Allen Test 02
May 2, 2017 | Author: Robin Alom | Category: N/A
Short Description
AIPMT Test...
Description
MAJOR TEST
18–02–2013
PRE-MEDICAL : ENTHUSIAST COURSE
HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS BEWARE OF NEGATIVE MARKING 1.
2.
A particle is moving along a circular path. The angular velocity, linear velocity, angular acceleration and centripetal acceleration of the particle at any instant respectively are; w , V , and a C . Which of the following relation is not correct ? (1) w V (2) w a C (3) w (4) V An engine blowing a whistle of 256 vibration/ second if approaching you with
3.
4.
5.
6.
1 20
th
IB5= 25µA 4
2.
1 20
th
(1) 256 Hz, 256 Hz (3) 256 Hz, 243.8 Hz 3.
(2) 269.5 Hz, 243.8 Hz (4) 243.8 Hz, 256 Hz U = 2.5x2 + 100
0.2 :-
4.
5.
(1) Yes, 2.5 sec. (2) Yes, 1.26 sec. (3) Yes, 5.2 sec. (4) No 47°C 320 27°C :(1) 310 Hz (2) 320 Hz (3) 330 Hz (4) 340 Hz :(1)
(2) (3)
6.
(4) 3 IB5= 25µA
IC 4
IB3= 15µA IB2= 10µA
(2) Saturation region (4) None of the above
IB4= 20µA IB3= 15µA IB2= 10µA IB= 5µA 1 3 IB=0A
2 1
VCE,sat
(1) Active region (3) Cut off region
(2) w a C (4) V
256
IB= 5µA 1 3 IB=0A
2 1
IB4= 20µA
a C w , V , ? (1) w V (3) w
the velocity
of sound. The frequency before and after crossing of engine as heard by you would be :(1) 256 Hz, 256 Hz (2) 269.5 Hz, 243.8 Hz (3) 256 Hz, 243.8 Hz (4) 243.8 Hz, 256 Hz The potential energy U of a particle is given by U = 2.5x2 + 100 joule. Is the motion simple harmonic. If the the mass of the particle is 0.2 Kg, what is its time period :(1) Yes, 2.5 sec. (2) Yes, 1.26 sec. (3) Yes, 5.2 sec. (4) No An organ pipe produces a fundamental frequency of 320 Hz at 47°C. At 27°C the fundamental frequency of pipe would be :(1) 310 Hz (2) 320 Hz (3) 330 Hz (4) 340 Hz The MI of disc is minimum about an axis :(1) coinciding with the diameter (2) Tangential to the rim and lying in the plane of disc (3) Passing through centre of mass and perpendicular to the plane of the disc (4) Any axis passing through centre of mass According to the graph point 3 of the curve is located in :IC
1.
VCE,sat
(1) (3)
(2) (4)
H
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1/32
MAJOR TEST
18–02–2013
TARGET : PRE-MEDICAL 2013 (NEET-UG) A solid sphere of mass M is rolling with a speed V on a horizontal surface and strikes a massless spring of force constant K. Then the maximum compression of spring is :-
7.
(1)
5MV2 3K
(2)
7.
7MV2 5K
MV 2 (4) None of the above K For a transistor connected in CE configuration = 40 voltage drop across 2K collector resistance is 2V. Calculates emitter current.
(3)
8.
40 mA (1) 1mA (2) 41 (3) 1.025 mA (4) 2 mA A central charge particle +q is surrounded by a square array of charged particles separated by either distance r or r/2 along the perimeter of the square. Find magnitude of net electrostatic force on the central particle :3q 5q +q +4q –6q +3q –2q +q +3q –6q +q +4q +5q +3q
9.
10.
11.
q2 q2 3q 2 q2 (1) (2) (3) (4) 8 0 r 2 16 0 r 2 4 0 r 2 0 r 2 For a CE transistor amplifier which of following parameter governs the operating point of the amplifier. (1) Operating value of VCE and IB (2) Biasing of transistor (3) Both (1) and (2) (4) None of the above There is a uniformly charge sphere with total charge +Q and radius R what will be value of electric flux paasing through a imaginary concentric sphere of radius R/2 :-
Q (1) 2 0 12.
Q (2) 4 0
Q (3) 8 0
2/32
(2) 2W
(3) 1 W
(4) None
(1)
5MV2 3K
(2)
7MV2 5K
(3)
MV 2 K
(4) None of the above
CE = 40 2K
2V
40 mA 41 (4) 2 mA
(1) 1mA
9.
(2)
(3) 1.025 mA +q
r 2r :+q –6q –2q +3q +4q
(1) 10.
11.
12.
3q
5q
+4q +3q
+q +5q
+3q
–6q +q
q2 q2 3q 2 q2 (2) (3) (4) 8 0 r 2 16 0 r 2 4 0 r 2 0 r 2
CE
(1)VCE IB (2) (3) (1) (2) (4) R +Q R/2 :Q (1) 2 0
(4) Zero
A force F = (3iˆ 4 ˆj) N displaces a particle by ˆ m in 3 sec. find the power :S (3ˆj 4k) (1) 4 W
8.
V M K :
Q (2) 4 0
Q (3) 8 0
(4) Zero
F = (3iˆ 4 ˆj) N 3 sec. ˆ m :S (3ˆj 4k) (1) 4 W (2) 2W (3) 1 W (4) None
Your Target is to secure Good Rank in Pre-Medical 2013
H
MAJOR TEST
18–02–2013
PRE-MEDICAL : ENTHUSIAST COURSE
R
13.
Sliding contact
x=0
R
13.
x = 10 cm
Sliding contact
x=0
x = 10 cm
R
R
i
14.
i
E When sliding contact is connected, at x = 0 then i = 10 A and at x = 10 cm it is i = 20 A. For what value of x, i will be 12 A:(1) x = 2 cm (2) x = 5 cm (3) x = 7.5 cm (4) x = 4 cm The binding energy of a deuterium nucleus is about 1.115 MeV per nucleon. Then the mass defect of the nucleus is about (1) 2.23 amu (2) 0.0024 amu (3) 2077 amu (4) none of the above
15.
E
x = 0 i = 10 A x = 10 cm i = 20 A x i 12 A :(1) x = 2 cm (3) x = 7.5 cm 14.
1.115 MeV (1) 2.23 amu (3) 2077 amu
(2) 0.0024 amu (4)
15. k1
d1
k1
d1
k2
d2
k2
d2
k2 k 1 :
Space between plates of a capacitor is filled two dielectric slab of dielectric constant k1 and k2. Its equivalent dielectric constant will be :-
2k1 k 2 (2) k k 1 2
k1 k 2 2 k1 k 2 (d1 d 2 ) (3) k d k d 1 2 2 1 (1)
16.
(2) x = 5 cm (4) x = 4 cm
(1)
k1 k 2 (4) k d k d 1 2 2 1
1 16 relative to living wood. The half life of C14 is 5700 years. The age of fossil wood is-
The ratio of C14 and C12 in a fossil wood is
(1) 1.14 × 104 years
(2) 2.28 × 104 years
(3) 9.12 × 104 years
(4) 18.24 × 104 years 5V
17.
B
2k1k 2 (2) k k 1 2
k1 k 2 2
k1 k 2 (d1 d 2 ) (3) k d k d 1 2 2 1 16.
k1 k 2 (4) k d k d 1 2 2 1
C14 C12
1 16
C14
5700 (1) 1.14 × 104 (2) 2.28 × 104 (3) 9.12 × 104 (4) 18.24 × 104 5V
17.
B
4
4
10V
10V
A
A
2V 2 In the circuit VB – VA is :(1) 3V (2) 1.5 V (3) 7 V
H
2V 2
VB – VA :(4) 13 V
(1) 3V
(2) 1.5 V
Your Target is to secure Good Rank in Pre-Medical 2013
(3) 7 V
(4) 13 V 3/32
MAJOR TEST
18–02–2013
TARGET : PRE-MEDICAL 2013 (NEET-UG) 18.
The rate of flow of a liquid through a capillary tube under a constant pressure head is Q. If the diameter of the tube is reduced to half and its length is doubled, then the new rate of flow of liquid will be (1)
19.
Q 4
(2)
Q 8
(3) 16 Q
(4)
18.
Q
Q 32
Find ratio of value of i before and after connecting the switch :R
(1) 19.
Q 4
(2)
Q 8
(1) 20.
21.
22.
23.
5 4
5 6
Q 32
R i
2R
R
(2)
(4)
i :
i E
(3) 16 Q
(3)
5 12
E
(4)
5 3
What is excess pressure inside the drop of mercury of radius 6.0 mm at room temp. [surface tension = 4.65 × 10 –1 Nm –1 ] 1atm = 1.01 × 105 Pa (1) 155 Pa (2) 310 Pa (3) 460 Pa (4) None of these A transparent cube of 0.21 m edge contains a small air bubble. Its apparent distance when viewed through one face of the cube is 0.10 m and when viewed from the opposite face is 0.04 m. The actual distance of the bubble from the second face of the cube is :(1) 0.06m (2) 0.17m (3) 0.05m (4) 0.04m The volume of a solid at 1 atmosphere pressure is 10 4 cm 3. If the pressure is increased to 51 atmosphere then percentage change in its volume will be (K = 1012 dyne/cm2) (1) 0.001% (2) 0.003% (3) 0.005% (4) 0.05% An unpolarized beam of light is incident on a glass surface at an angle of incidence equal to the polarizing angle of the glass. Read the following statements :(i) The reflected beam is completely polarized (ii) The refracted beam is partially polarized (iii) The angle between the reflected and the refracted beam is 90°. Which of the above statements is/are true ? (1) (i) only (2) (ii) only (3) (i) and (iii) (4) All the statements are true
(1) 20.
21.
5 4
2R
R
(2)
5 6
(3)
5 12
(4)
5 3
6.0 mm [ = 4.65 × 10–1 Nm–1]
1atm = 1.01 × 105 Pa (1) 155 Pa (2) 310 Pa (3) 460 Pa (4) None of these 0.21m
0.10 m 0.04 m :- (1) 0.06m 22.
(2) 0.17m
(3) 0.05m
(4) 0.04m
104 3 51 2 (K = 1012 ) (1) 0.001% (3) 0.005%
23.
(2) 0.003% (4) 0.05% (Unpolarized beam of light) :(i) (ii) (iii) 90° ? (1) (i) (2) (ii) (3) (i) (iii) (4)
Key Filling 4/32
Your Target is to secure Good Rank in Pre-Medical 2013
H
MAJOR TEST
PRE-MEDICAL : ENTHUSIAST COURSE
24.
Let V and E be the potential and the field repectively at a point. Which of the following assertions are correct ? (1) If V = 0, E must be zero. (2) If V 0, E cannot be zero. (3) If E 0, V cannot be zero (4) None of these
24.
25.
A prism of a certain angle deviates the red and blue rays by 8° and 12° respectively. Another prism of the same angle deviates the red and blue rays by 10° and 14° respectively. The prisms are small angled and made of different materials. The dispersive powers of the materials of the prisms are in the ratio :(1) 5 : 6 (2) 9 : 11 (3) 6 : 5 (4) 11 : 9 One kilowatt hour is a unit of :(1) Energy (2) Power (3) Electric charge (4) Electric current The aperture of the largest telescope in the world is 5m. If the separation between the moon and the earth is 4×105 km and the wavelength of visible light is 5000 Å, then the minimum separation between the objects on the surface of the moon which can be just resolved is approximately :(1) 1m (2) 10m (3) 50m (4) 200m The potential difference V and the current I flowing through an instrument in an ac circuit of frequency f are given by V = 5 cos t volts and I = 2 sin t amperes (where = 2f). The power dissipated in the instrument is :(1) Zero (2) 10W (3) 5 W (4) 2.5 W ˆ (Where K is a positive A force F K(yiˆ xj)
25.
26.
27.
28.
29.
30.
constant) acts on a particle moving in the xy plane. starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particle is :(1) –2Ka2 (2) 2 Ka2 (3) –Ka2 (4) Ka2 The period of oscillation of a bar magnet in a vibration magnetometer is 2 sec. The period of oscillation of a bar magnet whose magnetic moment is 4 times that of Ist magnet is :(1) 4 sec. (2) 1 sec. (3) 2 sec. (4) 0.5 sec.
H
26.
27.
V E ? (1) V = 0 E (2) V 0 E (3) E 0 V (4) 8° 12° 10° 14° :- (1) 5 : 6
(2) 9 : 11
(3) 6 : 5
(4) 11 : 9
(kwh), :(1) (2) (3) (4) 5m 5 4×10 km 5000 Å :(1) 1m
28.
18–02–2013
(2) 10m
(3) 50m
(4) 200m
ac f V I V = 5 cos t volts I = 2 sin t ( = 2f) :(1) (2) 10W (3) 5 W
29.
(4) 2.5 W ˆ x-y F K(yiˆ xj) ( K )
x- (a, 0) y- (a, a) F :(1) –2Ka2 30.
(2) 2 Ka2
(3) –Ka2
(4) Ka2
2 4 : (1) 4 (2) 1 (3) 2 (4) 0.5
Your Target is to secure Good Rank in Pre-Medical 2013
5/32
MAJOR TEST
18–02–2013
TARGET : PRE-MEDICAL 2013 (NEET-UG) 31.
A uniform chain of length and mass m overhangs from a smooth table so that
31.
2 rd part of it is on 3
2 rd 3
the table then velocity of chain when it completely slips off the table :-
32.
33.
34.
35.
2g
(2)
(3)
2 g 3
(4) None
Identify the paramagnetic substance :(1) Iron (2) Aluminium (3) Nickel (4) Hydrogen In a children's park, there is a slide which has a total length of 10 m and a heigyht of 8 m. A vertical ladder is provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is three tenth of his weight. The work done by the friction on th boy as he comes down is :(1) 0 J (2) +600 J (3) –600 J (4) +1600 J Lenz's law is consistent with law of conservation of :(1) Current (2) emf (3) Energy (4) All of the above Amount of work done to carry a block from A to B will be (Assume friction coefficient µ) (1) mgh B (2) µmg 2 h 2
36.
:-
2 2g 3
(1)
m
µ
32.
33.
34.
35.
2g
(2)
(3)
2 g 3
(4)
: (2) (1) (3) (4) children's park 10 m 8 m 200 N 3/10 :(2) +600 J (4) +1600 J
:(1)
(2)
(3)
(4)
A B ( µ ) (1) mgh B (2) µmg 2 h 2
36.
2 2g 3
(1)
(1) 0 J (3) –600 J
h
(3) µmg( + h) A (4) mg(h + µ) A flux of 1 mWb passes through a strip having an area A = 0.02 m2. The plane of the strip is at an angle of 60° to the direction of a uniform field B. The value of B is :-
m
m
µ
h
(3) µmg( + h) A (4) mg(h + µ) 1 mWb A = 0.02 m2 B 60° B : (1) 0.1 T
(1) 0.1 T
(2) 0.058 T
(2) 0.058 T
(3) 4.0 mT
(3) 4.0 mT (4) None of the above
(4) Use stop, look and go method in reading the question
6/32
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H
MAJOR TEST
18–02–2013
PRE-MEDICAL : ENTHUSIAST COURSE
37.
For the position (x) - time (t) graph shown of particle in one dimensional motion. Choose the incorrect alternatives from below :-
37.
(x) - (t) ?
X
X C
C
B
A
B
A D
D
o E
38.
t
o E
(1) Particle was released from rest at t = 0 (2) At C particle will reverse its direction of motion. (3) Average velocity for motion between B and D is positive (4) At E, velocity = 0 and acceleration > 0 There is a black spot on a body. If the body is heated and carried in dark room then it glows more. This can be explained on the basis of:-
(1) t = 0 (2) C (3) B D
38.
(1) Newton's law of cooling (2) Wein's law (3) Kirchhoff's law (4) Stefan's law 39.
40.
Which of the following statement is incorrect about friction ? (1) Limiting static friction is independent of area of contact. (2) Kinetic friction is independent of area of contact. (3) Kinetic friction is nearly independent of velocity of bodies. (4) Kinetic friction is self adjusting
39.
PV versus T graph of equal masses of H2, He and O2 is shown in fig. Choose the correct alternative:-
40.
(4) E , = 0 >0 (1) (2) (3) (4) ?
(1)
(2)
(3) (4)
PV-T H2,He O2 P
P
C
C
B
PV
B
PV
A
A O
T
T
(1) C corresponds to He, B to H2 and A to O2 (2) A corresponds to He, B to H2 and C to O2 (3) A corresponds to He, B to O2 and C to H2 (4) A corresponds to O2, B to He and C to H2
H
t
O
T
T
(1) He C, H2 B, O2 A (2) He A, H2 B, O2 C (3) He A, O2 B, H2 C (4) O2 A, He B, H2 C
Your Target is to secure Good Rank in Pre-Medical 2013
7/32
MAJOR TEST
18–02–2013
TARGET : PRE-MEDICAL 2013 (NEET-UG) 41.
42.
43.
If the given graph is possible in realistic situations, then y and x variables may represent respectively :y (1) acceleration and time (2) velocity and time (3) velocity and displacement o x (4) displacement and time A Carnot's engine used first an ideal monoatomic gas then an ideal diatomic gas. if the source and sink temperature are 411°C and 69°C respectivelly and the engine extracts 1000 J of heat in each cycle, then area enclosed by the PV diagram is (1) 100 J (2) 300 J (3) 500 J (4) 700 J If the coefficient of kinetic friction between the trolley and surface is 0.1, then tension in the string connecting masses is 2 [Take g = 10 m/s ]?
41.
y x :(1)
y
(2) (3) 42.
411°C 69°C 1000 J PV (1) 100 J
43.
(2) 300 J
(3) 500 J
2
[g = 10 m/s ]? 30 kg
6kg
(1) 48 N (2) 51 N (3) 53 N (4) 55 N A perfect gas goes from state A to another state B by absorbing 8 × 105 J of heat and doing 6.5 × 105 J of external work. It is now transferred between the same two states in another process in which it absorbs 105 J of heat. Then in the second process (1) Work done on the gas is 0.5 × 105 J (2) Work done by gas is 0.5 × 105 J (3) Work done on gas is 105 J (4) Work done by gas is 105 J Position-time graph of a body of mass 0.5 kg is shown. Time interval between two consecutive impulses and the magnitude of that impulse is ? x(m)
44.
45.
(2) 51 N
(3) 53 N
(4) 55 N
(A) (B) 8 × 105 J 6.5 × 105 J (A) (B) 10 5 J (1) 0.5 × 105 J (2) 0.5 × 105 J (3) 105 J (4) 105 J 0.5 kg ?
10 5
10 15 20
(1) 5 s, 4 N-s (3) 10 s, 2 N-s 8/32
(1) 48 N
x(m)
10 0
(4) 700 J
0.1
6kg
45.
x
(4)
30 kg
44.
o
25
t(s)
(2) 10 s, 4 N-s (4) 5 s, 2 N-s
0
5
10 15 20
(1) 5 s, 4 N-s (3) 10 s, 2 N-s
Your Target is to secure Good Rank in Pre-Medical 2013
25
t(s)
(2) 10 s, 4 N-s (4) 5 s, 2 N-s
H
MAJOR TEST
18–02–2013
PRE-MEDICAL : ENTHUSIAST COURSE
46.
47.
48.
49.
50.
Which is incorrect statement :(1) Formic acid gives fehling's test but acetic acid does not give. (2) Chloral reacts with water but acetone does not react. (3) Ethyl alcohol gives iodoform test but does not give DNP test. (4) Nitrobenzene is meta-directing towards sodamide. The correct order of the O–O bond length in O2,H2O2 and O3 is : (1) O2 > O3 > H2O2 (2) O3 > H2O2 > O2 (3) O2 > H2O2 > O3 (4) H2O2 > O3 > O2 Which is correct statement :– (1) Benzyl amine is more basic than acetanilide (2) Nitrobenzene is purified by steam-distillation method (3) Aniline does not prepare from Gabriel phthalimide reaction but it gives Hoffmann's isocyanide test (4) All the above Match list-I with list-II and select the correct answer:List-I (species) List-II(O–N–O angle) (i) 180º (A) NO2+ (ii) 132º (B) NO2 (C) NO2– (iii) 120º (iv) 115º (D) NO3– (v) 109º A B C D (1) v iv iii ii (2) v ii iv iii (3) i ii iv iii (4) i iv iii ii Which of the following reaction shows incorrect product :H3 C O C CN
47.
48.
49.
(2) (3) DNP (4) O2, H2O2 O3 O–O (1) O2 > O3 > H2O2
(2) O3 > H2O2 > O2
(3) O2 > H2O2 > O3
(4) H2O2 > O3 > O2
:– (1) (2) (3) (4) List-I (species) (A) NO2+ (B) NO2 (C) NO2– (D) NO3–
50.
List-II(O–N–O angle) (i) 180º (ii) 132º (iii) 120º (iv) 115º (v) 109º A B C D (1) v iv iii ii (2) v ii iv iii (3) i ii iv iii (4) i iv iii ii :H3 C O C CN
COOH (1) CO2
(2)
+
(2) H 3O
+
(2) H 3O
Ph
Ph Ph
OH
(3)
O
Cl (4) CH3–CH–O–CH2 CH3
HBr
Ph
OH
O
CH3–CH–Br +
Cl (4) CH3–CH–O–CH2
CH3
CH3
CH2OH
H
OCH3
3
MgBr
(1) CO2
(2)
(1) CH 3MgBr (2) H O +
OCH3
COOH
MgBr
(3)
(1)
OCH3
3
OCH3
:-
(1)
(1) CH 3MgBr (2) H O +
(1)
46.
Your Target is to secure Good Rank in Pre-Medical 2013
HBr
CH3–CH–Br + CH3 CH2OH
9/32
MAJOR TEST
18–02–2013
TARGET : PRE-MEDICAL 2013 (NEET-UG) 51.
Correct order of dipole moment is :-
Cl
OH NO2
(I)
51.
(II)
CH3
54.
55.
=
(1) I = II = III (3) I > II > III 52.
(2) I < II < III (4) II < III < I
NaOH COCl2 A B (CH3)2CH–C–NH2 Br2 O
=
(2) I < II < III (4) II < III < I
NaOH COCl2 (CH3)2CH–C–NH2 A B Br2 O
B is :-
B :-
(1) (CH3)2CH–N=C=O
(1) (CH3)2CH–N=C=O
(2) CH3–CH–CH3 NH2
(2) CH3–CH–CH3 NH2
(3) (CH3)2CH–NH–COCH3
(3) (CH3)2CH–NH–COCH3
(4) CH3–CH2–NHCOCH3
(4) CH3–CH2–NHCOCH3
–
Which of the following molecules or ions is not linear? (3) CS2 (4) ICl2+ (1) BeCl2 (2) ICl2–
53.
Which of the following is Borodine Hunsdieker reaction ?
54.
(1) BeCl2
(2) ICl2–
(3) CS2
(4) ICl2+
?
Acetone (1) C2H5Br + KI C2H5I + KBr
Acetone C2H5I + KBr (1) C2H5Br + KI
(2) 2C2H5Cl + Hg2F2 2C2H5F + Hg2Cl2
(2) 2C2H5Cl + Hg2F2 2C2H5F + Hg2Cl2
(3) C2H5ONa + CH3I C2H5OCH3 + NaI
(3) C2H5ONa + CH3I C2H5OCH3 + NaI
CCl 4 (4) CH3COOAg+Br CH3Br+CO2+ AgBr 2
CCl 4 CH3Br+CO2+ AgBr (4) CH3COOAg+Br 2
General electronic configuration of lanthanides is :1–14
(1) (n – 2) f
2 6 0–1
(n – 1)s p d ns
55.
(1) (n – 2) f1–14(n – 1)s2p6d0–1ns2
2
(2) (n – 2)f0–14(n –1)d0–1ns2
(2) (n – 2)f0–14(n –1)d0–1ns1–2
56.
Cl
CH 3
(III)
–
53.
(II)
CH 3
(1) I = II = III (3) I > II > III 52.
NO2
(I)
CH 3 (III)
Cl
OH
Cl
(3) (n – 2)f0–14(n – 1)d10ns2
(3) (n – 2)f0–14(n – 1)d10ns2
(4) (n – 2)d0–1(n – 1)f1–14ns2
(4) (n – 2)d0–1(n – 1)f1–14ns2
An example of antipyretic is :(1) (2) (3) (4)
Luminal Paracetamol Terpineol Iodoform
56.
: (1) (2) (3) (4)
10/32
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MAJOR TEST
18–02–2013
PRE-MEDICAL : ENTHUSIAST COURSE
57.
58.
59.
60.
For the process
X(g) + e– X–(g), H = x
and X–(g) X(g) + e–,
and X–(g) X(g) + e–,
H = y
H = y
Select correct alternate :-
(1) ionisation energy of X–(g) is y
(1) X–(g) y
(2) electron affinity of X(g) is x
(2) X(g) x
(3) electron affinity of X(g) is –y
(3) X(g) –y
(4) all are correct statements
(4)
When compound X is oxidised by acidified potassium dichromate, compound Y is formed. Compound Y on reduction with LiAlH4, gives X. (X) and (Y) respectively are :-
58.
X Y Y, LiAlH4 X X Y :(1) C2H5OH, CH3COOH
(1) C2H5OH, CH3COOH
(2) CH3COCH3, CH3COOH
(2) CH3COCH3, CH3COOH
(3) C2H5OH, CH3COCH3
(3) C2H5OH, CH3COCH3 (4) CH3CHO, CH3COCH3 Which is largest in size in aqueous solution ? (1) Li+ (2) Na+ (3) Cs+ (4) Rb+ In the following sequence of reactions – Mg Ether
HCHO
(4) CH3CHO, CH3COCH3 59.
(1) Li+ (3) Cs+
60.
H2 O
(2) Na+ (4) Rb+
–
CH3CH2–OH A B C D,
P I2 Mg A Ether B HCHO C H2O D, CH3CH2–OH
the compound D is :(1) Butanal (2) n–Butyl alcohol (3) n–Propyl alcohol (4) Propanal Carbogen is :(1) mixture of CO + CO2 (2) mixture of O2+CO2
D :(1) (2) n– (3) n– (4) (1) CO + CO2 (2) O2+CO2 (3) (4) CH2OH–(CHOH)4–CHO HIO4 :-
61.
(3) Pure form of carbon (4) unsaturated organic compound 62.
X(g) + e– X–(g), H = x
P I2
61.
57.
Glucose CH2OH–(CHOH)4–CHO, on oxidation with HIO4 gives :-
62.
(1) Six moles of HCOOH
(1) HCOOH 6
(2) Five moles of HCOOH + one mole of HCHO
(2) HCOOH 5 + HCHO
(3) Four moles of HCOOH + two moles of HCHO
(3) HCOOH 4 + HCHO 2
(4) Six moles of HCHO
(4) HCHO 6
H
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MAJOR TEST
18–02–2013
TARGET : PRE-MEDICAL 2013 (NEET-UG) 63.
Which one of the following arrangement does not truly represent the property indicated against it? (1) Br2 < Cl2 < F2 : Oxidising power
63.
(1) Br2 < Cl2 < F2 : Oxidising power (2) Br < Cl < F : Electronegativity
(2) Br < Cl < F : Electronegativity
(3) Br < F < Cl : Electron affinity
(3) Br < F < Cl : Electron affinity
(4) Br2 < Cl2 < F2 : Bond energy
(4) Br2 < Cl2 < F2 : Bond energy 64.
65.
66.
Which among the following is biodegradable pollutant :(1) Lead compounds (2) Pesticides (3) Domestic wastes (4) Mercuric salts A solution of sodium metal in liquid NH3 is:(1) strongly reducing (2) blue in colour (3) good conductor (4) all of the above CH3NH2 + CHCl3 + KOH Pungent smell compound + KCl + H2O. Pungent smelling compound is :(1) CH3–CN (2) CH3–NH–CH3 ⓛ
(3) CH3 N C 67.
68.
69.
64.
(3) 65.
66.
(4)
(1) (2) (3) (4) CH3NH2 + CHCl3 + KOH + KCl + H2O, :(1) CH3–CN
ⓛ
ⓛ
(4) CH3 N C
When orthoboric acid is heated to red heat the residue is :
(1) (2)
(2) CH3–NH–CH3
67.
(3) CH3 N C
ⓛ
(4) CH3 N C
(1) boron
(1)
(2) boron sesquioxide
(2)
(3) metaboric acid
(3)
(4) pyroboric acid Which of following gives effervescences of CO2 with NaHCO3 solution :(1) HCOOH (2) 2, 4, 6–Trinitrophenol (3) Both (1) & (2) (4) None of these
(4) NaHCO3 CO2 : (1) HCOOH (2) 2, 4, 6– (3) (1) (2) (4)
Which is not obtained when metal carbides react with H2O:(1) Al4C3 + H2O CH3–CH2–CH3 (2) CaC2 + H2OCHCH
68.
69.
(1) Al4C3 + H2O CH3–CH2–CH3 (2) CaC2 + H2OCHCH
(3) Mg4C3 + H2O CH3CCH
(3) Mg4C3 + H2O CH3CCH
(4) Be2C + H2O CH4
(4) Be2C + H2O CH4
Time Management is Life Management 12/32
Your Target is to secure Good Rank in Pre-Medical 2013
H
MAJOR TEST
18–02–2013
PRE-MEDICAL : ENTHUSIAST COURSE
70.
71.
72.
73.
74.
Match list-I and list-II and then select the correct answer from the codes given below the lists : List–I List–II [A] C6 H5CHO [a] Mesitylene [B] CH3COCHO [b] Paraldehyde [C] CH3COCH3 [c] Iodoform reaction [D] CH3CHO [d] Cannizzaro reaction Codes : A B C D (1) d c b a (3)
a c b d
70.
(2)
A B C D d b c a
Codes : A B C D (1) d c b a
(2)
A B C D d b c a
(4)
d c a b
(3)
(4)
d c a b
Which cannot be used to generate H2?
71.
a c b d
H2
(1) Al + NaOH
(2) Zn + NaOH
(1) Al + NaOH
(2) Zn + NaOH
(3) Mg + NaOH
(4) LiH + H2O
(3) Mg + NaOH
(4) LiH + H2O
Formaldehyde reacts with ammonia to give urotropine. The formula of urotropine is :
72.
:
(1) (CH2)6N4
(2) (CH2)4N3
(1) (CH2)6N4
(2) (CH2)4N3
(3) (CH2)6N6
(4) (CH2)3N3
(3) (CH2)6N6
(4) (CH2)3N3
Which does not exist in solid state :-
73.
(1) NaHCO3
(2) NaHSO3
(1) NaHCO3
(2) NaHSO3
(3) LiHCO3
(4) CaCO3
(3) LiHCO3
(4) CaCO3
Formaldehyde and acetaldehyde are readily distinguished by reaction with :
74.
(1) A solution of 2,4–dinitrophenylhydrazine (2) Fehling’s solution (3) Tollen’s reagent (4) Iodine and alkali 75.
I II : I II [A] C6 H5CHO [a] [B] CH3COCHO [b] [C] CH3COCH3 [c] [D] CH3CHO [d]
For the reversible reaction, N2(g) + 3H2(g)
2NH3(g)
75.
: (1) 2, 4– (2) (3) (4) , N2(g) + 3H2(g)
2NH3(g)
at 500ºC, the value of Kp is 1.44 × 10–5 when partial pressure is measured in atmosphere. The corresponding value of Kc with concentration in mol/L is :-
500ºC Kp 1.44 × 10–5 mol/L Kc :-
1.44 10 5 (1) (0.082 500) 2
1.44 10 5 (2) (8.314 773) 2
(1)
1.44 10 5 (3) (0.082 773) 2
1.44 10 5 (4) (0.082 773) 2
1.44 10 5 (3) (0.082 773) 2
H
1.44 10 5 (0.082 500) 2
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(2)
1.44 10 5 (8.314 773) 2
1.44 10 5 (4) (0.082 773) 2 13/32
MAJOR TEST
18–02–2013
TARGET : PRE-MEDICAL 2013 (NEET-UG) 76.
The entropy change during an isothermal expansion of an ideal gas from V 1 to V 2 at temperature T is given by :-
76.
T V1 V2 :-
(1) S = 0 (1) S = 0
(2) S = 2.303 R log10 V2/V1
(2) S = 2.303 R log10 V2/V1
(3) S = 2.303 RT log10 V2/V1
(3) S = 2.303 RT log10 V2/V1
(4) S = 2.303 R log10 V1/V2 77.
Ag+ + NH3
[Ag(NH3)]+; k1 = 3.5 × 10–3
[Ag(NH3)]+ + NH3
78.
79.
80.
Ag+ + NH3
[Ag(NH3)]+; k1 = 3.5 × 10–3
[Ag(NH3)]+ + NH3
[Ag(NH3)2]+; k2 = 1.7 × 10–3
[Ag(NH3)2]+; k2 = 1.7 × 10–3
[Ag(NH3)2]+ :-
(1) 6.08 × 10–6
(2) 6.08 × 106
(1) 6.08 × 10–6
(2) 6.08 × 106
(3) 6.08 × 10–9
(4) None o these
(3) 6.08 × 10–9
(4)
Entropy of universe, in the case of adiabatic expansion of a gas is :(1) Suniv = 0
(2) Suniv > 0
(3) Suniv < 0
(4) Suniv 0
A certain weak acid has a dissociation constant of 1.0 × 10–4. The equilibrium constant for its reaction with a strong base is :-
78.
79.
:(1) Suniv = 0 (3) Suniv < 0
(2) Suniv > 0 (4) Suniv 0
1.0 × 10–4 :
(1) 1.0 × 10–4
(2) 1.0 × 10–10
(1) 1.0 × 10–4
(2) 1.0 × 10–10
(3) 1.0 × 1010
(4) 1.0 × 1014
(3) 1.0 × 1010
(4) 1.0 × 1014
Which of the following arrangements of electrons is mostly likely to be stable ? (z 30 for this atom):4s
(1) 3d
3d
80.
3d
? (z 30 ) :3d
4s
4s
3d
(4)
Which of the following is a strongest acid :-
4s
(1)
(2)
4s
(3)
82.
77.
Then the formation constant of [Ag(NH3)2]+ is :-
3d
81.
(4) S = 2.303 R log10 V1/V2
:-
(2) HClO3
(1) HClO4
(2) HClO3
(3) H2SO4
(4) H2SO3
(3) H2SO4
(4) H2SO3
82.
:-
(1) 3p
(2) 3d
(1) 3p
(2) 3d
(3) 3s
(4) Equal
(3) 3s
(4)
14/32
4s
(4)
(1) HClO4
Which orbital gives an electron the greatest probablity of being found close to the nucleus:-
4s
3d
4s
(3) 81.
3d
(2)
Your Target is to secure Good Rank in Pre-Medical 2013
H
MAJOR TEST
PRE-MEDICAL : ENTHUSIAST COURSE
83.
The value of vander Waals' constant ‘a’ for the gases O2 N2, NH3 and CH4 are 1.360, 1.390, 4.170 and 2.253 L2 atm mol–2 respectively. The gas which can most easily be liquified is :(1) O2
84.
85.
86.
87.
(2) N2
89.
90.
(4) CH4
How many unpaired electrons are in gaseous Fe2+ ion in the ground state ? (1) 0 (2) 2 (3) 4 (4) 6 In the standardization of Na2S2O3, using K2Cr2O7 by iodometry, the equivalent weight of K2Cr2O7 is :(1) (Molecular weight)/2 (2) (Molecular weight)/6 (3) (Molecular weight)/3 (4) Same as molecular weight All of these sets of quantum numbers are permissible except :ms n l m1 (1) 1 0 0 +1/2 (2) 2 2 0 –1/2 (3) 31 1 –1/2 (4) 3 2 –1 +1/2 The oxidation number of iron in magnetite (Fe3O4) is :(1) +2 (2) +3 (3) Both of the above
88.
(3) NH3
83.
84.
85.
86.
(1) Particles only (2) Wave only (3) Both simultaneously (4) Sometimes waves and sometimes particles The largest number of molecules is in :-
‘a’ O2 , N2 , NH3 CH4 1.360, 1.390, 4.170 2 2.253 L atm mol–2 :-
(1) O2 (2) N2 (3) NH3 (4) CH4 2+ Fe ? (1) 0 (2) 2 (3) 4 (4) 6 Na2S2 O3 K2Cr2O7 K2Cr2O7 : (1) (Molecular weight)/2 (2) (Molecular weight)/6 (3) (Molecular weight)/3 (4) Same as molecular weight
:n 1 2 31 3
87.
l m1 ms (1) 0 0 +1/2 (2) 2 0 –1/2 (3) 1 –1/2 (4) 2 –1 +1/2 (Fe3 O4 ) :(1) +2 (2) +3 (3) (4) +4/3
(4) +4/3
According to Schrodinger model nature of electron in an atom is as :-
88.
89.
:(1) (2) (3) (4) :-
(1) 8g of hydrogen
(1) 8g
(2) 28 g of CO
(2) 28 g CO
(3) 92 g of C2H5OH
(3) C2H5OH 92 g
(4) 56 g of N2
(4) N2 56 g
When the same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide, the ratio of volumes of hydrogen evolved is :(1) 1 : 1 (2) 1 : 2 (3) 2 : 1 (4) 9 : 4
H
18–02–2013
90.
H2SO4 NaOH :(1) 1 : 1
(2) 1 : 2
(3) 2 : 1
(4) 9 : 4
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15/32
MAJOR TEST
TARGET : PRE-MEDICAL 2013 (NEET-UG) 91.
before how many years :-
(1) 320 million years
(2) 200 million years
(1) 320 million years
(2) 200 million years
(3) 100 million years
(4) 350 million years
(3) 100 million years
(4) 350 million years
During evolution see weads existed probably
91.
92.
Progressive degeneration of skeletal muscles mostly due to genetic disorder is called :(1) Myasthenia gravis (2) Muscular dystrophy (3) Tetany (4) Gout
92.
93.
During evolution the animal which evolved in to
93.
the first amphibian that lived on both land and water were:(1) Sauropsids (2) Synapsids (3) Lobefins (4) Therapsids 94. Match column-I with column-II :Column-I Column-II (number of bones) A. Cranial bones (i) 24 B. Ribs (ii) 26 C. Vertebral column (iii) 8 (iv) 12 (1) A - (i), B - (ii), C - (iii) (2) A - (iii), B - (i), C - (ii) (3) A - (iii), B - (iv), C - (ii) (4) A - (iv), B - (i), C - (iii) 95. During evolution which organism deposits and
95.
(3) Bryophyte (4) Pteridophytes
disappeared from the earth :(1) 100 million year (2) 65 million year (3) 200 million year (4) 165 million year
(2)
(3)
(4)
(1) (2) (3) (4)
(2) Dicotyledons
97.
(1)
-I -II : -I -II ( ) 24 A. (i) B. (ii) 26 C. (iii) 8
(1) Angiosperms
Ringworm is .......... disease :(1) helminth (2) fungal (3) bacterial (4) viral Before how many year dinosaur suddenly
: (1) (2) (3) (4)
94.
convert in to present days coal :-
96.
18–02–2013
96.
97.
A A A A
-
(iv) 12 (i), B - (ii), C - (iii) (iii), B - (i), C - (ii) (iii), B - (iv), C - (ii) (iv), B - (i), C - (iii)
(1) (2) (3) (4) .......... : (2) (1) (3) (4) (1) 100 (2) 65 (3) 200 (4) 165
16/32
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H
MAJOR TEST
18–02–2013
PRE-MEDICAL : ENTHUSIAST COURSE
Enters via contaminated food + water
98.
+
98.
cytotoxins, Slough villus cells
enterotoxin stimulates Adenylate cyclase Induction of diarrhoea due to loss of Cl– + water
+ (39°–40°C) :(1) (2)
Sustain high fever diagnosed by widal test (39°–40°C) above chart is related with :(1) Common cold
99.
(2) Typhoid (3) Haemophilus influenzae (4) Trichophyton Modern farmer's can increase the yield of Paddy upto 50% by the use of :(1) Cyanobacteria (2) Rhizobium (3) Cyanobacteria in Azolla pinnata (4) Farm yard manure
100. In which phase of meiosis, the chromosomes do undergoes some dispersion, but they do not reach the extremely extended state of the interphase nucleus ? (1) Prophase-I (2) Metaphase-I (3) Telophase-I (4) Prophase-II 101. First transgenic plant :(1) Potato
(2) Tomato
(3) Tobacco (4) Maize 102. Which one of the following cellular parts is correctly described ? (1) Ribosomes - Those on chloroplasts are larger (80s) while in the cytoplasm are smaller (70s) (2) Centrioles - Site for active RNA synthesis (3) Thylakoids - Flatted membranous sacs forming the grana of chloroplasts (4) Golgi body - Not surrounded by membrane 103. Which fish selectively feed on larva of mosquito:(1) Gambusia
(2) Rohu
(3) Clarias
(4) Exocoetus
H
(3) (4) 99.
50% :- (1) (2) (3) (4)
Cyanobacteria Rhizobium Cyanobacteria in Azolla pinnata
100. ? -I (2) -I (1) (3) -I (4) -II 101. (1) Potato (3) Tobacco
(2) Tomato (4) Maize
102.
? (1) - (80s) (70s) (2) - RNA (3) - (4) - 103. (1) (2) (3)
(4)
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ENTHUSIAST COURSE
TARGET : PRE-MEDICAL 2013 MAJOR TESTDATE # 09: 10 - 01 - 2010
NEET-UG (FULL Syllabus)
DATE : 18 - 02 - 2013
ANSWER KEY Q.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
A.
3
2
2
1
1
3
2
3
1
3
3
1
2
2
3
2
3
4
1
1
Q.
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
A.
1
3
3
4
3
1
3
1
3
2
2
2
3
3
4
1
3
3
4
4
Q.
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
A.
3
3
4
1
4
4
4
4
3
4
4
1
2
4
1
2
4
1
1
3
Q.
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
A.
2
2
4
3
4
4
2
3
1
4
3
1
3
4
4
2
1
2
3
1
Q.
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
A.
1
3
3
3
2
2
3
3
1
1
1
2
3
2
4
2
2
2
3
3
101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Q. A.
3
3
1
3
2
3
2
2
3
1
1
1
4
1
1
2
2
2
3
2
121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Q. A.
2
4
4
3
2
4
3
4
3
2
4
2
2
1
4
4
1
3
2
1
141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
Q. A.
3
3
1
4
1
3
4
2
3
2
4
3
4
3
3
1
4
3
4
3
161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
Q. A.
4
1
2
1
4
4
1
3
1
4
3
2
4
2
4
1
1
4
4
1
HINT – SHEET 2.
n2 = 256 ×
3.
Therefore time period
20 n1 = 256 × = 256 × = 269.5 Hz V 19 V 20 V
V V
V 20
= 256 ×
20 = 243.8 Hz 21
T=2
4.
The force is determined from the relation F=–
dU d =– (2.5x2 + 100) dx dx
F = – 5x newton The motion is simple harmonic because F –x compare the above relation with F = – Kx force constant K = 5
HS
N m
0 .2 2 = 1.26sec. 5 5
n1 T 15 1 n T 16 15 310 Hz 16
n1 = 320 5. 7.
MI of disc about diametric axis will be minimum. Sphere compresses the spring untill its all K.E. is converted to P.E. of spring
F GH
K2 1 1 2 MV r2 2
I JK
=
1 Kx2 2
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MAJOR TEST
TARGET : PRE-MEDICAL 2013 (NEET-UG)
FG H
1 2 MV2 1 2 5
for sphere
=
1 Kx2 2
16.
1 amu 931
4 10
For other prism,
2
1 × 1.115 amu 931
1 1 16 2
B R B R 12 8 2 10
or 1
mass defect corresponding BE 1.115 MeV is
For one prism,
1
mass defect corresponding BE 931 MeV is 1 amu mass defect corresponding BE 1 MeV is
50 10 5 × 100% = 0.005% 1011
= 25.
7 MV2 5K
x= 14.
IJ K
14 10 B R 14 10 2 12 2
1 43 2 101
or
1 12 6 2 10 5
t / 5700
p r4 Q= 8 4
27.
Q1 r1 2 = 32 Q2 r2 1
Q2 =
Q 32
P=
2T r
20.
or
0.10
So, =
i.e. d
x 0.04
x l
B=
P V V
1 d = RP r
r 4 10 5 10 3 1.22 50m RP 10 7
x
35.
B µ
A
h
mg sin + µ mg cos )x
V P % = × 100% V B
2/4
1 and if d is the RP
(0.21–x)
or 0.21 × 0.04 – x × 0.04 = x×0.10 On solving x = 0.06m 22.
D 5 1 10 7 –7 1.22 1.22 5 10 1.22
distance between objects on the surface of moon which is at a distance r from the telescope, = (d/r)
Real Distance Apparent Distance
0.21 x
RP
As limit of resolution =
2 4.65 10 1 = = 155 Pa 6 10 3 21.
4 1 12 3
or
t = 22800 years 18.
18–02–2013
h 1 Mg u .x x x
Mg (h + µ)
Your Target is to secure Good Rank in Pre-Medical 2013
HS
MAJOR TEST
18–02–2013
PRE-MEDICAL : ENTHUSIAST COURSE
37.
39.
NCERT, Part - I, Page No. 46,56,58,59 V = slope of x - t graph If sign of v changes, then direction reverses. if v, then a > 0 and if v, then a < 0 NCERT, Part - I, Page No. 101, Right column. PV =
40.
= –2 N-s Impulse = 2 N-s O 48.
C6H5–CH2–NH2 > Ph–NH–C–R Basic strength Aniline does not prepare by Gabriel phthalimide synthesis because this reaction is used to prepare aliphatic primary amine.
52.
(CH3)2CH–C–NH2
M RT Mw x
y = m
NaOH Br2
(CH3)2CH–NH2
COCl2
(CH3)2CH–N=C=O
O
1 slope M (M = const.) w 41.
42.
NCERT, Part - I, Page No. 57 two values of acceleration, velocity or displacement. at one particle time are not possible but two velocities are possible at one value of displacement.
60.
43.
W 684 342 1000 684
T 6kg
a Fk
a 6kg
T
60N
44.
45.
60 – T = 6a .....(i) T = ƒK = 30a T – 30 × 0.1 × 10 = 30a T – 30 = 5(6a) T – 30 = 5 (60 – T) (by eq. i) T – 30 = 300 – 5T 6T = 330 T = 55 N dUI = dUII QI – WI = QII – WII 8 × 105 – 6.5 × 105 = 105 – WII WII = – 0.5 × 105 J work done on the gas = 0.5 × 105 J Impulse = p = m (vƒ – vi)
CH3–CH2–I
66.
72. 74. 75. 77. 78. 79.
CHO CH–OH CH–OH CH–OH CH–OH CH2–OH
HS
CH3–CH2MgI
5 moles of HIO 4 required which gives 5 moles of formic acid & one mole of formaldehyde.
CH3NH2 + CHCl 3 + KOH
CH3–NC + KCl + H2 O (isocyanide test) 6HCHO + 4NH3 (CH2 )6 N4 [Urotropine] HCHO not give iodoform test while acetaldehyde gives. NCERT - 11th, Part -Ist, Page - 195 Ag+ + 2NH3 [Ag(NH3)2]+ ; k' = ? k' = k1 × k2 NCERT- 11th, Part-II, Page - 177 HA + OH¯ A¯ + H2O; k' = ? k' =
1 1 k 10 4 = = a = = 1010 kh kw / ka 10 14 kw
80. 83. 84.
NCERT- 11th, Part-I, Page - 58 Intermolecular attraction a NCERT- 11th, Part-I, Page - 58
85.
S2 O32 Cr2 O27 SO24 Cr 3
87.
FeO . Fe2O3, In FeO , In Fe 2 O3
2
10 10 = 0.5 5 5
Mg Ether
CH3–CH2–CH2–OH
62.
W = 500 J NCERT, Part - I, Page No. 102 and 103
P + I2
(i) HCHO (ii) H 2O
W T1 T2 = Q T 1 1
CH3–CH2–OH
Your Target is to secure Good Rank in Pre-Medical 2013
3
3/4
MAJOR TEST
TARGET : PRE-MEDICAL 2013 (NEET-UG) 90.
102. NCERT(XI)th , Page No. 136 104. NCERT(XI)th , Page No. 131 th
106. NCERT(XI) , Page No. 134 108. NCERT(XII)th , Page No. 84 110. 115. 121. 122.
126. NCERT Pg. # 321 Para-1
H2SO4 + Zn ZnSO4 + H2
2NaOH + Zn Na2ZnO2 + H2 91. NCERT (XII) : Pg. No. 138, Line 93. NCERT (XII) : Pg. No. 139, Line 95. NCERT (XII) : Pg. No. 140, Line 97. NCERT (XII) : Pg. No. 140, Line th 100. NCERT(XI) , Page No. 169
NCERT(XII)th , Page No. 82 NCERT-XI : Pg. No. 35. NCERT-XI : Pg. No. 19/20. NCERT Pg. # 333
18–02–2013
160. NCERT XI Page No. 72, 79 = = = =
6 1 1 6
162. NCERT XI Page No. 73, Ist Para 163. NCERT–XII : Pg. No. 23 (E), Pg. No. 24 (H). 164. NCERT XI Page No. 75, IInd Para 165. NCERT–XII : Pg. No. 34 (E), Pg. No. 36 (H). 166. NCERT XI Page No. 80 172. NCERT Page No. 178 173. NCERT Page no. # 53 174. NCERT Page No. 197 175. NCERT Page no. # 49 176. NCERT Page No. 176 178. NCERT Page No. 241 179. NCERT Page No. 218
124. NCERT Pg. # 324, Para - 2
4/4
Your Target is to secure Good Rank in Pre-Medical 2013
HS
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