(physics, chemistry and mathematics)cv v cc = + c15f2 = μ 13. a particle is moving 5 times as fast...

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Answers & Solutions

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JEE (MAIN)-2020 (Online) Phase-2

Important Instructions :

1. The test is of 3 hours duration.

2. The Test Booklet consists of 75 questions. The maximum marks are 300.

3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics

having 25 questions in each part of equal weightage. Each part has two sections.

(i) Section-I : This section contains 20 multiple choice questions which have only one correct answer. Each question

carries 4 marks for correct answer and –1 mark for wrong answer.

(ii) Section-II : This section contains 5 SA type questions. The answer to each of the questions is a

numerical value. Each question carries 4 marks for correct answer and there is no negative marking for

wrong answer.

(Physics, Chemistry and Mathematics)

02/09/2020

Evening


2

PART–A : PHYSICS

SECTION - I

Multiple Choice Questions: This section contains 20

multiple choice questions. Each question has 4

choices (1), (2), (3) and (4), out of which ONLY ONE

is correct.

Choose the correct answer :

1. In a plane electromagnetic wave, the directions

of electric field and magnetic field are

represented by k̂ and −ˆ ˆ2i 2 j , respectively.What is the unit vector along direction of

propagation of the wave?

(1) +1 ˆ ˆ(i 2 j)5

(2) +1 ˆˆ( j k)2

(3) +1 ˆ ˆ(i j)2

(4) +1 ˆ ˆ(2i j)5

Answer (3)

Sol. �C = � �E B×

= �(i j)

k2

−×� �

=i j

2

⎛ ⎞+⎜ ⎟⎝ ⎠

� �

2. In a Young’s double slit experiment, 16 fringes

are observed in a certain segment of the

screen when light of a wavelength 700 nm is

used. If the wavelength of light is changed to

400 nm, the number of fringes observed in the

same segment of the screen would be

(1) 28 (2) 24

(3) 30 (4) 18

Answer (1)

Sol. N1λ1 = N

2λ2

16 × 700 = N2 × 400

⇒ N2 = 28

3. When the temperature of a metal wire is

increased from 0°C to 10°C, its length

increases by 0.02%. The percentage change in

its mass density will be closest to

(1) 2.3 (2) 0.06

(3) 0.8 (4) 0.008

Answer (2)

Sol.M V

100 100V V

Δρ Δρ = ⇒ × = ×ρ

100 3 T 100Δρ × = αΔ ×ρ ...(i)

Given 42 10−Δ = ×��

⇒ αΔT = 2 × 10–4

⇒ α = 2 × 10–5

From (i), 100Δρ ×ρ

= 6 × 10–5 × 10 × 100

= 0.06

4. An ideal gas in a closed container is slowly

heated. As its temperature increases, which of

the following statements are true?

(A) the mean free path of the molecules

decreases.

(B) the mean collision time between the

molecules decreases.

(C) the mean free path remains unchanged.

(D) the mean collision time remains unchanged

(1) (C) and (D)

(2) (A) and (D)

(3) (B) and (C)

(4) (A) and (B)

Answer (3)

Sol. Mean free path is independent of temperature

and relaxation time decreases as temperature

increases.

5. A heat engine is involved with exchange of heat

of 1915 J, – 40J, + 125 J and –Q J, during one

cycle achieving an efficiency of 50.0%. The

value of Q is

(1) 980 J (2) 40 J

(3) 400 J (4) 640 J

Answer (1)

Sol. η = − Rg

Q1

Q

1 (40 Q)1

2 (1915 125)

+= −+

⇒ Q = 980 J


3

6. In a hydrogen atom the electron makes a

transition from (n + 1)th level to the nth level.

If n >> 1, the frequency of radiation emitted is

proportional to

(1)1

n(2) 2

1

n

(3) 31

n(4) 4

1

n

Answer (3)

Sol. 0 21

E En

= − ×

⇒ 0 32

E E ( n)n

Δ = − × × Δ

⇒ 0 31

h 2E 1n

ν = × ×

⇒3

1

nν ∝

7. If momentum (P), area (A) and time (T) are taken

to be the fundamental quantities then the

dimensional formula for energy is

(1)⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

1

–12P AT (2) [P2AT–2]

(3)⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

1

–12PA T (4) [PA–1T–2]

Answer (3)

Sol. Energy = Force × Distance

⇒ P[Energy] AT

= ×

= PT–1A1/2

8. The displacement time graph of a particle

executing S.H.M is given in figure (sketch is

schematic and not to scale)

OT

2T

4

3T

4

5T

4

T

4

time (s)

displacement

Which of the following statements is/are true

for this motion?

(A) The force is zero at =3T

t4

(B) The acceleration is maximum at t = T

(C) The speed is maximum at =T

t4

(D) The P.E. is equal to K.E. of the oscillation at

= Tt2

(1) (B), (C) and (D) (2) (A) and (D)

(3) (A), (B) and (C) (4) (A), (B) and (D)

Answer (3)

Sol. At 3T

t4

= , particle is at mean position

⇒ Force = 0

At t = T, particle is at extreme position

⇒ Acceleration is maximum

At T

t4

= , particle is at mean position

⇒ Speed is maximum

9. The figure shows a region of length ‘l’ with a

uniform magnetic field of 0.3 T in it and a proton

entering the region with velocity 4 × 105 ms–1

making an angle 60° with the field. If the proton

completes 10 revolution by the time it cross the

region shown, ‘l’ is close to (mass of proton =

1.67 × 10–27 kg, charge of the proton =

1.6 × 10–19 C)

60°

B

l

(1) 0.22 m (2) 0.11 m

(3) 0.88 m (4) 0.44 m

Answer (4)

Sol. (v0 cosθ) × ΔT = l

and, 2 m

T 10qB

πΔ = ×

∴27

5

19

1 10 2 1.67 104 10 l

2 1.6 10 0.3

−

−× π× ×× × × =

× ×

⇒ l = 0.44 m


4

10. An inductance coil has a reactance of 100 Ω.When an AC signal of frequency 1000 Hz is

applied to the coil, the applied voltage leads the

current by 45°. The self-inductance of the coil is

(1) 6.7 × 10–7 H (2) 5.5 × 10–5 H

(3) 1.1 × 10–1 H (4) 1.1 × 10–2 H

Answer (4)

Sol. Impedance of coil = 100 Ω

L45 R X 50 2φ = ° ⇒ = =

⇒ 50 2 L= ω×

⇒ 250 2L 1.1 10 H2 1000

−= = ×π×

Nearest value given is 1.1 × 10–2 H.

11. A wire carrying current I is bent in the shape

ABCDEFA as shown, where rectangle ABCDA

and ADEFA are perpendicular to each other. If

the sides of the rectangles are of lengths a and

b, then the magnitude and direction of

magnetic moment of the loop ABCDEFA is

(1)⎛ ⎞

+⎜ ⎟⎜ ⎟⎝ ⎠

ˆĵ 2kabl, along

5 5

(2)⎛ ⎞

+⎜ ⎟⎜ ⎟⎝ ⎠

ˆĵ k2 abl, along

2 2

(3)⎛ ⎞

+⎜ ⎟⎜ ⎟⎝ ⎠

ˆĵ kabl, along

2 2

(4)⎛ ⎞

+⎜ ⎟⎜ ⎟⎝ ⎠

ˆĵ 2k2 abl, along

5 5

Answer (2)

Sol.1 2

M M M= +� � �

1ˆM abI j=

�

2ˆM abI k=

�

ˆĵ kM 2abI

2 2

⎛ ⎞= +⎜ ⎟⎜ ⎟

⎝ ⎠

�

12. A 10 μF capacitor is fully charged to a potentialdifference of 50 V. After removing the source

voltage it is connected to an uncharged

capacitor in parallel. Now the potential

difference across them becomes 20 V. The

capacitance of the second capacitor is

(1) 20 μF (2) 10 μF

(3) 15 μF (4) 30 μF

Answer (3)

Sol.1

final

1 2

C VV

C C=

+

2C 15 F= μ

13. A particle is moving 5 times as fast as an

electron. The ratio of the de-Broglie wavelength

of the particle to that of the electron is 1.878 ×

10–4. The mass of the particle is close to

(1) 1.2 × 10–28 kg (2) 9.1 × 10–31 kg

(3) 4.8 × 10–27 kg (4) 9.7 × 10–28 kg

Answer (4)

Sol. 1 2 2

2 1 1

m v

m v

λ=

λ

e

1 –4

mm

5 1.878 10=

× ×

–289.7 10 kg= ×

14. A small point mass carrying some positive

charge on it, is released from the edge of a

table. There is a uniform electric field in this

region in the horizontal direction. Which of the

following options then correctly describe the

trajectory of the mass? (Curves are drawn

schematically and are not to scale).

x

y

E

(1)

x

y

(2)

x

y

(3)

x

y

(4)

x

y


5

Answer (2)

Sol. ˆ ˆF qEi mgj= +�

Since initial velocity is zero. It will move in

straight line.

15. The height ‘h’ at which the weight of a body will

be the same as that at the same depth ‘h’ from

the surface of the earth is (Radius of the earth

is R and effect of the rotation of the earth is

neglected)

(1)5 R -R

2(2)

3 R -R

2

(3)R

2(4)

5R -R

2

Answer (1)

Sol.( )2 2GM GM h

1–RRh R

⎛ ⎞= ⎜ ⎟⎝ ⎠+

( ) ( )23R h R R–h= +

Solving 5 –1

h R2

=

16. A potentiometer wire PQ of 1 m length is

connected to a standard cell E1. Another cell E

2

of emf 1.02 V is connected with a resistance ‘r’

and switch S (as shown in figure). With switch

S open, the null position is obtained at a

distance of 49 cm from Q. The potential

gradient in the potentiometer wire is

Q

G

S

r

J

E2

P

E1

(1) 0.04 V/cm (2) 0.01 V/cm

(3) 0.02 V/cm (4) 0.03 V/cm

Answer (3)

Sol. Potential gradient Potential drop

length=

1.02 V

100 – 49 cm=

V0.02

cm=

17. Two uniform circular discs are rotating

independently in the same direction around

their common axis passing through their

centres. The moment of inertia and angular

velocity of the first disc are 0.1 kg-m2 and 10

rad s–1 respectively while those for the second

one are 0.2 kg-m2 and 5 rad s–1 respectively. At

some instant they get stuck together and start

rotating as a single system about their common

axis with some angular speed. The kinetic

energy of the combined system is

(1)20

J3

(2)5J

3

(3)10

J3

(4)2J

3

Answer (1)

Sol. (I1 + I

2)ω = I

1ω1 + I

2ω2

0.1 10 0.2 5

0.1 0.2

× + ×ω =+

20rad/s

3=

( ) 21 21

K.E. I I2

= + ω

20J

3=

18. A charge Q is distributed over two concentric

conducting thin spherical shells radii r and

R (R > r). If the surface charge densities on the

two shells are equal, the electric potential at

the common centre is

r

R

(1)+

πε +2 20

1 (R r)Q

4 (R r )(2)

+πε +2 20

1 (R 2r)Q

4 2(R r )

(3)+

πε +2 20

1 (R r)Q

4 2(R r )(4)

+πε +2 20

1 (2R r)Q

4 (R r )

Answer (1)

Sol. Let charges on shells be q1 and q

2

q1 + q

2 = Q ....(i)

1 2

2 2

q q

4 r 4 R=

π π....(ii)


6

We get

2 2

1 22 2 2 2

r Rq Q, q Q

r R r R= =

+ +

1 2

0

q q1V

4 r R

⎛ ⎞= +⎜ ⎟πε ⎝ ⎠

( )( )2 20R r1

Q4 R r

+=

πε +

19. In the following digital circuit, what will be the

output at ‘Z’, when the input (A, B) are (1, 0),

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

A

B

Z

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

(3) 1, 0, 1, 1 (4) 0, 1, 0, 0

Answer (2)

Sol. Truth Table

A B W A.B X A B Y W.X Z W Y

1 0 1 1 0 0

0 0 1 0 0 0

1 1 0 1 0 1

0 1 1 1 1 0

= = + = = +

20. A capillary tube made of glass of radius 0.15

mm is dipped vertically in a beaker filled with

methylene iodide (surface tension = 0.05 Nm–1,

density = 667 kg m–3) which rises to height h in

the tube. It is observed that the two tangents

drawn from liquid-glass interfaces (from opp.

sides of the capillary) make an angle of 60°

with one another. Then h is close to (g = 10 ms–2)

(1) 0.049 m (2) 0.087 m

(3) 0.137 m (4) 0.172 m

Answer (2)

Sol. Angle of contact = 30°

30°

60°

2Tcosh

r g

θ=ρ –3

32 0.05

2

0.15 10 667 10

⎛ ⎞× ×⎜ ⎟⎜ ⎟

⎝ ⎠=× × ×

= 0.087 m

SECTION - II

Numerical Value Type Questions: This section

contains 5 questions. The answer to each question is

a NUMERICAL VALUE. For each question, enter the

correct numerical value (in decimal notation,

truncated/roundedoff to the second decimal place;

e.g. 06.25, 07.00, -00.33, -00.30, 30.27, -27.30) using

the mouse and the on-screen virtual numeric keypad

in the place designated to enter the answer.

21. A square shaped hole of side =a

l2

is carved

out at a distance =a

d2

from the centre ‘O’ of

a uniform circular disk of radius a. If the

distance of the centre of mass of the

remaining portion from O is a

–X

, value of X (to

the nearest integer) is __________.

Od

a

=l a /2

Answer (23)

Sol.

Let the density is σ

Then original mass m0 = πa2σ

Remaining mass 2

2 2a 4 1m a a

4 4

⎛ ⎞ π −⎛ ⎞′ = π − σ = σ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

Removed mass 2

am

4= ⋅σ

σ∴ π σ = ⋅σ× + π −2 2

2 a a aa (0) (4 1)r4 2 4

a a ar

2(4 1) 23.13 23

− − −⇒ = = ≈

π −


7

22. A wire of density 9 × 10–3 kg cm–3 is stretched

between two clamps 1 m apart. The resulting

strain in the wire is 4.9 × 10–4. The lowest

frequency of the transverse vibrations in the

wire is (Young’s modulus of wire = × 10Y 9 10Nm–2), (to the nearest integer), __________.

Answer (35)

Sol.m

m Al Al

⎛ ⎞= σ ∴ μ = = σ⎜ ⎟⎝ ⎠

Now,T

v =μ

andT

Y (strain)A

=

⇒T

vA

=σ

⇒ 10 4T

9 10 4.9 10A

−= × × ×

∴10 4

3

9 10 4.9 10v

9 10

−× × ×=×

⇒ v = 70 m/s

∴ 0v 70

f 35 Hz2L 2 1

= = =×

23. A particle of mass m is moving along the x-axis

with initial velocity û i . It collides elastically

with a particle of mass 10 m at rest and then

moves with half its initial kinetic energy

(see figure). If θ = θ1 2

sin nsin then value of n

is _________.m

10 m

θ2

θ1

10 mûim

Answer (10)

Sol. m

θ2

10 m

ûim x

y

θ1

Since energy of m reduced by half ∴ 1u

u2

=

and

221 1

1 1 u u10 m v m v

2 2 2 20× ⋅ = ⇒ =

Now, momentum in y direction will remain

conserved

∴ mu1sinθ

1 = 10mv

1sinθ

2

⇒ 1 2u u

sin 10 sin2 20

θ = θ

⇒1 2sin 10 sinθ = θ

24. An ideal cell of emf 10 V is connected in circuit

shown in figure. Each resistance is 2 Ω. Thepotential difference (in V) across the capacitor

when it is fully charged is ______.

10 V

R1

C

R5

R4

R3

R2

Answer (8)

Sol.

2

2

2 Am

2

2

A B

1 Am

3 A

10 V

2

After solving the circuit we got the final current

distribution as shown in the above diagram. So

potential difference between A and B is

0 + (2 × 3) + (2 × 1) = 8 volt

25. A light ray enters a solid glass sphere of

refractive index = 3μ at an angle of incidence60°. The ray is both reflected and refracted at

the farther surface of the sphere. The angle (in

degrees) between the reflected and refracted

rays at this surface is _____.

Answer (90)

Sol.

i′S

TR

rr

r

P60°

Q

O

QR is reflected ray and QS is refracted ray. QT

is tangent and OQ is normal.

Now, μ.sin60° = 3 sinr

⇒3

2sinr2

= ⇒ 1

sinr r 302

= ∴ = °

Yet again 3 sinr = sini′ ∴ i′ = 60°

∴ Angle between QR and QS ⇒ (∠RQS) = 90°


8

PART–B : CHEMISTRY

SECTION - I


multiple choice questions. Each question has 4

choices (1), (2), (3) and (4), out of which ONLY ONE

is correct.


1. Consider the reaction sequence given below:

BrOH

H O2

OH + Br

rate = k[t-BuBr]

...(1)

OH

C H OH2 5 H C2

CH3

CH3

+ HOH + Br

rate = k[t-BuBr][OH ]

...(2)

Which of the following statements is true?

(1) Changing the concentration of base will

have no effect on reaction (2)

(2) Changing the concentration of base will


(3) Changing the base from OH to OR will


(4) Doubling the concentration of base will

double the rate of both the reactions

Answer (2)

Sol. For reaction 1

rate = K [t-BuBr] ...(1)

for reaction 2

rate = K[t-BuBr][OH–] ...(2)

Reaction 1 is independent of concentration of

OH– where as reaction 2 is dependent on

concentration of OH–

Hence changing the concentration of base will

have no effect on reaction 1

2. Two elements A and B have similar chemical

properties. They don’t form solid

hydrogencarbonates, but react with nitrogen to

form nitrides. A and B, respectively, are

(1) Li and Mg

(2) Cs and Ba

(3) Na and Rb

(4) Na and Ca

Answer (1)

Sol. – Solid hydrogencarbonates are not formed

by lithium and magnesium.

– 6Li + N2 2Li

3N

– 3Mg + N2 Mg

3N

2

3. The major product obtained from

2

E elimination of 3-bromo-2-fluoropentane is

(1) CH CH – CH – CH = CH 3 2 2

Br

(2) CH – CH = CH – CH – CH3 3

F

(3) CH CH CH = C – F3 2

CH3

(4) CH – CH – C = CH – CH3 2 3

Br

Answer (3)

Sol. Base abstracts most acidic hydrogen and

departure of better leaving group takes place

simultaneously.

CH – CH – CHCHCH3 2 3

Br F

Base

E2

CH – CH – CH = C – CH3 2 3

F

4. Cast iron is used for the manufacture of

(1) wrought iron, pig iron and steel

(2) pig iron, scrap iron and steel

(3) wrought iron and pig iron

(4) wrought iron and steel

Answer (4)

Sol. Cast iron is used for the manufacture of

wrought iron and steel.

5. Three elements X, Y and Z are in the 3rd period

of the periodic table. The oxides of X, Y and Z,

respectively, are basic, amphoteric and acidic.

The correct order of the atomic numbers of X,

Y and Z is

(1) X < Z < Y

(2) Y < X < Z

(3) Z < Y < X

(4) X < Y < Z

Answer (4)


9

Sol. In periodic table in a period on moving from left

to right as atomic number increases the nature

of oxide follows the order basic, amphoteric

and acidic i.e., oxide of X is basic, oxide of Y

amphoteric oxide of Z is acidic

Hence atomic number follows the order

Z > Y > X

6. Arrange the following labelled hydrogens in

decreasing order of acidity

NO2

C C – H a

COO H b

O – H C

d H – O

(1) b > c > d > a

(2) b > a > c > d

(3) c > b > d > a

(4) c > b > a > d

Answer (1)

Sol. b > c > d > a

7. The number of subshells associated with n = 4

and m = –2 quantum numbers is

(1) 2

(2) 8

(3) 4

(4) 16

Answer (1)

Sol. 2 subshells are associated with n = 4 and

m = –2.

8. If you spill a chemical toilet cleaning liquid on

your hand, your first aid would be

(1) aqueous NaOH

(2) aqueous NaHCO3

(3) aqueous NH3

(4) vinegar

Answer (2)

Sol. Aqueous NaHCO3 is used as the first aid.

NaHCO3 is basic and not harmful.

9. Simplified absorption spectra of three

complexes ((i), (ii) and (iii)) of Mn+ ion are

provided below; their max

values are marked as

A, B and C respectively. The correct match

between the complexes and their max

values is

AB

C

Absorp

tion

max

Wavelength (nm)

max

max

(i) [M(NCS)6](–6 + n)

(ii) [MF6](–6 + n)

(iii) [M(NH3)

6]n+

(1) A-(i), B-(ii), C-(iii)

(2) A-(ii), B-(iii), C-(i)

(3) A-(ii), B-(i), C-(iii)

(4) A-(iii), B-(i), C-(ii)

Answer (4)

Sol. (max

)C

> (max

)B > (

max)A

From spectrochemical series

F– < NCS– < NH3

A - iii, B - i, C - ii

10. The size of a raw mango shrinks to a much

smaller size when kept in a concentrated salt

solution. Which one of the following processes

can explain this?

(1) Osmosis

(2) Reverse osmosis

(3) Diffusion

(4) Dialysis

Answer (1)

Sol. Osmosis can explain the given process. There

are many phenomena which we observed in

nature or at home. Raw mango shrivel when

pickled in brine.

The solvent molecules will flow through the

membrane from pure solvent to the solution.

This process of flow of the solvent is called

osmosis.


10

11. Match the type of interaction in column A with

the distance dependence of their interaction

energy in column B

A B

(i) ion-ion (a)1

r

(ii) dipole-dipole (b) 21

r

(iii) London dispersion (c) 31

r

(d) 61

r

(1) (I)-(a), (II)-(b), (III)-(d)

(2) (I)-(b), (II)-(d), (III)-(c)

(3) (I)-(a), (II)-(b), (III)-(c)

(4) (I)-(a), (II)-(c), (III)-(d)

Answer (4)

Sol. Ion-ion interaction energy 1

r

Dipole-dipole interaction energy 31

r

London dispersion 61

r

[Reference — NCERT (Page-137)]

12. The one that is not expected to show isomerism

is

(1) [Pt(NH3)

2Cl

2]

(2) [Ni(NH3)

4 (H

2O)

2]2+

(3) [Ni(en)3]2+

(4) [Ni(NH3)

2Cl

2]

Answer (4)

Sol. [Ni(NH3)

2Cl

2]

Ni2+ d8 sp3 hybridisation

Tetrahedral complex

[Ni(NH3)2Cl

2] cannot show isomerism.

[Pt(NH3)2Cl

2] dsp2 shows geometrical

isomerism

[Ni(NH3)4(H

2O)

2]2+ Octahedral show

geometrical isomerism

[Ni(en)3]2+ Octahedral shows optical

isomerism

13. The results given in the below table were

obtained during kinetic studies of the following

reaction

2 A + B C + D

–1 –1 –1 –1

–3

–2

–2

–2

–1

[A] / [B] / Initial rate /Experiment

molL molL molL min

I 0.1 0.1 6.00 10

II 0.1 0.2 2.40 10

III 0.2 0.1 1.20 10

IV X 0.2 7.20 10

V 0.3 Y 2.88 10

X and Y in the given table are respectively

(1) 0.4, 0.3

(2) 0.3, 0.4

(3) 0.4, 0.4

(4) 0.3, 0.3

Answer (2)

Sol. Rate = k[A]a [B]b

6 × 10–3 = k(0.1)a (0.1)b ...(1)

2.4 × 10–3 = k(0.1)a (0.2)b ...(2)

1.2 × 10–3 = k(0.2)a (0.1)b ...(3)

Solving eq (1), (2) and (3), we get

a = 1, b = 2

6 × 10–3 = k(0.1)1 (0.1)2 from (1)

k = 6

7.2 × 10–2 = 6(x)1 (0.2)2 x = 0.3

2.88 × 10–2 = 6(0.3)1 (y)2 y = 0.4

14. The molecular geometry of SF6 is octahedral.

What is the geometry of SF4 (including lone

pair(s) of electrons, if any)?

(1) Tetrahedral

(2) Trigonal bipyramidal

(3) Square planar

(4) Pyramidal

Answer (2)


11

Sol. SF4

Bond pair = 4

Lone pair = 1

Steric Number = 5

Hybridisation sp d3

S

F

F

F

F

Geometry Triagonal bipyramidal

Shape See saw

15. Amongst the following statements regarding

adsorption, those that are valid are

(a) H becomes less negative as adsorptionproceeds

(b) On a given adsorbent, ammonia is

adsorbed more than nitrogen gas

(c) On adsorption, the residual force acting

along the surface of the adsorbent

increases

(d) With increase in temperature, the

equilibrium concentration of adsorbate

increases

(1) (c) and (d)

(2) (a) and (b)

(3) (d) and (a)

(4) (b) and (c)

Answer (2)

Sol. Because of polar nature of NH3, it can be

easily liquified as well as easily adsorbed as

compared with non-polar N2 gas.

H becomes, less negative as the adsorptionproceeds.

16. Two compounds A and B with same molecular

formula (C3H

6O) undergo Grignard’s reaction

with methylmagnesium bromide to give

products C and D. Products C and D show

following chemical tests.

Test C D

Ceric

ammonium Positive Positive

nitrate Test

TurbidityTurbidity

obtainedLucas Test obtained

after fiveimmediately

minutes

Iodoform Test Positive Negative

C and D respectively are

(1) C = H C – CH – CH – CH3 2 3

OH

D = H C – C – OH3

CH3

CH3

;

(2) C = H C – CH – CH – CH – OH3 2 2 2

D = H C – CH – CH – CH 3 2 3

OH

;

(3) C = H C – CH – CH – CH – OH3 2 2 2

D = H C – C – OH3

CH3

CH3

;

(4) C = H C – C – OH;3

CH3

CH3

D = H C – CH – CH – CH 3 2 3

OH

Answer (1)

Sol.3 2 3

OH|

CH —CH —CH—CH can give positive CAN

test, iodoform test, and with Lucas reagent it

takes 5 minutes to give turbidity.

3

3

3

CH

|CH —C— OH

|CH

can not give iodoform test

but can give CAN test and Lucas test.

17. The shape / structure of [XeF5]– and XeO

3F

2,

respectively, are

(1) Pentagonal planar and trigonal bipyramidal

(2) Trigonal bipyramidal and pentagonal

planar

(3) Octahedral and square pyramidal

(4) Trigonal bipyramidal and trigonal

bipyramidal

Answer (1)

Sol. [XeF5]– is Pentagonal planar

XeO3F

2 is trigonal Pyramidal.


12

18. An organic compound ‘A’ (C9H

10O) when treated

with conc. HI undergoes cleavage to yield

compounds ‘B’ and ‘C’. ‘B’ gives yellow

precipitate with AgNO3 where as ‘C’

tautomerizes to ‘D’. ‘D’ gives positive iodoform

test. ‘A’ could be

(1) O – CH = CH – CH3

(2) O – CH = CH2H C3

(3) CH – O – CH = CH2 2

(4) O – CH – CH = CH2 2

Answer (3)

Sol. CH2

O CH CH2

HI

CH2

I CH2

CH OH

Tautomerise

CH3

CHO

+

can give yellow ppt with AgI

can give iodoform test

19. The major product of the following reaction is:

OH

NO2

CH3conc. HNO + conc.

3

H SO2 4

(1)

OH

NO2

H C3

O N2

(2)

OH

NO2

H C3

NO2

(3)

OH

NO2

H C3 NO

2

(4)

OH

NO2

H C3 NO

2

NO2

Answer (2)

Sol.

OH

H C3

NO2

conc. HNO3

+

conc. H SO2 4

OH

H C3

NO2

NO2

Position of electrophilic attack is directed by

the electron donating group present in ring

20. The correct observation in the following

reactions is :

Sucrose Glycosidic bond

Cleavage (Hydrolysis)

A + BSeliwanoff's

reagent?

(1) Gives no colour

(2) Formation of red colour

(3) Formation of violet colour

(4) Formation of blue colour

Answer (2)

Sol.

Seliwanoff’sreagent

Red colour will

be observed

HydrolysisSucrose Glucose Fructose

Ketose with seliwanoff’s reagent gives red

colour. It is a specific test for ketose.

SECTION - II

Numerical Value Type Questions: This section

contains 5 questions. The answer to each question is

a NUMERICAL VALUE. For each question, enter the

correct numerical value (in decimal notation,

truncated/roundedoff to the second decimal place;

e.g. 06.25, 07.00, -00.33, -00.30, 30.27, -27.30) using

the mouse and the on-screen virtual numeric keypad

in the place designated to enter the answer.


13

21. The oxidation states of transition metal atoms in

K2Cr

2O

7, KMnO

4 and K

2FeO

4, respectively, are

x, y and z. The sum of x, y and z is _______.

Answer (19)

Sol.1 x 2

2 72K Cr O

2 + 2x – 14 = 0 x = +6

y1 2

4KMnO

1 + y – 8 = 0 y = +7

1 z 2

2 4K FeO

2 + z – 8 = 0 z = +6

x + y + z = 6 + 7 + 6 = 19

22. The work function of sodium metal is

4.41 × 10–19 J. If photons of wavelength 300 nm

are incident on the metal, the kinetic energy of

the ejected electrons will be (h = 6.63 × 10–34 J s;

c = 3 × 108 m/s) ________ × 10–21 J.

Answer (222)

Sol. w, work function of sodium metal

= 4.41 × 10–19 J

, wavelength of incident light = 300 nm

= 3 × 10–7 m

According to photoelectric effect

hcw KE

34 8

19

7

6.63 10 3 104.41 10 KE

3 10

6.63 × 10–19 = 4.41 × 10–19 + KE

KE = 2.22 × 10–19 J = 222 × 10–21 J

23. The heat of combustion of ethanol into carbon

dioxide and water is – 327 kcal at constant

pressure. The heat evolved (in cal) at constant

volume and 27°C (if all gases behave ideally) is

(R = 2 cal mol–`1 K–1) ________.

Answer (–326400)

Sol. C2H

5OH (l) + 3O

2(g) 2CO

2(g) + 3H

2O(l)

H = –327 kcal; ng = – 1

H = U + ngRT

U = – 327 + 2 × 10–3 × 300

= – 326.4 kcal

= – 326400 cal

24. For the disproportionation reaction

2Cu+(aq) � Cu(s) + Cu2+(aq) at 298 K. ln K

(where K is the equilibrium constant) is

___________ × 10–1.

Given :

20Cu /Cu

E 0.16 V

0Cu /Cu

E 0.52 V

RT0.025

F

Answer (144)

Sol. 2 Cu+(aq) � Cu(s) + Cu2+(aq)

2

0

Cu /CuE 0.16 V ;

0

Cu /CuE 0.52 V

2

0 0 0cell

Cu /Cu Cu /CuE E – E

= 0.52 – 0.16 = 0.36 V

0cell

RTE ln K

nF

10.36ln K 14.4 144 10

0.025

25. The ratio of the mass percentages of ‘C & H’

and ‘C & O’ of a saturated acyclic organic

compound ‘X’ are 4 : 1 and 3 : 4 respectively.

Then, the moles of oxygen gas required for

complete combustion of two moles of organic

compound ‘X’ is ________.

Answer (5)

Sol. Let the masses of C, H and O in organic

compound X be x, y and z respectively

Mass Moles

C x 4 12 1

H y 1 3 3

O z16

316 1

Empirical formula : CH3O

Molecular formula : C2H

6O

2

2 6 2 2 2 2(X)

2C H O 5O 4CO 6H O

Number of moles of O2 required to oxidise

2 moles of (X) = 5.

��


14

PART–C : MATHEMATICS

SECTION - I


multiple choice questions. Each question has

4 choices (1), (2), (3) and (4), out of which ONLY ONE

is correct.


1. Let a, b, c R be all non-zero and satisfya3 + b3 + c3 = 2. If the matrix

a b c

A b c a

c a b

satisfies ATA = I, then a value of abc can be :

(1) 3 (2)13

(3)1

–3

(4)23

Answer (2*)

Sol. AAT = I |A| = 1

|A| = 3abc – (a3 + b3 + c3) = ±1

{a3 + b3 + c3 = 2}

3abc = 3 or 1

1

abc or 13

* For the data given in the question, there existno real numbers a, b and c; because

AAT = I a2 + b2 + c2 = 1 and ab + bc + ca = 0also a3 + b3 + c3 = 2 leads to no solutions.

2. The imaginary part of

1 1

2 23 2 –54 – 3 – 2 –54 can be :

(1) 6 (2) – 6

(3) –2 6 (4) 6

Answer (3)

Sol. Let 3 6 6 i a ib

a2 – b2 = 3 and ab 3 6 a2 + b2 = 15

So, a = ±3 and b = ± 6

3 + 6 6 i = ± 3 + 6 i

Similarly, 3 – 6 6 i = ± 3 – 6 i

Im 3 + 6 6 i – 3 – 6 6 i = ±2 63. Let A = {X = (x, y, z)T : PX = 0 and x2 + y2 + z2 =

1}, where

1 2 1

P 2 3 4 ,

1 9 1

then the set A :

(1) is an empty set.

(2) contains more than two elements.

(3) contains exactly two elements.

(4) is a singleton.

Answer (3)

Sol. det(P) = 0

So the system has infinitely many solutions.

All solution lies on the line of intersection ofplanes

x + 2y + z = 0, –2x + 3y – 4z = 0 and x + 9y – z =0

ˆˆ ˆi j kˆˆ ˆ1 2 1 = – 11i 2 j 7k

–2 3 –4

So, x = –11 , y = 2, z = 7

x2 + y2 + z2 = 1

2 = 1

174 = ±

1

174

Two values of gives two triplets of (x, y, z)

4. For some 0, ,

2

if the eccentricity of the

hyperbola, x2 – y2sec2 = 10 is 5 times theeccentricity of the ellipse, x2sec2 + y2 = 5, thenthe length of the latus rectum of the ellipse, is :

(1) 2 6 (2)2 5

3

(3)4 5

3(4) 30

Answer (3)


15

Sol.2 2 2

e2

x y cosE : 5 e 1 sin

1 1cos

2 22

H2

x yH : – 10 e = 1 + cos

1 cos

2 2H e1

e = 5 e 1 + cos = 5sin sin =3

Length of latus rectum of ellipse = 22a

b

22(5cos )=

5

= 2 4 5

2 53 3

5. The set of all possible values of in the interval(0, ) for which the points (1, 2) and (sin, cos)lie on the same side of the linex + y = 1 is :

(1) 0, 2

(2)3

,4 4

(3) 0, 4

(4)3

0,4

Answer (1)

Sol. Let f(x, y) = x + y – 1

f(1, 2).f(sin, cos) > 0

2 sin + cos – 1 0 > sin+ cos > 1

1

sin +4 2

3

+ ,4 4 4

0,2

6. Which of the following is a tautology?

(1) (~ p)(p q) q

(2) (~ q)(p q) q

(3) (p q)(q p)

(4) (q p)~ (p q)

Answer (1)

Sol. Truth table

P

T

T

F

F

q

T

F

T

F

~p

F

F

T

T

p q

T

T

T

F

(~p) (p q) q

T

T

T

T

(~p) (p q)

F

F

T

F

~p (p q) q be a tautology

7. The equation of the normal to the curve

y = (1 + x)2y + cos2(sin–1x) at x = 0 is :

(1) y = 4x + 2 (2) y + 4x = 2

(3) x + 4y = 8 (4) 2y + x = 4

Answer (3)

Sol. y = (1 + x)2y + cos2 (sin–1x) ...(1)

On differentiating both sides w.r.t. x we get

dydx

2y(1 + x)2y–1 + (1 + x)2y ln(1 + x).

12

sin 2sin xdy2

dx 1 x

...(2)

When x = 0 then y = (1 + 0)2y + cos2(0) = 2

When x = 0 and y = 2 then

dy4 0 0 4

dx

Slope of normal at x = 0 is 14

Equation of normal : 1

y 2 (x 0)4

x + 4y = 8

8. If the sum of first 11 terms of an A.P., a1, a2, a3,... is 0 (a1 0), then the sum of the A.P., a1, a3,a5, ..., a23 is ka1, where k is equal to :

(1)121

–10

(2)72

–5

(3)725

(4)12110

Answer (2)

Sol. Let common difference be d.

a1 + a2 + a3 + ... + a11 = 0

111

2a 10.d 02

a1 + 5d = 0

1a

d5

...(1)

Now a1 + a3 + a5 + ... + a23= a1 + (a1 + 2d) + (a1 + 4d) + ... + (a1 + 22d)

= 12a1 + 2d 11 12

2

= 11a

12 a 11.5

= 16

12 a5

= 172

a5


16

9. Let f(x) be a quadratic polynomial such thatf(–1) + f(2) = 0. If one of the roots of f(x) = 0 is3, then its other root lies in :

(1) (–1, 0) (2) (–3, –1)

(3) (0, 1) (4) (1, 3)

Answer (1)

Sol. Let f(x) = ax2 + bx + c

Let roots are 3 and

and f(–1) + f(2) = 0

4a + 2b + c + a – b + c = 0

5a + b + 2c = 0 ...(i)

f(3) = 0 9a + 3b + c = 0 ...(ii)From equation (i) and (ii)

a b c a b c1– 6 18 – 5 15 – 9 –5 13 6

f(x) = k(–5x2 + 13x + 6)

= –k(5x + 2)(x – 3)

Roots are 3 and 2

–5

2

–5

lies in interval (–1, 0)

10. Consider a region 2 2R = {(x, y) R : x y 2x}.

if a line y = divides the area of region R intotwo equal parts, then which of the following istrue?

(1) 23 – 8 + 8 = 0

(2) 3 3/2– 6 – 16 = 0

(3) 2 3/23 – 8 + 8 = 0

(4) 3 2– 6 + 16 = 0 Answer (3)

Sol. According to given condition

xx

y

O

y = x2

y = 2x

(2, 4)y =

y

4

0

y yy dy y dy

2 2

443/2 2 3/2 2

00

y y y y3 34 42 2

2 2

3/2 3/22 2 8 43 4 3 4

23/24 4

3 2 3

3/2 28 3 8

2 3/23 8 8 0 11. Let n > 2 be an integer. Suppose that there are

n Metro stations in a city located along acircular path. Each pair of stations isconnected by a straight track only. Further,each pair of nearest stations is connected byblue line, whereas all remaining pairs ofstations are connected by red line. If thenumber of red lines is 99 times the number ofblue lines, then the value of n is :

(1) 199 (2) 201

(3) 101 (4) 200

Answer (2)

Sol. Number of two consecutive stations = n

Number of two non-consecutive stations = 2Cn – n

Now, According to the question,

2Cn – n 99n

n n – 1

– 100n = 02

n – 1 – 200 = 0

n = 201

12. Let S be the sum of the first 9 terms of theseries :

{x + ka} + {x2 + (k + 2)a} + {x3 + (k + 4)a}+ {x4 + (k + 6)a} + ... where a 0 and x 1.

If 10x x+ 45a(x 1)

S ,x – 1

then k is equal to :

(1) –3 (2) 1

(3) –5 (4) 3

Answer (1)

Sol. Seires (x + ka) + (x2 + (k + 2)a) + ...... 9 terms

S = (x + x2 + x3 + ...... 9 terms) + a[k +(k + 2)+ (k + 4) + .... 9 terms]

9x x – 1 9

S = + 2ak + 8 × (+2a)x – 1 2

1010 x + 45a x – 1x – x 9ka + 72a

S = + = x – 1 1 x – 1

(given)


17

10 10x – x + 9a k + 8 x – 1 x – x + 45a x – 1

= x – 1 x – 1

9a(k + 8) = 45a

k + 8 = 5

k = – 3

13. Let EC denote the complement of an event E.Let E1, E2 and E3 be any pairwise independentevents with P(E1) > 0 and P(E1 E2 E3) = 0.

Then P( C2E C3E /E1) is equal to :

(1) C C3 2P E P E (2) C2 3P E P E(3) C3 2P E P E (4) C3 2P E P E

Answer (4)

Sol. Here

C CC C 1 2 32 3

1 1

P E E EE EP =

E P E

1 1 2 2 3 1 2 3

1

P E – P E E + P E E – P E E E=

P E

1 1 2 2 3

1

P E – P E E – P E E + 0=

P E

= 1 – P(E2) – P(E3)

= C C2 3 3 2P E – P E or P E – P E14. A plane passing through the point (3, 1, 1)

contains two lines whose direction ratios are1, –2, 2 and 2, 3, –1 respectively. If this planealso passes through the point (, –3, 5), then is equal to :

(1) 5 (2) 10

(3) –10 (4) –5

Answer (1)

Sol. The equation of plane passing through (3, 1, 1) is-

a(x – 3) + b (y – 1) + c(z – 1) = 0 ... (i)

This plane contains the lines having drs(1, –2, 2) & (2, 3, –1)

so,

a – 2b + 2c = 0

2a + 3b – c = 0 a b ca b c = = = =

2 – 6 – –1– 4 3 + 4–2 2 1 2 1 –2 a b c = = 23 –1 2 –1 3–4 5 7

So, equation of plane is

–4(x – 3) + 5(y – 1) + 7(z – 1) = 0

–4x + 12 + 5y – 5 + 7z – 7 = 0

–4x + 5y + 7z = 0

This also passes through (, –3, 5)

So, –4 – 15 + 35 = 0

–4 = –20

= 5

15. Let ƒ : R R be a function which satisfies

ƒ(x + y) = ƒ(x) + ƒ(y) x, y R . If ƒ(1) = 2 and

n 1

k 1

g(n) ƒ(k), n N then the value of n, for

which g(n) = 20, is :

(1) 20 (2) 9

(3) 5 (4) 4

Answer (3)

Sol. f(x + y) = f(x) + f(y), x, y R, f 1 2=

f(x) = 2x

Now, g(n) = n–1

k 1

f k

= f(1) + f(2) + f(3) + .......f(n – 1)

= 2 + 4 + 6 + ...... +2(n – 1)

= 2[1 + 2 + 3 + ..... +(n – 1)]

2n – 1 n= 2 × n – n2

So, n2 – n = 20 (given)

n2 – n –20 = 0

(n – 5)(n + 4) = 0

n 5

16.

1x

x 0lim tan x

4 is equal to :

(1) e2 (2) 1

(3) e (4) 2

Answer (1)

Sol.1/x

x 0

1 tanxlim

1 tanx

x 01 tanx (1 tanx) 1

lim1 tanx xe

x 02 tanx 1

lim21 tanx xe e


18

17. Let f : (–1, ) R be defined by f(0) = 1 and

e1

f(x) log (1 x), x 0.x

Then the function f :

(1) increases in (–1, )

(2) increases in (–1, 0) and decreases in (0, )

(3) decreases in (–1, 0) and increases in (0, )

(4) decreases in (–1, )

Answer (4)

Sol.2

xln(1 x)

1 xf (x)x

2x (1 x)ln(1 x)

0 x ( 1, ) 0(1 x)x

[ as g(x) = x – (1 + x)ln (1 + x) gives g(x) < g(0)for x (–1, 0) and g(x) < g(0) for x (0, )]

18. If the equation cos4 + sin4 + = 0 has realsolutions for , then lies in the interval

(1)

1–1, –

2(2)

3 5– , –

2 4

(3)

1 1– , –

2 4(4)

5– , – 1

4

Answer (1)

Sol. sin4 + cos4 = –

1 – 2sin2 cos2 = –

2(sin2 )

12

as sin22 [0, 1]

1

1,2

19. If a curve y = ƒ(x), passing through the point (1,2), is the solution of the differential equation,

2x2dy = (2xy + y2)dx, then 1

ƒ2

is equal to :

(1)e

11 log 2

(2)e

11 log 2

(3) 1 + loge2 (4)e

11 log 2

Answer (1)

Sol.2

2

dy 2xy ydx 2x

Put y = vx

2dv vv x v

dx 2

2dv dx

2xv

2

lnx cv

2x

lnx cy

(1, 2) c = –1

2xlnx 1

y

Hence, for 1

x2

1

y1 ln2

20. The area (in sq. units) of an equilateral triangleinscribed in the parabola y2 = 8x, with one of itsvertices on the vertex of this parabola, is :

(1) 64 3 (2) 256 3

(3) 128 3 (4) 192 3

Answer (4)

Let A = (2t2, 4t)

B (2t2, 4t) (by symmetry as equilateraltriangle)

Sol.

(0, 0)

A(2t , 4t)2

B(2t , –4t)2

O

for equilateral triangle (angle at O is 60º)

2

4t 1t 2 3

32t

Area 1

.8(2 3).2.24 192 32

SECTION - II

Numerical Value Type Questions: This sectioncontains 5 questions. The answer to each question isa NUMERICAL VALUE. For each question, enter thecorrect numerical value (in decimal notation,truncated/roundedoff to the second decimal place;e.g. 06.25, 07.00, -00.33, -00.30, 30.27, -27.30) usingthe mouse and the on-screen virtual numeric keypadin the place designated to enter the answer.

21. If

6

–1

k = 1

3 4y k cos coskx – sinkx ,

5 5 then

dy at x = 0 is

dx________.

Answer (91)


19

Sol.6

–1

k = 1

3 4y k cos coskx – sinkx

5 5

Let 3 4

cos and sin5 5

6

–1

k = 1

y k cos {coscoskx – sinsinkx}

6

–1

k = 1

k cos cos kx

6 6

2

k = 1 k = 1

k kx k x k

6 2k 1

dy 6 7 13k 91

dx 6

22. Let [t] denote the greatest integer less than or

equal to t. Then the value of 2

1| 2x – [3x]| dx

is ______.

Answer (1)

Sol. 2

1| 2x – [3x]| dx

21

3x x dx

21

x 3x dx

2 21 1

xdx 3x dx

13

0

4 13 3xdx

2

3 11

2 2

23. For a positive integer n, n1

1 +x

is expanded

in increasing powers of x. If three consecutivecoefficients in this expansion are in the ratio, 2: 5 : 12, then n is equal to ____.

Answer (118)

Sol. n n nr 1 r r 1C : C : C 2 : 5 : 12

n n

r r 1n n

r 1 r

C C5 12and

2 5C C

n r 1 5 n r 12

andr 2 r 1 5

2n – 7r + 2 = 0 and 5n – 17r – 12 = 0

Solving; x = 118, r = 34

24. If the variance of the terms in an increasingA.P., b1, b2, b3, ..., b11 is 90, then the commondifference of this A.P. is ________.

Answer (3)

Sol. Variance

211 112i i

i 1 i 1

b b

11 11

210 10

21 1

r 0 r 0

b rd b rd

11 11

2 21 1

10 11 10 11 2111b 2b d d

2 611

2

110 11

11b d2

11

22 2 21 1 1b 10b d 35d b 5d 10d Variance = 90

10d2 = 90

d = 3

25. Let the position vectors of points ‘A’ and ‘B’ be

ˆˆ ˆi + j+k and ˆˆ ˆ2i + j+3k, respectively. A point ‘P’

divides the line segment AB internally in theratio : 1 ( > 0). If O is the origin and

2OB.OP 3 OA×OP 6,

then is equal to

________.

Answer (0.8)

Sol. Let position vector of P is b a

OP1

P B(1, 1, 1) (2, 1, 3)

: 1

Given 2

OB OP 3 OA OP 6

2

b a b ab 3 a 6

1 1


20

2 2 2

2

a b b 3a b 6

1 1

ˆˆ ˆa b 2i j k

2

2

6 14 186

1 1

2

2

8 186 6

1 1

Let t1

18t2 – 8t = 0

4t 0,

9

4

1 9

4

0.85

Que & Sol_JEE(Main)-2020_Phase-2_02-09-2020_Evening_Physics_FinalQue & Sol_JEE(Main)-2020_Phase-2_02-09-2020_Evening_Chemistry_FinalQue & Sol_JEE(M)-2020_P-II (02-09-2020) Evening_Maths

(physics, chemistry and mathematics)cv v cc = + c15f2 = μ 13. a particle is moving 5 times as fast...

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