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Transcript of 5(9,(: $66,*10(17 -(( 0DLQ...-(( 0dlq 5hylhz $vvljqphqw $dndvk (gxfdwlrqdo 6huylfhv /lplwhg ±...
Aakash Educational Services Limited – Registered Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 [Page 1]
Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005, Ph.011-47623456
REVIEW ASSIGNMENT-2 (JEE Main) (PHYSICS)
Topics covered :
Complete syllabus of JEE(Main)
Choose the correct answer :
1. If 1 m/s2 is equal to x and 1 km/min2 is equal
to y then
(1) x < y (2) x = 3.6y
(3) y = 3.6x (4) x = 360y
2. A particle moves on x-axis such that its
position is given by
x = |t – T|m (Here t is time and T is a constant)
If the average velocity of the particle for first
6 s is zero then the distance covered by the
particle in first 3 s is
(1) 2 m (2) 6 m
(3) 3 m (4) 1.5 m
3. Loudness of sound at a distance x from a point
source of sound is L. At what distance from the
point source loudness is 2
L ?
(1) 2x
(2) 2x
(3) 4x
(4) It may be any of the above values
4. a, b and c represent maximum particle
velocity, maximum particle acceleration and
wave speed respectively for a harmonic
transverse wave travelling in negative y
direction on a string. Which of these equations
may be representing the wave?
(1) 2sin
6
a a bx t y
b b ac
(2) 2sin
a a bx t y
b b ac
(3) 2
sin6
a b bx t y
b a ac
(4) 2
sina b b
x t yb a ac
5. A body is projected from a height 2
R above
surface of earth with a minimum speed so that
it does not fall on the surface of earth. If mass
and radius of earth are M and R respectively,
then the speed of the body when it is closest
to earth is
(1) 3
2
GM
R (2)
6
7
GM
R
(3) 5
6
GM
R (4)
6
5
GM
R
6. A heavy rod of mass m, length , hangs
vertically downward from a rigid support as
shown in the figure. Young's modulus of
material of rod is Y and area of cross section
is A. The elastic potential energy of rod is
(1) 2 2 3
6
M g L
AY (2)
2 2 2
6
M g L
AY
(3) 2 2
3
M g L
AY (4)
2 2
6
M g L
AY
25/03/2020
JEE Main Review Assignment-2
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7. A block of mass 2.0 kg is moving on a
frictionless horizontal surface with a velocity v
m/s towards another block of equal mass kept
at rest. The spring constant of the spring fixed
at one end is 100 N/m. If maximum
compression of the spring is 10 cm, the
velocity v is
(1) 1 m/s (2) 0.5 m/s
(3) 0.25 m/s (4) 0.75 m/s
8. Two blocks of masses m1 and m2 are
connected by a light inextensible string
passing over a smooth fixed pulley of
negligible mass. Acceleration of centre of
mass of system is
(1) 1 2
1 2
m mg
m m
(2) 1 2
1 2
m mg
m m
(3) 1 2
1 2
2m mg
m m
(4) 2
1 2
1 2
m mg
m m
9. In shown figure, if mass A will go up by a
distance x, mass B will go
(1) 3x down (2) 4x down
(3) 2x down (4) 8x down
10. A cylinder whose inside diameter is 4.00 cm
contains air compressed by a piston of mass
m = 13.0 kg, which can slide freely in the
cylinder. The entire arrangement is immersed
in a water bath whose temperature can be
controlled. The system is initially in equilibrium
at temperature ti = 20°C. The initial height of
the piston above the bottom of the cylinder is
hi = 4.00 cm. The temperature of the water
bath is gradually increased to a final
temperature tf = 100°C. The height hf of the
piston is
(1) 10.18 cm (2) 5.09 cm (3) 2.03 cm (4) 20.36 cm
11. The above composite block is shown as a
2-dimensional figure below
The block is made up of materials of two
different thermal conductivities as shown. The equivalent thermal conductivity between face A and face B is given by
(1) 2 1
2
1
ln
k kk
k
(2) 2 1
2 1
k k
k k
(3) (k2 – k1) ln 2
1
k
k
(4) 2
12 1( )
k
kk k e
12. For a heat engine the temperature of the source is 127ºC. To have 60% efficiency the temperature of the sink is
(1) 160ºC (2) –200ºC (3) –113ºC (4) 113ºC 13. The two arms, of a vertical U-tube containing
a liquid, are maintained at different
temperatures, t1 and t2. The liquid columns in
the two arms have heights l1 and l2
respectively. The coefficient of volume expansion of the liquid is equal to (Initial temperature is assume to be zero)
(1) 1 2
2 1 1 2t t
l l
l l (2) 1 2
1 1 2 2t t
l l
l l
(3) 1 2
2 1 1 2t t
l l
l l (4) 1 2
1 1 2 2t t
l l
l l
Review Assignment-2 JEE Main
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14. The amount of work done by the agent to
change the configuration A into B is
(1) 2
04
Q
l (2)
2
08
Q
l
(3) 2
02
Q
l (4)
2
0
3
4
Q
l
15. A charged particle of specific charge
(charge/mass) is released from origin at time
t = 0, with velocity 0ˆ ˆ( )v v i j
in uniform
magnetic field 0ˆB B i
. Co-ordinates of the
particle at time 0
tB
are
(1) 0 0 0
0 0 0
2, ,
2
v v v
B B B
(2) 0
0
, 0, 02
v
B
(3) 0 0
0 0
20, ,
2
v v
B B
(4) 0 0
0 0
2, 0,
v v
B B
16. Consider a conducting ring of radius r placed
in a plane containing two parallel infinite wires
as shown. The wires carry equal and opposite
current i. Distance of the wires from centre of
the ring are also shown. For magnetic field at
the centre of the ring to be zero, the ring
should be carrying current
(1) 2
i
, anticlockwise (2)
2
i
, clockwise
(3) 6
i
, anticlockwise (4)
6
i
, clockwise
17. Two identical rings of identical radii and
resistance of 36 each are placed in such a
way that they cross each others centre C1 and
C2 as shown. Conducting joints are made at
intersection points A and B of the rings. An
ideal cell of emf 20 V is connected across AB.
The power delivered by cell is
(1) 80 W
(2) 100 W
(3) 120 W
(4) 200 W
18. The value of current passing through battery
immediately after (K) is closed, If initially all
capacitors are uncharged, will be
(1) 5
E
R
(2) 3
E
R
(3) 6
5
E
R
(4) Zero
19. Resistance of a wire at temperature °C is
given by relation 0 1b
R R a
. Here
R0 is the temperature at 0°C. The temperature
coefficient of the material of the wire at °C is
(1) 2
2
( )a b
(2) 2
2
( )
( )
a b
a b
(3) 2
( )
( )
a b
a a b
(4) 2
a b
a b
JEE Main Review Assignment-2
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20. A source of red light is placed at the bottom of two transparent media having refractive index
= 2 and 2 as shown in the figure. The
maximum area in which red is seen when observed from above is nearly
(1) 50.1 cm2 (2) 40 cm2
(3) 100 cm2 (4) 20 cm2
21. A ray of light falls on a prism of refractive index
2 at an angle i with the normal. What
should be the minimum value of i for the total internal reflection will take place at side AC is
(1) 30° (2) 45°
(3) 60° (4) 90°
22. If in the given diagram OO is the principal axis.
P is the point object and P is the image, then the mirror used and location of object (P) w.r.t. mirror is
(1) Concave mirror with P beyond centre of curvature
(2) Concave mirror with P between focus and pole
(3) Concave mirror with P between focus and centre of curvature
(4) Convex mirror with P at 2.5f on right,
where f is the focal length of the mirror
23. Select the correct alternative.
(1) In Young's double slit experiment (YDSE),
if one of the slits is covered, the intensity
at the centre becomes half
(2) If YDSE is performed under water instead
of air, fringe width will become 1
times of
the value in air
(3) If one of the slits is covered with a
transparent sheet, the interference pattern
shifts to the opposite side in which
transparent sheet is introduced
(4) All of these
24. A graph is plotted between rate of
disintegration and number of active nuclei.
What is half life of the radioactive substance?
(Here time is in second)
(1) 3 s (2) 1
s3
(3) 3 ln2s (4) ln2
s3
25. A nucleus of mass number A originally at rest
emits -particle with kinetic energy E. The
kinetic energy of the daughter nucleus is equal
to
(1) 4E
A (2)
4
( – 4)
E
A
(3) ( – 4)
4
A E (4)
( – 4)
AE
A
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REVIEW ASSIGNMENT-2 (JEE Main) (PHYSICS)
[ANSWERS]
1. (2)
2. (3)
3. (4)
4. (3)
5. (4)
6. (4)
7. (1)
8. (4)
9. (1)
10. (2)
11. (1)
12. (3)
13. (1)
14. (1)
15. (4)
16. (1)
17. (2)
18. (3)
19. (2)
20. (1)
21. (2)
22. (3)
23. (2)
24. (4)
25. (2)
25/03/2020
JEE Main Review Assignment-2
Aakash Educational Services Limited – Registered Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 [Page 6]
[ANSWERS]
1. Answer (2)
2
2
1km
10001m/s1
minute3600
x = 3.6 km/minute2
= 3.6y
2. Answer (3)
Above curve shows the position versus time curve.
3. Answer (4)
L = 10 log 0
I
I
If distance becomes n times, as I 2
1
r for point
source I becomes 2
1
n times
L= 10 log 2
0
I
n I
= 10 log 0
I
I – 10 log n2
= 10 log 0
I
I – 20 log n
= 2
L
20 log n = 2
L
n may be any value depending on L
4. Answer (3)
vmax = a = A, amax = b = A2, v = c
= ,b
a A =
2
,a b
kb v ac
x = Asin (t + ky + )
5. Answer (4)
The particle will revolve around earth in elliptical path as shown, when projected at A
By conservation of angular momentum,
11 2 2
33
2 2
vRv v R v
By conservation of energy,
2 21 2
1 2 1–
2 3 2
GMm GMmmv mv
R R
Solving v2 = 6
5
GM
R
6. Answer (4)
Taking small element of length dy at distance y
from lower end. Stress = 1
,2
M g dUY
L A dV stress
× strain.
21 (Stress)
Volume2
dUY
21(Stress)
2A dy
Y
2 2
6
M g LU
YA
7. Answer (1)
At maximum compression, velocity of both the block will same
2v = 2v1 + 2v1
1 2
vv
From conservation of energy
2 2
2 21 1 1 1
2 2 2 2 2 2
v vmv m m kx
v = 1 m/s
8. Answer (4)
Acceleration of both blocks
2 1
1 2
m ma g
m m
... (1)
21 2 1 2
cm
1 2 1 2
ˆ ˆ( ) ( )m a m a m a j m a ja
m m m m
1 2 1 2 2 1
1 2 1 2 1 2
ˆ ˆ( )m m m m m m
a j gjm m m m m m
2
1 2
1 2
ˆ( )m m
g jm m
Review Assignment-2 JEE Main
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9. Answer (1)
Point P will go up, pulley y will go down by x will
make 2x string free and length x due to
movement of Q upward.
10. Answer (2)
i f
i f
v v
T T i f
i f
h h
T T
ff i
i
Th h
T
373
4.00 5.09 cm293fh
11. Answer (1)
Thermal resistance of small element is
dR = 2
2 1 1
1
( )
ldx
k k x k lb
R = 2 1 12
12 1
( )ln
( )
k k l k lldR
k lk k b
k =l
AR = 2 1
2
1
ln
k kk
k
12. Answer (3)
2 2
1
601 1
100 (273 127)
T T
T
2 20.6 1 0.4400 400
T T
T2 = 160 K = –113°C
13. Answer (1)
2 1
2 11 1t t
l l
14. Answer (1)
2 2
2 – 22
kQ kQW U
l l
= 2 2
04
kQ Q
l l
15. Answer (4)
Time period = 2 m
Bq
2
B
0 2
Tt
B
path = helical
axis of helix = x-axis
x-coordinate = 0
0
v
B
z-coordinate = –zr 0
0
2v
B
So, 0 0
0 0
2, 0,
v v
B B
16. Answer (1)
Magnetic field due to wires at the centre of the
ring
= 0 0 0
2 (3 ) 2 (6 ) 4
i i i
r r r
directed down the plane of ring
Ring should carry anticlockwise current,
say i, such that
| |B
due to ring = | |B
due to wires
0 0
2 4
i i
r r
2
ii
(anticlockwise)
17. Answer (2)
AC1 = AC2 = C1C2 = radius
AC1B = 120
JEE Main Review Assignment-2
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1 1 1 1 1
24 12 12 24R
R = 4
2 2(20)
100 W4
VP
R
18. Answer (3)
net 3 2
R RR
5
6
R
6
5
Ei
R
19. Answer (2)
0 2
dR bR a
d
dR
Rd
0 22
2
0
( )
( )1
bR a
a bb a b
R a
20. Answer (1)
2 sin 2 sini r and 2 sin 1 sin2
r
1
sin 30 and 452
i i r
Radius of the circle on the upper surface is
3.46 tan30 2 tan 45 = 4 cm
Area = R2 = (4 × 10–2)2 = 16 × ×
10–4 cm2
= 50.1 cm2
21. Answer (2)
r1 aC
11
1sinr
r1 45° r1 = 45°
and r1 + r2 = 75°
r2 = 30°,
Also, 12
11 sin sin sin 2
2i r i
i = 45°
22. Answer (3)
A concave mirror with object placed between
focus and centre of curvature gives an inverted
and enlarged image.
23. Answer (2)
In any medium wavelength becomes 0
= fringe width = D
d
will become
1
times.
24. Answer (4)
–dN
Ndt
so slope = 1
tan30º
3
Half life ln2
3
25. Answer (2)
1 2
2 1
E m
E m
2
– 4
4
E A
E
E2 = 4
( – 4)E
A
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REVIEW ASSIGNMENT-2 (JEE Main) (CHEMISTRY)
Topics covered :
Complete syllabus of JEE (Main)
Choose the correct answer :
1. 0.250 g of a diacidic organic base required
15 ml of N/5 HCl for complete neutralisation.
The molecular mass of the base is
(1) 41.66 (2) 83.33
(3) 166.66 (4) 124.99
2. Thermite mixture contains
(1) 3 parts Fe2O3 and 2 parts Al by mass
(2) 3 parts Al2O3 and 4 parts Al by mass
(3) 1 part Fe2O3 and 12 parts Al by mass
(4) 3 parts Fe2O3 and 1 part Al by mass
3. Alkaline earth metal having minimum density is
(1) Mg (2) Be
(3) Ba (4) Ca
4.
(1)
(2)
(3)
(4)
5. The reactivity order of the following w.r.t.
electrophile is
I.
II.
III.
IV.
(1) I > II > III > IV
(2) II > I > III > IV
(3) II > III > I > IV
(4) I > III > II > IV
6. 15 mL of a gaseous sample believed to
contain C2H4 and C2H2 requires 45 mL of O2
for complete combustion. Determine
percentage composition of C2H2 in the
mixture.
(1) 15 mL (2) 7 mL
(3) 5 mL (4) Zero
7. Kinetic energy of the electron in 3rd shell of
Li+2 is
(1) –13.6 eV (2) 13.6 eV
(3) –27.2 eV (4) 27.2 eV
8. If x-axis is the internuclear axis then which of
the following is the correct combination for p-
d bond?
(1) px-dxy
(2) py-dyz
(3) 2z zp -d
(4) pz-dzx
25/03/2020
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9. Out of the following mentioned reactions, the
one which has highest rate is
(I)
(II)
(III)
(IV)
(1) I (2) II
(3) III (4) IV
10. Identify the non-narcotic analgesic among the
following.
(1) Morphine (2) Paracetamol
(3) Codeine (4) Heroin
11. A sample of ideal gas is compressed from
initial volume of 2V0 to V0 using three different
processes
(A) Reversible isothermal
(B) Reversible adiabatic
(C) Irreversible adiabatic under a constant
external pressure
Then
(1) Final temperature of gas will be highest at
the end of process (B).
(2) Magnitude of enthalpy change of sample
will be highest in isothermal process
(3) Final temperature of gas will be highest at
the end of process (C).
(4) Final pressure of gas will be highest at the
end of process (B).
12. FeC2O4 solution of some concentration is
titrated with KMnO4 in presence of HCl and in
presence of H2SO4 separately. If VHCI and
2 4H SOV represent volume of oxidising agent
consumed at the two cases respectively, then
which of the options is correct ? (Ignore the
reaction between FeC2O4 and Cl2)
(1) VHCI = 2 4H SOV
(2) VHCI > 2 4H SOV
(3) VHCI < 2 4H SOV but VHCI 2 4H SO
1V
2
(4) VHCI = 2 4H SO
1V
2
13. Vapour pressure of C6H6 and C7H8 mixture at
50°C is given by P (mm Hg) =180 XB+ 90,
where XB is the mole fraction of C6H6. A
solution is prepared by mixing 936 g benzene
and 736 g toluene and if the vapours over this
solution are removed and condensed into
liquid and again brought to the temperature of
50°C, what would be the new mole fraction of
C6H6 in the vapour state ?
(1) 0.93 (2) 1.93
(3) 1.00 (4) 2.93
14. 20 ml of KOH solution was titrated with 0.20 M
H2SO4 solution in a conductivity cell. The data
obtained were plotted to give the graph shown
below :
The concentration of the KOH solution was :
(1) 0.30 mol L–1 (2) 0.15 mol L–1
(3) 0.12 mol L–1 (4) 0.075 mol L–1
15. 84Po210 decays with particle to 82Pb206 with a
half-life of 138.4 days. If 1.0 g of 84Po210 is
placed in a sealed tube, how much helium will
accumulate in 69.2 days. Express the answer
in cm3 at STP.
(1) 11.25 cm3
(2) 21.25 cm3
(3) 31.25 cm3
(4) 41.25 cm3
Review Assignment-2 JEE Main
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16.
The products of ozonolysis are (considering
benzene does not undergo ozondysis)
(1)
(2)
(3)
(4)
17. Identify the final product
3 4
2 43 2 6 4 3
HNO MnO /HH SO
PhCH p O N C H CH (X)
2 52
3
P O(i) SOCl(ii) NH ,
(Y) (Z)
(1) p-O2N—C6H4—CO2H
(2) p-O2N—C6H4—CN
(3) p-O2N—C6H4—CONH2
(4) p-O2N—C6H4—CH2CN
18.
(1) HCHO & HCOCH(OCH3)2
(2) (CH3O)2CH – COOH & HCHO
(3) (CH3O)2CH – CHO + HCOOH + HCHO
(4) HCOOH & HCOCH(OCH3)2
19. When tin is treated with concentrated nitric
acid :
(1) It is converted into stannous nitrate
(2) It is converted into stannic nitrate
(3) It is converted into metastannic acid
(4) It becomes passive
20. Match column-I with column-II
Column-I (Chemical reaction)
I. 800 C/Pt3 2 24NH 5O 4NO 6H O
II. 2400 450 C/CuCI2 2 24HCI O 2CI 2H O
III. 2 5V O2 2 3450 500 C
2SO O 2SO
IV. Fe Mo2 2 3500 C
2N 3H 2NH
Column-II (Name of process)
(a) Contact process
(b) Ostwald’s process
(c) Deacon’s process
(d) Haber’s process
(1) I-a, II-b, III-d, IV-c
(2) I-b, II-c, III-a, IV-d
(3) I-a, II-d, III-c, IV-b
(4) I-a, II-c, III-b, IV-d
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Integer Value Correct Type Questions
21. The ratio of Fe+3 and Fe+2 ion per unit cell of
Fe3O4 in octahedral voids is _________.
22. 24.95 g of CuSO4·5H2O is dissolved in 171 g
of H2O. If the percentage lowering of vapour
pressure at 100°C is x, x is closest to integer
________ .
23. Which among the following species exhibit
geometrical isomerism?
[Co(NH3)4(H2O)2]Cl2, [Ni(H2O)6]Cl2,
[Co(H2O)5Cl]Cl, [Pt(NH3)2Cl2],
[Pt(en)Cl2], [Pt(en)2Cl2],
[PdCl2BrI], [Co(NH3)4Cl2],
[Co(en)2Br2]
24. Find the pH at isoelectric point of the following
-amino acid
Given : 1apK = 2,
2apK = 4 and 3apK = 8
25. Initially only PCl5 (g) is present at 6 atm. If at
equilibrium it is partially converted into PCl3
and Cl2 gases and total pressure becomes 8
atm, the value of Kp in atm will be ________.
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REVIEW ASSIGNMENT-2 (JEE Main) (CHEMISTRY)
[ANSWERS]
1. (3)
2. (4)
3. (4)
4. (3)
5. (1)
6. (4)
7. (2)
8. (4)
9. (4)
10. (2)
11. (3)
12. (2)
13. (1)
14. (1)
15. (3)
16. (2)
17. (2)
18. (3)
19. (3)
20. (2)
21. (1)
22. (2)
23. (6)
24. (3)
25. (1)
25/03/2020
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[SOLUTIONS]
1. Answer (3)
At the completion of reaction,
Molecular equivalent of acid = Molecular equivalent of base
0.25 1
2 1000 15Molecular weight 5
Molecular weight = 0.25 2 1000 5
15
= 166.66 g
2. Answer (4)
Fact.
3. Answer (4)
Density of Ca = 1.55 g cm3–, Be > Mg > Ca < Sr < Ba
Due to very large volume.
4. Answer (3)
5. Answer (1)
Activating power
–OCH3 > –CH3 > –Cl > –NO2
6. Answer (4)
2 4 2 2 2
x mL 3x mL
C H 3O 2CO 2H O
2 2 2 2 2
2.5(15 x) mL(15 x) m
5C H O 2CO H O
2
Total volume of O2
2.5(15 – x) + 3x = 45
x = 15
Volume of C2H2 in the sample = 0
7. Answer (2)
Kinetic energy of e–
– E = 2
2
13.6 313.6 eV
3
8. Answer (4)
9. Answer (4)
Toluene is more reactive than benzene and
formed carbocation as electrophile is best
favoured in so, reaction IV has most rate of
reaction.
10. Answer (2)
Fact.
11. Answer (3)
Work done on the system in an irreversible
adiabatic process is maximum. So, rise in
temperature of the system in such a process will
also be maximum.
12. Answer (2)
Part of KMnO4 will be used to oxidise HCl in
addition to oxidise FeC2O4. Therefore, vol of
KMnO4 used at the end point in presence of HCl
(VHCl) will be more than that used in presence of
H2SO4 2 4H SOV
13. Answer (1)
Vapour pressure of pure C6H6 B(P ) = 270 mm
Vapour pressure of pure toluene T(P ) 90 mm
B T
936 736n 12; n 8
78 92
XB = 0.6 and XT = 0.4
1 1B TP 0.6 270;P 0.4 90;
1 11 B TP P P 162 36 198 mm
1 1B T
162 36Y 0.82;Y 0.18
198 198
2BP 0.82 270;
2 2 2T 2 B TP 0.18 90; P P P
= 221.4 + 16.2
= 237.6 mm
2B
221.4Y 0.93
237.6
Review Assignment-2 JEE Main
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14. Answer (1)
Let the molarity of KOH be x mol/L
20 × x = (0.2 × 2) × 15; x = 0.30 mol/L
15. Answer (3)
No= initial number of moles of 0
1P
210
1t
2 half life of P0 = 138.4 days
n= no. of half lives in 69.2 days = 69.2 1
138.4 2
Nt, number of moles of p0 left at t = 69.2 days
12
t 0
1N N
2
No. of moles of He formed
= N0 – Nt
= 0 0
1 2 1N 1 N
2 2
= 0.293 N0
Volume of He gas formed at STP
= 0.293 22400
31.25cc210
16. Answer (2)
17. Answer (2)
18. Answer (3)
19. Answer (3)
20. Answer (2)
21. Answer (1)
Fe3O4
Fe+2 in 1
th4
of octahedral voids
Fe+3 in 1
th4
of octahedral voids + 1
th8
of
tetrahedral voids
JEE Main Review Assignment-2
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22. Answer (2)
Mass of CuSO4 in 25 g CuSO4·5H2O = 15.98 g
Mass of water = 196 – 15.98 = 180 g
o
15.982P 2 0.1 0.2159.5 0.019615.98 180 2 0.1 10 10.2P 2159.5 10
% = 0.0196 × 100 = 1.96 2
23. Answer (6)
[Ni(H2O)6]Cl2, [Co(H2O)5Cl]Cl, [Pt(en)Cl2] does
not show geometrical isomerism.
24. Answer (3)
The pH at isoelectric point of the given amino
acid is given by
p1 = 1 2a apK pK
2
= 2 4
2
= 3
25. Answer (1)
6 – p + 2p = 8
p = 2
Kp = 2 2
16 2
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REVIEW ASSIGNMENT-2 (JEE Main) (MATHEMATICS)
Topics covered :
Complete syllabus of JEE (Main)
Choose the correct answer :
1. The number of distinct rational numbers of the
form p/q, where , {1, 2, 3, 4, 5, 6}p q is
(1) 23 (2) 32
(3) 36 (4) 28
2. The point (1, 2) is one extremity of focal chord
of parabola 2 4y x . The length of this focal
chord is
(1) 2 (2) 4
(3) 6 (4) None of these
3. If [x] denotes the integral part of x, then domain
of the function
13 3 2( ) sin
( 1)( 2)( 3) 2
x xf x
x x x
is
(1) [0, 2) (2) [0, 2) – {1}
(3) (– , 3) – {1, 2} (4) None of these
4. If the function 3 2( ) 11 6f x ax bx x
satisfies conditions of Rolle’s theorem in [1, 3]
and
12 0,
3f then value of a and b are
respectively
(1) 1, –6 (2) –1, 6
(3) –2, 1 (4) –1, 1
2
5. If n arithmetic means are inserted between two
sets of numbers a, 2b and 2a, b, where
a, b R. Suppose that mth arithmetic mean
between these two sets of numbers is same,
then the ratio a : b equals
(1) 1:n m m (2) 1:n m n
(3) : 1m n m (4) : 1n n m
6. If 1 be a cube root of unity and
7(1 ) l m , then the value of l m
(1) 0 (2) 1
(3) 2 (4) –1
7. If a and b are chosen randomly from the set
consisting of numbers 1, 2, 3, 4, 5, 6 with
replacement. Then probability that
2/
0lim
2
xx x
x
a b= 6 is
(1) 1
3 (2)
1
4
(3) 1
9 (4)
2
9
8. From origin, chords are drawn to the circle
2 2 2 0x y y . The locus of the middle
points of these chords is
(1) 2 2 0x y y
(2) 2 2 0x y x
(3) 2 2 2 0x y x
(4) 2 2 0x y x y
25/03/2020
JEE Main Review Assignment-2
Aakash Educational Services Limited – Registered Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 [Page 2]
9. Identify the pair of functions which are not
identical.
(1) 21 1
tan cos ; x
y x yx
(2) 1 1tan cot ; y x y
x
(3) 2
sin arc tan ; 1
xy x y
x
(4) 1 1cos tan ; sin tany x y x
10. If z is a complex number satisfying
4 3 22 1 0,z z z z then the set of
possible values of | z | is
(1) {1, 2}
(2) {1}
(3) {1, 2, 3} (4) {1, 2, 3, 4}
11. Integral of 1 2cot (cot cosec )x x x with
respect to x is:
(1) 2lncos2
xc
(2) 2lnsin2
xc
(3) 1
lncos2 2
xc
(4) ln sin x ln(cosec x cot x) + c
12. 2
cos
0
cos sine d
(1) (2) 2
(3) 0 (4) 3
13. For each real number x such that – 1 < x < 1,
let A(x) be the matrix 1 1(1 )
1
xx
x
and
.1
x yz
xy
Then,
(1) A(z) = A(x) + A(y)
(2) A(z) = A(x) [A(y)]–1
(3) A(z) = A(x) A(y)
(4) A(z) = A(x) – A(y)
14. The length of the perpendicular from the origin
to the plane passing through the point a and
containing the line r b c is
(1) [ ]
| |
a b c
a b b c c a
(2) [ ]
| |
a b c
a b b c
(3) [ ]
| |
a b c
b c c a
(4) [ ]
| |
a b c
c a a b
15. If the standard deviation of 1 2, , ..... nx x x is
3.5, then the standard deviation of
1 22 3, 2 3,..., 2 3nx x x is
(1) –7 (2) –4
(3) 7 (4) 1.75
16. If 1( )y x is a solution of the differential
equation / ( ) 0,dy dx f x y then a solution of
the differential equation ( ) ( )dy
f x y r xdx
(1) 11
1( ) ( )
( )r x y x dx
y x
(2) 11
( )( )
( )
r xy x dx
y x
(3) 1( ) ( )r x y x dx
(4) None of these
17. The distance between the directrices of the
ellipse 22 24 8 16 3 10x y x y is K
then 2
K is
18. The number of distinct real roots of the
equation
cos sin sin
sin cos sin 0
sin sin cos
x x x
x x x
x x x
which lies
in the interval 4 4
x
is______
Review Assignment-2 JEE Main
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19. Given /2
0
1lnsin ln
2 2xdx
and
2/2
0
.ln2
sin 2
x kdx
x
then k = ______
20. The number of ordered pairs (x, y) where
, [0, 10]x y satisfying
22 sec1
sin sin .2 12
yx x
is 2K then K =
21. If (a, b] is the domain of the function
0.52 8
log2
xf x
x
then value of b – a
is___
22. In a triangle ABC, 5, 2,6
a b A
and
c1, c2 are the possible values of the third side.
Then |c1 – c2| is
23. 4 341 2
nn n n no n nC C C C C
4 6 4 93 ...n n n
n nC C C to n terms is equal
to (1 )nk then the value of k is
24. Let PN be the ordinate of a point P on the
hyperbola 2 2
2 21
97 79
x y and the tangent at P
meets the transverse axis in T. Then the value
of 2020
ON OT
is
(O being origin and [] denotes the greatest
integer function)
25. Find natural number n so that area bounded
by y = nx2, 21
2y nx and 2 4 3 0y y is
greatest.
JEE Main Review Assignment-2
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REVIEW ASSIGNMENT-2 (JEE Main) (MATHEMATICS)
[ANSWERS]
1. (1)
2. (2)
3. (2)
4. (1)
5. (3)
6. (3)
7. (3)
8. (1)
9. (4)
10. (2)
11. (2)
12. (2)
13. (3)
14. (3)
15. (3)
16. (1)
17. (8)
18. (2)
19. (2)
20. (8)
21. (4)
22. (04)
23. (2)
24. (4)
25. (1)
25/03/2020
Review Assignment-2 JEE Main
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[ANSWERS]
1. Answer (1)
1, 1, 2, 3, 4, 5, 6 6p q
2, 1, 3, 4, 5, 6 3 [ (2, 4), (2, 6)]p q
3, 1, 2, 4, 5, 6 4 [ (3, 6)]p q
4, 1, 3, 5, 6 3 [ (4, 6)]p q
5, 1, 2, 3, 4, 6 5p q
6, 1, 5 2p q
2. Answer (2)
The parabola 2 4 ,y x here 1a and focus is
(1, 0).
The focal chord is ASB. This is clearly latus
rectum of parabola, its value = 4
3. Answer (2)
For f(x) to be defined
(i) 3 2
1, 0, 12
x
3 2
1 2 2 3 2 42
xx
0 3 6 0 2x x ...(A)
(ii) 3 0x and 1, 2, 3x
x < 3 and 1, 2x ...(B)
From (A) and (B), we have
0 1x or 1 < x < 2
Domain 0,1 1,2 0,2 1
4. Answer (1)
f(1) = f(3)
a + b + 11 – 6 = 27a + 9b + 33 – 6
13a + 4b = –11
and 23 2 11f x ax bx ...(i)
2
1 12 3 2
3 3f a
1
2 2 11 03
b
1 4
3 43 3
a
1
2 2 11 03
b
...(ii)
From eqs. (i) and (ii), we get
a = 1, b = –6.
5. Answer (3)
Let 1 2, ..... nA A A be airthmetric means between
a and 2b , then 2
1mb a
A a mn
Again , let 1 2, ........ nB B B be arithmetic means
Between 2a and b then 2
21m
b aB a m
n
Now ,
2 2
21 1m m
b a b aA B a m a m
n n
1 1
b a a mm a
n b n m
6. Answer (3)
is cube root of unity.
21 0
21
Now if 7(1 ) l m
2 7( ) l m
14 l m
12 2 l m
3 4 2( ) l m
2 l m
1 l m
can comparison l = 1, m = 1
X
Y
B
A(1, 2)
S(1, 0)
JEE Main Review Assignment-2
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7. Answer (3)
2/
0lim 6
2
xx x
x
a b
0
1 1lim
6
x x
x
a b
x xe
log log 6a be
ab = 6
(a, b) = (1, 6), (6, 1), (2, 3), (3, 2)
Required probability = 4 1
6 6 9
8. Answer (1)
1T S
i.e., 2 21 1 1 1 1 1( ) 2xx yy y y x y y
Passes through (0, 0)
2 2 0x y y
9. Answer (4)
Conceptual
10. Answer (2)
The given equation is 2 2( 1)( 1) 0z z z .
2, , ,z i w w w being an imaginary cube root
of unity. Thus | | 1z .
11. Answer (2)
I = 21 2cosec cot 2cotx x x
= 2 2cosec 2cosec cot cotx x x x dx
= (cosec cot )x x dx
12. Answer (2)
Real part of 2
0
iee d
put ie z
Real part of 2
0
12
zedz
i z
.
13. Answer (3)
1
11( )
1 (1 )(1 )1
1
x y
xyx y xyA z A
xy x y x y
xy
( ). ( ) ( )A x A y A z
14. Answer (3)
Given plane passes through and a b
containing the line is [ ] 0AP AB c
( ) (( ) ) 0r a b a c
( ) [ ]r b c c a a b c
length of r from the origin [ ]
| |
a b c
b c c a
15. Answer (3)
We know that if ii
x Ad
h
then | |x dh .
In this case 3 / 2
2 31/ 2
ii
xx
.
So 1
2h .
Thus 1
2 3.5 7| |d xh
.
16. Answer (1)
( ). 0dy
f x ydx
( )dy
f x dxy
ln ( )y f x dx
( )
1( )f x dx
y x e Then for given equation I.F =
( )f x dxe
Hence Solution 1 1. ( ) ( ). ( )y y x r x y x dx
11
1( ). ( )
( )y r x y x dx
y x
17. Answer (8)
2 2
2 2 1 ( 3 10)( 2)
2 4
x yx y
1, , 0 ,
2h k z e
Perpendicular distance from (2, 0) to
3 10 0x y is a
aee
2 6 42
aa a
Distance between directrics = 2
16a
Ke
Review Assignment-2 JEE Main
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18. Answer (2)
The given equation can be identified as
2(cos sin ) (2sin cos ) 0x x x x
4
x
and 1 1tan
2x
19. Answer (2)
/2 /2/22 2 2
00 0
cos cot 2 cotI x ec xdx x x x xdx
/22
/2
00 0
0 2 lnsin 2 lnsintanx
xLt x x xdx
x
0
10 2 lnsin 2 ln ln2
2 2xLt x x
0 0
2
1.coslnsin sin 011/x x
xx xLt Ltx
x
20. Answer (8)
2
2 1 1 1 1sin sin sin
2 2 4 2x x x
and
22 sec(sec ) 1, 2 2yy
It is possible only when 21sin , sec 1
2x y
5 13 17, , ,
6 6 6 6x
0, , 2 , 3y
No. of ordered pairs = 16
21. Answer (4)
0.52 8
log 02
x
x
2 8 1
2
x
x
(2, 6] 4x b a
22. Answer (04)
cos 2 2 2
2 2 3 1 02
b c aA C C
bc
Now C1 + C2 = 2 3, C1C2 = – 1
So |C1 – C2| = 4
23. Answer (2)
Coefficient xn in 4 4 311 1
n nn noC x C x
4 62 1 ... 1 1
n n nn nnC x C x
Coefficient xn in 31 1 1 3
nn nx x
24. Answer (4)
ON OT = 97 cos 97sec = 972
97
42020
25. Answer (1)
3
1
2Area
y ydy
n n
is greatest when n is least
y = 1
y nx = 2
y nx = 1/2 2