2.63

0.8815 = 2.98 (Truncated value)

or = 2.99 (Round off)

Vidyamandir Classes


SECTION-3 1.(1) According to Bohr’s model radius of nth orbit 2n Angular momentum n of e in nth orbit

Kinetic Energy of e in nth orbit 2

1

n

Potential Energy 2

1

n

2.(3) 2

1K.E.

n

3-4 3.(4) 4.(3)

(I)

(II)

(III)

(IV)

(I) Q, R (II) P, S, U (III) Q, R, T (IV) Q, R

Vidyamandir Classes


PART-III MATHEMATICS

SECTION-1 1.(1,4) If P, Q and R are collinear

PQ PR 1

1 1

v

1

,1

v 0,1

2.(1, 3)

1/3

1

21

1

lim 541 1

n

rnn

r

r

n n

n ra

n

11/3

01

20

54

( )

x dx

dx

a x

1

( 1) 72

a a 8a or 9

3.(1, 2, 4) '( ) ( 1) ( 2) ( 5) F x x x x

Obviously 1 and 5 are the points local minima and 2 is a point of local maxima.

Claim: F(x) < 0 (0, 5) x .

Since 2 is a point of maxima, it is suffice to show that (2) 0F .

4 3 28 17( ) 10

4 3 2

x x xF x x and (2) 0F .

Therefore 2 is correct.

4.(2, 3, 4) 0

2

0

1 22 sin sin

2 2( )

12 sin

2

n

kn

k

k k

n nf n

kn

0

0

2 3cos cos

2 2

2( 1)1 cos

2

n

k

n

k

kn n

kn

3 1cos sin

2 2( 1) cos

2 sin2

1cos sin

2( 1)

sin2

n n

n nn

nn

n

nn

n

3( 1)cos cos

2 2( 1) 1

nn

n nn

( ) cos2

f nn

(i) lim ( ) lim cos 12n n

f nn

(ii) 1 1tan(cos ( (6))) (6) cos ; cos cos8 8 8

f f

Vidyamandir Classes


2

2tan ; tan 1

8 4 1

2 22 1 2 1 0

(iii) 1 1sin (7 cos ( (5))); (5) cos cos cos7 7 7

f f

sin 7 07

(iv) 3(4) cos

6 2f

5.(3, 4) 0

| | 0( ) | | lim 0

| |h

h hf x x x

h

20

| |lim does not existh

h h

h

0

sin sin( ) sin lim | | 0

| || |h

h hf x x h

hh

20

sinlim does not existh

h

h

2/3

2/3

0

( )( ) lim 0

| |h

hf x x

h

2/3

20

( )lim does not existh

h

h

0

| |( ) | | lim 0

| |h

hf x x

h

20

| |lim does not existh

h

h

6.(1, 2, 4) Clearly 1 TK K KP P P

As well as TA A where

2 1 3

1 0 2

3 2 1

A

Given 6

1

TK K

k

X P AP

6

1

T T TK K

k

X P A P 6

1

T TK K

k

X P AP [ ]TAs A A

TX X

Let

1

1

1

B

6

1

TK K

K

XB P A P B6

1

K KK

P AP B [ ]TK KAs P P

Vidyamandir Classes


6

1

Kk

XB P AB 1 2 6

6

... 3

6

P P P 30 30 B

1

( 30 ) 1 0

1

X I

As ( 30 ) 0 X I B has a non trivial solution

1

1 det( 30 ) 0

1

B X I

30X I is non invertible ( ) 3 6 18 rT X

7.(1, 2) 1det det( )R PQP

det detR Q

2 2 2

det 0 4 0 0 4 0 0 4 0

0 0 5 1 5 0 0 1

x x x x x x

R

x x x x

2

det 0 4 0 8

5

x x

R x R

x x

As 1

6 3 01

0 3 26

0 0 2

P

22 1

2 3 34 2

2 43 3

3 0 6

x xx

R x x

x x

As for 1x det 0R

So,

0

0 0

0

R

As

1 1

6

R a a

b b

or

1

( 6 ) 0 2, 3

R I a a b

b

PQ QP

1 1 1 2 2 1 1 1

0 2 2 0 4 0 0 4 0 0 2 2

0 0 3 6 6 0 0 3

x x x x

x x x x

Comparing 12a

2 4 2 2 x x x

Vidyamandir Classes


8.(2, 3, 4) 2

sin( ) ; 0

xf x x

x

2

4

cos( ) 2 sin'( )

x x x x

f xx 4

cos ( 2 tan )

x x x x

x 4

cos tan2

2

xx x

xx

In 1

2 , 22

n n the derivative goes from positive to negative. Therefore, it’s a point of maxima.

Similarly, in3

2 1, 22

n n there is a minima.

So, the first minima 1 :y 13

12

y and the first maxima 1 15

: 22

x x and 11tan

2 y

y

and 11tan( )

2 x

x

Since 1 1x y

1 1tan( ) tan( ) x y

1 1tan( ) tan ( 1) x y and 15

2 12

y

Notice the function tan t is increasing in 5

2,2

1 1 1 x y

Generalising: 1 n nx y similarly 1 1 n ny x

Since the function is differentiable, therefore the points of maxima and minima occur alternately.

Therefore, 1 1 n n nx y x

Now 1 1 1 1( ) ( ) n n n n n nx x x y y x using the foregoing proofs every quantity inside these brackets is

greater than 1. Therefore, 1 2 n nx x .

SECTION-2 1.(30) Maximum number of hats used of same colour may be two only.

i.e. RRGGB or BBGGR or BBRRG

Let it be RRGGB

Blue colour hat can be given in 5 ways and other four hats can be given in only two ways.

Total number of ways 3 5 2 30

2.(0.50)

/2

50

3 cos

sin cos

I d

/2

50

sin3

cos sin

I d

/2

40

12 3

cos sin

I d

/2 2

40

sec2 3

1 tan

I d

Let 2tan t

4

0

22 3

(1 )

t dt

It

1

2I

Vidyamandir Classes


3.(422) As A & B are independent events 6 ( ) ( ) ( ) n A B n A n B

6 5 31 2 1

62

6 4 32 1 2

63

64

65

( ) ( ) ( ) ( , )

1 6 1 6

1 3 2 180

2 6 2 15

2 4 3 180

3 6 3 20

4 6 4 15

5 6 5 6

total 422

n A B n A n B A B

C C C

C

C C C

C

C

C

4.(0) 10

1

0

1 7 7sec sec sec

4 12 2 12 2 2

k

k k

10

1

0

1 1sec

7 74cos sin

12 2 12 2

k k k

101

0

1 1sec

72sin

6

k k

10 1

1

0

1 ( 1)sec

2 1 / 2

k

k

1sec (1) 0

5.(6.2) 1 2

1

( 1)2 ( 1)2

2 0

2 4

n n

n n

n nn n n

n

0, 4, 1 4 n n

Now 1

1

0 1 1

n nnk k

k

C C

k n

541

0

316.2

5 5

k

k

C

6.(18) As , &a b c are co-planar

0 a b c Minimum value of ( ( )) c a b c = Minimum value of 2| |c

Now as given ( )

3 2| |

c a b

a b

2 2 2| | 6( 2 4) c

Minimum value of 2| | 18c

SECTION-3 1.(1) 2.(1)

( ) 0f x sin( cos ) 0x

cos ,x n n I cos 1,0,Ix

,2

nX n I

( ) 0 cos(2 sin ) 0 g x x

2 1

sin ,4

nx n I

1 3sin ,

4 4x

Vidyamandir Classes


1 11 3sin , sin ,

4 4

Z n n n I

If '( ) 0 cos( cos ) ( sin ) 0f x x x

cos (2 1) or2

x n x n

1

cos or2

x n x n 1

cos or2

x x n

2

2 ,2 , ,3 3

Y n n n n I

If '( ) 0g x

sin(2 sin ) (2 cos ) 0 x x sin or (2 1)2 2

nx x n

2

1sin 0 or (2 1)

2 2x x n

, (2 1) , ,2 6

W n n n n I

3.(2) 4.(1) Let P & C be the centre of circle 2 3&C C

2MN r OM OP PN

2 12 6r r

As centre of 3C is collinear with

&O P let it be 4

,3

t t

Now as 1C touch internally 3C

3 3 OC r

2 9OC

2 2169

9t t

9 9 12,

5 5 5t C

The equation of common chord of 1 2&C C will be given by 1 2 0S S

: 3 4 9XY x y

Now as XY is chord for 1 & isC ZW for 3 ,C we can find the length of chord by using the formula 2 22 ,r p

where r is radius of circle & p is distinct from centre on the chord.

24 24 6

,5 5

XY ZW

Area of 1 72 6

2 2 5

ZWMZN MN

Area of 1

( )2

ZMW ZW OM OQ 288 6

25

5

4

MZN

ZMW

Slope of tangent at M is 3

4

Equation of tangent on the circle 1C 3 15

4 4

y x and equation of tangent to 2 8x y for same

slope.

3 9 10

(2 )4 16 3

y

2019-SOLUTIONS-JEE ENTRANCE...

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