Question 1 1.33 / -0
Three players play a total of 9 games. In each game, one person wins and the other two lose; the winner gets 2 points and the losers get -1 each. The number of ways in which they can play all the 9 games and finish each with a zero score is
Solution
Let the three players are A, B and C.
Now, each player has score 0 after playing 9 games. It can happen only when each player wins 3 games and loses 6.
So,
A wins 3 games out of 9, so number of ways A win a game
B wins 3 games out of remaining 6
and, C wins 3 games out of remaining 3
So, required number of ways =
Question 2 1.33 / -0
If sin x + sin y = 7/5 and cos x + cos y = 1/5, then sin (x + y) equals
Solution
sin x + sin y =
------(1)
cos x + cos y =
------(2)
By (1)
2 + (2)
2 , we get
2 + 2 sin x sin y + 2 cos x cos y = 2
sin x sin y + cos x cos y = 0
cos (x - y) = 0
Therefore, x - y = 90°
By (1) x (2), we get
sin x cos x + sin x cos y + sin y cos x + sin y cos y =
sin (90 + y) cos x + sin (x + y) + sin (x - 90) cos y =
cos y cos x + sin (x + y) - cos x cos y =
sin (x + y) =
Question 3 1.33 / -0
What is the value of
? (Here, [t] denotes the integral part of the real number t.)
Solution
cos (
x) cos ([2x]
)dx
=
cos (
x) cos 0dx +
cos (
x) cos
dx
=
cos (
x)dx -
cos (
x)dx
=
=
=
Question 4 1.33 / -0
If m
1 and n are positive integers such that m
n
(here dk means d is a positive divisor of k), then
Solution
Let M = 2
x 1.3
x 2.5
x 3 ...., N = 2
y 1.3
y 2.5
y 3......x
i & y
i ∈
W
Question 5 1.33 / -0
Suppose two perpendicular tangents can be drawn from the origin to the circle x2 + y2 – 6x – 2py + 17 = 10, for some real p. Then |p| is equal to
Solution
Equation of chord of contact T = 0, with respect to origin is
– 3(x + 0) – p(y + 0) + 17 = 0
3x + py – 17 = 0
By homogenisation:
Therefore, for perpendicular, coefficient of x
2 + coefficient of y
2 = 0.
34 - 18 + 9 - p
2 = 0
p
2 = 25
Question 6 1.33 / -0
The mid-point of the domain of the function f(x) =
for real x is
Question 7 1.33 / -0
Let [x] and {x} be the integer part and fractional part of a real number x, respectively. The value of the integral
is
Solution
=
= (1 + 2 + 3 + 4)
= 10.
= 10.
= 5
Question 8 1.33 / -0
Suppose a1 , a2 , a3 , ....., a2012 are integers arranged in a circle. Each number is equal to the average of its two adjacent numbers. If the sum of all even indexed numbers is 3018, then what is the sum of all the numbers?
Solution
Now, a
2 + a
4 + ........ + a
2012 = 3018 ...............(1)
2a
2 + 2a
4 + ............ + 2a
2012 = 6036
a
1 + a
3 + a
3 + a
5 + ........... + a
2011 + a
1 = 6036
2(a
1 + a
3 + ..........+ a
2011 ) = 3018 ...............(2)
By adding (1) and (2), we get
a
1 + a
2 + a
3 + ......... + a
2012 = 6036
Question 9 1.33 / -0
Let I, ω and ω2 be the cube roots of unity. The least possible degree of a polynomial, with real coefficients, having 2ω2 , 3 + 4ω, 3 + 4ω2 and 5 - ω - ω2 as roots is:
Solution
Roots
,
,
, 5 -
-
Now,
= 5 -
= 5 + (1) = 6
If
is a root, then 2
has to be a root too.
Least number of total roots = 5
Least possible degree = 5
Question 10 1.33 / -0
The product (1 + tan1°)(1 + tan2°)(1 + tan3°)...(1 + tan45°) equals:
Question 11 1.33 / -0
Let I
n =
, where n is a non-negative integer. Then, I
2011 + 2011 I
2010 is equal to:
Question 12 1.33 / -0
The number of natural numbers n in the interval [1005, 2010] for which the polynomial 1 + x + x2 + x3 + ... + xn - 1 divides the polynomial 1 + x2 + x3 + x4 + ... + x2010 is:
Solution
1 + x
2 + x
4 + ... + x
2010 =
=
=
This is divisible by 1 + x + x
2 + ... + x
n - 1 ,
if
n - 1 = 502
Or n = 503
Question 13 1.33 / -0
Consider the cubic equation x3 + ax2 + bx + c = 0, where a, b and c are real numbers. Which of the following statements is correct?
Solution
f(x) = x
3 + ax
2 + bx + c
f'(x) = 3x
2 + 2ax + b
D = 4a
2 - 4.3.b = 4(a
2 - 3b)
If a
2 < 2b
a
2 < 3b
D < 0
f'(x) = 0 has no real roots.
Hence, f(x) = 0 has 1 real and two imaginary roots.
Question 14 1.33 / -0
Let f (x) = cos 5x + A cos 4x + B cos 3x + C cos 2x + D cos x + E and T = f(0) - f(
) + f(
) - f(
) +..........+ f(
) - f(
). Then, T
Solution
Clearly,
(Every term contains cosine)
and
.
Then,
T contains B and D terms.
Question 15 1.33 / -0
Let f: (2,
)
N be defined by f(x) = The largest prime factor of [x]. Then,
is equal to
Question 16 1.33 / -0
The sum of (12 – 1 + 1)(1!) + (22 – 2 + 1)(2!) + …. + (n2 – n + 1)(n!) is
Solution
(r
2 - r + 1) r!
T
r = [(r + 1) (r - 1) - (r - 2)] r!
T
r = (r - 1) (r + 1)! - (r - 2) (r!)
but r = 1, 2, ......., n
sum = 1 + (n + 1)! (n - 1)
Question 17 1.33 / -0
If a and b are positive real numbers such that the lines ax + 9y = 5 and 4x + by = 3 are parallel, then the least possible value of a + b is
Question 18 1.33 / -0
Let f: R
R be a function such that
f(x) = M > 0. Then, which of the following is false?
Solution
From option 3, L.H.S. =
= 1 × (0) × M = 0
R.H.S.
Question 19 1.33 / -0
The area bounded by the curve y = cos x, the line joining
and (0, 2), and the line joining
and (0, 2) is:
Question 20 1.33 / -0
For an integer n, let Sn = {n + 1, n + 2, ... n + 18}. Which of the following is true for all n
10?
Solution
n + 1, n + 2, ... n + 18
(1) False, if n = 19
(3) False if n = 15
16 to 33
20, 25, 30
only three multiples of 5
(4) Number of odd integers in S
n = 9
Every third odd integer is the multiple of 3.
So, maximum prime numbers = 6
Question 21 1.33 / -0
The five sides of a regular pentagon are represented by vectors
and
, in cyclic order as shown. Corresponding vertices are represented by
and
, drawn from the centre of the pentagon.
Then,
= ?
Solution
From the following figure,
Angle between every vector is same.
So,
Question 22 1.33 / -0
In the hydrogen spectrum, the ratio of the wavelength for Lyman -radiation to Balmer- radiation is
Question 23 1.33 / -0
A point electric dipole placed at the origin has a potential given by
, where θ is the angle made by the position vector with the direction of dipole. Then,
Solution
Electric field at each point of OA obtained is perpendicular to it and opposite to direction of the dipole moment.
Question 24 1.33 / -0
A stream of charged particles enters into the region with crossed electric and magnetic fields as shown in the figure. On the other side is a screen with a hole, that is right on the original path of the particles.
Then,
Solution
Force of positive charge = Electric force + Magnetic force
Question 25 1.33 / -0
Seven identical coins are rigidly arranged on a flat table in the pattern shown below so that each coin touches its neighbours. Each coin is a thin disc of mass m and radius r. Note that the moment of inertia of an individual coin about an axis passing through centre and perpendicular to the plane of the coin is
.
The moment of inertia of the system of seven coins about an axis that passes through the point P (the centre of the coin positioned directly to the right of the central coin) and perpendicular to the plane of the coins is
Solution
First we mark coins with same number that have same distance from point P.
Now,
Question 26 1.33 / -0
A steady current I flows through a wire of radius r, length L and resistivity
. The current produces heat in the wire. The rate of heat loss in a wire is proportional to its surface area. The steady temperature of the wire is independent of
Solution
It is a direct concept of fuse wire.
Or we can say that
I
2 R = Rate of heat loss through radiation
And R
Now, from above,
We can say that
is independent of L.
Question 27 1.33 / -0
A particle is acted upon by a force given by F = -αx3 - βx4 , where α and β are positive constants. At the point x = 0, the particle is
Solution
As F = -
At x = 0, F = 0
For x > 0, F < 0 and for x < 0, F > 0
So, the particle is in stable equilibrium as the particle has a tendency to move towards equilibrium position when disturbed from it.
Question 28 1.33 / -0
A parent nucleus X is decaying into daughter nucleus Y, which in turn decays to Z. The half lives of X and Y are 4000 years and 20 years, respectively. In a certain sample, it is found that the number of Y nuclei hardly changes with time. If the number of X nuclei in the sample is 4 × 1020 , the number of Y nuclei present in it is
Solution
Rate of decay of X = Rate of decay of Y
As we know
4 × 10
20 ×
N
2 = 2 × 10
7
Question 29 1.33 / -0
A comet (assumed to be in an elliptical orbit around the sun) is at a distance of 0.4 AU from the sun at the perihelion. If the time period of the comet is 125 years, what is the arphelion distance? AU : Astronomical Unit
Solution
r =
Also,
by Kepler's law of time-periods (T
1 , r
1 are taken for earth)
Therefore,
Solving, we get
y = 49.6 AU
Question 30 1.33 / -0
A point source of light is placed at the bottom of a vessel which is filled with water of refractive index μ to a height h. If a floating opaque disc has to be placed exactly above it so that the source is invisible from above, the radius of the disc should be
Solution
Radius of the disc should be such that rays beyond it get totally internally reflected.
For this,
, where C is the critical angle of incidence
Or
Also,
Therefore,
In limiting case,
Solving, we get
Question 31 1.33 / -0
The flat face of a plano-convex lens of focal length 10 cm is silvered. A point source placed 30 cm in front of the curved surface will produce a
Question 32 1.33 / -0
A charged particle of charge q and mass m gets deflected through an angle θ upon passing through a square region of side 'a' which contains a uniform magnetic field B normal to its plane. Assuming that the particle entered the square at right angles to one side, what is the speed of the particle?
Question 33 1.33 / -0
A container with rigid walls is covered with perfectly insulating material. The container is divided into two parts by a partition. One part contains a gas while the other is fully evacuated (vacuum). The partition is suddenly removed. The gas rushes to fill the entire volume and comes to equilibrium after a little time. If the gas is not ideal, the initial
Solution
Expansion is against vacuum. Therefore,
The container is insulated. Therefore,
Using the first law of thermodynamics, we get
0 = 0 +
(From above, we can say that U
initial is equal to u
final ).
Question 34 1.33 / -0
A plane polarised light passed through successive polarisers, which are rotated by 30° with respect to each other in the clockwise direction. Neglecting absorption by the polarisers and given that the first polariser's axis is parallel to the plane of polarisation of the incident light, the intensity of light at the exit of the fifth polariser is closest to
Solution
=
= 30% of I
0
Question 35 1.33 / -0
An electron collides with free molecules initially in its ground state. The collision leaves the molecules in an excited state that is metastable and does not decay to the ground state by radiation. Let K be the sum of the initial kinetic energies of the electron and the molecule and vector P the sum of their initial momenta. Let K' and vector P' represent the same physical quantities after the collision. Then,
Solution
Momentum is conserved during collision.
So,
=
'.
But, a small amount of kinetic energy is stored in excited state.
So, K' < K.
Question 36 1.33 / -0
Quantum Hall Resistance RH is a fundamental constant with dimensions of resistance. If h is the Planck's constant and e is the electron charge, then the dimension of RH is the same as
Solution
R
H =
=
.........................(1)
Energy =
=
Power =
P =
..........................(2)
and
.................(3)
From above equations, we get
Question 37 1.33 / -0
The equation of state of n moles of a non-ideal gas can be approximated by the equation
, where a and b are constants characteristic of the gas.
Which of the following can represent the equation of a quasistatic adiabatic equation for this gas (Assume that C
V , the molar heat capacity at constant volume, is independent of temperature)?
Solution
For the real gas, we have
Comparing it with ideal gas, the effective pressure is
and the effective volume is
.
For an ideal gas, the adiabatic process can be written as
Using
, the equation for the adiabatic process can be written as
Hence, for the real gas, the equation for the adiabatic process can be written as
Or
Question 38 1.33 / -0
The circuit shown has been connected for a long time. The voltage across the capacitor is
Solution
In steady state, current through capacitor is zero.
Hence, voltage across 2 k
= Voltage across capacitor
Voltage across 2 k
=
× 6 = 4 V
Question 39 1.33 / -0
An electron enters a chamber in which a uniform magnetic field is present as shown. Ignore gravity.
During its motion inside the chamber,
Solution
As force working on the electron inside the magnetic field will be perpendicular to the displacement, work done will be zero.
Question 40 1.33 / -0
A point particle of mass 0.5 kg is moving along the x-axis under a force described by the potential energy V shown below. It is projected towards the right from the origin with a speed v.
What is the minimum value of v for which the particle will escape infinitely far away from the origin?
Solution
From energy, 2 ) + 0 = 0 + 12 = 4
Question 41 1.33 / -0
In the following reactions
,
the major product X is
Solution
First step - Ozonolysis reaction
Second step - Intramolecular aldol reaction
Question 42 1.33 / -0
The solvent used for carrying out a Grignard reaction is
Solution
Diethyl ether is a good solvent for the formation of Grignard reagents because ether is aprotic in nature, while water, chloroform, ethyl acetate or alcohol would protonate and destroy the Grignard reagent. This is because the carbon atom in Grignard reagent is highly nucleophilic. This would form a hydrocarbon on reaction with water, alcohol etc.
Question 43 1.33 / -0
In one component second order reaction, if the concentration of the reactant is reduced to half, then the rate of the reaction
Solution
Suppose the reaction is 2A
Product.
According to rate law, rate R = k[A]
2 According to the question,
[
]
⇒
= 4
⇒
=
Hence, the final rate of reaction is reduced to one-fourth of the initial rate of reaction.
Question 44 1.33 / -0
The half-life of a first order reaction is 30 min. The time required for 75% completion of the same reaction is
Solution
t
1/2 = 30 minutes (Given)
For first order reaction,
K =
K =
= 0.0231 min
-1 Now, time required for the completion of 75% of the reaction is as follows.
Let the initial concentration = [A
o ]
Final concentration [A] = [A
o ] -
=
t = 60.01 ~ 60 minutes
Question 45 1.33 / -0
In the reaction of benzene with an electrophile E+ , the structure of the intermediate σ-complex can be represented as
Solution
During the electrophilic substitution reaction of benzene, the electrophile E
+ attacks the π-electron cloud of aromatic ring and forms σ-bond with carbon, creating a positive charge on the ring. This results in the formation of a σ-complex or a carbocation or arenium ion which gets stabilised by resonance.
Question 46 1.33 / -0
The isoelectronic pair of ions is:
Solution
Isoelectronic species have the same number of electrons. As 25 Mn2+ and 26 Fe3+ both have 23 electrons, both are an isoelectronic pair.
Question 47 1.33 / -0
For a tetrahedral complex [MCl4]2- , the spin-only magnetic moment is 3.83 BM. The element M is:
Solution
It is a high-spin complex as Cl- is a weak field ligand. In [CoCl4 ]2- , the oxidation state of Co is +2, in which 3 unpaired electrons are present; this gives the spin-only magnetic moment equal to 3.83 BM.
Question 48 1.33 / -0
The major product of the following reaction is:
Question 49 1.33 / -0
The energy of a photon of wavelength λ = 1 metre is: (Planck's constant = 6.625 × 10-34 Js, speed of light = 3 × 108 m/s)
Solution
According to Planck's quantum theory:
= 1.988 × 10
-25 J
Question 50 1.33 / -0
The most stable conformation of n-butane is:
Solution
The most stable conformation of butane is the one in which the two terminal methyl groups are the farthest removed from each other, i.e. the anti conformation, as shown below:
This is because dihedral angle between CH
3 group is 180°.
Question 51 1.33 / -0
From equations 1 and 2:
CO
2 CO +
O
2 [K
1 = 9.1 × 10
-13 at 1000°C] (eq. 1)
H
2 O
H
2 +
O
2 [K
2 = 7.1 × 10
-12 at 1000°C] (eq. 2),
The equilibrium constant for the reaction CO
2 + H
2 CO + H
2 O at the same temperature is:
Solution
CO
2 CO +
O
2 [K
1 = 9.1 × 10
-13 at 1000°C] (eq. 1)
H
2 O
H
2 +
O
2 [K
2 = 7.1 × 10
-12 at 1000°C] (eq. 2),
To get the required equation: CO
2 + H
2 CO + H
2 O, subtract (2) from (1).
Therefore, K' =
= 0.128
Question 52 1.33 / -0
The aromatic carbocation among the following is:
Question 53 1.33 / -0
At 298 K, the ratio of osmotic pressures of two solutions of a substance with concentrations of 0.01 M and 0.001 M, respectively is
Solution
at const. T.
Question 54 1.33 / -0
When the size of a spherical nanoparticle decreases from 30 nm to 10 nm, the surface area/volume ratio becomes
Solution
⇒ d
1 = 30 nm d
2 = 10 nm
Question 55 1.33 / -0
The standard Gibbs free energy change (ΔG° in kJ mol- 1 ) in a Daniel cell (E°cell = 1.1 V), when 2 moles of Zn(s) is oxidised at 298 K, is closest to
Solution
The cell reaction for Daniel cell is
Zn + Cu
+2 Zn
+2 + Cu
2Zn + 2Cu
+2 2Zn
+2 + 2Cu
For 2 moles of Zn, n = 4.
ΔG° = -nFE°
Cell = -4 × 96,500 × 1.1 = -424.6 KJ/mol
Question 56 1.33 / -0
The reaction
is known as
Solution
Reimer-Tiemann reaction is an organic reaction used to convert a phenol to an
o- hydroxy benzaldehyde, using chloroform and a base.
Question 57 1.33 / -0
Ethyl acetate reacts with NH2 NHCONH2 to form
Question 58 1.33 / -0
Hydrolysis of BCl3 gives 'X', which on treatment with sodium carbonate produces 'Y'. 'X' and 'Y' respectively are
Question 59 1.33 / -0
The number of peptide bonds in the compound is:
Solution
A peptide bond is a chemical bond formed between two molecules when the carboxyl group of one molecule reacts with the amino group of the other molecule, releasing a molecule of water (H
2 O). It is represented as -CO-NH-.
Question 60 1.33 / -0
The order of S
N 1 reactivity in aqueous acetic acid solution for the given compounds is:
Question 61 1.33 / -0
Glycolysis is the
Solution
Glycolysis is the metabolic pathway that converts glucose into two molecules of pyruvate. The process occurs in the cytoplasm of the cell.
Question 62 1.33 / -0
This cell organelle consists of two granule-like centrioles and is found in animal cells only. It helps in cell division. It is called
Solution
Centrosome is an organelle near the nucleus of a cell which contains the centrioles (in animal cells) and from which, the spindle fibres develop in cell division.
Question 63 1.33 / -0
The natural source of Ti plasmid is
Solution
A Ti or tumour inducing plasmid is a plasmid that often is a part of the genetic equipment that Agrobacterium tumefaciens and Agrobacterium rhizogenes (bacteria) used to transduce their genetic material to plants.
Question 64 1.33 / -0
The probability of having a girl child with blood group O when parents have blood group A and B is
Solution
Out of the four offsprings, one has blood group O. Hence, the chances are 25%.
Question 65 1.33 / -0
In which kind of rocks are fossils most commonly found?
Solution
Fossils are more common in some kinds of sedimentary rocks than others. Fossils are most common in limestones. That is because most limestones consist partly or mostly of the shells of organisms.
Question 66 1.33 / -0
Human chromosomes undergo structural changes during the cell cycle. Chromosomal structure can be best visualised if a chromosome is isolated from a cell at
Solution
Metaphase is the stage in which the chromosomes can be best visualised. Metaphase is a stage of mitosis or M phase of the cell cycle.
Question 67 1.33 / -0
DNA mutations that do not cause any functional change in the protein product are known as
Solution
Silent mutations are mutations in DNA that do not significantly alter the phenotype of the organism in which they occur. Silent mutations can occur in non-coding regions (outside of genes or within introns), or they may occur within exons. These mutations do not cause any functional change in the protein product.
Question 68 1.33 / -0
A reflex action is a quick involuntary response to stimulus. Which of the following is an example of BOTH, unconditioned and conditioned reflex?
Solution
A conditioned reflex is an automatic response established by training to an ordinarily neutral stimulus. An unconditioned reflex is a reflex that is inborn or dependent on physiological maturation rather than on learning. Salivation in response to the aroma of food is an example of both.
Question 69 1.33 / -0
A horse has 64 chromosomes and a donkey has 62. Mules result from crossing a horse and a donkey. State which of the following is INCORRECT.
Solution
Mules are a hybrid of two species – a female horse and a male donkey – so they end up with an odd number of chromosomes. A horse has 64 chromosomes and a donkey has 62. A mule inherits 63. Mules can be either male or female, but, because of the odd number of chromosomes, they can't reproduce.
Question 70 1.33 / -0
Fruit wrapped in paper ripens faster than when kept in open air because
Solution
Fruits produce ethylene on ripening. Ethylene serves as a ripening hormone. The fruit wrapped in paper ripens faster as ethylene produced is retained better.
Question 71 1.33 / -0
Out of the following combinations of cell biological processes, which one is associated with embryogenesis?
Solution
Embryogenesis is the process by which the embryo forms and develops. Embryogenesis starts with the fertilisation of the egg cell (ovum) by a sperm cell (spermatozoon). Once fertilised, the ovum is referred to as a zygote, a single diploid cell. The zygote undergoes mitotic divisions with no significant growth (a process known as cleavage) and cellular differentiation, leading to development of a multicellular embryo.
Question 72 1.33 / -0
The distance between two consecutive DNA base pairs is 0.34 nm. If the length of a chromosome is 1 mm, the number of base pairs in the chromosome is approximately
Solution
The total length of double helix DNA in an organism can be calculated by multiplying the total number of base pairs with distance between two consecutive base pairs.-9 m6 bp
Question 73 1.33 / -0
If the sequence of bases in DNA is 5'-ATGTATCTCAAT-3', then the sequence of bases in its transcript will be
Solution
Base pairing is according to Chargaff rule. According to it, a hydrogen bonds with T by two H bonds. G is hydrogen bonded to C by 3 H bonds during DNA replication. When RNA synthesis occurs from DNA, T is replaced by U.
Question 74 1.33 / -0
Carbon dioxide in the blood is mostly carried
Solution
Some of the carbon dioxide is transported dissolved in the plasma. Some carbon dioxide is transported as carbaminohaemoglobin. However, most carbon dioxide is transported as bicarbonate.
Question 75 1.33 / -0
Modern evolutionary theory consists of the concepts of Darwin modified by knowledge concerning
Solution
Modern evolutionary theory consists of the concepts of Darwin modified by knowledge concerning the process of natural selection. This explains that adaptability has genetic basis, which proves fitness of an organism and nature selects organism for its fitness and allows to produce its progeny in large numbers.
Question 76 1.33 / -0
Crossing over occurs at which of the following stages of meiosis I?
Solution
Pachytene is the third stage of the prophase of meiosis, following zygotene, during which the paired chromosomes shorten and thicken, the two chromatids of each separate and exchange of segments between chromatids may occur.
Question 77 1.33 / -0
If DNA codons are ATG GAA, then insertion of thymine after the first codon results in
Solution
A nonsense mutation is a point mutation in a sequence of DNA that results in a premature stop codon. Entry of T causes formation of stop codon.
Question 78 1.33 / -0
According to Mendel, ______________ segregate and _______________ assort independently.
Solution
According to the law of segregation, during gamete formation, the alleles for each gene segregate from each other so that each gamete carries only one allele for each gene. According to the law of independent assortment, genes for different traits can assort or segregate independently during the formation of gametes.
Question 79 1.33 / -0
Stroke could be prevented/treated with
Solution
Blood thinners do not actually thin blood, but they can stop blood clots from forming or growing larger. Hence, they prevent cardiovascular accidents, which can cause stroke.
Question 80 1.33 / -0
In a large isolated population, alleles p and q at a locus are at Hardy-Weinberg equilibrium. The frequencies are p = 0.6 and q = 0.4. The proportion of the heterogeneous genotype in the population is
Solution
In Hardy-Weinberg equation (p² + 2pq + q² = 1), p2 represents the frequency of the homozygous genotype AA, q2 represents the frequency of the homozygous genotype aa and 2pq represents the frequency of the heterozygous genotype Aa.
Question 81 1.33 / -0
The lengths of the sides and the diagonal of an isosceles trapezium form a two-element set {a, b}. If a > b, then a/b equals
Solution
=
-------(i)
=
=
------(ii)
From (i) & (ii):
=
Let
= x
3 - 2x - 1 = 0
= (x + 1)(x
2 - x - 1) = 0
= x = -1 or x =
=
as x cannot be negative
Therefore, x =
Question 82 1.33 / -0
Consider all the natural numbers whose decimal expansion has only the even digits 0, 2, 4, 6, 8. Suppose these are arranged in an increasing order. If an denotes the n-th number in this sequence, then limn ➝ ∞ (log an log n) equals
Question 83 1.33 / -0
The sum of all x ∈ [0, π], which satisfy the equation sin x +
cos x = sin
2 (x +
) is
Solution
sin x +
cos x = sin
2 sin x +
cos x =
sin x +
cos x =
+
sin 2x
2 sin x + cos x = 1 + sin 2x
2 sin x + cos x = 1 + 2 sin x cos x
2 sin x(1 - cos x) - 1(1 - cos x) = 0
(1 - cos x)(2 sin x - 1) = 0
cos x = 1 or sin x =
x = 0 and x =
,
Sum of roots = 0 +
+
=
Question 84 1.33 / -0
A man tosses a coin 10 times, scoring 1 point for each heads and 2 points for each tails. Let P(K) be the probability of scoring at least K points. The largest value of K such that P(K) >
is
Solution
P(K) = P(at least K points)
x
1 + x
2 + x
3 + ...... + x
10 = k
Coefficient x
k in (x
1 + x
2 )
10 = x
10 (1 + x)
10 Coefficient x
k in (1 + x)
10 =
10 C
k - 10 k
10
Now, P(K) >
C
0 +
10 C
1 + .......... +
10 C
K-10 > 29
1 + 10 + 45 + 120 + 210 + 252 + ..... > 512
C
0 . +
10 C
1 +
10 C
2 +
10 C
3 +
10 C
4 +
10 C
5 > 512
K - 10 = 5
K = 15
Question 85 1.33 / -0
The following figure shows the graph of a differentiable function y = f(x) on the interval [a, b] (not containing 0)
Let g(X) = f(x)/x, which of the following is a possible graph of y = g(x)?
Question 86 1.33 / -0
Let ABC be a triangle and P be a point inside ABC such that
. The ratio of the area of triangle ABC to that of APC is
Question 87 1.33 / -0
The smallest possible positive slope of a line whose y-intercept is 5 and which has a common point with the ellipse 9x2 + 16y2 = 144 is
Question 88 1.33 / -0
The maximum possible value of x2 + y2 - 4x - 6y, where x and y are real and are subject to the condition |x + y| + |x - y| = 4 is
Solution
|x + y| + |x - y| = 4 represents a square.
x
2 + y
2 - 4x - 6y = (x - 2)
2 + (y - 3)
2 - 13
A point on square ABCD which is at maximum distance from (2, 3) is C (-2, -2).
So, the required maximum value is 28.
Question 89 1.33 / -0
In a triangle ABC, let G denote its centroid and let M and N be points in the interiors of the segments AB and AC, respectively such that M, G and N are collinear. If r denotes the ratio of the area of triangle AMN to the area of ABC, then
Question 90 1.33 / -0
A box contains coupons labelled 1, 2, ..., 100. Five coupons are picked at random, one after another without replacement. Let the numbers on the coupons be x1 , x2 , ..., x5 . What is the probability that x1 > x2 > x3 and x3 < x4 < x5 ?
Solution
Suppose 1, 2, 3, 4 and 5 are the selected coupons.
Then, from question, place of 1 is fixed and it is at x
3 .
And, total number of arrangements of 5 is 2, either x
1 or x
5 .
Then, possible number of arrangements of 2, 3 and 4 are
Therefore, probability of selecting coupons =
=
Question 91 1.33 / -0
For what value of the resistor X, will the equivalent resistance of the two circuits shown be the same?
Solution
When we compare the figures, we see that
x =
Since resistance cannot be negative, so
x = 2R
Question 92 1.33 / -0
Two small blocks slide without losing contact with the surface along two frictionless tracks 1 and 2, starting at the same initial speed v. Track 1 is perfectly horizontal, while track 2 has a dip in the middle, as shown.
Which block reaches the finish line first?
[Hint: Use velocity time graph to solve]
Solution
Block on track 2 reaches the finish line first.
Time taken in covering the regions 1 and 5 will be the same for both blocks. Due to gravity, V increases in region 2 for track 2 and it will cover the region 3 with increased speed. in region 4 speed will decrease to V again it will travel the region 5 with speed V.
So now we can understand that block 2 reaches the finish line first.
Question 93 1.33 / -0
A stream of photons having energy 3 eV each impinges on a potassium surface. The work function of potassium is 2.3 eV. The emerging photo-electrons are slowed down by a copper plate placed 5 mm away. If the potential difference between the two metal plates is 1 V, then the maximum distance the electrons can move away from the potassium surface before being turned back is
Solution
Kmax = 3 - 2.3 = 0.7 eV
Question 94 1.33 / -0
Two blocks (1 and 2) of equal mass m are connected by an ideal string (see figure below) over a frictionless pulley. The blocks are attached to the ground by springs having spring constants k
1 and k
2 ,
such that k
1 > k
2 .
Initially, both the springs are unstretched. The block 1 is slowly pulled down a distance x and released. Just after the release, the possible value of the magnitudes of the acceleration of the blocks a
1 and a
2 can be:
Solution
Case-1: If T is not zero T + k
1 x - mg = ma...(1)
k
2 x + mg - T = ma...(2)
From above,
T + k
1 x - mg = k
2 x + mg - T
2T = (k
2 - k
1 )x + 2mg
T =
a =
Case-2: If T is 0, then a
1 =
a
2 =
Question 95 1.33 / -0
Consider the infinite ladder circuit shown below:
For what angular frequency 'ω' will the circuit behave like a pure inductance?
Solution
Let the equivalent impedance of the circuit be Z.
So, Z =
Now,
Therefore, Z =
On solving, we get
Z =
For Z to be purely inductive,
or
Question 96 1.33 / -0
Two identical particles of mass m and charge q are shot at each other from a very great distance with an initial speed v. The distance of closest approach of these charges is
Solution
Using conservation of energy:
At closest approach, they will come to rest for a moment and Initial Kinetic energy = Final potential energy.
Question 97 1.33 / -0
One mole of an ideal gas at initial temperature T undergoes a quasi-static process during which the volume V is doubled. During the process, the internal energy U obeys the equation U = aV3 , where a is a constant. The work done during this process is
Solution
n = 1 mole
Internal energy U = aV
3 ⇒
.....................(1)
and
PV = nRT .............................................(2)
Now, from eq. (1) and (2)
P =
and we also know that
W =
..........................(4)
Now from equations 2, 3 and 4
Question 98 1.33 / -0
Two batteries V
1 and V
2 are connected to three resistors as shown below. If V
1 = 2 V and V
2 = 0 V, then the current I = 3 mA. If V
1 = 0 V and V
2 = 4 V, current I = 4 mA. Now, if V
1 = 10 V and V
2 = 10 V, then the current I will be
Solution
We know that, V
eq =
;
and R
eq I =
Now, V
eq is changing and R
eq and R remains same for given cases.
V
eq =
[V
1 = 2, V
2 = 0]
V
eq =
For case 2 V
eq =
[V
1 = 0, V
2 = 4]
Now, from both the cases,
...............................................(1)
For final case V
eq =
From equation (1), we get
I' = 25 mA
Question 99 1.33 / -0
A singly ionised helium atom in an excited state (n = 4) emits a photon of energy 2.6 eV. Given that the ground state energy of hydrogen atom is –13.6 eV, the energy (Et ) and quantum number (n) of the resulting state are respectively,
Question 100 1.33 / -0
A block of mass m slides from rest at a height H on a frictionless inclined plane as shown in the figure. It travels a distance d across a rough horizontal surface with coefficient of kinetic friction μ and compresses a spring k by a distance x before coming to rest momentarily. Then, the spring extends and the block travels back, attaining a final height h. Then,
Solution
With the force, balance
Question 101 1.33 / -0
Consider the following electrochemical cell:
Zn(s) + 2Ag
+ (0.04 M)
Zn
2+ (0.28 M) + 2Ag(s)
If E°
cell = 2.57 V, then e.m.f of the cell at 298 K is
Solution
Zn + 2Ag
+ Zn
2+ + 2Ag
(0.04M) (0.28M)
According to Nernst equation:
E
cell = E
o -
log
E
cell = 2.57 -
log
[∵ At 298 K, [Ag] = [Zn] = 1]
Here, n = 2
By solving the equation, we get
E
cell 2.50 V
Question 102 1.33 / -0
234 Th90 gets converted to 206 Pb82 through a series of radioactive decay processes. The number of alpha and beta particles lost in this transformation, respectively, are
Solution
By comparing mass numbers,
234 = 206 + 4x + 0y
x = 7
By comparing atomic numbers,
90 = 82 + 2x - 1y
y = 82 + 2(7) - 90
y = 6
Hence, number of alpha particles (x) = 7
And, number of beta particles (y) = 6
Question 103 1.33 / -0
Phenol on treatment with dil. HNO3 gives two products P and Q. P is steam volatile but Q is not. P and Q are respectively
Solution
Phenol on treatment with dil. HNO
3 gives
o- nitrophenol and
p- nitrophenol.
o- nitrophenol ('P') has intramolecular H-bonding. Hence, it is steam volatile, whereas "Q" is
p- nitrophenol which has intermolecular H-bonding.
Question 104 1.33 / -0
Three moles of an ideal gas expand reversibly under isothermal conditions from 2 L to 20 L at 300 K. The amount of heat-change (in kJ/mol) in the process is
Solution
⇒ W = -3R x 300 ln 10
⇒
= -17.2 kJ/mol
q = -W = 17.2 kJ/mol (For an isothermal process, ΔE = 0)
Question 105 1.33 / -0
For a reaction A
B, ΔH° = 7.5 mol
-1 and ΔS° = 2.5 J mol
-1 . The value of ΔG° and the temperature at which the reaction reaches equilibrium are respectively
Solution
The change in Gibbs free energy is given as:
At equilibrium,
Therefore,
Question 106 1.33 / -0
The Crystal Field Stabilisation Energy (CPSE) and the spin-only magnetic moment in Bohr Magneton (BM) for the complex K3 [Fe(CN)6 ] respectively are
Solution
Let the oxidation state of the central metal atom (Fe) be 'x' in complex K
3 [Fe(CN)
6 ].
Therefore, 3(+1) + x + 6(-1) = 0
x = +3
Since there are six ligands around iron, it is an octahedral complex with a splitting energy of +
Δ
o and −
Δ
o from the free-ion energy of the metal. As CN
- is a strong field ligand, it forms a low spin complex as shown below:
And crystal field splitting energy is:
CFSE = (2 + 2 + 1) ×
= -2
The spin-only magnetic moment is:
Question 107 1.33 / -0
The electron in hydrogen atom is in the first bohr orbit (n = 1). The ratio of transition energies, E(n = 1
n = 3) to E(n = 1
n = 2) is
Question 108 1.33 / -0
The crystal field stabilisation energies(CFSE) of high spin and low spin d6 metal complexes in terms of Δo , respectively are
Question 109 1.33 / -0
In 108 g of water, 18 g of a non-volatile compound is dissolved. At 100°C, the vapour pressure of the solution is 750 mm Hg. Assuming that the compound does not undergo association or dissociation, the molar mass of the compound in g mol- 1 is
Question 110 1.33 / -0
The amount of Na2 S2 O3 .5H2 O required to completely reduce 100 mL of 0.25 N iodine solution is
Solution
N1 V1 = N2 V2
Question 111 1.33 / -0
Gregor Mendel showed that unit factors exist in pairs and exhibit a dominant-recessive relationship. These unit factors, in modern terminology, are called
Solution
Each of the two alternative forms of a gene that arise by mutation and are found at the same place on a chromosome are called alleles. These exist in pairs and exhibit a dominant-recessive relationship.
Question 112 1.33 / -0
Greatest proportion of photosynthesis in the world is carried out by
Solution
It is estimated that marine plants produce between 70 and 80 percent of the oxygen in the atmosphere. Nearly all marine plants are single-celled, photosynthetic algae.
Question 113 1.33 / -0
In cattle, the coat colours red and white are two dominant traits which express equally in F1 to produce roan (red and white colours in equal proportion). If F1 progeny are self-bred, the resulting progeny in F2 will have phenotypic ratio (red : roan : white)
Solution
Let the red colour be represented by R and white by W (as both are dominant). Parental generation: RR X WWF1 generation will give roan colour ---RW
Question 114 1.33 / -0
The following sequence contains the open reading frame of polypeptide. How many amino acids will the polypeptide consist of?
Solution
An ORF is a continuous stretch of codons that contains a start codon (usually AUG) and a stop codon (usually UAA, UAG or UGA). An ATG codon within the ORF (not necessarily the first) may indicate where translation starts.
Question 115 1.33 / -0
The amino acid sequences of a bacterial protein and a human protein carrying out similar functions are found to be 60% identical. However, the DNA sequences of the genes coding for these proteins are only 45% identical. This is possible because
Solution
The condition given can be explained by Wobble hypothesis. The Wobble Hypothesis explains why multiple codons can code for a single amino acid. One tRNA molecule (with one amino acid attached) can recognise and bind to more than one codon, due to the less-precise base pairs that can arise between the 3rd base of the codon and the base at the 1st position on the anti-codon.
Question 116 1.33 / -0
Which of the following graphs accurately represents the insulin levels (Y-axis) in a body as a function of time (X-axis) after eating sugar and bread/roti?
Solution
Sucrose is the table sugar made up of glucose and fructose. It is instantly broken down and causes insulin to be released in a short time, whereas roti is a complex carbohydrate which needs digestion. This increases the insulin releasing time.
Question 117 1.33 / -0
Although blood flows through large arteries at high pressure, when the blood reaches small capillaries the pressure decreases because
Solution
The total cross-sectional area of capillaries arising from an artery is much greater than that of the artery. This leads to lowering of the blood pressure when it reaches the capillary network.
Question 118 1.33 / -0
Nocturnal animals have retinas that contain
Solution
Nocturnal animals have retinas that contain a high percentage of rods to increase sensitivity to low light conditions.
Question 119 1.33 / -0
From an early amphibian embryo, the cells that would give rise to skin in adults were transplanted into the developing brain region of another embryo. The transplanted cells developed into brain tissue in the recipient embryo. What do you infer from this experiment?
Solution
Differentiation of cells in the embryo is brought about by both internal cellular factors as well as extracellular factors, that act on the cell from the outside.
Question 120 1.33 / -0
In some species, individuals forego reproduction and help bring up another individual's offspring. Such altruistic behaviour cannot be explained by which of the following?
Solution
Individuals are predicted to behave more altruistically and less competitively toward their relatives, because they share a relatively high proportion of their genes (e.g., one-half for siblings and one-eighth for cousins). Consequently, by helping a relative reproduce, an individual passes its genes to the next generation, increasing their Darwinian fitness. Altruistic behaviour toward relatives may at some later time lead to increased competition between relatives, reducing or even completely removing the net selective advantage of altruism. Hence, option (4) is not justified.
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