(physics, chemistry and mathematics) · 2020-01-14 · 3. there are three parts in the question...

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

Evening

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

2

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. A uniform sphere of mass 500 g rolls withoutslipping on a plane horizontal surface with itscentre moving at a speed of 5.00 cm/s. Itskinetic energy is

(1) 6.25 × 10–4 J

(2) 1.13 × 10–3 J

(3) 8.75 × 10–4 J

(4) 8.75 × 10–3 J

Answer (3)

Sol. v =

= v/R

5

100m/s

KE =

2

2 21 2 vmR mR

2 5 R

2

2

2 4

1 7 v 7 1 25KE mR

2 5 10 2R 10

KE = –43510 Joule

4

KE = 8.75 × 10–4 Joule

2. In the given circuit, value of Y is

1

0

Y

(1) Toggles between 0 and 1

(2) 0

(3) 1

(4) Will not execute

Answer (2)

Sol.

A =1

B = 0

Y=01

10

0

1

Y AB.A AB A AB A

Y = 0 + 0 = 0

3. Consider a mixture of n moles of helium gas and2n moles of oxygen gas (molecules taken to berigid) as an ideal gas. Its C

p/C

v value will be

(1) 40/27 (2) 19/13

(3) 67/45 (4) 23/15

Answer (2)

Sol.

1 2

1 2

1 P 2 P

1 v 2 v

N C N C

N C N C

5 7n. R 2n. R

19nR 22 23 5 2(13nR)

n. R 2n. R2 2

p

v mixture

C 19

C 13

4. An object is gradually moving away from thefocal point of a concave mirror along the axisof the mirror. The graphical representation ofthe magnitude of linear magnification (m)versus distance of the object from the mirror(x) is correctly given by

(Graphs are drawn schematically and are not toscale)

(1)

m

1

f 2f

x (2)

m

1

f 2f

x

(3)

m

1

f 2f

x (4)

m

1

f 2f

x

Answer (4)


3

Sol. At focus magnification is

And at x = 2f, magnification is 1.

m

f 2f x(0, 0)

5. A transverse wave travels on a taut steel wirewith a velocity of v when tension in it is 2.06 ×104 N. When the tension is changed to T, thevelocity changed to v/2. The value of T is closeto

(1) 30.5 × 104 N (2) 2.50 × 104 N

(3) 5.15 × 103 N (4) 10.2 × 102 N

Answer (3)

Sol. T

v

1 1

2 2

v T

v T

4

2

v 2.06 102

v T

4

4

2

2.06 10T 0.515 10 N

4

3

2T 5.15 10 N

6. A galvanometer having a coil resistance 100 gives a full scale deflection when a current of1 mA is passed through it. What is the value ofthe resistance which can convert thisgalvanometer into a voltmeter giving full scaledeflection for a potential difference of 10 V?

(1) 10 k (2) 9.9 k

(3) 8.9 k (4) 7.9 k

Answer (2)

Sol. V = 10 volt

When 1 mA current will be flowing

10 = 1 × 10–3 (100 + R0)

10000 – 100 = 9900 = R0

R0 = 9.9 k

7. A Carnot engine having an efficiency of 1

10 is

being used as a refrigerator. If the work done onthe refrigerator is 10 J, the amount of heatabsorbed from the reservoir at lowertemperature is :

(1) 90 J (2) 1 J

(3) 99 J (4) 100 J

Answer (1)

Sol.

1

Q

2

Q

w

1 2

1 1

Q Q 1 w

Q 10 Q

1

1 w

10 Q

Q1 = w × 10 = 100 J

So, Q1 – Q

2 = w

100 – 10 = Q2 = 90 J

8.

L R

SE

As shown in the figure, a battery of emf E isconnected to an inductor L and resistance R inseries. The switch is closed at t = 0. The totalcharge that flows from the battery, betweent = 0 and t = t

C (t

C is the time constant of the

circuit) is :

(1)2

EL

R(2)

2

ER

eL

(3) 2

11

e

EL

R(4)

2

EL

eR

Answer (4)

Sol. ct /tE

i (1 e )R

q

R L

Θ C

Lt

R

C–t /tE

dq (1 e )dtR

C

C

tt /t

C 0

Eq t t e

R

C

C C

tEq t t

R e

E Lq

R Re

2

ELq

R e


4

9.

M

N

O

5 m

5 m

Two liquids of densities 1 and

2 (

2 = 2

1) are

filled up behind a square wall of side 10 m asshown in figure. Each liquid has a height of 5 m.The ratio of the forces due to these liquidsexerted on upper part MN to that at the lowerpart NO is (Assume that the liquids are notmixing)

(1) 1/4

(2) 1/2

(3) 2/3

(4) 1/3

Answer (1)

Sol. P1

= 0

F1

P2 = g5

F2

P = 15 g3

1 2

1

(P P )F A

2

2 3

2

(P P )F A

2

1

2

F 5 g 5 1

F 20 g 20 4

10. A particle of mass m is dropped from a height habove the ground. At the same time anotherparticle of the same mass is thrown verticallyupwards from the ground with a speed of

2gh . If they collide head-on completely

inelastically, the time taken for the combined

mass to reach the ground, in units of h

g is

(1)1

2(2)

3

4

(3)1

2(4)

3

2

Answer (4)

Sol.

Time of collision

0

h ht

2g2gh

2

1 0

1 1 h hs gt g ·

2 2 2g 4

h

4

3h

4

s1

s2

2

3hs

4

Speed of (A) just before collision v1

0

ghgt

2

And speed of (B) just before collision v2

gh2gh

2

After collision velocity of centres of mass

cm

gh ghm 2gh m

2 2V 0

2m

So from there, time of fall ‘t’

23h 1gt

4 2

3 ht

2 g

11. A simple pendulum is being used to determinethe value of gravitational acceleration g at acertain place. The length of the pendulum is25.0 cm and a stopwatch with 1 s resolutionmeasures the time taken for 40 oscillation to be50 s. The accuracy in g is

(1) 3.40% (2) 2.40%

(3) 5.40% (4) 4.40%

Answer (4)

Sol. l = 25.0 cm

Time of 40 oscillation is 50 sec

2

2

4 l g l 2 Tg

g l TT

g 0.1 12

g 25.0 50

g100 4.4%

g


5

12. A particle of mass m and charge q is releasedfrom rest in a uniform electric field. If there isno other force on the particle, the dependenceof its speed v on the distance x travelled by it iscorrectly given by (graphs are schematic andnot drawn to scale)

(1)

v

x

(2)v

x

(3)v

x

(4)v

x

Answer (4)

Sol.

x

v

2 2qEv · x

m

13. A capacitor is made of two square plates eachof side ‘a’ making a very small angle betweenthem, as shown in figure. The capacitance willbe close to

V1

V2

d

a

(1)

2

0a 3 a

1d 2d

(2)

2

0a a

1d 2d

(3)

2

0a a

1d d

(4)

2

0a a

1d 4d

Answer (2)

Sol.

dC

ax = 0

y = tan x

adx 60

dCd xtan

x a

eq 0

x 0

dxC dc a

xtan d

[By Binomial expansion]

aa 2

0 0eq

0 0

a ax tan x tanC 1 dx x

d d d d2

2 20 0

eq

a aa tan aC 1 1

d 2d d 2d

14. Consider two charged metallic spheres S1 and

S2 of radii R

1 and R

2, respectively. The electric

fields E1 (on S

1) and E

2 (on S

2) on their surfaces

are such that E1/E

2 = R

1/R

2. Then the ratio

V1(on S

1)/V

2(on S

2) of the electrostatic

potentials on each sphere is

(1) (R1/R

2)2 (2) (R

2/R

1)

(3)

3

1

2

R

R(4) R

1/R

2

Answer (1)

Sol. Let 1 and

2 are the charge densities of two

sphere. Then,

1 1 2 2

1 0

0 0

R RE andE

3 3

∵1 1 1 1

2 2 2 2

E R R

E R R

1 =

2 =

So,

2 2

1 2

1 2

0 0

R RV andV

3 3

2

1 1

2 2

V R

V R


6

15. A plane electromagnetic wave of frequency25 GHz is propagating in vacuum along thez-direction. At a particular point in space andtime, the magnetic field is given by

��

8 ˆB 5 10 j T . The corresponding electric field

��

E is (speed of light c = 3 × 108 ms–1)

(1) 16 ˆ1.66 10 i V/m

(2) 16 ˆ1.66 10 i V/m

(3) ˆ15 i V/m

(4) ˆ15 i V/m

Answer (4)

Sol. E B C

��

|E| = |B| C

��

|E| = 5 × 10–8 × 3 × 108 = 15 N/C

ˆE 15 i V/m��

16. In a double-slit experiment, at a certain pointon the screen the path difference between the

two interfering waves is 1th

8 of a wavelength.

The ratio of the intensity of light at that point tothat at the centre of a bright fringe is

(1) 0.568

(2) 0.760

(3) 0.853

(4) 0.672

Answer (3)

Sol. x = /8

Phase difference = 4

2 2

0 0I I cos I cos

2 8

0

I0.853

I

17. A very long wire ABDMNDC is shown in figurecarrying current I. AB and BC parts arestraight, long and at right angle. At D wireforms a circular turm DMND of radius R. AB,BC parts are tangential to circular turn at Nand D. Magnetic field at the centre of circle is

A

BCD

N

M

(1) 0I( 1)

2 R(2)

0I

2R

(3)

0I 1

2 R 2(4)

0I 1

2 R 2

Answer (3)

Sol.

i

1

2

3

4i

ii

� � ��

0 0 1 0 2 0 3 0 4B (B ) (B ) (B ) (B )

0 0 0I I I1 1

1 14 R 2R 4 R2 2

�

�

��

0

0

I 1B

2 R 2

18. As shown in fig. when a spherical cavity(centred at O) of radius 1 is cut out of auniform sphere of radius R (centred at C), thecentre of mass of remaining (shaded) part ofsphere is at G, i.e. on the surface of the cavity.R can be determined by the equation

R

1

CG O

(1) (R2 + R + 1) (2 – R) = 1

(2) (R2 – R + 1) (2 – R) = 1

(3) (R2 – R – 1) (2 – R) = 1

(4) (R2 + R – 1) (2 – R) = 1

Answer (1)


7

Sol.R

30

4M R

3

3cavity

4M (1)

3

3 3(Remaining)

4 4M R (1)

3 3

3 3 3 34 4 4 4

R (1) 2 (1) R R3 3 3 3

R4 – 2R3 + 1 = 0 ∵ R 1

R3 – R2 – R – 1 = 0 (R2 + R + 1) (2 – R) = 1

19. A particle moves such that its position vector

�

ˆ ˆr(t) cos t i sin t j where is a constantand t is time. Then which of the following

statements is true for the velocity �

v(t) and

acceleration ��

a(t) of the particle

(1)� ��

v and a both are perpendicular to �

r

(2)�

v is perpendicular to � ��

r and a is directedtowards the origin

(3)� ��

v and a both are parallel to �

r

(4)�

v is perpendicular to � ��

r and a is directedaway from the origin

Answer (2)

Sol. �

ˆ ˆr cos t i sin t j

�

� dr ˆ ˆv ( sin t i cos t j)dt

�

��

2dv ˆ ˆa (cos t i sin t j)dt

��

2a r a is antiparallel to r

Also � �

v r 0 � �

v r

Actually particle is in state of uniform circularmotion.

20. An electron (mass m) with initial velocity

�

0 0ˆ ˆv v i v j is in an electric field

��

0ˆE E k .

If 0 is initial de-Broglie wavelength of

electron, its de-Broglie wavelength at time t isgiven by

(1)

0

2 2 2

0

2 2

0

e E t1

m v

(2)

0

2 2 2

2 2

0

2

e E t1

m v

(3)

0

2 2 2

2 2

0

e E t1

2m v

(4)

0

2 2 2

2 2

0

e E t2

m v

Answer (3)

Sol. 0

0 0

eE ˆˆ ˆv v i v j t km

2

2 0

0

eE t|v| 2v

m

�

Initialy, 0

0

h

mv 2

Now,

2

2 0

0

h

eE tm 2v

m

20

0

0

1

eE t1

2mv

=

0

2 2 2

0

2 2

0

e E t1

2m v

SECTION - II

Numerical Value Type Questions: This sectioncontains 5 questions. The answer to each of thequestions is a numerical value. Each question carries4 marks for correct answer and there is no negativemarking for wrong answer.

21. Three containers C1, C

2 and C

3 have water at

different temperatures. The table below showsthe final temperature T when differentamounts of water (given in liters) are takenfrom each container and mixed (assume noloss of heat during the process)

� �

� �

� �

� � �

1 2 3C C C T

1 2 60 C

1 2 30 C

2 1 60 C

1 1 1

The value of (in °C to the nearest integer)is _______.

Answer (50.00)


8

Sol. Let C1 is at

1 ; C

2 is at

2 and C

3 is at

3

then

ms (1 – 60) = 2ms (60 –

2)

1 – 60 = 120 – 2

2

1 = 180 – 2

2...(i)

and ms(2 – 30) = 2ms(30 –

3)

2 = 90 – 2

3...(ii)

and 2ms (1 – 60) = ms (60 –

3)

21 – 120 = 60 –

3

21 +

3 = 180 ...(iii)

Adding them together 3(1 +

2 +

3) = 9

= 50°C

22. An asteroid is moving directly towards thecentre of the earth. When at a distance of10 R (R is the radius of the earth) from theearths centre, it has a speed of 12 km/s.Neglecting the effect of earths atmosphere,what will be the speed of the asteroid when ithits the surface of the earth (escape velocityfrom the earth is 11.2 km/s)? Give your answerto the nearest integer in kilometer/s _______.

Answer (16.00)

Sol. 2 2E E

0

GM m GM m1 1mV mV

10R 2 R 2

2 2

E 0GM V1 V

1R 10 2 2

2 2

0

9V V gR

5

2

0

9V V gR 16km/s

5

23. The series combination of two batteries, bothof the same emf 10 V, but different internalresistance of 20 and 5 , is connected tothe parallel combination of two resistors 30 and R . The voltage difference across thebattery of internal resistance 20 is zero, thevalue of R (in ) is _______.

Answer (30.00)

Sol. 10V 10

R

10V 10V

R

30

5 20 5 20

30RR

30 R

20

IR 25

Also

20 2010

R 25 R + 25 = 40

30RR 15

30 R 30 + R = 2R

R = 30

24. The first member of the Balmer series ofhydrogen atom has a wavelength of 6561 Å.The wavelength of the second member of theBalmer series (in nm) is _______.

Answer (486.00)

Sol.

0

hC 1 1E E

4 9...(i)

So,

0

hC 1 1E

4 16...(ii)

5 16

9 4 3

5 4 656.1

nm 486 nm9 3

25. A ball is dropped from the to of a 100 m high

tower on a planet. In the last 1

2s before hitting

the ground, it covers a distance of 19 m.Acceleration due to gravity (in ms–2) near thesurface on that planet is _______.

Answer (08.00)

Sol. 21h gt

2

2200 gt ...(i)

200t

g

Also

21 1g t 81

2 2

21

g t 81 22

...(ii)

1 81 2t

2 g

1 1

( 200 81 2) g 2(10 2 9 2)2 g

g 2 2

g = 8 m/s2


9

CHEMISTRY

SECTION - I

Multiple Choice Questions: This section contains20 multiple choice questions. Each question has4 choices (1), (2), (3) and (4), out of which ONLY ONE

is correct.

Choose the correct answer :

1. Among the compounds A and B with molecularformula C

9H

18O

3, A is having higher boiling point

the B. The possible structures of A and B are

(1) A = H CO3

OCH3

OCH3

B = HO OH

HO

(2) A = HO OH

HO

B = HOOH

OH

(3)

B = HOOH

OH

A = H CO3

OCH3

OCH3

(4) A = HO OH

HO

B = H CO3

OCH3

OCH3

Answer (4)

Sol. In (A), Intermolecular H-bonding is possiblewhile in (B) there is no inter-molecular H-bonding. So A is having higher boiling pointthan B.

2. Which of the following compounds is likely toshow both Frenkel and Schottky defects in itscrystalline form ?

(1) ZnS (2) CsCI

(3) AgBr (4) KBr

Answer (3)

Sol. AgBr shows both, Frenkel as well as Schottkydefects.

3. Among the reactions (a) - (d), the reaction(s)that does/do not occur in the blast furnaceduring the extraction of iron is/are

(a) CaO + SiO2 CaSiO

3

(b) 3Fe2O

3 + CO 2Fe

3O

4 + CO

2

(c) FeO + SiO2 FeSiO

3

(d) 2

1FeO Fe + O

2

(1) (c) and (d)

(2) (d)

(3) (a)

(4) (a) and (d)

Answer (1)

Sol. (c) FeO + SiO2 FeSiO

3

(d) 2

1FeO Fe + O

2

Reactions (c) and (d) do not occur in the blast

furnace in the metallurgy of iron.

4. The increasing order of the atomic radii of the

following elements is

(a) C (b) O (c) F

(d) CI (e) Br

(1) (d) < (c) < (b) < (a) < (e)

(2) (b) < (c) < (d) < (a) < (e)

(3) (c) < (b) < (a) < (d) < (e)

(4) (a) < (b) < (c) < (d) < (e)

Answer (3)


10

Sol. C N O F

Size decreasesCl

Br

I

Size increases

Correct increasing order of atomic radii is

F < O < C < Cl < Br

5. The radius of the second Bohr orbit, in termsof the Bohr radius, a

0, in Li2+ is

(1) 04a

3(2) 0

4a

9

(3) 02a

3(4) 0

2a

9

Answer (1)

Sol.2n

r = 0.529 ÅZ

Bohr’s radius for hydrogen atom (a0) = 0.529 Å

Bohr’s radius of Li+2 ion for n = 2

0

2n

= aZ

04a

=3

6. The correct order of the calculated spin-onlymagnetic moments of complexes (A)to (D) is

(A) Ni(CO)4

(B) [Ni(H2O)

6]Cl

2

(C) Na2[Ni(CN)

4] (D) PdCl

2(PPh

3)2

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

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

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

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

Answer (1)

Sol.

Number of unpaired electrons

(A) Ni(CO)4

Ni = 3d 84s2(SFL) 0 0

(B) [Ni(H2O)

6]Cl

2Ni+2 = 3d 8(WFL) 2 8 BM

(C) Na2[Ni(CN)

4] Ni+2 = 3d8(SFL) 0 0

(D) PdCl2(PPh

3)2

Pd+2 = 4d 8 0 0

Correct order of the calculated spin onlymagnetic moments of complexes A to D is

(A) � (C) � (D) < B

7. The major product [B] in the followingsequence of reactions is

CH – C = CH – CH CH3 2 3

CH(CH )3 2

(i) B H2 6

(ii) H O , OH2 2

–

[A]

2 4dil.H SO

[B]

(1) CH – C – CH CH CH3 2 2 3

C

H C3

CH3

(2) CH = C – CH CH CH2 2 2 3

CH(CH )3 2

(3) CH – C = CH–CH CH3 2 3

CH(CH )3 2

(4) CH – CH – CH = CH – CH3 3

CH(CH )3 2

Answer (1)

Sol. CH – C = CH – CH – CH3 2 3

CH

CH3

CH3

(i) B H2 6

(ii) H O , OH2 2

–

CH – CH – CH – CH – CH3 2 3

CH

CH3

CH3

OH

(A)

dil H SO , 2 4

C = CCH

3

CH – CH – CH2 2 3

CH3

CH3

(B)

8. Hydrogen has three isotopes (A), (B) and (C). Ifthe number of neutron(s) in (A), (B) and (C)respectively, are (x), (y) and (z), the sum of (x),(y) and (z) is

(1) 4 (2) 2

(3) 3 (4) 1

Answer (3)

Sol. Isotopes of hydrogen and the number ofneutrons present in them are

Number of neutrons

1 2 3H H (D) H (T)

0 1 2(x) (y) (z)

1 1 1

Total number of neutrons in three isotopes ofhydrogen = 0 + 1 + 2 = 3


11

9. The major product in the following reaction is

O

CH3

+ H O3

+

(1)

O

CH3HO

(2)

O

CH3

OH

(3)

OH

CH3

OH

(4)

OH

CH3

+

Answer (4)

Sol. OH

CH3

+

O

CH3

H O3

+

Product is aromatic

10. For the following Assertion and Reason, thecorrect option is

Assertion : The pH of water increases withincrease in temperature.

Reason : The dissociation of water into H+

and OH– is an exothermicreaction.

(1) Both assertion and reason are false

(2) Assertion is not true, but reason is true

(3) Both assertion and reason are true, and thereason is the correct explanation for theassertion

(4) Both assertion and reason are true, but thereason is not the correct explanation for theassertion

Answer (1)

Sol. Both assertion and reason are incorrect

11. A metal (A) on heating in nitrogen gas givescompound B. B on treatment with H

2O gives a

colourless gas which when passed throughCuSO

4 solution gives a dark blue-violet

coloured solution. A and B respectively, are

(1) Mg and Mg3N

2(2) Na and Na

3N

(3) Mg and Mg(NO3)2

(4) Na and NaNO3

Answer (1)

Sol. 3Mg + N2

Mg N

3 2

H O2

(A) (B)

2NH + 3Mg(OH)3 2

(Colourless gas)

CuSO4

[Cu(NH ) ]SO3 4 4

dark blue coloured

1

2

1

2

12. Preparation of Bakelite proceeds via reactions

(1) Electrophilic substitution and dehydration

(2) Electrophilic addition and dehydration

(3) Nucleophilic addition and dehydration

(4) Condensation and elimination

Answer (1)

Sol.Formation of Bakelite

Electrophilic substitution reaction of phenolwith formaldehyde followed by dehydration

13. Consider the following plots of rate constant

versus 1

T for four different reactions. Which of

the following orders is correct for the activationenergies of these reactions?

d ca

b

log k

1 / T

(1) Eb > E

a > E

d > E

c(2) E

c > E

a > E

d > E

b

(3) Ea > E

c > E

d > E

b(4) E

b > E

d > E

c > E

a

Answer (2)

Sol.a

Elogk = logA –

2.303 RT

aE

Slope = –2.303R

So correct order of activation energies

Ec > E

a > E

d > E

b

14. For the following Assertion and Reason, thecorrect option is

Assertion : For hydrogenation reactions, thecatalytic activity increases fromGroup 5 to Group 11 metals withmaximum activity shown by Group7-9 elements.

Reason : The reactants are most stronglyadsorbed on group 7-9 elements.


12

(1) Both assertion and reason are true and thereason is the correct explanation for theassertion.

(2) Both assertion and reason are false.

(3) The assertion is true, but the reason is false.

(4) Both assertion and reason are true but thereason is not the correct explanation for theassertion.

Answer (3)

Sol. The assertion is true, but the reason is false

Refer : [NCERT Page No. 132]

Reactants must not get adsorbed so stronglythat they are immobilised and other reactantsare left with no space on the catalyst’s surfacefor adsorption.

15. Arrange the following bonds according to theiraverage bond energies in descending order

C – Cl, C – Br, C – F, C – I

(1) C – Cl > C – Br > C – I > C – F

(2) C – Br > C – I > C – Cl > C – F

(3) C – F > C – Cl > C – Br > C – I

(4) C – I > C – Br > C – Cl > C – F

Answer (3)

Sol. Generally, bond energy 1

bond length

So bond energy order is

C–F > C–Cl > C–Br > C–I

16. White phosphorus on reaction withconcentrated NaOH solution in an inertatmosphere of CO

2 gives phosphine and

compound (X). (X) on acidification with HClgives compound (Y). The basicity of compound(Y) is

(1) 3 (2) 2

(3) 4 (4) 1

Answer (4)

Sol.

acidification

with HCl

3 2(Y)

Basicity of H PO 1

17. Two monomers in maltose are

(1) -D-glucose and -D-glucose

(2) -D-glucose and -D-glucose

(3) -D-glucose and -D-galactose

(4) -D-glucose and -D-Fructose

Answer (1)

Sol.Maltose -D-glucose + -D-glucose Hydrolysis

H

HO

H

H OH

O

OOH

HH H

OH

H H

H HOH OH

O O

OH OH

H H

OH

Hydrolysis

CH OH2

CH OH2

CH OH2

H H HOH

(Maltose) -D-glucose(2 moles)

18. Kjeldahl’s method cannot be used to estimatenitrogen for which of the following compounds?

(1) CH3CH

2–CN (2) NH –C–NH

2 2

O

(3) C6H

5NO

2(4) C

6H

5NH

2

Answer (3)

Sol. Kjeldahl’s method is not applicable tocompounds containing nitrogen in nitro, azogroups and nitrogen present in ring (pyridine).

19. Among (a) – (d), the complexes that can displaygeometrical isomerism are

(a) [Pt(NH3)3Cl]+

(b) [Pt(NH3)Cl

5]–

(c) [Pt(NH3)2Cl(NO

2)]

(d) [Pt(NH3)4ClBr]2+

(1) (c) and (d) (2) (a) and (b)

(3) (b) and (c) (4) (d) and (a)

Answer (1)

Sol.2

3 2 2 3 4(c) (d)

[Pt(NH ) Cl(NO )] and [Pt(NH ) ClBr]

can display geometrical isomerism

Pt

NH3

Cl

Br

H N3

H N3

NH3

(cis)

Pt

Cl

NH3

NH3

H N3

H N3

Br

(trans)

PtCl

NO2

H N3

H N3

(cis)

PtCl

NH3

H N3

O2N

(trans)

(octahedralcomplex)

(square planarcomplex)

(c)

(d)


13

20. An unsaturated hydrocarbon X absorbs twohydrogen molecules on catalytic hydrogenation,and also gives following reaction

3 3 2

2

O [Ag(NH ) ]

Zn/H OX A

B(3-oxo-hexanedicarboxylic acid)

X will be

(1) (2)

(3)

O

(4)

Answer (2)

Sol.

O

(X)

O3

Zn/H O2

(A)

CHO

CHO

[Ag(NH ) ]3 2

+

OCOOH

COOH

(B)

12

3

4 5 6

SECTION - II

Numerical Value Type Questions: This sectioncontains 5 questions. The answer to each of thequestions is a numerical value. Each question carries4 marks for correct answer and there is no negativemarking for wrong answer.

21. At constant volume, 4 mol of an ideal gaswhen heated from 300 K to 500 K changes itsinternal energy by 5000 J. The molar heatcapacity at constant volume is __________.

Answer (6.25)

Sol. U = nCvT

5000 = 4 × Cv(500 – 300)

Cv = 6.25 JK–1mol–1

22. For an electrochemical cell

Sn(s)|Sn2+(aq, 1M)||Pb2+(aq, 1M)|Pb(s)

the ratio 2

2

[Sn ]

[Pb ]

when this cell attains

equilibrium is __________.

2

0

Sn |SnGiven : E 0.14V,

2

0

Pb |Pb

2.303RTE 0.13V, 0.06

F

Answer (2.15)

Sol.At equilibrium state o

cell cellE 0 E 0.01V

2 2Sn s Pb aq. Sn aq. Pb s

o

cell

0.06 [P]E E log

n [R]

2

2

0.06 [Sn ]0 0.01 log

2 [Pb ]

2

2

0.06 [Sn ]0.01 log

2 [Pb ]

2

2

1 [Sn ]log

3 [Pb ]

12

3

2

[Sn ]10 2.1544

[Pb ] = 2.15

23. NaClO3 is used, even in spacecrafts, to

produce O2. The daily consumption of pure O

2

by a person is 492 L at 1 atm, 300 K. Howmuch amount of NaClO

3, in grams, is required

to produce O2 for the daily consumption of a

person at 1 atm, 300 K? __________.

NaClO3(s) + Fe(s) O

2(g) + NaCl(s) + FeO(s)

R = 0.082 L atm mol–1 K–1

Answer (2130.00)

Sol.3 2

NaClO (s) Fe(s) NaCl(s) FeO(s) O (g)

moles of NaClO3 = moles of O

2

moles of

2

PV 1 492O 20 mol

RT 0.082 300

mass of NaClO3 = 20 × 106.5 = 2130 g

= 2130.00

24. In the following sequence of reactions themaximum number of atoms present inmolecule ‘C’ in one plane is __________.

3

3

CH Cl (1. eq.)Red hot

Cu tube Anhydrous AlClA B C

(A is a lowest molecular weight alkyne)

Answer (13.00)


14

Sol.Red hot Cu ubet

3 H–C C–H(Lowest molecular

weight alkyne)

(A)(B)

CH Cl(1 eq.)3

Anhyd AlCl3

C

HH

H

H H

H

(C)

H

H

Number of atoms in one plane = 13

25. Complexes (ML5) of metals Ni and Fe have ideal

square pyramidal and trigonal bipyramidalgeometries, respectively. The sum of the 90°,120° and 180° L-M-L angles in the twocomplexes is __________.

Answer (20.00)

Sol. M L

L

L

L

L

120° = 3, 90° = 6, 180° = 1 Total = 10

M

L

L

L

L

L

90° = 8, 180° = 2 Total = 10

Total number of 180°, 90° and 120° L-M-L bondangles = 10 + 10 = 20

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(physics, chemistry and mathematics) · 2020-01-14 · 3. there are three parts in the question...

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