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Space for Rough Work
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PEBPT2160815C0-1
PART-A
SECTION - I
Straight Objective Type
This section contains 28 multiple choice questions.
Each question has 4 choices (1), (2), (3) and (4) for
its answer, out of which ONLY ONE is correct.
1. A car travels from A to B at a speed of
20 km h�1 and returns at a speed of
30 km h�1. The average speed of the car for
the whole journey is :
(1) 5 km h�1
(2) 24 km h�1
(3) 25 km h�1
(4) 50 km h�1
2. A stone is thrown vertically upward with an
initial speed u from the top of a tower,
reaches the ground with a speed 3u. The
height of the tower is:
(1) 23u
g
(2) 24u
g
(3) 26u
g
(4) 29u
g
3. A body starts from rest and is uniformly
acclerated for 30 s. The distance travelled in
the first 10 s is x1, next 10 s is x2 and the last
10 s is x3. Then x1 : x2 : x3 is the same as
(1) 1 : 2 : 4
(2) 1 : 2 : 5
(3) 1 : 3 : 5
(4) 1 : 3 : 9
4. Two balls of equal masses are thrown
upward, along the same vertical line at an
interval of 2 seconds, with the same initial
velocity of 40 m/s. Then these collide at a
height of (Take g = 10 m/s2)
(1) 120 m
(2) 75 m
(3) 200 m
(4) 45 m
5. A balloon is moving upwards with velocity
10 ms�1. It releases a stone which comes
down to the ground in 11 s. The height of the
balloon from the ground at the moment when
the stone was dropped is :
(1) 495 m
(2) 592 m
(3) 460 m
(4) 500 m
Space for Rough Work
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6. The acceleration time plot for a particle
(starting from rest) moving on a straight line
is shown in figure. For given time interval,
choose the incorrect option :
(1) The particle has zero average acceleration
(2) The particle has never turned around.
(3) The particle has zero displacement
(4) The average speed in the interval 0 to 10s
is the same as the average speed in the
interval 10s to 20s.
7. Mark the incorrect statement for a particle
going on a straight line (x�position coordinate,
v�velocity, a�acceleration) :
(1) If the v and a have opposite sign, the
object is slowing down.
(2) If the x and v have opposite sign, the
particle is moving towards the origin.
(3) If the v is zero at an instant, the a should
also be zero at that instant.
(4) If the v is zero for a time interval, then a
is zero at every instant within the time
interval.
8. A particle is initially at rest, It is subjected to
a linear acceleration a, as shown in the
figure. The maximum speed attained by the
particle is
(1) 605 m/s (2) 110 m/s
(3) 55 m/s (4) 550 m/s
9. The coordinates of a moving particle at any
time t are given by x = t3 and y = t3. The
speed of the particle at time t is given by :
(1) 2 2
(2) 3t2 2 2
(3) t2 2 2
(4) 2 2
10. A particle is moving eastwards with a velocity
of 5 ms�1. In 10 second the velocity changes
to 5 ms�1 northwards. The average acceleration
in this time is :
(1) 1
2 ms�1 towards north-west
(2) 12
ms�2 towards north
(3) zero
(4) 12
ms�2 towards north-west.
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11. During projectile motion, acceleration of a
particle at the highest point of its trajectory is
(1) g
(2) zero
(3) less than g
(4) dependent upon projection velocity
12. The velocity of projection of a projectile is
(6 �i + 8 �j ) ms�1.. The horizontal range of the
projectile is : (g = 10 m/sec2)
(1) 4.9 m
(2) 9.6 m
(3) 19.6 m
(4) 14 m
13. A body is projected horizontally from the top
of a tower with initial velocity 18 ms�1. It hits
the ground at angle 45º. What is the vertical
component of velocity when it strikes the
ground?
(1) 18 2 ms�1
(2) 18 ms�1
(3) 9 2 ms�1
(4) 9 ms�1
14. A ball is horizontally projected with a speed v
from the top of a plane inclined at an angle
45º with the horizontal. How far from the
point of projection will the ball strike the
plane?
(1) 2v
g
(2) 22v
g
(3) 22v
g
(4) 22 2v
g
15. On an inclined plane of inclination 30º, a ball
is thrown at an angle of 60º with the
horizontal from the foot of the incline with a
velocity of 10 3 ms�1. If g = 10 ms�2, then
the time in which ball will hit the inclined
plane is -
(1) 1 sec.
(2) 6 sec.
(3) 2 sec.
(4) 4 sec.
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16. A particle moves in the xy plane with only an
x-component of acceleration of 2 ms�2. The
particle starts from the origin at t = 0 with an
initial velocity having an x-component of
8 ms�1 and y-component of �15 ms�1.
Velocity of particle after time t is :
(1) [(8 + 2t) �i � 15 �j ] m s�1
(2) zero
(3) 2t �i + 15 �j
(4) directed along z-axis.
17. A stone projected at an angle of 60º from the
ground level strikes at an angle of 30º on the
roof of a building of height �h�. Then the
speed of projection of the stone is :
(1) 2gh
(2) 6gh
(3) 3gh
(4) gh
18. A particle is projected at 60° to the horizontal
with a kinetic energy K. The kinetic energy at
the highest point is :
(1) K
(2) Zero
(3) K/4
(4) K/2
19. Shown in the figure are the position time
graph for two children going home from the
school. Which of the following statements
about their relative motion is true after both
of them started moving ?
Their relative velocity : (consider 1-D motion)
(1) first increases and then decreases
(2) first decreases and then increases
(3) is zero
(4) is non zero constant.
20. A boat which can move with a speed of
5 m/s relative to water crosses a river of
width 480 m flowing with a constant speed of
4 m/s. What is the time taken by the boat to
cross the river along the shortest path.
(1) 80 s
(2) 160 s
(3) 240 s
(4) 320 s
Space for Rough Work
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21. A man walks in rain with a velocity of
5 kmh�1. The rain drops strike at him at an
angle of 45° with the horizontal. Velocity of
the rain if it is falling vertically downward �
(1) 5 kmh�1 (2) 4 kmh�1
(3) 3 kmh�1 (4) 1 kmh�1
22. Two particles A and B move with velocities v1
and v2 respectively along the x & y axis. The
initial separation between them is �d� as
shown in the figure. Find the least distance
between them during their motion.
(1) 21
2 21 2
d.v
v v (2)
22
2 21 2
d.v
v v
(3) 1
2 21 2
d.v
v v (4) 2
2 21 2
d.v
v v
23. A body is thrown up in a lift with a velocity u
relative to the lift and the time of flight is
found to be � t �. The acceleration with which
the lift is moving up is :
(1) u gt
t
(2) 2u gt
t
(3) u gt
t
(4) 2u gt
t
24. Which figure represents the correct F.B.D. of rod of mass m as shown in figure :
(1) (2)
(3) (4) None of these
25. Two forces of 6N and 3N are acting on the
two blocks of 2kg and 1kg kept on
frictionless floor. What is the force exerted on
2kg block by 1kg block ?
(1) 1N (2) 2N
(3) 4N (4) 5N
26. A mass M is suspended by a rope from a rigid support at A as shown in figure. Another rope is tied at the end B, and it is pulled horizontally with a force F. If the rope AB makes an angle with the vertical in equilibrium, then the tension in the string AB is :
/////////////////////////////
M
B
A
F
(1) F sin (2) F/sin
(3) F cos (4) F/cos
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27. A body of mass 8 kg is hanging from another
body of mass 12 kg. The combination is
being pulled by a string with an acceleration
of 2.2 m s�2. The tension T1 and T2 will be
respectively : (use g = 9.8m/s2 )
(1) 200 N, 80 N
(2) 220 N, 90 N
(3) 240 N, 96 N
(4) 260 N, 96 N
28. A block is dragged on smooth plane with the
help of a rope which moves with velocity v.
The horizontal velocity of the block is :
//////
/////
/////
//////
/////
/
m
//////////////////
V
(1) v
(2) v
sin
(3) v sin
(4) v
cos
SECTION - II
Reasoning Type
This section contains 2 reasoning type questions.
Each question has 4 choices (1), (2), (3) and (4), out
of which ONLY ONE is correct.
29. STATEMENT-1 : A particle having negative
acceleration will slow down. STATEMENT-2 : Direction of the
acceleration is not dependent upon direction of the velocity.
(1) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1
(2) Statement-1 is true, statement-2 is true and statement-2 is not correct explanation for statement-1
(3) Statement-1 is true, statement-2 is false (4) Statement-1 is false, statement-2 is true 30. STATEMENT-1 : Two stones are
simultaneously projected from level ground from same point with same speeds but different angles with horizontal. Both stones move in same vertical plane. Then the two stones may collide in mid air.
STATEMENT-2 : For two stones projected simultaneously from same point with same speed at different angles with horizontal, their trajectories may intersect at some point.
(1) Statement-1 is true, statement-2 is true
and statement-2 is correct explanation
for statement-1
(2) Statement-1 is true, statement-2 is true
and statement-2 is not correct
explanation for statement-1
(3) Statement-1 is true, statement-2 is false
(4) Statement-1 is false, statement-2 is true
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PART � B
Atomic masses : [H = 1, D = 2, Li = 7, C = 12,
N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27,
Si = 28, P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40,
Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65,
As = 75, Br = 80, Ag = 108, I = 127, Ba = 137,
Hg = 200, Pb = 207]
Straight Objective Type
This section contains 30 multiple choice
questions. Each question has 4 choices (1),
(2), (3) and (4) for its answer, out of which
ONLY ONE is correct.
31. The ratio of the difference in energy between
the first and second Bohr orbit to that
between the second and third Bohr orbit in a
H-like species is
(1) 12
(2) 13
(3) 49
(4) 275
32. If the de-Broglie wavelength of an
electron revolving in 2nd orbit of
H-atom is x, then radius of that orbit is
given by :
(1) x
(2) 2x
(3) x
2
(4) Cannot be determined
33. The uncertainty in position and velocity of a
particle are 0.5 Å and 5.27 × 10�24 m/s
respectively. Then, the approximate mass of
the particle is :
(1) 0.1 Kg
(2) 0.2 Kg
(3) 0.3 Kg
(4) 0.4 Kg
34. The wavenumber of the spectral line of
shortest wavelength of Balmer series of He+
ion is : (R = Rydberg�s constant)
(1) R (2) 3R
(3) 4R (4) 4R/9
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35. The value of azimuthal quantum number of
an electron present in the orbital designated
with quantum numbers as n = 4, m = �3 may
be :
(1) 0 (2) 1
(3) 2 (4) 3
36. In a sample of H-atom electrons make
transition from 5th excited state to ground
state, producing all possible types of
photons, then number of lines in infrared
region are
(1) 4 (2) 5
(3) 6 (4) 3
37. Which of the following set of quantum
numbers is correct for an electron in 4f
orbital ?
(1) n = 4, l =3, m = +4, s = +1/2
(2) n = 4, l = 4, m = �4, s = �1/2
(3) n = 4, l = 3, m = +1, s = +1/2
(4) n = 3, l=2, m =�2, s = +1/2
38. The wave motion of an electron in a Bohr's
orbit of Hydrogen atom is as shown in
diagram. The orbit number is :
(1) 2 (2) 3
(3) 4 (4) 6
39. One atom of an element x weigh
6.643 × 10�23 g. Number of moles of atom in
20 kg is :
(1) 4
(2) 40
(3) 100
(4) 500
40. For reaction P + 2Q 3R, if reaction start
with 0.1 mole of Q then find out the mole of
R produed.
(1) 0.2
(2) 0.3
(3) 1.5
(4) 0.15
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41. 2K + 2 + 22 HNO3 2HO3 + 2KO3 +
22NO2 + 10H2O
If 3 mole of K & 2 moles 2 are reacted with
excess of HNO3. Volume of NO2 gas evolved
at NTP is
(1) 739.2 Lt
(2) 1075.2 Lt
(3) 44.8 Lt
(4) 67.2 Lt
42. A solution containing 0.1 mol of a metal
chloride MClx requires 500 ml of 0.8 M
AgNO3 solution for complete precipitation.
The value of x is
(1) 1
(2) 2
(3) 4
(4) 3
43. The atomic mass of an element is 27. if
valency is 3, the vapour density of the
volatile chloride will be:
(1) 66.75
(2) 6.675
(3) 667.5
(4) 81
44. A hydrocarbon contains 80% of carbon, then
the hydrocarbon is :
(1) CH4
(2) C2H5
(3) C2H6
(4) C2H2
45. 20 gm. CaCO3 on decomposition gives CO2
at STP
(1) 4.48 litre
(2) 22.4 litre
(3) 2.24 litre
(4) None of these
46. How many alcohols give immediate turbidity
with lucas reagent having molecular formula
(C5H12O)
(1) 1
(2) 2
(3) 3
(4) 4
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47. Which of the following compound can give
test with Tollen's reagent, and yellow
precipitate with iodine in NaOH.
(1) CH2=O
(2) CH3�CH=O
(3) CH3�CH2�CH=O
(4)
48. Which is incorrect match with respect to the
reagent used for lab test ?
(1) Carbohydrates �Naphthol (Molish
reagent)
(2) Nitro ethane Zn, NH4Cl and AgNO3
(Muliken Barker test)
(3) Phenol Anhydrous ZnCl2 + Conc.
HCl (Lucas Reagent)
(4) Benzoic acid NaHCO3
49. How many structural isomers of C5H10 give
bromine water test ?
(1) 1 (2) 3
(3) 5 (4) 10
50. How many structural isomeric ketones
having molecular formula (C5H10O) give
iodoform test ?
(1) 1
(2) 2
(3) 3
(4) 4
51. On oxidative ozonolysis of 3-Methylhex-3-
ene, two products A & B are formed.
A gives CO2 gas with sodium bicarbonate,
but B can not. The structures of A & B are
respectively :
(1) & CH3�CH2�COOH
(2) CH3�CH2�COOH & CH3�CH2�CH=O
(3) CH3�CH2�COOH &
(4) CH3�CH2�CH2�COOH &
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52. How many monochloro structure isomers will
produce when 3-Methylpentane reacts with
chlorine in presence of sunlight ?
(1) 2
(2) 4
(3) 6
(4) 3
53. Which of the following hydrocarbon can give
only acetone and CO2 on ozonolysis in
presence of Zinc ?
(1) CH3�CH=C=CH�CH3
(2) CH3�CH=CH�CH=C(CH3)2
(3) (CH3)2C=C=CH2
(4)
54. Which is incorrect order for atomic radii?
(1) Mg < Ca < Sr < Ba
(2) B < Al < Ga < In
(3) F < O < N < C
(4) F < Cl < Br < I
55. Which statement is correct for the ionization
energy of second period elements (Be, B, C,
N, O, F)
(1) Order of Ist and IInd ionization energy is
same for the above elements.
(2) Ist ionization energy is highest for Be
among the given elements.
(3) Ionization energy of N is higher than
Oxygen & Fluorine
(4) First Ionization energy of Be is higher
than Boron but lower than Carbon
56. The set representing the correct order of first
ionization potential is :
(1) K > Na > Li
(2) Be > Mg > Ca
(3) B > C > N
(4) Ge > Si > C
57. Element with atomic number 38, belongs to
(1) II A group and 5th period
(2) II A group and 2nd period
(3) V A group and 2nd period
(4) III A group and 5th period
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58. Which gas is released in the given reaction ?
(1) H2 gas
(2) SO2 gas
(3) CO2 gas
(4) NO2 gas
59. Which is the position isomers of
(1)
(2)
(3)
(4)
60. Which is the chain isomers of
(1)
(2)
(3)
(4)
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Space for Rough Work
PART � C
SECTION - I
Straight Objective Type
This section contains 27 multiple choice questions.
Each question has 4 choices (1), (2), (3) and (4) for
its answer, out of which ONLY ONE is correct.
61. If a1, a2, .......... an are positive number in
A.P. then 21
n1
aa
aa
+
32
n1
aa
aa
+........ +
n1�n
n1
aa
aa
is equal to:
(1) n + 1
(2) n � 1
(3) n
(4) None of these
62. log(0.3)(x � 1) < log(0.09)(x � 1) then x lies in the
interval
(1) (2, )
(2) (�2, �1) (2, )
(3) (1, 2)
(4) (�, 1) (2, 8)
63. If c�b2log
= a�c3log
= b�a5log
then 2a 3b 5c is
equal to :
(1) 1
(2) 10
(3) 15
(4) None of these
64. The value of x for which |x2 + 3x| + x2 � 2 0 is
(1) R � [0, 1)
(2) R
(3) R �
21
,32
�
(4) R � (� 2 , 2 )
65. If P(x) = ax2 + bx + c & Q(x) = � ax2 + dx + c,
where ac 0, then P(x).Q(x) = 0 has at least
(1) Two real roots
(2) Four imaginary roots
(3) Four real roots
(4) can not say
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Space for Rough Work
66. If roots of the equation
(a2 + b2)x2 � 2(ac + bd)x + c2 + d2 = 0 are
equal then :
(1) ad + bc = 0
(2) ab = dc
(3) ac = bd
(4) ad = bc
67. If a,b,c,d and x are distinct real numbers such
that (a2+b2+c2)x2 � 2(ab+bc+cd)x + b2+c2+d2 0
then a, b, c, d :
(1) are in A.P.
(2) are in G.P.
(3) are in H.P.
(4) satisfy ab = cd
68. Solution of
1x2�x2)32( + 1�x2�x2
)3�2( =4
2 3are
(1) 1 ± 3 , 1
(2) 1 ± 2 , 1
(3) 1 ± 3 , 2
(4) 1 ± 2 , ± 1
69. If a1, a2, a3 ....... a2k are in A.P., then
21a � 2
2a + 23a � 2
4a + ........ � 2k2a =
(1) 1�k2
k( 2
1a � 2k2a )
(2) 1�k
k2 ( 2
k2a �21a )
(3) 1k
k
( 21a + 2
k2a )
(4) None of these
70. If ratio of sum of m terms and n terms of an
A.P. be m2 : n2, then the ratio of its mth and
nth terms will be
(1) (2m + 1) : (2n + 1)
(2) m : n
(3) (2m � 1) : (2n � 1)
(4) none of these
71. If sum of the A.M. & G.M. of two positive
distinct numbers is equal to the difference
between the numbers then numbers are in
ratio :
(1) 1 : 3
(2) 1 : 6
(3) 9 : 1
(4) 1 : 12
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Space for Rough Work
72. Suppose x1, x2 be the roots of ax2 + bx + c = 0
and x3, x4 be the roots of px2 + qx + r = 0
and x1, x2, x3, x4 are in A.P. then common
difference of this A.P. is
(1) 21
pq
�ab
(2) 31
pq
�ab
(3) 41
pq
�ab
(4) None of these
73. Sum of positive roots of the equation
(x2 � 12x + 35)(x2 + 10x + 24) = 504 is
(1) 9
(2) 10
(3) 11
(4) 12
74. Solution of the equation x x�1x5 (64) = 2000
is (where x 2 and x is an integer)
(1) divisible by 2
(2) divisible by 3
(3) divisible by 5
(4) divisible by 6
75. If the equations x2 + ax + b = 0 and
x2 + bx + a = 0 have a common root, then the
value of a + b is
(1) 1
(2) 0
(3) � 1
(4) none of these
76. Roots of the quadratic equation
2 2x 4x 3 x 6x 8 0, R will
be
(1) always real
(2) real only when is positive
(3) real only when is negative
(4) always imaginary
77. The solution set of the inequality
(x + 1)2 > (x + 3) is
(1) {x : �3 < x < �1}
(2) {x : x > �1}
(3) {x : �3 x �2}
(4) {x : x > 1 or x < �2}
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MEBPT2160815C0-16
Space for Rough Work
78. The number of values of k for which
equation |x + 1| + |x � 1| = k has infinite
solution is
(1) greater than 2
(2) 2
(3) less than 2
(4) none of these
79. If |x| � |x�2| = 2 then number of prime
solution of this equation less than 10 are
(1) 4
(2) 3
(3) 5
(4) none of these
80. Common solution of the equation
2(x �3x 2)| x � 2 | = 1 and inequality x(x � 2) 0
is
(1) 1
(2) 2
(3) 3
(4) 1, 2
81. Number of positive integers which satisfy the
inequality 2(x � 2)(�x 1)(x 1)
(x 1)
0 are
(1) two
(2) one
(3) three
(4) infinite
82. Sum of the all possible solutions of
||x � 1| � 3| = 2 is
(1) 0
(2) 2
(3) 4
(4) 6
83. If {x} = 2.1 then complete set of values of x
is (where {.} denotes fractional part function).
(1)
(2) 0.1
(3) 2
(4) 0.9
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MEBPT2160815C0-17
Space for Rough Work
84. 3 3 3(171) � (123) � (48)
9 19 123 16 3 =
(1) 8
(2) 2
(3) �3
(4) 3
85. If Hn = 1 + 21
+ 31
+ ........+ n1
then value of
1 +23
+ 35
+ ........+ n
1�n2 is :
(1) 2n � Hn
(2) 2n + Hn
(3) Hn � 2n
(4) Hn + n
86. If positive numbers a, b, c, d are in HP then
(1) ab cd
(2) ac bd
(3) ad bc
(4) none of these
87. The value of 91/3 . 91/9. 91/27 ......upto , is
(1) 1
(2) 3
(3) 9
(4) None of these
SECTION - II
Reasoning Type
This section contains 3 reasoning type questions.
Each question has 4 choices (1), (2), (3) and (4), out
of which ONLY ONE is correct.
88. Let a, b, c, d, e are positive numbers
STATEMENT -1 : If a, b, c, d, e are in A.P.,
then bcde, acde, abde, abcd and abce are in
H.P.
STATEMENT -2 : If a
a�cb ,
bb�ac
,
cc�ba
are in A.P., then a, b, c are in H.P.
(1) Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation for
Statement-1.
(2) Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct
explanation for Statement-1
(3) Statement-1 is True, Statement-2 is False
(4) Statement-1 is False, Statement-2 is True
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MEBPT2160815C0-18
Space for Rough Work
89. STATEMENT -1 :
Equation x2(a) + x (a2 � 3a + 1) � a2 + 2a � 1
is an identity in 'a' if x = �1.
STATEMENT-2 : If all the coefficients of
equation are zero then equation is called an
identity.
(1) Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation for
Statement-1.
(2) Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct
explanation for Statement-1
(3) Statement-1 is True, Statement-2 is False
(4) Statement-1 is False, Statement-2 is True
90. STATEMENT- 1 : 12| 1� 3 | � 4 2 3
is a rational number.
STATEMENT - 2 : |x| = x x 0
�x x 0
(1) Statement -1 is True, Statement -2 is
True ; Statement -2 is a correct
explanation for Statement -1
(2) Statement-1 is True, Statement-2 is True
; Statement-2 is NOT a correct
explanation for Statement-1
(3) Statement -1 is True, Statement -2 is
False
(4) Statement -1 is False, Statement -2 is
True
dPps dk;Z ds fy, LFkku
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PEBPT2160815C0-1
Hkkx-A
[k.M- I
lh/ks oLrqfu"B izdkj
bl [k.M esa 28 cgq&fodYih iz'u gSaA izR;sd iz'u ds
4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. ,d dkj A ls B rd 20 fdeh-@?k.Vk ls xfr
djrh gS vkSj 30 fdeh-@?k.Vk ls okil ykSVdj
vk tkrh gSA iwjh ;k=kk ds nkSjku dkj dh vkSlr
pky D;k gksxh \
(1) 5 fdeh-@?k.Vk
(2) 24 fdeh-@?k.Vk
(3) 25 fdeh-@?k.Vk
(4) 50 fdeh-@?k.Vk
2. fdlh iRFkj dks feukj ls m/okZ/kj Å ij dh vksj
u çkjfEHkd osx ls Qsadus ij ;g tehu ij 3u
osx ls igq¡prk gS rks feukj dh Å ¡pkbZ gS:
(1) 23u
g (2)
24ug
(3) 26u
g (4)
29ug
3. ,d oLrq fojkekoLFkk ls çkjEHk gksdj 30 lSd.M
rd ,d leku Rojfr gksrh gSA ;fn igys
10 lSd.M esa r; nwjh x1] vxys 10 lSd.M eas
x2 rFkk vfUre 10 lSd.M esa x3 gS rks x1 : x2 : x3 dk
vuqikr gksxk &
(1) 1 : 2 : 4 (2) 1 : 2 : 5
(3) 1 : 3 : 5 (4) 1 : 3 : 9
4. leku nzO;eku dh nks xsanksa dks leku Å /okZ/kj js[kk
esa 2 lSd.M ds vUrjky esa ,d leku çkjfEHkd osx
40 eh-/lS- ls Å ij dh vksj Qsadk tkrk gS rks ;s
fdl Å ¡pkbZ ij Vdjk,xhA (g = 10 m/s2)
(1) 120 m (2) 75 m
(3) 200 m (4) 45 m
5. ,d xqCckjk 10 eh0@lS0 osx ls tehu ls Å ij dh
vksj tk jgk gSA blls ,d iRFkj fxjk;k tkrk gS]
tks tehu ij 11 lSd.M esa igq¡prk gSA tc iRFkj
fxjk;k x;k Fkk] rc xqCckjs dh tehu ls Å pk¡bZ D;k
Fkh\
(1) 495 m (2) 592 m
(3) 460 m (4) 500 m
dPps dk;Z ds fy, LFkku
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PEBPT2160815C0-2
6. lh/kh js[kk esa fLFkjkoLFkk ls xfreku d.k dk
Roj.k le; vkjs[k n'kkZ;k x;k gSA fn;s x;s
le;kUrjky ds fy, vlR; dFku gSA
(1) d.k dk vkSlr Roj.k 'kwU; gSA
(2) d.k dHkh Hkh okil ugha eqM+rk gSA
(3) d.k dk foLFkkiu 'kwU; gksxkA
(4) d.k dh 0 ls 10 lSd.M eas vkSlr pky ogh gksxh
tks 10 lSd.M ls 20 lSd.M ds nkSjku gSA
7. ,d d.k lh/kh js[kk esa xfr dj jgk gSA bl d.k ds
fy, xyr dFku igpkfu;s&(x�fLFkfr funsZ'kkad ,
v�osx, a�Roj.k gS) :
(1) vxj v vkSj a foijhr fn'kk esa gSa] rks d.k
dh pky de gks jgh gSA
(2) vxj x vkSj v foijhr fpUg ds gaS] rks d.k
ewy fcUnq dh vksj tk jgk gSA
(3) vxj fdlh {k.k v 'kwU; gS] rks ml {k.k ij a
Hkh 'kwU; gksxkA
(4) vxj fdlh le;kUrjky esa v 'kwU; gS] rks ml
le;kUrjky esa fdlh Hkh {k.k a Hkh 'kwU;
gksxkA
8. fojkekoLFkk ls çkjEHk gqvk ,d d.k] fu;r js[kh;
Roj.k a ls fp=kkuqlkj xfreku gSA d.k }kjk çkIr
vf/kdre pky gksxh&
(1) 605 m/s (2) 110 m/s
(3) 55 m/s (4) 550 m/s
9. xfr'khy d.k ds fdlh le; t ij funZs'kkad x = t3
o y = t3A le; t ij d.k dh pky gksxhA
(1) 2 2 (2) 3t2 2 2
(3) t2 2 2 (4) 2 2
10. ,d d.k 5 ms�1 pky ls iwoZ fn'kk esa xfr dj jgk
gSA 10 lSd.M ckn ;g mÙrj fn'kk esa 5 ms�1 pky
ls py jgk gksrk gSA rks bl le; esa d.k dk vkSlr
Roj.k D;k gksxkA
(1) 1
2 ms�1 mÙrj&if'pe dh vksj
(2) 12
ms�2 mÙrj dh vksj
(3) 'kwU;
(4) 12
ms�2 mÙrj&if'pe dh vksj
dPps dk;Z ds fy, LFkku
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PEBPT2160815C0-3
11. iz{ksI; xfr esa] iFk ds mPpre fcUnq ij d.k dk
Roj.k gksrk gS :
(1) g
(2) 'kwU;
(3) g ls de
(4) iz{ksi.k osx ij fuHkZj djrk gSA
12. ,d iz{ksI; dk iz{ksi.k osx (6 �i + 8 �j ) ms�1 gSA rks
iz{ksI; dh {kSfrt ijkl gksxh & (g = 10 m/sec2)
(1) 4.9 m
(2) 9.6 m
(3) 19.6 m
(4) 14 m
13. ,d oLrq ,d Å ¡ph bekjr ls 18 ms�1 ds
izkjfEHkd osx ls {kSfrt fn'kk esa iz{ksfir dh tkrh
gS ;g tehu ij 45º dks.k ij Vdjkrh gS rks
oLrq }kjk tehu ij Vdjkrs le; osx dk
Å /okZ/kj ?kVd gksxkA
(1) 18 2 ms�1
(2) 18 ms�1
(3) 9 2 ms�1
(4) 9 ms�1
14. ,d xsn dks 45º dks.k ij >qds urry ds Å ijh
fljs ls {kSfrt fn'kk esas v pky ls iz{ksfir fd;k
tkrk gSA rks iz{ksfir fcUnq ls xsan urry ij fdruh
nwj Vdjk,xh &
(1) 2v
g
(2) 22 v
g
(3) 22v
g
(4) 22 2 v
g
15. 30º mUu;u dks.k okys ,d urry ds vk/kkj ls
{kSfrt ls 60º ds dks.k ij ,d xsan dks
10 3 ms�1 ds osx ls Qsadrs gSaA ;fn g=10 ms�2
gS rks fdrus le; ckn xsan okil ur ry ls
Vdjk,xhA
(1) 1 sec.
(2) 6 sec.
(3) 2 sec.
(4) 4 sec.
dPps dk;Z ds fy, LFkku
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PEBPT2160815C0-4
16. x-y ry esa xfr'khy d.k] dsoy Roj.k dk
x�?kVd 2 ms�2 j[krk gSA d.k t = 0 ij ewy
fcUnq ls xfr izkjEHk djrk gSA izkjfEHkd osx dk
x-?kVd 8 ms�1 rFkk y-?kVd �15 ms�1 gSA rks
t le; i'pkr d.k dk osx gksxk &
(1) [(8 + 2t) �i � 15 �j ] m s�1
(2) 'kwU;
(3) 2t + 15 �j
(4) z-v{k ds vuqfn'k funsZf'kr
17. ,d iRFkj tehu ls 60º ds dks.k ij iz{ksfir
fd;k tkrk gS ,oa h Å ¡pkbZ dh ,d bekjr dh
Nr ij 30º ds dks.k ij Vdjkrk gSA rks iRFkj
dh iz{ksi.k pky gSA
(1) 2gh
(2) 6gh
(3) 3gh
(4) gh
18. ,d d.k dks {kSfrt ls 60° dk dks.k cukrs gq,
xfrt Å tkZ K ls iz{ksfir fd;k tkrk gSA mPpre
fcUnq ij xfrt Å tkZ gksxh %
(1) K (2) 'kwU;
(3) K/4 (4) K/2
19. ?kj ls Ldwy tkrs gq, cPpksa ds fLFkfr le; xzkQ
fp=k esa çnf'kZr gSA rks nksuks ds xfr izkjEHk djus ds
i'pkr~ buds lkis{k xfr ds lEcU/k esa dkSuls dFku
lR; gS? budk lkis{k osx% ¼ljy js[kh; xfr ekfu;s½
(1) igys c<+sxk ckn esa ?kVsxkA
(2) igys ?kVsxk ckn esa c<+sxkA
(3) 'kwU; gksxkA
(4) v'kwU; fu;rakd gksxkA
20. ikuh ds lkis{k 5 m/s dh pky ls xfreku ,d ukoa
4 m/s dh fu;r pky ls cgrh gqbZ 480 m pkSM+h
unh dks ikj djrh gS rks U;wure nwjh ls unh dks
ikj djus esa yxk le; Kkr djksA
(1) 80 s
(2) 160 s
(3) 240 s
(4) 320 s
dPps dk;Z ds fy, LFkku
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PEBPT2160815C0-5
21. ,d vkneh 5 kmh�1 osx ls xfr'khy gSA vkneh
dks c"kkZ dh cwans {kSfrt ls 45° ds dks.k ij vkrh
çrhr gksrh gS rks m/okZ/kj uhps fxjrh cwanksa dk osx
gksxk \
(1) 5 kmh�1 (2) 4 kmh�1
(3) 3 kmh�1 (4) 1 kmh�1
22. nks d.k A rFkk B Øe'k% v1 rFkk v2 osx ls x rFkk
y v{k esa xfreku gSA nksuksa d.kksa ds e/; izkjfEHkd
nwjh d fp=k esa n'kkZ;s vuqlkj gS rks xfr ds nkSjku
nksuksa ds e/; U;wure nwjh Kkr djksA
(1) 21
2 21 2
d.v
v v (2)
22
2 21 2
d.v
v v
(3) 1
2 21 2
d.v
v v (4) 2
2 21 2
d.v
v v
23. ,d oLrq dks Å ij dh rjQ xfr'khy fy¶V esa
fy¶V ds lkis{k Å ij dh rjQ u osx ls QSadk
tkrk gS vkSj oLrq dk mM~M;u dky � t � çkIr
gksrk gSA rks Å ij dh rjQ xfr'khy fy¶V dk
Roj.k gksxk %
(1) u gt
t
(2) 2u gt
t
(3) u gt
t
(4) 2u gt
t
24. fp=k esa iznf'kZr m nzO;eku dh NM+ dk eqDr oLrq
js[kkfp=k fuEu esa ls dkSuls fp=k }kjk fn;k tkrk gSA
(1)
(2)
(3)
(4) buesa ls dksbZ ugha
25. fp=kkuqlkj ?k"kZ.k jfgr lrg ij j[ks 2 kg o 1 kg ds
nks CykWdks ij nks cy 6N vkSj 3 N dk;Zjr gS] rks
1 kg ds CykWd }kjk 2kg ds CykWd ij vkjksfir cy
D;k gksxk \
(1) 1N (2) 2N
(3) 4N (4) 5N
dPps dk;Z ds fy, LFkku
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PEBPT2160815C0-6
26. fp=kkuqlkj ,d M nzO;eku dks jLlh dh lgk;rk
ls n< vk/kkj ij fcUnq A ls yVdk;k x;k gSA ,d nwljh jLlh fcUnq B ij ca/kh gS vkSj bldks {kSfrt fn'kk esa cy F ls [khapk tkrk gSA ;fn m/okZ/kj lkE;koLFkk eas jLlh AB Å /okZ/kj ls
dks.k cukrh gks rks jLlh AB esa ruko gksxk & /////////////////////////////
M
B
A
F
(1) F sin (2) F/sin (3) F cos (4) F/cos
27. ,d 8 kg nzO;eku dh oLrq ,d nwljh 12 kg dh oLrq ls yVdh gqbZ gSA bl la;kstu dks jLlh dh lgk;rk ls Å ij dh rjQ 2.2 m s�2 ds Roj.k ls [khapk tkrk gS rks ruko T1 o T2 Øe'k% gksaxs : (fn;k gS g = 9.8m/s2 )
(1) 200 N, 80 N (2) 220 N, 90 N (3) 240 N, 96 N (4) 260 N, 96 N 28. ,d CykWd dks fpdus ry ij v osx ls xfr'khy
jLlh }kjk fp=kkuqlkj [khapk tkrk gSA CykWd dk {kSfrt osx gksxk&
//////
/////
/////
//////
/////
/
m
//////////////////
V
(1) v (2) v
sin
(3) v sin (4) v
cos
[k.M - II
dkj.k&izdkj
bl [k.M esa 2 dkj.k ds ç'u gSA çR;sd ç'u ds 4 fodYi
(1), (2), (3) rFkk (4) gS] ftlesa ls flQZ ,d lgh gSA
29. oDrO;-1 % ,d d.k ftldk Roj.k _ .kkRed gS]
eafnr gksxkA
oDrO;-2 % Roj.k dh fn'kk osx dh fn'kk ij fuHkZj
ugha djrhA
(1) oDrO;&1 lR; gS] oDrO;&2 lR; gS ;
oDrO;&2, oDrO;&1 dk lgh Li"Vhdj.k gSA
(2) oDrO;&1 lR; gS] oDrO;&2 lR; gS ;
oDrO;&2, oDrO;&1 dk lgh Li"Vhdj.k ugha
gSA
(3) oDrO;&1 lR; gS] oDrO;&2 vlR; gSA
(4) oDrO;&1 vlR; gS] oDrO;&2 lR; gSA 30. oDrO;-1 : nks iRFkj tehu ls ,d lkFk ,d gh
fcUnq ls ,d gh pky ls ysfdu {kSfrt ls
fHkUu&fHkUu dks.kksa ij iz{ksfir fd;s tkrs gSaA nksuksa
iRFkj ,d gh Å /okZ/kj ry esa xfr djrs gSaA rks
nksuksa iRFkj chp gok esa Vdjk ldrs gSaA
oDrO;-2 : ,d gh fcUnq ls ,d lkFk leku pky ls
{kSfrt ls fHkUu&fHkUu dks.kksa ij iz{ksfir nks iRFkj ds
iz{ksI; iFk chp gok esa fdlh fcUnq ij dkV ldrs
gSaA
(1) oDrO;&1 lR; gS] oDrO;&2 lR; gS ;
oDrO;&2, oDrO;&1 dk lgh Li"Vhdj.k gSA
(2) oDrO;&1 lR; gS] oDrO;&2 lR; gS ;
oDrO;&2, oDrO;&1 dk lgh Li"Vhdj.k ugha
gSA
(3) oDrO;&1 lR; gS] oDrO;&2 vlR; gSA
(4) oDrO;&1 vlR; gS] oDrO;&2 lR; gSA
d Pps d k;Z d s fy, LFkku
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CEBPT2160815C0-7
PART � B
Atomic masses : [H = 1, D = 2, Li = 7, C = 12,
N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27,
Si = 28, P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40,
Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65,
As = 75, Br = 80, Ag = 108, I = 127, Ba = 137,
Hg = 200, Pb = 207]
lh/ks oLrqfu"B izdkj
bl [k.M esa 30 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4
fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d
lgh gSA
31. fdlh H ds leku Lih'kht esa çFke o f}rh; cksj
d{k rFkk f}rh; o rrh; cksj d{k ds Å tkZ vUrj
dk vuqikr gksxk µ
(1) 12
(2) 13
(3) 49
(4) 275
32. ;fn H-ijek.kq dh 2nd d{kk esa ifjØe.k djus
okys ,d bysDVªkWu dh Mh&czksXyh rjax}S/;Z x gS] rks
bl d{kk dh f=kT;k D;k gksxh :
(1) x
(2) 2x
(3) x
2
(4) Kkr ugha fd;k tk ldrk
33. ;fn fdlh ,d d.k dh fLFkfr rFkk osx esa
vfuf'pr~rk Øe'k% 0.5 Å rFkk 5.27 × 10�24 m/s
gSa] rks d.k dk nzO;eku yxHkx fdruk gksxk\
(1) 0.1 Kg (2) 0.2 Kg
(3) 0.3 Kg (4) 0.4 Kg
34. He+ vk;u dh ckej Js.kh esa izkIr U;wure rjax}S/;Z
dh LiSDVªeh js[kk dh rjaxla[;k D;k gksxh\
(R = fjMcxZ fu;rkad )
(1) R (2) 3R
(3) 4R (4) 4R/9
d Pps d k;Z d s fy, LFkku
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CEBPT2160815C0-8
35. DokaVe la[;k n = 4, m = �3 okys ,d d{kd esa
mifLFkr ,d bysDVªkWu ds fy, f}xa'kh DokaVe
la[;k dk eku D;k gks ldrk gS %
(1) 0
(2) 1
(3) 2
(4) 3
36. H-ijek.kq ds uewus esa bysDVªkWu 5th mÙksftr voLFkk
ls ewy voLFkk rd laØe.k dj lHkh laHko QksVksu
mRikfnr djrs gSa] rks vojDr {ks=k esa js[kkvksa dh
la[;k fuEu gSa &
(1) 4
(2) 5
(3) 6
(4) 3
37. 4f d{kd esa ,d by sDVªkWu ds fy, DokaVe la[;k
dk dkSulk leqPp; lgh gS ?
(1) n = 4, l =3, m = +4, s = +1/2
(2) n = 4, l = 4, m = �4, s = �1/2
(3) n = 4, l = 3, m = +1, s = +1/2
(4) n = 3, l=2, m =�2, s = +1/2
38. fuEu js[kkfp=k esa gkbMªkstu ijek.kq dh ,d cksgj
d{kk esa ,d bysDVªkWu dh rjax xfr n'kkZ;h x;h gSA
d{kk la[;k fuEu gS%
(1) 2
(2) 3
(3) 4
(4) 6
39. rRo x ds ,d ijek.kq dk Hkkj 6.643 × 10�23 xzke
gSA 20 kg esa ijek.kq ds eksyksa dh la[;k fuEu gS %
(1) 4
(2) 40
(3) 100
(4) 500
40. vfHkfØ;k P + 2Q 3R, ds fy, ;fn vfHkfØ;k
Q ds 0.1 eksy ysdj izkjEHk dh xbZ rc fufeZr
R ds eksy Kkr dhft,A
(1) 0.2 (2) 0.3
(3) 1.5 (4) 0.15
d Pps d k;Z d s fy, LFkku
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CEBPT2160815C0-9
41. 2K + 2 + 22 HNO3 2HO3 + 2KO3 +
22NO2 + 10H2O
;fn K ds 3 eksy rFkk 2 ds 2 eksy HNO3 ds
vkf/kD; ls fØ;k djrs gS] rks ekud rki o nkc
(NTP) ij NO2 dk izkIr vk;ru Kkr djksA
(1) 739.2 Lt
(2) 1075.2 Lt
(3) 44.8 Lt
(4) 67.2 Lt
42. /kkrq DyksjkbM MClx ds 0.1 eksy ;qDr ,d foy;u
ds iw.kZ :i ls vo{ksi.k ds fy, 0.8 M AgNO3
foy;u ds 500 ml vko';d gSA x dk eku fuEu gS%
(1) 1
(2) 2
(3) 4
(4) 3
43. ,d rRo dk ijek.kq Hkkj 27 gSA ;fn la;kstdrk 3
gS] rks ok"i'khy DyksjkbM dk ok"i ?kuRo gksxk :
(1) 66.75
(2) 6.675
(3) 667.5
(4) 81
44. ;fn ,d gkbMªksdkcZu esa 80% dkcZu gks] rks
gkbMªksdkcZu gksxk :
(1) CH4
(2) C2H5
(3) C2H6
(4) C2H2
45. 20 xzke CaCO3 dks fo?kfVr djus ij fdruh CO
2
STP ij çkIr gksrh gS\
(1) 4.48 yhVj
(2) 22.4 yhVj
(3) 2.24 yhVj
(4) buesa ls dksbZ ugha
46. (C5H
12O) v.kqlw=k okys fdrus ,YdksgkWy Y;wdkWl
vfHkdeZd ds lkFk rqjUr xanykiu nsrs gSa \
(1) 1
(2) 2
(3) 3
(4) 4
d Pps d k;Z d s fy, LFkku
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CEBPT2160815C0-10
47. fuEu esa ls dkSulk ;kSfxd VkWysu vfHkdeZd ds
lkFk ijh{k.k nsrk gS rFkk NaOH esa vk;ksMhu ds
lkFk ihyk vo{ksi nsrk gS \
(1) CH2=O
(2) CH3�CH=O
(3) CH3�CH
2�CH=O
(4)
48. iz;ksx'kkyk ijh{k.k ds fy, fuEu esa ls dkSu mlds
vfHkdeZd ds lkFk lgh lqesfyr ugha gSA
(1) dkcksZgkbMªsV �us¶Fkksy (eksfy'k vfHkdeZd )
(2) ukbVªks,Fksu Zn, NH4Cl rFkk AgNO
3
(eqfydu cdZj ijh{k.k)
(3) fQuksy futZy ZnCl2 + Conc. HCl
(Y;wdkWl vfHkdeZd )
(4) csUtksbd vEy NaHCO3
49. C5H
10 ds fdrus lajpukRed leko;oh czksehu ty
ijh{k.k nsrs gSa ?
(1) 1 (2) 3
(3) 5 (4) 10
50. (C5H
10O) v.kqlw=k okys fdrus lajpukRed
leko;oh dhVksu vk;ksMksQkWeZ ijh{k.k nsrs gSa?
(1) 1 (2) 2
(3) 3 (4) 4
51. 3-esfFkygsDl-3-bZu ds vkWDlhdkjh vkstksuhvi?kVu
ij nks mRikn A o B curs gSaA mRikn A lksfM;e
ckbdkcksZusV ds lkFk CO2 xSl nsrk gS fdUrq mRikn
B ughaA A rFkk B dh lajpuk,a Øe'k% gS :
(1) & CH3�CH
2�COOH
(2) CH3�CH
2�COOH & CH
3�CH
2�CH=O
(3) CH3�CH
2�COOH &
(4) CH3�CH
2�CH
2�COOH &
d Pps d k;Z d s fy, LFkku
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CEBPT2160815C0-11
52. tc 3-esfFkyisUVsu dh vfHkfØ;k lw;Z ds izdk'k dh
mifLFkfr esa Dyksjhu ds lkFk dh tkrh gS rks fdrus
eksuksDyksjks lajpuk leko;oh izkIr gksrs gSa\
(1) 2 (2) 4
(3) 6 (4) 3
53. dkSulk gkbMªksdkcZu ftad dh mifLFkfr esa
vkstksuhdj.k djus ij dsoy ,lhVksu ,oa CO2 nsrk
gS \
(1) CH3�CH=C=CH�CH
3
(2) CH3�CH=CH�CH=C(CH
3)
2
(3) (CH3)2C=C=CH
2
(4)
54. fuEu esa ls dkSulk ijekf.od f=kT;kvksa dk xyr
Øe gS \
(1) Mg < Ca < Sr < Ba
(2) B < Al < Ga < In
(3) F < O < N < C
(4) F < Cl < Br < I
55. f}rh; vkorZ rRoksa (Be, B, C, N, O, F) dh
vk;uu Å tkZ ds lEcU/k esa lgh dFku gS \
(1) mijksDr rRoksa ds fy, izFke rFkk f}rh;
vk;uu Å tkZvksa dk Øe leku gSA
(2) fn;s x;s rRoksa esa ls Be dh izFke vk;uu Å tkZ
lcls vf/kd gksrh gSA
(3) N dh vk;uu Å tkZ vkWDlhtu o ¶yksjhu ls
vf/kd gksrh gSA
(4) Be dh izFke vk;uu Å tkZ cksjkWu ls vf/kd
fdUrq dkcZu ls de gksrh gSA
56. fuEu lewg esa izFke vk;uu foHko dk lgh Øe
fuEu gS %
(1) K > Na > Li
(2) Be > Mg > Ca
(3) B > C > N
(4) Ge > Si > C
d Pps d k;Z d s fy, LFkku
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CEBPT2160815C0-12
57. vkorZ lkj.kh esa ijek.kq Øekad 38 okyk rRo
mifLFkr gksxkA
(1) II A lewg rFkk 5th vkorZ esa
(2) II A lewg rFkk 2nd vkorZ esa
(3) V A lewg rFkk 2nd vkorZ esa
(4) III A lewg rFkk 5th vkorZ esa
58. nh xbZ vfHkfØ;k esa dkSulh xSl eqDr gksrh gS \
(1) H2 gas
(2) SO2 gas
(3) CO2 gas
(4) NO2 gas
59. ;kSfxd dk fLFkfr leko;oh
dkSulk gS
(1)
(2)
(3)
(4)
60. ;kSfxd dk Ja[kyk leko;oh
dkSulk gS \
(1)
(2)
(3)
(4)
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MEBPT2160815C0-13
(dPps dk;Z ds fy, LFkku )
PART � C
[k.M - I lh/ks oLrqfu"B izdkj
bl [k.M esa 27 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4
fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
61. ;fn a1, a2, .......... an /kukRed la[;k,¡ lekUrj
Js.kh esa gS rc 21
n1
aa
aa
+
32
n1
aa
aa
+
.......... + n1�n
n1
aa
aa
cjkcj gS -
(1) n + 1
(2) n � 1
(3) n
(4) buesa ls dksbZ ugha
62. log(0.3)(x � 1) < log(0.09)(x � 1) gks x vUrjky esa
fLFkr gS -
(1) (2, )
(2) (�2, �1) (2, )
(3) (1, 2)
(4) (�, 1) (2, 8)
63. ;fn c�b2log
= a�c3log
= b�a5log
rc 2a 3b 5c cjkcj
gS -
(1) 1
(2) 10
(3) 15
(4) buesa ls dksbZ ugha
64. |x2 + 3x| + x2 � 2 0 ds fy, x dk eku gS -
(1) R � [0, 1)
(2) R
(3) R �
21
,32
�
(4) R � (� 2 , 2 )
65. ;fn P(x) = ax2 + bx + c vkSj Q(x) = � ax2 + dx + c
tgk¡ ac 0 rc P(x).Q(x) = 0 de ls de -
(1) nks okLrfod ewy j[krh gS
(2) pkj dkYifud ewy j[krk gS
(3) pkj okLrfod ewy j[krk gS
(4) dqN dgk ugh tk ldrk
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MEBPT2160815C0-14
(dPps dk;Z ds fy, LFkku )
66. ;fn lehdj.k
(a2 + b2)x2 � 2(ac + bd)x + c2 + d2 = 0 ls ewy
cjkcj gS rc -
(1) ad + bc = 0
(2) ab = dc
(3) ac = bd
(4) ad = bc
67. ;fn a, b, c, d vkSj x fofHkUu okLrfod la[;k,¡ bl
izdkj gS fd (a2 + b2 + c2)x2 � 2(ab + bc + cd)x
+ b2 + c2 + d2 0 rc a, b, c, d -
(1) lekUrj Js.kh es
(2) xq.kksÙkj Js.kh esa
(3) gjkRed Js.kh esa
(4) ab = cd dks larq"B djrk gSA
68. 1x2�x2)32( + 1�x2�x2
)3�2( =4
2 3 dk
gy gS -
(1) 1 ± 3 , 1
(2) 1 ± 2 , 1
(3) 1 ± 3 , 2
(4) 1 ± 2 , ± 1
69. a1, a2, a3 ....... a2k lekUrj Js.kh esa gS] rc
21a � 2
2a + 23a � 2
4a + ........ � 2k2a =
(1) 1�k2
k( 2
1a � 2k2a )
(2) 1�k
k2 ( 2
k2a �21a )
(3) 1k
k
( 21a + 2
k2a )
(4) buesa ls dksbZ ugh
70. ;fn lekUrj Js.kh ds m inksa vkSj n inksa ds
;ksxQyksa dk vuqikr m2 : n2 gS rc blds m osa
vkSj n osa inks dk vuqikr gksxk -
(1) (2m + 1) : (2n + 1)
(2) m : n
(3) (2m � 1) : (2n � 1)
(4) buesa ls dksbZ ugh
71. ;fn nks fofHkUu /kukRed la[;kvksa ds lekUrj ek/;
vkSj xq.kksÙkj ek/; dk ;ksxQy la[;kvksa ds vUrj
ds cjkcj gS rc la[;kvksa dk vuqikr gS -
(1) 1 : 3
(2) 1 : 6
(3) 9 : 1
(4) 1 : 12
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MEBPT2160815C0-15
(dPps dk;Z ds fy, LFkku )
72. ekuk fd lehdj.k ax2 + bx + c = 0 ds ewy x1
vkSj x2 vkSj lehdj.k px2 + qx + r = 0 ds ewy
x3, x4 gS rFkk x1, x2, x3, x4 lekUrj Js.kh esa gS rc
bl lekUrj Js.kh dk lkoZvUrj gS -
(1) 21
pq
�ab
(2) 31
pq
�ab
(3) 41
pq
�ab
(4) buesa ls dksbZ ugha
73. lehdj.k (x2 � 12x + 35)(x2 + 10x + 24) = 504
ds /kukRed ewyksa dk ;ksxQy gS -
(1) 9
(2) 10
(3) 11
(4) 12
74. lehdj.k x x�1x5 (64) = 2000 dk gy gS
(tgk¡ x 2 ,oa x iw.kk±d gS)
(1) 2 ls foHkkftr
(2) 3 ls foHkkftr
(3) 5 ls foHkkftr
(4) 6 ls foHkkftr
75. ;fn x2 + ax + b = 0 rFkk x2 + bx + a = 0
dk ,d mHk;fu"B ewy gS] rks a + b dk eku gS&
(1) 1
(2) 0
(3) � 1
(4) buesa ls dksbZ ugha
76. f}?kkr lehdj.k
2 2x 4x 3 x 6x 8 0, R ds
ewy gksxsa&
(1) lnSo okLrfod
(2) okLrfod dsoy tc /kukRed gSA
(3) okLrfod dsoy tc _ .kkRed gSA
(4) lnSo dkYifud
77. vlfedk (x + 1)2 > (x + 3) dk gy leqPp; gS&
(1) {x : �3 < x < �1}
(2) {x : x > �1}
(3) {x : �3 x �2}
(4) {x : x > 1 ;k x < �2}
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MEBPT2160815C0-16
(dPps dk;Z ds fy, LFkku )
78. lehdj.k |x + 1| + |x � 1| = k ds vuUr gy gksus
ds fy, k ds ekuksa dh la[;k gS&
(1) 2 ls vf/kd
(2) 2
(3) 2 ls de
(4) buesa ls dksbZ ugha
79. ;fn |x| � |x �2| = 2, rc bl lehdj.k ds] 10 ls
NksVs vHkkT; gyksa dh la[;k gS&
(1) 4
(2) 3
(3) 5
(4) buesa ls dksbZ ugha
80. lehdj.k 2(x �3x 2)| x � 2 |
= 1 ,oa vlfedk
x(x � 2) 0 dk mHk;fu"B gy gS&
(1) 1
(2) 2
(3) 3
(4) 1, 2
81. vlfedk 2(x � 2)(�x 1)(x 1)
(x 1)
0 dks larq"V
djus okys /kukRed iw.kk±dksa dh la[;k gS&
(1) nks
(2) ,d
(3) rhu
(4) vuUr
82. lehd j.k ||x � 1| � 3| = 2 d s lHkh
laHkkfor gy ksa d k ;ksxQ y gS&
(1) 0
(2) 2
(3) 4
(4) 6
83. ;fn {x} = 2.1 gS rks x ds lEi.wkZ ekuksa dk leqPp;
gS (tgk¡ {.} fHkUukRed HkkxQyu dks O;Dr djrk
gSA)
(1)
(2) 0.1
(3) 2
(4) 0.9
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MEBPT2160815C0-17
(dPps dk;Z ds fy, LFkku )
84. 3 3 3(171) � (123) � (48)
9 19 123 16 3 =
(1) 8
(2) 2
(3) �3
(4) 3
85. ;fn Hn = 1 + 21
+ 31
+ ........+ n1
gks] rks
1 +23
+ 35
+ ........+ n
1�n2dk eku gS &
(1) 2n � Hn
(2) 2n + Hn
(3) Hn � 2n
(4) Hn + n
86. ;fn /kukRed la[;k,sa a, b, c, d gjkRed Js<+h esa
gks] rks &
(1) ab cd
(2) ac bd
(3) ad bc
(4) buesa ls dksbZ ugh
87. 91/3 . 91/9. 91/27 ......rd] dk eku gS &
(1) 1
(2) 3
(3) 9
(4) buesa ls dksbZ ugh
[k.M- II
dkj.k çdkj
bl [k.M esa 3 dkj.k çdkj ds ç'u gSA çR;sd ç'u ds 4
fodYi (1), (2), (3),rFkk (4) gS, ftuesa ls flQZ ,d lgh gSA
88. ekuk fd a, b, c, d, e /kukRed la[;k,¡ gSA
oDrO;-1 : ;fn a, b, c, d, e lekUrj Js.kh esa gS rc
bcde, acde, abde, abcd vkSj abce gjkRed Js.kh
esa gSA
oDrO;-2 : ;fn a
a�cb ,
bb�ac
,
cc�ba lekUrj Js.kh esa gS rc a, b, c gjkRed
Js.kh esa gSA
(1) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2,
oDrO;-1 dk lgh Li"Vhdj.k gSA
(2) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2,
oDrO;-1 dk lgh Li"Vhdj.k ugha gSA
(3) oDrO;-1 lR; gS] oDrO;-2 vlR; gSA
(4) oDrO;-1 vlR; gS] oDrO;-2 lR; gSA
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MEBPT2160815C0-18
(dPps dk;Z ds fy, LFkku )
89. oDrO; -1 :
lehdj.k x2(a) + x (a2 � 3a + 1) � a2 + 2a � 1,
'a' esa ,d loZ lfedk gS ;fn x = �1 gSA
oDrO; -2 : ;fn fdlh lehdj.k ds lHkh xq.kkad
'kwU; gksa rks lehdj.k loZlfedk dgykrh gSA
(1) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2,
oDrO;-1 dk lgh Li"Vhdj.k gSA
(2) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2,
oDrO;-1 dk lgh Li"Vhdj.k ugha gSA
(3) oDrO;-1 lR; gS] oDrO;-2 vlR; gSA
(4) oDrO;-1 vlR; gS] oDrO;-2 lR; gSA
90. oDrO;-1 : 12| 1� 3 | � 4 2 3 ,d
vifjes; la[;k gS&
oDrO;-2 : |x| = x x 0
�x x 0
(1) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2,
oDrO;-1 dk lgh Li"Vhdj.k gSA
(2) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2,
oDrO;-1 dk lgh Li"Vhdj.k ugha gSA
(3) oDrO;-1 lR; gS] oDrO;-2 vlR; gSA
(4) oDrO;-1 vlR; gS] oDrO;-2 lR; gSA
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SOLEBPT2160815-1
PART TEST-2 (PT-2)
TARGET : JEE (MAIN)-2017
HINTS & SOLUTIONS ¼ladsr ,oa gy½
PART-A PHYSICS
1. A car travels from A ........................
,d dkj A ls B rd ........................ Sol. Suppose ekuk AB = x km Average speed
= Total distance cov ered
Total time taken
vkSlr pky = r; d h xb Zd qy nwjh
fy;k x;k dqy le;
= 2x
x x20 30
= 2
1 120 30
= 20 6020 30
= 24 km/h = 24 kmh�1
2. A stone is thrown ........................ fdlh iRFkj dks feukj ........................ Sol. I mothod � Let downward direction is taken
as +ve. initial vel is �ve = � u (say) From the equation; v2 � u2 = 2as we get
(3u)2 � (�u)2 = 2hg izFke fof/k � ekuk uhps dh vksj nwjh;ksa dks /kukRed
fy;k tkrk gS izkjfEHkd osx _ .kkRed gS = �u (sayekuk)
lehdj.k ls ; v2 � u2 = 2as ge izkIr djrs gS
(3u)2 � (�u)2 = 2hg
h = 24u
g
The stone is thrown vertically upward with an
initial velocity u from the top of a tower it
reaches the highest point and returns back
and reaches the top of tower with the same
velocity u vertically downward.
Now, from the equation, V2 = u2 + 2gh
feukj ds 'kh"kZ ls ,d iRFkj m/oZ Å ij dh vksj
izkjfEHkd osx u ls Qsadk x;k gS ;g mPpre fcUnq rd igq¡prk gS vkSj okil ykSVrk gS o feukj ds 'kh"kZ ij m/oZ uhps dh vksj leku pky ls igq¡prk gS
lehdj.k ls V2 = u2 + 2gh (3u)2 = u2 + 2 gh 2gh = 9u2 � u2
h = 28u
2g h =
24ug
3. A body starts from ........................ ,d oLrq fojkekoLFkk ls ........................ Sol. u = 0, Let acceleration ekuk Roj.k = a
Total time dqy le; t = 30 s X1 = distance travelled in the first 10 s.
X1 = igys 10 s esa r; dh xbZ nwjh
Using , S = ut + 12
at2, we get
S = ut + 12
at2 dk mi;ksx djds ge izkIr
djrs gSA
X1 = 0 + 12
a (10)2 , i.e., X1 = 50 a
Similarly, blh izdkj X2 = distance travelled in the next 10 s
X2 = vxys 10 s esa r; dh xbZ nwjh
So blfy,, X2 = (0 + 10a ) 10 + 12
a (10)2
So blfy,, X2 = 100 a + 50 a
or ;k, X2 = 150 a
and, X3 = distance travelled in the last 10 s
o, X3 = vfUre 10 s esa r; dh xbZ nwjh
Soblfy,, X3 = (10 a + 10 a) 10 + 12
a (10)2
or ;k, X3 = 200a + 50a
or ;k, X3 = 250a
Hence vr%, X1 : X2 : X3 = 50a : 150 a : 250a
= 1 : 3 : 5
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4. Two balls of equal ........................ leku nzO;eku dh nks xsanksa ........................ Sol.
f}rh; xsan
t=3S
u = 40 m/s , g = 10 m/s2 Let t be time taken by the first ball to reach
the highest point. ekuk izFke xsan }kjk mPpre fcUnq rd igq¡pus esa
fy;k x;k le; t gS V = u � gt ; 0 = 40 � 10 t ; t = 4 s From figure second ball will collide with first
ball after 3 second, there fore the height of collision point
= height gained by the second ball in 3 sec vr% Vdjkus okys fcUnq dh Å ¡pkbZ = f}rh; xsan }kjk 3 sec esa izkIr Å ¡pkbZ
= 40 (3) � 12
(10) (3)2 = 120 � 45 = 75 m 5. A balloon is moving ........................ ,d xqCckjk 10 eh0@lS0 ........................ Sol.
A
B
10 m/s
t = 11 secH
As pwafd s = ut + 12
at2
� H = 10 × 11 � 5 × (11)2 � H = 110 � 605 H = 495 m Aliter : oSdfYid fof/k
10ms�1
10ms�1
u=�10ms�1
t=11s
At the time of release, velocity of stone will
be same as that of balloon, hence eqDr djus ds le; iRFkj dk osx xqCckjs ds osx ds
rqY; gksxk vr% u = � 10 ms�1 , t = 11 s xqCckjs dh Å ¡pkbZ gksxh
h = ut + 12
gt2
= (� 10)×11 + 12
(10)(11)2 =�110 + 605 = 495 m
6. The acceleration time ........................ lh/kh js[kk esa fLFkjkoLFkk ........................ Sol.
aAvg vkSlr Roj.k = vt
= 020
= 0
From 0 to 20 time interval velocity of particle doesn't change it's direction.
Area under v�t curve is not zero. As the magnitude of area under v � t graph
from t = 0 to 10 is same as from t = 10 to 20, hence the average speed in both the intervals will be same.
0 ls 20 lsd.M vUrjky esa d.k dk osx viuh fn'kk ifjofrZr ugh djrkA
v�t vkjs[k ds vUrxZr {ks=kQy 'kwU; ugh gSA t = 0 ls 10 lsd.M rFkk t = 10 ls 20 lsd.M ds
fy, v � t vkjs[k ds vUrxZr {ks=kQy leku gSA vr% bu nksuks le; vUrjkyksa esa vkSlr pky leku gksxhA
7. Mark the incorrect statement ........................ ,d d.k lh/kh js[kk esa xfr dj ........................
Sol. If the velocity (u) and acceleration (a) have
opposite directions, then velocity (v) will decrease, therefore the object is slowing down.
If the position (x) and velocity (u) have opposite sign the position (x) reduces to become zero. hence the particle is moving towards the origin.
If a v 0
speed will increase. If velocity V = 0 , t1 < t < t2
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Hence; acceleration a = Vt
= 0 ;
t1 < t < t2 Therefore if the velocity is zero for a time
interval, the acceleration is zero at any instant within the time interval.
[acc, a = dvdt
v = u + at ]
Now , v = 0 a = 0 a = � u/t acceleration may not be zero when vel. 'V' = 0, 'c' is incorrect.
;fn osx a v 0
(u) rFkk Roj.k (a) foifjr fn'kkvksa esa gS rks vfUre osx (v) ?kVsxk vr% d.k /khek gks tk;sxkA
;fn fLFkfr (x) rFkk osx (v) foifjr fn'kkvksa esa gS rks fLFkfr ?kVdj 'kwU; gks tk;sxh vr% d.k ewy fcUnq dh vksj xfr dj jgk gSA
;fn rks pky c<sxh . ;fn osx V = 0 , t1 < t < t2
vr% Roj.k a = = 0 ; t1 < t t2
blfy, ;fn fdlh le; vUrjky esa osx 'kwU; gS rks ml le; vUrjky esa fdlh Hkh {k.k ij Roj.k 'kwU; gSA
[ a = dvdt
v = u + at ]
vc , v = 0 a = 0 a = � u/t Roj.k 'kwU; ugh Hkh gks ldrk gSA ;fn osx 'V' = 0, 'c'
xyr gSA 8. A particle is initially at ........................ fojkekoLFkk ls çkjEHk gqvk ........................
Sol. Area under acceleration-time graph gives the
change in velocity.
Roj.k≤ vkjs[k dk {ks=kQy] osx esa ifjorZu
nsrk gSA
Hence, vr% vmax = 12
× 10 × 11 = 55 m/s
Therefore, the correct option is (3)
vr% (3) lgh gSA
9. The coordinates of a ........................ xfr'khy d.k ds fdlh ........................ Sol. x = t3, y = t3
Vx = dxdt
= 3t2
Vy = dydt
= 3t2
Resultant velocity ifj.kkeh osx
V = 2 2x yv v
= 2 4 2 49 t 9 t
= 3t2 2 2a 10. A particle is moving........................ ,d d.k 5 ms�1 pky ........................ Sol. 1
�v �5i
2�v 5 j
2� �v 5 j 5i
v
= 5 2
a = v 5 2 1t 10 2
ms�2
W E
vN
1v 1v
2v
S
For direction, fn'kk ds fy,
tan = �55
= � 1
Average acceleration is 1
2 ms�2
towards north-west.
vkSlr Roj.k 1
2 ms�2 mÙkj if'pe dh
vksj gSA 11. During projectile motion........................ iz{ksI; xfr esa iFk ........................ Sol. Gravitational acceleration is constant near
the surface of the earth. iFoh dh lrg ds ikl xq:Roh; Roj.k fu;r jgrk
gSA
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12. The velocity of projection........................ ,d iz{ksI; dk iz{ksi.k........................ Sol. xu
= 6 �l + 8 �j
x�u 6l
uy = 8 �j
R = x y2u u
g =
2 6 810
= 9.6
13. A body is projected ........................ ,d oLrq ,d Å ¡ph ........................
Sol. tan45° = y
x
v
v
vy = vx = 18m/s Ans. 14. A ball is horizontally ........................ ,d xsn dks 45º dks.k........................
Sol. R = 2
2
vsin(2 ) � sin
gcos
Putting ;gka = 45º & = � 3
4 4
R =
2
2
v 3sin 2 � sin
4 4 41g
2
sx = 2 2v 2 1 v
�2 � 2 2g g2
(�ve sign indicates that the displacement is in �ve x direction)
(�ve fpUg ;g n'kkZrk gS fd foLFkkiu _ .kkRed x
fn'kk esa gksxk )
Range ijkl = 2v
2 2g
Ans "(4)"
Alternate II method
�4
& 4
R = 2
2
usin (2 ) � sin
g cos
= 2
2
usin � sin �
4 41g
2
R =22 2 u
g (along +ve x die.)
(+ve x fn'kk esa ) III Method
ux = u cos , T = 2 usingcos
, ax = g sin
; ay = � g cos
sx = ux t +12
axt2
= (u cos ) 2
2 u sin 1 2 u sin(gsin )
g cos 2 gcos
Let ;fn = = 45º So blfy,, sx
=2 2
2
u 2 1 1 1 2.2 ug
g 2 g2 2
= 22 u
[1 1]g2
sx = 2u
2 2g
Ans "(4)"
15. On an inclined plane ........................ 30º mUu;u dks.k okys ........................ Sol. u = 10m/s
Time of flight on the incline plane ur ry ij mM~M;u dky
60o
30o
u
T= 2u sing cos
given fn;k gS =30o & =30o & u =
10 3 m/s
T =o
o
2 10 3 sin30
10 cos 30
so vr% T= 2 sec .
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16. A particle moves in ........................ x-y ry esa xfr'khy d.k........................ Sol. ax = 2 m/s2 ; ay = 0 ux = 8 m/s uy = � 15 m/s.
V
= Vx�i + Vy
�j
Vy = uy + ay t Vy = � 15 m/s Vx = ux + ax t Vx = 8 + 2 t
V = [(8 + 2 t) �i � 15 �j ] m/s. Ans.
17. A stone projected at ........................ ,d iRFkj tehu ls 60º ........................ Sol. Let initial and final speeds of stone be u and v. ekuk iRFkj dh izkjfEHkd o vfUre pky u rFkk v gSA v2 = u2 � 2gh .........(1) and vkSj v cos 30° = u cos 60° ..........(2) solving 1 and 2 we get (1) o (2) dks gy djus ij u = 3gh
18. A particle is projected at 60º ........................ ,d d.k dks {kSfrt ls ........................
Sol. 2H
1mv
2 = K, 21
m(vcos60)2
= 21 v
m2 4
= 21 1mv
4 2
= K4
19. Shown in the figure are ........................ ?kj ls Ldwy tkrs gq, ........................ Sol. 1 = slope of C1 line = constant 2 = slope of C2 line = constant 1 � 2 0 but constant
1 = C1 js[kk dh <ky = fu;rkad 2 = C2 js[kk dh <ky = fu;rkad
1 � 2 0 ysfdu fu;rkad 20. A boat which can move ........................ ikuh ds lkis{k 5 m/s dh ........................
Sol. Vb = 2 25 4 = 3 m/s
t = 4803
= 160 s
21. A man walks in rain with ........................ ,d vkneh 5 kmh�1 osx ls ........................
Sol. r yV j
m�5i
r m y� �V � V (�5) i j
tan = 1 = y
5
so y = 5 km/hr
22. Two particles A and ........................ nks d.k A rFkk B Øe'k% ........................ Sol.
tan = 1
2
v
v
rmin = d sin
= 1
2 21 2
vd .
v v
23. A body is thrown up ........................ ,d oLrq dks Å ij dh ........................
Sol. We have, ge tkurs gSa] Srel = urelt + 12
arel t2
0 = ut � 12
(a + g) t2
a = 2ut� g =
2u gtt
24. Which figure represents........................ fp=k esa iznf'kZr m nzO;eku ........................ Sol. Force exerted by string is always along the
string and of pull type. When there is a contact between a point
and a surface the normal reaction is perpendicular to the surface and of push type.
Mksjh }kjk vkjksfir fd;k x;k cy ges'kk Mksjh ds vuqfn'k rFkk f[kapko ds :i esa gksrk gSA tc ,d fcUnq rFkk ,d lrg ds lEidZ gksrk gS rks vfHkyEc izfrfØ;k lrg ds yEcor~ rFkk ncko ds :i esa gksrk gSA
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25. Two forces of 6N and........................ fp=kkuqlkj ?k"kZ.k jfgr........................
Sol.
Both blocks are constrained to move with same acceleration.
nksuks CYkkWd leku Roj.k ls xfr djus ds fy, cfU/kr gSA
6 � N = 2a [Newtons II law for 2 kg block] [2 kg CykWd ds fy, U;wVu dk f}rh; fu;e] N � 3 = 1a [Newtons II law for 1 kg block]
[1 kg CYkkWd ds fy, U;wVu dk f}rh; fu;e] N = 4 Newton 26. A mass M is suspended ........................ fp=kkuqlkj ,d M nzO;eku ........................ Sol.
Point A is mass less so net force on it most
be zero otherwise it will have acceleration. fcUnq A nzO;ekujfgr gS blfy;s bl ij dqy cy
'kwU; gksxk vU;Fkk bldk Roj.k gksxkA F � Tsin = 0 [Equilibrium of A in
horizontal direction] [{kSfrt fn'kk esa A dh lkE;koLFkk]
T = F
sin
27. A body of mass 8 kg ........................ ,d 8 kg nzO;eku dh ........................ Sol.
T2 � 8g = 8a [Newton�s II law for 8 kg
block] [8 kg CykWd ds fy, U;wVu dk f}rh; fu;e] T2 = 8 × 2.2 + 8 × 9.8 = 96 N T1 � 12 g � T2 = 12 a [Newton�s II law for 12 kg
block] [12 kg CykWd ds fy, U;wVu dk f}rh; fu;e] T1 = 12 × 2.2 + 12 × 9.8 + 96 T1 = 240 N 28. A block is dragged on ........................ ,d CykWd dks fpdus ry ........................ Sol.
The length of string AB is constant. Mksjh AB dh yEckbZ fu;r gSA speed A and B along the string are same
u sin = V Mksjh ds vuqfn'k A rFkk B dh pky leku gS u
sin = V
u sin = V u = V
sin
29. STATEMENT-1 : A particle having ............... oDrO; 1 % ,d d.k ftldk ........................ Sol. Assertion is false , because direction of
velocity is not specified .It is not necessary that when acceleration is negative positive slow down .As when both velocity and acceleration are in opposite directions, then particle will increase its speed.
dFku vlR; gS D;ksfd osx dh fn'kk ugh crkbZ xbZ gSA ;g t:jh ugh gS fd tc Roj.k _ .kkRed gks rks d.k /khek gksA D;ksfd tc osx o Roj.k nksuks foijhr fn'kk es gksa rc d.k dh pky c<sxhA
30. STATEMENT-1 : Two stones........................ dFku-1 : nks iRFkj tehu ........................ Sol. Both the stones cannot meet (collide)
because their horizontal component of velocities are different. Hence statement I is false.
nksuks iRFkj vkil esa Vdjk ugha ldrs gS D;ksafd muds osxksa ds {kSfrt ?kVd vyx&vyx gSA vr% dFku-I vlR; gSA
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PART- B
CHEMISTRY
31. The ratio of the difference in energy ..................
fdlh H ds leku Lih'kht esa çFke ------------------------
Sol. Use : E1 � E2 / E2 � E3
32. If the de-Broglie wavelength ......... ... .. .
;fn H-ijek.kq dh 2nd d{kk esa ifjØe.k -------------------
Sol. 2r = n
Here (;gk¡), n = 2. 2r = 2x r = x
.
33. The uncertainty in position and ................
;fn fdlh ,d d.k dh fLFkfr rFkk osx ------------------
Sol. x × (mv) = h4
= 0.527 × 10�34
0.5 × 10�10 × m × 5.27 × 10�24 = 0.527 × 10�34
m = 0.2 Kg.
34. The wavenumber of the spectral line ............
He+ vk;u dh ckej Js.kh esa izkIr U;wure -----------------
Sol. Z = 2 n1 = 2 n2 =
= R (2)2 2 2
1 1
2
= R
35. The value of azimuthal quantum ..............
DokaVe la[;k n = 4, m = �3 okys ,d d{kd --------------
Sol. n = 4, = 0, 1, 2, 3
For m = �3, the only possible value of = 3
gy . n = 4, = 0, 1, 2, 3
m = �3 ds fy, dk laHkkfor eku = 3 gSA
36. In a sample of H-atom electrons ...............
H-ijek.kq ds uewus esa bysDVªkWu 5th mÙksftr ----------------
Sol. 65432n = 1
maximum number of lines produced = 6 (6 1)
2
= 15
out of these 15 lines,
5 lines belong to ultra violet region and 4 lines
are in visible region and rest are in infrared
region
Sol. 65432n = 1
mRikfnr js[kkvksa dh vf/kdre la[;k=6 (6 1)
2
= 15
15 js[kkvksa esa ls]
5 js[kk,¡ ijkcSaxuh {ks=k ls] 4 js[kk,¡ n'; {ks=k ls rFkk
'ks"k js[kk,¡ vojDr Js.kh ls gSaA
37. Which of the following set of quantum ...............
4f d{kd esa ,d by sDVªkWu ds fy, DokaVe --------------------
Sol. For 4 orbital electrons, n = 4
= 3 (because ) m = + 3, + 2, + 1, 0, �1, �2,
�3 s = + 1/2.
gy- 4 d{kh; bysDVªkWu ds fy;s] n = 4
= 3 (D;ksafd ) m = + 3, + 2, + 1, 0, �1, �2,
�3 s = + 1/2.
39. One atom of an element x weigh ..................
rRo x ds ,d ijek.kq dk Hkkj ...................
Sol. Atomic weight of an element
x = 6.643 × 10�23 × NA = 40
Number of moles of x = 20 1000
40
= 500
gy- ,d rRo dk ijek.kq Hkkj
x = 6.643 × 10�23 × NA = 40
x ds eksyksa dh la[;k = 20 1000
40
= 500
40. For reaction P + 2Q 3R, if reaction .................
vfHkfØ;k P + 2Q 3R, ds fy, ...............
Sol. P + 2Q 3R 0.1/2 = Rmole / 3
R = 0.15 mole
41. 2K + 2 + 22 HNO3 2HO3 + 2KO3 +
22NO2 + 10H2O
If 3 mole of K & 2 moles 2 ..................
;fn K ds 3 eksy rFkk 2 ds 2 eksy HNO3 --------------
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Sol. KI is limiting reagent
3 mole of KI will give 33 mole of NO2
according to stoichiometry.
Volume of NO2 at STP = 33 22.04
= 739.2 Lt
Sol. KI lhekUr vfHkdeZd gSaA
vfHkfØ;k dh jllehdj.kferh lsµ 3 eksy KI, }kjk
33 eksy NO2 nsrs gSA
Volume of NO2 at STP = 33 22.04
= 739.2 Lt
42. A solution containing 0.1 mol ................
/kkrq DyksjkbM MClx ds 0.1 eksy ;qDr -----------------
Sol. MClx + AgNO3 AgCl + MNO3
POAC on Ag
500
1000 × 0.8 = 1 × mole of AgCl
mole of AgCl = 0.4 ...........(1)
POAC on Cl
0.1 × x = 1 × mole of AgCl = 0.4........(1)
mole of AgCl = 0.1 x ....................(2)
put eq (2) in eq (1)
0.1 x = 0.4
x = 4
gy % MClx + AgNO3 AgCl + MNO3
Ag ij POAC
500
1000 × 0.8 = 1 × AgCl ds eksy
AgCl ds eksy = 0.4 ...........(1)
Cl ij POAC
0.1 × x = 1 × AgCl ds eksy = 0.4.......(1)
AgCl ds eksy = 0.1 x ....................(2)
leh- (1) esa (2) dks j[kus ij
0.1 x = 0.4
x = 4
43. The atomic mass of an element is 27 ...........
,d rRo dk ijek.kq Hkkj 27 -------------------
Sol. It may be AlCl3
V.D = 3AlClM.Wt
2
V.D = 133.5
2 = 66.75
gy % ;g AlCl3 gks ldrk gS\
ok-?k- = 3AlClM.Wt
2
ok- ?k- = 133.5
2 = 66.75
44. A hydrocarbon contains 80% .............
;fn ,d gkbMªksdkcZu esa 80% --------------------- Sol. ratiosimplest ratio
C 80 /12 6.6680 6.66 / 6.66 1 1
3H 20 / 6.66 320 20 /1 20
Hence hydrocarbon = n × CH3 (n = 1, 2, 3.....)
i.e. CH3, C2H6.................
gy %
C 80 /12 6.6680 16.66 / 6.66 1 1
3H 20 / 6.66 320 20 /1 20
lk/kkj.kre vuqikr vuqikr
vr % gkbMªkssdkcZu = n × CH3 (n = 1, 2, 3........)
i.e. CH3, C2H6.................
45. 20 gm. CaCO3 on decomposition ...............
20 xzke CaCO3 dks fo?kfVr -----------------
Sol. The decomposition of CaCO3 takes place as
below
3 256 gm 22.4 lit100 gm
CaCO CaO CO
100 gm CaCO3 at STP produce = 22.4 lit CO
2
20 gm CaCO3 at STP produce =
22.420
100
= 4.48 lit CO2
gy- 3 256 gm 22.4 lit100 gm
CaCO CaO CO
100 xzke CaCO3 nsrk gS 22.4 yhVj CO
2
20 xzke CaCO3 nsxk CO
2 =
22.420
100
= 4.48 yhVj
46. How many alcohols give immediate ............
(C5H
12O) v.kqlw=k okys fdrus ,YdksgkWy -------------
Sol.
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47. Which of the following compound ...............
fuEu esa ls dkSulk ;kSfxd VkWysu ---------------------
Sol. Aldehyde and Ketones having �C�CH3
O
group always gives +ve tollen's test.
,YMhgkbM rFkk dhVksu tks fd �C�CH3
O
lewg j[krs gS lnSo VkWysu ifj{k.k +ve nsrs gSA
48. Which is incorrect match with respect .............
iz;ksx'kkyk ijh{k.k ds fy, fuEu esa ls dkSu ----------
Sol. It is fact.
rF; gSA
49. How many structural isomers of C5H
10 ...........
C5H
10 ds fdrus lajpukRed leko;oh ----------------
Sol. C5H10
D.U. = 1
Unsaturated compound which having >C=C<
and �CC� always shown bromine water test
positive.
vlarIr ;kSfxd tks fd >C=C< rFkk �CC� j[krs
gS lnSo czksehu ty ifj{k.k /kukRed nsrs gSA
C�C�C�C=C
C�C�C=C�C
C=C�C�C
C
C�C=C�C
C
C�C�C=C
C 50. How many structural isomeric ketones ...............
(C5H
10O) v.kqlw=k okys fdrus lajpukRed --------------
Sol. ,
51. On oxidative ozonolysis of 3-Methylhex .............
3-esfFkygsDl -3-bZu ds vkWDlhdkjh -------------------
Sol.
CH3�CH2�C=CH�CH2�CH3
CH3
(i) O H O3 2�H O2 2
CH3�CH2�C=O
CH3
+ O=CH�CH2�CH3
CH3�CH2�C=O
CH3
+H2O2
+ CH3�CH2�COOH
52. How many monochloro structure ............
tc 3-esfFkyisUVsu dh vfHkfØ;k lw;Z ------------------
Sol.
53. Which of the following hydrocarbon ................
dkSulk gkbMªksdkcZu ftad dh mifLFkfr ---------------------
Sol. (1) CH3�CH=C=CH�CH3 (i) O H O3 2
(ii) Zn
CH3�CH=O + CO2
Acetaldehyde
,flVsfYMgkbM
(2) CH3�CH=CH�CH=C(CH3)2 (i) O H O3 2
(ii) Zn
CH3CH=O + CHO
CHO + (CH3)2C=O
Acetaldehyde glyoxal acetone
,flVsfYMgkbM XykbvkWDtsy ,flVksu
(3) (CH3)2C=C=CH2 (i) O H O3 2
(ii) Zn
(CH3)2C=O +
CO2 + H�CHO
Acetone Formaldehyd
,flVksu QkeZsfYMgkbM
(4) CH3�C=C=C�CH3
CH3 CH3
(i) O H O3 2(ii) Zn
CH3�C=O + CO2
CH3 Acetone
,flVksu
54. Which is incorrect order ..............
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SOLEBPT2160815-10
fuEu esa ls dkSulk ijekf.od f=kT;kvksa --------------------
Sol. Refer notes At. radii = B < Al ~� Ga < In
uksV~l ns[ksaA ijekf.od f=kT;k = B < Al ~� Ga < In
55. Which statement is correct for the ..............
f}rh; vkorZ rRoksa (Be, B, C, N, O, F) --------------
Sol. First ionization energy of Be(2s2) in higher than
Boron (3s22p1) but lower than carbon (2s22p2)
due to full field electronic configuration of
Be(2s2).
Be(2s2) izFke vk;uu Å tkZ B(3s22p1) ls vf/kd
gksrh gS ysfdu dkcZu (2s22p2) dh rqyuk esa de
gksrh gSA
D;ksafd Be(2s2) iw.kZiwfjr bysDVªksfud foU;kl
j[krk gSA
56. The set representing the correct.................
fuEu lewg esa izFke vk;uu foHko ---------------------
Sol. Down the group the effective nuclear charge
remains almost constant. But down the group
with increasing atomic number, the number of
shells increase and thereby atomic size
increases. As a result, the distance of valence
shell electron from nucleus increases, attraction
between them decreases and therefore
ionization energy decreases.
gy- oxZ esa uhps tkus ij izHkkoh ukfHkdh; vkos'k yxHkx
fLFkj jgrk gS] ijUrq oxZ esa uhps tkus ij ijek.kq
Øekad c<+us ij dks'kksa dh la[;k esa of) gksrh gS
vkSj ijek.kq dk vkdkj c<+rk gSA bl dkj.k]
ukfHkd ls la;ksth dks'k bysDVªkWu dh nwjh c<+rh gS
vkSj muds chp vkd"kZ.k ?kVrk gS] ftlds
ifj.kkeLo:i vk;uu Å tkZ ?kVrh gSA
57. Element with atomic number 38 ..............
vkorZ lkj.kh esa ijek.kq Øekad 38 ----------------------
Sol. Refer Periodic table.
vkoZr lkj.kh esa ns[ksaA
58. Which gas is released ................
nh xbZ vfHkfØ;k esa dkSulh xSl -------------------
Sol. SO3H NO2
OH
Na
SO3Na NO2
ONa
+ H2
59. Which is the position isomers ............
;kSfxd dk fLFkfr ----------------
Sol. (1) Identical
(2) Position
(3) Chain
(4) Homologous
(1) le:i
(2) fLFkrh
(3) Ja[kyk
(4) ltkrh;
60. Which is the chain isomers ..............
;kSfxd dk Ja[kyk --------------------
Sol. Only 1st is chain isomer while 2nd and 3rd are
positional isomers. 4th is functional isomer.
dsoy 1st Ja[kyk leko;oh gS tcfd 2nd rFkk 3rd
fLFkfr leko;oh gSA 4th fØ;kRed leko;oh gSA
PART- C
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SOLEBPT2160815-11
MATHEMATICS 61. If a1, a2, .......... an are.................... ;fn a1, a2, .......... an ....................
Sol. 21
n1
aa
aa
+
32
n1
aa
aa
+...... +
n1�n
n1
aa
aa
= 1a + na
1�nn
1�nn
n1�n23
23
3212
12
21 a�a
a�a
aa
1.....
a�a
a�a
aa
1
a�a
a�a
aa
1
= 1a + nad
]a�a.......a�aa�a[ 1�nn2312
= 1 n n 1[ a a ] [ a � a ]
d
=
d
a�a 1n
[as. a2 � a1 = a3 � a2 = a4 � a3 = .....d]
= d
d)1�n( = n � 1
62. log(0.3)(x � 1) < log(0.09) ....................
log(0.3)(x � 1) < log(0.09)(x � 1) gks x .................
Sol. log0.3(x � 1) < 2)3.0(log (x �1)
log0.3(x � 1) < 21
log0.3(x � 1)
log0.3(x � 1)2 < log0.3(x �1) (x � 1)2 > (x � 1) x2 � 3x + 2 > 0 (x � 1)(x � 2) > 0 xt(�, 1) (2, ) but ysfdu x � 1 > 0 x > 1 so blfy, x (2, )
63. If c�b2log
= a�c3log
= b�a5log
....................
;fn c�b2log
= a�c3log
= b�a5log
rc ....................
Sol. Let ekuk c�b2log
= a�c3log
= b�a5log
= k
log 2 = k(b � c) log 3 = k(c � a) log 5 = k(a � b) let 2a 3b 5c = x taking log both sides nksuksa rjQ log ysus ij a log 2 + b log 3 + c log 5 = log x a k(b � c) + bk(c � a) + ck(a � b) = log x k[ab � ac + bc � ab + ac � bc] = logx logx = 0 x = 1 2a 3b 5c = 1
64. The value of x, |x2 + 3x| ....................
x, |x2 + 3x| + x2 � 2 0 ds ....................
Sol. case fLFkfr- CasefLFkfr-
x2 + 3x 0 x2 + 3x 0
x (�, � 3] [0, ) x [�3, 0]
2x2 + 3x � 2 0 �x2 � 3x + x2 � 2 0
(2x � 1)(x + 2) 0 3x + 2 0
x (�, � 2]
,
21
x � 32
x (, �3]
,
21
x
32
�,3�
x
32
�,�
,
21
x R �
21
,32
�
65. If P(x) = ax2 + bx + c & ....................
;fn P(x) = ax2 + bx + c vkSj .................... Sol. P(x) = ax2 + bx + c D1 b2 � 4ac Q(x) = � ax2 + dx + c D2 d2 + 4ac D1 + D2 = b2 � 4ac + d2 + 4ac D1 + D2 = b2 + d2 0 D1 + D2 0 So at least one of D1 and D2 0 blfy, de ls de ,d D1 vkSj D2 0 So at least two real roots blfy, de ls de nks okLrfod ewy gSA 66. If roots of the equation.................... ;fn lehdj.k (a2 + b2)x2 ....................
Sol. D 4(ac + bd)2 � 4(a2 + b2)(c2 + d2) = 0 (ac + bd)2 = (a2 + b2)(c2 + d2) a2c2 + b2d2 + 2abcd = a2c2 + b2c2 + a2d2 + b2d2 b2c2 + a2d2 � 2abcd = 0 (bc � ad)2 = 0 bc = ad 67. If a, b, c, d and x are .................... ;fn a, b, c, d vkSj x fofHkUu .................... Sol. (a2 + b2 + c2)x2 � (ab + bc + cd)x + b2 + c2 + d2 0 (ax � b)2 + (bx � c)2 + (cx � d)2 0 ax � b = 0 bx � c = 0 cx � d = 0
x = ab
= bc
= cd
= k(common ratio lkoZvuqikr)
so a, b, c, d are in G.P.
blfy, a, b, c, d xq.kksÙkj Js.kh esa gSA
68. Solution of 1x2�x2)32( ....................
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SOLEBPT2160815-12
1x2�x2)32( + 1�x2�x2
)3�2( ....................
Sol. 1x2�x2)32( + 1�x2�x2
)3�2( = 3�2
4
1x2�x2)32( (2 � 3 ) + x2�x2
)3�2( = 4
x2�x2)32( (2 + 3 )(2 � 3 ) + x2�x2
)3�2(
= 4
x2�x2)32( + x2�x2
)3�2( = 4
let ekuk x2�x2)32( = y then rc
x2�x2)3�2( =
y1
y + y1
= 4 y = 2 ± 3
y2 � 4y + 1 = 0
x2�x2)32( = 2 + 3 or 2 � 3
x2 � 2x = ± 1 x2 � 2x + 1 = 0 (x � 1)2 = 0 &
x2 � 2x � 1 = 0 x = 2
82 = 1 ± 2
so x = 1 ± 2 , 1
69. If a1, a2, a3 ....... a2k........................
a1, a2, a3 ....... a2k lekUrj Js.kh esa........... Sol. a2 = a1 + d a2k = a2k�1 + d S = 2
1a � 22a + 2
3a � 24a + .......... 2
k2a
= ( 21a � 2
2a ) + ( 23a � 2
4a ) + ........ ( 21�ka � 2
k2a ) S = � d(a1 + a2) + (a3 + a4) + ....... (a2k�1 + a2k) S = � d[a1 + a2 + a3 + a4 + .......... a2k]
S = � d
2k2
[a1 + a2k]
S = � k
1�k2a�a 1k2 [a1 + a2k] a2k = a1 + (2k � 1)d
S = 1�k2
k( 2
1a � 2k2a ) d =
)1�k2(
a�a k2
70. If ratio of sum of m .................... ;fn lekUrj Js.kh ds m ....................
Sol. Sm = 2m
[2a + (m �1)d]
Sn = 2n
[2a + (n � 1)d]
n
m
SS
= ]d)1�n(a2[
2n
]d)1�m(a2[2m
= 2
2
n
m
d)1�n(a2d)1�m(a2
=
nm
; d
21�n
a
d2
1�ma
= nm
d)1�n(ad)1�m(a
=
1�n21�m2
n
m
TT
= 1�n21�m2
71. If sum of the A.M. & G.M. .................... ;fn nks fofHkUu /kukRed la[;k,¡ ....................
Sol. 2
ba + ab = a � b
a + b + 2 ab = 2a � 2b (a � 3b)2 = 4ab a2 + 9b2 � 10ab = 0 (a � 9b)(a � b) = 0 a = 9b as a � b 0
ba
= 19
72. Suppose x1, x2 be the .................... ekuk fd lehdj.k ax2 + bx + c = 0............
Sol. x1 + x2 = � ab
x1 = a
x3 + x4 = � pq
x2 = a + d
x3 = a + 2d x4 = a + 3d (x3 + x4) � (x1 + x2)
(a + 3d + a + 2d) � (a + a + d) = ab�
pq
4d = ab
� pq
d = 41
pq
�ab
73. Sum of positive roots .................... lehdj.k (x2 � 12x + 35) .................... Sol. (x2 � 12x + 35)(x2 + 10x + 24) = 504 (x � 5)(x � 7) (x + 6)(x + 4) = 504 (x � 7)(x + 6)(x � 5)(x + 4) = 504 (x2 � x � 42)(x2 � x � 20) = 504 x2 � x = t (t � 42)(t � 20) = 504 x2 � x = 6 t2 � 62t + 840 � 504 = 0 x2 � x � 6 = 0 t2 � 62t + 336 = 0 (x � 3)(x + 2) = 0 t2 � 56t � 6t + 336 = 0 x = 3 & � 2 (t � 6)(t � 56) = 0 x2 � x = 56 t = 6 or 56 x2 � x � 56 = 0 (x � 8)(x + 7) = 0 x = � 78 x1 + x2 = 8 + 3 = 1
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SOLEBPT2160815-13
74. Solution of the equation ....................
lehdj.k x x�1x5 (64) = 2000 ....................
Sol. x�1
x x5 .(64) = 53. 16
5x. 3(x�1)
x(4) = 53. 42
x = 3
75. If the equations ....................
;fn x2 + ax + b = 0 ....................
Sol. Let be a common root of
x2 + ax + b = 0 and x2 + bx + a = 0. Then,
ekuk x2 + ax + b = 0 vkSj x2 + bx + a = 0 dk
mHk;fu"B ewy gS] rks
2 + a + b = 0 andvkSj 2 + b + a = 0
= 1
Putting = 1, either of the two, we get
a + b = � 1.
= 1 j[kus ij, a + b = � 1.
76. Roots of the quadratic ....................
f}?kkr lehdj.k ....................
Sol. 2 2x 4x 3 x 6x 8 0 ,
2x 1 2x 2 3 3 8 0
Discriminant. foospd
2
D 4 2 3 4 1 3 8
2D 4 1
If R then D > 0 ;fn R rc D > 0
so root of given quadratic always real blfy, nh
xbZ f}?kkr lehdj.k ds ewy lnSo okLrfod gksxsaA
Answer A.
77. The solution set of ....................
vlfedk (x + 1)2 > (x + 3) ....................
Sol. (x + 1)2 > x + 3 or ;k x2 + x � 2 > 0
or ;k (x + 2)(x � 1) > 0
x < � 2 or ;k x > 1
78. The number of values....................
lehdj.k |x + 1| + |x � 1| = k ....................
Sol. |x + 1| + |x � 1| =
�2x , x �1
2 , �1 x 1
2x , x 1
for infinite solutionvuUr gy ds fy, k = 2
79. If |x| � |x�2| = 2 then ....................
;fn |x| � |x �2| = 2, rc 10 ....................
Sol. |x| = 2 + |x�2|
|a+b| = |a| + |b|
a.b 0
2(x � 2) 0 x 2
4 such solutions 4 gy gSA
80. Common solution of ....................
lehdj.k 2(x �3x 2)| x � 2 |
= 1 ,oa ....................
Sol. 2(x �3x 2)(x � 2)
= 1
CasefLFkfr-I x2 � 3x + 2 = 0 x = 1 or x = 2
(RejectedvLohdk;Z)
CasefLFkfr-II x � 2 = ±1 x = 2 ± 1
x = 3,1
andvkSj x(x � 2) 0 x [0, 2]
common solution mHk;fu"B gy x = 1
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SOLEBPT2160815-14
81. Number of positive integers ....................
vlfedk 2(x � 2)(�x 1)(x 1)
(x 1)
....................
Sol. 2(x � 2)(�x 1)(x 1)
(x 1)
0
2(x � 2)(x �1)(x 1)
(x 1)
0
x (�, � 1) [1, 2]
82. Sum of the all possible....................
lehdj.k ||x � 1| � 3| = 2 ds ....................
Sol. |x � 1| � 3 = 2 or ;k |x � 1| � 3 = � 2
|x � 1| = 5 or ;k |x � 1| = 1
x � 1 = ± 5 or ;k x � 1 = ±1
x = 6, � 4 or ;k x = 2, 0
sum ;ksxQy = 4
83. If {x} = 2.1 then complete....................
;fn {x} = 2.1 gS rks x ds ....................
Sol. {x} = 2.1 not possible lEHko ugha
as pwfd 0 {x} < 1
so blfy, x
84. 3 3 3(171) � (123) � (48)
9 19 123 16 3 ....................
Sol. Let a = 171, b = �123, c = � 48
then a + b + c = 0
a3 + b3 + c3 = 3abc
3 3 3(171) � (123) � (48)
9 19 123 16 3
= 3 171 (�123)(�48)
9 19 123 16 3
= 3
Hindi. ekuk a = 171, b = �123, c = � 48
rc a + b + c = 0
a3 + b3 + c3 = 3abc
3 3 3(171) � (123) � (48)
9 19 123 16 3
= 3 171 (�123)(�48)
9 19 123 16 3
= 3
85. If Hn = 1 + 21
+ 31
+ ........+ ....................
;fn Hn = 1 + 21
+ 31
+ ........+ ....................
Sol. Hn = 1 + 12
+ 13
+ ...... + 1n
1 + 32
+ 53
+ ....... + 2n 1
n
Here ;gk¡ Tn = 2n 1
n
Tn = 2 � 1n
T1 = 2 � 11
; T2 = 2 � 12
T3 = 2 � 13
Tn = 2 � 1n
Sn = T1 + T2 + T3 + ....... Tn
Sn = 2n � 1 1 1
1 .....2 3 n
Sn = 2n � Hn
86. If positive numbers .................... ;fn /kukRed la[;k,sa .................... Sol. HM between a and c = b
and GM = ac Also HM between b and d = c
and GM = bd But GM HM
ac b and bd c
ac bd bc ad bc
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SOLEBPT2160815-15
Hindi. a ,oa c ds e/; gjkRed ek/; = b
rFkk xq.kksÙkj ek/; = ac
blh izdkj b ,oa d dk gjkRed ek/; = c
,oa xq.kksÙkj ek/; = bd
ysfdu GM HM
ac b ,oa bd c
ac bd bc
ad bc
87. The value of 91/3 . 91/9. 91/27....................
91/3 . 91/9. 91/27 ......rd ....................
Sol. We have to find value of 91/3 . 91/9 . 91/27 .... = x
(say)
We can see that powers are in G.P. and |r| < 1
x =
13
11
39
= 91/2 = 3
Hindi. gesa 91/3 . 91/9 . 91/27 ......... = x (ekuk) dk eku
Kkr djuk gSA
?kkrs xq.kksÙkj Js<+h esa gS rFkk |r| < 1
x =
13
11
39
= 91/2 = 3
88. Let a, b, c, d, e ..................... Statement-1 : If a, b, c, d, e .................... ekuk fd a, b, c, d, e /kukRed .................... oDrO;-1 : ;fn a, b, c, d, e .................... Sol. a, b, c, d, e A.P.
a1
, b1
, c1
, d1
, e1
H.P.
bcde, acde, abde, abce, abcd multiplying by abcde ls xq.kk djus ij
bcde, acde, abde, abce, abcd are in H.P. gjkRed Js.kh esa gSA
Statement-1 is false dFku-1 vlR; gSA Statement-2 dFku-2
a
a�cb ,
bb�ac
, a b c
c
A.P.
a
a2�acb ,
bb2�bac
, a b c 2c
c
A.P.
a
acb � 2,
bbac � 2,
ccba � 2 A.P.
a
acb ,
bbac
, c
cba A.P.
a1
, b1
, c1
A.P. statement-2 is true dFku 2 lR; gSA
a, b , c H.P. 89. Statement 1 : Equation x2(a) + x .................... oDrO; -1 : lehdj.k x2(a) + x .................... Sol. Obvious Li"Vr%
90. Statement- 1 : 12| 1� 3 | � 4 2 3 .............
oDrO;-1 : 12| 1� 3 | � 4 2 3 ....................
Sol. 12| 1� 3 | � 4 2 3 = 3 � 1 � 3 1 = 2
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SOLEBPT2160815-16
PART TEST-2 (PT-2)
TARGET : JEE (MAIN)-2017
ANSWER KEY
CODE-0
PHYSICS
1. (2) 2. (2) 3. (3) 4. (2) 5. (1) 6. (3) 7. (3)
8. (3) 9. (2) 10. (1) 11. (1) 12. (2) 13. (2) 14. (4)
15. (3) 16. (1) 17. (3) 18. (3) 19. (4) 20. (2) 21. (1)
22. (3) 23. (2) 24. (3) 25. (3) 26. (2) 27. (3) 28. (2)
29. (4) 30. (4)
CHEMISTRY
31. (4) 32. (1) 33. (2) 34. (1) 35. (4) 36. (3) 37. (3)
38. (2) 39. (4) 40. (4) 41. (1) 42. (3) 43. (1) 44. (3)
45. (1) 46. (1) 47. (2) 48. (3) 49. (3) 50. (2) 51. (3)
52. (2) 53. (4) 54. (2) 55. (4) 56. (2) 57. (1) 58. (1)
59. (2) 60. (1)
MATHEMATICS
61. (2) 62. (1) 63. (1) 64. (3) 65. (1) 66. (4) 67. (2)
68. (2) 69. (1) 70. (3) 71. (3) 72. (3) 73. (3) 74. (2)
75. (3) 76. (1) 77. (4) 78. (3) 79. (1) 80. (1) 81. (1)
82. (3) 83. (1) 84. (4) 85. (1) 86. (3) 87. (2) 88. (4)
89. (4) 90. (1)
DATE : 16-08-2015 | CLASS-XI |COURSE : ALL INDIA TEST SERIES (VIKALP)