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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)
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 ˆE 60sin(1.6 10 x 48 10 t)k V /m
(2) ��
8 3 10 ˆE 3 10 sin(1.6 10 x 48 10 t)i V /m
(3) ��
3 10 ˆE 9sin(1.6 10 x 48 10 t)k V /m
(4) ��
8 3 10 ˆE 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 ˆE 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
JEE (MAIN)-2020 Phase-1 (7M)
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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
JEE (MAIN)-2020 Phase-1 (7M)
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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
JEE (MAIN)-2020 Phase-1 (7M)
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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
(2) 2.0 cm right and 0.9 cm above 1 kg mass
(3) 0.9 cm right and 2.0 cm above 1 kg mass
(4) 0.6 cm right and 2.0 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
JEE (MAIN)-2020 Phase-1 (7M)
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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
JEE (MAIN)-2020 Phase-1 (7M)
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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
ˆE y2
���
2
0
ˆ ˆE ( cos60 x sin60 y)2
0
1 3ˆ ˆx y
2 2 2
��� ��� ���
P 1 2
0
1 3ˆ ˆE 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
JEE (MAIN)-2020 Phase-1 (7M)
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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
� � �
JEE (MAIN)-2020 Phase-1 (7M)
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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)
JEE (MAIN)-2020 Phase-1 (7M)
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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)
JEE (MAIN)-2020 Phase-1 (7M)
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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)
JEE (MAIN)-2020 Phase-1 (7M)
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
JEE (MAIN)-2020 Phase-1 (7M)
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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