MET 2021 Sample Papers
MET 2021 Sample Papers
MET 2021 Sample Papers
SAMPLEPAPERS
(
PHYSI
CS,CHEMISTRY,
MATHS)
Physics
Q. 1 A Zener diode is connected to a battery and a load as shown below :
Option 1:
5 mA, 5 mA, 10 mA
Option 2:
15 mA, 7.5 mA, 7.5 mA
Option 3:
12.5 mA, 5 mA, 7.5 mA
Option 4:
12.5 mA, 7.5 mA, 5 mA
Correct Answer:
12.5 mA, 7.5 mA, 5 mA
Solution:
In the given gure
Q. 2 A car is moving with speed 30 m/s on a circular path of radius 500m. Its speed is increasing at the rate of .What is the acceleration of the
car?
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learned-
Centripetal acceleration -
When a body is moving in a uniform circular motion, a force is responsible to change direction of its velocity.This force acts towards the centre of circle and is called
centripetal force.Acceleration produced by this force is centripetal acceleration.
When a body is moving in circular motion,a centripetal acceleration acts along the radius and is directed towards the center of the crcular path and a
tangential acceleration along the tangent.
Q. 3 In nity number of the masses, each 1 kg , are placed along the X-axis at . The
magnitude of the resultant gravitational potential in terms of gravitational constant G at the origin is
Option 1:
2G
Option 2:
4G
Option 3:
8G
Option 4:
G/2
Correct Answer:
4G
Solution:
As we learn
Sum of mass
distances
- wherein
As we know
GP
As, m =1 kg
So, V= 4G
Q. 4 Equal amount of an ideal monoatomic gas at 400K is lled in two cylinders A and B. The piston A is free to move while that of B is held xed. The
same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 45 K. Then the rise in temperature of gas in
B is
Option 1:
100K
Option 2:
75K
Option 3:
55K
Option 4:
45K
Correct Answer:
75K
Solution:
The gas lled in monoatomic piston A is free to move i.e it is isobaric process. Piston B is xed i.e ir is isochoric process. If same amount of heat is given,
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
Q. 6 Find the value of pushing force P just to move the block. If the coe ecient of friction between block and the oor is given as .
Option 1:
55.6 N
Option 2:
60 N
Option 3:
35.5 N
Option 4:
50 N
Correct Answer:
55.6 N
Solution:
Use
F = Force
R = Reaction
So
Q. 7 Calculate the acceleration of the system as shown in the gure , assume all the surfaces are smooth
Option 1:
g/3
Option 2:
g/4
Option 3:
g/2
Option 4:
g
Correct Answer:
g/3
Solution:
By solving both the equation -
Q. 8 There are 5000 turnes of a wire in every metre length of a long solenoid . If 4 ampere current is owing in rthe solenoid the approximate value
of magnetic eld along its axis at one end will be
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learned
So,
Near one end
Q. 9 The line joining the earth's surface joining the points where the eld is horizontal is
Option 1:
Magnetic Meridian
Option 2:
Magnetic Axis
Option 3:
Magnetic Equator
Option 4:
Magnetic Line
Correct Answer:
Magnetic Equator
Solution:
At a place of Equator -
At magnetic equator, the magnetic eld is in horizontal direction.
Q. 10 A screw gauge gives the following reading when used to measure the diameter of a wire:
1mm of the main scale corresponds to 100 divisions of circular scale. The diameter of the wire is:
Option 1:
100 mm
Option 2:
50.05 mm
Option 3:
50.5 mm
Option 4:
99.5 mm
Correct Answer:
50.5 mm
Solution:
Least count of screw gauge = 0.01mm
= 50.5 mm
Option 2:
It depends on direction of projection of body.
Option 3:
It depends on height of location from where body is projected.
Option 4:
All of these.
Correct Answer:
It depends on height of location from where body is projected.
Solution:
Escape velocity ( in terms of radius of planet) -
Escape velocity
Radius of earth
For di erent height 'g' value would be di erent so it depends on height of location from where body is projected.
Q. 12
What is ratio of acceleration due to gravity at
Option 1:
0.983
Option 2:
0.446
Option 3:
0.892
Option 4:
1.32
Correct Answer:
0.446
Solution:
Variation in 'g' with height -
Radius of earth
Value of g decreases.
Radius of earth
at
at
Q. 13
3 mole of He is mixed with 2 mole of O2 . Then nd the value of for the mixture:
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
for Monoatomic gas
Q. 14 Two springs have a spring constant K1 and K2. These are the extended through a distance x1 and x2 respectively. If their elastic energies are
equal. Then is equal to:
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we learn
- wherein
x= elongation or compression of spring from natural position
U1 = U2
Option 1:
Friction
Option 2:
Inelastic deformation
Option 3:
Heat transfer due to nite temperature di erence between system and surrounding
Option 4:
All of these
Correct Answer:
All of these
Solution:
Condition of reversible process -
1) Complete absence of dissipative force. (Like friction, inelastic deformation etc.)
3) The temperature of system must not di er appreciably from surrounding. (Finite temperature di erence heat transfer will occur which is irreversible)
Q. 16 Two system starting from the same state A return to the same state A after performing a di erent path as shown in graph
Option 2:
Option 3:
Option 4:
None of these
Correct Answer:
Solution:
Non Cyclic Process -
The series of changes involved do not return the system back to its initial state.
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
Option 1:
10.5 Mev
Option 2:
13.4 Mev
Option 3:
23.6 Mev
Option 4:
9.7 Mev
Correct Answer:
23.6 Mev
Solution:
Q value -
is mass of product
So,
Q value -
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
Bragg's law -
= wavelength
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
Given that -
Option 1:
4.5 A
Option 2:
5.6 A
Option 3:
4.5 mA
Option 4:
5.6 mA
Correct Answer:
4.5 mA
Solution:
Knee voltage of P-N junction -
It is de ned as that forward voltage at which the current through the junction starts rising rapidly with increase in voltage .
Q. 22 Centre of the mass of a hollow cone of height "h" from the top of the cone is at a distance
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
Centre of mass lies at a distance of from apex of cone.
Q. 23 A ring is rolling an inclined plane of inclination The minimum value of the coe cient of friction between plane and body for pure rolling is
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we learned
- wherein
for ring
Q. 24 A time variable force is applied on block. Then nd work done by friction force for time interval t=0 to t=2 sec.
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we learned
as
at
and
So
so at, F = 0 at t = 2sec.
When
So
at t = 0 V = 0
Similarly
at t = 0 S = 0
so
Now,
So for t = 0 to t = 2 sec -- S = 0
So
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learned
Relation between RMS speed, average speed and most probable speed -
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
By rearranging
Q. 27 The magnitude of charge of electric dipole is 3.2 x 10-19 C and distance between them 2A0. Then the dipole moment is ( in C- m):
Option 1:
9 x10-5
Option 2:
12x 10-14
Option 3:
6.4 x 10-29
Option 4:
1.6 x 10-9
Correct Answer:
6.4 x 10-29
Solution:
Dipole moment M = q x l
= 6.4 x 10-29 C -m
Q. 28 what should be the height of transmitting antenna if TV telecast is to cover a radius 128 Km
Option 1:
1280 m.
Option 2:
1560 m.
Option 3:
640 m.
Option 4:
320 m.
Correct Answer:
1280 m.
Solution:
Range of transmitting antenna -
height of antenna
Radius of earth
d =1280 meter
Option 1:
static resistance
Option 2:
dynamic resistance
Option 3:
both (a) and (b)
Option 4:
None of the above
Correct Answer:
static resistance
Solution:
To determine resistance of given wire by plotting graph between V versus I -
So we can draw the graph by using Ohm's law and the slope of that graph gives the value of static resistance. -
V= IR
R= Resistance
I = Current
V= Voltage
Q. 30 If and are the instantaneous values of voltage and current then the apparent power of the
circuit is-
Option 1:
2.5 watt
Option 2:
5 watt
Option 3:
3 watt
Option 4:
2 watt
Correct Answer:
2.5 watt
Solution:
Q. 31 In the given gure, the potentiometer wire has a resistance of and length 4m. The balancing length AC for emf 0.5V is
Option 1:
2.2m
Option 2:
2m
Option 3:
4.4m
Option 4:
3.2m
Correct Answer:
2.2m
Solution:
l = 2.2 m
Option 1:
Phase di erence between voltage and current
Option 2:
Time di erence is
Option 3:
Power factor
Option 4:
None of the above
Correct Answer:
Power factor
Solution:
Power factor
Q. 33
Option 1:
Option 2:
Option 3:
Both (1) andd (2)
Option 4:
None of the above
Correct Answer:
Both (1) andd (2)
Solution:
This law is also known as kircho 's Voltage law
For loop A B C D E F A
For loop A B C D G H A
Q. 34 The length of an astronomical telescope of focal lenth of objective lens 'fo' and focal length of eye lens 'fe' for normal vision (relaxed eye) is
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
and
Option 1:
7
Option 2:
6
Option 3:
8
Option 4:
9
Correct Answer:
7
Solution:
Resultant amplitude of two wave -
- wherein
amplitude of wave 1
amplitude of wave 2
phase di erence
here
Q. 36
Light take 9sec. to cover a distance 'd' if the whole medium is lled with water then the time taken to cover the same distance is
Option 1:
12 sec
Option 2:
10 sec
Option 3:
8 sec
Option 4:
14 sec
Correct Answer:
12 sec
Solution:
Q. 37 If we want to reduce De - Brogile wavelength of electron from 10-9 m to 0.25 x 10-9 m then we have to increrase its velocity by:
Option 1:
twice of intial velocity
Option 2:
equal to intial velocity
Option 3:
thrice the itial velocity
Option 4:
four times the intial velocity
Correct Answer:
four times the intial velocity
Solution:
Q. 38 Two rods of same length are arranged as shown. The ends are maintained at temperature and . The temperature at the junction
point (where the area of cross-section changes suddenly) is -
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we learn
- wherein
Let be temperature of junction point and in series rate of heat ow is same.
temperature is
Option 1:
Option 2:
Option 3:
Option 4:
No neutral temperature is possible
Correct Answer:
No neutral temperature is possible
Solution:
At neuttral temperature
Since
So, No neutral temperature is possible because Neutral temperature can never be negative hence no θ is possible.
Q. 40 A 4 m long rod of radius 1cm which is xed from one end is given a twist of 0.4 radians . The shear strain developed in a rod is -
Option 1:
0.002
Option 2:
0.004
Option 3:
0.008
Option 4:
0.001
Correct Answer:
0.001
Solution:
As we learn
Shearing strain -
- wherein
by equating both
Q. 41 Within the elastic limit the ratio of stress and strain is de ned as -
Option 1:
Modulus of elasticity
Option 2:
Rigidity
Option 3:
Plastic coe cient
Option 4:
None of these
Correct Answer:
Modulus of elasticity
Solution:
The ratio of stress and strain up to elastic limit is known as Modulus of elasticity or Young's modulus. It is denoted by E.
Q. 42 The condition where the uid properties at a point in a system is time dependent is called -
Option 1:
Steady ow
Option 2:
Unsteady ow
Option 3:
Laminar ow
Option 4:
Turbulent ow
Correct Answer:
Unsteady ow
Solution:
As we learn
Unsteady ow -
Option 1:
Radio and microwave
Option 2:
Ultravoilet
Option 3:
Infrared
Option 4:
X-Rays
Correct Answer:
Radio and microwave
Solution:
Application of Radio and Microwaves -
Q. 44 The peak value of electric eld of light coming from sun is N/M then the average total density of the electromagnetic wave is
then the value of P is
Option 1:
16
Option 2:
8
Option 3:
32
Option 4:
64
Correct Answer:
32
Solution:
Energy density =
So, P = 32
Q. 45 Find the speed of sound in hydrogen gas at 300K.
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
Q. 46 A pressure wave is represented by . The frequency of the wave is
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
=
Q. 47 A 100 V carrier wave is made to vary between 160 V and 40 V by a modulating signal . What is the modulation index ?
Option 1:
0.4
Option 2:
0.5
Option 3:
0.6
Option 4:
0.3
Correct Answer:
0.6
Solution:
&
So,
Q. 48
Let where then the value of ' ' is given by :
Option 1:
Option 2:
Option 3:
Option 4:
None of these
Correct Answer:
Solution:
Option 1:
Surface tension
Option 2:
Kinetic Energy
Option 3:
Both (1) and (2)
Option 4:
None of these
Correct Answer:
Kinetic Energy
Solution:
As we know that the dimension of Work as well as that of energy is same i.e.,
,
Option 1:
Farad - Metre
Option 2:
Option 3:
Option 4:
Farad
Correct Answer:
Solution:
The standard SI unit for permittivity or absolute permitivity is farad per meter (F/m or F·m−1).
So the answer is -
Chemistry
Q. 1 The correct corresponding order of names of four aldoses with con guration given below
respectively, is:
Option 1:
L-Erythrose, L-Threose, L-Erythrose, D-Threose
Option 2:
D-Threose, D-Erythrose, L-Threose, L-Erythrose
Option 3:
L-Erythrose, L-Threose, D-Erythrose, D-Threose
Option 4:
D-Erythrose, D-Threose, L-Erythrose, L-Threose
Correct Answer:
D-Erythrose, D-Threose, L-Erythrose, L-Threose
Solution:
As we learnt,
Therefore, Option(4) is correct.
Option 1:
It is diamagnetic in nature.
Option 2:
It is a derivative of
Option 3:
oxidises to in acid medium.
Option 4:
It is the super oxide of sodium.
Correct Answer:
It is the super oxide of sodium.
Solution:
As we discussed in
Form oxides
- wherein
4Na+O2
2Ca +
Q. 3 Among the following complexes the one which shows zero crystal eld stabilization energy (CFSE):
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we learnt in
Option 2:
tetrahedral geometry and diamagnetic
Option 3:
square planar geometry and diamagnetic
Option 4:
tetrahedral geometry and paramagnetic
Correct Answer:
tetrahedral geometry and diamagnetic
Solution:
As we learnt in
Hybridisation -
d2sp3 - octahedral
sp3 - tetradedral
sp3 -
dsp2 -
Hence, the option number (2) is correct.
Option 1:
Ti
Option 2:
V
Option 3:
Cr
Option 4:
Mn
Correct Answer:
Mn
Solution:
As we learnt in
Generally, d1- d9 electronic con gurated metal complex show colour according to the
Q. 6 Which of the following liquid pairs shows a positive deviation from Raoult's law:
Option 1:
Water-nitric acid
Option 2:
Benzene-methanol
Option 3:
Water-hydrochloric acid
Option 4:
Acetone-chloroform
Correct Answer:
Benzene-methanol
Solution:
As we learned
Condition for positive deviation -
- wherein
In solution showing the positive type of deviation, the partial pressure of each component of the solution is greater than the vapour pressure as expected
according to Raoult’s law.
In a solution of methanol & benzene methanol molecules are held together due to hydrogen bonding as shown below.
On adding benzene, the benzene molecules get in between the molecule of methanol thus breaking the hydrogen bonds. As the resulting
solution has a weak intermolecular attraction, the escaping tendency of alcohol & benzene molecule from the solution increases. Consequently, the vapour
pressure of the solution is greater than the vapour pressure as expected from Raoult’s law.
Q. 7 The half life of a rst order reaction varies with temperature according to
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we learnt
= Activation Energy
T = Temperature
= Rate constant
- wherein
Option 2:
ethylene glycol
Option 3:
Benzene
Option 4:
Glucose
Correct Answer:
ethylene glycol
Solution:
As we learnt,
Antifreeze -
In cold countries, coolant water is mixed with Glycol making it as an freeze and does not freeze at low temperature.
- wherein
Option 2:
Trigonal bipyramidal
Option 3:
Octahedral shape
Option 4:
Tetrahedral shape
Correct Answer:
Octahedral shape
Solution:
As we have learned
Coordination Polyhedron -
The spatial arrangement of ligands around the central atom de nes a Coordination Polyhedron.
Option 2:
re ection
Option 3:
transmission
Option 4:
scattering of light
Correct Answer:
scattering of light
Solution:
As we learned
Dust particles along with water suspended in the air scatter blue light which reaches our eyes and the sky looks blue to us.
Q. 11 Which of the following enzymes are responsible for biological oxidation and reduction reaction?
Option 1:
Trypsin
Option 2:
Dehydrogenase
Option 3:
Lactose
Option 4:
Amylase
Correct Answer:
Dehydrogenase
Solution:
As we have learned
Oxidoreductase -
- wherein
Dehydrogenase belongs to the group of oxidoreductases that catalyse the oxidation and reduction reactions. These enzymes fall into six categories:
oxygenases, reductases, peroxidases, oxidases, hydroxylases, and dehydrogenases.
Therefore, Option(2) is correct.
Option 1:
Molases
Option 2:
Starch
Option 3:
Both a & b
Option 4:
Glucose
Correct Answer:
Glucose
Solution:
As we learned
Ethanol -
- wherein
Q. 13 The drugs which may interfere with the e ciency of oral contraceptives and increase the failure rates are the following except?
Option 1:
Barbiturates
Option 2:
Rifampicin
Option 3:
Ampicilin
Option 4:
Sulphonamides
Correct Answer:
Sulphonamides
Solution:
As we have learnt,
Sulpha drugs -
Synthetic chemotherapeutic drug which contain - group used for bacterial infection in human
- wherein
Therefore, Option(4) is correct.
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we learnt
or
Option 1:
Option 2:
Option 3:
Both 1 & 2
Option 4:
Correct Answer:
Solution:
As we learnt
Preparation of Alkanes - 1 -
Dihydrogen gas adds to alkenes and alkynes in the presence of nely divided catalysts like platinum, palladium or nickel to form alkanes. This process is
called hydrogenation. These metals adsorb dihydrogen gas on their surfaces and activate the hydrogen – hydrogen bond. Platinum and palladium catalyse
the reaction at room temperature but relatively higher temperature and pressure are required with nickel catalysts.
-
- wherein
Q. 16 Hybridisation of XeF6 is:
Option 1:
Sp3d
Option 2:
Sp3d2
Option 3:
Sp3d3
Option 4:
dsp2
Correct Answer:
Sp3d3
Solution:
XeF6 :
Hybridisation
Hybridisation = Sp3d3
Option 1:
also increases
Option 2:
remains same
Option 3:
decreases
Option 4:
rst increases then decreases
Correct Answer:
decreases
Solution:
As we have learnt,
Le Chatelier’s principle -
E ect of change in temperature
On increasing the temperature, equilibrium shifts to that direction which proceeds with the absorption of heat.
This is an endothermic reaction. Thus, on increasing the temperature, equilibrium shifts in the forward direction.
This is an exothermic reaction. Thus on increasing the temperature, equilibrium shifts in backward direction.
For example:
In this reaction, the product formed is HI and the release of 3000 calories of energy. Thus if temperature is increased then equilibrium will shift
backward and form the reactants.
When np is equal to nr there is no e ect of adding an inert gas either at constant volume or pressure.
When np ≠nr there is no e ect of adding an inert gas at constant volume.
When np ≠nr at constant pressure on adding inert gas equilibrium will shift towards more volume side. e.g., dissociation of ammonia will be more at
constant pressure by adding inert gas like argon (Ar).
The equilibrium constant for an exothermic reaction ( negative ) decreases as the temperature increases.
As the T is increased for an exothermic reaction more of the reactants are formed as the concentration of the reactants increases the value of equilibrium
constant decreases
Q. 18 In the estimation of Sulphur by Carius Method, 0.468g of organic compound gave 668mg . Calculate %S.
Option 1:
19.6
Option 2:
21.7
Option 3:
17.9
Option 4:
16.9
Correct Answer:
19.6
Solution:
As we learnt
- wherein
Q. 19 DDT is an:
Option 1:
Insecticide
Option 2:
fungicide
Option 3:
herbicide
Option 4:
None of these
Correct Answer:
Insecticide
Solution:
As we learnt
Insecticides -
Option 1:
A→R; B→P;C→S;D→Q
Option 2:
A→Q; B→S;C→P;D→R
Option 3:
A→R; B→S;C→P;D→Q
Option 4:
A→Q; B→P;C→S;D→R
Correct Answer:
A→R; B→S;C→P;D→Q
Solution:
Compound + 1% alkaline
KMNO4 Solution --------------> Pink Color Disappear
- wherein
Test of Ferric -
- wherein
- wherein
as we know
1. carbylamine test
2. sodium hydrogen carbonate test
Chloroxylenol
4.Bayer's test
Norethindrone
Therefore, Option(3) is correct.
Option 1:
Sublimation of dry ice
Option 2:
Dissolution of Iodine(s) in water
Option 3:
Synthesis of ammonia from N2 and H2
Option 4:
Dissociation of CaSO4 (s) to CaO(s) and SO3(g)
Correct Answer:
Synthesis of ammonia from N2 and H2
Solution:
As we learnt,
Also,
The decrease in moles of gas in the Haber ammonia synthesis drives the entropy change negative, making the reaction spontaneous only at low
temperatures. Thus higher T, which speeds up the reaction, also reduces its extent.
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
urea-formaldehyde resin -
Urea-formaldehyde (UF), also known as urea-methanal, so named for its common synthesis pathway and overall structure, is a non-
transparent thermosetting resin or polymer. It is produced from urea and formaldehyde. These resins are used in
adhesives nishes,particle-board, medium-density breboard (MDF), and moulded objects.
- wherein
This is a urea-formaldehyde polymer and its monomers are urea and formaldehyde.
Item - I Item-II
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
High density Polythene (HDP) -
- Polymerization of ethene in a hydrocarbon solvent under low pressure in presence of triethylaluminium and titanium tetrachloride (Ziegier. natta catalyst)
- wherein
Polyacrylonitrile -
- wherein
- Hompolymer
Nylon 6 -
- Obtained by heating caprolactam with water.
- Homopolymer
- wherein
--
(H.D.P)
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
Preparation of alkyl/ aryl halide by the electrophilic substitution reaction -
Benzene undergo electrophilic substitution reaction when treated with Cl2 in presence of lewis acid (FeX3 / AlX3).
- wherein
This is Intramolecular fridel craft acylation which provides a useful synthetic approach to cyclic ketones. The cyclization is well-suited to the preparation of six-
and particularly ve-membered rings
Q. 25 The major product of the following reaction is :
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
Oxidation of primary alcohol or an aldehyde -
Primary alcohol oxidizes in the presence of oxidising agent to give aldehyde which further oxidises to give the carboxylic acid.
- wherein
Option 1:
Calgen's Method
Option 2:
Clark's Method
Option 3:
Ion-exchange method
Option 4:
Synthetic Resins method
Correct Answer:
Clark's Method
Solution:
Clark's Method -
This method is used for removing the temporary hardness of the water.
Option 1:
penicillin G
Option 2:
ampicillin
Option 3:
amoxycillin
Option 4:
chloramphenicol
Correct Answer:
penicillin G
Solution:
- wherein
Eg: Penicillin G
Penicillin G is a narrow spectrum antibiotic. It is used for the speci c infection when the causative organism is known and will not kill as many of the normal
microorganisms as a broad spectrum antibiotic in the body.
Therefore, Option(1) is correct
Q. 28 The de-Broglie wavelength of a tennis ball of mass 60g moving with a velocity of 10 metres per second is approximately:
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
De-broglie wavelength
v its velocity
p its momentum
Bohr model and de-broglie principle :
Bohr model,
Debroglie,
Thus, we have:
Therefore, Option(1) is correct
Option 1:
Li, Na, K
Option 2:
Be, Mg, Cr
Option 3:
Ca, Sr, Ba
Option 4:
Cl, Br, I
Correct Answer:
Be, Mg, Cr
Solution:
Introduction of Periodic Table -
J.W. Dorbereiner pointed out that within a group of three elements having similar chemical and physical properties, the atomic weight of the middle element
is the mean of the other two. Some examples of such triads are given below. He also pointed out the triad - iron, cobalt and nickel in which the atomic
weights of the elements are almost the same.
Triads Elements Li Na K Ca Sr Ba S Se Te Cl Br I
Though it was the rst successful attempt to rationalize the problem, it could not be generalised or extended.
Drawback or Limitation: All the known elements could not be arranged as triads.
John Alexander Reina newland in England made the rst attempt to correlate the chemical properties of the elements with their atomic weight. According to
him -
1. If the elements are arranged in order to their increasing atomic weights, every eighth element had similar properties to rst one like the rst and eighth
note in music. For example
Sa Re Ga Ma Pa Dha Ni Sa
Li Be B C N O F Na
Na Mg Al Si P S Cl K
We know that according to Dobereiner's law of triads the atomic mass of the middle element of a triad is the arithmetic mean of the atomic masses of the
other two elements.
The group which does not follow Dobereiner’s law of triad is Be, Mg, Cr.
It is because Cr is not of the same group as Be and Mg are. Be and Mg belong to group second whereas Cr belongs to group sixth.
Moreover, the arithmetic mean of the atomic masses of the other two elements is not equal to the atomic mass of the middle element.
Arithmetic mean = 51.9+82=29.9. This arithmetic mean is not equal to atomic mass of central element.
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
Energy Level Diagram for Molecules -
The relative energy levels of atomic and molecular orbitals are typically shown in a molecular orbital diagram. As given in the gure below, for a diatomic
molecule, the atomic orbitals of one atom are shown on the left, and those of the other atom is shown on the right. Each horizontal line represents one
orbital that can hold two electrons. The molecular orbitals formed by the combination of the atomic orbitals are shown in the center. Dashed lines show
which of the atomic orbitals combine to form the molecular orbitals. For each pair of atomic orbitals that combine, one lower-energy (bonding) molecular
orbital and one higher-energy (antibonding) orbital result. Thus we can see that combining the six 2p atomic orbitals results in three bonding orbitals (one σ
and two π) and three antibonding orbitals (one σ* and two π*).
22
The molecular orbitals are lled in the same manner as atomic orbitals, using the Aufbau principle and Hund’s rule.
As we have learnt,
Q. 31 Calculate the (in m/sec) of if its density at 1 atm pressure and is 1.429g/l.
Option 1:
462.21
Option 2:
46.12
Option 3:
56.31
Option 4:
44.23
Correct Answer:
462.21
Solution:
Root Mean Square Speed urms
It is the square root of the mean of the square of the velocities of di erent molecules.
-
Now,
Therefore, Option(1) is correct
Option 1:
Temperature
Option 2:
Method of expressing activity or active mass
Option 3:
Both 1 and 2
Option 4:
Volume
Correct Answer:
Both 1 and 2
Solution:
Equilibrium constants are changed if you change the temperature of the system. Kc or Kp is constant at a constant temperature, but they vary as the
temperature changes.
The equilibrium constant K is determined by the activities of the components in the equilibrium expression.
The value of Kc and Kp can be di erent in magnitude as well as dimensions.
Option 1:
0.295
Option 2:
-0.295
Option 3:
-0.59
Option 4:
0.59
Correct Answer:
-0.295
Solution:
Therefore,option(2) is correct
Option 1:
to move electrons from the electrolyte
Option 2:
to maintain charge balance
Option 3:
for both (1) and (2)
Option 4:
To form precipitate.
Correct Answer:
to maintain charge balance
Solution:
The purpose of a salt bridge is not to move electrons from the electrolyte, rather to
maintain charge balance because the electrons are moving from one half cell to the
other.
Without the salt bridge the solution in the anode compartment would become positively charged and the solution in the cathode compartment would
become negatively charged, because of the charge imbalance, the electrode reaction would quickly come to halt, therefore it helps to maintain the ow of
electrons from the oxidation half cell to a reduction half cell.
Therefore, option(2) is correct
Q. 35 The rate of a reaction triples when temperature changes from 20oC to 50oC. Calculate the energy of activation for the reaction? (R = 8.314 J K-
1mol-1).
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learned,
Complex Reaction - Mechanism of Reaction -
The reactions, which occur in single step, are called simple or elementary reactions. For example:
An elementary reaction is an individual molecular event that involves breaking or making of chemical bonds. The overall reaction describes the
stoichiometry of the overall process but provides no information how the reaction occurs.
Complex Reaction
Important Facts:
The numbeer of reactant molecules taking part in an elementary step or in an elementary reaction is expressed as molecularity of that step of
molecularity of that reaction respectively.
For elementary reactions usually order of reaction and molecularity are same. Thus, it can be said that if order of a reaction for a change is fractional it
cannot be an elementary reaction.
-
Therefore, option(2) is correct
Option 1:
Sodium carbonate
Option 2:
Limestone
Option 3:
Sodium chloride
Option 4:
Silica
Correct Answer:
Limestone
Solution:
As we have learnt,
Process in Metallurgy -
There are various stages of metals before it comes into pure form. Hence we need to perform certain steps to extract a particular metal. We will discuss the
di erent processes which are required for extracting these metals.
1. Crushing and grinding: The rst process in metallurgy is the crushing of ores into a ne powder. This process is known as pulverization.
2. The concentration of ores: The process of removing impurities from ore is known as a concentration of minerals or ore dressing. The concentration
of ores can be done by the following methods
Hydraulic Washing
Magnetic Separation
Froth Floatation Method
Leaching
3. Roasting: In metallurgy, the process of heating a concentrated ore in the presence of oxygen is known as roasting. This process is applied in the case
of sul de ores.
4. Calcination: For ores containing carbonate or hydrated oxides, heating is done in the absence of air to melt the ores, and this process is known as
calcination.
5. Re ning: The impurities contained in the metal product of the roasting/reduction process are removed.
Basic ux like limestone(CaCO3) is used to remove acidic impurities such as SiO2. The process occurs as follows:
Option 1:
Five
Option 2:
One
Option 3:
Four
Option 4:
Two
Correct Answer:
One
Solution:
As we have learnt,
Water is neutral in nature. pH of the pure water is 7. It is a weak electrolyte and ionises into H+ and OH- ions.
It reacts with active metals and evolves hydrogen The reaction is exothermic in the case of alali and alkaline earth metals.
Reaction with non-metals: Water reacts with non-metals like uorine and chlorine as follows:
Action on non-metallic oxides: Acidic oxides combine with water to form acids viz:
Action on hydrides carbides, nitrides, phosphides: Water decomposes these compounds with the ration of hydrogen acetylene, ammonia,
phosphine respectively.
Hydrolysis: Many salts undergo hydrolysis with water.
Decomposition: Water containing either alkali or acid when electrolysed gets decomposed into H2 and O2.
Water of crystallisation: It combines with many salts during crystallisation to form hydrates. For example, CuSO4.5H2O, FeSO4.7H2O, etc.
Water as a catalyst: Water acts as a catalyst in many reactions. Perfectly dry gases generally do not react but the presence of moisture brings the
chemical change. Ammonia and hydrochloric acid gas combine only in presence of moisture.
In CuSO4.5H2O, Cu is cordinated with 4 water molecules. Now fth water molecule is hydrogen bonded and is deeply embedded in the crystal. It is not
cordinated. Only 4 water molecules are coordinated and fth is only hydrogen bonded.
Therefore, Option(2) is correct
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
Let us rst look at the structures of the given compounds
Therefore, Option(1) is correct.
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learnt,
Grignard Reagent - 1 -
All three types of monohydric alcohols can be prepared by the use of Grignard reagents. Grignard reagents form addition compounds by nucleophile attack
with aldehydes and ketones which on hydrolysis with dilute acid yields alcohol.
Mechanism
For example:
NaBH4 can only reduce keto groups. But LiAlH4 can reduce even anhydrides and esters. LiAlH4 is a very good reducing agent because (Al) atom present in ir is
more covalent than (B) atom in NaBH4. Therefore, Al has more tendency to gain the electrons, thus, it will try to keep the electrons to itself and hence H- will
go in a particular manner. Thus, LiAlH4 is better reducing agent than NaBH4.
Mechanism
The mechanism for LiAlH4 occurs in the following steps:
1. Deprotonation
2. Nucleophilic attack by the hydride ion
3. Nucleophilic attack by the hydride ion
4. Leaving group removal
5. Alkoxide is protonated
Therefore, Option(2) is correct.
Q. 40 The number of orbitals associated with quantum numbers
is:
Option 1:
25
Option 2:
11
Option 3:
15
Option 4:
50
Correct Answer:
25
Solution:
As we have learnt,
Quantum Numbers -
Quantum numbers:
They are the set of four numbers which explain the state of electron i.e., location, energy, type of orbital, orientation of orbital, etc. in an atom. Various
quantum numbers are as follows:
Principal quantum number(n):
It represents the principal shell of an atom. It can have integral values except zero like 1,2,3,.... Also denoted as K,L,M,.....etc.
Maximum number of electrons in a principal shell can be 2n2 where n is principal quantum number.
1. Value of l is 0
2. m has value=0
For ‘p’ subshell :
1. Value of l is 1
Thus, an electron can have only two possible values of this quantum number, either or respectively.
We have:
l = 0 to 4
Therefore, Option(1) is correct.
Q. 41 Two solutions A and B each of 100L was made by dissolving 4g of NaOH and 9.8g of H2SO4 in water, respectively. The pH of the resultant
solution obtained from mixing 40L of solution A and 10L of solution B is_________:
Option 1:
pH = 10.6
Option 2:
pH = 12.7
Option 3:
pH = 9.8
Option 4:
pH = 11
Correct Answer:
pH = 10.6
Solution:
As we have learnt,
pH of solution/mixture -
Mixture of Strong Bases:
Mixture of Strong Acid and Strong Base:
Now, 40L of NaOH solution and 10L of H2SO4 solution are added, thus we get:
Total volume = 50L
[OH-] = 4x10-4
pOH = -log[4x10-4]
Further, we know:
pH = 14 - 3.4
pH = 10.6
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learnt,
Alkylation and Acylation of Amines -
Alkylation
Amines undergo alkylation with RX and undergo complete methylation and this is called exhaustive methylation, but with Me2SO4 amines undergo
monomethylation. 1o and 2o amines are also methylated by heating HCHO and excess of HCOOH at 100oC. This reaction is known as Eschweiler-Clarke
methylation. The reaction occurs as follows:
Acylation
1o and 2o aliphatic and aromatic amines react with acid chlorides (RCOCl), anhydrides and esters by SN2 reaction is called acylation reaction. The reaction is
carried out in the presence of a base stronger than amine, such as pyridine, which removes HCl so formed and shifts the equilibrium to the product side. The
reaction occurs as follows:
-
Therefore, Option(1) is correct.
Option 1:
C>A>B
Option 2:
C>B>A
Option 3:
B>A>C
Option 4:
B>C>A
Correct Answer:
C>B>A
Solution:
As we have learnt,
Carbanions -
The carbon species carrying a negative charge on carbon atom is called carbanion. Carbon in carbanion is generally sp3 hybridised and its structure is
distorted tetrahedron as shown in the gure given below. Carbanions are also unstable and reactive species.
The resonance e ect is de ned as ‘the polarity produced in the molecule by the interaction of two π-bonds or between a π-bond and lone pair of electrons
present on an adjacent atom’. The e ect is transmitted through the chain. There are two types of resonance or mesomeric e ect designated as R or M e ect.
Positive Resonance E ect (+R e ect): In this e ect, the transfer of electrons is away from an atom or substituent group attached to the conjugated
system. This electron displacement makes certain positions in the molecule of high electron densities. This e ect in aniline is shown as:
Negative Resonance E ect (- R e ect): This e ect is observed when the transfer of electrons is towards the atom or substituent group attached to
the conjugated system. For example, in nitrobenzene, this electron displacement can be depicted as:
The atoms or substituent groups, which represent +R or –R electron displacement e ects are as follows :
The presence of alternate single and double bonds in an open-chain or cyclic system is termed as a conjugated system. These systems often show abnormal
behaviour. The examples are 1,3- butadiene, aniline and nitrobenzene etc. In such systems, the π-electrons are delocalised and the system develops polarity.
When a negative charge is delocalised with an electron-withdrawing group like (NO2) then stability increases.
(B) The negative charge is delocalised with the carbon of the alkene
Therefore, Option(2) is correct.
Q. 44 Consider the following reactions:
(a)
(b)
(c)
(d)
Option 2:
b and d
Option 3:
b, a and d
Option 4:
a and d
Correct Answer:
b and d
Solution:
As we have learnt,
Preparation of Aryl Halides -
The ortho and para isomers can be easily separated due to large di erence in their melting points. Reactions with iodine are reversible in nature and
require the presence of an oxidising agent (HNO3, HIO4) to oxidise the HI formed during iodination. Fluoro compounds are not prepared by this
method due to high reactivity of uorine.
1. Vinyl halides (C) do not give Fridel Craft's reaction because of unstable carbocation formation.
2. Aryl halides (A) do not give Fridel Craft's reaction with Benzene because of formation of unstable phenyl carbocation.
Therefore, Option(2) is correct.
Q. 45 The molarity of HNO3 in a sample which has density 1.4 g/mL and mass percentage of 63% is ____ (Molecular weight of HNO3 = 63)
Option 1:
14
Option 2:
15
Option 3:
12
Option 4:
10
Correct Answer:
14
Solution:
Mass of HNO3 is 63 g
Moles of HNO3 is 1
Q. 46 Which of the following compounds is likely to show both Frenkel abnd Schottky defects in its crystalline form ?
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learnt,
Imperfections in Solids -
Stoichiometric Defects
Those compounds in which the number of positive and negative ions are exactly in the ratio indicated by their chemical formula are called
stoichiometric compounds example, NaCl. These solids show following types of defects:
1. Vacancy Defect: When some of the lattice sites are vacant. the crystal is said to have a vacancy defect.
This results in a decrease in the density of the substance. This defect can also develop when a substance is heated.
2. Interstitial Defect: When some constituent particles (atoms or molecules) occupy an interstitial site, the crystal is said to have an interstitial
defect.
This defect increases the density of the substance. Vacancy and interstitial defects as explained above can be shown by non-ionic solids. Ionic
solids must always maintain electrical neutrality. Rather than simple vacancy or interstitial defects, they show these defects as Frenkel and
Schottky defects.
3. Frenkel Defect: This defect is shown by ionic solids. The smaller ion (usually cation) is dislocated from its normal site to an interstitial site.
It creates a vacancy defect at its Original site and an interstitial defect at its new location. Frenkel defect is also called dislocation defect. It does
not change the density of the solid. Frenkel defect is shown by ionic substance in which there is a large di erence in the size of ions, for example,
ZnS, AgCl, AgBr and Agl due to small size of Zn2+ and Ag+ ions.
4. Schottky Defect: It is basically a vacancy defect in ionic solids. In order to maintain electrical neutrality, the number of missing cations and
anions are equal.
Like simple vacancy defect, Schottky defect also decreases the density of the substance. Number of such defects in ionic solids is quite
signi cant. For example, in NaCl, there are approximately 106 Schottky pairs per cm3 at room temperature. In 1 cm3 there are about 1022 ions.
Thus, there is one Schottky defect per 1016 ions. Schottky defect is shown by ionic substances in which the cation and anion are of almost similar
sizes. For example, NaCl, KCl, CsCl and AgBr. It may be noted that AgBr shows both, Frenkel as well as Schottky defects.
Impurity Defects
If molten NaCl containing a little amount of SrCl2 is crystallised, some of the sites of Na+ ions are occupied by Sr2+. Each Sr2+ replaces two Na+ ions. It
occupies the site of one ion and the other site remains vacant. The cationic vacancies thus produced are equal in number to that of Sr2+ ions. Another
similar example is the solid solution of CdCl2 and AgCl.
Non-Stoichiometric Defects
There are many compounds in which the ratio of positive and negative ions present in the compound di ers from the required by ideal formula of the
compound. Such compounds are called Non-stoichiometric compounds. For example, VOx
In these compounds, a balance of positive and negative charges is maintained by having extra electrons or extra positive charge. These defects are of
following types:
A compound may have excess metal ion if a negative ion is absent from its lattice site, leaving a hole which is occupied by an electron to
maintain electrical neutrality.
1. The holes occupied by electrons are called F -centres and are responsible for the colour of the compound.
5. Greater the number of F-centres greater is the intensity of colour. This type of defects is found in a crystal which is likely to possess
Schottky defects.
Due to Cationic Vacancies
2. Electrical neutrality is maintained by the presence of an extra electron in the Interstitial site.
3. These types of defects are exhibited by the crystals which are likely to exhibit Frenkel defects. Example,— Yellow colour of ZnS.
Therefore, Option(1) is correct.
Q. 47 The solubility product of at is . The concentration of hydroxide ions in a saturated solution of will be
:
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
Q. 48 At constant volume, of an ideal gas when heated from to changes its internal energy by . The molar heat capacity
at constant volume is ______.
Option 1:
Option 2:
5.85
Option 3:
5.25
Option 4:
6.0
Correct Answer:
Solution:
As we have learnt,
Q. 49
Kjeldahl's method can be used for:
Option 1:
Option 2:
Option 3:
Azobenzene
Option 4:
Pyridine
Correct Answer:
Solution:
Kjeldahl's Method -
Kjeldahl's method is a method for the quantitative determination of nitrogen contained in organic substances as well as nitrogen contained in unorganic
compounds ammonia and ammonium.
But, Kjeldahl's method cannot be used for nitrogen determination of compounds having nitro group or azo-group or nitrogen present in rings as the nitrogen
of these compounds cannot be converted to ammonium sulphate under the condition of this method.
Q. 50 Which of the following sulphide is completely precipitated only when the acidic solution is made dilute?
Option 1:
HgS
Option 2:
PbS
Option 3:
CuS
Option 4:
CdS
Correct Answer:
CdS
Solution:
As we have learnt,
Con rmatory tests for the anions which react with concentrated sulphuric acid are given below in the Table:
Con rmatory tests for Cl–, Br–, I–, NO3– and C2O42-
-
CdS is only completely precipitated when the acidic solution is made dilute.
Therefore, Option(4) is correct.
Q. 51 If are natural numbers such that , then the minimum possible value of is
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
Q. 52 If are four distinct number chosen from the set then, the minimum value of is
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
Q. 53 Find out the pair of words from the options which exhibits the same relationship as shown by the question pair of words
Renegade: Traitor
Option 1:
Ghetto: Colony
Option 2:
Hit: Destroy
Option 3:
Exhilarate: Excruciate
Option 4:
Damp: Lively
Correct Answer:
Ghetto: Colony
Solution:
The question pair of words contains synonyms, similarly, ghetto and colony are also synonyms.
Q. 54 Find out the pair of words from the options which exhibits the same relationship as shown by the question pair of words:
Implore: Command
Option 1:
Brutal: Humane
Option 2:
Gregarious: Gracious
Option 3:
Stupendous: Staggering
Option 4:
Monotonous: Tedious
Correct Answer:
Brutal: Humane
Solution:
The question pair of words contains antonyms, similarly, brutal and humane are also antonyms.
Q. 55 Find out the pair of words from the options which exhibits the same relationship as shown by the question pair of words:
Haggle: Bargain
Option 1:
Razor: Obliterate
Option 2:
Attack: Exterminate
Option 3:
Obdurate: Obstinate
Option 4:
Impersonate: Paragon
Correct Answer:
Attack: Exterminate
Solution:
Haggling is done to bargain, similarly, attacking is done to exterminate or kill.
Q. 56 Find out the pair of words from the options which exhibits the same relationship as shown by the question pair of words:
Poignant: Marked
Option 1:
Powerful: Potent
Option 2:
Embellish: Admonish
Option 3:
Braggart: Flaunt
Option 4:
Sensational: Moribund
Correct Answer:
Powerful: Potent
Solution:
The question pair of words contains synonyms, similarly, powerful and potent are synonyms
Q. 57 Find out the pair of words from the options which exhibits the same relationship as shown by the question pair of words
Proselytize: Faith
Option 1:
Fly: Nation
Option 2:
Elevate: Exile
Option 3:
Relegate: Delegate
Option 4:
Retail: Retailer
Correct Answer:
Fly: Nation
Solution:
Proselytize is a verb which denotes moving into another faith; ying may denote moving into another nation
Q. 58 Find out the pair of words from the options which exhibits the same relationship as shown by the question pair of words:
Inordinate: Nondescript
Option 1:
Convoluted: Tortuous
Option 2:
Bawdy: Decent
Option 3:
Decrepit: Dilapidated
Option 4:
Mercenary: Vindictive
Correct Answer:
Bawdy: Decent
Solution:
The question pair of words contains antonyms, similarly, bawdy and decent are antonyms
Q. 59 Find out the pair of words from the options which exhibits the same relationship as shown by the question pair of words:
Plateau: Altitude
Option 1:
Meticulous: Victory
Option 2:
Fraternity: Disunity
Option 3:
Palpable: Sensible
Option 4:
Astonish: Dexterous
Correct Answer:
Meticulous: Victory
Solution:
By being on plateau (the raised part of land) the altitude is enhanced, similarly, by being meticulous, victory may be achieved.
Q. 60 Rearrange the following paragraph to answer the following question
(A) Therefore, it is an encouraging symbol, but we must watch against our rising pace of populace expansion.
(C) Even though this modi cation is sluggish and steady, it appears to be unswerving.
(D) The degree of success of this programme can be gauged when we examine the share of citizens underneath de ciency line.
Option 1:
A
Option 2:
B
Option 3:
C
Option 4:
D
Correct Answer:
C
Solution:
The correct arrangement of sentences is: BCAD
B- This is the opening line of the paragraph which establishes the idea of de ciency mitigation
C- C and B form a mandatory pair. Here ‘this modi cation refers to de ciency mitigation.
(A) Therefore, it is an encouraging symbol, but we must watch against our rising pace of populace expansion.
(C) Even though this modi cation is sluggish and steady, it appears to be unswerving.
(D) The degree of success of this programme can be gauged when we examine the share of citizens underneath de ciency line
Option 1:
A
Option 2:
B
Option 3:
C
Option 4:
D
Correct Answer:
D
Solution:
The correct arrangement of sentences is: BCAD
B- This is the opening line of the paragraph which establishes the idea of de ciency mitigation
C- C and B form a mandatory pair. Here ‘this modi cation refers to de ciency mitigation.
(A) Therefore, it is an encouraging symbol, but we must watch against our rising pace of populace expansion.
(C) Even though this modi cation is sluggish and steady, it appears to be unswerving.
(D) The degree of success of this programme can be gauged when we examine the share of citizens underneath de ciency line
Option 1:
A
Option 2:
B
Option 3:
C
Option 4:
D
Correct Answer:
A
Solution:
The correct arrangement of sentences is: BCAD
B- This is the opening line of the paragraph which establishes the idea of de ciency mitigation
C- C and B form a mandatory pair. Here ‘this modi cation refers to de ciency mitigation.
Maths
Q. 1 If is less than or equal to 500y and y is (0-1), If y is 0, then will be
Option 1:
equal to 0
Option 2:
less than 0
Option 3:
Option 4:
=500
Correct Answer:
equal to 0
Solution:
Solution of Linear Programming Problems -
As we learnt in
This method of solving a LPP graphically is based on the principle of extreme points theorem.
Q. 2 The points (x,3) satis es the inequality , -5x-2y 13, nd the smallest possible value of x?
Option 1:
-1.4
Option 2:
1.4
Option 3:
-3.8
Option 4:
3.8
Correct Answer:
-3.8
Solution:
Solution of Linear Programming Problems -
-
This method of solving a LPP graphically is based on the principle of extreme points theorem.
Q. 3
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learned
Option 1:
81/2
Option 2:
81/9
Option 3:
81
Option 4:
none of these
Correct Answer:
81/2
Solution:
As we have learned
If we have two functions intersection each other.First nd the point of intersection. Then integrate to nd area
- wherein
cut
Q. 5
If f (x) = Then
Option 1:
Option 2:
Option 3:
Option 4:
none of this
Correct Answer:
Solution:
As we have learned
ex:
- wherein
Let
Q. 6 In how many ways can 5 women and 3 men be seated in a row so that no two men are together
Option 1:
6 ! 5!
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learned
- wherein
Where
It can be done in 5 ! ways for such arrangement, 3 men can be seated only at X position
XWXWXWXWXWX
Q. 7 In expansion of
Option 1:
3
Option 2:
6
Option 3:
7
Option 4:
none of these
Correct Answer:
7
Solution:
As we have learned
Coe cient of x^{R} -
and
compare :
nd
- wherein
Take a in terms of x.
Q. 8
If
Option 1:
1
Option 2:
2
Option 3:
mulitple of 15
Option 4:
none of these
Correct Answer:
2
Solution:
As we have learned
Sum of the binomial coe cients of the odd term is equal to sum of the binomial coe cients of even term and each is equal to .
- wherein
Q. 9
If in ratio of
Option 1:
4
Option 2:
5
Option 3:
6
Option 4:
217
Correct Answer:
4
Solution:
As we have learned
If the three consecutive terms in the expansion of are in the ratio then the value of is given by
- wherein
Take term
Q. 10 What is the equation of an upward opening parabola with distance between its focus and directrix as 7 and passing through origin as vertex
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learned
- wherein
2a = 7
a = 7/2
Q. 11 What is the name of the line segment through the center and perpendicular to the major axis of an ellipse ?
Option 1:
Principal Axis
Option 2:
Minor Axis
Option 3:
Major Axis
Option 4:
Axis
Correct Answer:
Minor Axis
Solution:
As we have learned
Minor axis -
The line segment through the centre and perpendicular to the major axis.
- wherein
Q. 12 Write the equation of rectangular hyperbola in standard form with x-axis as transverse axis and origin as center with length of transverse axis as
8
Option 1:
Option 2:
Option 3:
Option 4:
none of these above
Correct Answer:
Solution:
As we have learned
Option 1:
3x+y +4 = 0
Option 2:
x+3y-4 = 0
Option 3:
y = 3x+4
Option 4:
none of these
Correct Answer:
3x+y +4 = 0
Solution:
As we have learned
- wherein
Option 1:
Option 2:
Option 3:
Option 4:
None of these
Correct Answer:
Solution:
As we learned
- wherein
denotes conjugate of z.
Q. 15 Modulus of 2 complex no's are 4 and 7 and modulus of sum of these complex numbers is 9, then
Option 1:
16
Option 2:
2
Option 3:
11
Option 4:
0
Correct Answer:
16
Solution:
As we learned
- wherein
denotes conjugate of z
So,
81=16+49+
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we learned
- wherein
Here,
&, Arg
Option 1:
Option 2:
Option 3:
Option 4:
None of these
Correct Answer:
Solution:
As we learned
- wherein
Arg
So,
Q. 18
What is the argument of zw, such that and .
Option 1:
Option 2:
Option 3:
Option 4:
None of these
Correct Answer:
Solution:
As we learned
Property of Argument of a Complex Number -
- wherein
and it is chosen such that Arg(z.w) lies in the principal value range of Argument.
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learned
Normal form -
- wherein
p is the length of the perpendicular segment from the origin and is the angle made by this perpendicular with +ve -axis.
Q. 20 For what value of angle between intersecting plane and vertical axis of adouble napped circular cone with as semi vertical angle , we get
ellipse ?
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learned
- wherein
Q. 21 Find the equation of circle whose center is in 2nd quadrant and radius is 3 units , Also , the circle touches both the axes
Option 1:
Option 2:
Option 3:
Option 4:
none of these
Correct Answer:
Solution:
As we have learned
Hence
C = (-3,3)
= (-g,-f)
r = 3
Q. 22 Find the parameter of a point (3,-4) on a circle whoese center is (2,-4 + ) and radisu is 2 units
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learned
Parametric form -
- wherein
x = 3 , y = -4
h = 2
Option 1:
Option 2:
Option 3:
Option 4:
None of these
Correct Answer:
Solution:
As we learned
Transformation of equation -
To nd equation whose roots are symmetrical functions of and , Where & are roots of some other equation.
- wherein
Take any of the roots to be equal to & calculate or accordingly in terms of & satisfy the given equation to get the required equation.
now,
Q. 24 Which of the following gives "if p then q " true :
Option 1:
p is false , q is false
Option 2:
p is false , q is true
Option 3:
p is true , q is true
Option 4:
all of the above
Correct Answer:
all of the above
Solution:
As we have learned
By assuming that p is true, prove that q must be true. By assuming that q is false, prove that p must be false.
Q. 25 What is the no. of ways to divide 100 persons into 20 equal groups?
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
The number of ways in which mn di erent things can be divided equally into m groups is .
Q. 26 What is the rank of word "AGAIN" in a dictionary with all possible arrangements of its letters?
Option 1:
7
Option 2:
12
Option 3:
8
Option 4:
15
Correct Answer:
7
Solution:
As we learnt
- wherein
Q. 27 What is the no. of positive integral solutions of : x+y+z=8 and zero values of x, y, z are included?
Option 1:
42
Option 2:
21
Option 3:
45
Option 4:
90
Correct Answer:
45
Solution:
As we have learnt in
Number of non-negative integral solutions of the equation x1 + x2 + x3 +......... + xr=n is .
- wherein
Where
Option 1:
9
Option 2:
45
Option 3:
10
Option 4:
None of these
Correct Answer:
10
Solution:
As we learned
- wherein
=9+1
=10
Q. 29
General Solution of the D.E
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we learnt
Put
Option 2:
Population of mice increase at constant rate
Option 3:
both (a) and (b)
Option 4:
None of these
Correct Answer:
both (a) and (b)
Solution:
As we learnt
- wherein
* Temperature Problems
* Dilution Problems
for (a)
for (b)
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we learnt
- wherein
P is the function of x alone
If=
General Solution
Option 1:
is continuous at
Option 2:
is discontinuous with a removable discontinuity at
Option 3:
is discontinuous with an irremovable discontinuity at
Option 4:
None of the above
Correct Answer:
is discontinuous with a removable discontinuity at
Solution:
As we have learnt,
Removable discontinuity -
- wherein
, ,
Q. 33
is equal to:
Option 1:
0
Option 2:
1
Option 3:
2
Option 4:
3
Correct Answer:
3
Solution:
As we learned
L - Hospital Rule -
- wherein
=
=
Q. 34
Find
Option 1:
1
Option 2:
Option 3:
Option 4:
Correct Answer:
1
Solution:
As we learned
so,
Applying Lim
=1
Q. 35 Find
Option 1:
3.1
Option 2:
3.11
Option 3:
3.007
Option 4:
3.012
Correct Answer:
3.007
Solution:
As we learn
Approximation -
- wherein
Let
So,
Q. 36 Find the local maximum value of ?
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we learn
- wherein
Option 1:
f is one-one and f is onto
Option 2:
f is many-one and g is onto
Option 3:
f is one-one and g is onto
Option 4:
g is one-one and g is onto
Correct Answer:
f is one-one and g is onto
Solution:
As we learnt
Option 1:
(-3,-9)
Option 2:
Option 3:
(1,3)
Option 4:
None of these
Correct Answer:
Solution:
As we learnt
We have x-3=y
and 2x=-x+y
x = y and x - x - 3 = 0
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we learned from
Independent events -
If A and B are independent events then probability of occurrence of A is not a ected by occurrence or non occurrence of event B.
and
so
Q. 40 What is the probability of getting a king in rst withdrawl a red card in second withdrawl from a deck of 52 cards without replacment ?
Option 1:
2/ 52
Option 2:
1/ 51
Option 3:
1/26
Option 4:
2/ 13
Correct Answer:
1/26
Solution:
As we learned
Independent events -
- wherein
Q. 41 What is the probability of getting a red ball from a bag 1 with 2 red and 4 black balls or from bag 2 with 3 red and 2 black balls, where the
probability of selecting bags is equal:
Option 1:
1/2
Option 2:
14/15
Option 3:
7/15
Option 4:
7/8
Correct Answer:
7/15
Solution:
We leaarned from
Let S be the sample space and E1, E2, ......En be n mutually exclusive and exhaustive events associated with a random experiment.
- wherein
Option 2:
4/243
Option 3:
4/81
Option 4:
1/81
Correct Answer:
4/243
Solution:
As we learned from
Binomial Distribution -
In a series of n independent trials if the probability of success P in each trial is same, then the probapility of r success is
- wherein
Q. 43 Find standard deviation of
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we learned
Standard Deviation -
If x1, x2...xn are n observations then square root of the arithmetic mean of
- wherein
Where
Q. 44 If mean of 30 class Xth students marks are 15 and mean of 20 class XIIth students marks is 18, then combined mean of marks of this 50 student
is
Option 1:
16.2
Option 2:
17.5
Option 3:
16
Option 4:
17
Correct Answer:
16.2
Solution:
As we learned
Combined Mean -
If x1 and x2 be the means of two related groups having n1 and n2 items respectively then the combined mean of both the groups is given by
-
Q. 45
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learnt
- wherein
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution:
As we have learnt
This shows the formulae for the conversion of half angles and their doubles.
Q. 47
Option 1:
Option 2:
Option 3:
Option 4:
None of these
Correct Answer:
Solution:
As we have learnt