AIEEE 2007 Practice Test 2
- Test duration is 2 hours.
Instructions
1. All questions are compulsory.
2. Every question carries 3 marks.
3. Do not write anything on answer sheet except in the marked area.
4. Every incorrect answer carries -1(negative) mark
5. The paper is of 2 hours duration
Syllabus
Physics - Heat, SHM and waves
Chemistry - Electrochemistry , Hydrocarbons,
Mathematics - Probability, Trigonometric equations and inverse trigonometric functions
Q1.
Two holes of unequal diameters d1 and d2 (d1 > d2) are cut in a metal sheet. If the sheet is heated,
(a) both d1 and d2 will decrease
(b) both d1 and d2 will increase
(c) d1 will increase, d2 will decrease
(d) d1 will decrease, d2 will increase
Q2.
In the previous question, the distance between the holes will
(a) increase
(b) decrease.
(c) remain constant
(d) may either increase or decrease depending on the positions of the holes on the sheet and on the
Q3.
A metal wire of length l and area of cross-section A is fixed between rigid supports at negligible
cooled, the tension in the wire will be
(a) proportional to l
(b) inversely proportional to l
(c) independent of l
(d) independent of A
ratio d1/d2
tension. If this is
Q4.
Two metal rods of the same length and area of cross-section are fixed end to end between rigid
supports. The
material of the rods have Yong modulii Y1 and Y2, and coefficients of linear expansion
a1 and a2. The junction between the
rods does not shift if the rods are cooled.
(a) Y1a1 = Y2a2
(b) Y1a2 = Y2a1
(c) Y1a12 = Y2a22
(d) Y12a1 = Y22a2
Q5.
Three rods of equal length are joined to form an equilateral triangle ABC. D is the midpoint of AB.
The coefficient
of linear expansion is a1 for AB, and a2 for AC and BC. If the distance DC remains
constant for small changes in
temperature,
(a) a1 = a2
(b) a1 = 2a2
(c) a1 = 4a2
(d) a1 =
Q6.
a2
When the temperature of a body increases from t to t + Dt, its moment of inertia increases from I to I
coefficient of linear expansion of the body is a. The ratio
+ DI. The
is equal to
(a)
(b)
(c) aDt
(d) 2aDt
Q7.
A horizontal tube, open at both ends, contains a column of liquid. The length of this liquid column
does not change
with temperature. Let g = coefficient of volume expansion of the liquid and a =
coefficient of linear expansion of the material
of the tube.
(a) g = a
(b) g = 2a
(c) g = 3a
(d) g = 0
Q8.
In a vertical U-tube containing a liquid, the two arms are maintained at different temperatures, t1 and
t2. The liquid
columns in the two arms have height l1 and l2 respectively. The coefficient of volume
expansion of the liquid is equal to
(a)
(b)
(c)
(d)
Q9.
A solid whose volume does not change with temperature floats in a liquid. For two different
temperatures t1 and t2
of the liquid, fractions f1 and f2 of the volume of the solid remain submerged in
the liquid. The coefficient of volume
expansion of the liquid is equal to
(a)
(b)
(c)
(d)
Q10.
A solid with coefficient of linear expansion a just float in a liquid whose coefficient of volume
system is heated, the solid will
(a) sink in all cases
(b) continue to float in all cases
expansion is g. If the
(c) sink if g > 3a
(d) sink if g < 3a
Q11.
A gas at absolute temperature 300 K has pressure = 4 x 10–1 N/m2.
Boltzmann constant = k = 1.38 x 10–23 J/K. The number of molecules per cm3 is of the order of
(a) 100
(b) 105
(c) 108
(d) 1011
Q12.
The root-mean-square (rms) speed of oxygen molecules (O2) at a certain absolute temperature is v.
temperature is doubled and the oxygen gas dissociates into atomic oxygen, the rms speed
would be
(a) v
If the
(b)
(c) 2 v
(d)
Q13.
The average translational kinetic energy of O2 (molar mass 32) at a particular temperature is 0.048
average translational kinetic energy of N2 (molar mass 28) molecules in eV at the same
temperature is
(a) 0.0015
(b) 0.003
(c) 0.048
(d) 0.768
eV. The
Q14.
gas is
A gas has volume V and pressure p. The total translational kinetic energy of all the molecules of the
(a)
pV only if the gas is monatomic
(b)
pV only if the gas is diatomic
(c) >
(d)
pV only if the gas is diatomic
pV in all cases
Q15.
A closed vessel is maintained at a constant temperature. It is first evacuated and then vapour is
continuously. The pressure of the vapour in the vessel
(a) increases continuously
(b) first increases and then remains constant
(c) first increases and then decreases
(d) none of the above
injected into it
Q16.
When an air bubble rises from the bottom to the surface of a lake, its radius becomes double. Find
the depth of the
lake, given that the atmospheric pressure is equal to the pressure due to a column of
water 10 m high. Assume constant
temperature and disregard surface tension
(a) 30 m
(b) 40 m
(c) 70 m
(d) 80 m
Q17.
Two containers of equal volume contain the same gas at pressures p1 and p2 and absolute
temperatures T1 and
T2 respectively. On joining the vessels, the gas reaches a common pressure p
and a common temperature T. The ratio p/T
is equal to
(a)
(b)
(c)
(d)
Q18.
Two identical containers joined by a small pipe initially contain the same gas at pressure p0 and
absolute
temperature T0. One container is now maintained at the same temperature while the other is
heated to 2T0. The common
pressure of the gases will be
(a)
(b)
(c)
(d) 2p0
Q19.
In the pressure question, let V0 be the volume of each container. All other details remain the same.
moles of gas in the container at temperature 2T0 will be
The number of
(a)
(b)
(c)
(d)
Q20.
A horizontal cylinder has two sections of unequal cross-sections in which two pistons can move
are joined by a string. Some gas is trapped between the pistons. If this gas is
heated, the pistons will
freely. The pistons
(a) mole to the left
(b) move to the right
(c) remain stationary
(d) either (a) or (b) depending on the initial pressure of the gas
Q21
A particle of mass 1 kg is moving in SHM with an amplitude 0.02 and a frequency of 60 Hz. The
acting on the particle is
(a) 144 p2
(b) 188 p2
(c) 288 p2
(d) None of these
Q22
at time
An instantaneous displacement of a simple harmonic oscillator is x = A cos (wt + p/4). Its speed will
maximum force
be maximum
(a) p/4 w
(b) p/2w
(c) p/w
(d) 2 p/w
Q23
Velocity of sound waves in air is 330 m/s. For a particular sound wave in air, a path difference of 40
equivalent to phase difference of 1.6 p. The frequency of this wave
(a) 165 Hz
(b) 150 Hz
(c) 660 Hz
(d) 330 Hz
Q24
The velocity of sound in air is 330 m/s. The rms velocity of air molecules (g = 1.4) is approximately
cm. is
equal to
(a) 400 m/s
(b) 471.4 m/s
(c) 231 m/s
(d) 462 m/s
Q25
Sound waves of length l traveling with velocity v in a medium enter into another medium in which their
v. The wavelength in 2nd medium is
(a) 4 l
(b) l
(c) l/4
(d) 16 l
Q26
The velocity of sound in a container of air at − 73 °C is 300 m/s. Its temp. of container were raised
would be the velocity of sound ?
(a) 300 m/s
(b) 300
(c) 300 /
(d) 600 m/s
velocity is 4
to 127 °C. What
m/s
m/s
Q27
If the intensities of two interfering waves be I1 and I2, the contrast between maximum and minimum
maximum, when
(a) I1 >> I2
(b) I1 << I2
(c) I1 = I2
(d) Either I1 or I2 is zero
Q28
Two periodic waves of intensities I1 and I2 pass through a region at the same time in the same
of the maximum and minimum intensities is
(a) 2(I1 + I2)
(b) I1 + I2
intensity is
direction. The sum
(c)
(d)
Q29
If two tuning forks A and B are sounded together, they produce 4 beats per sec. A is then slightly
and same no. of beats / sec are produced again. If frequency of A is 256, the
frequency of B would be
(a) 250
(b) 262
(c) 252
(d) 260
Q30
If the ratio of maximum to minimum intensity in beats is 49, then the ratio of amplitudes of two
trains is
(a) 7 : 1
(b) 4 : 3
(c) 49 : 1
(d) 16 : 9
Q31
A uniform wire of length 20 m and weighing 5 kg hangs vertically. If g = 10 m/s2, then the speed of
waves in the middle of the wire is
(a) 10 m/s
loaded with wax
progressive wave
transverse
(b) 10
m/s&
(c) 4 m/s
(d) 0
Q32
If there are six loops for 1 m length in transverse mode of Melde’s experiment., the no. of loops in
mode under otherwise identical conditions would be
(a) 3
(b) 6
(c) 12
(d) 8
Q33
The fundamental frequency of an open organ pipe is 300 hz. The first overtone of this pipe has same
first overtone of a closed pipe. If speed of sound is 330 m/s, then the length of closed
organ pipe is
longitudinal
frequency as
(a) 41 cm
(b) 37 cm
(c) 31 cm
(d) 80 cm
Q34
Tube A has both ends open while tube B has one end closed, otherwise they are identical. The ratio
fundamental frequency of tube A and B is
(a) 1 : 2
(b) 1: 4
(c) 2 : 1
(d) 4 : 1
Q35
When temperature increases, the frequency of a tuning fork
(a) increases
(b) decreases
(c) remains same
(d) increases or decreases depending upon the material
Q36
Two waves of wavelengths 99 cm and 100 cm both traveling with velocity 396 m/s are made to
number of beats produced by them per second are
(a) 1
(b) 2
(c) 4
(d) 8
Q37
of
y = 2 (cm) sin
interfere. The
, what is the maximum acceleration of the particle doing the SHM
(a)
(b)
(c)
(d)
Q38
A cylindrical tube open at both the ends, has a fundamental frequency ‘f’ in air. The tube is dipped
vertically in air.
The tube is dipped vertically in water so that half of it is in water. The fundamental
frequency of the air column in now
(a) f/2
(b) 3 f/4
(c) f
(d) 2f
Q39
A wave is represented by the equation y = a cos (kx − wt) is superimposed with another wave to
wave such that point x = 0 is a node. The equation for the other wave is
(a) a sin (kx + wt)
(b) − a cos (kx − wt)
(c) − a cos (kx + wt)
(d) − a sin (kx − wt)
Q40
An organ pipe open at one end is vibrating in first overtone and is in resonance with another pipe
ends and vibrating in third harmonic. The ratio of length of two pipes is
(a) 1 : 2
(b) 4 : 1
(c) 8 : 3
(d) 3 : 8
form a stationary
open at both
[ Chemistry ]
Q41.
In the reaction, H2(g) + I2(g)
2HI(g) the concentration of H2, I2 and HI at equilibrium are
28.0 moles are litres respectively. What will be the equilibrium constant (a) 30.61
8.0 , 3.0 and
(b) 32.66
(c) 29.40
(d) 20.90
Q42.
For a gas reaction, 3H2(g) + N2(g)
2NH3(g), the partial pressures of H2 and N2 are 0.4
and 0.8
atmosphere, respectively. The total pressure of the entire system is 2.8atmosphere. What will
be the value of Kp if all the
concentration are given in atmosphere ?
(a) 32 atm-2
(b) 20 atm-2
(c) 50 atm-2
(d) 80 atm-2
Q43
One mole of nitrogen and three moles of hydrogen are mixed in a 4 litre container. If 0.25 percent of
nitrogen is
coverd to ammonia by the following reaction N2(g) + 3H2(g)
2NH3(g). What
will be the equilibrium constant (Kc) in
concentration units ? What will be the value of K for the
following equilibrium -
N2(g) +
H2(g)
(a) 1.49 x 10-5 lit mol-1
(b) 2.22 x 10-10 lit2 mol-2
(c) 3.86 x 10-3 lit mol-1
(d) Question is incomplete
NH3(g).
Q44
When ethanol and acetic acid were mixed together in equilimolecular proportion 66.6% are
converted into ethyl
acetate. Calculate Kc . Also calculate quantity of ester produced if one mole of
acetic acid is treated with 0.5 mole and 4
mole of alcohol respectively.
(a) 4, 0.93, 0.43
(b) 0.93, 4, 0.43
(c) 0.43, 0.93, 4
(d) 4, 0.43, 0.93
Q45
One mole of ammonium carbonate dissociate as shown below at 500 K.
NH2COONH4(s)
2NH3(g) + CO2(g)
If the pressure exerted by the released gases is 3.0 atm, the value of Kp is.
(a) 7 atm
(b) 3 atm
(c) 4 atm
(d) 8 atm
Q46
Iron filling and water were placed in a 5 litre tank and sealed The tank was heated to 1273 K. Upon
analysis the
tank was found to contain 1.10 gram of hydrogen and 42.5 gm of water vapour. If the
reaction in the tank is represented by
3Fe(s) + 4H2O(g)
Fe3O4 (s) + 4H2(g)
the equilibrium constant will be (a) 2.949 x 103
(b) 6.490 x 103
(c) 4.940 x 103
(d) 3.200 x 103
Q47
At 700 K, the equilibrium constant KP, for the reaction
2SO3(g)
2SO2(g) + O2(g) is 1.8 x 10-3 kPa.
What is the numerical value of Kc for this reaction at the same temperature
(a) 3.09 x 10-7mole litre-1
(b) 9.03 x I 0-7mole litre-1
(c) 5.05 x 10-9 mole litre-1
(d) 5.05 x 10-5mole litre-1
Q48
The value of KC for the reaction N2(g) + 3H2(g)
2NH3(g) is 0.50 at 400 °C. What will
at 400 °C when concentration are expressed in mole litre-1 and pressure in
atmosphere (a) 1.64 x 10-4
(b) 2.80 x 10-6
(c) 2.80 x 10-4
(d) 1.64 x 10-6
Q49
The equilibrium constant for the reaction H2(g) + S(s)
respectively. The enthalpy of the reaction will be -
H2S(g) ; is 18.5 at 935 K and 9.25
be the value of KP
at 1000 K
(a) - 68000.05 J mol-1
(b) -71080.57 J mol-1
(c) -80071.75 J mol-1
(d) 57080.75 J mol-1
Q50
Inhibitor is (a) An activator
(b) Negative catalyst
(c) Catalyst for a catalyst
(d) Promoter
Q51
KP for the reaction A(g) + 2B(g)
3C(g) + D(g) ; is 0.05 atm. What will be its KC at 1000
K in terms of R -
(a)
(b)
(c)
R
(d) none of these
Q52
The vapour density of N2O4 at a certain temperature is 30. The percentage dissociation of N2O4 at
temperature is
(a) 55.5 %
(b) 60 %
(c) 70 %
(d) 53.3 %
Q53
If PCl5 is 80 % dissociation at 523 K. Calculate the vapour density of the equilibrium mixture at 523
(a) 75.9
(b) 57.9
(c) 97.5
(d) 95.7
Q54
Ammonium carbamate when heated to 200 °C gives a mixture of vapours
(NH2COONH4
2NH3 + CO2)
with a density 13.0. What is the degree of dissociation of ammonium carbamate (a) 1
(b) 2
(c) 3
(d) 4
Q55
The volume of a closed reaction vessel in which the equilibrium
2SO2(g) + O2(g)
2SO3(g) sets is halved, Now(a)
The rates of forward and backwards reactions will remain the same.
(b)
The equilibrium will not shift.
(c)
The equilibrium will shift to the right.
(d)
The rate of forward reaction will become double that of reverse reaction and the
this
K
equilibrium will shift to
the right.
Q56
H2(g) + I2(g)
2HI(g)
When 46g of I2 and 1g of H2 are heated at equilibrium at 450 °C, the equilibrium mixture contained
How many moles of I2 and HI are present at equilibrium (a) 0.0075 & 0.147 moles
(b) 0.0050 & 0.147 moles
(c) 0.0075 & 0.347 moles
(d) 0.0052 & 0.347 moles
Q57
A two litre flask contains 1.4 gm nitrogen and 1.0 gm hydrogen. The ratio of active mass of nitrogen
would be
(a) 1 : 3
(b) 1 : 5
(c) 1.4: 1
(d) 1 : 10
1.9 g of I2.
and hydrogen
Q58
In the reaction
A+B
C+D
The initial concentration of A is double the initial concentration of B. At equilibrium the concentration
found to be one third of the concentration of C. The value of equilibrium constant is
(a) 1.8
(b) 1.008
(c) 0.0028
(d) 0.08
Q59
of B was
The value of KC for the reaction :
A + 3B
2C at 400 °C . Calculate the value of KP
(a) 1.64 x 10-4
(b) 1.6 x 10-6
(c) 1.6 x 10-5
(d) 1.6 x 10-3
Q60
Two moles of ammonia was introduced in an evacuated vessel of 1litre capacity. At high temperature
undergoes partial dissociation according to the equation
2NH3(g)
N2(g) + 3H2(g)
At equilibrium the concentration of ammonia was found to be 1 mole. What is the value of ‘K’ ?
the gas
(a) 3/4 = 0.75 mol2
(b) 3/2 = 1.5 mol2
(c) 27/16 = 1.7 mol2
(d) 27/64 = 0.42 mol2
Q61.
Q62
Which sodium salt will be heated with soda lime to obtain propane –
(a) CH3 – CH2 – C – OΘNaÅ
(b) CH3 – CH2 – CH2 – C – OΘNaÅ
||
||
O
O
(c) CH3 – CH – C – OΘNaÅ
(d) 2nd and 3rd both
||
O
Which of the following compounds cannot be prepared by Wurtz reaction (a) CH3CH3
(b) CH3CH – CH
|
CH3
(c) (CH3)2CHCH3
(d) CH3CH2CH2CH3
(a) a, b
(b) b, c
(c) c, d
(d) none of these
Q63
Which of the following compounds liberate methane when treated with excess of methyl magnesium
dry ether
(a) CH3 - CH2 - CH2OH
(b) H3C - CH2 - C ≡ CH
(c) H3C - CH2 - CO2H
(d) H3C - CH2 - CHO
(a) a, b
(b) b, c
(c) a, b, c
(d) All above
Q64
Raney Nickel is a suitable catalyst for: (a) Bromination
(b) Dehydration
(c) Hydrogenation
iodide in
(d) Alkylation
Q65.
Correct order of boiling point is (a) n-Pentane < neohexane < isohexane < 3-methyl pentane
(b) neohexane < n-pentane < isohexane < 3-methyl pentane
(c) 3-methyl pentane < neohexane < n-pentane < isohexane
(d) n-pentane < isohexane < 3-methyl pentane < neohexane
Q66
In the complete combustion of Cn H2n + 2 , the number of oxygen rmoles required is
(a) n / 2O2
(b)
O2
(c)
Q67
O2
(d)
O2
What is X in the following sequence of reaction:
X
Z
(a) Methane
(b) Ethanoic acid
(c) Propane
(d) None of the above
CH4
Q68.
The number of isomers of C6H14 are:
(a) 4
(b) 5
(c) 6
(d) 7
Q69.
Which of the following reaction pairs constitutes the chain propagation step in chlorination of methyl
(a)
(c)
·CH3 + Cl2 ® CH3Cl + ·Cl
CH3Cl + ·Cl ® CH2Cl2 + ·H
CH3Cl + ·CI ® ·CH2Cl + HCl
·CH2Cl + ·CH2Cl ® CH2Cl + CH2Cl
(b)
(d)
chloride?
CH3Cl + ·Cl ® CH2Cl + HCl
·CH2Cl + Cl2 ® CH2Cl2 + ·Cl
·CH2Cl + Cl2 ® CH2Cl2 + ·Cl
·CH2Cl + ·Cl ® CH2Cl2
Q70
Kolbe's reaction is convenient for the preparation of:
(a) Methane
(b) Alkanes containing even number of carbon atoms
(c) Alkanes containing even as well as odd number of carbon atoms
(d) Alkanes containing odd number of carbon atoms
Q71.
For the formation of 27 gm of water, what volume of neopentane is required for the complete
(a) 5.6 lit.
(b) 11.2 lit.
(c) 33.6 lit
(d) 2.24 lit.
Q72.
Acetylene and ethylene reacts with alk. KMnO4 to give (a) Oxalic acid and formic acid
(b) Acetic acid and ethylene glycol
(c) Ethyl alcohol and ethylene glycol
(d) None
Q73.
Chloroform when heated with silver powder gives an alkyne. For the substitution of hydrogen atom
reaction must be carried out with (a) Cl2 at 0°C in the presence of ultra violet light
(b) NaOCl at 0°C in the presence of light and air
(c) Cl2 at 0°C in dark
combustion:
by chlorine the
(d) None of the above
Q74.
Acetylene on passing into excess of HOCl solution forms (a) Ethylene chlorohydrin
(b) Acetaldehyde
(c) Dichloroacetaldehyde
(d) Methyl Chloride
Q75.
10 ml of a certain hydrocarbon require 25 ml of oxygen for complete combustion and the volume of
produced is 20 ml. What is the formula of hydrocarbon (a) C2H2
(b) C2H4
(c) CH4
(d) C2H6
Q76.
Lindlar's catalyst consists of (a) Metallic nickel + nickel boride,
(b) Metallic platinum
(c) Metallic palladium deposited on calcium carbonate containing lead acetate and quinoline
(d) Sodium borohydride in ethanol.
Q77.
2-Butyne and 1-Butene show resemblance in all except (a) Both decolourise alkaline KMnO4
(b) Both turn bromine water colourless
(c) Both undergo addition reaction
(d) Both from white precipitate with Tollen's reagent
Q78.
Acetylene reacts with formaldehyde in the presence of sodium alkoxide to form mainly (a) CH2 = CH – CH2OH
(b) CH2OH – CH = CH2
CO2
(c)
(d)
Q79.
Acetylene reacts with 42% H2SO4 containing 1% HgSO4 to give:
(a) C2H5HSO4
(b) CH3CHO
(c) HCHO
(d) CH2 == CH2
Q80.
The alkene which on ozonolysis yields acetone is:
(a) CH2 == CH2
(b) CH3 –– CH == CH2
(c) (CH3)2C == C(CH3)2
(d) CH3 –– CH == CH – CH3
[ Mathematics ]
Q81
A car is parked by a driver amongst 25 cars in a row, not at either end. When he returns he finds
are empty. The probability that both the neighboring places of driver’s car are vacant
is
(a)
(b)
(c)
(d)
that 10 places
Q82
A is one of the six race horses which is to be ridden by one of the two jockeys B or C. It is 2 :1 that
B rides A in
which case all the horses are equally likely to win but if C rides, then A’s chances of
wining are trebled. The odds against
his winning are
(a) 13 : 5
(b) 5 : 13
(c) 8 : 5
(d) 5 : 8
Q83
A and B play by throwing a pair of dice alternately. A wins if he throws 6 before B throws 7. If A
their chances of wining the game are in the ratio
(a) 28 : 33
(b) 29 : 32
(c) 30 : 31
(d) none
Q84
Five coins whose faces are marked 2, 3, …. Are thrown. The probability of obtaining a total of 12
starts the game
is
(a)
(b)
(c)
(d)
Q85
The probability that a teacher will give an unannounced test during any class meeting is
twice, the probability that he will miss at least one test is
. If a
student is absent
Q86
Two cards are randomly selected from a deck of 52 playing cards. The probability that both the
than 3 and less than 8 is
cards are greater
(a)
(b)
(c)
(d)
(a)
(b)
(c)
(d)
Q87
is
Probabilities of teams A, B and C wining are
(a)
(b)
respectively. Probability that one of these
teams will win
(c)
(d) none
Q88
If two squares are chosen at random on a chess board, the probability that they have a side in
common is
(a)
(b)
(c)
(d) none
Q89
A single letter is selected at random from the word PROBABILITY . The probability that it is a
vowel is
(a)
(b)
(c)
(d) none
Q90
if P(A Ç B) =
, P(
Ç
)=
, P(A) = p, P(B) = 2p, then the value of p is given by
(a)
(b)
(c)
(d)
Q91
The probability that a leap year selected at random contains either 53 Sundays or 53 Mondays is
(a)
(b)
(c)
(d)
Q92
A bag contains 50 tickets numbered 1, 2, 3 , ……., 50 of which five are drawn at random and
ascending order of magnitude (
(a)
(b)
). The probability that x3 = 30 is
arranged in
(c)
(d) none
Q93
A box contains 10 mangoes out of which 4 are rotten. 2 mangoes are taken out together. If one of
to be good, the probability that the other is also good is
them is found
(a)
(b)
(c)
(d)
Q94
Three letters, to each of which corresponds an envelope, are placed in the envelopes at random. The
that all the letters are not placed in the right envelope is
(a) 1/6
(b) 5/6
(c) 1/3
(d) 2/3
Q95
Two athletes A and B participate in a race along with other athletes. If the chance of A wining the
that of B wining the same race is 1/8, then the chance that neither wins the race is
(a) 1/4
(b) 7/24
(c) 17/24
(d) 35/38
Q96
race is 1/6 and
Six coins are tossed simultaneously. The probability of getting at least 4 heads is
(a) 11/64
(b) 11/32
(c) 15/44
(d) 21/32
Q97
Dialing a telephone number an old man forgets the last two digits remembering that these are different
random. The probability that the number is dialed correctly is
(a) 1/45
(b) 1/90
(c) 1/100
(d) none
Q98
If A and B are two independent events, the probability that both A and B occur is 1/8 and the
neither of them occur is 3/8. The probability of the occurrence of A is
(a) 1/2
(b) 1/3
(c) 1/4
(d) 1/5
Q99
is
probability
If three vertices of a regular hexagon are chosen at random, then the chance that they form an
dialed at
probability that
equilateral triangle
(a) 1/3
(b) 1/5
(c) 1/10
(d) 1/2
Q100
is
10 apples are distributed at random among 6 persons. The probability that at least one of them will
(a)
receive none
(b)
(c)
(d) none
Q101.
If (tan–1 x)2 + (cot–1 x)2 =
(a) – 1
(b) 1
(c) 0
(d) none of these
Q102.
cos–1 [cos (2 cot–1 (
, then x equals
))] is equal to
(a)
(b)
(c) p/4
(d) 3p/4
Q103.
The equation
(a) no solution
(b) unique solution
(c) infinite number of solution
(d) None of these
has
Q104.
Two angles of a triangle are cot–1 2 and cot–1 3. Then the third angle
(a) p/4
(b) 3p/4
(c) p/6
(d) p/3
Q105.
Complete set of values of x satisfying [tan–1 x] + [cot–1 x] = 2, where [.] denotes the greatest
(a) (cot 3, cot 2]
(b) (cot 3, – tan 1]
(c) (cot 3, 0)
(d) None of these
Q106.
Complete solution set of the equation [cot–1 x] + 2[tan–1 x] = 0, where [.] denotes the greatest
equal to
(a) (0, cot 1)
(b) (0, tan 1)
(c) (tan 1, ¥)
(d) (cot 1, tan 1)
Q107.
The trigonometric equation sin–1 x = 2 sin–1 a, has a solution
(a)
(b) all real values of a
(c)
(d)
Q108.
, then sin x is equal
for:
integer function, is
integer function, is
(a)
(b)
(c) tan a
(d)
Q109.
is equal to:
(a)
(b)
(c)
(d)
Q110.
The value of
is:
(a)
(b)
(c)
(d) none of these
Q111.
The number of real solutions of
(a) 0
(b) 1
(c) 2
(d) infinite
is:
Q112.
If
(a) 1/2
(b) 1
(c) – 1/2
(d) – 1
Q113.
The value of x for which sin (cot–1 (1 + x)) = cos (tan–1 x) is
(a) 1/2
(b) 1
(c) 0
(d) – 1/2
Q114.
The principal value of
for 0 < |x| <
is
, then x equals:
(a) –
(b)
(c)
(d)
(e) none of these
Q115.
The period of the function f(x) = sin4 x + cos4 x is
(a) p
(b)
(c) 2p
(d) none of these
Q116.
The value of
(a) 1
is
(b)
(c)
(d) 2
Q117.
If
then sin q is
(a)
but not
(b)
or
(c)
but not
(d) none of these
Q118.
If sin (a + b) = 1, sin (a – b) =
(a) 1
(b) – 1
(c) zero
(d) none of these
Q119.
Number of solutions of the equation tan x + sec x = 2 cos x lying in the interval [0, 2p] is&
(a) 0
(b) 1
(c) 2
(d) 3
Q120.
In a triangle PQR, ÐR = p/2. If tan (P/2) and tan (Q/2) are the roots of the equation ax2 + bx + c =
(a) a + b = c
(b) b + c = a
(c) a + c = b
(d) b = c
, then tan (a + 2b) tan (2a + b) is equal to:
0 (a ¹ 0) then
Answ ers
Organic Chemistry Fact Sheet
Organic Chemistry Quick Fact Sheet
Acidic power
Order
Why ?
1.
III < II < I
Farther the (–I) group (Cl), lesser the
acidic strength
2.
I < II < III
Farther the (+I) group, greater the
acidic power
3.
II < I < III
—CH3 is electron donating and
— NO2 is electron attracting
4.
II < I < III
—CH3 is electron repelling;
decreases acidic strength of phenol
III < I < II
— OCH3 group contains +M effect
and decreases acidic poer.
I < II < III
— NO2 is electron attracting; III is
more resonance stabilised than I and
also than II. In I, only inductive
effect is operative.
I > II > III
sp2 hybridised carbon of I, II are
more el-ectronegative hence acid
strength is inc-reased. Benzylic
(C6H5CH2) is more stab-ilised than
allylic (CH2==CHCH2).
8.
I > II > III > IV
Effect of one —COOH on the other
decr-eases as its distance between
them increases, (COOH)2 is
maximum acidic.
9.
II < III < I
—NO2 is electron attracting (–I
effect)
10.
III < II < I
—OH shows electron withdrawing
nature at o - and m - and electron
repelling at p -, o - isomer due to
intramolecular bonding in salicylate
ion is stronger than m - isomer
11.
III < II < I
—do—
12.
I < III < II
—NH2 is electron donating.
5.
6.
7.
Section B
Basic power
1.
Order
I > III > II
Why ?
lone pair on N is not used in
resonance of -electrons in I. In II
lone pair of the ring is itself used in
delocalisation while that of outside
ring in III.
I > II > III > IV
—OCH3 is strong electron donating
group. This is due to ortho effect, all
the aniline are less basic than psubstituted aniline due to steric
hindrance.
3.
I > II > III > IV
I (hyper conugation and induction) II
(induction) IV (ortho effect), ortho
effect normally decreases basic
nature.
4.
II > I > III
In II there is sp3 hybridised C, In I,
sp2. NO2 is electron withdrawing.
III > I > II
lone pair on N is used in
delocalisation of -electrons in
aromatic amines while cyclohexyl is
electron repelling (III); in II, lone
pair on N is used by two benzene
ring.
6.
I > II > III > IV
NO2 is electron-withdrawing, thus
nitro-anilines are less basic than
aniline. IV is less basic than III
because —NO2 is closer and exerts a
stronger inductive effect.
7.
III > I > II
phenyl and —COCH3 are
electronwithdra-wing and —C6H5 <
COCH3
8.
I < II < III
Electron donating nature of C2H5 >
CH3 So more basic strength.
2.
5.
9.
I < II < III
ortho effect in I.
10.
I < II < III
ortho effect in (I)
General series
Order
Why ?
There is intermolecular H-bonding I.
III has weak force of attraction and
is most volatile.
1.
I > II > III
2. B.P. of o, m, p-nitro phenol
o<m<p
Intramolecular H-bonding in oisomer makes it more volatile.
I > II > IV > III
—CHO group is easily oxidised
compared to keto group due to
redusing hydrogen.
I > II > IV > III
—do—
I < II < III < IV
Aldehydes are more hydrated than
ketones. Halide makes C of carbonyl
group more electropositive.
I > II > III
CH3 group decreases +ve charge on
C hence nucleophilic attack.
Reactivity of ... with Tollen’s reagent
3.
4.
Reactivity of ... with Fehling’s solution
Extent of hydration of
5.
Electrophilic nature of ........ for nucleophilic
attack
6.
7.
Reactivity of isomeric 1°, 2°, 3° butyl halide
towards elimination (E1 or E2)
3° < 2° < 1°
Dehydration of
8.
due to stability of intermediate
carbocation
IV < I < II < III
Alcohol leading to increase in
conjugation due to dehydration is
more easily dehydration is more
easily dehydrated. IV is vinylic,
hence least.
I < II < III < IV
< V < VI
Substituted alkenes are more
stable.More the alkyl groups are
attached to the doubly bonded
carbon atom more
is the stability.
9. Stability of
Stability of
10.
I < III < II
II is more substituted than III (More
hyperconjugation more stability)
III > II > I > IV
IV is vinylic while in conjugative, II
allylic.
I < IV < II < III
III is 3° allylic and II is 1° allylic
3° < 2° < 1°
More the stability of intermediate,
greater the reactivity of chemical
reaction.
Stability of
11.
Stability of
12.
Dehydration of
13.
1°, 2°, 3° isomeric butyl alcohol
14.
Boiling points of
II > I > III
I, II have H-bonding but
electronegativity of O > N hence Hbonding in II > I
Formation of
15.
I > II > III > IV
(easiest I)
greater the stability, easier the
formation of perticular species.
I < II < III < IV
< V < VI
Vinyl < methyl 1° < 2° < 3° < allylic
I < II < III ~ IV
< V < VI < VII
< VIII
If acid is strong, its conjugate base is
weak and greater the leaving
tendency.
I > II > III > IV
>V
As the size of the substituents on the
—C increases, the tetrahedrally
bonded interme. diate becomes more
crowded and these slower the rate.
16. Reactivity of C—H bond (abstraction of H)
Leaving nature (tendency) of ... in SN
reaction.
17.
Rate of esterification of the following acids
with MeOH
18.
19.
Relative reactivity of ... with electrophile in
SE reaction
I > II > IV
> III > V
—CH3 is o-, p-directing and
responsible for activation.
Relative reactivity of these compunds with
electrophile inSE reaction
20.
II > I > III > IV
—CH3 is o-, p-directing due to
activation while —COOH is mdirecting and deactivating group.
21. Relative reactivity of ... with electrophile in
II > I > IV > II
As the number of sp3 hybridised C
atoms separating the ring from the
positively charged substituent
increases, deactivating effect
decreases due to less
electronegativity.
SE reaction.
is best able to donate electrons
there by giving a very stable
Activating effects of the following o, pdirectors.
22.
II > I > III
Relative reactivity of ... towards SN1 reaction
23.
II > I > III
Relative reactivity of ... towards SN1 and SN2
reaction
24.
25.
Relative reactivity of ... with E+ (electrophile)
in SE reaction.
26.
Order of SN2 reactivity of alkoxide
nucleophiles
SN1 :
III > II > I
SN2 :
II < II < I
uncharged intermediate. In
cross conjugation diminished its
ability to donate electrons to an
arenium ion.
Intermediates are benzylic cations.
So CH3O(electron repelling) gives
greater stability through
delocalisation while NO2 (electron
attracting) decreases stability.
SN1 : 1° < 2° < 3° alkyl halide
SN2 : 3° < 2° < 1° alkyl halide
II > I > III
—NO2 deactivates benzene ring for
SE
I < IV < V < III
< II
SN2 reactivity is suseptible to steric
hindrance by the nucleophile as well
as by the size of alkyl group.
IMPORTANT ORDER AND FACTS OF ORGANIC CHEMISTRY
1. RCOCl > RCOOCOR > RCOOR > RCONH2 Nucleophilic substitution reaction.
2. HI > HBr > HCl > RCOOH > C6H5OH > H2O > CH CH > NH3 (Acidic nature).
3. CCl3CHO > HCHO > CH3CHO > CH3COCH3 Nucleophilic addition reaction.
4. CH2 = CH2 > CH CH > C6H6 Electrophilic addition reaction.
Electrophilic Substitution
Reaction
Nucleophilic Substitution Reaction
Nucleophilic Substitution Reaction
10. (CH3)2C = C(CH3)2 > CH3 - CH = C -(CH3)2 > CH3 - CH = CH - CH3 > CH3 - CH = CH2 (Stability)
(Heat of Hydration)
13. NI3 > NBr3 > NCl3 > NF3 (Basic strength)
14. Br2 > Cl2 > I2 (Selectivity for halogenation)
15. Halogenation of alkenes by cyclic halonium state, so anti attack takes place.
16. Hydroboration followed by oxidation is always anti markownikoff’s addition due to steric effect.
17. Oximercuration - demercuration is m.K. addition of water because some carbocation character in cyclic
mercurium state.
18. CHCl3 in the presence of strong bases forms biradical : CCl2 which undergo addition with double or
triple bonds.
19. When conjugated diene reacts with alkene or alkyne it is known as diel’s elder synthesis.
20. Ozonolysis of cyclo alkene forms one mole dialdehyde while ozonolysis of cyclo alkadiene forms two
moles of dialdehyde.
21. Ozonolysis with (CH3)2S is known as reductive ozonolysis.
22. Hydration of alkyne occur’s in HgSO4 and dil H2SO4.
24. Cis-2-butene reacts with Br2 to forms dl( ) pair of enantiomers of 2,3-dibromobutane while in case of
trans-2-butene forms meso-2,3-dibromo butane due to anti addition always.
25. Haloform test given by species with CH3CO-group but not in case of A.A.E. and tert. Butyl alcohol.
26. Chloral reacts with chloro benzene in con. H2SO4 to form insectiside DDT.
27. NBS is used for free radical allylation.
28. Rate for SN1 reaction is 3° > 2° > 1° in protic polar solvent.
29. Rate for SN2 reaction is 1° > 2° > 3° in polar aprotic solvent like DMSO, DMF, HMPT.
30. Chemical reactions like Hoffmann carbylamine and Reimer Tiemann’s reaction active species is
biradical CCl2.
31. If cyclo 1,3-penta diene reacts with CHCl3 and potassium tert. butoxide to form chlorobenzene.
32. Alkyl halides reacts with AgCN to form isocyanides due to ambident nature of nucleophile, other
ambident nucleophiles are
and SO3-2.
33. In dehydration of alcohols active species is carbocation so rearrangement occurs like hydride shift or
alkyl shift.
34. Dehydration of cyclobutyl methyl alcohol ring expansion takes place, formation of cyclo pentene occurs.
35. In esterification where acid reacts with alcohol to form ester, - OH given by acid while - H by alcohol
36.Ether’s reacts with HI to form alcohol and halide where fission of lower ether by SN2 mechanism while
higher ethers like ter. butyl methyl ether or alkyl methyl ether by SN1 mechanism.
37. Quantitative estimation of ethers is done by ziesal’s method.
38. If unsym. cyclic ether undergo fission it depends upon medium weather it is acid or basic like in acidic
medium some character of carbocation so nucleophile goes to carbon where more alkyl groups are there
while reverse in basic medium due to steric factor.
39. Aldehydes are reducing agent while ketones are not.
40. Aldehydes and ketones are separated by tollen’s reagent.
41. Carbonyl and noncarbonyl are seprated by sodium bi sulphite and bradye’s reagent.
42. As the size of alkyl group increases steric hindrance comes into play, reactivity towards nucleophilic
addition decreases.
43. Aldehydes with hydrogen atom in the presence of dil base undergo enolization and form
to give aldol product.
44. If there is two-CHO group with
carbonian
hydrogen atom to form cyclic intra aldol product.
45. Aldehydes without -hydrogen atom in the presence of con. alkali to form each molecule of acid and
alcohol by hydride active species.
46. Glyoxal reacts with con. KOH to form glycolate ion by Intra Cannizaro’s.
47. Pinacol pinacolone type reactions involve protonation, deprotonation and alkyl shift.
48. Aldol, Perkin, reformatsky and knovengel’s reactions are classified as carbanian active reaction.
49. In Beckmann’s rearrangement migration of group which is anti to-OH group takes place.
50. Beckmann’s rearrangent is a reaction of oximes in the presence of H2SO4 or PCl5 to produce N-alkyl
amide derivatives.
51. Migratory attitude of alkyl group in Pinacol-pinacolone, beckmann’s and bayer villegar oxidation is
- C6H5 > (CH3)3C - > (CH3)2CH - > C2H5 - > CH352. Cyclo hexanoneoxime on beckmann’s reaction gives caprolectum which on reaction with
polymer nylon-6
53. 2-methyl propanal even contains
to give
-hydrogen atom but does not give aldol reaction.
54. Diphenyl glyoxal reacts with con. KOH to form salt of benzillic acid.
55. Propanone in the presence of dry HCl gas by enolic intermediate to form diacetone alcohol undergo
heating form mesityl oxide. If this again reacts with propanone to form phoron.
56. Benzaldehyde reacts with alc. KCN to form Benzoin which on oxidation form benzil.
57. By wittig reaction carbonyl compounds are converted into E-Z form of alkene.
58. Benzaldehyde do not respond to benedict’s and fehling’s solution due to less redusing power of aromatic
aldehyde.
59. HCOOH respond to oxidising agent due to presence of - CHO group.
60. HI > H2SO4 > HNO3 > RCOOH > H2CO3 > C6H5OH > H2O > CH CH > NH3 > CH3 - C CH > CH2 =
CH2 > H2 this is decreasing acidic nature.
61. Acids with
-hydrogen atom when reacts with halogen in the presence of P to form
-haloacid (HVZ).
62. Carboxyllic acid on reaction with PCl5, CH2N2 and H2O to form higher acid or next homologue.
63. For reactivity of acid derivatives use funda weaker the base better the leaving group.
64. Acid amide on reaction with PCl5 to form alkane nitrile.
65. Anhydride on reaction with carbonyl compound in the presence of base (carbanian) forms
unsaturated carboxyllic acid (perkin reaction)
, -
66. Amide on reaction with Br2 and alkali to form primary amine of lower homologue. Intermediate species
is nitrene which undergo intra rearrangement to form RNH2.
67. Ester’s with -hydrogen atom in the presence of strong base to form carbanian undergo nucleophilic
substitution reaction forms -keto ester for example ethyl acetate in the presence of pot. ter. Butoxide form
aceto acetic ester (AAE). reaction is known as clesen’s ester condensation of four types
(a) Simple clesen’s ester condensation.
(b) CROSS clesen’s ester condensation.
(c) Intra CEC (Dieckmann’s condensation).
(d) mixed clesen ester condensation.
68. For a compound to be aromatic it must be cyclic, planar and obey (4n + 2) electron rule.
69. Cyclooctatetraene is non aromatic compound while pyrrole, pyredene, furan, cyclopentadieneylanion all
are aromatic.
70. In aromatic electrophilic substitution reaction there is no hydrogen isotopic effect except sulphonation
and iodination.
71. m- directing groups like nitrobenzene and benzaldehyde cannot undergo fridal craft reaction.
72. Phenol is less acidic than general carboxylic acid, cannot react with NaHCO3.
73. Anilene is more reactive than phenol towards electrophilic substitution reaction because less energy
difference between nitrogen and carbon.
74. Benzene diazonium chloride reacts with phenol or anilene to form azo compounds.
75. C6H5NH2 > C6H5OH > C6H5OR > C6H5Br > C6H5NO2 (Reactivity towards electrophilic substitution
reaction.
AIEEE Minor Test #1
Ques 1 -12 [1. 5 marks] / Ques 13- 25 [3 marks] / Ques 26- 40 x 4. 5 marks
-ve marks are one third
Test duration 2 hour
Q1. The triangle with vertices (1, 5); (–3, 1) and (3, –5) is
(a)
isosceles
(b)
equilateral
(c)
right angled
(d)
None of these
Q2. If the points (4, – 4), (– 4, 4) and (x, y) form an equilateral triangle then
(a)
(b)
(c)
(d)
None of these
Q3. If (–4, 6), (2, 3) and (2, –5) are vertices of a triangle, then its incentre is
(a)
(–1, 2)
(b)
(2, –1)
(c)
(1, 2)
(d)
(2, 1)
Q4.Circumcentre of a triangle whose vertex are (0, 0), (4, 0) and (0, 6) is
(a)
(b)
(0, 0)
(c)
(2, 3)
(d)
(4, 6)
(d)
(0, 0)
(d)
37.5
Q5.Orthocentre of a triangle whose vertex are (8, –2), (2, –2) and (8, 6) is
(a)
(8, –2)
(b)
(8, 6)
(c)
Q6.The area of a triangle with vertices (3, 8); (–4, 2) and (5, –1) is
(a)
40.5
(b)
36.5
(c)
3.75
Q7.If D, E, F are mid points of the sides AB, BC and CA of a triangle formed by the points A(5, -1) B(-7, 6) and C(1, 3), then area of
DDEF is
(a)
2/5
(b)
5/2
(c)
5
(d)
10
Q8. The point (4, 1) undergoes two successive transformations
(i) Reflection about the line y = x
(ii) Translation through a distance 2 units along the positive direction of x – axis
The final position of the point is given by the coordinates
(a)
(4, 3)
(b)
(3, 4)
(c)
(7/2, 7/2)
(d)
Q9.If A(c, 0) and B(– c, 0) are two points, then the locus of a point P which moves such that
PA2+ PB2 = AB2 is
(a)
x2 – y2 =c2
(b)
y2 = 4cx
(c)
x2 + y2 = c2
(1, 4)
(d)
None of these
Q10. Let A(2, 3) and B(–4, 5) are two fixed points. A point P moves in such a way that DPAB = 12 sq. units, then its locus is
(a)
x2 + 6xy + 9y2 + 22 x + 66y – 23 = 0
(b)
x2 + 6xy + 9y2 + 22 x + 66y + 23 = 0
(c)
x2 + 6xy + 9y2 – 22 x – 66y – 23 = 0
(d)
None of these
Q11.If sum of square of distances of a point from axes is 4, then its locus is
(a)
x+y=2
(b)
x2 + y2 = 16
(c)
x+y=4
(d)
x2 + y2 = 4
Q12.The extremities of diagonal of a right-angled triangle are (2, 0) and (0, 2), then locus of its third vertex is
(a)
x2 + y2 – 2x – 2y = 0
(b)
x2 + y2 + 2x – 2y = 0
(c)
x2 + y2 – 2x + 2y = 0
(d)
x2 + y2 + 2x + 2y = 0
Q13.Keeping coordinate axes parallel, the origin is shifted to a point (1, –2), then transformed equation of
x2 + y2 = 2 is
(a)
(c)
x2 + y2 + 2x – 4y + 3 = 0
x2 + y2 – 2x – 4y + 3 = 0
(b)
x2 + y2 – 2x + 4y + 3 = 0
(d)
x2 + y2 – 2x + 4y + 3 = 0
Q14.To remove xy term from the second degree equation 5x2 + 8xy + 5y2 + 3x + 2y + 5 = 0, the coordinates axes are rotated through an
angle q, then q equals
(a)
p/2
(b)
p/4
(c)
3p/8
(d)
p/8
Q15.The ratio in which the line y – x + 2 = 0 divides the line joining (3, – 1) and (8, 9) is
(a)
2:3
(b)
3:2
(c)
–2:3
(d)
–3:2
Q16.The area of the triangle, formed by the straight lines 7x – 2y + 10 = 0, 7x + 2y – 10 = 0 and 9x + y + 2 = 0, is
(a)
(b)
(c)
(d)
None of these
Q17.Two vertices of a triangle are (3, – 1) and (– 2, 3) and its orthocenter is origin, the coordinates of the third vertex are
(a)
(b)
(c)
(d)
Q18.The equation of the internal bisector of ÐBAC of DABC with vertices A(5, 2), B(2, 3) and C(6, 5) is
(a)
2x + y + 12 = 0
(b)
x + 2y – 12 = 0
(c)
2x + y – 12 = 0
(d)
Q19.The equation of the straight line upon which the length of perpendicular from origin is
angle of 75° with the positive direction of x – axis, is
(a)
(b)
(c)
(d)
None of these
units and this perpendicular makes an
None of these
Q20.The image of the point (– 8, 12) with respect to the line mirror 4x + 7y + 13 = 0 is
(a)
(16, – 2)
(b)
(– 16, 2)
(c)
(16, 2)
(– 16, – 2)
(d)
Q21.The equation of the straight line passing through the point of intersection of lines 3x – 4y – 7 = 0 and 12x – 5y – 13 = 0 and
perpendicular to the line 2x – 3y + 5 = 0 is
(a)
33x + 22y + 13 = 0
(b)
33x + 22y – 13 = 0
(c)
33x – 22y + 13 = 0
(d)
None of these
Q22.If the family of lines x(a + 2b) + y(a + 3b) = a + b passes through the point for all values of a and b, then the coordinates of the
points are
(a)
(2, 1)
(b)
(2, – 1)
(c)
(– 2, 1)
(d)
None of these
Q23.The value of k so that the lines 2x – 3y + k = 0, 3x – 4y – 13 = 0 and 8x – 11y – 33 = 0 are concurrent, is
(a)
7
(b)
–7
(c)
5
(d)
–5
Q24.Let P be the image of the point (– 3, 2) with respect to x-axis. Keeping the origin as same, the coordinate axes are rotated through
an angle 60° in the clockwise sense. The coordinates of point P with respect to the new axes are
(a)
(b)
(c)
(d)
Q25.If for a variable line
drawn from origin to this is
None of these
, the condition a–2 + b–2 = c–2 (c is a constant) is satisfied, then the locus of foot of the perpendicular
(d)
x2 – y2 = c2
(d)
none of these
(d)
none of these
Q29.Let R be reflexive relation on a finite set A having n elements, and let there be m ordered pairs in R. Then
(a)
m³n
(b)
m£n
(c)
m=n
(d)
none of these
(a)
(b)
x2
+
y2
x2 + y2 = 2c2
(c)
x2 + y2 = c2
=
Q26.If A and B are two sets, then A  (A È B)¢ equals
(a)
A
(b)
B
(c)
f
Q27.Which of the following is an empty set ?
(a)The set of prime numbers which are even
(b)The solution set of the equation
(c)(A x B) Ç (B x A), where A and B are disjoint
(d)The set of real which satisfy x2 + ix + i – 1 = 0
=0;xÎR
Q28.If sets A and B are defined as : A = {(x, y) : y = , x ¹ 0, x Î R} , B = {(x, y) : y =-x, x Î R}, then
(a)
AÇB=A
(b)
AÇB=B
(c)
AÇB=f
Q30.f(x) = | sin x | has an inverse if its domain is :
(a)
[ 0, p ]
(b)
Q31.Consider the following equations
(1) A – B = A – (A Ç B)
(2) A = (A Ç B) È (A – B)
(2) A – (B È C) = (A – B) È (A - C)
(c)
(d)
none of these
Which of these is / are correct
(a)
1 and 3
(b)
2 only
(c)
2 and 3
(d)
1 and 2
Q32.If a is the set of the divisors of the number 15, B is the set of prime numbers smaller than 10 and C is the set of even numbers
smaller than 9, then (A È C) Ç B is the set
(a)
{1, 3, 5}
(b)
{1, 2, 3}
(c)
{2, 3, 5}
(d)
{2, 5}
Q33.x, y and z are rational numbers. Consider the following statements in this regard
(1)
x+3=y+zÞx=y
(2)
Which of the above statement(s) is / are correct ?
(a)
1 alone
(b)
2 alone
(c)
xz = yz Þ x = y
Both 1 and 2
(d)
Neither 1 nor 2
Q34.If second term of an AP is 2 and 7th term is 22, then sum of 9 terms is
(a)
126
(b)
– 126
(c)
90
(d)
252
Q35.If Sn denotes the sum of the first n terms of an AP and S2n = 3Sn, then
(a)
7
(b)
6
(c)
8
(d)
10
, then the ratio of the 7th term and 5th term of the AP is
Q36.The sum of r terms of an AP is denoted by Sr and
(a)
(b)
(c)
(d)
Q37.A square is drawn by joining the mid points of the given square a third square in the same way and this process continues
indefinitely. If a side of the first square is 16 cm, then the sum of the areas of all the squares
(a)
128 sq cm
(b)
256 sq cm
(c)
512 sq cm
(d)
1024 sq cm
Q38.If the pth term of an AP is
(a)
0
and qth term is
then the sum of the first pq terms is
(b)
(c)
(pq – 1)
(d)
(pq + 1)
Q39.If the AM between pth and qth terms of an AP be equal to AM between rth and sth terms of the AP, then
(a)
p+s=q+r
(b)
p+q=r+s
(c)
p+r=q+s
(d)
p+q+r+s=0
Q40.The value of 5 + 55 + 555 + …… to n terms is
(a)
10n + 1 – 10
(b)
(c)
(d)
[10n + 1 – 1 + 9n]
(10n + 1 – 10)
Answ ers
Question
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Answer
c
c
a
c
a
d
b
b
c
c
d
a
a
b
a
a
b
c
a
d
a
b
b
a
c
Question
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Answer
c
c
c
a
b
d
a
d
a
b
d
c
c
b
b
ATOMIC STRUCTURE
Introduction
The most special thing about this chapter is that the no. of scientists involved in developed of this concept is
remarkably large. Starting with heucippus of Miletus in 440 BC it extends to 2000's. And the research is still
going on. Even the great scientist EINSTEIN had his hands in developing this extraordinary concept.
Atomic No.: Total no. of protons present in nucleus.
Mass No.: Total no. of protons & Neutrons in neucleus.
Atomic No.(Z): no. of electrons (when atom is neutral).
Mass No.(A): no. of neutrons + no. of protons.
Representation
Ilustration: Calculate no. of protons, neutrons & electrons in
.
Ans: Z = 35, A = 80
No. of protons = Z = 35
No. of neutrons = A - Z = 80 - 35 = 45
As atom is neutral. No. of electrons = No. of protons = 35.
Isotopes: Such atoms of same element having same atomic no. but different mass no.s are c/d isotopes.
Protiun (
). Denterium (
) & Tritium (
) are isotopes.
Isobars: Atoms of different elementswhich have same mass no.s but different atomic No.
,
,
are isobars.
Isotones: Atomes of different elements which contain same no. of neutrons c/d isotones.
,
are isotones because they have same no. of neutrons.
subatomic particle like electron at high speed varies.
m=
m0 restmass
c velocity of light
Photoelectric effect: When radiations with certain minimum frequency (v0) strike surface of metal electrons are
ejected from surface of metal. This is c/d photoelectric effect & electrons are called photoelectrons.
Threshold frequency: It is min. frequency (v0) below which no electrons are ejected.
Work function: Min. energy required to eject electron (h
0).
h = h 0 + mv2
Total energy = Workfunction + Kinetic energy.
Planck's Quantam Theory:
(i) Radiant energy is emitted or absorbed not continuously but discontinuously in form of small discrete packets
of energy. Each packet c/d 'quanta'. In case of hight energy 'quanta' is 'photon'.
(ii) Energy of each quantum is directly proportional to frequency of radiation.
E=h
h Planck's Constant.
h = 6.626 x 10-34 J/sec.
Body will be some whole no. quanta.
E = nh
n integer.
Illustration: Calculate KE of electrons ejected when yellow light of frequency 5.2 x 1014 s-1 on a metal whose
threshold frequency is 4 x 1014 s-1.
Ans: h
h
=h
-h
0
+
mv2
= mv2
mv2 = h(1.2 x 1014)
-34
= 6.625 x 10 x 1.2 x 1014
= 7.95 x 10-20 J
0
Bohr's Model:
Postulate:
(1) As long as electrons occupy definite energy level, it does not radiate out energy. Emission or absorption of
energy occures only when electrons jumps from one level to other.
E = En2 - En1 = h
If n2 > n1 emission epectra.
If n2 < n1 absorption spectra.
n1
n2
Initial state.
Final state.
(2) Angular momentum of electron in closed is always quantized i.e. integer multiple of
Angular momentum = n
mvr = n
.
.
Dumb Question: Why angular momentum is mvr ?
Ans: Angular momentum = Moment of Inertia x angular velocity.
=
Note: In CGS unit,
MKS unit,
= mvr
=1
= 9 x 109 Nm2c-2.
BOHR MODEL & X-RAYS
Physics: Modern Physics
Neils Bohr gave forward his model of atom with answeres most of the flaws that Rutherford model had. This model introduced
quantization principal and this model is regarded as first quantum mechanical model. We will explore in detail, this structure of atom.
We will also look into X - ray formation which is a converse process of photoclectric effect and will examine Moreley's law in pantium.
THE BOHR THEORY OF HYDROGE ATOM (AND HYDROGEN LIKE ATOM)
Bohr gave following postulates for electron in hydrogen atom :• An electron in an atom could resolve in certain stable orbits without the emission of radiant energy.
• An electron resolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of
angular of electron is QUANTAIZED.
L=
, where n
. Thus
principal quantum number.
• An electron might a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a single
photon is amitted having energy equal to the energy difference between the initial and final slater.
The frequency of emitted photon is given by
h = Ei - Ef
Dumb Question:- Are these equations valid for all atoms ?
Ans:- No, equations are valid only for single electron system, i.e. systems where only one electron is present. Ideally this equation
applies only to Hydrogen atom, Singly ionized helium He +, and doubly ionized Lithium Li+ +.
In Bohr's model, radius of nth orbit is given by rn =
and speed of electron is given by Vn =
if values of constant are evaluate then,
rn = 0.529 n2/z
and Vn = 2.19 X 106
m/sec.
where n
Principal quantum number/shell number.
z
atomic number of nucleus involved.
Why ??
from Bohr's postulate, angular momentum of electron in nth orbit is L =
also from mechanics L = mvnrn
mvnrn =
............................................... (1)
for electron to resolve in its orbit, Electrostatic force is providing the required centripetal force.
Solving (1) & (2) simultaneously yields the required result.
BOHR MODEL & X-RAYS
Physics: Modern Physics
Bohr Energy levels:- The energy of a revolving electron is :(i) Kinetic Energy, due to its revolution in the orbit.
(ii) Electrostatic Potential Energy.
The total energy Eu is the sum of kinetic and potential energies.
Eu = Ku + Vu = if the value of constants are placed then
Eu =
Dumb Question:- What is the significance of total energy of e- as a negative quantity?
Ans:- Total energy as negative quantity sindicates that it is held in a stable state and energy needs to be supplied to it if electron has to
be pulled out.
Dumb Question:- Where has the zero of potential energy been let ?
Ans:- The zero of electrostatic potential energy has been setup at infinity i.e. outside the bounds of atom.
If an electron moves from ni orbit to lower nf orbit then it lose, energy which is emitted in the form of radiations and the wavelength of
such radiation is given by formula.
, Where R = Rydberg's constant = 1.097 x 107 m- 1
Why ??
energy of an orbit is given by En = from Bohr's postulate when an electron falls from higher orbit to lower orbit then energy of emitted radiation is given by
E = Ei - Ef =
if R =
(constant)
then
Few named transitions :-
Lymen Series
n = 2, 3, 4,...............
Balmen Series
n = 3, 4, 5,...............
Paschen Series
n = 4, 5, 6,...............
Brackett Series
n = 5, 6, 7,...............
Pfund Series
n = 6, 7, 8,...............
(In all these cases electron is moving down from nth orbit to concerned orbit)
• Total number of emission lines from same excited state n 1 to another state n2 (< n1) is given by
BOHR MODEL & X-RAYS
Physics: Modern Physics
few handy difinations:IONIZATION POTENTIAL:- The minimum energy needed to ionize an atom is called ionization energy, and the potential defference
through which an electron should be accelerated to acquire this energy is called ionization potential. The ioniozation energy of hydrogen
atom is ground state is 13.6 ev and ioniozation potentiation is 13.6 v.
BINDING ENERGY:- Binding enrgy of a system is defined as energy liberated when its constituents are brought from infinity to form the
system. For hydrogen atom binding energy is same as its ionization energy.
EXCITATION ENERGY:- The energy required to take an atom from its ground state to an excited state is ncalled excitation energy of
that excited state, and the potential.
ILLUSTRATION:- Find the longest and shortest wavelength in the Lyman Series for Lydrogen. In what region of the electromagnetic
spectrum does each series lie ?
Solution:- The transition equation for Lyman Series is given by,
, n = 2, 3, ........................
The longest wavelength is corresponding to n = 0
= 1.099 x 107(1 - 1/4) = 0.823 x 1s7
= 1.254 x 10- 7m = 1215
The shortest wavelength corresponds to n =
= 1.097 x 107(1 min
= 0.911 x 10-7 m = 911
)
AS can be seen both, the weavelength lie in ultraviolet (UV) region of electromagnetic spectrum.
X-RAYS
Electromagnetic radiations with wavelength from 0.1A 0 to 100A0 falls into the category of X - rays, however the boundaries of this
catygory are not sharp.
• The process of production of X-rays is reserve of photoelectric efffect. Here electromagnetic radiations are emitted by the
bombordment of high speed thermionically emitted electrons on the metal surface.
X-rays produced are of two types:(a) CONTIMUOUS X-RAYS:- When electrons of vary high kinetic energy accelarates in an electric field then according to Maxwells
theory, it radiates energy in the form of electromagnetics waves X-rays produced in this fashion are called continuous X-rays (also
called Bremsstrahlung).
• Continuous X-rays produced at a given accelerating potential V very in wavelength, but none has wavelength shorten than a certain
value
,V
min.
potential aeron which electron is accelerated.
Dumb Question:- What is the significance of this minimum wavelength of the X-Ray ?
Ans:- This minimum wavelength corresponds to the maximum energy of the X-ray produced. This condition occures when striking
electron looses all its energy and this energy gets converted into X-rays. This for it
Energy of X-ray = total kinetic energy of electron = energy gained when accelerated across potential barrier
Note That the maximum wavelength
0 , this is the case when striking electron looses very small fraction of its energy.
CHARACTERISTICS X-RAYS:- The sharp lines superimposed on the continuous spectrum are known as characteristic X-rays.
because they are characteristic of the target material.
s
• Characteristic X-rays emission occures when a banbarding electron that collides with a target atom has sufficient energy to remove on
inner shell electron from the atom.
• The vacancy created in the shell is filled when an electron from a higher level drops down into it. This transition is of a photon whose
energy equals the difference in energy between the two levels.
Naming of characteristic X-rays:- Let the incoming electron dislodge an atomic electron from innermost shell - the K shell. If the vacancy
is filled by an electron dropping from the next higher shell - the L shell emission is said to be of Kd series. If vacancy is filled by an
electron dropping from M shell,
Dumb Question:- Why do
Ans:-
represents transition from higher shell then the kd line,
lower wavelength as
line is produced and Jo an.
occur earlier than Kd in the plot ?
reperesents transition over larger energy range, thus would have
. thats why
occurs earlier than Kd.
Electric Capacitance And Capacitor
INTRODUCTION
The capacitance of any conductor is its capacity to store charge. As the quantity of charge on conductor
is increased its potential also increases. Beyond a limit even if a small additional charge is given then it
starts leaking.
Hence Q  V or Q = CV.
Where C is called capacitance of the conductor. It depends upon geometrical shape and size of the
conductor.
If V = 1 volt, then Q = C, hence capacitance of any conductor will be numerically equivalent to that
charge which makes its potential equal to 1 volt.
Unit of capacitance is ‘Farad’:
1 Farad = 1 coulomb/1 volt.
= coulomb/ (joule/coulomb).
= coulomb2/joule.
= (ampere x second) 2 / ( Newton x meter).
3) Dimensions of capacitance:
In practice 1 Farad is very large capacity therefore its small units are as following:
1) 1 micro-farad = 10-6 farad = 1f.
2) 1 nano-farad = 10-9 farad = 1f.
3) 1 pico-farad or micro-micro farad = 10-12 farad = 1f = 1f.
4)
Capacitance of spherical conductor:
For a spherical conductor of radius R,
Potential is
Fig (1)
1)
.
2)  C  R.
3) C (capacitance) does not depend upon the charge given.
5) Energy stored in a charged conductor:
6) Redistribution of charge:
Fig (2)
Electric Capacitance And Capacitor
INTRODUCTION
Before contact:
Q = Q1+Q2
= V1C1+V2C2
In contact or after contact
Q11 = C1V
Q21 = C2V
And
7) Energy loss in redistribution of charge:
1) If V1 = V2, means two conductors have the same potential then energy loss is zero otherwise
2)
V1>V2 or V2>V1; E will always be time positive. Hence, there will be loss of energy in the
form of heat.
8) Capacitor (Parallel plate capacitor):
It consist of two plates placed close together and is insulated from one another by air or same dielectric
medium. One of the plates is given a charge and other is grounded. This device can store charge or
electrical energy.
9) Different types of grouping of capacitors and the effective capacitance:
1) Series Combination:
Fig (3)
Remember
V = V1 + V2 + V3 + --------------
2) Parallel Combination:
Fig (4)
Remember:
Q = Q1 + Q2 + Q3 + ----------------C = C1 + C2 + C3 + ----------------10) If n parallel plates are connected as shown in the figure below and C1 is the capacitance between
two adjacent plates then total capacitance will be C
C = (n-1) C1.
Fig (5)
Electric Capacitance And Capacitor
INTRODUCTION
11) Capacitance of spherical capacitor:
(A) When outer sphere is earthed.
Fig (6)
(B) When inner sphere is earthed:
Fig (7)
12) Capacitance of cylindrical capacitor:
Fig (8)
Electric Capacitance And Capacitor
INTRODUCTION
13) Capacitance per unit length between two parallel wires of radius r and separated by distance ‘d’:
Fig (9)
14) The energy stored in a fully charged capacitor:
E2 = Energy density.
Where E is electric field in a capacitor.
15) Charging and discharging of capacitors:
(a)
Fig (10)
(b)
Fig (11)
VC = E at t=0.
(c)
Fig (12)
(d)
Fig (13)
HEAT TRANSFER
Introduction
Heat Transfer helps us to understand various aspects for the transfer of heat through various materials(solids, liquid and gasses)
through various known process such as conduction, convection, radiation etc.
So, it finds a great use in engineering applications.
Heat is transferred from higher temperature to lower temperature. [secondary information]
There are three different ways in which heat can be transferred:
1. Conduction:- it is process by which heat can be transferred in solids.
2. Convection:- it is aprocess by which heat is transferred in fluides (gaseous &liquid)
3. Radiation:- it is a process in which the heat is transferred in the form of electromagnetic waves without yhe aid of any material
medium.
Thermal conductivity:
In solids, heat is transferred through conduction. We will study conduction of heat through a solid bar in the following sector.
Consider a solid bar of thickness d and area of cross-section A. The left side of bar is maintained at QA temperature and right side at QB
Let us suppose QAQB
Heat flows from higher temperature to lower temperature after some time, the temp of each section becomes constant with time. This is
known as steady state.
In steady state, if
Q Amount of heat crosses through any cross section in time
T, then
and
or,
K: coefficient of thermal conductivity.
For a small thickness Dx along the direction of heat and whose thickness dx is small and temperature difference is
Q
[Secondary information]
Quantity
called the temperature gradient and minus sign indicates that
is negative along the direction of heat flow.
[Secondary Information]
We can compose the equation with the Ohm's Law
V1 - V2 = I.R
I=
Similarly
hence
is called thermal resistance rth.
Rth =
Accretion of Ice [Secondary Information]:
Consider a layer of ice of thickness x has air temp - Q0c and is 00c.
HEAT TRANSFER
Introduction
Considering unit cross -section area of ice , if alayer of thickness dx given in time dt, The heat given by thin layer.
= mass x Latent heat
= 1.dx.
.L.
= density of ice
L = Latent heat of fusion of ice.
The quantity of heat is conducted upward through the ice layer in time dt
dx.
.L. =
Time taken
Dumb Question:
Q1: Pieces of iron and glass are heated to same temperature. which one will be more hot ? why ?
A: Pieces of iron wil be hotter. Because of higher thermal conductivity.
Q: Cooking utensils are provided wth wooden handes why ?
A: Wood is pure conductor of heat. So hot utensils can be easily handled.
Q: In winter, metallic handles of wooden door appear colder. Why ?
A: Because they are good conductor of heat.
Q: Why is ice packed in gunny bags or sawdust.
A: Air trapped in saw dust prevent transfer of heat from the surroundings, to the ice.
Illustration:
Q: A cubical ice box of thermocole has each side = 30cm and a thickness of 5cm. 4kg of ice is put in the box. If outside temp is 450c
and the coeff. of thermal conductivity = 0.01 J s-1 m-1 0c-1 Calculate the mass of ice left after 6 hrs. Take latent heat of fusion of ice =
335 x 103 J/kg.
Solution: l = 0.3 m
Thickness = 0.05 m
Total surface area = 6l2
= 0.54 m2
T = 450c
T = 6 hrs
Q = mL = KA
m = KA
= 0.313 kg
wt of ice left = 3.687 kg
Radation:
Absorptive Power a:
Absorptive power of a body is defined as the fraction of the incident radiation that is absorbed by the body.
Absorptive Power a =
Emmisive Power E:
The emmisive power denotes the energy radiated per unit time per unit area of the surface.
Emmisitivity E:
The emmisitivity of a surface is the ratio of the Emmisive power of the surface to the emmisive power of the black body at that temp.
E=
Black Body:
A perfectly black body is the one which absorbs completely all the radiations of whatever wavelength incident on it.
Kirchoff's Law:
It states that the ratio of the Emmisive power to the Absorptive power for radiation of a given wavelength is the same for all bodies at
the same temperature.
[constant]
[Secondary Information]
Everything is continuously emitting radient energy in all direction at a rate depending only on the nature of the surface and its
temperature and also it is absorbing radient energy from all surrounding bodies at arate depending on its surface and temperature of
the surrounding bodies.
HEAT TRANSFER
Introduction
Stefan's Law of Radiation:
The total radient energy emitted E per unit time by a black body of surface A is propertional to the fourth power of its absolute
temperature.
E T4
4
or E =
AT
= Stefan's const
For abody which is not black body
E=
AT4
= emmisitivity of te Black Body.
Note: Emmisitivity and absorptive power have the same value.
Net loss of thermal Energy[secondary information]:
If abody of surface area A is kept at absolute temp T in asurrounding of temperature T0(T0 > T) Then the energy emitted by the body
per unit time.
E=
AT4
And the energy absorbed per unit time by the body
E0 =
AT04
Net loss of thermal energy per unit time.
E = E - E0 =
A[T4 - T04]
Newton's Law of Cooling:
For a small temperature difference between a body and its surroundings, the rate of cooling of the body is dirctly propertional to the
temperature difference. If abody of temperature T and surface area A is kept in a surrounding temperature T 0(T0 < T). Then net loss of
thermal energy per unit time.
=
A(T0 - T00)
If the temperature difference is small
T = T0 +
T)4] =
=>
A[(T0 +
=>
AT04[1 + 4
T
A[T04 (1 +
)4 - T04]
+ higher powers of
- 1]
3
=4
AT0
T
Now, Rate of loss of heat at temperature T
mc
=-4
AT03[T - T0]
= - K[T - T0]
K=
(T - T0)
WIENS LAW:
At a const temperature T, when wavelength
is increased , the energy emitted E, first increases, reaches a maximumand then
decreases i.e at a particulartemperatuure the spectral radiancy
is a maximum at a particular wavelength
As the temperature increases , the maximum radiancy of energy occurs at shorter wavelength i.e
b is a constant.
This is called wiens Displacement Law.
Dumb Question:
mt = b
m.
Q: At a particular wavelength the bodies A, B, C, have their emmisive Power in Ratio 1:2:3. what is the ratio of the absorptive power ?
A: 1:2:3 became
= constant