JEE Main – 2020 | 8th January 2020 (Evening Shift)...

Vidyamandir Classes

VMC | JEE Main-2020 1 Solutions |8th January Evening

SOLUTIONS

JEE Main – 2020 | 8th January 2020 (Evening Shift) PHYSICS

SECTION – 1

1.(4) 3 2 1R R i.e. 4 32 1 0R R

1R can’t be solution so,

Remaining part is 3 2 1 1 0R R R R

i.e. R must satisfy 3 2 1 0R R R

2.(2) 00 2h

mv

00 0

eE tV V i V j km

02 2 2

2 20

12

hm v e E t

m v

3.(1) 10 1

100mA

R

.

9 9R . K 4.(1)

22 2

0 0 0 02

0

1 122

a adx a a aa a a ac lnd x d d d dd

5.(2) It is uniform circular motion. 6.(4) 2pI I I I cos

2

8 4.

12 12pI I

4mI I

112 0 85

2Ratio .

Vidyamandir Classes


7.(3) T V

5

32

2 06 10 5 15 104

.T . N

8.(3)

9.(4) 2

242T g

g T

0.1 12 225 50

g Tg t T

100 0.4 4 4.4%gg

10.(4) 2 47 8 75 1010

K.E. mv . J

So, option is (4)

11.(3)

1

1 2

2 5 50 15 2 5 50 4

v

D

F g. .F g g .

12.(4) 1 t /i eR

0

20

11 1t / /E E E E ELQ e dt eR R R e R e eR

13.(1)

E i

15E BC

14. (4) 32

1 1 1 1 12

2 2 22 2

kQ / R R Q RR Q RkQ / R

21 1 1

22 2

kQ / R RRkQ / R

15.(2) 2 2V ax

Vidyamandir Classes


16.(2) Total time to land (using C.O.M)

22 1 2 62 2 2 2

ghh h.T gT Tg

Time to collide 1

2 2h ht

ggh

So, time for combined mass to land 2 6 1 2 12 2 6 3

2 22 2 2 4T t

17.(4) 3 52 132 2

3 6v

R Rn. n RC mixn

196pRC mix ;

1913

Y

18.(4) fm

x f

f: mag of focal length

m = mag of magnification.

19.(2) 1

10H L

H

Q QQ

1 100

10 HH

W QQ

10H LQ Q

100 10 90L LQ Q

20.(4) 0 0 02 1

4 2 22 2outI I I

BR R R

SECTION – 2

21.(8.00) 211002

gt

21 181 ( )2 2

g t 10 59 0.5

t tt

2200 200 8

25g

t

22.(486) For Balmer series, Transition happen to 2n from higher orbits. For transition 3 2n to n

3 21

hcE E

2 2

1 113 66561Å3 2

hc. ……..(i)

For transition 4n to 2n

4 22

hcE E

2 22

1 113 64 2

hc.

…….(ii)

i / ii

2

1 1209 4

1 16561 2716 4

220 6561 4860Å 48627

nm

Vidyamandir Classes


23.(50) Let containers have temperature 1 2T ,T and 3T C respectively. For case: 1 1 21 2 0w w w wc T T c T T

1 23 2 0T T T

1 22 180 60T T ....... i common temperature T

For case II

2 31 2 0w w w wc T T c T T

2 33 2 0T T T

2 32 90 30T T common temperature T in this csse …….(ii)

For case III:

1 32 1 0w w wc T T c T T

1 33 2 0T T T

1 32 180 60T T ......... iii common temperature T inthis case

For Case: -IV

1 2 31 1 1 0w w w w w wc T c T c T

1 2 33 0T T T ; 1 2 33

T T T

1 2 33 450i ii iii T T T C 1 2 3 450 503 9

T T T C

24.(16) From energy conservation i i f fU K U K

2 21 110 2 2

e ei f

GM m GM mm v mvR R

2 22 2 2 22 1 1 91 1 11 2 1210 10 10

ef i e i

GMv v v v .R

256 896 16 02fv . . km / s

25.(30) Current through the batteries;

10 10 20

30 3020 5 2530 30

iR RR R

P.d. across 20 internal resistance battery;

110 20 02

i i Or 20 1

30 22530

RR

3040 2530

RR

30 15 30R R ; 30R

Vidyamandir Classes


CHEMISTRY SECTION – 1

1.(4)

2.(2) AgBr shows both Frenkel and Schottky defects.

3.(1) 2 3

2

FeO SiO FeSiO

1FeO Fe O2

These reactions don’t occur in the blast furnace during extraction of Fe

4.(4) Kjeldahl’s method can’t be used for Nitro compounds.

5.(4) Ea/RTK Ae

aElog k log A2.303RT

; aEslope2.303R

In the given graph, greater the slope, more is the aE Order c a d bE E E E

6.(1) Three isotopes of hydrogen are: Protinum Deuterium Tritium 1

1H 21 H 3

1 H

No. of neutrons in 11H 0

No. of neutrons in 21 H 1

No. of neutrons in 31 H 2

Total no. of neutrons 3

7.(4) Catalytic hydrogenation depends upon the extent of adsorption. Group 7-9 elements exhibit maximum adsorption property The reactants must get adsorbed reasonably strongly on to the catalyst to become active. However, they must not

get adsorbed so strongly that they are immobilized and other reactants are left with no space on the catalyst’s surface for adsorption.

Hence, assertion is true but reason is false.

8.(2)

Vidyamandir Classes


9.(1) has greater boiling point than than because of extensive –H bonding.

10.(3) Greater the bond length, lesser is the bond energy. Order of bond length: C I C Br C Cl C F Order of bond energy: C F C Cl C Br C I

11.(1) More the no. of shells, greater is the atomic radius. Size of Br Size of Cl While moving along the period, the atomic size decreases due to increase in effective nuclear charge.

Size: C O F Overall order: Br Cl C O F

12.(4) Compound Geometry Geometrical isomerism

3 3Pt NH Cl

Square planar ×

3 5Pt NH Cl

Octahedral ×

3 22Pt NH ClNO Square Planar

23 2Pt NH ClBr

Octahedral

13.(3) Bohr’s radius 2

0naZ

For 2Li , n 2and z 3 04av3

14(4) In 4Ni CO , CO is SFL pairing occurs ; 0

Similarly for 24Ni CN

and 2 3 2PdCl PPh , under the effect of SFL, all e s are paired, hence 0

In 2

2 2 26 6Ni H O Cl or Ni H O

,

2H O is a weak field ligand,

2 8Ni : d configuration

Vidyamandir Classes


15.(1)

16.(4)

17.(3) 2H O H OH Dissociation of 2H O is an endothermic reaction.

On increasing temperature, H ion concentration increases. pH decreases.

Vidyamandir Classes


18.(3)

Basicity of 3 2H PO 1 (one replacable H ion per molecule)

19.(2)

Maltose is a disaccharide of two D glucose monomers.

20.(2) 3 23

2

Ag NHOZN/H OX A B

SECTION – 2 21.(20.00)

No. of 90° angles 8 No. of 90°angles 6 No. of 120° angles 0 No. of 120°angles 3

No. of 180° angles 2 No. of 180°angles 1 Total 20

Vidyamandir Classes


22(13.00) A CH CH

Max no of atoms in one plane 13

23.(2130.00) 3 s s 2 g s sNaClO Fe O NaCl FeO

No. of moles of 2O no. of moles of 3NaClO

No. of moles of 2PV 1 492O n 20RT 0.082 300

Molar mass of 3NaClO 23 35.5 48 106.5

Mass of 3NaClO required 20 106.5 2130g

Ans. 2130

24.(2.15) 2 2Sn s | Sn aq,1M || Pb aq,1M | Pb s

Overall cell reaction 2 2

s aq aq sSn Pb Sn Pb

2

2

Sn0.06E E log2 Pb

E 0.13 0.14 0.01V

At equilibrium, E 0

2

2

Sn0 0.01 0.03log

Pb

2

2

Sn0.01 0.03log

Pb

2

2

Sn1 log3 Pb

2

1/32

Sn10 2.15

Pb

25.(6.25) vU nC T

v500 4 C 500 300

1 1v

5000C 6.25JK mol4 200

Vidyamandir Classes


MATHEMATICS

SECTION – 1

1.(4)

2

2

1 21

22 3

1

xx ,

xf xx

, x ,x

2

2

2

2

(1 ) 0 (1, 2)(1 )

( )2(1 ) 0 [2,3)1

x xx

f xx xx

So, the function is decreasing in (1, 2) and [2, 3) therefore, range is 2 1 3 4

5 2 4 5, ,

2.(4 ) As given 10 20020

i

i

xx x

Similarly,

2 21

22 21100 4 2080

20 20

ix xx x

Final mean after correction 200 9 11 202

20 20

Final variance after correction

22 222080 9 11 202 2120

10 1 3 9920 20 20

. .

3.(2)

0

0

10x

x

t sin t dt

limx

;

0

100

1x

x sin xlim

4.(3) 0b c b a b c a

c a b

; a.c a.a b.a

; 0 6 4 ; 3

2

3

2c a i j k ; 3

2c.b a.b i j k i j k ;

34 3

2c.b

94

2c.b

; 9 1

42 2

c.b

5.(3) Given curve is 2 22 3 0x xy y

2 23 3 0x xy xy y

3 3 0x x y y x y

3 0x y x y

Equation of line to 0y x and passing through 2 2,

2 1 2 4y x x y

Its distance from origin 0 0 4

2 22

Vidyamandir Classes


6.(2) Given 2

3 21 2 9 12 4

dxI

x x x

Say 3 22 9 12 4f x x x x

26 18 12f x x x

1 9 2 8f and f

f x decreases from 1 2x to x

So 3 2

1

2 9 12 4x x x

Increases from 1 2x ,to x

3 2

1 1 1

3 2 22 9 12 4x x x

2

3 21

1

3 2 22 9 12 4

dx dxdx

x x x

So 2 1 1

9 8I ,

7.(4) 2 2 1 09 4 0 1

A ,I

0A I characteristic equation

2 2

09 4

2 4 18 0

2 6 8 18 0

2 6 10 0

2 6 10 0A A I

16 10 0A I A

110 6A A I

8.(1) w ; clearly 3 21 and1 0

100 100

2 3

0 01 101k k

k ka w w , b b

202

2

11

1

wa w

w

2021 1

11 1

w wa

w w

Hence equation having roots 1 and 101

2 102 101 0x x

Vidyamandir Classes


9.(Bonus) For constant function Which is continuous and differentiable in [0, 1] Option 2, 3, 4 are false

( ) (1) ( )1

f c f f dC

by LMVT ( , 1)d c

and we can’t compare ( ) and ( )f d f c None option is correct

10.(1) Given equation is 3 3 1 2 3 1 3 2x x x x

Let 3 0x t ,t

2 2 1 2t t t t

2 2y t t

1 2y t t

Point of intersections of the graphs of these two functions will be solution (for 0t ) 1B will intersect at only one point

2B will not intersect

For 3B

2 2 1 2t t t t

2 2 2 3t t t

2 3 5 0t t Roots are imaginary So solution set is a singleton

11.(3) 2 2 3 2x, y R : x y x

2 3 2x x

2 2 3 0x x 2 3 3 0x x x

3 1x ,x

Required area 11 1 3

2 2 2

3 3 3

3 2 3 2 33x

x x dx x x dx x x

1

3 1 9 9 93

32

3

12.(1) Given system of equations 2 2 5x y z

2 3 5 8x y z

4 6 10x y z

2 22 3 5 04 6

218 5 2 20 12 2 2 12 0

2 6 16 0

Vidyamandir Classes


2 8and

Now, 1

5 2 28 3 5

10 2 6D

1 0D

Thus for 2 , given set of equations has no solution.

13.(2) Given family of curves is 2 4x b y b , b R .(i) Differentiating both sides w.r.t x

2 42

dy xx b b

dx dy dx ……..(ii)

On substituting the value of ‘b’ from (ii) to (i)

2 42 2

x xx y

dy dydx dx

We get 2

2dy dy

x y xdx dx

14.(2) Mid point of point and its image is 2 1 4

3 3 3, ,

Direction ratios of the normal 10 10 10

3 3 3, ,

So, the equation of plane is

10 2 10 1 10 4

03 3 3 3 3 3

x y z

1x y z

1 11, , ratio for equation of the plane

15.(3) 1

2P A B

2

25

P A P B P A B ……..(i)

1

2P A P B P A B …….(ii)

(i) – (ii)

2 1

5 2P A B

1

10P A B

1

10P A B

16.(1) ~ p q p q

~ ~ p q ~ p ~ q

Vidyamandir Classes


17.(4) Let the equation of the hyperbola is 2 2

2 136

x y

b

Since it passes through 10 16,

22

100 2561 144

36b

b

So the equation of the hyperbola is 2 2

136 144

x y

Equation of normal 36 14410 36 16 144

x y

/ /

2 5 100x y

18.(2) 1

920

a d ……….(i)

1

1910

a d …….(ii)

1 1

1010 20

d ; 1

1020

d ; 1

200d

Now 9 1

200 20a ;

1 9

20 200a ;

1

200a

200200 2 1 201 1

199 100 1002 200 200 200 2

S

19.(3) 1 2L x y

Since y mx c is to 2x y So y mx c becomes 0y x c y x c

Now 0y x c is tangent to 2 23 1x y

3 0

1 3 22

cc

2 6 9 2c c ; 2 6 7 0c c

20.(4) 6 6 2 32 2 6 6 6 4 2 6 2 2 6 2

0 2 4 61 1 2 1 1 1x x x x C x C x x C x x C x

6 6 4 6 4 2 6 4 22 15 15 15 30 15 6 3 1x x x x x x x x x

6 4 2 6 4 22 32 48 18 1 16 96 36 2x x x x x x

96 36 96 36 132; ;

SECTION – 2 21.(2454) Given word is EXAMINATION 2A, 2I, 2N, 1E, 1X, 1M, 1T, 1O

Case (i) two pairs 32

4

2 2

!C

! !

Case (ii) one pair two diff 3 71 2

4

2

!C C

!

Case (iii) all diff 84 4C !

Total no of words 18 756 1680 2454

Vidyamandir Classes


22.(3) Let 3 2 23 2f x ax bx cx d , f x ax bx c ; 6 2f x ax b 10a b c d …(i)

6a b c d …….(ii) 3 2 0a b c ……(iii) 6 2 0a b ….(iv) So, 1 3 9 5a ,b , c ,d

Also, for local minima 0f x 23 6 9 0x x 1 3x , So local minima at 3x

23(1) 2 1 1 2 1

01 2 7 2 10 2

sin cosand , ,

cos

2

2 1

71 2 1

sin

cos

1

07 2

sin,

cos

;

1

7tan

Now 21 1 2 1

2 10

sin

10

10 2sin , ,

110

sin , 1

3tan ;

2 3 2 5 32 1 8 41 3

9 9

/ /tan

Now

1 32 7 42 11 31 2 1

7 4

tan tantan

tan ,tan

24.(504)

7

1

1 2 1

4n

n n n

7

3 2

1

12 3

4 nn n n

; 21 7 8 7 8 15 7 8

2 34 2 6 2

25.(0.5) Area of 4OPQ

Slope of OP 2

t

2

2

0 0 11 1

0 1 42 4

1 11

4 2

t

t t

318

8t

3 64 4t t So 1

2m

JEE Main – 2020 | 8th January 2020 (Evening Shift)...

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