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Answers & Solutions

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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)

07/01/2020

Morning

JEE (MAIN)-2020 Phase-1 (7M)

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.

A parallel plate capacitor has plates of area A

separated by distance ‘d’ between them. It is

filled with a dielectric which has a dielectric

constant that varies as k(x) = K(1 + x) where

‘x’ is the distance measured from one of the

plates. If (d) << 1, the total capacitance of the

system is best given by the expression

(1)

0AK

(1 d)d

(2)

2 20

A K d1

d 2

(3)

0

AK d1

d 2(4)

2

0A K d

1d 2

Answer (3)

Sol. k K 1 x

0

el

K 1 x AC

dxx

d

el 00

1 1 dxd

C C KA 1 x

d0

0

1 1ln 1 x

C KA

0

1 1ln 1 d

C KA

2 2

0

1 dd

KA 2

0

d d1

KA 2

0 0KA KA dC C 1

d d 2d 1

2

2. If the magnetic field in a plane

electromagnetic wave is given by

��

8 3 10 ˆB 3 10 sin(1.6 10 x 48 10 t) j T , then

what will be expression for electric field?

(1) ��

3 10 Ê 60sin(1.6 10 x 48 10 t)k V /m

(2) ��

8 3 10 Ê 3 10 sin(1.6 10 x 48 10 t)i V /m

(3) ��

3 10 Ê 9sin(1.6 10 x 48 10 t)k V /m

(4) ��

8 3 10 Ê 3 10 sin(1.6 10 x 48 10 t) jV /m

Answer (3)

Sol. 8 3 10 ˆB 3 10 sin 1.6 10 x 48 10 t j T ��

8 80 0E B C 3 10 3 10 9 V/m

3 10 Ê 9sin 1.6 10 x 48 10 t k V /m ��

3. A litre of dry air at STP expands adiabatically to

a volume of 3 litres. If = 1.40, the work done

by air is (31.4 = 4.6555) [Take air to be an ideal

gas]

(1) 90.5 J

(2) 60.7 J

(3) 48 J

(4) 100.8 J

Answer (1)

Sol. V1 = 1 litre, P

1 = 1 atm

V2 = 3 litre, = 1.40,

2 12 1P V P V

1.4

2 1

1 1P P atm

3 4.6555

5 –3

1 1 2 2

11 1 3 1.01325 10 10

P V P V 4.6555W

1 0.4

= 90.1 J

Closest value of W = 90.5 J


3

4. The radius of gyration of a uniform rod of

length l, about an axis passing through a point

l

4 away from the centre of the rod, and

perpendicular to it, is:

(1)7

l48

(2)1l

8

(3)1l

4(4)

3l

8

Answer (1)

Sol.2 2 2

Ml l 7MlI M

12 16 48

22 7Ml

MK48

7K l

48

5. A stellite of mass m is launched vertically

upwards with an initial speed u from the

surface of the earth. After it reaches height

R(R = radius of the earth), it ejects a rocket of

mass m

10 so that subsequently the satellite

moves in a circular orbit. The kinetic energy of

the rocket is (G is the gravitational constant; M

is the mass of the earth):

(1)

2 119 GM5m u

200 R(2)

2

3m 5GMu

8 6R

(3)

2

m 2GMu

20 3R(4)

2m 113 GMu

20 200 R

Answer (1)

Sol.mu

2R

v

2 21 GMm 1 GMmmu mv

2 R 2 2R

2 21 GMmm v u

2 2R

2 GMv u

R ...(i)

0 rad

GM m vv v 10 v

m2R

10

29m GM m GMv v 81

10 2R 10 2R

2rocket

1 m GM GMKE (u )100 81

2 10 R 2R

2m GM 81 GM100 u

20 R 200 R

2 119 GM5m u

200 R

6. A long solenoid of radius R carries a time (t) -

dependent current I(t) = I0t(1 – t). A ring of

radius 2R is placed coaxially near its middle.

During the time interval 0 t 1, the induced

current (IR) and the induced EMF(V

R) in the ring

change as:

(1) Direction of IR

remains unchanged and VR

is zero at t = 0.25

(2) Direction of IR

remains unchanged and VR

is maximum at t = 0.5

(3) At t = 0.5 direction of IR reverses and V

R is

zero

(4) At t = 0.25 direction of IR reverses and V

R is

maximum

Answer (3)

Sol.

2R

R

×

×

×

×

×

×

20 0I I t I t

20

nI R

d

dt

20 0 0n R I 2I t

10 at t s

2


4

7. Speed of transverse wave on a straight wire

(mass 6.0 g, length 60 cm and area of cross-

section 1.0 mm2) is 90 ms–1. If the Young’s

modulus of wire is 16 × 1011 Nm–2, the extension

of wire over its natural length is:

(1) 0.01 mm (2) 0.02 mm

(3) 0.04 mm (4) 0.03 mm

Answer (4)

Sol. l = 60 cm, m = 6 g, A = 1 mm2, v = 90 m/s

2T mv

v l Tm l

2Tl mv l

LYA l YA

3 24

11 6

6 10 903 10 m

16 10 10

= 0.03 mm

8. A 60 HP electric motor lifts an elevator having a

maximum total load capacity of 2000 kg. If the

frictional force on the elevator is 4000 N, the

speed of the elevator at full load is close to : (1

HP = 746 W, g = 10 ms–2)

(1) 1.5 ms–1 (2) 1.9 ms–1

(3) 1.7 ms–1 (4) 2.0 ms–1

Answer (2)

Sol. totalF Mg friction

= 2000 × 10 + 4000

= 20,000 + 4000 = 24000 N

P = F × v

60 × 746 = 24000 × v

v 1.86 m/s 1.9 m/s

9. The time period of revolution of electron in its

ground state orbit in a hydrogen atom is 1.6 ×

10–16 s. The frequency of revolution of the

electron in its first excited state (in s–1) is:

(1) 1.6 × 1014 (2) 7.8 × 1014

(3) 5.6 × 1012 (4) 6.2 × 1015

Answer (2)

Sol. T n3

T2 = 8T = 8 × 1.6 × 10–16 s

16

2

1 1

T 8 1.6 10

147.8 10 Hz

10. Two moles of an ideal gas with P

V

C 5

C 3 are

mixed with 3 moles of another ideal gas with

P

V

C 4

C 3. The value of

P

V

C

Cfor the mixture is :

(1) 1.47 (2) 1.42

(3) 1.50 (4) 1.45

Answer (2)

Sol.1 1 22

1 2 m

n n nn

1 1 1

∵

m

3 2 5

4 5 11 1

3 3

m

9 2 3 5

1 2 1

m

51

12

m

171.42

12

11. The current I1 (in A) flowing through 1 resistor

in the following circuit is :

1 V

2

1

1 2

I1

(1) 0.5 (2) 0.4

(3) 0.25 (4) 0.2

Answer (4)

Sol. eq

(2 0.5) 2 2.5 2 10R

4.5 4.5 9

1 9I A

10 10

9

1

1 2 9I

2 2 2.5 10

1

1I A = 0.2 A

5


5

12.

r

m

m

As shown in the figure, a bob of mass m is tied

by a massless string whose other end portion is

wound on a fly wheel (disc) of radius r and mass

m. When released from rest the bob starts

falling vertically. When it has covered a distance

of h, the angular speed of the wheel will be:

(1)3

r2gh

(2)3

r4gh

(3)1 4gh

r 3(4)

1 2gh

r 3

Answer (3)

Sol.2 21 1

mgh mv I2 2

2

2 21 1 mrm( r)

2 2 2

2 23mgh m r

4

2

4gh 1 4gh

r 33r

13. Visible light of wavelength 6000 × 10–8 cm falls

normally on a single slit and produces a

diffraction pattern. It is found that the second

diffraction minimum is at 60° from the central

maximum. If the first minimum is produced at

1, then

1 is close to

(1) 25° (2) 30°

(3) 20° (4) 45°

Answer (1)

Sol. 2

dsin 2 (for 2nd minima)

3d 2

2... (i)

For first minima,

1

dsin

1

3sin 0.43

d 4

1 1

30 , 25

14. Three point particles of masses 1.0 kg, 1.5 kg

and 2.5 kg are placed at three corners of a

right angle triangle of sides 4.0 cm, 3.0 cm and

5.0 cm as shown in the figure. The center of

mass of system is at a point

2.5 kg

1.5 kg1.0 kg3 cm

4 cm 5 cm

(1) 1.5 cm right and 1.2 cm above 1 kg mass




Answer (3)

Sol.2.5 kg

1.5 kg1.0 kg 3 cm

4 cm 5 cm

(0, 4)

(0, 0) (3, 0)

cm

1 0 1.5 3 2.5 0 1.5 3X 0.9 cm

1 1.5 2.5 5

cm

1 0 1.5 0 2.5 4 2.5 4Y 2 cm

1 1.5 2.5 5

15. If we need a magnification of 375 from a

compound microscope of tube length 150 mm

and an objective of focal length 5 mm, the focal

length of the eye-piece, should be close to

(1) 2 mm (2) 33 mm

(3) 22 mm (4) 12 mm

Answer (3)

Sol.

�

0 e

L Dm 1

f f

e

150 250375 1

5 f

e

25012.5 1

f

e

250f 21.7

11.5

ef 22mm


6

16. Consider a circular coil of wire carrying

constant current I, forming a magnetic dipole.

The magnetic flux through an infinite plane that

contains the circular coil and excluding the

circular coil area is given by i. The magnetic

flux through the area of the circular coil area is

given by 0

. Which of the following option is

correct?

(1) i 0

(2) i 0

(3) i 0

(4) i 0

Answer (1)

Sol.

Magnetic field lines form close loop.

i 0

17. A polarizer-analyser set is adjusted such that

the intensity of light coming out of the analyser

is just 10% of the original intensity. Assuming

the polarizer - analyser set does not absorb any

light, the angle by which the analyser need to

be rotated further to reduced the output

intensity to be zero, is

(1) 71.6° (2) 45°

(3) 90° (4) 18.4°

Answer (4)

Sol.

2

0I I cos

20

0

II cos

10

21cos

10

1cos 0.316

10

71.6

90 90 71.6 18.4

18. Which of the following gives a reversible

operation?

(1)

(2)

(3)

(4)

Answer (1)

Sol. NOT gate is reversible gate because the input

signal can be recovered.

19. A LCR circuit behaves like a damped harmonic

oscillator. Comparing it with a physical spring-

mass damped oscillator having damping

constant ‘b’ the correct equivalence would be

(1) 1L m,C ,R b

k

(2) L k,C b,R m

(3) 1 1 1L ,C ,R

b m k

(4) L m,C k,R b

Answer (1)

Sol.

vkf

m

LC

R

In damped oscillation,

2

2

md xkx bv

dt

2

2

md x dxb kx 0dtdt

... (i)

In LCR circuit, q Ldi

iR 0C dt

2

2

d q dq qL R 0

dt Cdt

1m L,b R,k

C


7

20. Two infinite planes each with uniform surface

charge density +are kept in such a way that

the angle between them is 30°. The electric

field in the region shown between them is given

by

30°+

+ y

x

(1) 0

x̂ˆ1+ 3 y

2 2

(2)

0

x̂ˆ1+ 3 y –

2 2

(3)

0

ˆ3 xˆ1– y –

2 2 2

(4)

0

ˆ3 xˆ1+ y+

2 2

Answer (3)

Sol.

30°60°

+

+

y

x

E1

E2

��

1

0

Ê y2

��

2

0

ˆ Ê ( cos60 x sin60 y)2

0

1 3ˆ ˆx y

2 2 2

��

P 1 2

0

1 3ˆ Ê E E x 1 y

2 2 2

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. A particle (m = 1 kg) slides down a frictionless

track (AOC) starting from rest at a point A

(height 2 m). After reaching C, the particle

continues to move freely in air as a projectile.

When it reaching its highest point P (height

1 m), the kinetic energy of the particle (in J) is :

(figure drawn is schematic and not to scale;

take g = 10 ms–2)

HeightA

P

2 m

O

C

Answer (10.00)

Sol. KE mg h

1 10 1 10 J

22. A Carnot engine operates between two

reservoirs of temperatures 900 K and 300 K.

The engine performs 1200 J of work per cycle.

The heat energy (in J) delivered by the engine

to the low temperature reservoir, in a cycle, is

_________ .

Answer (600.00)

Sol. 2

1 1

T W1

T Q

1

1 12001

3 Q

1

2 1200

3 Q

1

Q 1800

2

1Q (1800) 600 J

3


8

23. A loop ABCDEFA of straight edges has six

corner points A(0, 0, 0), B(5, 0, 0), C(5, 5, 0),

D(0, 5, 0), E(0,5, 5) and F(0, 0, 5). The magnetic

field in this region is ��

ˆˆB 3 i 4k T. The quantity

of flux through the loop ABCDEFA (in Wb) is

_________.

Answer (175.00)

Sol.

y

C

xA B

D

E

Fz

��

ˆˆB 3i 4k T

3 25 25 4 175 Wb

24. A non-isotropic solid metal cube has

coefficients of linear expansion as: 5 × 10–5/°C

along the x-axis and 5 × 10–6/°C along the y and

the z-axis. If the coefficent of volume expansion

of the solid is C × 10–6/°C then the value of C is

________.

Answer (60.00)

Sol. V lbh

�

�

V b h

V b h

5 6 6

5 10 5 10 5 10

660 10 / C

25. A beam of electromagnetic radiation of intensity

6.4 × 10–5W/cm2 is comprised of wavelength, = 310 nm. It falls normally on a metal (work

function = 2 eV) of surface area of 1 cm2. If

one in 103 photons ejects an electron, total

number of electrons ejected in 1 s is 10x.

(hc = 1240 eVnm, 1 eV = 1.6 × 10–19 J), then x is

_______.

Answer (11.00)

Sol. 0

hc 1240E PV 4eV

310

5

1 14

19

6.4 10N s 10 per sec

4 1.6 10

Photoelectrons/s = 11

3

N10

10

� � �


9

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. In comparison to the zeolite process for the

removal of permanent hardness, the synthetic

resins method is

(1) Less efficient as it exchanges only anions

(2) More efficient as it can exchange only

cations

(3) More efficient as it can exchange both

cations as well as anions

(4) Less efficient as the resins cannot be

regenerated

Answer (3)

Sol.Nowadays hard water is softened by using

synthetic resins method. This method is more

efficient than zeolite process as it can

exchange both cations as well as anions.

2. The atomic radius of Ag is closest to

(1) Cu (2) Au

(3) Hg (4) Ni

Answer (2)

Sol.Ag and Au have similar atomic radius.

3. Given that the standard potentials (E°) of

Cu2+/Cu and Cu+/ Cu are 0.34 V and 0.522 V

respectively, the E° of Cu2+/Cu+ is:

(1) +0.158 V (2) –0.158 V

(3) –0.182 V (4) 0.182 V

Answer (1)

Sol.2+ – °

1

+ – °

2

Cu + 2e Cu, G = –2F(0.34) (i)

Cu + e Cu, G = –F(0.522) (ii)

...

...

Subtract (ii) from (i)

– 2+ + ° 0

3Cu +e Cu , G = –F(E )

– ° ° °

1 2 3G G G

°–FE = –2F(0.34) +F(0.522)

°

E = 0.68 – 0.522 = 0.158 V

4. At 35°C, the vapour pressure of CS2 is 512 mm

Hg and that of acetone is 344 mm Hg. A

solution of CS2 in acetone has a total vapour

pressure of 600 mm Hg. The false statement

amongst the following is

(1) Raoult’s law is not obeyed by this system

(2) A mixture of 100 mL CS2 and 100 mL

acetone has a volume < 200 mL

(3) Heat must be absorbed in order to produce

the solution at 35°C

(4) CS2 and acetone are less attracted to each

other than to themselves

Answer (2)

Sol. Since the vapour pressure of the solution is

greater than individual vapour pressure of both

pure components, the solution shows a

positive deviation from Raoult’s law.

5. Match the following

(i) Riboflavin (a) Beriberi

(ii) Thiamine (b) Scurvy

(iii) Pyridoxine (c) Cheilosis

(iv) Ascorbic acid (d) Convulsions

(1) (i)-(c), (ii)-(d),(iii)-(a), (iv)-(b)

(2) (i)-(c), (ii)-(a),(iii)-(d), (iv)-(b)

(3) (i)-(d), (ii)-(b),(iii)-(a), (iv)-(c)

(4) (i)-(a), (ii)-(d),(iii)-(c), (iv)-(b)

Answer (2)

Sol.Thiamine (vitamin B1) : Beriberi

Riboflavin (vitamin B2) : Cheilosis

Pyridoxine (vitamin B6) : Convulsions

Ascorbic acid (vitamin C) : Scurvy

6. The increasing order of pKb for the following

compounds will be

(1) (B) < (C) < (A) (2) (B) < (A) < (C)

(3) (C) < (A) < (B) (4) (A) < (B) < (C)

Answer (2)


10

Sol. pKb = –logK

b

So, as Kb increases, pK

b decreases

NH – CH = NH2

(A)

(B)

CH – NH – CH3 3

(C)

N

N NH

Kb : (B) > (A) > (C)

pKb : (B) < (A) < (C)

7. Oxidation number of potassium in K2O, K

2O

2

and KO2, respectively, is

(1) +1, +2 and +4 (2) +2, +1 and +1

2

(3) +1, +4 and +2 (4) +1, +1 and +1

Answer (4)

Sol.Alkali metals show an oxidation state of +1 in

their compounds.

8. Consider the following reaction:

———

N + Na SO3N Cl

2

CH3

CH3

–

OH'X '

The product ‘X’ is used

(1) In acid base titration as an indicator

(2) In protein estimation as an alternative to

ninhydrin

(3) In laboratory test for phenols

(4) As food grade colourant

Answer (1)

Sol.

N

N

Me

Me

Me

Me

N= N SO Na3

——

—

——

OH–

SO Na3 Cl N

–

2

——

++–

+–

+

Formed product is methyl orange and it is used

as an indicator in acid base titrations.

9. The relative strength of interionic/

intermolecular forces in decreasing order is :

(1) ion-ion > ion-dipole > dipole-dipole

(2) ion-dipole > dipole-dipole > ion-ion

(3) ion-dipole > ion-ion > dipole-dipole

(4) dipole-dipole > ion-dipole > ion-ion

Answer (1)

Sol.Ionic interactions are stronger as compared to

van der waal interactions.

So, correct order is

ion-ion > ion-dipole > dipole-dipole

10. What is the product of following reaction?

4

3

2 3

(i) NaBH

(ii)PBr

(iii)Mg/ether

(iv)CO /H O

Hex – 3 – ynal ?

(1) COOH

(2)COOH

(3) COOH

(4)COOH

Answer (4)

Sol.O

NaBH4 OH

PBr3

Br

Mg/Ether

MgBr

CO2

O

O

H+

COOH

H

11. A solution of m-chloroaniline, m-chlorophenol

and m-chlorobenzoic acid in ethyl acetate was

extracted initially with a saturated solution of

NaHCO3 to give fraction A. The left over

organic phase was extracted with dilute NaOH

solution to give fraction B. The final organic

layer was labelled as fraction C. Fractions A, B

and C, contain respectively :

(1) m-chloroaniline, m-chlorobenzoic acid and

m-chlorophenol

(2) m-chlorophenol, m-chlorobenzoic acid and

m-chloroaniline

(3) m-chlorobenzoic acid, m-chlorophenol and

m-chloroaniline

(4) m-chlorobenzoic acid, m-chloroaniline and

m-chlorophenol

Answer (3)


11

Sol.m-chlorobenzoic acid dissolves in aq NaHCO3

while m-chlorophenol dissolves in aq NaOH

12. Amongst the following statements, that which

was not proposed by Dalton was :

(1) All the atoms of a given element have

identical properties including identical

mass. Atoms of different elements differ in

mass

(2) Matter consists of indivisible atoms.

(3) Chemical reactions involve reorganization

of atoms. These are neither created nor

destroyed in a chemical reaction.

(4) When gases combine or reproduced in a

chemical reaction they do so in a simple

ratio by volume provided all gases are at

the same T & P.

Answer (4)

Sol.“When gases combine or reproduced in a

chemical reaction, they do so in a simple ratio

by volume provided all gases are at the same

T & P”

This is not a postulate of Dalton’s theory

13. The purest form of commercial iron is

(1) Scrap iron and pig iron

(2) Cast iron

(3) Wrought iron

(4) Pig iron

Answer (3)

Sol.Purest form of commercial iron is wrought iron

14. Consider the following reactions :

(a) 2 4conc.H SO

3 33CH CCH OH CH

(b) alc.KOH

3 32CH CHCH Br CH

(c)

○–

3 3CH O K

3 32CH CHCH Br CH

(d) 3 22

CH C–CH –CHO

|

OH

Which of these reaction(s) will not produce

Saytzeff product ?

(1) (a), (c) and (d) (2) (c) only

(3) (b) and (d) (4) (d) only

Answer (2)

Sol.Tertiary butoxide is a bulky base and it extracts

the least hindered hydrogen

CH — CH — CH

3

Br

CH3

CH3

tBuOCH — CH = CH

2

CH3

CH3

15. The electron gain enthalpy (in kJ/mol) of

fluorine, chlorine, bromine and iodine,

respectively, are :

(1) –296, –325, –333 and –349

(2) –333, –325, –349 and –296

(3) –349, –333, –325 and –296

(4) –333, –349, –325 and –296

Answer (4)

Sol.Electron gain enthalpy is most negative for

chlorine followed by fluorine and bromine and

least negative for iodine among given

elements.

So, F(–333 kJ mol–1)

Cl(–349 kJ mol–1)

Br(–325 kJ mol–1)

I(–296 kJ mol–1)

16. The IUPAC name of the complex [Pt(NH3)2

Cl(NH2CH

3)]Cl is :

(1) Diamminechlorido(methanamine)platinum

(II) chloride

(2) Diammine(methanamine)chloridoplatinum

(II) Chloride

(3) Bisammine(methanamine)chloridoplatinum

(II) chloride

(4) Diamminechlorido(aminomethane)platinum

(II) chloride

Answer (1)

Sol.The IUPAC name is Diamminechlorido

(methanamine)platinum(II) chloride

17. The theory that can completely/properly

explain the nature of bonding in [Ni(CO)4] is :

(1) Crystal field theory

(2) Werner’s theory

(3) Valence bond theory

(4) Molecular orbital theory

Answer (4)


12

Sol.CFT assumes that the ligands are point

charges and it follows that anionic ligands

should exert the greatest splitting effect.

Further it does not take into account the

covalent character of bonding which exists

between the metal and the carbon atom of the

carbonyl. The M – C bond is formed by the

donation of lone pair of electrons on the

carbonyl carbon into a vacant orbital of the

metal and the M-C bond is formed by the

donation of a pair of electrons from a filled

d-orbital of metal into vacant antibonding

orbital of carbon monoxide.

So, MOT explains the bonding in Ni(CO)4.

18. The number of orbitals associated with

quantum numbers n = 5, ms =

1

2 is :

(1) 15 (2) 50

(3) 25 (4) 11

Answer (3)

Sol.The number of orbitals possible in a shell with

principal quantum number ‘n’ is ‘n2’.

19. 1-methylethylene oxide when treated with an

excess of HBr produces:

(1)Br

CH3

(2)Br

Br

(3) Br

Br

CH3

(4)Br

CH3

Answer (3)

Sol.H

+

O O

H

+

Br–

Br

OH

HBr

Br

Br

20. The dipole moments of CCl4, CHCl

3 and CH

4

are in the order :

(1) CCl4 < CH

4 < CHCl

3

(2) CHCl3 < CH

4 = CCl

4

(3) CH4 = CCl

4 < CHCl

3

(4) CH4 < CCl

4 < CHCl

3

Answer (3)

Sol.CHCl3 is polar while CH

4 and CCl

4 are non-

polar. So, dipole moment order is :

CHCl3 > CH

4 = CCl

4

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. Two solutions, A and B, each of 100 L was

made by dissolving 4 g of NaOH and 9.8 g of

H2SO

4 in water, respectively. The pH of the

resultant solution obtained from mixing 40 L of

solution A and 10 L of solution B is________.

Answer (10.60)

Sol.Molarity of H2SO

4 =

0.1

100= 10–3 M

Molarity of NaOH = 0.1

100= 10–3 M

Since, NaOH is present in excess

So, [OH–] = –3 –340 10 – 2 10 10

50

= –3210

5

pOH = 3 – log2 + log5 = 3 – 0.3 + 0.7 = 3.4

pH = 14 – 3.4 = 10.6


13

22. During the nuclear explosion, one of the

products is 90Sr with half life of 6.93 years. If

1 g of 90Sr was absorbed in the bones of a

newly born baby in place of Ca, how much

time, in years, is required to reduce it by 90%

if it is not lost metabolically _________.

Answer (23.03)

Sol.Decay constant () = �n2

half life =

0.693

6.93

= 0.1 yr–1

Now (t90%

) = ln

100

10

t90%

= 10 ln 10

= 10 × 2.303

= 23.03 yr

23. The number of chiral carbons in

chloramphenicol is ___________.

Answer (02.00)

Sol. The structure of chloramphenicol is

O N2 CH — CH — CH OH2

*

*

NH — C — CHCl 2

O

OH

It contains 2 chiral centres

24. For the reaction

A(l) 2B(g)

U = 2.1 kcal, S = 20 cal K–1 at 300 K.

Hence G in kcal is ___________.

Answer (–02.70)

Sol. ��⇀� ↽�� gA 2B g , n = 2

H = U + (ng) RT

= –32.1 2 2 300 10 kcal

= 3.3 kcal

G = H – TS

= 3.3 – 300 × 20 × 10–3

= 3.3 – 6

= – 2.7 kcal.

25. Chlorine reacts with hot and concentrated

NaOH and produces compounds (X) and (Y).

Compound (X) gives white precipitate with

silver nitrate solution. The average bond order

between Cl and O atoms in (Y) is _________.

Answer (01.67)

Sol.

Conc.

2 3 23Cl + 6NaOH 5NaCl +NaClO + 3H O

Ag+ forms precipitate with Cl– i.e. AgCl.

AgClO3 is soluble.

X is NaCl

Y is NaClO3

Now, structure of –

3ClO is

O

Cl

O O

–

Bond order is 5

3 i.e. 1.67

��

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(Physics, Chemistry and Mathematics)...Jul 01, 2020 · There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics ... (Physics, Chemistry