THOMAStutorials - Meritstoresingly ionised ions in one spectrograph and doubly ionized ions in the...
Transcript of THOMAStutorials - Meritstoresingly ionised ions in one spectrograph and doubly ionized ions in the...
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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 60Β° 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 π between wheels.
Its centre of gravity is 0.62 π 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 π? (Road
surface is horizontal)
a) 7.64 π/π b) 6.28 π/π
c) 10.84 π/π d) 11.23 π/π
5. The force acting on a body moving along π₯-axis
varies with the position of the particle as
shown in the fig
The body is in stable equilibrium at
a) π₯ = π₯1 b) π₯ = π₯2
c) Both π₯1 and π₯2 d) Neither π₯1 nor π₯2
6. An inclined plane makes an angle of 30Β° 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 π,
are placed a distance π 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) βπΊπ
π
c) β2πΊπ
π
d) β4πΊπ
π
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 π£
and π is the mass per unit length of the water
jet. What is the rate at which kinetic energy is
imparted to water
a) 1
2ππ£3
b) ππ£3
c) 1
2ππ£2
d) 1
2π2π£2
10. On heating, the temperature at which
water has minimum volume is
a) 0β b) 4β c) 4K d) 100β
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 7β. The gas is
(π = 8.3 J molβ1Kβ1)
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 (πΆπ) and gas
constant (π )
a) πΆπ =
π
2 b) πΆπ = π
F
x x2 x1
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c) πΆπ = 2π d) πΆπ = 3π
13. A mass of 4 kg suspended from a spring of
force constant 800 ππβ1 executes simple
harmonic oscillations. If the total energy of the
oscillator is 4π½, the maximum acceleration (in
ππ β2) of the mass is
a) 5 b) 15 c) 45 d) 20
14. A wave is represented by the equation
π¦ = 7 sin{π(2π‘ β 2π₯)} where π₯ is in πππ‘πππ
and π‘ in seconds. The velocity of the wave is
a) 1 π/π b) 2 π/π c) 5 π/π d) 10 π/π
15. If the electric field given by (5οΏ½ΜοΏ½ + 4οΏ½ΜοΏ½ + 9οΏ½ΜοΏ½), 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 πΎ and X respectively, then a
material suitable for use as a dielectric in a
capacitor must have
a) High πΎ and high π b) High πΎ and low π
c) Low πΎ and high π d) Low πΎ and low π
17. In a copper voltmeter, the mass deposited in
30 s is π gram. If the current-time graph is as
shown in figure, the electrochemical
equivalent of copper, in gCβ1 is
a) 0.1 π b) 0.6 π c)
π
2 d) π
18. A lead-acid battery of a car has an emf of 12 V.
If the internal resistance of the battery is 0.5 Ξ©,
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 π€π/π2. If a
proton having mass = 1.7 Γ 10β27ππ and
charge = 1.6 Γ 10β19πΆ moves in this field
vertically downward with energy 5 πππ, then
the force acting on it will be
a) 7.4 Γ 1012 π b) 7.4 Γ 10β12 π
c) 7.4 Γ 1019 π d) 7.4 Γ 10β19 π
20. If the magnetic flux is expressed in π€ππππ, then
magnetic induction can be expressed in
a) πππππ/π2 b) πππππ/π
c) πππππ-π d) πππππ-π2
21. In a transformer, the number of turns in
primary coil and secondary coil are 5 and 4
respectively. If 240 π 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 Asβ1. 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 π. The ratio of the
wavelength of incident and refracted waves is
a) 1 βΆ π b) 1 βΆ π2 c) π βΆ 1 d) 1 βΆ 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 Thomsonβs
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 β«. The radius of
the orbit of the electron in the third excited
state of π»π+ will be
a) 8 β« b) 4 β« c) 0.5 β« d) 0.25 β«
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 kmβ2?
Radius of the earth is 6.4 Γ 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 β N bond weakens
b) In O2+, O β 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) β 2N2O5(g), βππ»
= β111 kJ.
If N2O2(π ) is formed instead of N2O5(g) in
the above reaction, the βππ» value will be
(Given, βπ» of sublimation for N2O2is 54
kJ molβ1)
a) -165 kJ b) +54 kJ c) +219 kJ d) -219 kJ
37. The ionisation constant of acetic acid is
1.8 Γ 10β5 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 hallβs process
a) Bauxite ore is fused with Na2CO3
b)
Bauxite ore is heated with NaOH solution at
50β
c) Bauxite ore is heated with NaHCO3
d)
Bauxite ore is fused with coke and heated at
1800β in a current of nitrogen
42. Following reaction,
(CH3)3CBr + H2O β (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 π or phenol with π. π
and π 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
0.52β. The boiling point of 1.0 modal aqueous
KCl solution (assuming complete dissociation
of KCl), therefore, should be
a) 98.96β b) 100.52β
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c) 101.04β d) 107.01β
47. The cell reaction is spontaneous, when
a) πΈredΒ° is negative b) πΈred
Β° is positive
c) ΞπΊΒ° is negative d) ΞπΊΒ° is positive
48. What is the order of a reaction which has a rate
expression rate = π[π΄]3/2[π΅]β1?
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 π-
block elements because
a) It is low melting
b) π-orbital is complete
c) It is a soft metal
d) Two electrons are present in the outermost
orbit
53. Which one doesnβt have Ο βbond?
a) Grignard reagent
b) Dibenzene chromium
c) Zeiseβs salt
d) Ferrocene
54. In the chemical reactions,
The compounds βAβ and βBβ 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 βS;C= Polythene.
Arrange these in increasing order of their
intermolecular force (lower to higher).
a) π΄ < π΅ < πΆ b) π΄ < πΆ < π΅
c) π΅ < πΆ < π΄ d) π΅ < πΆ < π΅
60. A drug that is antipyretic as well as analgesic is
a) Chloropromazine hydrochloride
b) ππππ-acetamidophenol
c) Chloroquin
d) Penicillin
61. Let π΄ and π΅ be two sets, then (π΄ βͺ π΅)β² βͺ
(π΄β² β© π΅)is equal to
a) π΄β² b) π΄
c) π΅β² d) None of these
62. If π is an equivalence relation on a set π΄, then
π β1 is
a) Reflexive only
b) Symmetric but not transitive
c) Equivalence
d) None of the above
63. If
cos(ΞΈ β Ξ±) = π, sin(ΞΈ β π½) = π, then cos2(πΌ β
π½) + 2ππ sin(πΌ β π½) is equal to
a) 4π2π2 b) π2 β π2
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c) π2 + π2 d) βπ2π2
64. If π is a positive integer, then π3 + 2π is
divisible by
a) 2 b) 6 c) 15 d) 3
65. If Ξ± is a cube root of unit and is not real, then
Ξ±3π+1 + Ξ±3π+3 + Ξ±3π+5 has the value
a) β1 b) 0 c) 1 d) 3
66. If π₯2 + 6π₯ β 27 < 0 and π₯2 β 3π₯ β 4 < 0, then
a) π₯ > 3 b) π₯ < 4
c) 3 < π₯ < 4 d) π₯ =
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 Γ 4!
c) (26 β 1)(4!) d) None of these
68. β 10π=0 20πΆπis equal to
a) 219
+1
2 20πΆ10
b) 219
c) 20πΆ10
d) None of
these
69. If {ππ} is a sequence with π0 = π and
ππ β ππβ1 = πππβ1 for π β₯ 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 2π₯ β 3π¦ = 4 from the
point (1, 1) measured parallel to the line
π₯ + π¦ = 1, is
a) β2 b) 5/β2 c) 1/β2 d) 6
71. Let (Ξ±, Ξ²) be a point from which two
perpendicular tangents can be drawn to the
ellipse 4π₯2 + 5π¦2 = 20. If πΉ = 4Ξ± + 3Ξ², then
a) β15 β€ πΉ β€ 15
b) πΉ β₯ 0
c) β5 β€ πΉ β€ 20
d) πΉ β€ β5β5 or πΉ β₯ 5β5
72. If π1 = limπ₯β2+(π₯ + [π₯]), π2 = limπ₯β2β(2π₯ β
[π₯]) and π3 = limπ₯βπ/2cosπ₯
(π₯βπ/2), then
a) π1 < π2 < π3 b) π2 < π3 < π1
c) π3 < π2 < π1 d) π1 < π3 < π2
73. The switching function for the following
network is
a) (π β§ π β¨ π) β§ π‘ b) (π β§ π β¨ π) β¨ π‘
c) π β¨ π β§ π β¨ π‘ 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 β π΄π΅πΆ is given by β= π2 β
(π β π)2, then tan (π΄
2) is equal to
a) β1
b) 0
c) 1
4
d) 1
2
77. In an equilateral triangle, π : π: π1 is equal to
a) 1:1:1 b) 1:2:3 c) 2:1:3 d) 3:2:4
78. If π₯β1 + π¦ + π¦β1 + π₯ = 0, thenπy
dπ₯ is equal to
a)
1
(1 + π₯)2 b) β
1
(1 + π₯)2
c)
1
1 + π₯2 d)
1
1 β π₯2
79. If tan ΞΈ + tan (π
3+ ΞΈ) + tan (β
π
3+ ΞΈ) =
π tan3ΞΈ, then π is equal to
a) 1 3β b) 1
c) 3 d) None of these
80.
If [1 π₯ 1] [
1 2 30 5 10 3 2
] [π₯1β2]=0, then the value of π₯
is
a) 0
b) 2
3
c) 5
4
d) β4
5
81.
|π β π + π β π β π + π 1
π + π + 2π β π + π + 2π 23π 3π 3
| is
a) 6ππ b) ππ c) 12ππ d) 2ππ
82. For the function π(π₯) =logπ(1+π₯)+logπ(1βπ₯)
π₯ to
be continuous at = 0, the value of π(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 π‘ is given by
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a) π¦ = πππ‘ + π, for some constants π β€ 0 and
π β₯ 0
b) π¦ = ππππ‘, for some constants π β₯ 0 and
π β€ 0
c) π¦ = πππ‘ + π, for some constants π β€ 0 and
π β₯ 0
d) π¦ = π πππ‘ , for some constants π β₯ 0 and
π β€ 0
84. The value of β« π2π₯(2 sin3π₯ + 3 cos 3π₯)ππ₯ is
a) π2π₯ sin 3π₯ + π b) π2π₯ cos 3π₯ + π
c) π2π₯ + π d) π2π₯ (2 sin 3π₯) + π
85. If a function π(π₯) satisfies πβ²(π₯) = π(π₯)
Then, the value of
β« π(π₯)π(π₯)ππ₯π
π is
a)
1
2[(π(π))
2
β (πΉ(π))2]
b)
1
2[(π(π))
2
+ (π(π))2]
c) 1
2[π(π) β π(π)]2
d) None of these
86. The area bounded by π¦ = 2 β |2 β π₯| and
π¦ =3
|π₯| is
a) 4 + 3 log 3
2 sq unit b)
4 β 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 ππ¦
ππ₯+
π¦
(1βπ₯)βπ₯= 1 β βπ₯ is
a) 1ββπ₯1+βπ₯
b) 1+βπ₯
1ββπ₯ c)
1βπ₯
1+π₯ d) β
π₯
1ββπ₯
88. If οΏ½βοΏ½ Γ π = π and π Γ π = οΏ½βοΏ½ , then
a) |οΏ½βοΏ½ |= 1, |π |= |π | b) |π |= 1, |οΏ½βοΏ½ | = 1
c) |π |= 2, |π | = 2|οΏ½βοΏ½ | d) |π |= 1, |π |= |οΏ½βοΏ½ |
89. If π is the image of the point π(2, 3, 4) under
the reflection in the plane
π₯ β 2π¦ + 5π§ = 6, then the equation of the line
ππ is
a)
π₯ β 2
β1=π¦ β 3
2
=π§ β 4
5
b)
π₯ β 2
1=π¦ β 3
β2
=π§ β 4
5
c)
π₯ β 2
β1=π¦ β 3
β2
=π§ β 4
5
d)
π₯ β 2
1=π¦ β 3
2
=π§ β 4
5
90. π§ = 30π₯ + 20π¦, π₯ + π¦ β€ 8, π₯ + 2π¦ β₯ 4, 6π₯ +
4π¦ β₯ 12, π₯ β₯ 0, π¦ β₯ 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, π =1
2πβπΏπΆ
Or βπΏπΆ =1
2ππ= time
Thus, βπΏπΆ has the dimension of time.
2 (b)
Only directions of displacement and velocity gets
changed, acceleration is always directed vertically
downward
3 (c)
π΄π₯ = 50, ΞΈ = 60Β°
Then tanΞΈ = π΄π¦/π΄π₯ or π΄π¦ = π΄π₯ tanΞΈ
Or π΄π¦ = 50 tan60Β° = 50 Γ β3 = 87 N
4 (c)
π£ = βπ π π = β0.8 Γ 9.8 Γ 15 = 10.84 π/π
5 (b)
When particle moves away from the origin then at
position π₯ = π₯1 force is zero and at π₯ > π₯1, force is
positive (repulsive in nature) so particle moves
further and does not return back to original
position
π. π. the equilibrium is not stable
Similarly at position π₯ = π₯2 force is zero and at
π₯ > π₯2, force is negative (attractive in nature)
So particle return back to original position π. π. the
equilibrium is stable
6 (a)
π =g sinΞΈ
1 + πΌ/ππ2=g sin30Β°
1 +2
5
=5
7g Γ
1
2=5g
14
7 (d)
Gravitational potential of π΄ at π = βπΊπ
π/2= β
2πΊπ
π
For π΅, potential at π = βπΊπ
π/2= β
2πΊπ
π
β΄ Total potential = β4πΊπ
π
8 (b)
π =πΉπ
π΄βπ
π, πΉ πππ π are constants.
β΄ βπ2β1=π1π2=4
8=1
2
Or βπ2 =βπ1
2=0.1
2mm = 0.5 mm
9 (a)
ππ
ππ‘=π
ππ‘(1
2ππ£2) =
π£2
2.ππ
ππ‘=π£2
2(ππ
ππΓππ
ππ‘)
βππ
ππ‘=1
2ππ£2 Γ
ππ
ππ‘=1
2ππ£3
10 (b)
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Water has maximum density at 4β so at this
temperature, it has minimum volume.
11 (a)
For adiabatic process,
dQ=0
So, ππ = βΞπ
β ππΆπππ = +146 Γ 103J
β πππ
2Γ 7 = 146 Γ 103
[π β Degree of freedom]
β 103 Γ π Γ 8.3 Γ 7
2= 146 Γ 103
π = 5.02 β 5
So, it is a diatomic gas.
12 (d)
From πΆπ =1
2ππ =
1
2Γ 6π = 3π
13 (d)
Here, π = 4ππ; π = 800ππβ1; πΈ = 4π½
In SHM, total energy is πΈ =1
2ππ΄2
where π΄ is the amplitude of oscillation
β΄ 4 =1
2Γ 800 Γ π΄2
π΄2 =8
800=
1
100
β π΄ =1
10π = 0.1π
Maximum acceleration, πmax = π2π΄
=π
ππ΄ [β΅ π = β
π
π]
=800ππβ1
4ππΓ 0.1π = 20ππ β2
14 (a)
π£ =π
π=2π
2π= 1 π/π
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
β΄ π¬. ππ¨ = Ο = (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
π =50+100+50
3=200
3mA
π§ =π
ππ‘=
3π
200 Γ 10β3 Γ 30=π
2
19 (b)
πΉ = ππ£π΅ and πΎ =1
2ππ£2 β πΉ = ππ΅β
2π
π
= 1.6 Γ 10β19 Γ 1.5β2 Γ 5 Γ 106 Γ 1.6 Γ 10β19
1.7 Γ 10β27
= 7.344 Γ 10β12π
20 (a)
Flux = π΅ Γ π΄; β΄ π΅ =πΉππ’π₯
π΄= π€ππππ/π2
21 (a) ππ ππ=ππππ βππππ =4
5
22 (a)
π = πΏππΌ
ππ‘= 2 Γ 10β3 = 2V
23 (d)
The wavelength order of the given types of waves
are given below
Waves Wavelength Range (in meter)
Gamma rays 10β14 β 10β10
IR-rays 7 Γ 10β7 = 10β3
UV-rays 10β9 β 4 Γ 10β7
Microwave 10β4 β 100
Hence, statements (A) and (D) are correct.
24 (c)
π β1
πβπ1π2=π2π1=π
1
25 (d)
Huygenβs theory explains propagation of
wavefront
26 (c)
Using π2 = π (π
π) π¦; where π =
π΅2πΏπ·
πΈ
For parabolas to coincide in the two photographs,
the ππ
π should be same for the two cases
Thus,π΅12πΏπ·π
πΈ1π1=π΅22πΏπ·(2π)
πΈ2π2
βπ1π2
= (π΅1π΅2)2
Γ (πΈ2πΈ1) Γ
1
2=9
4Γ2
1Γ1
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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P a g e | 9
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 ππ = π0π2
π; where π0 = Radius of the Bohr
orbit in the ground state atom. So for π»π+ third
excited state π = 4, π = 2, π0 = 0.5β« β π4 = 0.5 Γ42
2= 4β«
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 ππ, 50 Hz.
30 (a)
β = 150 m,π = 6.4 Γ 106 m
Average population density = 103 km = 103 Γ
(103)β2
= 10β3 m
Distance up to which the transmission would be
view
π = β2βπ
Total area over which transmission could be
= ππ2 = 2πβπ
Population covered = 10β3 Γ 2πβπ
= 10β3 Γ 2 Γ 3.00 Γ 150 Γ 6.4
Γ 10β6
= 60.288 lakhs
31 (d)
Ag2CO3276g
βΆ2Ag216g
+ CO2 +1
2O2
As 276 g of Ag2CO3 will give = 216g of Ag
So, 2.76 g of Ag2CO3 will give =2.76Γ216
276= 2.16g
32 (d)
πΈ1 β πΈ2 = 1312 Γ π2 [1
12β1
22]
πΈ1 β πΈ2 = 1312 Γ π2 [3
4] β¦ (i)
πΈ2 β πΈ3 = 1312 Γ π2 [1
22β1
32]
πΈ2 β πΈ3 = 1312 Γ π2 [5
36] β¦ (ii)
From Eqs. (i) and (ii) πΈ1 β πΈ2πΈ2 β πΈ3
=3 Γ 36
4 Γ 5=27
5
33 (b)
Electron affinity is defined as, βThe energy
released when an extra electron is added to a
neutral gaseous atom.β
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 = π π3π (Trigonal pyramidal)
IF7 = π π3π3(Pentagonal bipyramidal)
H3O+ = π π3 (Pyramidal)
ClO2 = π π2 (Angular) bond length are shorter
than single bond due to resonance.
NH4+ = π π3(Tetrahedral)
35 (d)
For bcc lattice, the coordination number is 8
36 (d)
4NO2(g) + O2(g) β 2N2O5(g),β111 kJ
2N2O5(g)β2N2O5(π ); βπ π»=(β54Γ2)kJ
4NO2(g)+O2(g)β2N2O5(π ); βππ»=β219 kJ
Note : βπ» of sublimation of N2O5(π ) is +54 kJ
molβ1.
Thus, for reverse process, it is β54 kJ molβ1.
37 (b)
CH3COOH β CH3COOβ + H+
πΆ 0 0
πΆ(1 β Ξ±) πΆΞ± πΆΞ±
πΎ =πΆΞ± β πΆΞ±
πΆ(1 β Ξ±)=
πΆΞ±2
(1 β Ξ±)
If 1 >>> πΆ then πΎ = πΆΞ±2
πΆ = 1.8 Γ 10β5
(0.02)2= 0.045 M
38 (c)
Na2S2O3,
2(+1) + 2π₯ + 3(β2) = 0
2 + 2π₯ β 6 = 0
π₯ = +2
Na2S4O6
2(+1) + 4(π₯) + 6(β2) = 0
2 + 4π₯ β 12 = 0
4π₯ = +10
π₯ = +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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P a g e | 10
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
1000Γ 15 = 0.3 L = 300 mL.
40 (b)
(i) The alkali metal superoxides contain O2β 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
CO32β + 2H2O β H2CO3 + 2OH
β
Thus, basic nature of carbonates in aqueous
solution is due to anionic hydrolysis.
41 (a)
In Hallβs process
Al2O3 β 2H2O+ Na2CO3
βΆ 2NaAlO2 + CO2 + 2H2O
2NaAlO2 + 3H2O+ CO2 333K β 2Al(OH)3
β + Na2CO3
2Al(OH)3 1473K β Al2O3 + 3H2O
42 (c)
(CH3)3CBr + H2O β (CH3)3C β OH + HBr
Br is subsituted by βππ»β(nucleophile)
ππ1(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 ππ, 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 Γ1
8= 1
No. of O atoms present at the centre of edges
= 12 Γ1
4= 3
No. of W atoms present at the centre of cube = 1
Formula of the compound = NaWO3
46 (c)
βππ = ππ ππ = 0.52 Γ 1 Γ 2 = 1.04
β΄ ππ = π + βππ = 100 + 1.04 = 101.04β
47 (c)
β³ πΊ = β³ π» β π β³ π
For a spontaneous cell reaction, β³π» should be
negative and β³ π should be positive. Hence, β³ πΊ
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 β CaSiO3
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acidic impurity basic flux slag
51 (a)
Chlorine heptaoxide (Cl2O7) is the anhydride of
perchloric acid.
2HCIO4 ββ Cl2O7 + H2O
53 (a)
Grignard reagent is a Ο-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 (Fentonβs 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 πΌ-H atom is
absent.
Formaldehyde (HCHO)and benzaldehyde
(C6H5CHO) both due to the absence of πΌ-H atom
undergo Cannizaro reaction.
57 (b)
There are four π½- hydrogens, in this quaternary
ammonium salt.
On heating quaternary ammonium salt gives
Hofmann elimination (abstraction of most acidic
hydrogen which is π½1).
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
Cπ H2π+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 β S < ππππ¦π‘βπππ < ππ¦πππ 66
(π΅)(πΆ)(π΄)
61 (a)
From Venn-Eulerβs Diagram it is clear that
(π΄ βͺ π΅)β² βͺ (π΄β² β© π΅) = π΄β²
62 (c)
Since, inverse of an equivalent relation is also an
equivalent relation.
β΄ π β1 is an equivalent relation.
63 (c)
Now, sin(πΌ β Ξ²) = sin(ΞΈ β Ξ² β (ΞΈ β πΌ))
= sin(ΞΈ β Ξ²) = cos(ΞΈ β Ξ±)
β cos(ΞΈ β Ξ²) sin(ΞΈ β Ξ±)
= ππ β β1 β π2β1β π2
and cos (Ξ± β Ξ²) = cos(ΞΈ β Ξ² β (ΞΈ β Ξ±))
= cos(ΞΈ β Ξ²) cos(ΞΈ β Ξ±) + sin(ΞΈ β Ξ²) sin(ΞΈ β Ξ±)
= πβ1 β π2 + πβ1 β π2
β΄ cos2(Ξ± β Ξ²) + 2ππ sin(Ξ± β Ξ²)
= (πβ1 β π2 + πβ1 β π2)2+ 2ππ(ππ
β β1 β π2β1 β π2)
= π2 + π2
64 (d)
Let π(π) = π3 + 2π
βΉ π(1) = 1 + 2 = 3
βΉ π(2) = 8 + 4 = 12
βΉ π(3) = 27 + 6 = 33
Here, we see that all these number are divisible by
3
65 (b)
Since, Ξ± is an imaginary cube root of unity. Let it
be Ο, then Ξ±3π+1 + Ξ±3π+3 + Ξ±3π+5 = (Ο)3π+1 +
(Ο)3π+3 + (Ο)3π+5
= Ο+ 1 + Ο5
= Ο + 1 + Ο2 = 0
66 (c)
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We have, π₯2 + 6π₯ β 27 > 0
β (π₯ + 9)(π₯ β 3) > 0 β π₯ < β9 or π₯ > 3
β π₯ β (ββ,β9) βͺ (3,β) β¦.(i)
And π₯2 β 3π₯ β 4 < 0
β (π₯ β 4)(π₯ + 1) < 0
β β1 < π₯ < 4 β¦.(i)
From relations (i) and (ii), we get
3 < π₯ < 4
67 (a)
Required number of selections
3πΆ1 Γ4 πΆ1 Γ
2 πΆ1(6πΆ3+
6πΆ2+ 6πΆ1+
6πΆ0)
= 3 Γ 4 Γ 2(20 + 15 + 6 + 1) = 42(4!)
68 (a)
β
10
π=0
20πΆπ = 20πΆ0 +
20πΆ1 + 20πΆ2 +β―+
20πΆ10
On putting π₯ = 1and π = 20in (1 + π₯)π
= ππΆ0 + ππΆ1π₯ +
ππΆ2π₯2 +β―+ ππΆππ₯
π
We get
220 = 2( 20πΆ0 + 20πΆ1 +
20πΆ2 +β―+ 20πΆ9)
+ 20πΆ10
βΉ 219 = ( 20πΆ0 + 20πΆ1 +
20πΆ2 +β―+ 20πΆ9)
+1
2 20πΆ10
βΉ 219 = 20πΆ0 + 20πΆ1 +
20πΆ2 +β―+ 20πΆ10
β1
2 20πΆ10
βΉ 20πΆ0 + 20πΆ1+. . . +
20πΆ10 = 219 +
1
2 20πΆ10
69 (b)
Given, π0 = π and ππ β ππβ1 = πππβ1
β ππ = ππβ1(π + 1)
For π = 1, π1 = π0(π + 1) = π(π + 1)
π = 2, π2 = π1(π + 1) = π(π + 1)2
π = 3, π3 = π2(π + 1) = π(π + 1)3
This shows that the sequence is a geometric
progression.
70 (a)
β΅ The slope of line π₯ + π¦ = 1 is β1
β΄ It makes an angle of 135Β° with π₯-axis
The equation of line passing through (1, 1) and
making an angle of 135Β° is π₯ β 1
cos 135Β°=
π¦ β 1
sin135Β°= π
βπ₯ β 1
β1
β2
=π¦ β 11
β2
= π
Coordinates of any point on this line are
(1 βπ
β2, 1 +
π
β2)
If this point lies on 2π₯ β 3π¦ = 4, then
2 (1 βπ
β2) β 3(1 +
π
β2) = 4
β 2β2π
β2β 3 β
3π
β2= 4
β5π
β2= β5
β π = β2 (neglect negative sign)
71 (a)
(Ξ±, Ξ²) lies on the director circle of the ellipse ππ, on
π₯2 + π¦2 = 9
So, we can assume
Ξ± = 3 cos ΞΈ , Ξ² = 3 sinΞΈ
β΄ πΉ = 12 cos ΞΈ + 9 sinΞΈ = 3(4 cos ΞΈ + 3 sinΞΈ)
β β15 β€ πΉ β€ 15
72 (c)
π1 = limπ₯β2+
(π₯ + [π₯])
= limββ02 + β + [2 + β] = 4
π2 = limπ₯β2β
(2π₯ β [π₯])
= limββ0{2(2 β β) β [2 β β]}
= limββ0{2(2 β β) β 1} = 3
π3 = limπ₯β
π
2
cos π₯
π₯ βπ
2
= limπ₯β
π
2
β sinπ₯ = β1
[by LβHospitalβs rule]
Thus, π3 < π2 < π1
73 (b)
The switching function for the given network is
(π β§ π β¨ π) β¨ π‘
74 (b)
Since, there are 19 observations. So, the middle
term is 10th
After including 8 and 32, ππ, 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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β΄Required probability= π(7) + π(8)
= 4πΆ1(0.05)(0.5)3 + (0.5)4
= 0.0250 + 0.0625
= 0.0875
76 (c)
Given, β= π2 β (π β π)2
= (π + π β π)(π β π + π)
= 2(π β π) β 2(π β π)
βπ (π β π)(π β π)(π β π) = 4(π β π)(π β π)
β 1
4= β
(π β π)(π β π)
π (π β π)= tan
π΄
2
β΄ tanπ΄
2=1
4
77 (c)
Let each side of equilateral triangle = π
β΄ β=β3
4π2, π =
3π
2
Now, π =β
5=β3
4π2 β
2
3π=
π
2β3
π =πππ
4β=
π3
β3π2=π
β3
π1 =β
π β π=β3
4π2 β
2
π=β3
2π
β΄ π : π1: π1 =π
β3:π
2β3:β3
2π
= 2: 1: 3
78 (b)
Given, π₯β1 + π¦ = βπ¦β1 + π₯ β¦(i)
On squaring both sides, we get
π₯2(1 + π¦) = π¦2(1 + π₯)
β (π₯ β π¦)(π₯ + π¦) + π₯π¦(π₯ β π¦) = 0
β (π₯ β π¦)(π₯ + π¦ + π₯π¦) = 0
π₯ β π¦ β 0 because it does not satisfy the Eq. (i).
β΄ π₯ + π¦ + π₯π¦ = 0 β π¦ = βπ₯
1 + π₯
β ππ¦
ππ₯= β
(1 + π₯)(1) β π₯(1)
(1 + π₯)2= β
1
(1 + π₯)2
79 (c)
tan ΞΈ + tan (π
3+ ΞΈ) + tan (β
π
3+ ΞΈ) = π tan3ΞΈ
β tanΞΈ +β3 + tanΞΈ
1 β β3 tanΞΈ+tan ΞΈ β β3
1 + β3 tan ΞΈ= a tan 3ΞΈ
β tanΞΈ +8 tanΞΈ
1 β 3 tan2 ΞΈ= π tan 3ΞΈ
β3(3 tan ΞΈ β tan3 ΞΈ)
1 β 3 tan2 ΞΈ= π tan 3ΞΈ
β 3 tan3ΞΈ = π tan3ΞΈ
β π = 3
80 (c)
Given, [1 π₯ 1] [1 2 30 5 10 3 2
] [π₯1β2] = 0
β [1 π₯ 1] [π₯ + 2 β 60 + 5 β 20 + 3 β 4
] = 0
β [1 π₯ 1] [π₯ β 43β1
] = 0
β π₯ β 4 + 3π₯ β 1 = 0 β π₯ =5
4
81 (c)
|π β π + π β π β π + π 1π + π + 2π β π + π + 2π 23π 3π 3
|
|2π β 2π 0
π + π + 2π β π + π + 2π 23π 3π 3
|
[using π 1 β π 1 + π 2 β π 3]
= 2π(β3π + 3π + 6π β 6π) + 2π(3π + 3π + 6π
β 6π)
= 12ππ
82 (b)
Since, the function π(π₯) is continuous
β΄ π(0) =RHL π(π₯) =LHLπ(π₯)
Now, RHL π(π) = limββ0
log(1+0+β)+log(1β0ββ)
0+β
= limββ0
log(1 + β) + log(1 β β)
β
= limββ0
1
1+ββ
1
1ββ
1= 0
[by L βHospitalβs rule]
β΄ π(0) =RHL π(π₯) = 0
83 (b)
Given, ππ¦
ππ‘β π¦,π€βπππ y is the position of village
β1
π¦ππ¦ = π ππ‘
β log π¦ = log π + ππ‘ [on integrating]
β logπ¦
π= ππ‘ β π¦ = ππππ‘
84 (a)
β«π2π₯(2 sin 3π₯ + 3 cos 3π₯)ππ₯
= 2β«π2π₯ sin 3π₯ + ππ₯ + 3β«π2π₯ cos 3π₯ ππ₯
= π2π₯ sin 3π₯ β 3β«π2π₯ cos3π₯ ππ₯
+ 3β«π2π₯ cos 3π₯ ππ₯
= π2π₯ sin 3π₯ + π
85 (a)
β΅ πβ²(π₯) = π(π₯)
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β β« π(π₯)π(π₯)ππ₯ = β« π(π₯)πβ²(π₯)ππ₯π
π
π
π
= [(π(π₯))
2
2]
π
π
=1
2[(π(π))
2β (π(π))
2]
86 (b)
Required area = β« (2 β 12 β π₯)ππ₯ β β«3
|π₯|
3
β3
3
β3ππ₯
= β« π₯ ππ₯ +β« (4 β π₯)ππ₯ β3
2
2
β3
β«3
π₯ππ₯
3
β3
= [π₯2
2]β3
2
+ [4π₯ βπ₯2
2]2
3
β [3 log π₯]β33
=1
2[4 β 3] + [12 β
9
2β (8 β 2)]
β 3[log 3 β logβ3]
=1
2+3
2β 3 log
3
β3=4
2β3
2log 3
=4 β 3 log 3
2 sq unit
87 (b)
Given, ππ¦
ππ₯+
π¦
(1βπ₯)βπ₯= 1 β βπ₯
β΄ IF = πβ«
1
(1βπ₯)βπ₯ ππ₯
Put βπ₯ = π‘
β 1
2βπ₯ ππ₯ = ππ‘
β΄ IF = πβ«2
1βπ‘2 ππ‘
= π2
2log(
1+π‘
1βπ‘) =
1+π‘
1βπ‘=1+βπ₯
1ββπ₯
88 (d)
We have, οΏ½βοΏ½ Γ π = π
β π is perpendicular to οΏ½βοΏ½ and π and π Γ π = οΏ½βοΏ½ .
β οΏ½βοΏ½ is perpendicular to π and π .
β οΏ½βοΏ½ , π , π are mutually perpendicular.
Again οΏ½βοΏ½ Γ π = π
β |οΏ½βοΏ½ Γ π | = |π |
= |οΏ½βοΏ½ ||π | β sin90Β° = |π |
β |οΏ½βοΏ½ ||π | = |π | ...(i)
Also, π Γ π = |οΏ½βοΏ½ |
|π ||π | β sin90Β° = |οΏ½βοΏ½ |
|π ||π | = |οΏ½βοΏ½ | β¦.(ii)
From Eqs. (i) and (ii), we get
|π |2|π | = |π |
β΄ |π |2= 1 (β΅ |π | β 0)
β |π | = 1
β |οΏ½βοΏ½ | = |π |
89 (b)
Since, point π is the image of π, therefore ππ
perpendicular to the plane
π₯ β 2π¦ + 5π§ = 6
β΄ Required equation of line is π₯ β 2
1=π¦ β 3
β2=π§ β 4
5
90 (d)
Feasible region is π΄π΅πΆπ·πΉπ΄ and π§ = 30π₯ + 20π¦
Now, at π΄(4, 0), π§ = 30 Γ 4 + 0 = 120
π΅(8, 0), π§ = 30 Γ 8 + 0 = 240
πΆ(0, 8), π§ = 0 + 20 Γ 8 = 160
π·(0, 3), π§ = 0 + 20 Γ 3 = 60
And πΉ (1,3
2) , π§ = 30 Γ 1 + 20 Γ
3
2= 60
It is clear that minimum value of π§ is 60 at points
π·(0, 3) and πΉ (1,3
2)
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