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THOMAStutorials
JEE (FINAL) Date: TEST NO: 33 Time: 03 HRS PCM MARKS: 360
1. The square root of the product of inductance
and capacitance has the dimension of
a) Length b) Mass
c) Time d) No dimension
2. Consider the acceleration, velocity and
displacement of a tennis ball as it falls to the
ground and bounces back. Directions of which
of these changes in the process
a) Velocity only
b) Displacement and velocity
c) Acceleration, velocity and displacement
d) Displacement and acceleration
3. A force is inclined at 60° to the horizontal. If its
rectangular component in the horizontal
direction is 50 N, then magnitude of the force
in the vertical direction is
a) 25 N
b) 75 N
c) 87 N
d) 100 N
4. A motor car has a width 1.1 𝑚 between wheels.
Its centre of gravity is 0.62 𝑚 above the ground
and the coefficient of friction between the
wheels and the road is 0.8. What is the
maximum possible speed, if the centre of
gravity inscribes a circle of radius 15 𝑚? (Road
surface is horizontal)
a) 7.64 𝑚/𝑠 b) 6.28 𝑚/𝑠
c) 10.84 𝑚/𝑠 d) 11.23 𝑚/𝑠
5. The force acting on a body moving along 𝑥-axis
varies with the position of the particle as
shown in the fig
The body is in stable equilibrium at
a) 𝑥 = 𝑥1 b) 𝑥 = 𝑥2
c) Both 𝑥1 and 𝑥2 d) Neither 𝑥1 nor 𝑥2
6. An inclined plane makes an angle of 30° with
the horizontal. A solid sphere rolling down the
inclined plane from rest without slipping has a
linear acceleration equal to
a) 5 g/14 b) 5 g/4
c) 2 g/3 d) g/3
7. Two small and heavy spheres, each of mass 𝑀,
are placed a distance 𝑟 apart on a horizontal
surface. The gravitational potential at the mid-
point on the line joining the centre of the
spheres is
a) Zero
b) −𝐺𝑀
𝑟
c) −2𝐺𝑀
𝑟
d) −4𝐺𝑀
𝑟
8. A wire whose cross-section is 4 mm2 is
stretched by 0.1 mm by a certain weight. How
far will a wire of the same material and length
stretch if its cross-sectional area is 8 mm2 and
the same weight is attached ?
a) 0.1 mm b) 0.05 mm
c) 0.025 mm d) 0.012 mm
9. An engine pumps water continuously through
a hose. Water leaves the hose with a velocity 𝑣
and 𝑚 is the mass per unit length of the water
jet. What is the rate at which kinetic energy is
imparted to water
a) 1
2𝑚𝑣3
b) 𝑚𝑣3
c) 1
2𝑚𝑣2
d) 1
2𝑚2𝑣2
10. On heating, the temperature at which
water has minimum volume is
a) 0℃ b) 4℃ c) 4K d) 100℃
11. The work of 146 kJ is performed in order to
compress one kilo mole of a gas adiabatically
and in this process the temperature of the gas
increases by 7℃. The gas is
(𝑅 = 8.3 J mol−1K−1)
a) Diatomic
b) Triatomic
c) A mixture of monoatomic and diatomic
d) Monoatomic
12. For a gas molecule with 6 degrees of
freedom the law of equipartition of energy
gives the following relation between the
molecular specific heat (𝐶𝑉) and gas
constant (𝑅)
a) 𝐶𝑉 =
𝑅
2 b) 𝐶𝑉 = 𝑅
F
x x2 x1
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c) 𝐶𝑉 = 2𝑅 d) 𝐶𝑉 = 3𝑅
13. A mass of 4 kg suspended from a spring of
force constant 800 𝑁𝑚−1 executes simple
harmonic oscillations. If the total energy of the
oscillator is 4𝐽, the maximum acceleration (in
𝑚𝑠−2) of the mass is
a) 5 b) 15 c) 45 d) 20
14. A wave is represented by the equation
𝑦 = 7 sin{𝜋(2𝑡 − 2𝑥)} where 𝑥 is in 𝑚𝑒𝑡𝑟𝑒𝑠
and 𝑡 in seconds. The velocity of the wave is
a) 1 𝑚/𝑠 b) 2 𝑚/𝑠 c) 5 𝑚/𝑠 d) 10 𝑚/𝑠
15. If the electric field given by (5�̂� + 4�̂� + 9�̂�), the
electric flux through a surface of area 20 unit
lying in the Y-Z plane will be
a) 100 unit b) 80unit c) 180 unit d) 20 unit
16. If dielectric constant and dielectric strength be
denoted by 𝐾 and X respectively, then a
material suitable for use as a dielectric in a
capacitor must have
a) High 𝐾 and high 𝑋 b) High 𝐾 and low 𝑋
c) Low 𝐾 and high 𝑋 d) Low 𝐾 and low 𝑋
17. In a copper voltmeter, the mass deposited in
30 s is 𝑚 gram. If the current-time graph is as
shown in figure, the electrochemical
equivalent of copper, in gC−1 is
a) 0.1 𝑚 b) 0.6 𝑚 c)
𝑚
2 d) 𝑚
18. A lead-acid battery of a car has an emf of 12 V.
If the internal resistance of the battery is 0.5 Ω,
the maximum current that can be drawn from
the battery will be
a) 30 A b) 20 A c) 6 A d) 24 A
19. A uniform magnetic field B is acting from south
to north and is of magnitude 1.5 𝑤𝑏/𝑚2. If a
proton having mass = 1.7 × 10−27𝑘𝑔 and
charge = 1.6 × 10−19𝐶 moves in this field
vertically downward with energy 5 𝑀𝑒𝑉, then
the force acting on it will be
a) 7.4 × 1012 𝑁 b) 7.4 × 10−12 𝑁
c) 7.4 × 1019 𝑁 d) 7.4 × 10−19 𝑁
20. If the magnetic flux is expressed in 𝑤𝑒𝑏𝑒𝑟, then
magnetic induction can be expressed in
a) 𝑊𝑒𝑏𝑒𝑟/𝑚2 b) 𝑊𝑒𝑏𝑒𝑟/𝑚
c) 𝑊𝑒𝑏𝑒𝑟-𝑚 d) 𝑊𝑒𝑏𝑒𝑟-𝑚2
21. In a transformer, the number of turns in
primary coil and secondary coil are 5 and 4
respectively. If 240 𝑉 is applied on the primary
coil, then the ratio of current in primary and
secondary coil is
a) 4 : 5 b) 5 : 4 c) 5 : 9 d) 9 : 5
22. For a coil having L = 2 mH, current flows at the
rate of 103 As−1. The emf induced is
a) 2 V
b) 1 V
c) 4 V
d) 3 V
23. A. The wavelength of microwaves is greater
than that of UV-rays.
B. The wavelength of IR rays is lesser than that
of UV-rays.
C. The wavelength of microwaves is lesser than
that of IR-rays.
D. Gamma rays have shortest wavelength in
the Electromagnetic Spectrum.
Of the above statements
a) A and B are true
b) B and C are true
c) C and D are true
d) A and D are true
24. Monochromatic light is refracted from air into
the glass of refractive index 𝜇. The ratio of the
wavelength of incident and refracted waves is
a) 1 ∶ 𝜇 b) 1 ∶ 𝜇2 c) 𝜇 ∶ 1 d) 1 ∶ 1
25. Huygens wave theory allows us to know
a) The wavelength of the wave
b) The velocity of the wave
c) The amplitude of the wave
d) The propagation of wave fronts
26. In order to coincide the parabolas formed by
singly ionised ions in one spectrograph and
doubly ionized ions in the other Thomson’s
mass spectrograph, the electric field and
magnetic fields are kept in the ratios 1:2 and
3:2 respectively. Then the ratio of masses of
the ions is
a) 3 :4 b) 1 :3
c) 9 :4 d) None of these
27. Which of the following atoms has the lowest
ionization potential?
a) N714 b) Cs55
133 c) Ar1840 d) O8
16
28. The radius of the Bohr orbit in the ground
state of hydrogen atom is 0.5 Å. The radius of
the orbit of the electron in the third excited
state of 𝐻𝑒+ will be
a) 8 Å b) 4 Å c) 0.5 Å d) 0.25 Å
29. In the half wave rectifier circuit operating
from 50 Hz mains frequency, the fundamental
t (s)
i (mA)
100
10 20 30
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frequency in the rippel would be
a) 25 Hz b) 50 Hz c) 70.7 Hz d) 100 Hz
30. A TV tower has a height 150 m. What is the
total population covered by the TV tower, if the
population density around the TV tower is
103 km−2?
Radius of the earth is 6.4 × 106 m.
a) 60.288 lakhs b) 40.192 lakhs
c) 100 lakhs d) 20.228 lakhs
31. 276 g of silver carbonate on being strongly
heated yields a residue weighing
a) 3.54 g b) 3.0 g c) 1.36 g d) 2.16 g
32. The ratio of the difference in energy between
the first and the second Bohr orbit to that
between the second and the third Bohr orbit is
a) 1
2
b) 1
3
c) 4
9
d) 27
5
33. Highest electron affinity among the following is
a) Fluorine b) Chlorine
c) Sulphur d) Xenon
34. N2 and O2 are converted into N2+ and O2
+
respectively.
Which of the following is not correct?
a) In N2+, the N – N bond weakens
b) In O2+, O – O bond order increases
c) In O2+, paramagnetism decreases
d) N2+ becomes diamagnetic
35. Potassium crystallizes in a bcc lattice, hence
the coordination number of potassium metal is
a) 0 b) 4 c) 6 d) 8
36. Consider the reaction,
4NO2(g) + O2(g) → 2N2O5(g), ∆𝑟𝐻
= −111 kJ.
If N2O2(𝑠) is formed instead of N2O5(g) in
the above reaction, the ∆𝑟𝐻 value will be
(Given, ∆𝐻 of sublimation for N2O2is 54
kJ mol−1)
a) -165 kJ b) +54 kJ c) +219 kJ d) -219 kJ
37. The ionisation constant of acetic acid is
1.8 × 10−5 at what concentration it will be
dissociated to 2%?
a) 0.025 M b) 0.045 M
c) 0.240 M d) 0.082 M
38. Oxidation state of sulphur in Na2S2O3 and
Na2S4O6
a) 4 and 6 b) 3 and 5
c) 2 and 2.5 d) 6 and 6
39. The volume of oxygen liberated from 15mL of
20 volume H2O2 is
a) 250mL b) 300mL c) 150mL d) 200mL
40. Which of the following statements are correct
for alkali metal compounds?
(i) Superoxides are paramagnetic in nature.
(ii) The basic strength of hydroxides increases
down the group.
(iii) The conductivity of chlorides in their
aqueous solutions decreases down the group.
(iv) The basic nature of carbonates in aqueous
solutions is due to cationic hydrolysis.
a) (i), (ii), and (iii) only
b) (i), and (ii), only
c) (ii), (iii) and (iv) only
d) (iii) and (iv) only
41. In purification of bauxite by hall’s process
a) Bauxite ore is fused with Na2CO3
b)
Bauxite ore is heated with NaOH solution at
50℃
c) Bauxite ore is heated with NaHCO3
d)
Bauxite ore is fused with coke and heated at
1800℃ in a current of nitrogen
42. Following reaction,
(CH3)3CBr + H2O → (CH3)3COH + HBr
is an example of
a) Elimination reaction
b) Free radical substitution
c) Nucleophilic substitution
d) Electrophilic substitution
43. Benzene can be obtained by heating either
benzoic acid with 𝑋 or phenol with 𝑌. 𝑋
and 𝑌 are respectively
a) Zinc dust and soda lime
b) Soda lime and zinc dust
c) Zinc dust and sodium hydroxide
d) Soda lime and copper
44. Depletion of ozone layer over Antarctica takes
place
a) In November
b) In the months of September and October
c) In the months of October and November
d) In summers
45. If a crystal lattice of a compound, each corner
of a cube is enjoyed by sodium, each edge of a
cube has oxygen and centre of a cube is
enjoyed by tungsten (W), then give its formula
a) Na2WO4 b) NaWO3 c) Na3WO3 d) Na2WO3
46. The modal elevation constant of water is
0.52℃. The boiling point of 1.0 modal aqueous
KCl solution (assuming complete dissociation
of KCl), therefore, should be
a) 98.96℃ b) 100.52℃
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c) 101.04℃ d) 107.01℃
47. The cell reaction is spontaneous, when
a) 𝐸red° is negative b) 𝐸red
° is positive
c) Δ𝐺° is negative d) Δ𝐺° is positive
48. What is the order of a reaction which has a rate
expression rate = 𝑘[𝐴]3/2[𝐵]−1?
a) 3
2
b) Zero
c) 1
2
d) None of these
49. Lyophilic sols are
a) Irreversible sols
b) They are prepared from inorganic
compounds
c) Coagulated by adding electrolytes
d) Self-stabilising
50. When a metal is to be extracted from its ore, if
the gangue associated with the ore is silica,
then
a) A basic flux is needed
b) An acidic flux is needed
c) Both basic and acidic flux are needed
d) Neither of them is needed
51. Which of the following is anhydride of
perchloric acid?
a) Cl2O7 b) Cl2O5 c) Cl2O3 d) HCIO
52. Zinc does not show variable valency like 𝑑-
block elements because
a) It is low melting
b) 𝑑-orbital is complete
c) It is a soft metal
d) Two electrons are present in the outermost
orbit
53. Which one doesn’t have π −bond?
a) Grignard reagent
b) Dibenzene chromium
c) Zeise’s salt
d) Ferrocene
54. In the chemical reactions,
The compounds ‘A’ and ‘B’ respectively are
a) Nitrobenzene and fluorobenzene
b) Phenol and benzene
c) Benzene diazonium chloride and
fluorobenzene
d) Nitrobenzene and chlorobenzene
55. Mild oxidation of glycerol with H2O2/FeSO4
gives
a) Glyceraldehyde
b) Dihydroxy acetone
c) Both (a) and (b)
d) None of the above
56. Cannizaro reaction is performed by
a) Formaldehyde
b) Formaldehyde and acetaldehyde
c) Benzaldehyde
d) Formaldehyde and benzaldehyde
57.
The alkene formed as a major product in the
above elimination reaction is
a)
b) CH2 = CH2
c)
d)
58. An example for a saturated fatty acid, presents
in nature is
a) Oleic acid b) Linoleic acid
c) Linolenic acid d) Palmitic acid
59. Given the polymers,
A = Nylon 6.6; B=Buna –S;C= Polythene.
Arrange these in increasing order of their
intermolecular force (lower to higher).
a) 𝐴 < 𝐵 < 𝐶 b) 𝐴 < 𝐶 < 𝐵
c) 𝐵 < 𝐶 < 𝐴 d) 𝐵 < 𝐶 < 𝐵
60. A drug that is antipyretic as well as analgesic is
a) Chloropromazine hydrochloride
b) 𝑝𝑎𝑟𝑎-acetamidophenol
c) Chloroquin
d) Penicillin
61. Let 𝐴 and 𝐵 be two sets, then (𝐴 ∪ 𝐵)′ ∪
(𝐴′ ∩ 𝐵)is equal to
a) 𝐴′ b) 𝐴
c) 𝐵′ d) None of these
62. If 𝑅 is an equivalence relation on a set 𝐴, then
𝑅−1 is
a) Reflexive only
b) Symmetric but not transitive
c) Equivalence
d) None of the above
63. If
cos(θ − α) = 𝑎, sin(θ − 𝛽) = 𝑏, then cos2(𝛼 −
𝛽) + 2𝑎𝑏 sin(𝛼 − 𝛽) is equal to
a) 4𝑎2𝑏2 b) 𝑎2 − 𝑏2
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c) 𝑎2 + 𝑏2 d) −𝑎2𝑏2
64. If 𝑛 is a positive integer, then 𝑛3 + 2𝑛 is
divisible by
a) 2 b) 6 c) 15 d) 3
65. If α is a cube root of unit and is not real, then
α3𝑛+1 + α3𝑛+3 + α3𝑛+5 has the value
a) −1 b) 0 c) 1 d) 3
66. If 𝑥2 + 6𝑥 − 27 < 0 and 𝑥2 − 3𝑥 − 4 < 0, then
a) 𝑥 > 3 b) 𝑥 < 4
c) 3 < 𝑥 < 4 d) 𝑥 =
7
2
67. A bag contains 3 black, 4 white and 2 red balls,
all the balls being different. The number of
selections of at most 6 balls containing balls of
all the colours, is
a) 42(4!) b) 26 × 4!
c) (26 − 1)(4!) d) None of these
68. ∑ 10𝑘=0 20𝐶𝑘is equal to
a) 219
+1
2 20𝐶10
b) 219
c) 20𝐶10
d) None of
these
69. If {𝑎𝑛} is a sequence with 𝑎0 = 𝑝 and
𝑎𝑛 − 𝑎𝑛−1 = 𝑟𝑎𝑛−1 for 𝑛 ≥ 1, then the
terms of the sequence are in
a) An arithmetic progression
b) A geometric progression
c) A harmonic progression
d) An arithmetic-geometric progression
70. The distance of the line 2𝑥 − 3𝑦 = 4 from the
point (1, 1) measured parallel to the line
𝑥 + 𝑦 = 1, is
a) √2 b) 5/√2 c) 1/√2 d) 6
71. Let (α, β) be a point from which two
perpendicular tangents can be drawn to the
ellipse 4𝑥2 + 5𝑦2 = 20. If 𝐹 = 4α + 3β, then
a) −15 ≤ 𝐹 ≤ 15
b) 𝐹 ≥ 0
c) −5 ≤ 𝐹 ≤ 20
d) 𝐹 ≤ −5√5 or 𝐹 ≥ 5√5
72. If 𝑙1 = lim𝑥→2+(𝑥 + [𝑥]), 𝑙2 = lim𝑥→2−(2𝑥 −
[𝑥]) and 𝑙3 = lim𝑥→𝜋/2cos𝑥
(𝑥−𝜋/2), then
a) 𝑙1 < 𝑙2 < 𝑙3 b) 𝑙2 < 𝑙3 < 𝑙1
c) 𝑙3 < 𝑙2 < 𝑙1 d) 𝑙1 < 𝑙3 < 𝑙2
73. The switching function for the following
network is
a) (𝑝 ∧ 𝑞 ∨ 𝑟) ∧ 𝑡 b) (𝑝 ∧ 𝑞 ∨ 𝑟) ∨ 𝑡
c) 𝑝 ∨ 𝑟 ∧ 𝑞 ∨ 𝑡 d) None of these
74. The median of 19 observations of a group is
30. If two observations with values 8 and 32
are further included, then the median of the
new group of 21 observations will be
a) 28 b) 30 c) 32 d) 34
75. India plays a two ODI matches each with
Australia and Pakistan. The probability of India
getting points 0,1,2 are 0.45, 0.05, 0.50. The
probability of India getting at least 7 points in
the series is
a) 0.00875 b) 0.875
c) 0.0875 d) None of these
76. If the area of a ∆ 𝐴𝐵𝐶 is given by ∆= 𝑎2 −
(𝑏 − 𝑐)2, then tan (𝐴
2) is equal to
a) −1
b) 0
c) 1
4
d) 1
2
77. In an equilateral triangle, 𝑅: 𝑟: 𝑟1 is equal to
a) 1:1:1 b) 1:2:3 c) 2:1:3 d) 3:2:4
78. If 𝑥√1 + 𝑦 + 𝑦√1 + 𝑥 = 0, then𝑑y
d𝑥 is equal to
a)
1
(1 + 𝑥)2 b) −
1
(1 + 𝑥)2
c)
1
1 + 𝑥2 d)
1
1 − 𝑥2
79. If tan θ + tan (𝜋
3+ θ) + tan (−
𝜋
3+ θ) =
𝑎 tan3θ, then 𝑎 is equal to
a) 1 3⁄ b) 1
c) 3 d) None of these
80.
If [1 𝑥 1] [
1 2 30 5 10 3 2
] [𝑥1−2]=0, then the value of 𝑥
is
a) 0
b) 2
3
c) 5
4
d) −4
5
81.
|𝑎 − 𝑏 + 𝑐 – 𝑎 − 𝑏 + 𝑐 1
𝑎 + 𝑏 + 2𝑐 – 𝑎 + 𝑏 + 2𝑐 23𝑐 3𝑐 3
| is
a) 6𝑎𝑏 b) 𝑎𝑏 c) 12𝑎𝑏 d) 2𝑎𝑏
82. For the function 𝑓(𝑥) =log𝑒(1+𝑥)+log𝑒(1−𝑥)
𝑥 to
be continuous at = 0, the value of 𝑓(0) is
a) -1 b) 0 c) -2 d) 2
83. Let y be the number of people in a village at
time t. Assume that the rate of change of the
population is proportional to the number of
people in the village at any time and further
assume that the population never increase in
time. Then, the population of the village at any
fixed 𝑡 is given by
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a) 𝑦 = 𝑒𝑘𝑡 + 𝑐, for some constants 𝑐 ≤ 0 and
𝑘 ≥ 0
b) 𝑦 = 𝑐𝑒𝑘𝑡, for some constants 𝑐 ≥ 0 and
𝑘 ≤ 0
c) 𝑦 = 𝑒𝑐𝑡 + 𝑘, for some constants 𝑐 ≤ 0 and
𝑘 ≥ 0
d) 𝑦 = 𝑘 𝑒𝑐𝑡 , for some constants 𝑐 ≥ 0 and
𝑘 ≤ 0
84. The value of ∫ 𝑒2𝑥(2 sin3𝑥 + 3 cos 3𝑥)𝑑𝑥 is
a) 𝑒2𝑥 sin 3𝑥 + 𝑐 b) 𝑒2𝑥 cos 3𝑥 + 𝑐
c) 𝑒2𝑥 + 𝑐 d) 𝑒2𝑥 (2 sin 3𝑥) + 𝑐
85. If a function 𝑓(𝑥) satisfies 𝑓′(𝑥) = 𝑔(𝑥)
Then, the value of
∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥𝑏
𝑎 is
a)
1
2[(𝑓(𝑏))
2
− (𝐹(𝑎))2]
b)
1
2[(𝑓(𝑏))
2
+ (𝑓(𝑎))2]
c) 1
2[𝑓(𝑏) − 𝑓(𝑎)]2
d) None of these
86. The area bounded by 𝑦 = 2 − |2 − 𝑥| and
𝑦 =3
|𝑥| is
a) 4 + 3 log 3
2 sq unit b)
4 − 3 log 3
2 sq unit
c) 3
2log 3 sq unit d)
1
2+ log 3 sq unit
87. The integrating factor of the differential
equation 𝑑𝑦
𝑑𝑥+
𝑦
(1−𝑥)√𝑥= 1 − √𝑥 is
a) 1−√𝑥1+√𝑥
b) 1+√𝑥
1−√𝑥 c)
1−𝑥
1+𝑥 d) √
𝑥
1−√𝑥
88. If �⃗� × 𝐛 = 𝐜 and 𝐛 × 𝐜 = �⃗� , then
a) |�⃗� |= 1, |𝐛 |= |𝐜 | b) |𝐜 |= 1, |�⃗� | = 1
c) |𝐛 |= 2, |𝐛 | = 2|�⃗� | d) |𝐛 |= 1, |𝐜 |= |�⃗� |
89. If 𝑄 is the image of the point 𝑃(2, 3, 4) under
the reflection in the plane
𝑥 − 2𝑦 + 5𝑧 = 6, then the equation of the line
𝑃𝑄 is
a)
𝑥 − 2
−1=𝑦 − 3
2
=𝑧 − 4
5
b)
𝑥 − 2
1=𝑦 − 3
−2
=𝑧 − 4
5
c)
𝑥 − 2
−1=𝑦 − 3
−2
=𝑧 − 4
5
d)
𝑥 − 2
1=𝑦 − 3
2
=𝑧 − 4
5
90. 𝑧 = 30𝑥 + 20𝑦, 𝑥 + 𝑦 ≤ 8, 𝑥 + 2𝑦 ≥ 4, 6𝑥 +
4𝑦 ≥ 12, 𝑥 ≥ 0, 𝑦 ≥ 0 has
a) Unique solution
b) Infinity many solution
c) Minimum at (4, 0)
d) Minimum 60 at point (0, 3)
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THOMAStutorials
JEE (FINAL) Date: TEST NO: 33 Time: 03 HRS PCM MARKS: 360
: ANSWER KEY : 1) c 2) b 3) c 4) c 5) b 6) a 7) d
8) b 9) a 10) b 11) a 12) d 13) d 14) a
15) a 16) a 17) c 18) d 19) b 20) a 21) a
22) a 23) d 24) c 25) d 26) c 27) b 28) b
29) b 30) a 31) d 32) d 33) b 34) d 35) d
36) d 37) b 38) c 39) b 40) b 41) a 42) c
43) b 44) b 45) b 46) c 47) c 48) c 49) d
50) a 51) a 52) b 53) a 54) c 55) c 56) d
57) b 58) d 59) c 60) b 61) a 62) c 63) c
64) d 65) b 66) c 67) a 68) a 69) b 70) a
71) a 72) c 73) b 74) b 75) c 76) c 77) c
78) b 79) c 80) c 81) c 82) b 83) b 84) a
85) a 86) b 87) b 88) d 89) b 90) d
: HINTS AND SOLUTIONS : 1 (c)
We know, 𝑓 =1
2𝜋√𝐿𝐶
Or √𝐿𝐶 =1
2𝜋𝑓= time
Thus, √𝐿𝐶 has the dimension of time.
2 (b)
Only directions of displacement and velocity gets
changed, acceleration is always directed vertically
downward
3 (c)
𝐴𝑥 = 50, θ = 60°
Then tanθ = 𝐴𝑦/𝐴𝑥 or 𝐴𝑦 = 𝐴𝑥 tanθ
Or 𝐴𝑦 = 50 tan60° = 50 × √3 = 87 N
4 (c)
𝑣 = √𝜇 𝑔 𝑟 = √0.8 × 9.8 × 15 = 10.84 𝑚/𝑠
5 (b)
When particle moves away from the origin then at
position 𝑥 = 𝑥1 force is zero and at 𝑥 > 𝑥1, force is
positive (repulsive in nature) so particle moves
further and does not return back to original
position
𝑖. 𝑒. the equilibrium is not stable
Similarly at position 𝑥 = 𝑥2 force is zero and at
𝑥 > 𝑥2, force is negative (attractive in nature)
So particle return back to original position 𝑖. 𝑒. the
equilibrium is stable
6 (a)
𝑎 =g sinθ
1 + 𝐼/𝑚𝑟2=g sin30°
1 +2
5
=5
7g ×
1
2=5g
14
7 (d)
Gravitational potential of 𝐴 at 𝑂 = −𝐺𝑀
𝑟/2= −
2𝐺𝑀
𝑟
For 𝐵, potential at 𝑂 = −𝐺𝑀
𝑟/2= −
2𝐺𝑀
𝑟
∴ Total potential = −4𝐺𝑀
𝑟
8 (b)
𝑌 =𝐹𝑙
𝐴∆𝑙
𝑌, 𝐹 𝑎𝑛𝑑 𝑙 are constants.
∴ ∆𝑙2∆1=𝑎1𝑎2=4
8=1
2
Or ∆𝑙2 =∆𝑙1
2=0.1
2mm = 0.5 mm
9 (a)
𝑑𝑘
𝑑𝑡=𝑑
𝑑𝑡(1
2𝑀𝑣2) =
𝑣2
2.𝑑𝑀
𝑑𝑡=𝑣2
2(𝑑𝑀
𝑑𝑙×𝑑𝑙
𝑑𝑡)
⇒𝑑𝑘
𝑑𝑡=1
2𝑚𝑣2 ×
𝑑𝑙
𝑑𝑡=1
2𝑚𝑣3
10 (b)
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Water has maximum density at 4℃ so at this
temperature, it has minimum volume.
11 (a)
For adiabatic process,
dQ=0
So, 𝑑𝑈 = −Δ𝑊
⇒ 𝑛𝐶𝑉𝑑𝑇 = +146 × 103J
⇒ 𝑛𝑓𝑅
2× 7 = 146 × 103
[𝑓 → Degree of freedom]
⇒ 103 × 𝑓 × 8.3 × 7
2= 146 × 103
𝑓 = 5.02 ≈ 5
So, it is a diatomic gas.
12 (d)
From 𝐶𝑉 =1
2𝑓𝑅 =
1
2× 6𝑅 = 3𝑅
13 (d)
Here, 𝑚 = 4𝑘𝑔; 𝑘 = 800𝑁𝑚−1; 𝐸 = 4𝐽
In SHM, total energy is 𝐸 =1
2𝑘𝐴2
where 𝐴 is the amplitude of oscillation
∴ 4 =1
2× 800 × 𝐴2
𝐴2 =8
800=
1
100
⇒ 𝐴 =1
10𝑚 = 0.1𝑚
Maximum acceleration, 𝑎max = 𝜔2𝐴
=𝑘
𝑚𝐴 [∵ 𝜔 = √
𝑘
𝑚]
=800𝑁𝑚−1
4𝑘𝑔× 0.1𝑚 = 20𝑚𝑠−2
14 (a)
𝑣 =𝜔
𝑘=2𝜋
2𝜋= 1 𝑚/𝑠
15 (a)
Electric flux is equal to the product of an area
element and the perpendicular component of E.
As the surface is lying in Y-Z plane
∴ 𝑬. 𝑑𝑨 = ϕ = (5)(20)
= 100 unit.
16 (a)
The material suitable for use as dielectric must
have high dielectric strength X and large dielectric
constant K.
17 (c)
Average current
𝑖 =50+100+50
3=200
3mA
𝑧 =𝑚
𝑖𝑡=
3𝑚
200 × 10−3 × 30=𝑚
2
19 (b)
𝐹 = 𝑞𝑣𝐵 and 𝐾 =1
2𝑚𝑣2 ⇒ 𝐹 = 𝑞𝐵√
2𝑘
𝑚
= 1.6 × 10−19 × 1.5√2 × 5 × 106 × 1.6 × 10−19
1.7 × 10−27
= 7.344 × 10−12𝑁
20 (a)
Flux = 𝐵 × 𝐴; ∴ 𝐵 =𝐹𝑙𝑢𝑥
𝐴= 𝑤𝑒𝑏𝑒𝑟/𝑚2
21 (a) 𝑁𝑠𝑁𝑝=𝑖𝑝𝑖𝑠⇒𝑖𝑝𝑖𝑠=4
5
22 (a)
𝑒 = 𝐿𝑑𝐼
𝑑𝑡= 2 × 10−3 = 2V
23 (d)
The wavelength order of the given types of waves
are given below
Waves Wavelength Range (in meter)
Gamma rays 10−14 − 10−10
IR-rays 7 × 10−7 = 10−3
UV-rays 10−9 − 4 × 10−7
Microwave 10−4 − 100
Hence, statements (A) and (D) are correct.
24 (c)
𝜆 ∝1
𝜇⇒𝜆1𝜆2=𝜇2𝜇1=𝜇
1
25 (d)
Huygen’s theory explains propagation of
wavefront
26 (c)
Using 𝑍2 = 𝑘 (𝑞
𝑚) 𝑦; where 𝑘 =
𝐵2𝐿𝐷
𝐸
For parabolas to coincide in the two photographs,
the 𝑘𝑞
𝑚 should be same for the two cases
Thus,𝐵12𝐿𝐷𝑒
𝐸1𝑚1=𝐵22𝐿𝐷(2𝑒)
𝐸2𝑚2
⇒𝑚1𝑚2
= (𝐵1𝐵2)2
× (𝐸2𝐸1) ×
1
2=9
4×2
1×1
2=9
4
27 (b)
As 55Cs133 has larger size among the four atoms
given, thus, electrons present in the outermost
orbit will be away from the nucleus and the
electrostatic force experienced by electrons due
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to nucleus will be minimum. Therefore, the
energy required to liberate electrons from outer
orbit will be minimum in case of55Cs133.
28 (b)
By using 𝑟𝑛 = 𝑟0𝑛2
𝑍; where 𝑟0 = Radius of the Bohr
orbit in the ground state atom. So for 𝐻𝑒+ third
excited state 𝑛 = 4, 𝑍 = 2, 𝑟0 = 0.5Å ⇒ 𝑟4 = 0.5 ×42
2= 4Å
29 (b)
In half wave rectifier, we get the output only in
one half cycle of input AC therefore, the frequency
of the ripple of the output is same as that of input
AC 𝑖𝑒, 50 Hz.
30 (a)
ℎ = 150 m,𝑅 = 6.4 × 106 m
Average population density = 103 km = 103 ×
(103)−2
= 10−3 m
Distance up to which the transmission would be
view
𝑑 = √2ℎ𝑅
Total area over which transmission could be
= 𝜋𝑑2 = 2𝜋ℎ𝑅
Population covered = 10−3 × 2𝜋ℎ𝑅
= 10−3 × 2 × 3.00 × 150 × 6.4
× 10−6
= 60.288 lakhs
31 (d)
Ag2CO3276g
⟶2Ag216g
+ CO2 +1
2O2
As 276 g of Ag2CO3 will give = 216g of Ag
So, 2.76 g of Ag2CO3 will give =2.76×216
276= 2.16g
32 (d)
𝐸1 − 𝐸2 = 1312 × 𝑍2 [1
12−1
22]
𝐸1 − 𝐸2 = 1312 × 𝑍2 [3
4] … (i)
𝐸2 − 𝐸3 = 1312 × 𝑍2 [1
22−1
32]
𝐸2 − 𝐸3 = 1312 × 𝑍2 [5
36] … (ii)
From Eqs. (i) and (ii) 𝐸1 − 𝐸2𝐸2 − 𝐸3
=3 × 36
4 × 5=27
5
33 (b)
Electron affinity is defined as, “The energy
released when an extra electron is added to a
neutral gaseous atom.”
Electron affinity of F=332.6 kJ/mol
Electron affinity of Cl=348.5 kJ/mol
Electron affinity of S=200.7 kJ/mol
Electron affinity of O=140.9 kJ/mol
Highest electron affinity among fluorine, chlorine,
sulphur and oxygen, is of chlorine.
The low value of electron affinity of fluorine than
chlorine is probably due to small size of fluorine
atom i.e., electron density is high which hinders
the addition of an extra electron.
34 (d)
PCl5 = 𝑠𝑝3𝑑 (Trigonal pyramidal)
IF7 = 𝑠𝑝3𝑑3(Pentagonal bipyramidal)
H3O+ = 𝑠𝑝3 (Pyramidal)
ClO2 = 𝑠𝑝2 (Angular) bond length are shorter
than single bond due to resonance.
NH4+ = 𝑠𝑝3(Tetrahedral)
35 (d)
For bcc lattice, the coordination number is 8
36 (d)
4NO2(g) + O2(g) → 2N2O5(g),−111 kJ
2N2O5(g)→2N2O5(𝑠); ∆𝑠𝐻=(−54×2)kJ
4NO2(g)+O2(g)→2N2O5(𝑠); ∆𝑟𝐻=−219 kJ
Note : ∆𝐻 of sublimation of N2O5(𝑠) is +54 kJ
mol−1.
Thus, for reverse process, it is −54 kJ mol−1.
37 (b)
CH3COOH ⇌ CH3COO− + H+
𝐶 0 0
𝐶(1 − α) 𝐶α 𝐶α
𝐾 =𝐶α ∙ 𝐶α
𝐶(1 − α)=
𝐶α2
(1 − α)
If 1 >>> 𝐶 then 𝐾 = 𝐶α2
𝐶 = 1.8 × 10−5
(0.02)2= 0.045 M
38 (c)
Na2S2O3,
2(+1) + 2𝑥 + 3(−2) = 0
2 + 2𝑥 − 6 = 0
𝑥 = +2
Na2S4O6
2(+1) + 4(𝑥) + 6(−2) = 0
2 + 4𝑥 − 12 = 0
4𝑥 = +10
𝑥 = +2.5
39 (b)
Quantity of H2O2 = 15 mL and volume of
H2O2 = 20
We know that 20 volume of H2O2 means 1 L of
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this solution will give 20 L of oxygen at NTP.
Since, oxygen liberated from 1000mL (1L) of
H2O2 = 20 L, therefore, oxygen liberated from
15mL of H2O2
=20
1000× 15 = 0.3 L = 300 mL.
40 (b)
(i) The alkali metal superoxides contain O2− ion,
which has an unpaired electron, hence they are
paramagnetic in nature.
(ii) The basic character of alkali metal hydroxides
increases on moving down the group.
(iii) The conductivity of alkali metal chlorides in
their aqueous solution increases on moving down
the group because in aqueous solution alkali
metal chlorides ionize to give alkali metal ions.
On moving down the group the size of alkali metal
ion increases, thus degree of hydration decreases,
due to this reason their conductivity in aqueous
solution increases on moving down the group.
(iv) DIAGRAM
CO32− + 2H2O → H2CO3 + 2OH
−
Thus, basic nature of carbonates in aqueous
solution is due to anionic hydrolysis.
41 (a)
In Hall’s process
Al2O3 ∙ 2H2O+ Na2CO3
⟶ 2NaAlO2 + CO2 + 2H2O
2NaAlO2 + 3H2O+ CO2 333K → 2Al(OH)3
↓ + Na2CO3
2Al(OH)3 1473K → Al2O3 + 3H2O
42 (c)
(CH3)3CBr + H2O → (CH3)3C − OH + HBr
Br is subsituted by –𝑂𝐻−(nucleophile)
𝑆𝑁1(unimolecular nuclerophilic substitution
reaction)
43 (b)
Benzene can be obtained by heating benzoic
acid with sodalime.
Benzene can also be obtained by heating
phenol with zinc dust.
44 (b)
During spring season 𝑖𝑒, in the month of
September and October, the sunlight returs to the
Antarctica and breaks up the clouds and
photolysis HOCl and Cl2
These free radical again reacts with ozone
molecules and leads to ozone depletion
45 (b)
No. of Na atoms present at each corner
= 8 ×1
8= 1
No. of O atoms present at the centre of edges
= 12 ×1
4= 3
No. of W atoms present at the centre of cube = 1
Formula of the compound = NaWO3
46 (c)
∆𝑇𝑏 = 𝑖𝑚 𝑘𝑏 = 0.52 × 1 × 2 = 1.04
∴ 𝑇𝑏 = 𝑇 + ∆𝑇𝑏 = 100 + 1.04 = 101.04℃
47 (c)
△ 𝐺 = △ 𝐻 − 𝑇 △ 𝑆
For a spontaneous cell reaction, △𝐻 should be
negative and △ 𝑆 should be positive. Hence, △ 𝐺
should be negative.
49 (d)
Lyophilic sols are self stabilizing because these
sols are reversible and are highly hydrated in the
solution.
50 (a)
SiO2 + CaO → CaSiO3
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acidic impurity basic flux slag
51 (a)
Chlorine heptaoxide (Cl2O7) is the anhydride of
perchloric acid.
2HCIO4 ∆→ Cl2O7 + H2O
53 (a)
Grignard reagent is a σ-bonded organometallic
compound because all the bonds present in the
reagent are single bonds.
55 (c)
With mild oxidising agent like bromine water or
H2O2 in the presence of FeSO4 (Fenton’s reagent),
glycerol is oxidised to a mixture of glyceraldehyde
and dihydroxy acetone
56 (d)
Cannizaro reaction is given by only those
aldehydes and ketones in which 𝛼-H atom is
absent.
Formaldehyde (HCHO)and benzaldehyde
(C6H5CHO) both due to the absence of 𝛼-H atom
undergo Cannizaro reaction.
57 (b)
There are four 𝛽- hydrogens, in this quaternary
ammonium salt.
On heating quaternary ammonium salt gives
Hofmann elimination (abstraction of most acidic
hydrogen which is 𝛽1).
Hence, major product is CH2 = CH2. (Least
substituted alkene).
58 (d)
Palmitic acid = C15H31COOH
Saturated monocarboxylic acids form a
homologous series which has a general formula
C𝑛 H2𝑛+1 COOH. Out of all the options only
palmitic acid follows this .
59 (c)
Buna-S is a elastomer, thus has weakest
intermolecular forces. Nylon 66, is a fibre, thus
has strong intermolecular forces like H-bonding.
Polythene is a thermoplastic polymers,thus the
intermolecular force present in polythene are in
between elastomer and fibres. Thus, the order of
intermolecular force of these polymers is
Buna − S < 𝑃𝑜𝑙𝑦𝑡ℎ𝑒𝑛𝑒 < 𝑁𝑦𝑙𝑜𝑛 66
(𝐵)(𝐶)(𝐴)
61 (a)
From Venn-Euler’s Diagram it is clear that
(𝐴 ∪ 𝐵)′ ∪ (𝐴′ ∩ 𝐵) = 𝐴′
62 (c)
Since, inverse of an equivalent relation is also an
equivalent relation.
∴ 𝑅−1 is an equivalent relation.
63 (c)
Now, sin(𝛼 − β) = sin(θ − β − (θ − 𝛼))
= sin(θ − β) = cos(θ − α)
− cos(θ − β) sin(θ − α)
= 𝑏𝑎 − √1 − 𝑏2√1− 𝑎2
and cos (α − β) = cos(θ − β − (θ − α))
= cos(θ − β) cos(θ − α) + sin(θ − β) sin(θ − α)
= 𝑎√1 − 𝑏2 + 𝑏√1 − 𝑎2
∴ cos2(α − β) + 2𝑎𝑏 sin(α − β)
= (𝑎√1 − 𝑏2 + 𝑏√1 − 𝑎2)2+ 2𝑎𝑏(𝑎𝑏
− √1 − 𝑎2√1 − 𝑏2)
= 𝑎2 + 𝑏2
64 (d)
Let 𝑃(𝑛) = 𝑛3 + 2𝑛
⟹ 𝑃(1) = 1 + 2 = 3
⟹ 𝑃(2) = 8 + 4 = 12
⟹ 𝑃(3) = 27 + 6 = 33
Here, we see that all these number are divisible by
3
65 (b)
Since, α is an imaginary cube root of unity. Let it
be ω, then α3𝑛+1 + α3𝑛+3 + α3𝑛+5 = (ω)3𝑛+1 +
(ω)3𝑛+3 + (ω)3𝑛+5
= ω+ 1 + ω5
= ω + 1 + ω2 = 0
66 (c)
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We have, 𝑥2 + 6𝑥 − 27 > 0
⇒ (𝑥 + 9)(𝑥 − 3) > 0 ⇒ 𝑥 < −9 or 𝑥 > 3
⇒ 𝑥 ∈ (−∞,−9) ∪ (3,∞) ….(i)
And 𝑥2 − 3𝑥 − 4 < 0
⇒ (𝑥 − 4)(𝑥 + 1) < 0
⇒ −1 < 𝑥 < 4 ….(i)
From relations (i) and (ii), we get
3 < 𝑥 < 4
67 (a)
Required number of selections
3𝐶1 ×4 𝐶1 ×
2 𝐶1(6𝐶3+
6𝐶2+ 6𝐶1+
6𝐶0)
= 3 × 4 × 2(20 + 15 + 6 + 1) = 42(4!)
68 (a)
∑
10
𝑘=0
20𝐶𝑘 = 20𝐶0 +
20𝐶1 + 20𝐶2 +⋯+
20𝐶10
On putting 𝑥 = 1and 𝑛 = 20in (1 + 𝑥)𝑛
= 𝑛𝐶0 + 𝑛𝐶1𝑥 +
𝑛𝐶2𝑥2 +⋯+ 𝑛𝐶𝑛𝑥
𝑛
We get
220 = 2( 20𝐶0 + 20𝐶1 +
20𝐶2 +⋯+ 20𝐶9)
+ 20𝐶10
⟹ 219 = ( 20𝐶0 + 20𝐶1 +
20𝐶2 +⋯+ 20𝐶9)
+1
2 20𝐶10
⟹ 219 = 20𝐶0 + 20𝐶1 +
20𝐶2 +⋯+ 20𝐶10
−1
2 20𝐶10
⟹ 20𝐶0 + 20𝐶1+. . . +
20𝐶10 = 219 +
1
2 20𝐶10
69 (b)
Given, 𝑎0 = 𝑝 and 𝑎𝑛 − 𝑎𝑛−1 = 𝑟𝑎𝑛−1
⇒ 𝑎𝑛 = 𝑎𝑛−1(𝑟 + 1)
For 𝑛 = 1, 𝑎1 = 𝑎0(𝑟 + 1) = 𝑝(𝑟 + 1)
𝑛 = 2, 𝑎2 = 𝑎1(𝑟 + 1) = 𝑝(𝑟 + 1)2
𝑛 = 3, 𝑎3 = 𝑎2(𝑟 + 1) = 𝑝(𝑟 + 1)3
This shows that the sequence is a geometric
progression.
70 (a)
∵ The slope of line 𝑥 + 𝑦 = 1 is −1
∴ It makes an angle of 135° with 𝑥-axis
The equation of line passing through (1, 1) and
making an angle of 135° is 𝑥 − 1
cos 135°=
𝑦 − 1
sin135°= 𝑟
⇒𝑥 − 1
−1
√2
=𝑦 − 11
√2
= 𝑟
Coordinates of any point on this line are
(1 −𝑟
√2, 1 +
𝑟
√2)
If this point lies on 2𝑥 − 3𝑦 = 4, then
2 (1 −𝑟
√2) − 3(1 +
𝑟
√2) = 4
⇒ 2−2𝑟
√2− 3 −
3𝑟
√2= 4
⇒5𝑟
√2= −5
⇒ 𝑟 = √2 (neglect negative sign)
71 (a)
(α, β) lies on the director circle of the ellipse 𝑖𝑒, on
𝑥2 + 𝑦2 = 9
So, we can assume
α = 3 cos θ , β = 3 sinθ
∴ 𝐹 = 12 cos θ + 9 sinθ = 3(4 cos θ + 3 sinθ)
⇒ −15 ≤ 𝐹 ≤ 15
72 (c)
𝑙1 = lim𝑥→2+
(𝑥 + [𝑥])
= limℎ→02 + ℎ + [2 + ℎ] = 4
𝑙2 = lim𝑥→2−
(2𝑥 − [𝑥])
= limℎ→0{2(2 − ℎ) − [2 − ℎ]}
= limℎ→0{2(2 − ℎ) − 1} = 3
𝑙3 = lim𝑥→
𝜋
2
cos 𝑥
𝑥 −𝜋
2
= lim𝑥→
𝜋
2
− sin𝑥 = −1
[by L’Hospital’s rule]
Thus, 𝑙3 < 𝑙2 < 𝑙1
73 (b)
The switching function for the given network is
(𝑝 ∧ 𝑞 ∨ 𝑟) ∨ 𝑡
74 (b)
Since, there are 19 observations. So, the middle
term is 10th
After including 8 and 32, 𝑖𝑒, 8 will come before 30
and 32 will come after 30
Here, new median will remain 30
75 (c)
Maximum points in four matches can be 8 only.
Therefore, at least 7 points means 7 or 8 points
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P a g e | 13
∴Required probability= 𝑃(7) + 𝑃(8)
= 4𝐶1(0.05)(0.5)3 + (0.5)4
= 0.0250 + 0.0625
= 0.0875
76 (c)
Given, ∆= 𝑎2 − (𝑏 − 𝑐)2
= (𝑎 + 𝑏 − 𝑐)(𝑎 − 𝑏 + 𝑐)
= 2(𝑠 − 𝑐) ∙ 2(𝑠 − 𝑏)
√𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) = 4(𝑠 − 𝑏)(𝑠 − 𝑐)
⇒ 1
4= √
(𝑠 − 𝑏)(𝑠 − 𝑐)
𝑠(𝑠 − 𝑎)= tan
𝐴
2
∴ tan𝐴
2=1
4
77 (c)
Let each side of equilateral triangle = 𝑎
∴ ∆=√3
4𝑎2, 𝑆 =
3𝑎
2
Now, 𝑟 =∆
5=√3
4𝑎2 ∙
2
3𝑎=
𝑎
2√3
𝑅 =𝑎𝑏𝑐
4∆=
𝑎3
√3𝑎2=𝑎
√3
𝑟1 =∆
𝑠 − 𝑎=√3
4𝑎2 ∙
2
𝑎=√3
2𝑎
∴ 𝑅: 𝑟1: 𝑟1 =𝑎
√3:𝑎
2√3:√3
2𝑎
= 2: 1: 3
78 (b)
Given, 𝑥√1 + 𝑦 = −𝑦√1 + 𝑥 …(i)
On squaring both sides, we get
𝑥2(1 + 𝑦) = 𝑦2(1 + 𝑥)
⇒ (𝑥 − 𝑦)(𝑥 + 𝑦) + 𝑥𝑦(𝑥 − 𝑦) = 0
⇒ (𝑥 − 𝑦)(𝑥 + 𝑦 + 𝑥𝑦) = 0
𝑥 − 𝑦 ≠ 0 because it does not satisfy the Eq. (i).
∴ 𝑥 + 𝑦 + 𝑥𝑦 = 0 ⇒ 𝑦 = −𝑥
1 + 𝑥
⇒ 𝑑𝑦
𝑑𝑥= −
(1 + 𝑥)(1) − 𝑥(1)
(1 + 𝑥)2= −
1
(1 + 𝑥)2
79 (c)
tan θ + tan (𝜋
3+ θ) + tan (−
𝜋
3+ θ) = 𝑎 tan3θ
⇒ tanθ +√3 + tanθ
1 − √3 tanθ+tan θ − √3
1 + √3 tan θ= a tan 3θ
⇒ tanθ +8 tanθ
1 − 3 tan2 θ= 𝑎 tan 3θ
⇒3(3 tan θ − tan3 θ)
1 − 3 tan2 θ= 𝑎 tan 3θ
⇒ 3 tan3θ = 𝑎 tan3θ
⇒ 𝑎 = 3
80 (c)
Given, [1 𝑥 1] [1 2 30 5 10 3 2
] [𝑥1−2] = 0
⇒ [1 𝑥 1] [𝑥 + 2 − 60 + 5 − 20 + 3 − 4
] = 0
⇒ [1 𝑥 1] [𝑥 − 43−1
] = 0
⇒ 𝑥 − 4 + 3𝑥 − 1 = 0 ⇒ 𝑥 =5
4
81 (c)
|𝑎 − 𝑏 + 𝑐 – 𝑎 − 𝑏 + 𝑐 1𝑎 + 𝑏 + 2𝑐 – 𝑎 + 𝑏 + 2𝑐 23𝑐 3𝑐 3
|
|2𝑎 − 2𝑎 0
𝑎 + 𝑏 + 2𝑐 – 𝑎 + 𝑏 + 2𝑐 23𝑐 3𝑐 3
|
[using 𝑅1 → 𝑅1 + 𝑅2 − 𝑅3]
= 2𝑎(−3𝑎 + 3𝑏 + 6𝑐 − 6𝑐) + 2𝑎(3𝑎 + 3𝑏 + 6𝑐
− 6𝑐)
= 12𝑎𝑏
82 (b)
Since, the function 𝑓(𝑥) is continuous
∴ 𝑓(0) =RHL 𝑓(𝑥) =LHL𝑓(𝑥)
Now, RHL 𝑓(𝑋) = limℎ→0
log(1+0+ℎ)+log(1−0−ℎ)
0+ℎ
= limℎ→0
log(1 + ℎ) + log(1 − ℎ)
ℎ
= limℎ→0
1
1+ℎ−
1
1−ℎ
1= 0
[by L ‘Hospital’s rule]
∴ 𝑓(0) =RHL 𝑓(𝑥) = 0
83 (b)
Given, 𝑑𝑦
𝑑𝑡∝ 𝑦,𝑤ℎ𝑒𝑟𝑒 y is the position of village
⇒1
𝑦𝑑𝑦 = 𝑘 𝑑𝑡
⇒ log 𝑦 = log 𝑐 + 𝑘𝑡 [on integrating]
⇒ log𝑦
𝑐= 𝑘𝑡 ⇒ 𝑦 = 𝑐𝑒𝑘𝑡
84 (a)
∫𝑒2𝑥(2 sin 3𝑥 + 3 cos 3𝑥)𝑑𝑥
= 2∫𝑒2𝑥 sin 3𝑥 + 𝑑𝑥 + 3∫𝑒2𝑥 cos 3𝑥 𝑑𝑥
= 𝑒2𝑥 sin 3𝑥 − 3∫𝑒2𝑥 cos3𝑥 𝑑𝑥
+ 3∫𝑒2𝑥 cos 3𝑥 𝑑𝑥
= 𝑒2𝑥 sin 3𝑥 + 𝑐
85 (a)
∵ 𝑓′(𝑥) = 𝑔(𝑥)
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⇒ ∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑓′(𝑥)𝑑𝑥𝑏
𝑎
𝑏
𝑎
= [(𝑓(𝑥))
2
2]
𝑎
𝑏
=1
2[(𝑓(𝑏))
2− (𝑓(𝑎))
2]
86 (b)
Required area = ∫ (2 − 12 − 𝑥)𝑑𝑥 − ∫3
|𝑥|
3
√3
3
√3𝑑𝑥
= ∫ 𝑥 𝑑𝑥 +∫ (4 − 𝑥)𝑑𝑥 −3
2
2
√3
∫3
𝑥𝑑𝑥
3
√3
= [𝑥2
2]√3
2
+ [4𝑥 −𝑥2
2]2
3
− [3 log 𝑥]√33
=1
2[4 − 3] + [12 −
9
2− (8 − 2)]
− 3[log 3 − log√3]
=1
2+3
2− 3 log
3
√3=4
2−3
2log 3
=4 − 3 log 3
2 sq unit
87 (b)
Given, 𝑑𝑦
𝑑𝑥+
𝑦
(1−𝑥)√𝑥= 1 − √𝑥
∴ IF = 𝑒∫
1
(1−𝑥)√𝑥 𝑑𝑥
Put √𝑥 = 𝑡
⇒ 1
2√𝑥 𝑑𝑥 = 𝑑𝑡
∴ IF = 𝑒∫2
1−𝑡2 𝑑𝑡
= 𝑒2
2log(
1+𝑡
1−𝑡) =
1+𝑡
1−𝑡=1+√𝑥
1−√𝑥
88 (d)
We have, �⃗� × 𝐛 = 𝐜
⇒ 𝐜 is perpendicular to �⃗� and 𝐛 and 𝐛 × 𝐜 = �⃗� .
⇒ �⃗� is perpendicular to 𝐛 and 𝐜 .
⇒ �⃗� , 𝐛 , 𝐜 are mutually perpendicular.
Again �⃗� × 𝐛 = 𝐜
⇒ |�⃗� × 𝐛 | = |𝐜 |
= |�⃗� ||𝐛 | ∙ sin90° = |𝐜 |
⇒ |�⃗� ||𝐛 | = |𝐜 | ...(i)
Also, 𝐛 × 𝐜 = |�⃗� |
|𝐛 ||𝐜 | ∙ sin90° = |�⃗� |
|𝐛 ||𝐜 | = |�⃗� | ….(ii)
From Eqs. (i) and (ii), we get
|𝐛 |2|𝐜 | = |𝐜 |
∴ |𝐛 |2= 1 (∵ |𝐜 | ≠ 0)
⇒ |𝐛 | = 1
⇒ |�⃗� | = |𝐜 |
89 (b)
Since, point 𝑄 is the image of 𝑃, therefore 𝑃𝑄
perpendicular to the plane
𝑥 − 2𝑦 + 5𝑧 = 6
∴ Required equation of line is 𝑥 − 2
1=𝑦 − 3
−2=𝑧 − 4
5
90 (d)
Feasible region is 𝐴𝐵𝐶𝐷𝐹𝐴 and 𝑧 = 30𝑥 + 20𝑦
Now, at 𝐴(4, 0), 𝑧 = 30 × 4 + 0 = 120
𝐵(8, 0), 𝑧 = 30 × 8 + 0 = 240
𝐶(0, 8), 𝑧 = 0 + 20 × 8 = 160
𝐷(0, 3), 𝑧 = 0 + 20 × 3 = 60
And 𝐹 (1,3
2) , 𝑧 = 30 × 1 + 20 ×
3
2= 60
It is clear that minimum value of 𝑧 is 60 at points
𝐷(0, 3) and 𝐹 (1,3
2)
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