Vidyamandir Classes · 2020. 1. 17. · Barfoed’s test detects monosccharides. It is based on...

Vidyamandir Classes

VMC | JEE Main-2020 1 Solutions |9th January Evening

SOLUTIONS

JEE Main – 2020 | 9th January 2020 (Evening)

PHYSICS

SECTION – 1

1.(1) ( ) ( )

9 90.3

10 5 10 5 30= = = =

+ + +

VI A

R net

2.(Bonus)

1 1 1 280.6324 280.6A B C+ + = = (Rounded off to 1 decimal)



4 4 4 280.691 281A B C+ + = = (Rounded off to 0 decimal)

(In addition or subtraction, the final result should retain as many decimal places as are there in the

number with least decimal places.) No option gives correct relation

3.(3) ( )3

4

10 10 10 100.55

1 13.6 31.84 18.25 18.25340 40 10

314 10

−

−

= = = = = =−

−

VI

X

( ).52cos 314I t

4.(3) Mean free time 2

mean free path 1

32rms

M

V RTd= =

2 21 1 2

2 2 1

40 0.1 2 1001.09

10 0.07 7 49

M d

M d

= = = =

No option matches. But closest is 1.83

5.(1) 22 21 2

1 1 1 = −

Rz

n n

71 1 1 89 9 8 8 1.1 10

1 9 9

= − = = =

R R R

7 91 10 100 10

11.48 1.1 8.8

− − = =

nm

6.(2) 3 32 42

3 3 + = rT r g r d g

( )

32 32 23 2

= − =−

TrT r d g r

d g

7.(3) At the topmost point velocity before collision = cos3

u

by conservation of linear momentum

cos / 3 2 + =mu mu mv

Vidyamandir Classes


3

4=

uv

Height = 2 2sin / 3

2

u

g ;

23

8=

uH

g

Range = 2H

vg

23 2 3

4 8

=

u u

g g

23 3

8=

u

g

8.(3) 1 2

2 1

= =

AFll

Ay A

Energy density = 1

stress strin2

1 2 1

2 1 2

=

E AF L

E A F L

2 1

1 2

1

4

=

A

A; 2

1

1

2=

A

A

9.(3)

2

2

0

2

2

0

+

= =

+

L

cm L

xx a b dx

x dm LX

dm xa b dx

L

2 4 22

2

0

3

2

0

22 4 2 4 42 4

3

3 3 33

+ + + +

= = = =+ + + +

L

cm L

x bx L b a baL bLa a L LL

Xb b a bbx aL L aax

L

3 2

4 3

+ =

+ am

a bX L

a b

10.(3) 1

2

14

1= = = =

e

ee

vGMv n

R v

11.(4) 1

420 6

496 7+= =n

n

f

f

3

6 6 540420

2 2 6 10−= =

T

L L ;

243 3 3 10420 9 10

= =

L L ; ( )

962.1

42= =L m

12.(4) 2

2

= =mv

qvB evBr

2

=mv

eBr

; 0

2 2

= =

e nIReBrv

m m

13.(1) ( )2 + =m l x kx

2

2

=

−

m lx

k m

2

2( )

= x m

k ml k

Vidyamandir Classes


14.(1) =h

mv

0

= + = =

eEv at at t

m

= =h h m

mat meE t

=h

e E t

2

1−=

dy h

dt eE t

15.(3) sin = = MB MB = ( )iAB

= k k = (i A B)

( )

2

22 2

2= =

I MRT

k i R B =

12 2 2

2

4

2

MR

i R B ;

2=

MT

iB

16.(4) B is ⊥ to direction of propagation of e.m wave ( )0 cos2

−= −

i jB B t kr

17.(3) Loss in G.P.E of 1m = Gain in K.E. of blocks

+ Gain in G.P.E of 2m + Gain in rotational K.E. of cylinder

2 2 2

2 1 1 2

1 1 1

2 2 2− = + + amm gh m gh m v m v I

( ) ( ) ( )2 2 2

2 1 1 2

1 1 1

2 2 2− = + + m m gh m R m R I

( ) ( )2 2 22 1 1 21

2− = + + m m gh m R m R I ;

( )

( )2 1

21 2

2 −=

+ +

m m gh

m m R I

18.(3) Light flux = 0 1 cos2

− = 0 0

71 0.34

2 4 2

− =

= 0.17

critical angle 17%= = ci

3

sin sin4

= =ci

9 7 2.645

cos 1 0.66116 4 4

= − = = =

19.(4) As acceleration is constant 21

2= +y y yS u t a t

2132 4

2= t ; t = 4s

21

2= +x x xS u t a t =

13 4 6 16

2 + = 12 48+ ; 60m=xS

20.(1) =AQ VC open circuit due to reverse biased diode

/−= = =t RCB

VC VCQ VCe

e e due to forward biased diode

Vidyamandir Classes


SECTION – 2

21.(1818,1819)

1.4 1

4 1 4 1 11 1 2 2 2 1

2

VTV T V T T

V

−

− − = =

2/51(16)T=

2/53 22 2(300)(16) 1818.8T T= = =

22.(750) 15 500 10 = = D D D

nd d d

750 = nm

23.(40) Potential difference across resistors = 12 – 8 = 4V

4 1

400 100I A= =

P.d across each Zener diode 8

42

V= =

Power dissipated 31

(4) 40 10 40100

VI m−

= = = =

24.(–48) 2ˆ ˆ4 ( 1)E xj y j= − +

For ABCD: ( ) 0dA dA K E dA= =

1 0 =

For BCGF : ˆ( )dA dA j=

22ˆ ˆ ˆ[4 ( 1) ].( )xi y j dAi = − +

4 4(3) 12xdA dA A= = =

12(4) 48= =

1 2 48 − = −

25.(40) 100

=−

R l

S l

25

75 3= =

R SR

S

2

= =

l lR

A d ;

2 2

22 2 23

4

= = = =

ll S

R Rd d

100

=

−

R l

S l ;

2

3 100=

−

l

l; 200 2 3− =l l ; 200 5= l ; 40 = l

CHEMISTRY

SECTION – 1

1.(2) The amount of oxygen required by bacteria to break down the inorganic matter present in a certain

volume of a sample of water is called biochemical oxygen demand (BOD)

2.(1) 2 2 2Zn 2NaOH Na ZnO H+ → +

2 2Zn 2HCl ZnCl H+ → +

According to stoichiometry in both the reactions, equal number of mole of 2H are evolved

Vidyamandir Classes


3.(1) ( ) 33 S 3S

Ca OH Ca 3OH+ −→ +

( )( )3

spK s 3s=

31 46 10 27s− − = ;

1

4316S 1027

− =

OH 3s− =

=

1

43163 1027

−

= ( )1

31 418 10 M−

4.(4) In Benzene, total six 2sp hybrid carbon atoms are present. Each carbon atom has 3 2sp hybrid orbitals.

Therefore, total 2sp hybrid orbital are 18 in Benzene.

5.(2)

6.(4) dq

ST

= ; T

0

ncdTS

T=

7.(4) All carbohydrates – Monosaccharides, disaccharides, and polysaccharides should give a positive

reaction.

Barfoed’s test detects monosccharides. It is based on reduction of copper (II) acetate to copper (I) oxide

which forms brick red precipitate.

Biuret test detects presence of peptide bonds. Copper (II) ion forms mauve colored complexes in an

alkaline solution

8.(3)

In this reaction, major product is chiral

9.(1) 3 3 3 3 4 3 3 6 3B N H Cl LiBH B N H LiCl BCl+ → + +

(A)

( )3 3 3 3 3 3 3 3 3B N H Cl 3MeMgBr B N H CH 3MgBrCl+ → +

(A) (C)

10.(3) Resonance form of 2Cl CH CH NO− = − is more stable than resonance form of any other given

compounds. Hence double bond character in C Cl− bond is maximum and bond length is minimum

Vidyamandir Classes


11.(2)

12.(3) Distilled water show least conductivity due to less number of ions to flow in the solution

13.(4) Complex (I) 2Cr + weak field ligand

32gt

1eg

s 24 B.M =

Complex (II) 2Fe + strong field ligand

6 02gt eg ; s 0 =

Complex (III) 3Fe + strong field ligand

5 02gt eg ; s 3 B.M =

Complex (IV) 2CO + weak field ligand

5 22gt eg

s 15 B.M =

14.(1) eqP 11

k 1.83 2R 6

= = =

15.(3)

Vidyamandir Classes


16.(3)

17.(3) Statements (a), (c) & (d) are correct.

Option (a): Because of small size lithium has very high hydration energy.

Option (b): LiCl is soluble in pyridine because of covalent character. (Incorrect statement)

Option (c): Lithium unlike other alkali metals forms no ethynide on reaction with ethyne.

Option (d): Lithium and magnesium react slowly with water because of high enthalpy of atomisation.

18.(1) Theory based

19.(1) Basic strength depends upon availability of lone pairs. Greater the resonance of lone pairs lesser is the

basic strength.

20.(4) According to given data of I.E, the element must belong to group I and is monovalent & form hydroxide

of type M(OH)

( ) 21 mole

1mole

M OH HCl MCl H O+ → +

2 4 2 4 21 mole 1

mole2

2MOH H SO M SO H O+ → +

SECTION – 2

21.(66.66 to 66.67)

A 12 4

% carbon 100 66.6712 4 8 16

= =

+ +

22.( 3.98 to 3.99)

2

1 1 2

k Ea 1 1ln

k R T T

= −

60 Ea 100

ln40 8.314 400 300

=

3

Ea ln 8.314 12002

=

= 3984 J/mol = 3.984 kJ/mol

Vidyamandir Classes


23.(10) 3

610.3 10ppm 101030

−= = 10

24.(12) Chromate ion

Dichromate ion

Total number of Cr & O bonds is 12.

25.(2.18) fk 2.0=

m = 0.5 m

f fT k m = = 0.5 2

Tinital = 272 K

n = 0.1 mol

V = 1 3atm

gasnRT 0.1 0.08 272

Pv 1

= = = 2.176 atm

After releasing piston 1 1 2 2P V P V= , 22.176 1 1 V =

32V 2.176 dm= 32.18dm

MATHEMATICS

SECTION – 1

1.(4) ( )

2 1

3 2

4 3

+ + +

= + + +

+ + +

x a x x

f x x b x x

x c x x

Apply 1 1 3 22 + −R R R R

( )

1 0 0

3 2

4 3

= + + +

+ + +

f x x b x x

x c x x

( ) ( )( )2

3 2 4 1= + − + + =x x x

( ) ( )1 50 1= =f x f

2.(Bonus) 2sin sin 2x = −

2cos cos2y = −

2cos 2 2

2sin 2sin 2

dxcos

d

dy

d

= + −

= − +

sin 2 sin 2cos(3 / 2)sin / 2

cos cos2 2sin(3 / 2)sin( / 2)

dy

dx

−= =

−

3cot

2

dy

dx

=

Vidyamandir Classes


22

2

3 3cos .

2 2

d y dec

dxdx

= −

2

2

3 1 31

2 4 8x

d y

dx=

− = − =

3.(1)

1616

1

1

cos sin

−

+

=

r r

r r

xT C

x

Put 16 2 0 8− = =r r

9T is independent of x

( )

816 16

9 8 88 8 8

1 2

sin cos sin 2= =

T C C

When 16 81 8, , 2 at8 4 4

= =

L C

When 8

162 8 8

2, , at

16 8 81

2

= =

L C = 16 8 48 2 . 2C

16 8 4

82

16 81 8

2 .216

2= =

CL

L C

4.(1) ( )21 1, 2 2 , 42 2

− − =

t t t

Other end of focal chord is at t = 2

( ) ( )( ) ( )22 2 , 4 2 8, 8B B equation of tangent at B is

( )2

2 2 2 2 8= + − =y x y x

5.(1) 2sec

sec2 tan 2

+ d =

2

2

2 2

sec

1 tan 2 tan

1 tan 1 tan

+ +

− −

d = ( )2sec 1 tan

1 tan

−

+ d

Put tan = t

2sec =d dt

1 2

11 1

− = − +

+ +

tdt dt

t t = ( )2log 1− + + +t t C = ( )tan 2log 1 tan− + + +C

1 = − and ( ) 1 tan = + f

6.(4) Refer to the figure

Required area = Are of trapezium PRQS

3

22

1

2

1

2

− −

x dx

=

33 2

1

2

1 3 1 1 3 1 11

2 2 2 2 3 2

− + − − −

x = 3 1

4 3−

Vidyamandir Classes


7.(3) Put =y vx

= +dy dv

v xdx dx

2

2 2 2 21+ = =

+ +

xdv vx vv

dx x v x v

2

3

1 1+ −=

vdv dx

xv

3

1 1 1 + = −

dv dx

v xv

3

1 1 1− + =

dv dxv xv

2

1 1log log

2

−+ = − +e ev x C

v

2

2log

2

−= − +e

xy C

y

Put x = 1, y = 1

we get 1

2= −C

2 2 12 log

2

− = − −

ex y y ( )

2 2 1 2log= + ex y y

Put y = e

( )2 2 3=x e 3= x e 3=x e

8.(3) ( ) ( )2 =F x x g x

( ) ( )1 1 0 = =F g …(i) ( ) ( )1

1

1 0

= =

g f t dt

Now, ( ) ( ) ( )22 = +F x xg x x f x

( )1 0 3 3 0 = + = F

F(x) has a local minima at x = 1

9.(3) ( ) 26 5 1= + =P x K K

1

1,6

= −K

1=−K (rejected) ( )( ), 0P x 1

6=K

Now ( ) 22 5 3 = +P x K K = 3 5 23

6 36 36+ =

10.(4) ( ) ; 2, 2= −A x x

( ) ; , 1 5,= − − B x x

( 2, 1 = − −A B

( ) ), 2 5, = − A B

( )1, 2− = −A B

( ), 2 5,− = − − B A

11.(3) ( )( ) =f g x x ( )( ) ( ) 1 =f g x g x

Putting x = a, we get ( )( ) ( ) 1 =f g a g a

( ) 5 1 =f b ( )1

5 =f b

Vidyamandir Classes


12.(1) Let G.P is a, ar, 2ar , ………

100

2 4 2002 1

1

...... 200+=

= + + = nn

a ar ar ar

( )200

2

2

1200

1

−=

−

rar

r …(i)

100

3 1992

1

.... 100=

= + + = nn

a ar ar ar

( )200

2

1100

1

−=

−

rar

r …(ii)

Dividing (i) & (ii), we get

r = 2

adding we get ,

2 3 200 201... 300+ + + =a a a a

( )1 2 200... 300+ + =r a a a 200

1

300 300150

2=

= = = nn

ar

13.(1) We have

7 6 2

3 4 2

1 2 6

−

− −

= ( ) ( ) ( )7 20 6 20 2 10 0− + + =

so infinite solution exists

Now, Equation (i) + 3 eq (iii), we get

10 20 0− =x z x = 2z

14.(2) ( )

2

2

1cos

1 tan= =

− − x …(i)

2

2 2

1 1 1sin

1 cos sin= = =

− y

y

By (1) & (2)

21

1 sin− = = xy

( )1 1− =y x

15.(4) 0

4lim→

=

xx A

x

0

4 4lim→

− = x

x Ax x

0

4lim 4→

− =

xx x A

x 4 0 4− = =A A

Now ( ) ( )2 sin = f x x x is discontinuous when 2x is integer but x is not integer

At 5 9 3= + = = x A continuous

4 2= = = x A continuous

21 25 5= + = = x A continuous

1 5= + = x A discontinuous

Vidyamandir Classes


16.(1) Let , ,= + z x iy x y R

4+ =x y

( ) ( )2 2

= +z x y = 2 2+x y = ( )

22+ −x y x y

Now by A.M, G.M inequality

2

+

x yx y

2 x y 4x y

0 4 x y

8 2 0− − x y ( )2

8 16 2 16− + + − x y x y 2 2 4 z

z can’t be equal to 7

17.(Bonus)

10

(10)! 10! 10! 4!(4)! 4!

2!3!5! 2!3!5! 2!2!2!3!3!2!

4

+ +

15

17 945

2

=

None option is correct.

18.(2) 22 52 , = = =

b b

a a a

2

2

5=

b

aa

2 5=b a …(i)

2+ = b …(ii)

10=− …(iii)

=b

a is a root of 2 2 10 0− − =x bx

2 2 22 10 0− − =b ab a

2 25 10 10 0− − =a a a

21 5,

4 4= =a b

( ) ( )22 2 2 + = + − = ( )

22 20+b = 24 20 25+ =b

19.(1) 4 2

23 3

= =b b

We have, 2 2 2= +y mx a m b

Is equation of tangent of slope m comparing with 4

6 3= − +

xy

2 2 21 16,

6 9= − + =m a m b

2 4 16

36 3 9+ =

a

2 16=a 2

21= −

be

a =

4 111

3 16 12− =

Vidyamandir Classes


20.(3)

p q q p q ( )→ p p q

T T F F F

T F T T T

F T F F T

F F T F T

SECTION – 2

21.(3) P is on line 1 ; ( )1, 3, 1− −A

P is on line 2; ( )3, 2, 1− −B

2 5 2 ;

2 4 3

= − −

i j k

AB DR of line 1 < 2, 4, 3 >

23 10 2= + − −n i j k

eqn of plane : ( ) ( ) ( )23 1 10 3 2 1 0+ − − − + =x y z

23 10 2 51 0− − + =x y z

Other plane is 23 10 2 48 0− − + =x y z

disc between them = 3

633 K = 3

22.(30) . 10=b c

cos 103

=b c

15. 10

2 =c 4 =c

Also, ( ). 0 =a b c

( ) ( ) sin2

= a b c a b c = 3 . 4sin 1 30

3

=b

23.(36) The circles are ( )22 4+ − =x y k & ( )

2 23 1− + =x y

Towards each other if 1 2 1 2= C C r r

Where ( ) ( )1 20, 4 , 3, 0C C

1 2 1 or 1= + −C C k k ( )5 1 or 1 5= + − =k k 16 or 36= =k k Maximum value of k = 36

24.(51) ( )25 25 25

25 25 25

0 0 0

4 1 4= = =

+ = + r r rr r r

r C r C C

= 25

24 251

0

4 25 2−=

+ rr

C = 24 25100 2 2+ = ( )25 252 50 1 51.2+ = 51=k

25.(14) First common term = 23

Common difference = 7 4 28 =

Last term 407

( )23 1 28 407+ − n ( )28 1 384− n

13.71 1 +n 14.71n n = 14

Vidyamandir Classes · 2020. 1. 17. · Barfoed’s test detects monosccharides. It is based on...

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Transcript of Vidyamandir Classes · 2020. 1. 17. · Barfoed’s test detects monosccharides. It is based on...