(Physics, Chemistry and Mathematics) · 2020. 5. 9. · d4g g1 R9 5R d 9 15. For a concave lens of...

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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)

05/09/2020

Morning


2

PART–A : PHYSICS

SECTION - I

Multiple Choice Questions: This section contains 20multiple choice questions. Each question has 4choices (1), (2), (3) and (4), out of which ONLY ONE

is correct.

Choose the correct answer :

1. An electrical power line, having a totalresistance of 2 , delivers 1 kW at 220 V. Theefficiency of the transmission line isapproximately

(1) 85% (2) 96%

(3) 72% (4) 91%

Answer (2)

Sol. I = P (1000)

V 220

PLoss

= 2

2

trans

1000I R 2

220

efficiency = Loss

1000100

1000 P

� 96%

2. A bullet of mass 5 g, travelling with a speed of210 m/s, strikes a fixed wooden target. One halfof its kinetic energy is converted into heat inthe bullet while the other half is converted intoheat in the wood. The rise of temperature ofthe bullet if the specific heat of its material is0.030 cal/(g – °C) (1 cal = 4.2 × 107 ergs) closeto

(1) 83.3°C (2) 87.5°C

(3) 38.4°C (4) 119.2°C

Answer (2)

Sol. MS T = 21 1

Mv2 2

T = 2v

4S

= 2

7

(210,00)

4 0.030 4.2 10

� 87.5°C

3. An electron is constrained to move alongthe y-axis with a speed of 0.1 c (c is thespeed of light) in the presence ofelectromagnetic wave, whose electric

field is 7 2ˆE 30j sin(1.5 10 t 5 10 x) V/m ��

.

The maximum magnetic force experienced bythe electron will be

(given c = 3 × 108 ms–1 and electron charge =1.6 × 10–19 C)

(1) 4.8 × 10–19 N (2) 2.4 × 10–18 N

(3) 3.2 × 10–18 N (4) 1.6 × 10–19 N

Answer (1)

Sol. maxmF = (q)VB0

0

0

E

B = c 00

EB

c

maxmF =

19 0E

(1.6 10 )(0.1c)c

� 4.8 × 10–19 N

4. A wheel is rotating freely with an angular speed on a shaft. The moment of inertia of thewheel is I and the moment of inertia of theshaft is negligible. Another wheel of moment ofinertia 3I initially at rest is suddenly coupled tothe same shaft. The resultant fractional loss inthe kinetic energy of the system is

(1)5

6(2)

1

4

(3) 0 (4)3

4

Answer (4)

Sol. I = 4I

= 4

(K)Loss

= 2

21 1I (4I)2 2 4

Fractional loss = 2Loss 11

( K) 1K I

K 2

= 3

4


3

5. Activities of three radioactive substances A, Band C are represented by the curves A, B andC, in the figure. Then their half-lives

1 1 12 2 2

T A : T B : T C are in the ratio

6

4

2

05 10

t

(yrs)

C B

A

ln R

(1) 2 : 1 : 3 (2) 4 : 3 : 1

(3) 2 : 1 : 1 (4) 3 : 2 : 1

Answer (1)

Sol. R = N0e–t

lnR = lnN0 – t

Slope of curves = –

also 1t

2

ln2

½ ½ ½

6for A slope

10

6for B slope T (A):T (B):T (C) 2:1:3

5

2for C slope

5

6. A helicopter rises from rest on the groundvertically upwards with a constant accelerationg. A food packet is dropped from the helicopterwhen it is at a height h. The time taken by thepacket to reach the ground is close to [g is theacceleration due to gravity]

(1)h

t = 3.4g

(2)h

t =1.8g

(3)2h

t = 3g

(4)2 h

t = 3 g

Answer (1)

Sol. t0 = 2h

g

V0 =

2hg= 2gh

g

t1 = time to reach top = 0

v 2h=

g g

H = h + h = 2h

t2 = time of fall =

2× 2h h2

g g

Total time = t1 + t

2

= h2 2g

= h

3.4g

7. A hollow spherical shell at outer radius R floatsjust submerged under the water surface. Theinner radius of the shell is r. If the specific

gravity of the shell material is 27

8 w.r.t water,

the value of r is

(1)2R

3(2)

4R

9

(3)1R

3(4)

8R

9

Answer (4)

Sol. 3 3 3w m4 4R g = R – r g3 3

R3 = 3 3 27R – r8

r =

1

319R

27

8R

9

8. Assume that the displacement (s) of air isproportional to the pressure difference (p)created by a sound wave. Displacement (s)further depends on the speed of sound (v),density of air () and the frequency (f). Ifp~10 Pa, v~300 m/s, ~1 kg/m3 and f~1000 Hz,then s will be of the order of (take themultiplicative constant to be 1)

(1)3

mm100

(2) 10 mm

(3) 1 mm (4)1

mm10

Answer (1)


4

Sol. ∵

0 0P B × S

v

S =

P × v

B =

2

P × v

v 2

S P

vf

P

Svf

= 10

1 300 1000

= 1

mm30

3mm

100

9. Two capacitors of capacitances C and 2C arecharged to potential differences V and 2V,respectively. These are then connected inparallel in such a manner that the positiveterminal of one is connected to the negativeterminal of the other. The final energy of thisconfiguration is

(1) 23CV

2(2) 2

9CV

2

(3) Zero (4) 225

CV6

Answer (1)

Sol. Vcommon

= 4CV – CV

3C = V

+ –

+–

C, V

2C, 2V

Ceq

= C + 2C = 3C

Uf = 2 21 3× 3C × V = CV

2 2

10. With increasing biasing voltage of a photodiode,the photocurrent magnitude

(1) Increases initially and after attainingcertain value, it decreases

(2) Increases linearly

(3) Increases initially and saturates finally

(4) Remains constant

Answer (3)

Sol. In photodiode, photocurrent increases withincreasing biasing voltage and then becomessaturated.

I

IS

V

11. A physical quantity z depends on four

observables a, b, c and d, as

2

2 3

3

a bz .

c d

The

percentages of error in the measurement of a,b, c and d are 2%, 1.5%, 4% and 2.5%respectively. The percentage of error in z is

(1) 13.5% (2) 14.5%

(3) 16.5% (4) 12.25%

Answer (2)

Sol.z A 2 b 1 c d

2 3z A 3 b 2 c d

= 2 1

2 2 1.5 4 3 2.53 2

= 14.5%

12. Three different processes that can occur in anideal monoatomic gas are shown in the P vs Vdiagram. The paths are labelled as A B,A C and A D. The change in internalenergies during these process are taken asE

AB, E

AC and E

AD and the work done as W

AB,

WAC

and WAD

.

The correct relation between these parametersare

T > T1 2

T1

B

C

D

A

V

P

T2

(1) EAB

< EAC

< EAD

, WAB

> 0, WAC

> WAD

(2) EAB

= EAC

= EAD

, WAB

> 0, WAC

= 0, WAD

> 0

(3) EAB

> EAC

> EAD

, WAB

< WAC

< WAD

(4) EAB

= EAC

< EAD

, WAB

> 0, WAC

= 0, WAD

< 0

Answer (None of the option)


5

Sol. For all process

EAB

= EAC

= EAD

WAB

> 0

WAC

= 0

WAD

< 0

None of the option matches.

13. A solid sphere of radius R carries a chargeQ + q distributed uniformly over its volume. Avery small point like piece of it of mass m getsdetached from the bottom of the sphere andfalls down vertically under gravity. This piececarries charge q. If it acquires a speed v whenit has fallen through a vertical height y (seefigure), then : (assume the remaining portion tobe spherical).

Q R

y

q

v

(1) 2

0

qQv y g

4 R(R y)m

(2) 23

0

qQ Rv 2y g

4 (R y) m

(3) 22

0

qQv y g

4 R ym

(4) 2

0

qQv 2y g

4 R(R y)m

Answer (4)

Sol.2

elec.

1mv mgy U

2

elec.

1 1U kQq

R (R y)

20

qQv 2y g

4 R(R y)m

14. The value of the acceleration due to gravity is

g1 at a height

Rh

2 (R = radius of the earth)

from the surface of the earth. It is again equalto g

1 at a depth d below the surface of the

earth. The ratio d

R

equals

(1)5

9(2)

1

3

(3)7

9(4)

4

9

Answer (1)

Sol.2

R GM 4gg h

2 93R

2

d 4gg 1

R 9

5Rd

9

15. For a concave lens of focal length f, the relationbetween object and image distances u and v,respectively, from its pole can best berepresented by (u = v is the reference line)

(1)

f

v

u =

v

f u

(2)

f

v

u =

v

f u

(3)

f

v

u =

v

f u

(4)

f

v

u =

v

f u

Answer (4)

Sol. Theoretical


6

16. In a resonance tube experiment when the tubeis filled with water up to a height of 17.0 cmfrom bottom, it resonates with a given tuningfork. When the water level is raised the nextresonance with the same tuning fork occurs ata height of 24.5 cm. If the velocity of sound inair is 330 m/s, the tuning fork frequency is

(1) 2200 Hz

(2) 3300 Hz

(3) 1100 Hz

(4) 550 Hz

Answer (1)

Sol.

h

L

1L h n

2 4

2L h (n 1)

2 4

2 1

h h 24.5 17.0 7.5 cm2

= 15 cm

u = f

f = 2200 Hz

17. A galvanometer of resistance G is convertedinto a voltmeter of range 0 – 1 V by connectinga resistance R

1 in series with it. The additional

resistance that should be connected in serieswith R

1 to increase the range of the voltmeter

to 0 – 2 V will be

(1) G

(2) R1

(3) R1 + G

(4) R1 – G

Answer (3)

Sol. ig(G + R

1) = 1

ig(G + R

1 + R) = 2

R = G + R1

18. A balloon is moving up in air vertically above apoint A on the ground. When it is at a height h

1,

a girl standing at a distance d (point B) from A(see figure) sees it at an angle 45° with respectto the vertical. When the balloon climbs up afurther height h

2, it is seen at an angle 60° with

respect to the vertical if the girl moves furtherby a distance 2.464 d (point C). Then the heighth

2 is (given tan30° = 0.5774)

h1

h2

BA d 2.464 d

45° 60°

C

(1) d (2) 0.732 d

(3) 1.464 d (4) 0.464 d

Answer (1)

Sol. 1h

tan45d

1 2h h

tan30d 2.464d

h2 = d

19. Number of molecules in a volume of 4 cm3 ofa perfect monoatomic gas at some temperatureT and at a pressure of 2 cm of mercury is closeto ? (Given, mean kinetic energy of a molecule(at T) is 4 × 10–14 erg, g = 980 cm/s2, density ofmercury = 13.6 g/cm3)

(1) 5.8 × 1018

(2) 4.0 × 1016

(3) 5.8 × 1016

(4) 4.0 × 1018

Answer (4)

Sol. PV = NkT

3E kT

2

3PVN (P h g)

2E

= 4 × 1018


7

20. A square loop of side 2a, and carrying currentI, is kept in XZ plane with its centre at origin. Along wire carrying the same current I is placedparallel to the z-axis and passing through thepoint (0, b, 0), (b > > a). The magnitude of thetorque on the loop about z-axis is given by

(1)

2 2

02 I a

b

(2)

2 2

0I a

2 b

(3)

2 3

0

2

I a

2 b

(4)

2 3

0

2

2 I a

b

Answer (1)

Sol.

X

Y

I

F1

F2

Z

2a

2

0

1

IF 2a toward wire

2 r

2

0

2

IF 2a away from wire

2 r

F1

F2

2a

b

Torque about z-axis

= F1 a cos + F

2 a cos

= 2

0I 2a b

2 a2 r r

= 2 2

0

2 2

2 I a b

(a b )

= 2 2

02 I a

for (b a)b

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. A force � ˆˆ ˆF i 2 j 3k N acts at a point ˆˆ ˆ4i 3 j k m. Then the magnitude of torqueabout the point ˆˆ ˆi 2 j k m will be x N m.The value of x is ____ .

Answer (195.00)

Sol. �

��

r f

ˆ ˆˆ ˆ ˆ ˆr 4i 3i k i 2 j k �

ˆˆr 3i j 2k �

� ˆ ˆˆ ˆ ˆ ˆ3i j 2k i 2 j 3k � ˆ ˆˆ ˆ ˆ ˆ6k 9 j k 3i 2i 4i

� ˆˆ ˆ7i 11j 5k

( ) 195

22. Two concentric circular coils, C1 and C

2 are

placed in the XY plane. C1 has 500 turns, and a

radius of 1 cm. C2 has 200 turns and radius of20 cm. C

2 carries a time dependent current I(t)

= (5t2 – 2t + 3) A where t is in s. The emfinduced in C1 (in mV), at the instant t = 1 s is

4.

x The value of x is _____ .

Answer (05.00)

Sol. I = 5t2 – 2t + 3

dI10t 2

dt

At (t = 1 sec) dI

8 A/sdt

0200 I 100 500

2 20 100 100

0200 100 500d dI

edt 2 20 100 100 dt

d 8 4e m volt m volt

dt 10 5

x = 5


8

��

23. A particle of mass 200 MeV/c2 collides with ahydrogen atom at rest. Soon after the collisionthe particle comes to rest, and the atom recoilsand goes to its first excited state. The initial

kinetic energy of the particle (in eV) is N.

4 The

value of N is

(Given the mass of the hydrogen atom to be1 GeV/c2) ______ .

Answer (51.00)

Sol. M0 = 200 MeV/c2 ; m = 1 GeV/c2

Initial velocity = v0

Mv0 = mv 0

Mv v

m

2 2 0

0

E 31 1Mv mv

2 2 4

0

( E 13.6 eV)

2

0 0

1 M 3Mv 1 E

2 m 4

2

0 0 0

1 3 10 1 10Mv E 3 E

2 4 8 4 8

51 51

eV4 4

N = 51

24. A compound microscope consists of anobjective lens of focal length 1 cm and aneyepiece of focal length 5 cm with a separationof 10 cm.

The distance between an object and theobjective lens, at which the strain on the eye is

minimum is n

40 cm. The value of n is _____.

Answer (50.00)

Sol. f = 1 cm0 f = 5 cme

10 cm

Image by objective is formed at the focus ofeye-piece

For objective, v = 5, u, f = 1 cm

1 1 1 1 1

15 u 1 5 u

5| u | cm4

50

|u | cm40

n = 50

25. A beam of electrons of energy E scatters froma target having atomic spacing of 1 Å. The firstmaximum intensity occurs at = 60°. Then E(in eV) is ______.

(Planck constant h = 6.64 × 10–34 Js, 1 eV =1.6 × 10–19 J, electron mass m = 9.1 × 10–31 kg)

Answer (50.47)

Sol.21mv E

2

P 2Em h

2Em

For first maxima,

2d sin =

10 3 h2 10

2 2Em

10 2(10 h)2Em

3

10 2

e

(10 h)E (eV)

6m e

E = 50.47 eV


9

PART–B : CHEMISTRY

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.


1. In the sixth period, the orbitals that are filled

are

(1) 6s, 4f, 5d, 6p (2) 6s, 5d, 5f, 6p

(3) 6s, 6p, 6d, 6f (4) 6s, 5f, 6d, 6p

Answer (1)

Sol. Filling of electrons in orbitals in any period

takes place as:

ns, (n – 2)f (n – 1)d np

if possible if possible

for sixth period n = 6,

orbitals that are filled are 6s, 4f, 5d and 6p

2. The values of the crystal field stabilization

energies for a high spin d6 metal ion in

octahedral and tetrahedral fields, respectively,

are

(1) –1.6 0 and –0.4

t(2) –2.4

0 and –0.6

t

(3) –0.4 0

and –0.27 t(4) –0.4

0 and –0.6

t

Answer (4)

Sol. Crystal field stabilization energy (CFSE) for

high spin d6 metal ion

In octahedral

d6 eg

t2g

n = 2eg

2gtn 4

CFSE = 2g gt e 0–0.4n 0.6n

= [– 0.4 × 4 + 0.6 × 2]0

= – 0.4

0

In tetrahedral fields

d6 t2g

eg

t2g

CFSE = 2eg t g t

–0.6 n 0.4n

= (– 0.6 × 3 + 0.4 × 3)t

= (–1.8 + 1.2)t

= – 0.6 t

3. The equation that represents the water-gas

shift reaction is

(1) 1270 K4 2 2NiCH (g) H O(g) CO(g) 3H (g)

(2) 673 K

2 2 2CatalystCO(g) H O(g) CO (g) H (g)

(3) 1273 K2 2 22C(s) O (g) 4N (g) 2CO(g) 4N (g)

(4) 1270 K2 2C(s) H O(g) CO(g) H (g)

Answer (2)

Sol. In water gas shift reaction, Hydrogen gas is

produced economically by the reaction of

carbon monoxide with water vapour at 673 K in

presence of iron, cromium and copper zinc

catalyst

Catalyst2 2 2CO H O(g) CO H (g)

4. The condition that indicates a polluted

environment is

(1) 0.03% of CO2 in the atmosphere

(2) pH of rain water to be 5.6

(3) eutrophication

(4) BOD value of 5 ppm

Answer (3)

Sol. Eutrophication is a process by which

environment gain nutrients for habitats and

cause algal bloom, harmful algal produce toxins

that are harmful and create dead zones for fish

and polluting the environment.

5. If a person is suffering from the deficiency of

nor-adrenaline, what kind of drug can be

suggested?

(1) Analgesic (2) Antidepressant

(3) Anti-inflammatory (4) Antihistamine

Answer (2)

Sol. Noradrenaline is an organic chemical that

functions in brain and body as hormone and

neurotransmitter.

Antidepressant drugs can be suggested for the

deficiency of noradrenaline.


10

6. An Ellingham diagram provides information

about

(1) the temperature dependence of the

standard Gibbs energies of formation of

some metal oxides

(2) the pressure dependence of the standard

electrode potentials of reduction reactions

involved in the extraction of metals

(3) the conditions of pH and potential under

which a species is thermodynamically

stable

(4) the kinetics of the reduction process

Answer (1)

Sol. An Ellingham diagram provides information

about, the temperature dependence of the

standard gibbs energies of formation of some

metal oxides.

7. Which of the following is not an essential amino

acid?

(1) Tyrosine (2) Valine

(3) Lysine (4) Leucine

Answer (1)

Sol. Tyrosine is not an essential amino acid.

8. The correct electronic configuration and spin-

only magnetic moment (BM) of Gd3+ (Z = 64),

respectively, are

(1) [Xe] 5f7 and 8.9 (2) [Xe] 4f7 and 7.9

(3) [Xe] 5f7 and 7.9 (4) [Xe] 4f7 and 8.9

Answer (2)

Sol. Gd3+ (Z = 64) = [Xe] 4f7

n(n 2) 7(7 2) 7.9B.M.

9. The most appropriate reagent for conversion of

C2H

5CN into CH

3CH

2CH

2NH

2 is

(1) NaBH4

(2) CaH2

(3) Na(CN)BH3

(4) LiAlH4

Answer (4)

Sol. C2H

5CN 4LiAlH C2H5CH2NH2

10. The structure of PCl5 in the solid state is

(1) tetrahedral [PCl4]+ and octahedral [PCl

6]–

(2) square pyramidal

(3) trigonal bipyramidal

(4) square planar [PCl4]+ and octahedral [PCl

6]–

Answer (1)

Sol. PCl5 in solid state exist as [PCl

4]+ [PCl

6]–

[PCl4]+ is tetrahedral

[PCl6]– is octahedral

11. The increasing order of the acidity of the

-hydrogen of the following compounds is

O O O

Ph Ph

(A) (B)

O

OM e

(C)

O

NMe2

(D)

(1) (C) < (A) < (B) < (D)

(2) (B) < (C) < (A) < (D)

(3) (A) < (C) < (D) < (B)

(4) (D) < (C) < (A) < (B)

Answer (4)

Sol.

O O

Ph Ph

Activemethylene

(highly acidic)

>

O

>

O

OMe

O

NMe2

>

Lower – M effect as compared to —C— group

O

12. Which of the following derivatives of alcohols is

unstable in an aqueous base?

(1) RO Me

O

(2) RO—CMe3

(3)

ORO

(4) RO

Answer (1)

Sol.

O

RO Me

Ester

OH /H O–

2

HydrolysisROH +

O

Me O

So, Ester is unstable in an aqueous basic

solution and undergoes hydrolysis to give

alcohol and carboxylate.


11

13. The potential energy curve for the H2 molecule

as a function of internuclear distance is

(1) Energy

Internuclear distance

(2) Energy


(3) Energy


(4) Energy


Answer (3)

Sol.

Energy


Bond energy

(minimum)

14. The increasing order of basicity of the

following compounds is

NN

H

N

H

N

H

N

(A) (B) (C) (D)

(1) (B) < (A) < (D) < (C)

(2) (D) < (A) < (B) < (C)

(3) (B) < (A) < (C) < (D)

(4) (A) < (B) < (C) < (D)

Answer (3)

Sol.

NN

H

N

H

N

H

N

sp2

sp3

lone pairinvolvedin reso.

Most basic(because of ResonatingStructure after accepting H )+

hyb.

N N

H

N

H

N

H

N

(A)(B)(C)(D)

> > >

15. Identify the correct molecular picture showing

what happens at the critical micellar

concentration (CMC) of an aqueous solution of

a surfactant ( polar head; non-polar tail;

• water).

• • • •• • • •

•• • •• •

• • ••

• ••

•• •

• •• •

• •• ••

• •••

•(A) (B) (C) (D)

(1) (C) (2) (A)

(3) (D) (4) (B)

Answer (3)

Sol.Polar part interactswith water• •• ••

• •••

•

At CMC, micelle formation takes place.

16. A flask contains a mixture of compounds A and

B. Both compounds decompose by first-order

kinetics. The half-lives for A and B are 300 s

and 180 s, respectively. lf the concentrations of

A and B are equal initially, the time required for

the concentration of A to be four times that of

B (in s) is : (Use In 2 = 0.693)

(1) 120 (2) 300

(3) 180 (4) 900

Answer (4)

Sol. ∵ A = A0e–kt

1/2

ln2k

t

ln2t

3000

ln2t

1800

4B A e

B B e

[given A0 = B

0, A = 4B]

1 1ln2 t

180 3004 e

t = 900 sec.


12

17. In the following reaction sequence the major

products A and B are:

+ O

O

O

anhydrous

AlCl3A

1. Zn – Hg/HCl

2. H PO3 4B

(1)

O

A =

CO H2

; B =

O

(2)

O

A =

CO H2

; B =

O

(3)

O

A =

CO H2

; B =

(4)

O

A =

CO H2

; B =

O

Answer (1)

Sol. +O:

O

O

AlCl3

O

O

HO

(A)

Zn – Hg/HCl

O

HO

O

H PO3 4

(B)

18. Consider the following reaction :

N2O

4(g) � 2NO

2 (g); H0 = + 58 kJ

For each of the following cases (a, b), the

direction in which the equilibrium shifts is :

(a) Temperature is decreased

(b) Pressure is increased by adding N2 at

constant T.

(1) (a) Towards product, (b) towards reactant

(2) (a) Towards reactant, (b) no change

(3) (a) Towards reactant, (b) towards product

(4) (a) Towards product, (b) no change

Answer (2)

Sol. ∵ Given reaction is endothermic

On decreasing temperature backwardreaction will be favoured.

On adding N2, pressure is increased at

constant T, and volume would also be constant

so no change is observed.

19. A diatomic molecule X2 has a body-centred

cubic (bcc) structure with a cell edge of

300 pm. The density of the molecule is 6.17 g

cm–3. The number of molecules present in

200 g of X2 is :

(Avogadro constant (NA) = 6 × 1023 mol–1)

(1) 40 NA

(2) 4 NA

(3) 2 NA

(4) 8 NA

Answer (2)

Sol.3

A

Z Md

a N

8 3 23

2 M6.17

(3 10 ) 6 10

(For BCC Z = 2)

M = 50 g/mol

in 200 g, 4 moles of X2 is present

20. The difference between radii of 3rd and 4th

orbits of Li2+ is R1. The difference between the

radii of 3rd and 4th orbits of He+ is R2. Ratio

R1 : R

2 is

(1) 3 : 2

(2) 8 : 3

(3) 2 : 3

(4) 3 : 8

Answer (3)

Sol.2

n 0

nr a

Z

2

n

nr

Z

2

1 He

2 Li

ZR 2

R Z 3


13

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. The minimum number of moles of O2 required

for complete combustion of 1 mole of propane

and 2 moles of butane is _______.

Answer (18)

Sol. C3H

8(g) + 5O

2(g) 3CO

2(g) + 4H

2O(l)

C4H

10(g) + 2

13O (g)

2 4CO

2(g) + 5H

2O(l)

No. of moles of O2 required to oxidise 1 mole of

propane and 2 moles of butane = 5 + 2 × 13

2

= 18

22. The total number of coordination sites in

ethylenediaminetetraacetate (EDTA4–) is ____.

Answer (6.00)

Sol. [EDTA]4– is ethylenediaminetetraacetate anion.

It is a hexadentate ligand.

O

O – C – CH2

O – C – CH2

O

N – CH – CH – N 2 2

CH – C – O2

–

CH – C – O2

–

O

O

It has six co-ordination sites.

23. The number of chiral carbon(s) present in

peptide, I�e-Arg-Pro, is _____.

Answer (4)

Sol. The amino acids present in the given tripeptide

I�e–Arg–Pro are isoleucine, arginine and proline

CH —CH —CH—CH—23

COOH* *

CH3

NH2

(I e)�

HN—CH—COOH*

(Pro)

HN==C—NH—(CH ) —CH—COOH2 3

*

NH2

NH2

(Arg)

Number of chiral carbons present in the given

tripeptide is 4.

24. An oxidation-reduction reaction in which

3 electrons are transferred has a G0 of

17.37 kJ mol–1 at 25°C. The value of ocellE

(in V)

is _____ × 10–2.

(1 F = 96,500 C mol–1)

Answer (–6.00)

Sol. G° = 17.37 kJ ; n = 3

G° = –nFEocell

o –2

cell

17.37 1000E –0.06 6.00 10

3 96500

25. A soft drink was bottled with a partial pressure

of CO2 of 3 bar over the liquid at room

temperature. The partial pressure of CO2 over

the solution approaches a value of 30 bar when

44 g of CO2 is dissolved in 1 kg of water at room

temperature. The approximate pH of the soft

drink is ______ × 10–1.

(First dissociation constant of H2CO

3 =

4.0 × 10–7; log 2 = 0.3; density of the soft

drink = 1 g mL–1)

Answer (37)

Sol. At 30 bar pressure mass of CO2 in 1 kg water

= 44 gm

At 3 bar pressure mass of CO2 in 1 kg water

= 4.4 gm

Moles of CO2 in 1 kg water = 0.1

2 3 3H CO H HCO

0.1(1– ) 0.1 0.1

�

31

2 3

[H ][HCO ]Ka

[H CO ]

2–7 20.1

4 10 0.11–

�

= 2 × 10–3

[H+] = 0.1 = 2 × 10–4 ; pH = 3.7

= 37 × 10–1

� � �


14

PART–C : MATHEMATICS

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.


1. If x x2x x x (e e )e 2e e 1 e dx x x(e e )g(x)e c,

where c is a constant of

integration, then g(0) is equal to :

(1) e2 (2) 1

(3) 2 (4) e

Answer (3)

Sol. x –x2x x –x (e e )e 2e – e – 1 e dx

Let xdt

e t, dxt

1t

2 t1 dtt 2t – – 1 et t

12 tt

2

(t – 1)(t 1)1 e dt

t

1 1t tt t

2I

II

11–t 1 e dt e dt

t

1 1 1t t t

t t t(t 1) e – e dt e dt

x –xx (e e )(e 1) e c

So, g(x) = 1 + ex and g(0) = 2

2. Let R. The system of linear equations

2x1 – 4x2 + x3 = 1

x1 – 6x2 + x3 = 2

x1 – 10x2 + 4x3 = 3

is inconsistent for :

(1) exactly two values of .

(2) exactly one positive value of .

(3) every value of .

(4) exactly one negative value of .

Answer (4)

Sol. 2

2 –4

1 –6 1 0 3 – 7 – 12 0

–10 4

2

3 or –3

Adding first two equations, we get

1 2 33x – 10x ( 1)x 3

and the last equation is x1 –10x2 + 4x3 = 3

So, for = 3 there will be infinitely many

solutions and for 2

–3

there will be no

solution (i.e. equations will be inconsistent).

3. If the function

21

2

xk (x ) 1,f(x)

xk cosx,

is twice

differentiable, then the ordered pair (k1, k2) isequal to :

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

, 12

(3) (1, 1) (4) 1, 12

Answer (2)

Sol.

21

2

k (x – ) – 1 ; xf(x)

k cos x ; x

1

2

2k (x – ) ; xf (x)

–k sinx ; x

and 1

2

2k ; xf (x)

–k cosx ; x

f(x) is twice differentiable at x = , then

(i) – 2 2x xlim f(x) lim f(x) –1 –k k 1

(ii) – 2 1 1x x

1lim f (x) lim f (x) k 2k k

2


15

4. The negation of the Boolean expression x ~yis equivalent to :

(1) (x ~ y) (~ x y) (2) (x y) (~ x ~ y)

(3) (x y) (~ x ~ y) (4) (~ x y) (~ x ~ y)

Answer (2)

Sol. p : x y (x y) ( y x )

( x y) (y x )

(x y) (x y)

Negation of p is p (x y) (x y)

(x y) ( x y)

5. If the co-ordinates of two points A and B are

( 7, 0) and ( 7, 0) respectively and P is any

point on the conic, 9x2 + 16y2 = 144, thenPA + PB is equal to :

(1) 9 (2) 16

(3) 6 (4) 8

Answer (4)

Sol.2 2x y

E : 116 9

7, 0 are the foci of given ellipse. So forany point P on it; PA + PB = 2a

PA + PB = 2(4) = 8

6. If the common tangent to the parabolas,y2 = 4x and x2 = 4y also touches the circle,x2 + y2 = c2, then c is equal to :

(1)12

(2)14

(3)1

2 2(4)

1

2

Answer (4)

Sol. Equation tangent to parabola y2 = 4x

with given slope m is :

1y mx

m …(i)

Line (i) is tangent to x2 = 4y.

24

x 4mxm

mx2 – 4m2x – 4 = 0

For tangent : 16m4 + 16m = 0

16m (m3 + 1) = 0

m = 0, –1

Equation tangent : x + y + 1 = 0

It is tangent to circle x2 + y2 = c2

1

c2

7. If y = y(x) is the solution of the differential

equation x

x5 + e dy · + e = 02 + y dx

satisfying y(0) = 1,

then a value of y(loge13) is :

(1) –1 (2) 2

(3) 0 (4) 1

Answer (1)

Sol.x

x5 + e dy · + e = 02 + y dx

x

x

dy edx

2 y 5 e

ln|2 + y| + ln|5 + ex| = ln C

y(0) = 1 ln C = ln 18

|(2 + y) · (5 + ex)| = 18

When x = ln 13 then |(2 + y)·18| = 18

2 + y = ±1

y = –1, –3

y(ln 13) = –1

8. The value of 2

sinx–

2

1dx is :

1 + e

(1)4

(2)

(3)32

(4)2

Answer (4)


16

Sol.2

sinx–

2

1I dx

1 + e

2

sinx sinx0

1 1dx

1 e 1 e

sinx2

sinx0

1 edx

1 e

2

9. If S is the sum of the first 10 terms of theseries

1 –1 –1 –11 1 1 1tan + tan + tan + tan +...,3 7 13 21

then tan(S) is equal to :

(1)6

–5

(2)511

(3)1011

(4)56

Answer (4)

Sol. 1 1 11 1 1

S tan tan tan .........3 7 13

1 1 11 1tan tan tan1 1 2 1 2 3

1.........

1 3 4

1 12 1 3 2tan tan1 2 1 1 3 2

1 14 3 11 10tan .......... tan1 3 4 1 11 10

= (tan–12 – tan–11) + (tan–1 3 – tan–12) +

(tan–14 – tan–13) + ..... + (tan–111 – tan–110)

= tan–111 – tan–11

1 11 1tan1 11 1

1 5tan6

tan(S) = 56

10. If (a, b, c) is the image of the point (1, 2, –3) inthe line,

x + 1 y – 3 z = ,

2 –2 –1 then a + b + c is equal to :

(1) 2 (2) 1

(3) 3 (4) –1

Answer (1)

Sol. Equation of line : x 1 y 3 z say2 2 1

a point on line L is

= Q(2 – 1, –2 + 3, –)

D·RS of PQ = < 2 – 2, –2 + 1, – + 3 >

PQ is perpendicular to line L

2(2 –2) –2 (–2 + 1) –1 (– + 3) = 0

4 – 4 + 4 – 2 + – 3 = 0

9 – 9 = 0

= 1

Coordinate of foot of = Q = (1, 1, –1)

Coordinate of image R = (1, 0, 1) = (a, b, c)

a + b + c = 2

11. If 32 sin2–1, 14 and 34–2 sin2 are the first threeterms of an A.P. for some , then the sixth termof this A.P is :

(1) 65 (2) 78

(3) 81 (4) 66

Answer (4)

Sol. Given 32sin2 – 1, 14, 34 – 2sin2 are in A.P.

So 32sin2 – 1 + 34 – 2sin2 = 28

2sin2

2sin2

3 8128

3 3

t 81

283 t {Put 32sin2 = t}

t2 – 84t + 243 = 0 t = 81, t = 3

When t = 81, when t = 3


17

sin2 = 2 (Not possible) 2sin2 = 1

1sin2

2

26

12

So, I term a = 3° = 1, d = 14 – 1 = 13

Now, T6 = a + 5d = 1 + 65 = 66

12. The mean and variance of 7 observations are 8and 16, respectively. If five observations are 2,4, 10, 12, 14, then the absolute difference ofthe remaining two observations is :

(1) 2 (2) 4

(3) 3 (4) 1

Answer (1)

Sol. Let two remaining observations are x, y

So 2 4 10 12 14 x y

x 87

(given)

x + y = 14 …(1)

Now also 22

2 i ix x 16N N

(given)

2 24 16 100 144 196 x y64 16

7

460 + x2 + y2 = (16 + 64) × 7

x2 + y2 = 100 …(2)

Now (x + y)2 = x2 + y2 + 2xy xy = 48 …(3)

Now (x – y)2 = (x + y)2 – 4xy = 196 – 192 = 4

x – y = 2 |x – y| = 2

13. If the four complex numbers z, z , z –2Ree zand z – 2Re(z) represent the vertices of asquare of side 4 units in the Argand plane, then|z| is equal to :

(1) 4 2

(2) 2

(3) 2 2

(4) 4

Answer (3)

Sol. D(z – 2Re(z))

A(z)B( )z

C(z – 2Re(z))

Let z = x + iy

Length of side of square = 4 units

| z z | 4 | 2iy | 4

| y | 2

Also | z (z 2Re(z)) | 4

|2Re(z)| = 4 |2x| = 4 |x| = 2

Now 2 2| z | x y

4 4 2 2 14. If is the positive root of the equation, p(x) = x2

– x – 2 = 0, then x

1 cos(p(x))lim

x 4

is equal to :

(1)1

2(2)

12

(3)3

2(4) 3

2

Answer (3)

Sol. Eq. P(x) = x2 – x – 2 = 0 x = 2, –1 = 2

Now 2

x 2

1 cos(x x 2)lim

x 2

2

x 2

x x 22 sin

2lim

x 2

22

2x 2

sin(x x 2)2 (x x 2)2lim

2(x 2)x x 22

2

2x 2 x 2

x x 2sin

21 (x 2)(x 1)lim lim

(x 2)2 x x 22

1 31 3

2 2


18

15. If the minimum and the maximum values of the

function f : , R,4 2

defined byy

2 2

2 2

sin 1 sin 1

f( ) cos 1 cos 1

12 10 2

are m and M respectively, then the ordered pair(m, M) is equal to :

(1) 0, 2 2 (2) (0, 4)(3) (–4, 4) (4) (–4, 0)

Answer (4)

Sol.

2 2

2 2

sin 1 sin 1

f( ) cos 1 cos 1

12 10 2

f() = 4cos2 (after finding determinant)

f() = –8sin2 < 0 ,4 2

So f is decreasing function

So minf( ) f 4 cos 4 m2

maxf( ) f 4 cos 04 2

= M

So (m, M) = (–4, 0)

16. A survey shows that 73% of the personsworking in an office like coffee, whereas 65%like tea. If x denotes the percentage of them,who like both coffee and tea, then x cannot be :

(1) 63 (2) 36

(3) 38 (4) 54

Answer (2)

Sol. n(C) = 73, n(T) = 65

65 n (C T) 65 + 73 – 100

65 x 38

x 36

17. If 10 9 1 8 2 9 10 112 2 3 2 3 .... 2 3 3 S – 2 ,then S is equal to :

(1) 112 3 (2) 11 123 – 2

(3)11

103 22

(4) 311

Answer (4)

Sol. LHS is G.P of common ratio 32

1110

11

32 1

2S 2

31–

2

11 11

1110 11

3 2

22 S 2

12

S = 311

18. The product of the roots of the equation29x – 18 x 5 0 , is :

(1)259

(2)2581

(3)59

(4)5

27Answer (2)

Sol. Let |x| = t we have

9t2 – 18t + 5 = 0

9t2 – 15t – 3t + 5 = 0

(3t – 1)(3t – 5) = 0

1 5 1 5

t or x or 3 3 3 3

Roots are 1 5

and 3 3

Product = 2581

19. If the point P on the curve, 4x2 + 5y2 = 20 isfarthest from the point Q(0, –4), then PQ2 isequal to :

(1) 29 (2) 48

(3) 21 (4) 36

Answer (4)

Sol. Curve is 2 2x y

C 15 4

Let a point on curve be 5 cos , 2sin

PQ2 = 2 25 cos (–4 – 2 sin ) = 2 25cos 4 sin 16 16 sin


19

PQ2 = 21 + 16sin – sin2

= 21 + 64 – (sin – 8)2

=85 – (sin – 8)2

For PQ2 to be maximum sin = 1

2maxPQ = 85 – 49 = 36

20. If the volume of a parallelopiped,whose coterminus edges are given by

the vectors a i j nk, b 2i 4 j nk and

c i nj 3k (n 0), is 158 cu.units, then :

(1) n = 7 (2) b c 10

(3) n = 9 (4) a c 17

Answer (2)

Sol. Volume of parallelopiped = a b c

1 1 n

2 4 –n 158

1 n 3

(12 + n2) – 1 (6 + n) + n (2n – 4) = 158

3n2 – 5n – 152 = 0

3n2 – 24n + 19n – 152 = 0

3n(n – 8) + 19 (n – 8) = 0

n = 8

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆa i j 8k, b 2i 4 j – 8k and c i 8 j 3k

a c 1 8 24 33

b c 2 32 – 24 10

SECTION - II

Numerical Value Type Questions: This sectioncontains 5 questions. The answer to each questionis a NUMERICAL VALUE. For each question, enterthe correct numerical value (in decimal notation,truncated/rounded off 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 numerickeypad in the place designated to enter the answer.

21. Four fair dice are thrown independently27 times. Then the expected number of times,at least two dice show up a three or a five,is ______.

Answer (11)

Sol. Probability of getting at least two 3s or 5s inone trial =

2 2 3 44 4 4

2 3 41 2 1 2 1

C C C3 3 3 3 3

= 433 11

273

E(x) = np = 11

27 1127

22. If the line, 2x – y + 3 = 0 is at a distance 1

5

and 2

5 from the lines 4x – 2y + = 0 and

6x – 3y + = 0, respectively, then the sum of allpossible values of and is ______.

Answer (30)

Sol. L1 : 2x – y + 3 = 0

L1 : 4x – 2y + = 0

L1 : 6x – 3y + = 0

Distance between L1 and L2;

6 1 6 2

2 5 5

= 4, 8

Distance between L1 and L3;

9 2 9 6

3 5 5

= 15, 3

Sum of all values = 4 + 8 + 15 + 3 = 30

23. The natural number m, for which the coefficient

of x in the binomial expansion of 22

m2

1x

x

is

1540, is _________.

Answer (13)

Sol.r

22 m 22–rr 1 r 2

1T C (x )

x

22 22m mr 2r

r 1 rT C x

22m – mr – 2r = 1 and 22 rC 1540

22 3C 1540 r 3 or 19

Now, for r = 3; 22m – 3m – 6 = 1

19m = 7 7

m19

(not acceptable)

for r = 19; 22m – 19m = 39

m = 13


20

24. The number of words, with or without meaning,

that can be formed by taking 4 letters at a time

from the letters of the word ‘SYLLABUS’ such

that two letters are distinct and two letters are

alike, is _______.

Answer (240)

Sol. LLSSYABU

For two alike and two distinct letters, select any

one pair from LL, SS in 2C1 ways

Now from rest, select any 2 in 5C2 ways and

they can be arranged in 4!2!

wayss

Required number of ways = 2 51 24!

C C2!

= 240

25. Let x

ƒ(x) x ,2

for –10 < x < 10, where [t]

denotes the greatest integer function. Then thenumber of points of discontinuity of ƒ is equalto ______.

Answer (8)

Sol.x

f(x) x2

may be discontinuous where x2

is

an integer.

So possible points of discontinuity are;

x = ± 2, ± 4, ± 6, ± 8 and 0

but at x = 0

x 0 x 0lim f(x) 0 f(0) lim f(x)

So f(x) will be discontinuous at x = ± 2, ± 4, ± 6and ± 8

Que & Ans_JEE(Main)-2020_Phase-2_05-09-2020_Morning_Physics_CompileAns & Sol_JEE(Main)-2020_Phase-2_05-09-2020_Morning_ChemistryQue & Sol_JEE(M)-2020_P-II (05-09-2020) Morning_Maths

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