(Physics, Chemistry and Mathematics)...(Physics, Chemistry and Mathematics) 08/01/2020 Morning JEE...

Time : 3 hrs. M.M. : 300

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Answers & Solutions

forforforforfor

JEE (MAIN)-2020 (Phase-1)

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 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)

08/01/2020

Morning

JEE (MAIN)-2020 Phase-1 (8M)

2

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. When photon of energy 4.0 eV strikes the

surface of a metal A, the ejected

photoelectrons have maximum kinetic energy

TA

eV and de-Broglie wavelength A

.The

maximum kinetic energy of photoelectrons

liberated from another metal B by photon of

energy 4.50 eV is TB

= (TA

– 1.5) eV. If the

de-Broglie waelength of these photoelectrons

B

= 2A, then the work function of metal B is :

(1) 1.5 eV (2) 4 eV

(3) 3 eV (4) 2 eV

Answer (2)

Sol. de-Broglie wavelength (),

h

mv p 2m(KE)

h

2mKE

A AB

B A A

T 1.5K(as given)

K T

Also,

A

B

1

2

On solving TA = 2 eV

KB = T

A – 1.5 = 0.5 eV

Work function of metal B is

B = E

B – K

B

= 4.5 – 0.5 = 4 eV

For A, A

= EA

– TA

= 2 eV

2. The length of a potentiometer wire is 1200 cm

and it carries a current of 60 mA. For a cell of

emf 5 V and internal resistance of 20 , the

null point on it is found to be at 1000 cm. The

resistance of whole wire is :

(1) 80 (2) 100

(3) 60 (4) 120

Answer (2)

Sol.

Potential gradient for the potentiometer wire

‘AB’ is

� �

mvm

AB

dV 60 R

d

AP AP

dV 60 RV 1000mV

d 1200

��

VAP

= 50 R mV

Also, VAP

= 5 V (for balance point at P)

3

5R 100

50 10

3. Effective capacitance of parallel combination

of two capacitors C1 and C

2 is 10 F. When

these capacitors are individually connected to

a voltage source of 1 V, the energy stored in

the capacitor C2 is 4 times that of C

1. If these

capacitors are connected in series, their

effective capacitance will be :

(1) 1.6 F

(2) 3.2 F

(3) 4.2 F

(4) 8.4 F

Answer (1)

Sol. In parallel combination, C1 + C2 = 10 F

When connected across 1 V battery, then

1 1

2 2

U C1 1

U 4 C 4

C2 = 8 F C

1 = 2 F

In series combination,

1 2

equivalent1 2

C C 16C 1.6 F

C C 10


3

4. A particle of mass m is fixed to one end of a

light spring having force constant k and

unstretched length l. The other end is fixed. The

system is given an angular speed about the

fixed end of the spring such that it rotates in a

circle in gravity free space. Then the stretch in

the spring is

(1)

2ml

k m(2)

2

2

ml

k m

(3)

2ml

k m(4)

2

2

ml

k m

Answer (4)

Sol.

m

( + x)�0

kx

At elongated position (x),

Fradial

= mr2

kx = m(� + x)2

2

2

mx

k m

�

5. Boolean relation at the output stage-Y for the

following circuit is :

(1) A B (2) A B

(3) A + B (4) A B

Answer (1)

Sol.A

B5 V

Output (Y)

5 V

Diode OR Gate

Transistor NOT Gate

OR + NOT NOR Gate

Y A B A B

6. Consider a solid sphere of radius R and mass

density

2

0 2

r(r) 1 ,

R 0 < r R. The

minimum density of a liquid in which it will float

is :

(1)0

2

5

(2)0

3

(3)0

5

(4)0

2

3

Answer (1)

Sol. For minimum density of liquid, solid sphere has

to float (completely immersed) in the liquid.

mg = FB (Also V

immersed = V

total)

or �

34dV R

3

�

R 22 3

0 2

0

r 44 1 r dr R

3R

�

R3 5

3

0 20

r r 44 R

3 35R

�

3304 R 2 4

R3 5 3

�02

5

7. Proton with kinetic energy of 1 MeV moves

from south to north. It gets an acceleration of

1012 m/s2 by an applied magnetic field (west to

east). The value of magnetic field :

(Rest mass of proton is 1.6 × 10–27 kg)

(1) 0.071 mT

(2) 0.71 mT

(3) 71 mT

(4) 7.1 mT

Answer (2)


4

Sol.

Magnetic Force F = qvB

� qvBa

m perpendicular to velocity.

Also

62K 2 e 10

vm m

6qvB eB 2 e 10a

m m m

3

19 212 3

27

1.6 1010 2 10 B

1.67 10

31

B 10 T 0.71 mT (approx)2

�

8*. The dimension of stopping potential V0 in

photoelectric effect in units of Planck’s

constant ‘h’, speed of light ‘c’ and Gravitational

constant ‘G’ and ampere A is :

(1) h1/3G2/3c1/3A–1

(2) h2/3c5/3G1/3A–1

(3) h–2/3c–1/3G4/3A–1

(4) h2G3/2c1/3A–1

Answer (No option is correct) Bonus

Sol. Stopping potential (V) hxIyGzCr

Here, h = plank’s constant = [ML2T–1]

I = current = [A]

G = Gravitational constant = [M–1L3T–2]

and c = speed of light = [LT–1]

and V = potential = [ML2T–3A–1]

[ML2T–3A–1] = [ML2T–1]x [A]y [M–1L3T–2]z [LT–1]r

Comparing dimension of M, L, T, A, we get

y = –1, x = 0, z = – 1, r = 5

0 1 1 5v h I G c

9. A thermodynamic cycle xyzx is shown on a

V-T diagram

The P-V diagram that best describes this cycle

is : (Diagrams are shcematic and not to scale)

(1)

(2)

(3)

(4)

Answer (1)

Sol.

From the corresponding V-T graph

Process xy Isobaric expansion,

Process yz Isochoric (Pressure decreases)

Process z – x Isothermal compression

Therefore, corresponding PV graph is


5

10. Consider two solid spheres of radii R1 = 1 m,

R2 = 2 m and masses M

1 and M

2, respectively.

The gravitational field due to sphere and

are shown. The value of 1

2

M

M is

(1)1

6

(2)1

2

(3)2

3

(4)1

3

Answer (1)

Sol. From the diagram

Gravitation field at the surface

2

GmE

r

11 2

1

GmE

r

and 22 2

2

GmE

r

2

1 12

2 1 2

E mr

E r m

2

1

2

m2 2

3 1 m

1

2

m 1

m 6

11. The plot that depicts the behavior of the mean

free time (time between two successive

collisions) for the molecules of an ideal gas, as

a function of temperature (T), qualitatively, is

(Graphs are schematic and not drawn to scale)

(1) (2)

(3) (4)

Answer (2)

Sol. Relaxation time mean free path( )

speed

1

v

or 1

T

Graph between 1

v/sT

is a straight line,

12. The graph which depicts the results of

Rutherford gold foil experiment with -particle is

: Scattering angle

Y : Number of scattered -particles detected

(Plots are schematic and not to scale)

(1) (2)

(3) (4)

Answer (4)


6

Sol. In 1911, Ernest Rutherford publish a formula.

Which indicates that number of particles (Y)

that would be deflected by an angle ‘’ due to

scattering is

( )

4

KY

sin2

where K = constant

Corresponding graph.

13. Three charged particles A, B and C with

charges –4q, 2q and –2q are present on the

circumference of a circle of radius d. The

charged particles A, C and centre O of the

circle formed an equilateral triangle as shown

in figure. Electric field at O along x-direction is

(1) 2

0

3q

d(2) 2

0

3 3q

4 d

(3) 2

0

3q

4 d(4) 2

0

2 3q

d

Answer (1)

Sol.

Electric field due to charge +2q at centre O –

� ��

1 20

1 2q 3 i jE

4 2d

Due to –2q

� ��

2 20

1 2q 3 i jE

4 2d

Due to –4q

� ��

3 20

1 4q 3 i jE

4 2d

Net electric field at point O

� � � ��

0 1 2 3 2

0

3 qE E E E i

d

14. In finding the electric field using Gauss law the

formula

enc

0

qE

A

��

is applicable. In the formula

0

is permittivity of free space, A is the area of

Gaussian surface and qenc is charge enclosed

by the Gaussian surface. This equation can be

used in which of the following situation?

(1) Only when the Gaussian surface is an

equipotential surface and E��

is constant on

the surface.

(2) Only when E��

= constant on the surface.

(3) Only when the Gaussian surface is an

equipotential surface.

(4) For any choice of Gaussian surface.

Answer (1)

Sol. By Gauss law

in

0

QE dA

��

�

If in

0

Q

EA

�

�

or in

0

Q

E A��

then E || A��

Surface is equipotential too.


7

15. The coordinates of centre of mass of a uniform

flag shaped lamina (thin flat plate) of mass

4 kg. (The coordinates of the same are shown

in figure) are

(0, 3) (2, 3)

(2, 2)(1, 2)

(0, 0) (1, 0)

(1) (0.75 m, 1.75 m) (2) (1.25 m, 1.50 m)

(3) (1 m, 1.75 m) (4) (0.75 m, 0.75 m)

Answer (1)

Sol. (0, 3) (2, 3)

(2, 2)(1, 2)

(0, 0) (1, 0)

(1, 3)

C2

C1

For given Lamina

A1 = 1, C1 = (1.5, 2.5)

A2

= 3, C2 = (0.5, 1.5)

cm

1.5 1.5X 0.75

4

cm

2.5 4.5Y 1.75

4

Coordinate of centre of mass

(0.75, 1.75)

16. At time t = 0 magnetic field of 1000 Gauss is

passing perpendicularly through the area

defined by the closed loop shown in the figure.

If the magnetic field reduces linearly to 500

Gauss, in the next 5 s, then induced EMF in the

loop is

16 cm

2 cm4 cm

(1) 48 V (2) 36 V

(3) 56 V (4) 28 V

Answer (3)

Sol. Using faraday law

Induced EMF d dB

Adt dt

4 2dB 1000 50010 10 T/sec

dt 5

16 cm

2 cm4 cm

Area = ( 16 × 4 – 2 × Area of triangle) cm2

21cm64 2 2 4

2

= 56 × 10–4 m2

induced

= 56 × 10–6 V = 56 V

17. The critical angle of medium for a specific

wavelength, if the medium has relative

permittivity 3 and relative permeability 4

3 for

this wavelength, will be

(1) 60°

(2) 45°

(3) 15°

(4) 30°

Answer (4)

Sol. For relative permittivity = 3, = 30

For relative permeability 4,

3

0

4

3

= 40

0

0 0 v 1

c 2

c

1sin

2

c = 30°


8

18. The magnifying power of a telescope with tube

length 60 cm is 5. What is the focal length of its

eye piece?

(1) 30 cm (2) 10 cm

(3) 20 cm (4) 40 cm

Answer (2)

Sol.

fo

fe

Eye-piece

Objective

For telescope

Tube length (L) = fo + f

e

and magnification (m) e

o

f

f

where fo and f

e are focal length of objective

and eyepiece

fo + f

e = 60

and fe = 5f

o

fo = 50 cm

fe = 10 cm

19. Consider a uniform rod of mass M = 4 m and

length l pivoted about its centre. A mass m

moving with velocity v making angle 4

to

the rod’s long axis collides with one end of the

rod and sticks to it. The angular speed of the

rod-mass system just after the collision is

(1)4 v

7 l(2)

3 v

7 l

(3)3 v

l7 2(4)

3 2 v

7 l

Answer (4)

Sol.

P 45°

v

l

2 2

About point P, angular momentum (L) of thesystem

Linitial

mv l

22

2 2

final

(4m)l mlL

12 4

As Linitial

= Lfinal

3 2v

7 l

20. A leak proof cylinder of length 1 m, made of a

metal which has very low coefficient of

expansion is floating vertically in water at 0°C

such that its height above the water surface is

20 cm. When the temperature of water is

increased to 4°C, the height of the cylinder

above the water surface becomes 21 cm. The

density of water at T = 4°C, relative to the

density at T = 0°C is close to

(1) 1.04

(2) 1.03

(3) 1.26

(4) 1.01

Answer (4)

Sol. Law of floatation

bodyi

liquid

V

V

In given case

4 C1 water

0 C2 water

h

h body

constant

4 C

0 C

80100 201.01

79100 21

SECTION - II

Numerical Value Type Questions: This section

contains 5 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.

21. Four resistances of 15 , 12 , 4 and 10 respectively in cyclic order to form

Wheatstone’s network. The resistance that is

to be connected in parallel with the resistance

of 10 to balance the network is _________ .

Answer (10)


9

Sol. 12 15

4 10

R

G

Wheatstone bridge balance condition

1 3

2 4

R R

R R

15 15

10R4

10 R

10R 15 4

510 R 12

2R = 10 + R

R = 10

22. A one metre long (both ends open) organ pipe

is kept in a gas that has double the density of

air at STP. Assuming the speed of sound in air

at STP in 300 m/s, the frequency difference

between the fundamental and second

harmonic of this pipe is _______ Hz.

Answer (106)

vair

= 300 m/s

gasair

B 300v 150 2 m/s

2 2

th

n harmonics

nvf

2L (open organ pipe)

gas

1 0

v 150 2f f 75 2 Hz.

2L 2 1

= 106 Hz (approx)

23*. A body A, of mass m = 0.1 kg has an initial

velocity of ˆ3i ms–1. It collides elastically with

another body, B of the same mass which has

an initial velocity of ˆ5 j ms–1. After collision, A

moves with a velocity ˆ ˆv 4 i j�

. The energy

of B after collision is written as x

J10

. The value

of x is ________.

Answer (01) Bonus

Sol. By conservation of momentum

A B A Bu u v v� � � �

(Masses are equal)

Bˆ ˆ ˆ ˆ ˆ ˆv (3i 5 j) (4i 4 j) j i

�

22

B B

1 1KE mv 0.1 2

2 2

1J

10

xJ

10

x = 1

24*. A particle is moving along the x-axis with

its coordinate with time ‘t’ given by x(t) = 10 +

8t – 3t2. Another particle is moving along the

y-axis with its coordinate as a function of time

given by y(t) = 5 – 8t3. At t = 1 s, the speed of

the second particle as measured in the frame

of the first particle is given as v . Then v

(in m/s) is ________.

Answer (580) Bonus

Sol. For particle ‘A’ For particle ‘B’

XA = – 3t2 + 8t + 10 Y

B = 5 – 8t3

A

ˆV (8 6t)i�

2

BˆV 24t j

�

A

ˆa 6i�

B

ˆa 48t j�

at t = 1 sec

A B

ˆ ˆv 2i, v 24 j� �

� � �

B/A A Bˆ ˆv v v 2i 24 j

Speed of B w.r.t. A, �

B/Av 4 576 580

2v 580 m/s

25. A point object in air is in front of the curved

surface of a plano-convex lens. The radius of

curvature of the curved surface is 30 cm and

the refractive index of the lens material is 1.5,

then the focal length of the lens (in cm) is

______.

Answer (60)

Sol. Lens-maker formula 1 2

1 11( 1)

R Rf

for plano-convex lens.

R1 then R

2 = – R

R 30f 60 cm.

1 1.5 1


10

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.

Choose the correct answer :

1. The complex that can show fac-and mer-

isomers is

(1) [CoCl2(en)

2] (2) [Co(NH

3)

3(NO

2)

3]

(3) [Pt(NH3)

2Cl

2] (4) [Co(NH

3)

4Cl

2]+

Answer (2)

Sol. [Co(NH3)3(NO

2)3] will show fac and mer isomers

2. As per Hardy-Schulze formulation, the

flocculation values of the following for ferric

hydroxide sol are in the order

(1) AlCl3 > K

3[Fe(CN)

6] > K

2CrO

4 > KBr = KNO

3

(2) K3 [Fe(CN)

6] < K

2CrO

4 < KBr = KNO

3 = AlCl

3

(3) K3[Fe(CN)

6] > AlCl

3 > K

2CrO

4 > KBr > KNO

3

(4) K3[Fe(CN)

6] < K

2CrO

4 < AlCl

3 < KBr < KNO

3

Answer (2)

Sol. Fe(OH)3 is a positive sol. Its coagulation will be

caused by the anion of the electrolyte.

The flocculation value is inversely proportional

to coagulation power or valency of the anion.

The correct order of flocculation value is

K3 [Fe(CN)

6] < K

2CrO

4 < KBr = KNO

3 = AlCl

3

3. The decreasing order of reactivity towards

dehydrohalogenation (E1) reaction of the

following compounds is

(A) Cl (B) Cl

(C)

Cl

(D)

Cl

(1) B > A > D > C

(2) B > D > A > C

(3) B > D > C > A

(4) D > B > C > A

Answer (4)

Sol. Dehydrohalogenation of the given halides by E1

mechanism is decided by the stability of

carbocation formed in the first step. The

correct decreasing order of the given halides

towards dehydrohalogenation by E1 mechanism

is

Cl

Cl

(D) (B)

> >

(C) (A)

Cl

Cl>

4. For the Balmer series in the spectrum of H

atom,

H 2 2

1 2

1 1R ,

n n the correct statements

among (I) to (VI) are

(I) As wavelength decreases, the lines in the

series converge

(II) The integer n1 is equal to 2

(III) The lines of longest wavelength

corresponds to n2 = 3

(IV)The ionization energy of hydrogen can be

calculated from wave number of these lines

(1) (I), (II), (III)

(2) (II), (III), (IV)

(3) (I), (III), (IV)

(4) (I), (II), (IV)

Answer (1)

Sol. In Balmer series of H-atom, the electronic

transitions take place from higher orbits to 2nd

orbit and the longest wavelength will

correspond to transition from 3rd orbit to 2nd

orbit.

n1 = 2 and n

2 = 3 for longest wavelength.

As wavelength decreases the lines in the

Balmer series converge. The correct

statements are (I), (II) and (III).


11

5. The first ionization energy (in kJ/mol) of Na, Mg,

Al and Si respectively, are

(1) 786, 737, 577, 496 (2) 496, 577, 786, 737

(3) 496, 737, 577, 786 (4) 496, 577, 737, 786

Answer (3)

Sol. Ionisation energy of elements belonging to

period III in general increases as we move from

left to right with the exception of Group-2 and

Group-15 elements due to their stable

configuration. The increasing order of first

ionisation energy of the given elements is

Na < Al < Mg < Si

Ionisation energy of the given metals are

Na : 496 kJ/mol ; Al : 577 kJ/mol

Mg : 737 kJ/mol ; Si : 786 kJ/mol

6. The predominant intermolecular forces present

in ethyl acetate, a liquid, are

(1) Dipole-dipole and hydrogen bonding

(2) London dispersion and dipole-dipole

(3) Hydrogen bonding and London dispersion

(4) London dispersion, dipole-dipole and

hydrogen bonding

Answer (2)

Sol. The intermolecular forces present in liquid

ethyl acetate are

(i) Dipole-dipole interaction

(ii) London dispersion

7. A graph of vapour pressure and temperature for

three different liquids X, Y, and Z is shown below

Va

po

ur

pre

ss

ure

(mm

Hg

) 800

500

400

200

0 293 313 333 353

X Y Z

Temp

The following inferences are made

(A) X has higher intermolecular interactions

compared to Y.

(B) X has lower intermolecular interactions

compared to Y

(C) Z has lower intermolecular interactions

compared to Y.

The correct inferences is/are

(1) (B) (2) (C)

(3) (A) and (C) (4) (A)

Answer (1)

Sol. Vapour pressure of a liquid at a given

temperature is inversely proportional to

intermolecular force of attraction. At the same

temperature, vapour pressure of X is higher

than that of Y.

Therefore (X) has lower intermolecular

interactions compared to Y. Statement (B) is

correct.

8. The stoichiometry and solubility product of a

salt with the solubility curve given below is,

respectively

[Y]/mM 3

2

1

[X]/mM

1 2 3

(1) X2Y, 2 × 10–9 M3 (2) XY

2, 4 × 10–9 M3

(3) XY2, 1 × 10–9 M3 (4) XY, 2 × 10–6 M3

Answer (2)

Sol. From the given curve,

if [X] = 1 mM then [Y] = 2 mM

Salt is XY2

Ksp

= [X][Y]2 = (10–3)(2 × 10–3)2 = 4 × 10–9 M3

9. When gypsum is heated to 393 K, it forms

(1) Anhydrous CaSO4

(2) Dead burnt plaster

(3) CaSO4 5H

2O

(4) CaSO4 0.5H

2O

Answer (4)

Sol. Gypsum on heating to 393 K forms plaster of

Paris

393K

4 2 4 2 2CaSO 2H O CaSO 0.5H O 1.5H O

10. Arrange the following compounds in increasing

order of C – OH bond length

methanol, phenol, p-ethoxyphenol

(1) phenol < p-ethoxyphenol < methanol

(2) methanol < phenol < p-ethoxyphenol

(3) methanol < p-ethoxyphenol < phenol

(4) phenol < methanol < p-ethoxyphenol

Answer (1)


12

Sol. C-OH bond length in methanol, phenol and

p-ethoxyphenol is least in phenol due to

resonance and maximum in methanol due to

lack of resonance whereas it will have some

intermediate value in p-ethoxyphenol.

correct increasing order is

phenol < p-ethoxyphenol < methanol

11. The rate of a certain biochemical reaction at

physiological temperature (T) occurs 106 times

faster with enzyme than without. The change in

the activation energy upon adding enzyme is

(1) – 6RT (2) + 6RT

(3) + 6(2.303)RT (4) – 6(2.303)RT

Answer (4)

Sol. The rate constant of a reaction is given by

aE /RT

k Ae

The rate constant in presence of catalyst is

given by

aE /RT

k Ae

a a(E E )/RTk

ek

a a(E E )/RT6

10 e

6 a a(E E )

ln 10RT

a aE E 6(2.303)RT

12. The strength of an aqueous NaOH solution is

most accurately determined by titrating

(Note : consider that an appropriate indicator is

used)

(1) Aq. NaOH in a pipette and aqueous oxalic

acid in a burette

(2) Aq. NaOH in a burette and aqueous oxalic

acid in a conical flask

(3) Aq. NaOH in a burette and concentrated

H2SO

4 in a conical flask

(4) Aq. NaOH in a volumetric flask and

concentrated H2SO

4 in a conical flask

Answer (2)

Sol. In the titration of acid with a strong base like

NaOH, the oxalic acid is taken in a conical flask

and NaOH is taken in a burette.

13. The most suitable reagent for the given

conversion is

CONH2

HO C2

CN

C ==O

CH3

?

CONH2

HOH C2

CN

COCH3

(1) LiAIH4

(2) NaBH4

(3) H2/Pd (4) B

2H

6

Answer (4)

Sol. Diborane selectively reduces carboxylic acid to

alcohol in preference to other functional groups

like amide, carbonyl group and cyanide group.

14. Which of the following statement is not true for

glucose?

(1) Glucose reacts with hydroxylamine to form

oxime

(2) The pentaacetate of glucose does not react

with hydroxylamine to give oxime

(3) Glucose exists in two crystalline forms

and (4) Glucose gives Schiff’s test for aldehyde

Answer (4)

Sol. Glucose exists is two anomeric forms and .

It forms oxime with NH2OH and its

pentaacetate does not react with NH2OH

because its anomeric OH group is converted

into acetate group. But glucose does not give

Schiff test for aldehyde

15. The third ionization enthalphy is minimum for

(1) Mn (2) Fe

(3) Co (4) Ni

Answer (2)

Sol. The electronic configurations of the given

metals and in their +3 state are :

Mn : 3d 5 4s2 Mn3+ : 3d 4

Fe : 3d 6 4s2 Fe3+ : 3d 5

Co : 3d7 4s2 Co3+ : 3d 6

Ni : 3d 8 4s2 Ni3+ : 3d 7

Since Fe3+ has stable configuration of 3d5, the

third ionization energy of Fe is minimum.


13

16. Among the gases (a) - (e), the gases that cause

greenhouse effect are

(a) CO2

(b) H2O

(c) CFCs (d) O2

(e) O3

(1) (a), (b), (c) and (d)

(2) (a), (c), (d) and (e)

(3) (a), (b), (c) and (e)

(4) (a) and (d)

Answer (3)

Sol. Green house gases = CO2 , H

2O (Vapour), CFCs

and O3

17. The major product of the following reaction is

OHH C3

H C3H C

3

dil. H SO2 4

(1)

CH3

CH3

OHH C3

(2)

H C3

H C3

OH

OH

CH3

OH

(3)

H C3

H C3

OHOH

CH3

HO

(4)

CH3

OH

CH3H C3

Answer (4)

Sol.

+

H O2

–H+

OH

OH

dil H SO2 4

– H O2CH2

+

18. A flask contains a mixture of isohexane and

3-methylpentane. One of the liquids boils at

63°C while the other boils at 60°C. What is the

best way to separate the two liquids and which

one will be distilled out first?

(1) Simple distillation, isohexane

(2) Fractional distillation, isohexane

(3) Simple distillation, 3-methylpentane

(4) Fractional distillation, 3-methylpentane

Answer (2)

Sol. Two volatile liquids having their boiling points

close to each other can be separated by

fractional distillation. Out of the two given

liquids, isohexane has lower boiling point and

hence will distill out first.

19. The major products A and B in the following

reactions are

CN Peroxide

Heat[A]

[A] + B

(1)CN

ACN

and B

(2)CN

and B

CN

A

(3)CN CN

and B A

(4)CN

CNA and B

Answer (4)

Sol. Peroxide generates a radical that abstracts

H-atom from the C-atom adjacent to CN group

to give more stable radical

CN Peroxide

CN

[A]

[A] attacks 1-pentene to give 2° radical that

picks up H-atom to give [B]

CN CN+

[B]


14

20. The number of bonds between sulphur and

oxygen atoms in 28

2S O and the number of

bonds between sulphur and sulphur atoms in

rhombic sulphur, respectively, are

(1) 4 and 6 (2) 8 and 8

(3) 8 and 6 (4) 4 and 8

Answer (2)

Sol. S O2 8

2– – –

O — S — O — O — S — O

O

O

O

O

Number of S — O bonds in 28

2S O is 8

S8

S

S

S

S

S

SS

S

Number of S — S bonds in rhombic sulphur is 8.

SECTION - II

Numerical Value Type Questions: This section

contains 5 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.

21. The magnitude of work done by a gas that

undergoes reversible expansion along the path

ABC shown in the figure is _____________.

4 6 8 10 12

4

6

8

10B

C

(2, 2)

Pressure (Pa)

Volume

(m )3

A

Answer (48.00)

Sol. Work done by a gas that undergoes a reversible

expansion along the path ABC is given by

8 12

B

C

2

P (Pa)

Volume(m )

3

A

2

8

1W = 6×6 + × 4 ×6

2 = 48.00 J

22. The volume (in mL) of 0.125 M AgNO3

required

to quantitatively precipitate chloride ions in

0.3 g of [Co(NH3)

6]CI

3 is _____________.

M[Co(NH3)6]CI

3 = 267.46 g/mol

MAgNO3 = 169.87 g/mol

Answer (26.92)

Sol. Number of moles of the given complex,

0.3=267.46

Number of moles of CI– ions = 0.3× 3

267.46

Moles of Ag+ ions needed to ppt. CI– = 0.3× 3

267.46

Let the volume of 0.125 M AgNO3 needed by V ml.

0.125×V 0.3×3=

1000 267.46

V = 26.92 ml

23. The number of chiral centres in penicillin is

_____________.

Answer (3.00)

Sol. General structure of pencillin is

R — C — NH

O

O

H H

* *

N *

S CH3

CH3

COOH

H

The number of chiral centres in pencillin is 3.

24. What would be the electrode potential for the

given half cell reaction at pH = 5? _________.

– 0

2 2 red2H O O + 4H + 4e ; E = 1.23 V

(R = 8.314 J mol–1K–1; Temp = 298 K; oxygen

under std. atm. pressure of 1 bar)

Answer (–0.93)

Sol. The given half cell reaction is

+ –

2 22H O O + 4H + 4e

ox

E = –1.23 V

4+°ox ox 2

0.0591E =E – log Po H

4


15

At PO2 = 1 bar and [H+] = 10–5 M

–20

ox

0.0591E = –1.23 – log10

4

= –1.23 + 0.2955 = –0.93 V

25. Ferrous sulphate heptahydrate is used to fortify

foods with iron. The amount (in grams) of the

salt required to achieve 10 ppm of iron in

100 kg of wheat is_____________.

Atomic weight : Fe = 55.85; S = 32.00; O = 16.00

Answer (4.97)

Sol. Mass of iron needed in 100 kg wheat = 5

6

10×10

10

= 1.0 gm

Molecular mass of FeSO4.7H

2O is 277.85

55.85 gm iron is present in 277.85 gm of salt

1 gm iron is present in 277.85

= 4.97 gm.55.85

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(Physics, Chemistry and Mathematics)...(Physics, Chemistry and Mathematics) 08/01/2020 Morning JEE...

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