Question 1 1.33 / -0
The positive integer k for which
is a maximum is
Solution
T
k =
and T
k - 1 =
till K = 10
T
k > T
k - 1 Let K = 11
T
11 < T
10 Hence, T
k is maximum at k = 10.
Question 2 1.33 / -0
Let the line 2x + 3y = 18 intersect the y-axis at B. Suppose C (. ≠ B) with coordinates (a, b) is a point on the line such that PB = PC, where P = (10, 10). Then, 8a + 2b equals
Solution
PB = PC
---- (i)
Also, (a, b) lies on the line: 2x + 3y = 18
Then, 2a + 3b = 18
a = 9 -
Using equation (i):
13b
2 - 68b - 60 = 0
13b
2 - 78b + 10b - 60 = 0
13b(b - 6) + 10(b - 6) = 0
(13b + 10) (b - 6) = 0
Then, b = 6 or b =
Therefore, when b = 6. then a = 9 -
= 0 (Not possible)
When b =
, then a = 9 +
= 9 +
=
Therefore, 8a + 2b = 8
=
Question 3 1.33 / -0
The shortest distance from (0, 3) to the parabola y2 - 4x is
Solution
Equation of parabola: y
2 - 4x = 0
(2y)y' = 4
y' =
Slope of any tangent to the parabola =
and, slope of normal on the parabola =
So, equation of normal on the point (t
2 , 2t):
y - 2t = - t(x - t
2 )
tx + y - t
3 - 2t = 0
Since we know that perpendicular distance is the shortest distance between two points, then from question:
Point (0,3) lie on the normal
3 - t
3 - 2t = 0
t
3 + 2t - 3 = 0
(t - 1)(t
2 + t + 3) = 0
t
2 + t + 3 > 0
Hence, t = 1
Point on parabola (t
2 , 2t)
(1, 2)
Hence, distance of point (0, 3) from parabola =
=
Question 4 1.33 / -0
A vector which bisects the angle between
and
is
Solution
and
and
Therefore, the vector which bisects the angle is 13
.
Question 5 1.33 / -0
In the real number system, the equation
has
Question 6 1.33 / -0
Let f(x) =
, then
Solution
f(x) =
=
= cot a
Question 7 1.33 / -0
The value
is
Solution
I
1 =
I
1 =
I
1 =
Here I
1 =
=
Question 8 1.33 / -0
Let H be the orthocentre of an acute-angled triangle ABC and O be its circumcentre. Then,
Question 9 1.33 / -0
Suppose the sides of a triangle form a geometric progression with common ratio r. Then, r lies in the interval of:
Solution
From property of triangle:
a + ar > ar
2 r
2 - r - 1<0
r
---(1)
ar
2 + ar > a
r
2 + r - 1 > 0
r
R -
---(2)
ar
2 + a > ar;
r
2 - r + 1 > 0 always true
Solving (1) and (2),
Question 10 1.33 / -0
Let P be an arbitrary point on the ellipse
= 1, a > b > 0. Suppose F
1 and F
2 are the foci of the ellipse. The locus of the centroid of the triangle PF
1 F
2 as P moves on the ellipse is
Question 11 1.33 / -0
The figure shown below is the graph of the derivative of some function y = f'(x):
Then,
Solution
f'(a) = f'(b) = f'(c) = 0- ) < 0 f'(a+ ) > 0- ) < 0 f'(c+ ) > 0] minima at a and c- ) > 0 f'(b+ ) < 0] maximum at b
Question 12 1.33 / -0
Let
and
be vectors in R
3 and
be a unit vector in the xy-plane. Then, the maximum possible value of |
| is:
Solution
=
Let
s.t. a
2 + b
2 = 1
a =
; b =
.
=
=
Max. value =
Question 13 1.33 / -0
Suppose n is a natural number such that
= 18
, where i is the square root of -1. Then, n is
Question 14 1.33 / -0
In a rectangle ABCD, the coordinates of A and B are (1, 2) and (3, 6), respectively and some diameter of the circumscribing circle of ABCD has equation 2x – y + 4 = 0. Then, the area of rectangle is
Solution
Slope of AB =
= 2
Slope of BC =
Distance between two lines 2x - y + 4 = 0 and 2x - y = 0 is equal to
.
Area of rectangle =
Question 15 1.33 / -0
If f(x) = (2011 + x)
n , where x is a real variable and n is a positive integer, then the value of f(0) + f'(0) +
+ ........ +
is
Question 16 1.33 / -0
Two players play the following game: A writes 3, 5 and 6 on three different cards, B writes 8, 9 and 10 on three different cards. Both draw randomly two cards from their collections. Then, A computes the product of two numbers he/she has drawn and B computes the sum of two numbers he/she has drawn. The player getting the larger number wins. What is the probability that A wins?
Solution
For A to win, A can draw either 3 and 6 or 5 and 6. If A draws 3 and 6, then B can draw only 8 and 9.
Probability =
If A draws 5 and 6, then B can draw any two.
Probability =
Total probability of winning of A =
Question 17 1.33 / -0
Let n be a positive integer such that2 log2 log2 log2 log2 (n) < 0 < log2 log2 log2 log2 (n).
Solution
log2 log2 log2 log2 log2 (n) < 0 < log2 log2 log2 log2 (n)16
Question 18 1.33 / -0
Let ABC be an acute-angled triangle and let D be the midpoint of BC. If AB = AD, then
equals
Solution
Using M-N rule:
3 cotB = cotC
Question 19 1.33 / -0
For real X with -10 ≤ X ≤ 10, define
, where for a real number r, we denote by [r] the largest integer less than or equal to r. The number of points of discontinuity of f in the interval (-10, 10) is
Solution
Let r be an integer in (-10, 10).
Now, LHL =
=
=
=
---(1)
And
=
=
---(2)
f(r) =
=
---(3)
From (1), (2) and (3), f(x) is continuous at all integers.
Question 20 1.33 / -0
Let
be unit vectors in the xy-plane, one each in the interior of the four quadrants. Which of the following statements is necessarily true?
Solution
In this, options 2, 3 and 4 are not possible.
Question 21 1.33 / -0
The molecules of air in the room that you are sitting in are all experiencing the force of gravity trending to bring them down. The molecules are also frequently and randomly undergoing collisions, which tend to oppose the effect of fall under gravity. The density of air is nearly uniform throughout the room because
Solution
The potential energy due to gravity (mgh) is much lesser than thermal energy (kT) of particles. The force working on particles f = mg is negligible with respect to forces due to collisions. So, density of air remains nearly unchanged.
Question 22 1.33 / -0
A charge Q is spread non-uniformly on the surface of a hollow sphere of radius R such that the charge density is given by
, where
is the usual polar angle. The potential at the centre of the sphere is
Solution
We know that potential inside a hollow sphere is constant and is given by
.
Sphere is hollow. So, potential inside the sphere will be the same as that on the surface.
Question 23 1.33 / -0
The moment of inertia of a solid disc made of a thin metal of radius R and mass M about one of its diameters is given by
. What will be the moment of inertia about this axis if the disc is folded in half about the diameter?
Solution
We know, I
As there is no change in mass, so there is no change in moment of inertia.
Question 24 1.33 / -0
A source of frequency f is emitting sound waves. If temperature of the medium increases, then
Solution
For a wave equation
wavelength
and velocity v =
As temperature increases, velocity also increase (since v =
)
So, v increases, k decreases and wavelength increases.
Question 25 1.33 / -0
The capacitor of capacitance C in the circuit shown is fully charged initially. Resistance is R.
After the switch S is closed, the time taken to reduce the stored energy in the capacitor to half its initial value is:
Solution
As we know, charge Q is given by:
Q = Q
o e
-t/RC Now,
t = RC
Question 26 1.33 / -0
A loop carrying current I has the shape of a regular polygon of n sides. If R is the distance from the centre to any vertex, then the magnitude of the magnetic induction vector B at the centre of the loop is:
Solution
B
net = n × B
1 Now, calculation for B
1: B
1 =
=
B
1 =
B
net =
Question 27 1.33 / -0
A body is executing simple harmonic motion of amplitude a and period T about the equilibrium position x = 0. Large numbers of snapshots are taken at random of this body in motion. The probability of the body being found in a very small interval x to x + |dx| is highest at
Solution
The probability of being found is maximum where speed is minimum. Because the availability of body will be maximum at extreme positions for the same interval where speed is minimum.
Question 28 1.33 / -0
A particle released from rest is falling through a thick fluid under gravity. The fluid exerts a resistive force on the particle proportional to the square of its speed. Which one of the following graphs best depicts the variation of its speed v with time t?
Solution
Suppose the ball is moving with speed v at anytime t.
So,
Hence, m
We can say that speed of the particle will increase and become constant when mg = kv
2 .
⇒
Question 29 1.33 / -0
Two masses m1 and m2 connected by a spring of spring constant k rest on a frictionless surface. If the masses are pulled apart and let go, the time period of oscillation is
Solution
Let the masses be slightly displayed by x
1 and x
2 from this equilibrium position in opposite direction so net stretch in spring is x = x
1 + x
2 . Because of this, a restoring force kx will act on each mass and therefore, equation for m
1 and m
2 will be
And
Therefore,
Replacing values of
and
from acceleration equations, we get
Put
, where m is the reduced mass of the system
Therefore,
or
and
Therefore,
Question 30 1.33 / -0
A charge +q is placed somewhere inside the cavity of a thick conducting spherical shell of inner radius R1 and outer radius R2 . If a charge Q is placed at a distance r > R2 from the centre of the shell, then the electric field in the hollow cavity
Solution
For a conductor, electric field inside its cavity is only due to inside charge and not due to outside charge.
Question 31 1.33 / -0
A book is resting on shelf that is undergoing vertical simple harmonic oscillations with an amplitude of 2.5 cm. What is the minimum frequency of oscillation of the shelf from which the book will lose contact with the shelf? (Assume that g = 10 m/s2 )
Solution
The book will lose contact with the shelf when a = g.
Now, |a| =
Therefore, g =
(
Amplitude)
Also,
Therefore, f =
Replacing g = 10 m/s
2 and A = 2.5 × 10
-2 m, we get f = 3.18 Hz
Question 32 1.33 / -0
In Young's double slit experiment, the distance between the two slits is 0.1 mm, the distance between the slits and the screen is 1 m and the wavelength of the light used is 600 nm. The intensity at a point on the screen is 75% of the maximum intensity. What is the smallest distance of this point from the central fringe?
Solution
d = 0.1 mm, D = 1 m,
= 600 nm
I
p = 75% of maximum or I
p = 3I
0 Here, I
0 is the intensity of a single wave.
Now, I
p = 3I
0 =
Therefore,
=
Also,
Now,
Hence,
Or y = 1.0 mm
Question 33 1.33 / -0
Two plates, each of the mass m, are connected by a mass-less spring as shown.
A weight W is put on the upper plate, which compresses the spring further. When W is removed, the entire assembly jumps up. The minimum weight W needed for the assembly to jump up when the weight is removed is just more than
Solution
For lower block's +ve lift,
Using Work Energy theorem, we get
Therefore h =
Before lifting the weight, Resultant force on the upper block = 0
W + mg = kh
W + mg = 3mg
W = 2mg
Question 34 1.33 / -0
Jet aircrafts fly at altitudes above 30,000 ft, where the air is very cold at -40°C and the pressure is 0.28 atm. The cabin is maintained at 1 atm pressure by means of a compressor, which exchanges air from outside adiabatically. In order to have a comfortable cabin temperature of 25°C, we will require in addition
Question 35 1.33 / -0
For a domestic AC supply of 220 V at 50 cycles per second, the potential difference between the terminals of a two pin electric outlet in a room is given by
Solution
R.M.S value = 220 V
Peak value V
0 = 220
ω = 2πn
= 2π × 50 = 100π
We know that,
V = V
o cos ωt
V(t) = 220
cos (100 πt)
Question 36 1.33 / -0
Due to transitions among its first three energy levels, hydrogenic atom emits radiation at three discrete wavelengths λ1 , λ2 and λ3 , where λ1 < λ2 < λ3 . Then,
Question 37 1.33 / -0
Consider the circuit shown in the figure below:
All the resistors are identical. The charge stored in the capacitor, once it is fully charged, is
Question 38 1.33 / -0
An engine moving away from a vertical cliff blows a horn at a frequency f. Its speed is 0.5% of the speed of sound in air. The frequency of the reflected sound received at the engine is
Question 39 1.33 / -0
The bulk modulus of a gas is defined as B = –VdP/dV. For an adiabatic process, the variation of B is proportional to Pn . For an idea gas, n is
Solution
As we know, for adiabatic process,
On differentiating the equation w.r.t. V, we get
= B (According to the question)
So, B is proportional to p
1 .
Hence, n = 1
Question 40 1.33 / -0
A plank is resting on a horizontal ground in the northern hemisphere of the Earth at 45° latitude. Let the angular speed of the Earth be
and its radius r
e . The magnitude of the frictional force on the plank will be:
Solution
⇒ F
r = mω
2 r cos 45°;
where, r = r
e cos 45°
⇒
Question 41 1.33 / -0
Assuring ideal behaviour, the ratio of kinetic energies of 3 g of H2 and 4 g of O2 at any temperature is
Solution
Number of moles of H
2 =
Number of moles of O
2 =
Kinetic energy of 'n' moles of gas =
So,
= 12 : 1
Question 42 1.33 / -0
The ion that is isoelectronic with CO is
Solution
Isoelectronic species are those species that have the same number of electrons.- has 6 + 7 + 1 = 14 electrons- are isoelectronic species.
Question 43 1.33 / -0
Oxalic acid, when treated with potassium permanganate in the presence of an acid, produces
Solution
Reaction between oxalic acid and potassium permanganate is a redox reaction and occurs in the presence of acid (a source of H
+ ) and heat. So, it is an endothermic reaction. Potassium permanganate and the acid release oxygen, which combines with oxalic acid to form carbon dioxide and water. The chemical reaction involved in this process is given below:
Question 44 1.33 / -0
The order of basicity of the following compounds is:
Solution
The electronegativity of an sp
2 hybridised N-atom is greater than that of an sp
3 hybrdised N-atom and basic strength is inversely proportional to the electronegativity.
(I)
N-atom is sp
2 hybridised and the lone pair of N takes part in resonance with conjugated double bond, so it is not easily available on N for the protonation. Hence, it has the least basic strength.
(II)
N-atom is sp
2 hybridised and the lone pair is not taking part in resonance, so it is easily available for the protonation.
(III)
N-atom is sp
3 hybridised, but the availability of lone pair on N decreases due to the -I effect of the oxygen atom that has a higher electronegativity.
(IV)
N-atom is sp
3 hybridised, and there is no extra effect, so availability of lone pair on N increases.
Hence, the above order is justified.
Question 45 1.33 / -0
The molecule with non-zero dipole moment is:
Solution
BCl
3 , BeCl
2 and CCl
4 have zero dipole moment due to symmetric structures, whereas shape of NCl
3 is pyramidal due to the presence of lone pair.
Thus, NCl
3 has non-zero dipole moment.
Question 46 1.33 / -0
The major final product in the following reaction is:
CH3CH2CN
Solution
Nitriles are moderate electrophiles, which react with nucleophilic Grignard reagent to give an imine salt as an intermediate and acidic hydrolysis of this imine salt leads to give respective ketone.
Question 47 1.33 / -0
The oxidation state of cobalt in the following molecule is:
Solution
In metal carbonyls, the oxidation state of metal is equal to zero because carbonyls are neutral ligands.
Question 48 1.33 / -0
The C-O bond length in CO, CO2 and CO3 2- follows the order:
Solution
The given chemical species are:
As we know,
Bond length
Hence, the increasing order of bond lengths is
Question 49 1.33 / -0
Among the following, the π-acid ligand is
Solution
The π-acid ligand donates electron density into a metal d-orbital from a π-symmetry bonding orbital between the carbon atoms. Out of the given, CN
- accepts electrons from metal ion in its vacant
-ABMO.
Question 50 1.33 / -0
Friedel-Crafts Acylation is
Solution
Friedel-Crafts Acylation is an organic reaction which is used to convert an aryl compound and an acyl halide or anhydride to an aryl ketone by using a Lewis acid catalyst (such as AlCl3 ).
Question 51 1.33 / -0
Increasing the temperature increases the rate of reaction but does not increase the
Solution
As the temperature increases, the molecules move faster and, therefore, collide more frequently. The molecules also carry more kinetic energy. Thus, the proportion of collisions that can overcome the activation energy for the reaction increases with temperature.
Question 52 1.33 / -0
The enantiomeric pair among the following four structures is:
Solution
Enantiomers are chiral molecules that are mirror images of one another and are non-superimposable. There is no chiral centre in molecules II and III, while in case of I and IV, chiral centre is present. Thus, molecules I and IV are mirror images and are not superimposable on each other.
Question 53 1.33 / -0
Extraction of silver is achieved by initial complexation of the ore (argentite) with 'X', followed by reduction with 'Y'. X and Y, respectively are
Solution
The extraction of silver from the ore-argentite (Ag
2 S) is known as cyanide process. In this process, the ore is crushed, concentrated and treated with sodium cyanide (X) solution. This reaction forms sodium argento cyanide Na[Ag(CN)
2 ].
Ag
2 S + 4NaCN ⇌ 2Na[Ag(CN)
2 ] + Na
2 S
The solution of sodium argento cyanide combines with zinc dust (Y) and forms sodium tetra cyanozicate and precipitated silver.
Zn + 2Na[Ag(CN)
2 ]
Na
2 [Zn(CN)
4 ] + 2Ag
Question 54 1.33 / -0
Two possible stereoisomers for
are which of the following?
Solution
The given compound is chiral and the possible stereoisomers are enatiomer to each other are as shown below.
Question 55 1.33 / -0
The value of the limiting molar conductivity (∧°) for NaCl, HCl and NaOAc are 126.4, 425.9 and 91.0 S cm2 mol-1 , respectively. For HOAc, ∧° in S cm2 mol-1 is
Solution
Kohlrausch law of independent migration of ions is2 mol-1
Question 56 1.33 / -0
In a zero-order reaction, if the initial concentration of the reactant is doubled, the time required for half the reactant to be consumed
Solution
For Zero order reaction.
, when t = t
1/2 , [A
t ] = [A
o ]/2
Initial concentration
So, if the initial concentration of the reactant is doubled, the time required for half the reactant also doubled.
Question 57 1.33 / -0
For a zero order reaction with rate constant k, the slope of the plot of reactant concentration against time is
Solution
For a zero order reaction:
[R
t ] = [R
o ] - kt;
Slope = -k, and intercept [R
o ]
Question 58 1.33 / -0
The reaction of methyl ketone with Cl2 /excess OH- gives the following major product:
Question 59 1.33 / -0
D-Glucose upon treatment with bromine water gives
Solution
On reaction with mild oxidising agent like bromine water, D-Glucose is oxidised to gluconic acid which is monocarboxylic acid with six carbon atoms. It is a confirmatory test, that Glucose has only one carbonyl group.
Question 60 1.33 / -0
The number of ions produced in water by dissolution of the complex having the empirical formula COCl3 .4NH3 is
Solution
The cordination number of Co is 6. Two of the chloride ions are bound to cobalt in [Co(NH
3 )
4 Cl
2 ]Cl to balance both primary and secondary valencies.
Thus, only two ions are formed when this compound dissolves in water as shown below:
Question 61 1.33 / -0
In the organism muscle, oxygen is carried by
Solution
Myoglobin is an iron and oxygen binding protein found in the muscle tissue of vertebrates and in almost all mammals.
Question 62 1.33 / -0
The chromosomal attachment of the spindle microtubule is
Solution
Centromere is the region of a chromosome to which the microtubules of the spindle attach, via the kinetochore, during cell division.
Question 63 1.33 / -0
ELISA, the standard screening test for HIV, detects
Solution
Enzyme-linked immunosorbent assay, also called ELISA, is a test that detects and measures antibodies in blood. This test can be used to determine if one has antibodies related to a certain infectious condition.
Question 64 1.33 / -0
The abnormal development of which of the following lymphoid organs results in the most severe immunodeficiency?
Solution
The thymus is a specialised primary lymphoid organ of the immune system. Within the thymus, T-cells or T-lymphocytes mature. T-cells are critical to the adaptive immune system, where the body adapts specifically to foreign invaders.
Question 65 1.33 / -0
Which of the following classes of immunoglobulins can trigger the complement cascade?
Solution
IgM antibodies are the best because they have more antigen-binding sites. They can achieve binding of two adjacent antigens by single IgM molecule. Hence, complement activation will be maximum in this case.
Question 66 1.33 / -0
Some animals excrete uric acid in urine (uricotelic) as it requires very little water. This is an adaptation to conserve water loss. Which animals among the following are most likely to be uricotelic?
Solution
Uricotelic organisms (terrestrial arthropods (including insects), lizards, snakes, and birds) excrete uric acid. Excretion of uric acid requires very less water.
Question 67 1.33 / -0
Which of the following is true for TCA cycle in eukaryotes?
Solution
The citric acid cycle, also known as the tricarboxylic acid (TCA) cycle or the Krebs cycle, is a series of chemical reactions used by all aerobic organisms to release stored energy through the oxidation of acetyl-CoA derived from carbohydrates, fats, and proteins into carbon dioxide and chemical energy in the form of adenosine triphosphate (ATP). It occurs in the mitochondria. The NADH and FADH2 produced during TCA cycle are oxidised in ETC and energy is released.
Question 68 1.33 / -0
In Dengue virus infection, patients often develop haemorrhagic fever due to internal bleeding. This happens due to the reduction of
Solution
In Dengue fever, the number of platelets are highly reduced. Platelet count below 20,000 has higher chances of developing bleeding complication as seen in Dengue haemorrhagic fever.
Question 69 1.33 / -0
In 1670, Robert Boyle conducted an experiment wherein he placed a viper (a poisonous snake) in a chamber and rapidly reduced the pressure in that chamber. Which of the following would be true?
Solution
The first recorded experimental work related to decompression was conducted by Robert Boyle, who subjected experimental animals to reduced ambient pressure by use of a primitive vacuum pump. This led to the formation of gas bubbles in the tissues of the snake.
Question 70 1.33 / -0
Sometimes urea is fed to ruminants to improve their health. It works by
Solution
Urea is included in the animal diet as it helps in the growth of microbiota in the gut. These microbiota help in the breakdown of cellulose.
Question 71 1.33 / -0
Athletes often experience muscle cramps. Which of the following statements is true about muscle cramps?
Solution
Muscle cramps occur due to formation of lactic acid. Under the anaerobic conditions in the muscles (during rigorous exercise), pyruvate is converted into lactic acid in the cytoplasm.
Question 72 1.33 / -0
Restriction endonucleases are enzymes that cleave DNA molecules into smaller fragments. Which type of bond do they act on?
Solution
In DNA and RNA, the phosphodiester bond is the linkage between the 3' carbon atom of one sugar molecule and the 5' carbon atom of another, deoxyribose in DNA and ribose in RNA. Hence, restriction endonucleases break down this bond present in nucleic acids.
Question 73 1.33 / -0
A person has difficulty in breathing at higher altitudes because oxygen
Solution
At a high altitude, the atmospheric pressure is lower than the sea level pressure. Hence, the partial pressure of oxygen is less than that of carbon dioxide. Therefore, a person feels difficulty in breathing at higher altitude.
Question 74 1.33 / -0
A protein with 100 amino acid residues has been translated based on triplet genetic code. Had the genetic code been quadruplet, the gene that codes for the protein would have been
Solution
According to the triplet code, 1 amino acid is coded by a set of three bases called codon.
Question 75 1.33 / -0
A water molecule can be formed from a maximum of _____ hydrogen bonds.
Solution
Water is unique because its oxygen atom has two lone pairs and two hydrogen atoms, meaning that the total number of bonds of a water molecule is up to four.
Question 76 1.33 / -0
The process of cell death involving DNA cleavage in cells is known as
Solution
Apoptosis is a process of programmed cell death that occurs in multicellular organisms. Biochemical events lead to characteristic cell changes (morphology) and death.
Question 77 1.33 / -0
Nucleolus is an organelle that is responsible for the production of
Solution
Nucleolus is the principal site for the formation of ribosomal RNAs. Nucleolus is the centre for the formation of ribosomes.
Question 78 1.33 / -0
Fission yeasts are
Solution
Schizosaccharomyces pombe , also called "fission yeast", is a species of yeast used in traditional brewing and as a model organism in molecular and cell biology. It is a unicellular eukaryote, whose cells are rod-shaped.
Question 79 1.33 / -0
Morphogenetic movements take place predominantly during which of the following embryonic stages?
Solution
Morphogenetic movements occur during gastrulation and serve not only to generate shape but also to ensure that cells receive the right signals at the right time.
Question 80 1.33 / -0
Part of epidermis that keeps out unwanted particles is called
Solution
The hair, called cilia, moves back and forth to help move particles out of our body.
Question 81 1.33 / -0
Let a, b, c be the sides of a triangle. If t denotes the expression (a2 + b2 + c2 )/(ab + bc + ca), then the set of all possible values of t is
Solution
(a - b)
2 |a - b| < c -----(i)
|b - c| < a -----(ii)
|c - a| < b -----(iii)
Squaring and adding:
a
2 + b
2 + c
2 < 2ab + 2bc + 2ca
So, b
Question 82 1.33 / -0
Let A denote the area bounded by the curve y = 1/x and the lines y = 0, x = 1, x = 10, let B = 1 +
+
...... +
, and let C =
+
...... +
. Then
Solution
A =
=
B = 1 + 0.5 + 0.33 + 0.25 + 0.20 + 0.16 + 0.14 + 0.12 + 0.11
2.81
C =
So, C < A < B and
Since 2A < B + C
A - C < B - A
Question 83 1.33 / -0
The coefficient of x
2012 in
is
Solution
Coeff. Of x
2012 = (1 + x)
2 (1 - x
4 )
-1 = (1 + 2x + x
2 )(1 - x
4 )
-1 Coeff. Of x
2012 + 2Coeff of x
2011 + Coeff of x
2010 in the expansion of (1 - x
4 )
-1 x
2011 and x
2010 not possible in (1 - x
4 )
-1 = only coeff. Of x
2012 in the expassion of (1 - x
4 )
-1 1 + 503-1 C
503 = 1
Question 84 1.33 / -0
The value of the integral
, where a > 0, is
Solution
I =
(a > 0) ------------------(1)
I =
I =
----------------------------(2)
(1) + (2)
2I =
2I = 2
I =
I =
=
Question 85 1.33 / -0
Let r be a real number and n ∈ N be such that the polynomial 2x2 + 2x + 1 divides the polynomial (x + 1)n - r. Then (n, r) can be
Solution
2x
2 + 2x + 1 = 0
x satisfies (x + 1)
n - r = 0
- r = 0
- r = 0
= r
= r
RHS = real
LHS = real only when n = multiple of 4
n = 4000
r =
Question 86 1.33 / -0
Let f: R
R be a continuous function satisfying f(x) = x +
f(t) dt, for all x ∈ R. The number of elements in the set S = { x ∈ R ; f(x) = 0 }, is
Solution
f '(x) = 1 + f(x)
log (1 + f(x)) = dx
f(x) = e
x - 1
f(x) = 0
e
x = 1
x = 0 (One solution)
Question 87 1.33 / -0
The minimum value of n for which
< 1.01
Solution
Suppose 1
2 + 3
2 + ...... + (2n - 1)
2 = x and
c =
(4r
2 - 4r + 1) =
(4n
2 - 1)
then
< 1.01
< 2.01
< 2.01
⇒
< 2.01
⇒ 4n + 1 < (2.01)(2n - 1)
⇒ 4n + 1 < 4.02n - 2.01
3.01 < 0.02 n
< n
150.5 < n
Hence minimum value of n = 151
Question 88 1.33 / -0
Let f(x) = x12 – x9 + x4 – x + 1. Which of the following is true?
Solution
f(x) = x
9 (x
3 - 1) + x(x
3 - 1) + 1 is positive for x
1 or x
0.
1 - x + x
4 - x
9 + x
12 is positive for x
(0, 1).
f(x) is always positive.
Hence, f takes only positive value.
Question 89 1.33 / -0
Let a, b, c and d be real numbers such that
for every natural number n. Then, |a| + |b| + |c| + |d| is equal to
Solution
n
4 (12 - 3a) - n
3 (4b + 6a) - n
2 (6c + 6b + 3a) - n(6c + 2b + 12d) = 0
12 - 3a = 0, 4b + 6a = 0, 6c + 6b + 3a = 0 and 6c + 2b + 12d = 0
a = 4, b = -6, c = 4 and d = -1
|a| + |b| + |c| + |d| = 15
Question 90 1.33 / -0
Suppose that the Earth is a sphere of radius 6400 kilometers. What is the height from the Earth's surface from where exactly a fourth of the Earth's surface is visible?
Question 91 1.33 / -0
A plano-convex lens made of material of refractive index
with radius of curvature R is silvered on the curved side. How far away from the lens-mirror must you place a point object so that the image coincides with the object?
Question 92 1.33 / -0
A vehicle is moving with speed v on a curved road of radius r. The coefficient of friction between the vehicle and the road is μ. The angle θ of banking needed is given by
Question 93 1.33 / -0
A small asteroid is orbiting around the sun in a circular orbit of radius r0 with speed V0 . A rocket is launched from the asteroid with speed V = α V0 , where V is the speed relative to the sun. The highest value of α for which the rocket will remain bound to the solar system is (ignoring gravity due to the asteroid and effect of other planets)
Solution
As we know that
....(1)
Mechanical energy of asteroid =
For rocket to remain bounded to solar system, mechanical energy should be less than zero.
.....(2)
From equations (1) and (2) and V from the question, by putting the value of V, we get
Question 94 1.33 / -0
As shown in the figure below, a cube is formed with ten identical resistances R (thick lines) and two shorting wires (dotted lines) along the arms AC and BD.
Resistance between points A and B is
Solution
This figure can be simplified into two Wheatstone bridges parallel as follows:
As we know that in a Wheatstone Bridge, for one bridge R
AB = R.
But there are two in parallel, so R
AB = R/2
Question 95 1.33 / -0
A certain planet completes one rotation about its axis in time T. The weight of an object placed at the equator on the planet's surface is a fraction f (f is close to unity) of its weight recorded at a latitude of 600 . The density of the planet (assumed to be a uniform sphere) is given by
Solution
Also, T =
Or T =
, when V is replaced by
-
f =
Solving, we get
or
=
Also, T =
Therefore,
or
Question 96 1.33 / -0
An ideal gas with heat capacity at constant volume Cv undergoes a quasistatic process described by PVα in a P-V diagram, where α is a constant. The heat capacity of the gas during this process is given by
Solution
Direct formula is to be used.
Question 97 1.33 / -0
The surface of a planet is found to be uniformly charged. When a particle of mass m and no charge is thrown at an angle from the surface of the planet, it has a parabolic trajectory as in projectile motion with horizontal range L. A particle of mass m and charge q, with the same initial conditions has a range L/2. The range of particle of mass m and charge 2q with the same initial conditions is
Solution
For uncharged particle,
----(i)
Range for particle of mass m and charge q,
----(ii)
From (i) and (ii)
⇒ mg = qE
Range of particle of mass m and charge 2q
R =
Question 98 1.33 / -0
At a distance L from a uniformly charged long wire, a charged particle is thrown radially outward with a velocity u in the direction perpendicular to the wire. When the particle reaches a distance 2L from the wire, its speed is found to be
. The magnitude of the velocity, when it is a distance 4L away from the wire, is (ignore gravity)
Solution
Energy conversation at A and B
q
⇒
..............................................(1)
Energy conservation at A and C
⇒
⇒
⇒
Using equation (1),
⇒
⇒
Question 99 1.33 / -0
A small boy is throwing a ball towards a wall 6 m in front of him. He releases the ball at a height of 1.4 m from the ground. The ball bounces from the wall at a height of 3 m, rebounds from the ground and reaches the boy's hand exactly at the point of release. Assuming the two bounces (one from the wall and the other from the ground) to be perfectly elastic, how far ahead of the boy did the ball bounce from the ground?
Question 100 1.33 / -0
A solid sphere rolls without slipping, first horizontally and then, up to a point X at height h on an inclined plane before rolling down, as shown.
The initial horizontal speed of the sphere is
Question 101 1.33 / -0
A cylinder of cooking gas in a household contains 11.6 kg of butane. The thermochemical reaction for the combustion of butane is:
2C
4 H
10 (g) + 13O
2 (g)
8CO
2 (g) + 10H
2 O(l); ΔH = -2658 kJ/mol
If the household needs 15,000 KJ of energy per day, the cooking gas cylinder will last for about
Solution
2C
4 H
10 + 13O
2 8CO
2 + 10H
2 O
ΔH = -2658 kJ/mol
Butane present in cylinder = 11.6 kg = 11,600 g =
mol
As combustion of 1 mol of C
4 H
10 gives 2658 kJ of energy, therefore
Combustion of 11,600/58 mol of C
4 H
10 gives =
kJ = 5,31,600 kJ energy
Energy consumed in 1 day = 15,000 kJ
So, 5,31,600 kJ of energy will be consumed in
Question 102 1.33 / -0
Reactions and their equilibrium constants are given below:
The equilibrium constant, K for the reaction
is
Solution
Equilibrium constant for given equation
K =
.........................(v)
By multiplying the right hand side of equation (i), (ii) and (iii), we get right hand side of equation (v). It means K = K
1 K
2 K
3 .
Question 103 1.33 / -0
In the DNA of E. Coli, the mole ratio of adenine to cytosine is 0.7. If the number of moles of adenine in the DNA is 3,50,000, then the number of moles of guanine is equal to
Question 104 1.33 / -0
The
H for vaporisation of a liquid is 20 kJ/mol. Assuming ideal behaviour, the change in internal energy for the vaporisation of 1 mole of the liquid at 60°C and 1 bar is close to
Solution
20 =
+ 8.314 × 10
-3 × 333
= 20 - 2.76 = 17.23 kJ/mol
Question 105 1.33 / -0
Two isomeric compounds I and II are heated with HBr:
The products obtained are:
Solution
Since phenols do not react with hydrogen halides to give aryl halides, the reaction of isomeric compounds is given as follows:
(1) Alcohols on treatment with HBr undergo a nucleophilic substitution reaction to generate an alkyl halide and water.
(2) Ethers, R-O-Ar, are cleaved by the strong acids HI or HBr in a nucleophilic substitution reaction. The products are the phenol and the appropriate alkyl halide. Depending on the structure of the alkyl group, the reaction can be S
N 1 or S
N 2 .
Question 106 1.33 / -0
Identify the cyclic silicate ion given in the figure below:
Solution
The basic building block, the structural unit of silicate minerals is the SiO4 4- tetrahedron, which can link together by sharing corners resulting in Si-O-Si covalent bonds. The given structure of cyclic silicate ion is [Si6 O18 ]12- , which has 6 tetrahedra in a ring.
Question 107 1.33 / -0
oxidises
(i) oxalate ion in acidic medium at 333 K and
(ii) HCl.
For balanced chemical equations, the ratios [
] in (i) and [
] in (ii), respectively are
Solution
The chemical reaction of
with oxalate ions in acidic medium is
Therefore, the ratio is
The chemical reaction of
with HCl is
∴ ratio is
Question 108 1.33 / -0
Optically active (S) -α – methoxypropionaldehyde on reaction with MeMgX gave a mixture of alcohols. The major diastereomer 'P' on treatment with Mel/K
2 CO
3 gave an optically inactive compound. P is
Question 109 1.33 / -0
The maximum number of isomers that can result from monobromination of 2-methyl-2-pentene with N-bromosuccinimide in boiling CCl4 is
Question 110 1.33 / -0
Aqueous solution of metallic nitrate X reacts with NH4 OH to form Y, which dissolves in excess NH4 OH. The resulting complex is reduced by acetaldehyde to deposit the metal. X and Y, respectively are
Question 111 1.33 / -0
Which of the following statements is true for meiosis?
Solution
Meiosis is a type of cell division that reduces the number of chromosomes in the parent cell by half and produces four gamete cells. In meiosis, DNA replication is followed by two rounds of cell division to produce four daughter cells, each with half the number of chromosomes.
Question 112 1.33 / -0
The effect of consumption of excess protein by normal individual would result in
Solution
If plenty of carbohydrates are consumed, body converts the extra protein to fatty acids and stores them in the adipose tissue.
Question 113 1.33 / -0
Persons suffering from hypertension (high blood pressure) are advised a low-salt diet because
Solution
Too much sodium in the system causes the body to retain (hold onto) water. This puts an extra burden on the heart and blood vessels. In some people, this may lead to or raise high blood pressure.
Question 114 1.33 / -0
Vibrio cholerae causes cholera in humans. Ganga water was once used successfully to combat the infection. The possible reason could be
Solution
Bacteriophage is a virus that can kill bacteria. Cholera is caused by the bacteria called Vibrio cholerae .
Question 115 1.33 / -0
In a diploid organism, there are three different alleles for a particular gene. Out of these three alleles, one is recessive and the other two alleles exhibit co-dominance. How many phenotypes are possible with this set of alleles?
Solution
As two alleles are co-dominant and one recessive, the recessive allele will show a corresponding dominant allele. Hence, a cross between the parental generation will be: ABC X aBC.
Question 116 1.33 / -0
Four species of birds have different egg colours: [1] White with no markings, [2] Pale brown with no markings, [3] Grey-brown with dark streaks and spots, [4] Pale blue with dark blue-green spots. Based on egg colour, which species is most likely to nest in a deep tree hole?
Solution
The eggs of hole-nesting birds are generally white or pale blue so that the parent birds can easily locate them and avoid breaking them. Egg camouflage is less important because the egg is usually well hidden within the nest. Birds that lay their eggs in the open and on the ground rely on their eggs being very well camouflaged, and so usually have brown or speckled eggs.
Question 117 1.33 / -0
Rice has a diploid genome with 2n = 24. If crossing-over is stopped in a rice plant and then selfed seeds are collected, will all the offspring be genetically identical to the parent plant?
Solution
All the plants will have the same genetic makeup as crossing over introduces variation.
Question 118 1.33 / -0
Increasing the number of measurements of an experimental variable will
Solution
Different samples drawn from that same population would in general have different values of the sample mean, so there is a distribution of sampled means (with its own mean and variance).
Question 119 1.33 / -0
Watson and Crick model of DNA is
Solution
Watson-Crick structure of DNA.
Question 120 1.33 / -0
A scientist has cloned an 8 kb fragment of a mouse gene into the Eco RI site of a vector of 6 kb size. The cloned DNA has no other Eco RI site within. Digestion of the cloned DNA is shown below.
Which of the following sets of DNA fragments generated by digestion with both Eco RI and Bam HI as shown in (iii) is from the gene?
Solution
Restriction map as per data is shown above.
Thus, 1 kb, 3 kb and 4 kb are fragments of gene.
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