Vidyamandir Classes - Amazon Web Services

Vidyamandir Classes

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

SOLUTIONS

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

SECTION – 1

1.(3) Mean free path

rmsV

2

Number of moleculesvolume of inter action per unit volume

32

v tRTd v t nvM

23 2M

RT d nv

;

T

Correct answer (3)

2.(3) 1 2 10C C F

2 22 1

11 4 12 2

C C

2 14C C

1 14 10C C F

1 2C F

2 8C F

Equation 1 2

1 2

2 8 1.62 8

C C FC C

Correct answer (3)

3.(3) 0 60ef f ; 0 5e

ff

60 5e

e

ff

; 10 ef cm

4.(1) 27 2 6 191 1.6 10 1 10 1.6 102

v

2 142 10v ; 72 10v

19 7

1227

1.6 10 2 10101.6 10

B

310

2B

]

30.7 10B T Correct answer is (1)

Vidyamandir Classes


5.(1)

2( )m l x kx

2 2m l m x kx

2

2m l x

k m

Correct answer is (1)

6.(4) For x to y ; V T PV RT This means P is constant Also volume is increasing Only graph with pressure constant in process xy is (4). Correct answer is (4)

7.(3) By conservation of angular momentum about hinge

224sin 45

2 12 2l ml lmv m

2 21

2 3 42mvl ml ml

7122 2

v l

122 2 7

vl

3 27v

l


8.(2) 2 2(2 ) (4 )2 cos30 cos30k q k qi id d

2 20 0

4 1 3 4 1 32 24 4

q qi id d

20

3q id


9.(1) In Ruther ford gold foil experiment, angle of scattering is negligible for most of f particle, and very few scatter through large angle and extremely small number retrace its path.

1Y

(Note exactly but close to)


Vidyamandir Classes


10.(3) 2( )

hKE m

1KE

A B

B A

KEKE

1 1.52

A

A

TT 2AT eV

2 1.5 0.5BKE eV

4.5 0.5 4B eV Correct answer is (3)

11.(3)

Potential gradient 51000 1200

PV

6PV V

And 36 100

60 10P

PVRI

12.(3) Magnitude of electric field is constant and the surface is equipotential.

13.(1) 1V

2r rn

1sin2

c ; c = 30°

14.(3) 0(80) Cmg A g

4(79) Cmg A g

4

0

80 1.0179

C

C

15.(2) 2

0 21 rR

0 r R

mg = B

2 34(4 )3Lr dr R

2

2 30 2

41 43L

r r dr RR

4 3 5

2 30 02 2

00

44 43 35

RR

Lr r rr dr RR R

025 L

Vidyamandir Classes


16.(2) First part of figure shown OR gate and Second part of figure shown NOT gate So pY = OR + NOT = NOR gate

Y A B A B

17.(3) d AdBdt dt

4 4(1000 500).(16 4 4 2) 10 105

8 650056 10 56 10

5V

18.(1)

51 12 2

2cm

i jj ir

; 3 74 4cmr i j

19.(Bonus)

0x y z tV kh c G A

2 3 1 2 1 1 1 3 2 2[ ] [ ] [ ] [ ] [ ]x y tML T A ML T LT M L T A 1t ….(1)

2 1x …(2) 2 32 2x y ….(3)

22 3x y …(4) (3) + (4) : 2 1x …(5) (2) + (5) : 2 0 0x x

2 1, 5y No option is correct

20.(4) 223

2Gm

122

1Gm

2

1

3 12 4

mm

; 1

2

16

mm

SECTION – 2

21.(580) For particle 1 210 8 3x t t 1

ˆ(8 6 )V t i

For particle 2, 35 8y t

22

ˆ( 24 )V t j

At 2 1

ˆ ˆ1, 24 , 2t V j V i

21 2 1ˆ ˆ24 2V V V j i

2 221| | (24) (2) 580V

Vidyamandir Classes


22.(1) ˆ ˆ ˆ ˆ(0.1)(3 ) (0.1)(5 ) (0.1)4( ) (0.1) Bi j i j V

ˆ ˆ ˆ ˆ ˆ ˆ3 5 4 4BV j j i j i j

21 1| | 2, (0.1)( 2)2 10B BV K

23.(10)

10 12 15 410

RR

On solving 10 R

24.(60) 1 2

1 1 1( 1)f R R

1R

2 30 R cm

1 1 1(1.5 1)30f

1 0.530f

60 f cm

25.(106.05-106.07)

v

pipe

air

v /2 1v / 2

airpipe

vv2

pipe( 1)v2n

nf

l

pipe1 0

v 3002 2 2

f fl

Vidyamandir Classes


CHEMISTRY SECTION – 1

1.(1) 3 3 2 3 3 3[Co(NH ) (NO ) ] [Ma b ] type If three donor atoms of the same ligands occupy adjacent positions at the corners of an octahedral, we

have the facial (fac) isomer. When the positions are around the meridian of the octahedron, we get the meridional (mer) isomer.

2.(2) Ethyl acetate is polar molecule, so dipole-dipole and London dispersion forces are present in it.

3.(3)

4.(4) When gypsum is heated to 393 K, it forms hemihydrate of calcium sulphate (plaster of paris).

5.(1) Diborane reduces carboxylic acid to 1°-alcohols.

6.(1) 1E reaction proceeds with Carbocation intermediate, therefore greater the stability of carbocation, faster is the rate of reaction.

So rate of

7.(4) +2 618Fe [Ar] 4s°3d

So, it will achieve 53d (stable configuration) after removal of 1 electron from 2Fe .

8.(4)

( 22 8S O ) (rhombic sulphur)

(Number of S — O bonds = 8) (Number of S — S bonds = 8)

9.(2) According to Hardy-Schulze rule,

Coagulation value or flocculation value 1Coagulation power

Example In the Coagulation of the positive sol, the flocculating power is in the order– 4 3 2

6 4 4[Fe(CN) ] PO SO Cl

10.(2) 3CH OH No resonance

In p-Ethoxyphenol, due to +R effect of 2 5—O — C H group, resonance will be less as compared to phenol. So, C — OH bond length will be Phenol < p-ethoxyphenol < methanol

Vidyamandir Classes


11.(4) Ea /RTK Ae …(1) a cat/RT( E )610 K Ae …(2) Dividing equation (2) by (1),

a a cat /RTE (E )610 e a a catE (E )6 ln10RT

a a a catE E (E ) 6 ln10 RT a cat a(E ) E 6 2.303 RT

12.(4) H 2 21 2

1 1v Rn n

As wavelength decrease, the lines in the series converge. For Balmer series, 1n 2 For Balmer series, the lines of longest wavelength corresponds to 2n 3.

13.(2) Glucose exists in two crystalline forms and Glucose does not give the schiff’s test for aldehyde Glucose penta-acetate does not react with hydroxylamine Glucose combines with hydroxylamine to form a monoxime. 14.(3) At a particular temperature as intermolecular forces of attraction increases, vapour presence decrease. So intermolecular forces of attraction of X < Y < Z.

15.(2) The gases that cause green-house effect are 2 4 3—CO , CH , O , chlorofluorocarbon (CFCs) and water vapour.

16.(2) Oxalic acid is a primary standard solution while 2 4H SO is a secondary standard solution.

17.(2)

18.(4) Vapours of the liquid with higher boiling point condense before the vapours of the liquid with lower boiling point. Hence isohexane will be distilled out first. If the difference in boiling points of two liquids is not much, simple distillation cannot be used to separate them.

19.(4)

The correct answer would be 6 2XY,ksp 2 10 M

As only this expression of ksp will satisfy all the points on curve.

20.(4) Correct order of 1st ionization energy will be– Na < Al < Mg < Si

Vidyamandir Classes


SECTION – 2

21.(–0.93) 2 22H O O 4H 4e Re d 1.23V

4cell cell

0.059E E log [H ]4

5 4cell

0.059E 1.23 log(10 )4

200.0591.23 log104

0.0591.23 ( 20)4

1.23 0.2950.935V

22.(4.95 to 4.97)

Concentration in ppm 6mass of solute 10Mass of solution

63

mass of Fe (in gm)10 10100 10 gm

Mass of Fe = 1 gm Molecular weight of 4 2FeSO .7H O 278 gm / mole 56 gm of Fe is present in 278 gm of salt

56 gm of Fe is present in 278 4.9656

gm of salt

23.(3.00)

(Structure of pencillin)

24.(48.00) Work done = Area under the curve

1 (6 10) 62

= 48 J

25.(26.80 to 27.00)

3 6 3 3 3 6 3 30.3 gm x ml of 0.125 M

[Co(NH ) ]Cl 3AgNO 3AgCl [Co(NH ) ](NO )

0.3 x3 0.125267.46 1000

x 26.91 ml

Vidyamandir Classes


MATHEMATICS

SECTION – 1 1.(1) Rolle’s theorem will be applicable if f (b) = f(u)

2 23 4ln ln7 3 7 4

9 163 4

; 12

2 12 12/( ) ln ln7 7

x x xf xx

22

2 2

1217 ( 12)( ) 0712 ( 12)

x xxf xx x x

2 12x or 12c

22 21 2( ) ( 12)

( 12) ( 12)d xf x xdx x x x x

; 2 1( 12)

24 12f

2.(3) 1 12 2 3 3( )

2 2

x x x xf x

Minimum at x = 0 Minimum at x = 0 Hence f (x) is minimum at x = 0

min2 2 1 1( ) 3

2f x

3.(4) Volume of parallelepiped [ ] 1u v w

1 11 1 3 12 1 1

(–2 5 ) 1 2 or 4

2w i j k

cos u w

u w

2 4

2u i j k 4u i j k

2 1 2cos6 6

2 1 4cos18 6

56

76 3

Vidyamandir Classes


4.(1)

21/2

20

3 2lim7 2

x

x

xx

2

2 201 3 2lim 1

7 2exx

x x

2

2 201 4lim

7 2xx

x xe

2

21ee

5.(1)

Product of roots 2 2 45, sum of roots = 2b or

We can say z will lies on 2 2 45x y …(1)

Also, 1 2 10z z x iy

2 2( 1) 40x y

2 2 2 1 40x y x …(2) Solving (1) and (2) 45 + 2x + 1 = 40 x = –3 b = –2x = 6 2 30b b 6.(2) ( ( ))P P Q Q ( (~ ))P P Q Q

(( ~ ) ( )F

P P P Q Q

( )P Q Q (~ ( ))P Q Q ~ ~P Q Q ~ P T = T (tautology)

7.(3)

Q(b, b)

Vidyamandir Classes


Area 1 1( )2 2

OQR b b

b = 1 2

1 :C y ax

y a x for x > 0

22 :C x ay

2xy

a

Area bounded by 1C and 2 2.C Area bounded by 1C and 2C in (0, 1)

12 2

0 0

2a x xa x dx a x dx

a a

2

3/22 2 123 3 3 3

aa a aa

2 4 2

3 3 3a a

a ; 3 4 2a a a

3 2 3( 2) 16a a ; 6 312 4 0a a

8.(3) Mean ( ) 20x 1 2 10 200x x x S.D = 2 Let observation are 1 2 3 10, ,x x x x New observations are 1 2 10, – ,px q px q px q

New mean 1 2 10( ) ( ) ( )10

px q px q px q

1 2 10( ) 1010

p x x x q 200 10

10p q

120 (20)2

p q

20 10p q …(1)

Old S.D. 2 2( ) ( 20)

10 10i ix x x

New S.D. 2(( ) (20 ))

10ipx q p q

2 2( 20) 1

10 2ip x

old S.D.

2( 20)1

2 10ix

; 2 12

p

12

p or 12

20 10p q ; q = 20p – 10

120 10 202

or 120 10 02

Vidyamandir Classes


9.(Bonus)

2 21 1

dy dx

y x

1 1sin ( ) sin ( )y x C 1 32 2

y

3 6 2C C

1 1sin ( ) sin ( )2

y x

12

x

1sin ( )4 2

y Not Possible

10.(2) 3 6 2/3cos

sin (1 sin )x dx

x x 3 4 6 2/3cos

sin sin (cos 1)x dx

x x ec x

5

6 2/3cos cos cot

(cos 1)ec x ecx x dxI

ec x

61 cosec x t

56cos ( cos cot )ec x ecx x dx dt

2/36dtIt

1/336

t 6 1/31 (1 cos )2

e x 6 1/3

21 (1 sin )2 sin

xx

Therefore 21( ) cos2

f x ec x and =3

1 4( /3) 3 22 3

f

11.(1) 3 8 33 1 1

x y z

…(1)

3 8 3a i j k

3p i j k

3 7 63 2 4

x y z

…(2)

3 7 6b i j k

3 2 4q i j k

Shortest distance between lines is ( ).( )a b p qdp q

(6 15 3 ).( 6 15 3 )

6 15 3i j k i j k

i j k

36 225 9

36 225 9

270 3 30

Vidyamandir Classes


12.(2) x + 2y + 3z = 1 3 4 5x y z 4 4 4 x y z

Here 1 2 33 4 5 04 4 4

D

1

1 2 34 5 4 2 4 2(2 2)4 4

D

2

1 1 33 5 8 4 8 4(2 2)4 4

D

3

1 2 13 4 4 2 4 2(2 2)4 4

D

For inconsistent system D = 0 and atleast one of 1 2 3, ,D D D should be non zero. 2 2 0 Now check option

13.(2) Because A and B are two independent events therefore.

1 2( / ) ( ) 13 3

P A B P A

1( / ) ( )3

P A B P A ; 1( / ) ( )3

P A B P A

14.(4) (1, 1), (0,2), ( , )A B P x y Area ( ) 5PAB

1

1 1 1 1 52

0 2 1

x y

1[ ( 3) (1) 1(2)] 52

x y

3 2 10x y ; 3 8x y or –12

Also, 3 4x y By solving we get 2, 3

15.(4) Let a point on the parabola is 2(2 , )t t

23th 3

2ht …(1)

2 2

3tk …(2)

From (1) and (2)

293 2

4hk

Locus of P(h, k) is 29 12 8x y

Vidyamandir Classes


16.(4) 2 2

2 28 8( )

18 8

x x

x xyf x

Using componendo – Dividendo

2

21 2.81 2.8

x

xyy

4 181

x yy

8

14 log1

yxy

18

1 1log ( )4 1

yx f yy

18

1 1( ) log4 1

xf xx

17.(2) 1( ) cos sinf x x x

For 0,2

x

1( ) cos ( sin )f x y x x

1[ cos sin ]x x

1sin sin2

x x

1sin sin2

x x

2

y x x

2 02

y x f(x) is increasing

2 0y ( )f x is increasing

For ,02

x

1( ) cos ( sin )f x x x

1sin sin2

x x

2

x x

( ) 2 02

f x x

( ) 2 0f x

f(x) is increasing ( )f x is decreasing

18.(3) 11 (sin ( ))2

dy d f xdx dx

12 sin ( ( )) , ( 3) 06

y f x C y C

( 3)6

y

Vidyamandir Classes


19.(4) Ellipse 2 2

11/2 1x y

Let 1 cos ,sin2

P

Also P satisfies y = mx

1sin cos2

m

tan2

m …(1)

Equation of normal at P

112 1

cos sin 2

x y

It satisfies 1 ,03 2

and (0, )

1 16cos 2

and 1sin 2

1cos3

sin 2

sin 2tan 6cos 1/3

…(2)

From (1) and (2)

62

m

Also if 1cos3

tan 2 2 6 2 2 ; 23

20.(2) 199

19!9! 10!

a C

2010

20! 20 19!10! 10! 10 9! 10!

b C

2110

21! 21 20 19!11! 10! 11 10 9! 10!

c C

10 11 101 20 21 20a b c

111 2 42a b c

11 22 42a b c

SECTION – 2

21.(8) 2 332 ( 10) 22

x a x a

For real roots 0D

2 33( 10) 4 2 2 02

a a

2 100 20 132 16 0a a a 2 4 32 0a a ( 8)( 4) 0a a

( , 4] [8, )a

least 8a

Vidyamandir Classes


22.(672) Sum of diagonal elements of 2 2 2 2 2 2 2 2 211 12 13 21 22 23 31 32 333TAA a a a a a a a a a

Therefore out of these 9 terms, 6 terms should be zero and remaining 3 terms are either +1 or –1 Number of total possible matrices 9

6 2 2 2C = 672

23.(4) Let 1 1( )P x y

2 21 1 13 10 0y x y …(1)

2 23 10 0y x y Differentiate curve with respect to x

2 6 0dy dyy xdx dx

62 1

dy xdx y

Slope of normal at 11 1

1

(2 1)( )6yP x y

x

Equation of normal at 1 1( , )P x y

11 1

1

(2 1) ( )6yy y x x

x

It satisfies (0, 3/2)

11

3 2 12 6

yy

1 19 6 2 1y y 1 1y From equation 1 2x

Slope of tangent at 11 1

1

6( , ) 42 1

xP x yy

|m| = 4

24.(490) 5 red, 4 black, 3 white There are 4 possible cases (1) 0 red + 4 other 7

4 35C

(2) 1 red + 3 other 5 71 3 175C C

(3) 2 red + 2 other 5 72 2 210C C

(4) 3 red + 1 other 5 73 1 70C C

Total ways = 490

25.(1540) 20

1(1 2 3 )

kk

20

1

1 ( 1)2 k

k k

20 20

2

1 1

1 12 2k k

k k

1 20 21 41 1 20 212 6 2 2

= 1540

Vidyamandir Classes - Amazon Web Services

Documents

Transcript of Vidyamandir Classes - Amazon Web Services