Question 1 1.33 / -0
Suppose the sequence a1 , a2 , a3 ....... is an arithmetic progression of distinct numbers such that the sequence a1 , a2 , a4 , a8 ........ is a geometric progression. The common ratio of the geometric progression is
Solution
Given: a
1 , a
2 , a
3 , ................ A.P. and a
1 , a
2 , a
4 , a
8 , ......... G.P.
Let common difference of A.P. be d.
a
2 = a
1 + d
a
4 = a
1 + 3d
a
8 = a
1 + 7d
a
1 2 + d
2 + 2a
1 d = a
1 2 + 3a
1 d
d
2 = a
1 d (
)
d = a
1 ......... (i)
Hence,
(Using (i))
r = 2
Question 2 1.33 / -0
Consider an ellipse with foci at (5, 15) and (21, 15). If the x-axis is a tangent to the ellipse, then the length of its major axis equals
Solution
Mid point of PP' =
L = (13, 15)
Therefore, point A will be (13, 0).
By property, PA + PA' = 2a
= 17 cm
PA' =
= 17 cm
Since in ellipse, 2a = PA + PA'
2a = 17 + 17
2a = 34 cm
Hence, the length of major axis = 2a = 34 cm.
Question 3 1.33 / -0
Define a function f : R
R by
Suppose f(x) is differentiable on R. Then
Solution
f(x) is differentiable on R.
So, it will be continuous on R.
Continuity at x = 0
LHL
Put x = 0 - h, then h
RHL
Put x = 0 + h, then h -> 0
Value of f(x) at x = 0
f(0) = b
as f(x) is continuous at x= 0
Therefore, LHL = RHL = f(0)
0 = b = b
b = 0
Differentiability at x= 0
LHD
RHD
as f(x) is differentiable at x = 0, LHD = RHD
a = 1
Question 4 1.33 / -0
The area bounded by the parabolas y = x2 and y = 1 - x2 equals
Solution
y = x
2 & y = 1 - x
2 Point of intersection of graphs:
x
2 = 1 - x
2 2x
2 = 1
x =
Therefore, points of intersection are
and
.
Area under graph:
=
=
=
= 2 ×
=
Question 5 1.33 / -0
Three children, each accompanied by a guardian, seek admission in a school. The principal wants to interview all the 6 persons one after the other, subject to the condition that no child is interviewed before its guardian. In how many ways can this be done?
Solution
Number of ways =
=
= 90 ways
Question 6 1.33 / -0
Consider the conic ex2 + πy2 - 2e2 x - 2π2 y + e3 + π3 = πe. Suppose P is any point on the conic and S1 , S2 are the foci of the conic, then the maximum value of (PS1 + PS2 ) is
Question 7 1.33 / -0
Let f : R
R be the function f(x) = (x – a
1 )(x - a
2 ) + (x – a
2 )(x – a
3 ) + (x – x
3 )(x – x
1 ) with a
1 , a
2 , a
3 ∈ R. Then f(x) ≥ 0, if and only if
Solution
f(x) = 3x
2 - 2(a
1 + a
2 + a
3 )x + a
1 a
2 + a
2 a
3 + a
3 a
1 0
Therefore, D
0.
4(a
1 + a
2 + a
3 )
2 - 4.3(a
1 a
2 + a
2 a
3 + a
3 a
1 )
0
a
1 2 + a
2 2 + a
2 3 - a
1 a
2 - a
2 a
3 - a
3 a
1 0
[(a
1 - a
2 )
2 + (a
2 - a
3 )
2 + (a
3 - a
1 )
2 ]
0
Therefore, a
1 = a
2 = a
3
Question 8 1.33 / -0
A purse contains 4 copper coins and 3 silver coins. A second purse contains 6 copper coins and 4 silver coins. A purse is chosen randomly and a coin is taken out of it. What is the probability that it is a copper coin?
Solution
Required probability =
=
Question 9 1.33 / -0
Let A
denote the matrix, where i
2 = 1, and let I
denote the identity matrix.
Then, I + A + A
2 + ... + A
2010 is equal to
Solution
A =
, A
2 =
; A
3 =
; A
4 = I =
I + A + A
2 + A
3 =
I + A + A
2 + A
3 + ... + A
2010 = (I + A + A
2 + A
3 ) + A
4 (I + A + A
2 + A
3 ) + ... + A
2008 (I + A + A
2 )
=
Question 10 1.33 / -0
Let ABC be an equilateral triangle. Let KLMN be a rectangle with K and L on BC, M on AC, and N on AB. Suppose AN/NB = 2 and the area of triangle BKN is 6. Then, the area of the triangle ABC is:
Solution
⇒
, BK = z =
⇒
Area of
=
= 12 +
Question 11 1.33 / -0
The roots of (x - 41)49 + (x - 49)41 + (x - 2009)2009 = 0 are
Solution
(x - 41)49 + (x - 49)41 + (x - 2009)2009 = 049 + (x - 49)41 + (x - 2009)2009 48 + 41(x - 49)40 + 2009(x - 2009)48 > 0
Question 12 1.33 / -0
Three vertices are chosen randomly from the seven vertices of a regular 7-sided polygon. The probability that they form the vertices of an isosceles triangle is:
Question 13 1.33 / -0
Suppose loga b + logb a = c. The smallest possible integer value of c for all a, b > 1 is
Question 14 1.33 / -0
Let A = (4, 0), B = (0, 12) be two points in the plane. The locus of a point C such that the area of triangle ABC is 18 sq. units is
Question 15 1.33 / -0
Which of the following intervals is a possible domain of the function f(x) = log{x} [x] + log[x] {x}, where [x] is the greatest integer not exceeding x and {x} = x – [x]?
Solution
is the only option satisfying the domain condition.
Question 16 1.33 / -0
The value of
is
Question 17 1.33 / -0
The sum of non-real roots of the polynomial equation x3 + 3x2 + 3x + 3 = 0
Solution
f(x) = x
3 + 3x
2 + 3x + 3 = 0
f'(x) = 3x
2 + 6x + 3 = 3(x + 1)
2 0
f(x) is an increasing function.
Now, f(-3) = -6 < 0
f(-2) = 1 > 0
Therefore, real root
lies between -3 and 2.
= -3;
⇒
⇒
⇒
Question 18 1.33 / -0
Consider a triangle ABC in the xy-plane with vertices A = (0, 0), B = (1, 1) and C = (9, 1). If the line x = a divides the triangle into two parts of equal area, then a equals
Solution
Area of PQC =
× area of ABC
⇒
Question 19 1.33 / -0
Let f(x) =
and g(x) = cos x. Which of the following statements is/are true?
1. Domain of f((g(x))
2 ) = Domain of f(g(x))
2. Domain of f(g(x)) + g(f(x)) = Domain of g(f(x))
3. Domain of f(g(x)) = Domain of g(f(x))
4. Domain of g((f(x))
3 ) = Domain of f(g(x))
Solution
As 2 - cosx - cos
2 x
0
(cosx + 2)(cosx - 1)
0
⇒ -2
cosx
1
The domain of g(f(x)) is [-2, 1] as
.
f(g(x)) =
For domain of f(g(x)),
⇒
⇒
⇒
The domain of f(g(x)
2 ) is the domain of g(f(x)), i.e. [-2, 1].
Hence, option 2 is correct.
Question 20 1.33 / -0
Let n ≥ 3. A list of numbers 0 < X
1 < X
2 < ... X
n has mean
and standard deviation
. A new list of numbers is made as follows:
Y
1 = 0, Y
2 = X
(2) ..... X
(n - 1) , y
n = X
1 + X
n The mean and the standard deviation of the new list are
and
. Which of the following is necessarily true?
Question 21 1.33 / -0
The relation Cp - Cv = R (Cp, Cv ; Molar specific heats at constant pressure, volume) is exactly true for
Solution
For ideal gas,
and
Now,
From here,
Derivation of all these depends on PV = nRT (Ideal Gas Law)
Question 22 1.33 / -0
A progressive wave travelling in positive x-direction given by y = a cos(kx - ωt) meets a denser surface at x = 0 and t = 0. The reflected waves are then given by
Solution
Wave travelling in positive x-direction is given by y = a cos(kx - ωt)
Question 23 1.33 / -0
A small rectangular loop of wire in the plane of the paper is moved with uniform speed across a limited region of uniform speed across a limited region of uniform magnetic field, perpendicular to the plane of the paper.
Which of the following graphs would best represent the variation of the electric current I in the wire with time t?
Solution
As the block enters in the magnetic field, rate of change in flux will be constant. So, constant current will be produced.
Question 24 1.33 / -0
Consider a one dimensional potential V(x) as shown in the figure below. A classical particle of mass m moves under its influence and has total energy E as shown.
The motion is
Solution
In the given graph, initial potential energy is maximum. So, K.E. is minimum (since total energy remains constant E).
Question 25 1.33 / -0
An ideal monatomic gas expands to twice its volume. If the process is isothermal, the magnitude of work done by the gas is Wi . If the process is adiabatic, the magnitude of work done by the gas is We . Then, which of the following is true?
Solution
We know that area under the curve in PV graph is called work done.
Since area under isothermal curve is greater than that under adiabatic curve in PV graph,
W
i > W
a > 0
Question 26 1.33 / -0
In a Young's double slit experiment the intensity of light at each slit is I
0 . Interference pattern is observed along a direction parallel to the line S
1 S
2 on screen S.
The minimum, maximum, and the intensity averaged over the entire screen are respectively:
Solution
Now,
I
max = 4I
0 I
min = 0
I
avg = 2I
0
Question 27 1.33 / -0
In one model of the electron, the electron of mass m
e is thought to be a uniformly charged shell of radius R and total charge e, whose electrostatic energy E is equivalent to its mass m
e via Einstein's mass energy relation E = m
e C
2 . In this model, R is approximately (m
e = 9.1 × 10
-31 kg, c = 3 × 10
8 m/s
-1 ,
πε
0 = 9 × 10
9 Farads m
-1 , magnitude of the electron charge = 16 × 10
-19 C)
Solution
From the question, we can say that
By putting the values in the formula,
= 9.1 × 10
-31 × (3 × 10
8 )
2 R =
⇒ 1.4 × 10
-15 m
Question 28 1.33 / -0
A ball of mass m suspended from a rigid support by an inextensible massless string is released from a height h above its lowest point. At its lowest point, it collides elastically with a block of mass 2m at rest on a frictionless surface. Neglect the dimensions of the ball and the block. After the collision, the ball rises to a maximum height of
Solution
Velocity of the ball after collision,
Hence, after collision, maximum height,
h
max =
h
max =
Question 29 1.33 / -0
A pen of mass 'm' is lying on a piece of paper of mass M placed on a rough table. If the coefficient of friction between the pen and paper, and the paper and table are μ1 and μ2 , respectively, then the minimum horizontal force with which the paper has to be pulled for the pen to start slipping is given by
Solution
For pen to start slipping, maximum horizontal force on it is f
1 =
Therefore, a =
is the maximum common acceleration for both pen and paper.
F.B.D. for both pen and paper:
f
1 =
and N
1 = W
1 Therefore,
N
2 = N
1 + W
1 F - f
1 - f
2 = Ma
Also,
Therefore, F = f
1 + f
2 + Ma
Therefore,
Question 30 1.33 / -0
Consider a uniform spherical volume charge distribution of radius R. Which of the following graphs correctly represents the magnitude of the electric field E at a distance r from the centre of the sphere?
Question 31 1.33 / -0
A nucleus has a half life of 30 minutes. At 3 p.m, its decay rate was measured as 1,20,000 counts/sec. What will be the decay rate at 5 p.m?
Solution
Given: T = 30 minutes,
= 1,20,000 counts/sec
After each half life, activity is reduced to half. Therefore, after n half lives, activity reduces to
.
Also,
at 5 p.m. will be equal to activity remaining after four half lives.
i.e
of the initial activity
of the initial activity
Question 32 1.33 / -0
Velocity of sound measured at a given temperature in oxygen and hydrogen is in the ratio:
Question 33 1.33 / -0
A narrow but tall cabin is falling freely near the Earth's surface. Inside the cabin, two small stones A and B are released from rest (relative to the cabin). Initially, A is much above the centre of mass and B is much below the centre of mass of the cabin. A close observation of the motion of A and B will reveal that
Solution
Since cabin is tall, gravitation acceleration is different for particles A and B.
(Acceleration is given by g =
)
Because A and B are on different height. g
B > g
A so that acceleration of A is less than of acceleration B.
So, a
B > a
CM > a
A .
A moves slowly upward and B moves slowly downward relative to the cabin.
Question 34 1.33 / -0
Two rods, one made of copper and the other steel of the same length and cross sectional area are joined together. (The thermal conductivity of copper is 385 Js-1 .m-1 . K-1 and steel is 50 J.s-1 .m-1 .K-1 ). If the copper end is held at 100ºC and the steel end is held at 0ºC, what is the junction temperature (assuming no other heat losses)?
Solution
As we know,
Let the temperature of the junction to be T, In steady state the rate of heat flow through the copper rod = heat flow through the steel rod.
K
1 (T - 100) = K
2 ( -T)
T(K
1 + K
2 ) = 100K
1 °C
Question 35 1.33 / -0
In a Young's double slit set-up, light from a laser source falls on a pair of very narrow slits separated by 1.0 micrometre and bright fringes separated by 1.0 millimeter are observed on a distant screen. If the frequency of the laser light is doubled, what will be the separation of the bright fringes?
Solution
Bright fringe =
If f is doubled,
becomes halved. Therefore,
becomes half.
=
= 0.5 mm
Question 36 1.33 / -0
Given below are three schematic graphs of potential energy V(r) versus distance r for three atomic particles: electron (e), proton (p+) and neutron (n), in the presence of a nucleus at the origin O. The radius of the nucleus is r
o . The scale on the V-axis may not be the same for all figures. The correct pairing of each graph with the corresponding atomic particle is
Solution
Outside the nucleus, electric potential decreases and e is negatively charged.
Therefore, its PE is negative even outside the nucleus, where nuclear attractive force is negligible.
Outside the nucleus, neutron will not experience electric force as it is neutral. So, no potential energy is associated with it outside the nucleus.
Question 37 1.33 / -0
Consider an initially neutral hollow conducting spherical shell with inner radius r and outer radius 2r. A point charge +Q is now placed inside the shell at a distance r/2 from the centre. The shell is then grounded by connecting the outer surface to the earth. P is an external point at a distance 2r from the point charge +Q on the line passing through the centre and the point charge +Q as shown in the figure.
The magnitude of the force on a test charge +q placed at P will be
Solution
Charge on outer most surface is zero and force is directly proportional to the charge.
Question 38 1.33 / -0
In a photocell circuit, the stopping potential, V
0 , is a measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency v of incident light:
The values of Planck's constant and the work function as determined from the graph are (Taking the magnitude of electronic charge to be e = 1.6 × 10
-19 C)
Solution
V =
Hence, from the graph,
= 2 eV
h =
Question 39 1.33 / -0
Two species of radioactive atoms are mixed in equal number. The disintegration of the first species is
and of the second is
/3. After a long time, the mixture will behave as a species with mean life of approximately
Solution
Average life of species is given by:
Here, dN
1 =
dN
2 =
Now, average life =
Integrating, we get
Average life =
Question 40 1.33 / -0
Young-Laplace law states that the excess pressure inside a soap bubble of radius R is given by
P = 4
/R, where
is the coefficient of surface tension of the soap. The number E
0 is a dimensionless number that is used to describe the shape of bubbles rising through a surrounding fluid. It is a combination of g, the acceleration due to gravity,
, the density of the surrounding fluid,
and a characteristic length scale L which could be the radius of the bubble. A possible expression for E
0 is
Question 41 1.33 / -0
Which of the following gases has the slowest rate of diffusion?
Solution
According to Graham's law, rate of diffusion
Molar mass of O
2 = 32 g,
Molar mass of H
2 = 2 g,
Molar mass of CO
2 = 12 + 32 = 44 g,
Molar mass of CH
4 = 12 + 4 = 16 g
Hence, due to highest molar mass of CO
2 , its rate of diffusion is the slowest.
Question 42 1.33 / -0
If the ratio of the heat capacities Cp /Cv for one mole of a gas is 1.67, then the gas is
Solution
The values of C
p /C
v for different nature of gases.
= 1.66 (For monoatomic gas)
= 1.40 (For diatomic gas)
= 1.33 (For non - linear triatomic gas)
Hence, the given ratio of the heat capacities C
p /C
v for one mole of gas is for monoatomic gas. Out of the following, only He is monoatomic in nature. Therefore, correct answer is helium.
Question 43 1.33 / -0
What are the respective hybridisations of Ni centre in [Ni(PPh3 )2 Cl2 ] and [NiCl4 ]2- ?
Solution
dsp
2 hybridisation
PPh
3 is a strong ligand. Hence, pairing of electrons takes place.
sp
3 hybridisation
Cl
- is a weak ligand. Hence, pairing of electrons does not take place.
Question 44 1.33 / -0
Which of the following compounds are aromatic?
Solution
Both II and IV are aromatic compounds as both have close system of conjugated double bonds and follow Huckel's (4n + 2)πe- rule.
Question 45 1.33 / -0
Among the following, the species with the highest bond order is:
Solution
According to molecular orbital theory:
Bond order =
Bond order =
Bond order =
Bond order =
Hence, among the given species, O
+ 2 has the highest bond order, i.e. 2.5.
Question 46 1.33 / -0
Typical electronic energy gaps in molecules are about 1.0 eV. In terms of temperature, the gap is closest to
Solution
1 eV = 1.6 × 10
-19 J
To convert Joule into Kelvin, Boltzmann constant is required.
Kelvin (K) = (1 eV/k
B ) [Where, k
B = 1.38 × 10
-23 J/K]
Or K =
= 11594.20 ≈ 10000 or 10
4 K
Question 47 1.33 / -0
The major product of the following reaction is:
Question 48 1.33 / -0
The reaction that gives the above molecule as the major product is
Solution
The given compound is formed as the major product during Williamson Synthesis reaction. In this reaction, an alkyl halide is allowed to react with sodium alkoxide followed by S
N 2 mechanism to give respective ethers.
Question 49 1.33 / -0
The number of isomers of [Co(dien)Cl3] is
Solution
[Co(dien)Cl3 ] is an MA3 B3 type complex. Therefore, it exhibits geometrical isomerism and exists in the cis and trans forms which in this case may also be referred to as fac and mer isomers.
Question 50 1.33 / -0
The shape of the molecule ClF3 is:
Solution
ClF
3 has five electron pairs, which are arranged in a trigonal bipyramidal shape with a 175° F-Cl-F bond angle, due to the presence of two equatorial lone pairs making the final structure T-shaped.
Question 51 1.33 / -0
A concentrated solution of copper sulphate, which is dark blue in colour, is mixed at room temperature with a dilute solution of copper sulphate, which is light blue. For this process,
Solution
On dilution, entropy change is positive, while there is no change in enthalpy.
Question 52 1.33 / -0
The correct structure of PCl3 F2 is:
Solution
According to VSEPR Theory, the correct structure of PCl
3 F
2 is as follows:
Question 53 1.33 / -0
The hybridisations of [Ni(CO)4 ] and [Cr(H2 O)6 ]3+ , respectively are
Solution
In [Ni(CO)
4 ], carbonyl is a neutral ligand. Therefore, oxidation state of Ni is zero.
So, Ni(0) = [Ar]3d
8 , 4s
2 CO is a strong field ligand. Therefore, pairing of electrons takes place as shown below.
In case of [Cr(H
2 O)
6 ]
3+ , H
2 O is a neutral ligand. Thus, oxidation state of Cr = +3.
Electronic configuration of Cr
3+ = [Ar]3d
3 H
2 O is a weak field ligand. Therefore, pairing of electrons does not take place.
Question 54 1.33 / -0
In i - iv,
the compound that does
not undergo polymerisation under radical initiation is
Solution
The radical formed as an intermediate in compound (iv) is unstable. Therefore, it does not undergo polymerisation under radical initiation. In all other cases, resonance stabilised radicals are formed.
Question 55 1.33 / -0
For transforming
,
the reagent used is
Solution
H
3 PO
2 a mild reducing agent and is used to remove the diazonium group with hydrogen.
Question 56 1.33 / -0
The spin-only magnetic moments of [Mn(CN)
Solution
[Mn(CN)
6 ]
4- - Oxidation state of Mn is 'x'
x + 6(-1) = -4
x = +2
Electronic configuration of Mn
+2 3d
5 4s
0 4p
CN
- is a strong ligand. So, it creates back pairing of (n - 1)d orbitals configuration.
So, unpaired electrons = 1
B.M
1.73 B.M
And, in [MnBr
4 ]
2- Br is a weak ligans. So, no back pairing effect on (n - 1) orbital. So, unpaired electrons = 5
Question 57 1.33 / -0
Among the following, the set of isoelectronic ions is:
Solution
Na+ , Mg2+ , F- and O2- are isoelectronic as all are 10 electron species.
Question 58 1.33 / -0
For the reaction
, the concentration of A decreases from 0.06 to 0.03 mol/L and that of B rises from 0 to 0.06 mol/L at equilibrium. The values of n and equilibrium constant for the reaction respectively are
Solution
A
nB
t = 0 0.06 0
t = t 0.03 0.06
⇒ αc = 0.03
⇒ nαc = 0.06
n = 2
Question 59 1.33 / -0
The entropy change in the isothermal reversible expansion of 2 moles of an ideal gas from 10 to 100 L at 300 K is
Question 60 1.33 / -0
The angle of incidence of X-ray of wavelength 3
which produces a second offer diffraction beam from the (100) plane in a simple cubic lattice with interlayer spacing d = 6
is 30°. The angle of incidence that produces a first order diffracted beam from the (200) plane is
Question 61 1.33 / -0
During photosynthesis, the chemical conversion of water is called
Solution
Photolysis is a chemical reaction in which a chemical compound is broken down by photons. Photolysis of water results in the release of oxygen (as a by-product) and hydrogen.
Question 62 1.33 / -0
During development, unspecified cells become cells having unique functions. This process is called
Solution
Cellular differentiation is the process where a cell changes from one cell type to another. Most commonly, the cell changes to a more specialised type.
Question 63 1.33 / -0
Fertilisation in humans usually takes place in
Solution
Fertilisation is the fusion of male and female gametes to produce a zygote. It usually takes place in the fallopian tube.
Question 64 1.33 / -0
One difference between blood and lymph is that blood
Solution
Blood contains red blood cells, white blood cells and platelets. Lymph is a whitish and clear liquid. It is generally similar to blood plasma, except that it does not contain red blood cells.
Question 65 1.33 / -0
The disorders that arise when the immune system destroys self cells are called autoimmune disorders. Which of the following would be classified under this?
Solution
Rheumatoid arthritis is an autoimmune disease that causes chronic inflammation of the joints and other areas of the body.
Question 66 1.33 / -0
Transfer RNA (tRNA)
Solution
The Cloverleaf model of tRNA is a model that depicts the molecular structure of tRNA. The model revealed that the chain of tRNA consists of two ends, sometimes called "business ends" and three arms. Two of the arms have a loop, D-loop (dihydro U loop) and Tψ-loop with a ribosome recognition site.
Question 67 1.33 / -0
In mammals, the hormones secreted by the pituitary, the master gland, are regulated by
Solution
The hormones of the pituitary gland help regulate the functions of other endocrine glands. The hypothalamus sends signals to the pituitary to release or inhibit pituitary hormone production.
Question 68 1.33 / -0
Vitamin A deficiency leads to night-blindness. Which of the following is the reason for the disease?
Solution
Night blindness, also called nyctalopia, means a person cannot see well at night or in poor light. Vitamin A deficiency is one of the reasons for night blindness. In this condition, the rhodopsin pigment of the rod cells becomes defective.
Question 69 1.33 / -0
Ribonucleic Acids (RNA) that catalyse enzymatic reactions are called ribozymes. Which one of the following acts as a ribozyme?
Solution
Within the ribosome, ribozymes function as part of the large sub-unit ribosomal RNA to link amino acids during protein synthesis. They also participate in a variety of RNA processing reactions, including RNA splicing, viral replication, and transfer RNA biosynthesis.
Question 70 1.33 / -0
Pathfinding by ants is by means of
Solution
Pheromones help ants keep track of food and also direct each other through their environment. Hence, it is the chemical signal between the ants that helps them track their path.
Question 71 1.33 / -0
Bacteriochlorophylls are photosynthetic pigments found in phototrophic bacteria. Their function is distinct from the plant chlorophyll in that they
Solution
Bacteriochlorophylls are photosynthetic pigments that occur in various phototrophic bacteria. None of the anoxygenic phototrophic bacteria has the ability to use water as an electron donor (or cannot oxidise water), and thus, perform photosynthesis using sulphide, hydrogen or organic substrates. Therefore, photosynthesis by these bacteria does not involve oxygen.
Question 72 1.33 / -0
If you dip a sack full of paddy seeds in water overnight and then keep it out for a couple of days, it feels warm. What generates this heat?
Solution
The conditions mentioned are appropriate for increasing the rate of respiration in the seeds. The heat released is due to the respiration of the seeds.
Question 73 1.33 / -0
The major constituents of neurofilaments are
Solution
Neurofilaments (NF) are the 10 nanometer or intermediate filaments found in neurons. They are a major component of the neuronal cytoskeleton and are believed to function primarily to provide structural support for the axon and to regulate axon diameter.
Question 74 1.33 / -0
In which of the following phases of the cell cycle are sister chromatids available as template for repair?
Solution
M phase is the mitotic phase. The sister chromatid form is present in the M phase.
Question 75 1.33 / -0
Saline drip is given to a cholera patient because
Solution
Saline drip is given to a cholera patient because sodium ions help in the retention of water in the body. This helps to prevent dehydration.
Question 76 1.33 / -0
Puffs in the polytene chromosomes of Drosophila melanogaster salivary glands represent
Solution
Polytene chromosomes are oversised chromosomes, which have developed from standard chromosomes and are commonly found in the salivary glands of Drosophila melanogaster . These represent transcriptionally active genes.
Question 77 1.33 / -0
Human foetal haemoglobin differs from the adult haemoglobin in that it
Solution
HbF has more affinity for oxygen than adult haemoglobin. This property helps in extracting oxygen from the mother.
Question 78 1.33 / -0
Which of the following techniques is used for the detection of proteins?
Solution
The western blotting (sometimes called the protein immunoblot) is a widely used analytical technique used in molecular biology, immunogenetics and other molecular biology disciplines to detect specific proteins in a sample of tissue homogenate or extract.
Question 79 1.33 / -0
Chlorofluorocarbons (CFCs) are believed to be associated with cancer because
Solution
CFCs destroy ozone layer. This allows the penetration of harmful UV rays to the Earth. UV rays damage DNA by causing thymine dimerisation.
Question 80 1.33 / -0
Which of the following statements about nitrogenase is correct?
Solution
The nitrogenase enzyme complex (the nitrogen fixing enzyme) is sensitive to O2 and inactivates the enzyme.
Question 81 1.33 / -0
Let p(x) = a0 + a1 x + . . . + an xn . If p(-2) = -15, p(-1) = 1, p(0) = 7, p(1) = 9, p(2) = 13 and p(3) = 25, then the smallest possible value of n is
Solution
p(x) = a0 + a1 x + .........+ an xn 0 = 70 + a1 +a2 + ......... + an = 90 - a1 + 2a2 - ........... = 10 + 2a1 + 4a2 ......... = 130 - 2a1 + 4a2 ......... = -150 + a2 + ......] = 100 + a2 + a4 = 5 -----(1)2 + a4 = 52 + a4 = -2 ------(2)0 + 4a2 + ........) = -20 + 4a2 + 16a4 = -12 + 16a4 = -8 -----(2)0 + 3a1 + 9a2 + ........ = 250 + 3a1 + 9a2 + 27a3 + 81a4 + 243a5 = 252 + 4a4 = -82 + 16a4 = -84 = 0 and a2 = -2
Question 82 1.33 / -0
The range of the function f(x) = (sin x)
sin x defined on (0,
) is
Solution
f(x) = (sin x)
sin x f(x) = e
sin x log sin x Maximum value of sin x log (sin x) is 0.
Therefore, maximum value of (sin x)
sin x is e
0 = 1
Minimum value of sin x log (sin x) is
.
Therefore, minimum value of (sin x)
sin x is
.
Hence, range of f(x) =
Question 83 1.33 / -0
Suppose a, b, c are real numbers and each of the equations x2 + 2ax + b2 = 0 and x2 + 2bx + c2 = 0 has two distinct real roots. Then, the equation x2 + 2cx + a2 = 0 has
Solution
D
1 = 4a
2 - 4b
2 > 0
4(a
2 - b
2 ) > 0
a
2 - b
2 > 0 ---------(1)
D
2 = 4b
2 - 4c
2 > 0
b
2 - c
2 > 0 ----------(2)
Now, D = 4c
2 - 4a
2 = 4(c
2 - b
2 + b
2 - a
2 )
= -4(b
2 - c
2 + a
2 - b
2 ) = -4(D
1 + D
2 )
= Negative
Therefore, the equation has no real roots.
Question 84 1.33 / -0
Define a sequence (a
n ) by a
1 = 5, a
n = a
1 a
2 .... a
n = 1 + 4 for n > 1. Then,
Question 85 1.33 / -0
Arrange the expansion of
in decreasing powers of x. Suppose the coefficient of the first three terms form an arithmetic progression. Find the number of terms in the expansion having integer powers of x.
Solution
⇒
⇒
⇒ n - 1 =
⇒ n = 8
⇒
⇒ Integers r = 0, 4, 8
Question 86 1.33 / -0
Let V1 be the volume of a given right circular cone with O as the centre of the base and A as its apex. Let V2 be the maximum volume of the right circular cone inscribed in the given cone whose apex is O and whose base is parallel to the base of the given code. The ratio V2 /V1 is
Solution
ΔABC and ΔAOP are similar
⇒ h =
V
2 =
r
2 (H - h) =
r
2 H (1 -
) =
(r
2 -
)
⇒
= 0 ⇒ r =
Thus, V
2 is maximum at r =
V
1 =
Therefore,
Question 87 1.33 / -0
Let A and B be any two n × n matrices such that the following conditions hold: AB = BA and there exists a positive integers k and l such that Ak = I (the identity matrix) and Bl = 0 (the zero matrix). Then,
Solution
A
k = I and B
l = 0
(det(B) = 0)
det(AB) = 0
Question 88 1.33 / -0
Let f(x) = x
3 + ax
2 + bx + c, where a, b and c are real numbers. If f(x) has a local minimum at x = 1 and a local maximum at x = -
and f(2) = 0, then
f(x)
dx equals
Solution
From question, x = 1 and
are solutions of the function f'(x).
f'(x) =
f(x) =
f(2) = 8 - 4 - 2 +
= 0
= -2
f(x) =
Question 89 1.33 / -0
Let f(x) be a non-constant polynomial with real coefficients such that f(
) = 100 and f(x) ≤ 100 for all real x. Which of the following statements is
not necessarily true?
Solution
Cooefficient of highest degree term must be negative because if it is positive, then x
,
and it is not possible, since f(x) ≤ 100.
Now, graph will be like
at least two real roots will be there. If x
, then f(x) < 100. It is not true always, as the graph can be like this also.
Now, let the highest coefficient it can have be 49.
Then, f(
) = 49 +
+
+ .......
But, the sum cannot be equal to 100 and less than 100.
Question 90 1.33 / -0
Let f(x) = 1 +
. How many real roots does f(x) = 0 have?
Solution
Since f(x) = 1 +
f'(x) = 1 +
f'(x) > 0 for all x > -1
f'(-1) =
and f'(-2) =
Since f'(-2)
f'(-1) < 0
Then, x
0 ∈ (-2, -1) such that f'(x
0 ) = 0
Therefore, minimum value of f(x) lies on x = x
0 f(x
0 ) = 1 +
> 0 for all x
0 ∈ (-2, -1)
So, no real solution exists for f(x).
Hence, number of real solutions = 0
Question 91 1.33 / -0
A spherical cavity of radius r is carved out of a uniform solid sphere of radius R as shown in the figure.
The distance of the centre of mass of the resulting body from that of the solid sphere is given by
Solution
Mass of sphere of radius r
m =
Since
,
Then,
X
cm =
Because
So,
X
cm =
X
cm =
=
Question 92 1.33 / -0
A cubical box of side a sitting on a rough table-top is pushed horizontally with a gradually increasing force until the box moves. If the force is applied at a height from the table-top which is grater than critical height H, the box topples first. If it is applied at a height less than H, the box starts sliding first. Then the coefficient of friction between the box and the table-top is
Solution
When height is greater than H:
Balance of torque about point A
F × H = mg
------ (1)
When height is less than H, Force balance
F =
-------(2)
From eq. (1) and (2)
Question 93 1.33 / -0
The total energy of a black body radiation source is collected for five minutes and used to heat water. The temperature of the water increases from 10.0°C to 11.0°C. The absolute temperature of the black body is doubled and its surface area halved, and the experiment is repeated for the same time. Which of the following statements would be most nearly correct?
Solution
We know that
Now, from given values
So, the temperature of water would increase from 10°C to a final temperature of 18°C.
Question 94 1.33 / -0
On a bright sunny day, a diver of height h stands at the bottom of a lake of depth H. Looking upward, he can see objects outside the lake in a circular region of radius R. Beyond this circle, he sees the images of objects lying on the floor of the lake. If refractive index of water is 4/3, then the value of R is
Solution
From the figure:
Now, tan C =
So, R =
Question 95 1.33 / -0
A ball is dropped vertically from a height of h onto a hard surface. If the ball rebounds from the surface with a fraction r of the speed with which it strikes the latter on each impact, what is the net distance travelled by the ball up to the 10th impact?
Solution
Total distance = (
+ r
2 + r
4 + .... upto 10
th terms) - h
=
(1 + r
2 + r
4 + ... + 10
th term) - h
Also,
Therefore, total distance =
Or total distance =
Question 96 1.33 / -0
A particle of mass m undergoes oscillations about x = 0 in a potential given by
, where V
0 , k and a are constants. If the amplitude of oscillation is much smaller than a, the time period is given by
Solution
E =
Since x < < a,
This resembles F = -k x
Hence,
Question 97 1.33 / -0
An isolated sphere of radius R contains uniform volume distribution of positive charge. Which of the curves shown below correctly illustrate the dependence of the magnitude of the electric field of the sphere as a function of the distance r from its centre?
Question 98 1.33 / -0
A material is embedded between two glass plates. Refractive index n of the material varies with thickness as shown below. The maximum incident angle (in degrees) on the material for which beam will pass through the material is
Solution
1.5 × sin C = 1.2 sin r
sin r =
for total internal reflection
Therefore, sin r = 1
C = 53°
Question 99 1.33 / -0
A bullet of mass m is fired horizontally into a large sphere of mass M and radius R resting on a smooth horizontal table.
The bullet hits the sphere at a height h from the table and sticks to its surface. If the sphere starts rolling without slipping immediately on impact, then
Solution
By linear momentum conservation,
mc = (m + M)v
o ⇒
⇒
⇒
⇒
Question 100 1.33 / -0
A ball is rolling without slipping in a spherical shallow bowl (radius R) as shown in the figure and is executing simple harmonic motion. If the radius of the ball is doubled, the period of oscillation
Question 101 1.33 / -0
If the pH of a mixture of 10 ml of 0.1 M NH4 OH and 10 ml of 1 M NH4 Cl solution is 8, then the pKb value of NH4 OH is closest to
Solution
Millimoles of NH
4 OH in 10 ml of 0.1 M NH
4 OH solution = M × n-factor × v = 0.1 × 1 × 10 = 1
Millimoles of NH
4 Cl in 10 ml of 1 M NH
4 Cl solution = M × n-factor × v = 1 × 1 × 10 = 10
pOH = pK
b + log
6 = pK
b + log
6 = pK
b + log 10
pK
b = 6 - 1
pK
b = 5
Question 102 1.33 / -0
The rate constant for the reaction COCI
2 (g)
CO(g) + CI
2 (g) is given by
In k (min
-1 ) =
+ 31.33. The temperature at which the rate of this reaction will be doubled from that at 25°C is
Solution
In k =
2.303 log k =
Suppose k
1 and T
1 are rate constant and temperature, respectively, in case-I and k
2 and T
2 are rate constant and temperature, respectively, in case-II:
So, 2.303 log k
1 =
------(1)
2.303 log k
2 =
--------(2)
Subtracting equation (2) from equation (1):
2.303 log
= -11,067
2.303 log
= -11,067
[As k
2 = 2k
1 ]
2.303 × (-0.3010) =
By solving the above equation, we get
T
2 T
2 31°C
Question 103 1.33 / -0
The final major product obtained in the following sequence of reactions is:
Question 104 1.33 / -0
The inter-planar spacing between the (2 2 1) planes of a cubic lattice of length 450 pm is
Question 105 1.33 / -0
2.52 g of oxalic acid dihydrate was dissolved in 100 mL of water, 10 mL of this solution was diluted to 500 mL. The normality of the final solution and the amount of oxalic acid (in g) in the solution respectively are
Solution
N
1 V
1 = N
2 V
2 N
1 =
= 0.4
V
1 = 10 mL
V
2 = 500 mL
N
2 =
= 0.008 N
0.008 =
w =
w = 0.252 g
Question 106 1.33 / -0
A metal with an atomic radius of 141.4 pm crystallises in the face centred cubic structure. The volume of the unit in pm3 is
Solution
For an FCC unit cell,
a =
Hence,
= 6.4 × 10
7 pm
3
Question 107 1.33 / -0
XeF6 hydrolysis to give an oxide. The structure of XeF6 and the oxide, respectively is
Question 108 1.33 / -0
In the reaction sequence,
The major products X and Y respectively, are
Solution
Due to intramolecular hydrogen bonding, reaction is favoured at ortho position.
Question 109 1.33 / -0
The major product obtained in the reaction of aniline with acetic anhydride is
Solution
Aniline is a primary amine and it is basic in nature. Acetic anhydride is an anhydride of acetic acid, which is a source of acyl group. Aniline reacts with acetic anhydride by nucleophilic substitution reaction.
Question 110 1.33 / -0
Consider the equilibria (1) and (2) with equilibrium constants K
1 and K
2 , respectively.
K
1 and K
2 are related as
Question 111 1.33 / -0
The mode of action of penicillin is as follows:
Solution
β-Lactam antibiotics are a broad class of antibiotics that includes penicillin derivatives, cephalosporins, monobactams, and carbapenems. β-Lactam antibiotics are bacteriocidal and act by inhibiting the synthesis of the peptidoglycan layer of bacterial cell walls.
Question 112 1.33 / -0
Male offspring of which of the following couples has the highest chance of haemophilia?
Question 113 1.33 / -0
When hydrogen peroxide is applied on the wound as a disinfectant, there is frothing at the site of injury, which is due to the presence of an enzyme in the skin that uses hydrogen peroxide as a substrate to produce
Solution
This is because blood and most living cells contain the enzyme catalase, which attacks hydrogen peroxide and converts it into water and nascent oxygen. The frothing is caused due to the release of nascent oxygen.
Question 114 1.33 / -0
Which one of the following is true about enzyme catalysis?
Solution
Enzymes lower the activation energy of the reaction. This makes the reaction occur at a higher rate. The enzyme is recovered without any change.
Question 115 1.33 / -0
Which of the following statements is true about bacterial symbionts in insects?
Solution
Heritable bacterial endosymbionts are very common in insects. One important feature of all these symbionts is that most of them are exclusively or predominantly vertically (and usually maternally) transmitted, and thus heritable. Not all bacterial symbionts influence the host's interactions with other species directly, and many of these bacterial species affect the reproductive system of their hosts by either causing female-biased sex ratios or impacting on the ability of pairs of individuals to produce offspring.
Question 116 1.33 / -0
Which of the following correctly represents the results of an enzymatic reaction?
Solution
Enzyme combines with the substrate resulting in the formation of products. Enzyme is not used in the reaction and recovered as it is.
Question 117 1.33 / -0
Which of the following sequence of events gives rise to flaccid guard cells and stomatal closure at night?
Solution
The sequence followed is: 2 At night, photosynthesis stops and glucose gets consumed. This leads to low osmotic pressure and a low pH as respiration starts in the plant cells. This increases the carbon dioxide concentration in the plant cell. This leads to closure of stomata.
Question 118 1.33 / -0
Selection of lysine auxotroph from a mixed population of bacteria can be done by growing the bacterial population in the presence of
Solution
Auxotrophy is the inability of an organism to synthesise a particular organic compound required for its growth. An auxotroph is an organism that displays this characteristic. A lysine auxotroph will grow only in a medium containing lysine.
Question 119 1.33 / -0
109 bacteria were spread on an agar plate containing penicillin. After incubation overnight at 370 > C, 10 bacterial colonies were observed on the plate. The colonies that are likely to be resistant to penicillin can be tested by
Solution
The colonies resistant to penicillin will grow on agar plates containing penicillin. Hence, option (3) is correct.
Question 120 1.33 / -0
The figure below demonstrates the growth curves of two organisms A and B growing in the same area. What kind of relation exists between A and B?
Solution
The relationship between A and B is of competition. Competition is harmful for both species A and B as illustrated by the graph.
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