JEE MAIN MODEL - HINTS AND SOLUTIONS IIT, MEDICAL & ENGINEERING Entrance COACHING Now @ MUKKAM: BHABHA Institute of Sciences 2 nd floor, Karayil Tower, Near Bend Pipe Bridge, Orphanage Road, Mukkam, PIN - 673602 Ph : 8943300201 , 8943300204.
JEE MAIN MODEL - HINTS AND SOLUTIONS
IIT, MEDICAL & ENGINEERING
Entrance COACHING
Now
@
MUKKAM:
BHABHA Institute of Sciences
2nd floor, Karayil Tower,
Near Bend Pipe Bridge, Orphanage Road,
Mukkam, PIN - 673602
Ph : 8943300201 , 8943300204.
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PART : A — PHYSICS
ALL THE GRAPHS/DIAGRAMS GIVEN ARE
SCHEMATIC AND NOT DRAWN TO SCALE
01. The diameter of a cylinder is measured using a
Vernier callipers with no zero error. It is found that
the zero of the Vernier scale lies between 5.10 cm
and 5.15 cm of the main scale. The Vernier scale has
50 divisions equivalent to 2.45 cm. The 24th division
of the Vernier scale exactly coincides with one of the
main scale divisions. The diameter of the cylinder is
(1) 5.112 cm (2) 5.124 cm
(3) 5.136 cm (4) 5.148 cm
Sol : Answer (2)
Reading = M.S.R + No of division of V.S
matching the main scale division
(1MSD – 1VSD)
50
45.205.02410.5
= 5.124 cm option (2) is correct.
02. The velocity - displacement graph of a particle
moving along a straight line is shown
The most suitable acceleration-displacement graph
will be
(1) (2)
(3) (4)
Sol : Answer (1)
03. At what angle the vector BA
and BA
must
act, so that the resultant is 22 BA
(1) cos-1
22
22
BA
BA (2) cos
-1
22
22
BA
BA
(3) cos-1
22
22
2 BA
BA
(4) cos-1
22
22
2 AB
BA
Sol : Answer (4)
04. A uniform cylinder of radius R is spinned about its
axis to the angular velocity 0 and then placed into a
corner (see figure). The coefficient of friction
between the corner walls and the cylinder is equal to
. How many turns does the cylinder accomplish
before it stops?
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(1)
18
1 22
0
g
R (2)
14
1 22
0
g
R
(3)
12
1 22
0
g
R (4)
16
1 22
0
g
R
Sol : Answer (1)
05. A particle which is constrained to move along x-axis,
is subjected to a force in same direction which varies
with distance x of the particle from the origin as
F(x) = - kx + ax3. Here k and a are positive constant.
For x 0, the functional form of the potential energy
U(x) of the particle is:
(1) (2)
(3) (4)
Sol : Answer (4)
From the given function we can see that F = 0 at
x = 0 i.e., slope of U – x graph is zero at , x = 0.
Therefore, the most appropriate option is (4)
06. A particle of mass ‘m’ oscillates along the horizontal
diameter AB inside a smooth spherical shell of radius
R. At any instant K.E. of the particle is K. Then force
applied by particle on the shell at this instant is:
(1) R
K (2)
R
K2
(3) R
K3 (4)
R
K
2
Sol : Answer (3)
sin2
mgR
mvN
By conservation of energy
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2
2
1sin mvmgR
07. The magnitudes of gravitational field at distances r1
and r2 from the centre of a uniform sphere of radius R
and mass m are Eg1 and Eg2, respectively. Then,
[a] 2
1
2
1
r
r
E
E
g
g if r1 < R and r2 < R
[b] 2
1
2
2
2
1
r
r
E
E
g
g if r1 > R and r2 > R
[c] 2
1
2
1
r
r
E
E
g
g if r1 > R and r2 > R
[d] 2
1
2
2
2
1
r
r
E
E
g
g if r1 < R and r2 < R
(1) [a] and [b] are true (2) [b] and [c] are true
(3) [b] and [d] are true (4) [a] and [d] are true
Sol : Answer (1)
08. A metallic cube as a bulk modulus ‘B ’and a density
‘ ’ . A pressure of ‘P ’ is applied uniformly from all
sides of the cube. The increase in density is
(1) PB
B
(2)
B
PB
(3) PB
B
(4)
PB
B
Sol : Answer (1)
= V
M , B =
V
PV
= VV
M
=
V
M
V
V1
1
=
B
P1
1 =
PB
B
09. A bubble is rising from bottom of a lake to the top
slowly such that its diameter gets doubled. If a
barometer placed on the bank of lake reads ‘h’, then
depth of lake is (Given is the relative density of
mercury in barometer)
(1) 5h (2) 4h
(3) 7h (4) 2h
Sol : Answer (3)
The rising of bubble is isothermal
P1V1= P2V2
VPVPgH aw 3
2 2)(
VPa 8
gH w = aP7
g
gh
g
PH
w
Hg
w
a
77
h7
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10. A vessel having area of cross-section ‘A’contains a
liquid up to a height ‘h’ . At the bottom of the vessel,
there is a small hole having area of cross-section ‘a’.
Then the time taken for the liquid level to fall from
height ‘H1’ to ‘H2’ is given by
212)1( HHg (2) 21
2HH
ga
A
(3) 212
HHg
a
A
(4) gH2
Sol : Answer (2)
:
.2. 21 HH
ga
At
11. Steam at 1000
C is passed into 1.1 kg of water
contained in a calorimeter of water equivalent 0.02
kg at 25 0C till the temperature of the calorimeter
and its contents rises to 80 0C.The mass of the steam
condensed (in kg) is
(1) 0.130 (2) 0.065 (3) 0.260 (4) 0.135
Sol : Answer (1)
12. Two rods, one of aluminium and the other made of
steel, having initial lengths l1 and l2 are connected
together to form a single rod of length l1 + l2 .The
coefficients of linear expansion for aluminium and
steel are a and s , respectively . If the length of
each rod increases by the same amount when their
temperature are raised by t 0C , then find the ratio
21
1
ll
l
,
(1) a
s
(2)
s
a
(3) as
a
(4)
as
s
Sol : Answer (4)
13. When an ideal diatomic gas is heated at constant
pressure, the fraction of the heat energy supplied
which increases the internal energy of the gas is
(1) 5
2 (2)
5
3
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(3) 7
3 (4)
7
5
Sol : Answer (4)
14. Which of the following graphs correctly represents
the variation of V
dPdV
with P for an ideal gas
at constant temperature?
(1) (2)
(3) (4)
Sol : Answer (1)
15. In a given forced field, the potential energy of
a particle is given as a function of its x – coordinate
as x
k
x
kxU 2
2
1 , where k1 and k2 are positive
constants. Find the period of small oscillations of the
particle about its equilibrium position in the field.
(1) 4
2
3
14
k
mk (2)
4
2
3
18
2 k
mk
(3) 4
2
3
124
k
mk (4)
4
2
3
122
k
mk
Sol : Answer (3)
16. A uniform rope having mass m hangs vertically
from a rigid support. A transverse wave pulse is
produced at the lower end. The speed (v) of the wave
pulse varies with height (h) from the lower end as
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(1) (2)
(3) (4)
Sol : Answer (2)
17. A thin semi-circular ring of radius r has a positive
charge q distributed uniformly over it. The net field
E
at the centre O is
(1) jr
q ˆ2 2
0
2 (2) j
r
q ˆ4 2
0
2
(3) jr
q ˆ4 2
0
2 (4) j
r
q ˆ2 2
0
2
Sol : Answer (4)
18. Three capacitors C1, C2 and C3 are connected as
shown in the figure given below to a battery of V
volt. If the capacitor C3 breaks down electrically, the
change in total charge on the combination of
capacitors is
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(1)
321
321 1
CCC
CVCC
(2)
321
2121 1
CCC
CCVCC
(3)
321
321 1
CCC
CVCC
(4)
321
221 1
CCC
CVCC
Sol : Answer (1)
19. For the given circuit,
If internal resistance of cell is 1.5 , then
(1) VP – VQ = 0 (2) VP – VQ = 4 V
(3) VP – VQ = - 4 V (4) VP – VQ = - 2.5V
Sol : Answer (4)
20. A current I flows around a closed path in the
horizontal plane of the circle as shown in the figure
given below. The path consists of eight arcs with
alternating radii r and 2r. Each segment of arc
subtends equal angle at the common centre P. The
magnetic field produced by current path at point P is
(1) r
I0.8
3 ; perpendicular to the plane of the paper
and directed inwards
(2) r
I0.8
3 ; perpendicular to the plane of the paper
and outwards
(3) r
I0.8
1 ; perpendicular to the plane of the paper
and inwards
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(4) r
I0.8
1 ; perpendicular to the plane of the paper
and outwards
Sol : Answer (1)
21. A domain in ferromagnetic iron is in the form of a
cube of side length 1 m. Estimate the number of
iron atoms in the domain. The molecular mass of
iron is 55 g/mol and its density is 7.9 g/cm3.
(1) 8.65 10-10
atoms (2) 8 10-13
atoms
(3) 8 105 atoms (4) 8.65 10
10atoms
Sol : Answer (4)
22. An LCR circuit is equivalent to a damped pendulum.
In an LCR circuit the capacitor is charged to Q0 and
then connected to the L and R as shown below
If a student plots graphs of the square of maximum
charge (Qmax2) on the capacitor with time (t) for two
different values L1 and L2 (L1 > L2) of L then which of
the following represents this graph correctly?
(1) (2)
(3) (4)
Sol : Answer (1)
23. An infinitesimally small bar magnet of dipole
moment M is pointing and moving with a speed v in
the x direction. A small closed circular conducting
loop of radius ‘a’ and negligible self inductance lies
in the x - y plane with its centre at x = 0 and its axis
conciding with x axis. If x = 2a , the emf induced in
the loop is
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2
0
16
3)1(
a
Mv
2
0
32
3)2(
a
Mv
2
0
8
1)3(
a
Mv
2
0
16
1)4(
a
Mv
Sol : Answer (2)
24. An electromagnetic radiation represented by
ttaE 0coscos1 falls over lithium surface
with work function of 2.39 eV. Maximum kinetic
energy of emitted photoelectrons will be
(Given, a = 4 N/C, 14106 rads
-1 and
15
0 106.3 rads-1
)
(1) 0 (2) 0.37 eV
(3) 0.70 eV (4) No photoemission occurs
Sol : Answer (2)
25. In the situation as shown in the figure, the focal
length of the thin plane concave lens is 20 cm. A
point object P is at distance of 20 cm from the lens.
Find the velocity of image formed by plane mirror
at the instant as shown in the figure.
(1) ji ˆ3ˆ2 cms-1
(2) j3 cms-1
(3) ji ˆ3ˆ2 cms-1
(4) j3 cms-1
Sol : Answer (2)
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26. In an interference pattern of Young’s double still
experiment, we observe the 12th
order maxima for
wavelength 600 nm at a point on the screen. What
order will be visible at the same point, if the source
is replaced by light of wavelength 480 nm?
(1) 4th (2) 10
th
(3) 15th (4) 20
th
Sol : Answer (3)
27. An electromagnetic wave of frequency
= 3.0 MHz passes from vacuum into a dielectric
medium with permittivity = 4.0. Then
(1) wavelength is doubled and the frequency
remains uncharged
(2) wavelength is doubled and frequency becomes
half
(3) wavelength is halved and frequency remains
uncharged
(4) wavelength and frequency both remains
uncharged
Sol : Answer (3)
28. A diatomic molecules is made of two masses m1 and
m2 which are separated by a distance r. If we
calculate its rotational energy by applying Bohr’s
rule of angular momentum quantisation, its energy
will be given by (n is an integer)
(1)
22
2
2
1
222
21
2 rmm
hnmm (2)
2
21
22
2 rmm
hn
(3) 2
21
222
rmm
hn
(4)
2
21
22
21
2 rmm
hnmm
Sol : Answer (4)
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29. Half-life of a neutron is about 693 s and mass of
neutron is 1.6 10-27
kg. A beam of 10000 high
speed neutrons with kinetic energy 0.08 eV is fired
in space. Number of neutrons that possibly decay in
travelling through a distance of 4 km will be
(1) 10 (2) 100
(3) 1000 (4) 10000
Sol : Answer (1)
30. For the given logic circuit
Which of these is correct?
(1) y = 0 for x1 = x3 = 0 and x2 = 1
(2) y = 0 for x1 = x2 = x3 = 0
(3) y = 1 for x1 = x2 = x3 = 1
(4) y = 1 for x1 = x2 = 1 and x3 = 0
Sol : Answer (1)
when x1 = x3 = 0 and x2 = 1, inputs to AND gate
are 1 and 1 so, y = 0 due to inversion of NOT gate
PART : B — CHEMISTRY
31. A sample of a hydrate of barium chloride weighing
of 61 g was heated until all the water of hydration is
removed. The dried sample weighed 52g. The
formula of the hydrated salt is
(atomic mass Ba = 137 amu)
(1) BaCl2. H2O (2) BaCl2.2H2O
(3) BaCl2.3H2O (4) BaCl2.4H2O
Sol : Answer (2)
Weight of hydrated BaCl2 = 61 g
Weight of anhydrous BaCl2 = 52 g
Loss in Mass = 61 – 52 = 9g
Mass of H2O removed = 9g
Moles of H2O removed = 5.018
9
Molecular mass of BaCl2 = 208
% of H2O in the hydrated BaCl2
= %75.1410061
9
10018208
1875.14
x
x
x = 2
Formula of the hydrated salt is BaCl2. 2H2O
32. At very high pressure, the compressibility
factor of one mole of a gas is given by:
(1) RT
Pb1 (2)
RT
Pb
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(3) RT
Pb1 (4)
)(1
VRT
b
Sol : Answer (1)
For 1 mole of gas )(2V
aP (V – b) = RT
At very high pressure, P >2V
a so
2V
a is
negligible.
P ( V – b) = RT
PV - Pb = RT
RT
Pb
RT
PVZ 1
33. The ionisation enthalpy of hydrogen atom is
1.312 x 106 J Mol
-1. The energy required to excite
the electron in the atom from n = 1 to n = 2 is:
(1) 9.84 x 105J Mol
-1 (2) 8.51 x 10
5J Mol
-1
(3) 6.56 x 10
5J Mol
-1 (4) 7.56 x 10
5J Mol
-1.
Sol : Answer (1)
IE = 1EE
1.312 x 106 = 0 - E1
16
1 10312.1 MolJE
4
10312.1
2
10312.1 6
2
6
2
E
(Energy of electron in second orbit n = 2)
Energy required when an electron makes
transition from n = 1 to n = 2
)10312.1(4
10312.1 66
12
EEE
= 9.84 x 105 J Mol
-1
34. The hybridization of orbitals of N atom in
NO3- , NO2
+ and NH4
+ are respectively
(1) Sp, Sp2, Sp
3 (2) Sp, SP
3, SP
2
(3) SP2, Sp, Sp
3 (4) Sp
2, Sp
3, Sp
Sol : Answer (2)
35. In a fuel cell methanol is used as fuel and oxygen gas
is used as oxidizer. The reaction is:
CH3OH( )(2)()(2
3) 222 OHgCOgO
At 298 K, G of formation for CH3OH( ),
H2O( ) and CO2(g) are -166.2, -237.2 and
-394.4 kJ/Mol respectively. If standard enthalpy of
combustion of Methanol is -726 kJ/Mol. Efficiency
of the fuel cell will be
(1) 80% (2) 87%
(3) 90% (4) 97%
Sol : Answer (4)
1726 MolKJH
MolKJOHCHG f /2.166)( 3
MolKJOHG f /2.237)( 2
MolKJCOG f /4.394)( 2
)2.166()]2.237(24.394[ rG
= -702.6 KJ /Mol
% Efficiency = 100
H
G
=1726
/6.702
MolKJ
MolKJ=96.77%
36. The equilibrium constant at 298 K for a reaction
A + B C + D is 100. If the Initial
concentration of all the four species were 1M each,
then equilibrium concentration of D will be:
(1) 0.182 (2) 0.818
(3) 1.818 (4) 1.182
Sol : Answer (3)
A + B C + D
Initial conc 1 1 1 1
Equilibrium conc 1-x 1- x 1+x 1+x
Now 2
2
1
1
x
xKc
2
2
1
1100
x
x
x
x
1
110
10 – 10x = 1+ x = 9
x = 0.81
i e [D] at equilibrium = 1+0.81 = 1.818 M
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37. The increasing order of the ionic radii of the given
isoelectronic species is:
(1) S2-
, Cl-, Ca
2+, K
+ (2) Ca
2+, K
+, Cl
-, S
2-
(4) K+, S
2-, Ca
2+, Cl
- (3) Cl
-, Ca
2+, K
+, S
2-
Sol : Answer (2)
38. In a face centered cubic lattice, atom A occupies the
corner positions and atom B occupies the face centre
points. If one atom of B is missing from one face
centre point, the formula of the compound is:
(1) A2B5 (2) A2B
(3) AB2 (4) A2B3
Sol : Answer (1)
Share of A from corners = 8 x 1/8 = 1
Share of B, with one face vacant = 5/2
A : B = 1: 5/2 = 2:5
Formula is A2B5
39. In 0.2 molal aqueous solution of a weak acid HX, the
degree of ionization is 0.3.Taking Kf of water is
1.85, the freezing point of the solution will be nearest
to:
(1) - 0.480oC (2) -0.360
oC
(3) -0.260oC (4) + 0.480
oC
Sol : Answer (1)
HX H+ + X
-
1 Mole 0 0
1-0.3 0.3 0.3
i = ( 1- 0.3) + 0.3 + 0.3 = 1.3
Tf = i. Kf . M = 1.3 x 1.85 x 0.2 = 0.48
Freezing point = 0oC – 0.48
oC = -0.48
0C
40. Consider the following cell reaction:
2Fe (s) + O2(g) + 4H+(aq) 2Fe
2+(aq)+ 2H2O( )
Eo = 1.67 V At [Fe
2+] = 10
-3M, Po2 = 0.1atm and
pH = 3 the cell potential at 25oC is:
(1) 1.47 V (2) 1.77 V
(3) 1.87 V (4) 1.57V
Sol : Answer (4)
2Fe Fe2+
n = 4
pH=3 [H+] = 1 x 10
-3M
2Fe (s) + O2(g) + 4H+(aq)
2Fe2+
(aq)+ 2H2O( ) Eo = 1.67 V
7
43
23
4
2
22
10)10(1.0
)101(
][
][
HPO
FeQ
Qn
EE o log059.0
VE 57.110log
4
059.067.1 7
41. A first order reaction goes as follows
K1 = 1.26 x 10-4
s-1
K2 = 3.8 x 10-5
s-1
The percentage of B in the mixture of B and C
is likely to be
(1) 80% (2) 76.83%
(3) 92% (4) 68%
Sol : Answer (2)
%83.76108.31026.1
1026.1%
54
4
BOf
42. Coagulation of 90 ml of a negative sol requires 10ml
of 0.5 M NaCl. The coagulation value of NaCl is
(1) 25 (2) 50
(3) 75 (4) 100
Sol : Answer (2)
10ml of 0.5 M NaCl = 10 x 0.5 millimoles of NaCl =
5 millimoles of NaCl.
100 ml of total sol requires
NaCl = 5 millimoles
1000 ml (ie 1 litre) sol requires
NaCl = 50 millimoles
43. In the context of Hall Heroult process for the
extraction of Al, which of the following statements is
false?
(1) Na3AlF6 serves as the electrolyte
(2) CO and CO2 are produced in this process
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(3) Al2O3 is mixed with CaF2 which lowers the M.P.
of the mixture and brings conductivity.
(4) A13+
is reduced at the cathode to form Al
Sol : Answer (1)
44. Which is thermodynamically the most stable
allotropic form of phosphorus?
(1) Red (2) White
(3) Black (4) Yellow
Sol : Answer (3)
45. How many moles of KMnO4 are required to oxidise
one mole of ferrous oxalate FeC2O4 in acidic
medium?
(1) 5
2 (2)
5
1
(3) 4
5 (4)
5
3
Sol : Answer (4)
3MnO4- + 24 H
+ + 5FeC2O4 3Mn
2+ +
12 H2O + 5Fe3 + 10 CO2
5 moles of FeC2O4 = 3 Moles of KMnO4
1 mole of FeC2O4 = 5
3moles of KMnO4
46. Low spin complex of d6 cation in an octahedral field
will have the following energy in total.
(1) Po
5
12 (2) Po 3
5
12
(3) Po 25
2
(4) Po
5
2
Sol : Answer (2)
For low spin complex d6 is t2g
222 eg
o.
Total energy will be o + energy of 3 pair of
electrons
Net o + 3p = -6 x 0.4 o + 3p
= Po 35
12
47. In the reaction
Ag2O + H2O22Ag + H2O + O2
H2O2 act as
(1) reducing agent (2) Oxidising agent
(3) Bleaching agent (4) None of these
Sol : Answer (1)
48. The least stable carbonate of alkali metal is:
(1) Na2CO3 (2) Li2CO3
(3) K2CO3 (4) Cs2CO3
Sol : Answer (2)
Li2CO3 is unstable to heat, Li+ being very small in
size polarises a large CO32-
ion leading to the
formation of more stable Ci2O and CO2
49. Silica is soluble in:
(1) HCl (2) HNO3
(3) H2SO4 (4) HF
Sol : Answer (4)
HF is the only acid in which silica (SiO2) is soluble
due to the formation of Silicon tetra fluoride as
SiO2 + 4HFSiF4 + 2H2O
50. A hydrocarbon contains 20% hydrogen and 80% of
carbon. The empirical formula is:
(1) CH4 (2) CH3
(3) CH2 (4) CH
Sol : Answer (2)
Elemen
t
Percentag
e
Atomi
c mass
Relativ
e
number
of
atoms
Simples
t ratio
H 20 1 20 366.6
20
C 80 12 6.66 166.6
66.6
Empirical formula = CH3
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51. CaC2 +H2O BA HgSOSOH 442 /
(1) C2H2 and CH3CHO (2) CH4 and HCOOH
(3) C2H4 and CH3COOH (4) C2H2 and CH3COOH
Sol : Answer (1)
CaC2 + 2H2O CH CH + Ca(OH)2
CHOCH
OHCHCHOHCHCH mTautomeris
HgSO
SOH
3
224
42
52. BABrCHCH HKMnOKOHaq /.
234
DCKOH
BrNH
23 D is:
(1) CH3Br (2) CH3CONH2
(3) CH3NH2 (4) CHBr2
Sol : Answer (3)
23233
/
23
.
23
23
4
NHCHCONHCHCOOHCH
OHCHCHBrCHCH
KOH
BrNH
HKMnOKOHaq
53. ZYXPhenolOHKMnO
AlClanhy
ClCHdustZn /
.
4
3
3
The product ‘z’ is:
(1) Toluene (2) Benzaldehyde
(3) Benzoic acid (4) Benzene
Sol : Answer (3)
54. In the following compounds
(1) iii > iv> i > ii (2) i > iv > iii > ii
(3) ii > i > iii > iv (4) iv > iii > i > ii
Sol : Answer (4)
Electron withdrawing group stabilizes the phenoxid
anion.
Electron withdrawing effect is maximum at ortho and
para position than meta position.
55.
What is the major product ‘A’
(1) (2)
(3) (4)
Sol : Answer (3)
Ring expansion
56. Which one is most reactive towards nuceloplilic
addition reaction?
Sol : Answer (4)
Electron withdrawing groups increase the reactivity
towards nuclephilic addition.
57. When CH3CONH2 is heated with P2O5, the product
is:
(1) CH3COOH (2) CH3COONH4
(3) CH3CN (4) CH3COCl’
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Sol : Answer (3)
OHNCCHNHCCHD
OP
Oll
232352
58. CBANHHCNHPClHNO 332
252
recognise the compound ‘C’ from the following
(1) Propane nitrile (2) Methyl amine
(3) Ethyl amine (4) Acetamide
Sol : Answer (3)
22323
23223
3
32
NHCHCHClCHCH
OHCHCHNHCHCH
NH
PClHNO
59. Polysaccharides have which of the following
linkage?
(1) Glycosidic linkage (2) H-bond
(3) Peptide linkage (4) No linkage
Sol : Answer (1)
In polysaccharides different monosaccharides units
are linked to one another by glycosidic linkage
60. Salol is
(1) Acetyl salicylic acid (2) Phenyl salicylate
(3) Methyl salicylate (4) None of these
Sol : Answer (2)
Phenyl salicylate is known as salol. It is used as an
antiseptic. It is prepared by the reaction of salicylic
acid with phenol.
PART : C — MATHEMATICS
61. The principal value of tan-1
(cot 4
43) is:
(1) 4
3 (2)
4
3
(3) 4
(4)
4
Sol : Answer (4)
4
43cottan 1
=
44
44cottan 1
=
411cottan 1
=
4222cottan 1
=
4cottan 1
= 1tan 1
= 4
62. The distance of the point (1, -2, 4) from the plane
passing through the point (1, 2, 2) and perpendicular
to the planes x – y + 2z = 3 and 2x – 2y + z = 0 is:
(1) 2 (2) 2
(3) 2 2 (4) 2
1
Sol : Answer (3)
Normal vector =
122
211
ˆˆˆ
kji
=
22ˆ41ˆ41ˆ kji
= kji ˆ0ˆ3ˆ3
Equation of the plane is 02313 yx
0933 yx
Distance = 222 033
92313
= 18
963
= 222
4
23
12
23
12
63. If the vector a kji ˆˆˆ , kjbi ˆˆˆ and kcji ˆˆˆ ,
are coplanar, then the value of
cba 1
1
1
1
1
1
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(1) -1 (2) 2
1
(3) 1/2 (4) 1
Sol : Answer (4)
0
11
11
11
c
b
a
Taking
C2 R2 – R1,
C3 R3 – R2
0
101
111
01
c
bb
aa
0111]011[ bcacba
0111111 bacacba
0111111 bacacba
Dividing by cba 111
01
1
1
1
1
cba
a
011
1
1
1
11
cba
a
11
1
1
1
1
1
cba
64. Let a
and b
be two unit vector such that
3ba
. If )(32 babac
then 2 c
=
(1) 55 (2) 51
(3) 43 (4) 37
Sol : Answer (1)
3ba
32
ba
3ˆˆ.ˆ2ˆ
22 bbaa
31cosˆˆ21 22 ba
01cos1121
3cos22
1cos2
2
1cos
3
babac
32
baababbababac ˆˆ.ˆ6ˆˆ.12ˆ.ˆ4ˆˆ9ˆ4ˆˆ2
222
= 002
14
3sin11941 2
24
395
24
275
= 4
277
= 4
2728
= 4
55
2
55ˆ c
55ˆ2 c
65. If A =
13
14then the determinant of the matrix
A2019
– 2A2018
- A2017
is:
(1) 25 (2) -25
(3) 75 (4) -75
Sol : Answer (1)
13
14A
201720182019 2 AAA
IAAA 222017
IAAA 222017
IAAA 222017
515
5201
2017
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25251
66. The domain of the function f(x) =
2
1sin 1
xis
(1) [-3, 3] (2) [-1,1]
(3) ),1[]1,( (4) ),3(]3,(
Sol : Answer (2)
22
1sin0 1
x
1
2
10
x
210 x
11 x
11 x
11 x
Domain = 1,1
67. If 4)1
1(
2
bax
x
xxlt
x, then
(1) a = 1, b = 4 (2) a = 1, b = - 4
(3) a = 2, b = -3 (4) a = 2, b = 3
Sol : Answer (2)
41
12
bax
x
xxlt
x
41
112
x
xbaxxxlt
x
41
11 22
x
bxaxaxxxlt
x
41
11 2
x
xbaxalt
x
101 aa
41
1
x
xbalt
x
41
1
ba
4
4
31
3
41
b
b
b
ba
ba
68. The value of )2sin(
2/
2
x
dttlt
x
x
=
(1) (2) 2/
(3) 4/ (4) 8/
Sol : Answer (3)
0
02sin
2
2
x
tdt
lt
x
x
RHLx
xlt
x
..22cos
2
420cos
2
69. For xxfRx sin2log)(, and g(x) = f(f(x))
then
(1) g1(0) = cos (log 2)
(2) g1(0) = - cos (log 2)
(3) g is differentiable at x = 0 and g1(0) = -sin (log 2)
(4) g is not differentiable at x = 0
Sol : Answer (1)
xxf
xfxffxg
xffxg
sin2log
'''
xx
xxf cos
sin2log
sin2log'
0cos0sin2log
0sin2log0'
f
= - 1
2logcos2logsin2log
2logsin2log2log1
f
2logcos1
2logcos
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0'0'0' fffg
0'2log' ff
2logcos12logcos
70. For ),2
5,0(
x define f(x) = dtttx
sin0 then f
has
(1) Local maximum at and 2
(2) Local minimum at and 2
(3) Local minimum at and Local maximum at 2
(4) Local maximum at and Local minimum at 2
Sol : Answer (4)
dtttxf
x
sin0
x
xxxxf
xxxf
2
1sincos''
sin'
25,0,sin2,
,00sin
0sin0'
xx
xx
xxxf
02
1sincos11
f
0222
12sin2cos2211
f
71. The integral
dx
xx
xx335
912
)1(
52
(1) cxx
x
235
10
)1(2
(2) cxx
x
235
5
)1(2
(3) cxx
x
235
10
)1(2
(4) cxx
x
235
5
)1(
Sol : Answer (1)
dx
xx
xx
335
912
1
52
= dx
xxx
xx
3525
912
1
52
=
dxxxx
xx
35215
912
1
52
=
dxxx
xx
352
93
1
52
[put , txx 521
dtdxxx
dx
dtxx
63
63
52
52
]
=
3t
dt =
c
xxc
t
252
2
12
1
2
c
xxc
t
252212
1
2
1
= c
xx
2
52
1112
1
= c
x
xx
2
5
35 12
1
=
cxx
x
235
10
12
72. The value of
3log
2log 2
2
)6sin(logsin
sindx
xx
xx2
(1) )2/3log(2
1 (2) )2/3log(
4
1
(3) log(3/2) (4) 1/6 log(3/2)
Sol : Answer (2)
3log
2log 22
2
6logsinsin
sindx
xx
xx
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[Put tx 2
3log3log
2log2log
2
2
tx
tx
dtdxx
dx
dtx
]
3log
2log 26logsinsin
sin dt
tt
t
3log
2log 6logsinsin
sin
2
1dt
tt
t
2
2log3log
2
1
b
a
ab
xbafxf
dxxf
2
2
3log
4
1
3. The area of the region {(x, y): y2
}142 xyandx
is:
(1) 7/32 (2) 5/64
(3) 9/32 (4) 15/64
Sol : Answer (3)
142, 2 xyandxyyx
xy 22 (1)
14 xy (2)
Solving (1) and (2)
8
1
2
1
01812
012128
012816
011016
021816
214
2
2
2
2
xorx
xx
xxx
xxx
xx
xxx
xx
2
1,
8
1,3,
2
1
74. If a curve y = f(x) passes through the point (1, -1) and
satisfies the differential equations y (1 + xy)dx = x dy
then f(-1/2) =
(1) -4/5 (2) 2/5
(3) 4/5 (4) -2/5
Sol : Answer (3)
xdydxxyy 1
1) - , 1 ( through pass This
2
2
2
2
2
cx
y
x
xdxy
xd
xdxy
xd
xdxy
xdyydx
dxxyxdyydx
xdydxxyydx
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5
4
5
82
8
5
2
1
2
1
8
1
2
1
2
1
2
1
2
2
1
2
11
2
1
1
1
2
y
y
y
yx
x
y
x
c
c
75. The probability of a man hitting a target is 2/5. He
fires at the target k times (k, a given number). Then
the minimum k so that the probability of hitting the
target at least once is more than 7/10 is:
(1) 3 (2) 5
(3) 2 (4) 4
Sol : Answer (1)
P [hitting at least 1 time ] = 1 –P [ hitting no time]
=
k
5
31
10
3
5
3
10
7
5
31
k
k
From options k =3
76. The value of
4/3
4/ cos1
x
dx
(1) 2 (2) 4
(3) -2 (4) 0
Sol : Answer (1)
4
3
4cos1
x
dxI
= 4
3
4 2
2cos2
x
dx
= 4
3
4
2
2sec
2
1
dxx
=
4
3
42
12
tan
2
1
x
4
3
4
2tan
x
8
tan8
3tan
21212
77. The integral
dxxxxxxx
xx2523235
22
)coscossinsincos(sin
cossin
(1) cx
3cot1
1 (2) c
x
3cot1
1
(3) cx
)tan1(3
13
(4) cx
)tan1(3
13
Sol : Answer (4)
dx
xxxxxx
xx2523235
22
)coscossinsincos(sin
cossin
=
dxxx
xx233
22
)cos(sin
cossin
Dividing by cos6 x
Then = dx
x
xx
)1(tan
sectan3
22
[ Put
dx
dtxx
tx
22
3
sectan3
1tan
]
dx
t
dt
)(
32
c
t
t
3
1
13
1 1
c
x
)1(tan3
13
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78. If 5(tan2x – cos
2 x) = 2 cos 2x + 9, then the value of
cos 4x is:
(1) 1/3 (2) 2/9
(3) -7/9 (4) -3/5
Sol : Answer (3)
92cos2costan5 22 xxx
91cos22coscos
sin5 22
2
2
xx
x
x
7cos4
cos
coscos15 2
2
42
x
x
xx
xxxx 2442 cos7cos4cos5cos55
05cos12cos9 24 xx
18
18014412cos 2
x
= 18
32412
18
1812
18
30
18
6 or
3
5
3
1 or
9
71
9
2
13
12
12cos24cos
3
11
3
121cos22cos
3
1cos
2
2
2
2
xx
xx
x
79. If , C are the distinct roots of the equation
x2 – x + 1 = 0, then 107101
(1) 1 (2) 2
(3) -1 (4) 0
Sol : Answer (1)
2,
112
2
214101
1072101107101
80. Let R = { (3, 3), (6, 6), (9, 9), (12, 12), (6, 12),
(3, 9), (3, 12), (3, 6) } be a relation on the set
A = { 3, 6, 9 12}. The relation is:
(1) An equivalence relation
(2) Reflexive and symmetric only
(3) Reflexive and transitive only
(4) Reflexive only
Sol : Answer (3)
81. Let z, w be complex numbers such that 0 wiz
and arg (zw) = , then arg (z) =
(1) 4
5 (2)
2
(3) 4
3 (4)
4
Sol : Answer (3)
0 wiz
iwz
wiz
wiz
argarg
4
3arg
2
3arg2
arg2
argarg
z
z
z
wi
82. If twice the 11th term of an AP is equal to 7 times of
its 21st term, then its 25
th term is equal to:
(1) 24 (2) 120
(3) 0 (4) 48
Sol : Answer (3)
Let a is the first term of AP and d is common difference
∴ 11th term will be a + 10d
21st term will be a + 20d
We need to find 25th term which is a + 24d
Here given that 2 times of 11th term is equals to
7 times the 21st term
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2( a + 10d) = 7( a + 20d)
2a + 20d = 7a + 140d
5a + 120d = 0 or
a + 24d =0 (which is 25th term)
∴ 25th term of given AP = 0
83. The number of arrangements of the letters of the
word ‘BANANA’ in which two N’s do not appear
adjacently is:
(1) 40 (2) 60
(3) 80 (4) 100
Sol : Answer (1)
1804
720
!2!2
!6
two N’s come together 602
120
!2
!5
!2
!2
required 12060180
84. If x4 occurs in the r
th term in the expression of
(x4 + 1/x
3)
15 then r =
(1) 7 (2) 8
(3) 9 (4) 10
Sol : Answer (3)
r
r
rx
xCt
3
154
41
115
rxC 760
415
8
756
4760
r
r
thr 9181
85. In ,ABC if (a + b + c) ( a – b + c) = 3ac, then
(1) 060B (2)
030B
(3) 060C (4)
090 CA
Sol : Answer (1)
accbacba 3
0
222
222
222
22
60
2
1cos
2
1
2
32
3
B
B
ac
bca
acbca
acbcaca
acbca
86. The number of integral values of m, for which the x
co-ordinate of the point of intersection of the lines
3x + 4y = 9 and y =mx + 1 is also an integer is:
(1) 2 (2) 0
(3) 4 (4) 1
Sol : Answer (1)
Solving, 1,943 mxyyx we get m
x43
5
x is an integer if 3 + 4 m = 1, -1, 5, -5
4
8,
4
2,
4
4,
4
2 m
so m has two integral values.
87. The number of common tangents to the circles
x2 + y
2 – 4x – 6y–12 = 0 and
x2 + y
2 + 6x + 18y + 26= 0 is:
(1) 1 (2) 2
(3) 3 (4) 4
Sol : Answer (3)
88. The point of intersection of the normals to the
parabola y2 = 4x at the ends of its latus rectum is:
(1) (0,2) (2) (3, 0)
(3) (0,3) (4) (2,0)
Sol : Answer (2)
End of the latus rectum are
12,21,1 2 ttt
Equation of the normal at tt 2,2 is
22 tttxy
Equation of the normal for 1t are
3xy and 3 xy
They intersect at (3, 0) Or
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89. Let the equation of two ellipses be E1 : 123
22
yx
and E2 : 116 2
22
b
yx, if the product of their
eccentricities is 1/2, then the length of the minor axis
of E2 is:
(1) 8 (2) 9
(3) 4 (4) 12
Sol : Answer (3)
e1 e2 = 2
1
2
1
2
2
1
1 a
c
a
c
2c1 c2 = a1a2 …………………….(1)
2
1
1
16
123
bc
c
2 216 b = 43
4(16 - b2) = 48
64 - 48 = 4b2
16 = 4 b2
b = 2
2b = 4
90. If the standard deviation of the number 2, 3 a and 11
is 3.5, then which of the following is true
(1) 3a2 – 32 a + 84 = 0 (2) 3a
2 – 34a + 91 = 0
(3) 3a2 – 23a + 44 = 0 (4) 3a
2 - 26a + 55 = 0
Sol : Answer (1)
As we learnt in ,
Standard Deviation -
If x1, x2...xn are n observations then square root of the
arithmetic mean of
Where x is mean