Question 1 1.33 / -0
Let p(x) = a
0 + a
1 x + ….... + a
n x
(n) be a non-zero polynomial with integer coefficients. If p(
) = 0, then the smallest possible value of n is
Solution
x =
x
4 + 49 - 14x
2 = 128
x
4 - 14x
2 - 79 = 0
So, smallest possible value of n in x
n is 4.
Question 2 1.33 / -0
If cosec2 (α + β) - sin2 (β - α) + sin2 (2α - β) = cos2 (α - β) where α, β ∈ (0, π/2), then sin (α - β) is equal to
Solution
cosec
2 - sin
2 + sin
2 = cos
2 cosec
2 + sin
2 =
Minimum value of
is 1 and maximum value of
is 1.
Therefore, they will be equal for the value 1.
α + β = π/2 ---- (i) and 2α - β = 0 ---- (ii)
By adding (i) & (ii):
3α = π/2
α = π/6
β = π/3
Therefore, sin (α - β) = sin (π/6 - π/3) = -sin (π/6) =
Question 3 1.33 / -0
Ten trucks numbered 1 to 10, are carrying packets of sugar. Each packet weighs either 999 gm or 1000 gm and each truck carries only the packets of equal weight. The combined weight of 1 packet selected from the first truck, 2 packets from the second, 4 packets from the third, and so on, and 29 packets from the tenth truck is 1,022,870 gm. The trucks that have the lighter bags are
Solution
We can find the answer through options as the sum of weights of packet taken from trucks is 1,022,870 gm and its unit digit is 0. The trucks that have heavier bags have unit digit 0. So, in the trucks that have lighter bags, the sum of weights of bags must have unit digit 0.1 bags and total weight = 21 × 999 gm = 1998 gm7 bags and total weight = 27 × 999 = 128 × 999 gm = 127,872 gm
Question 4 1.33 / -0
An envelope has space for at most 3 stamps. If you are given three stamps of denomination 1 and 3 of denomination a, (a > 1), the least positive integer for which there is no stamp value is
Solution
So, using 1 stamp of value 1, the stamp value is 1.
Question 5 1.33 / -0
The maximum value M of 3x + 5x – 9x + 15x – 25x , as x varies over reals, satisfies
Solution
M = 3
x + 5
x – 9
x + 15
x – 25
x Suppose, a = 3
x , then a
2 = 3
2x = 9
x And b = 5
x , then b
2 = 5
2x = 25
x and 15
x = (3
x )(5
x ) = ab
So, M =
[2 - (a - 1)
2 - (b - 1)
2 - (a - b)
2 ]
So, maximum value of M is possible only when (a - 1)
2 + (b - 1)
2 + (a - b)
2 = 0
Therefore, maximum value of M =
× 2 = 1
So value of M is lies between 0 < M < 2
Question 6 1.33 / -0
The value of tan 81° – tan 63° – tan 27° + tan 9° is
Question 7 1.33 / -0
Find the value of the following function:
Solution
I =
I =
Since sin (x
3 ) + x
5 is an odd function,
then
= 0
Therefore, I =
= 0 + 2
2012 = 4024
Question 8 1.33 / -0
The number of ordered pairs (m, n), where m, n ∈ {1, 2, 3, ......, 50}, such that 6m + 9n is a multiple of 5 is
Solution
6m + 9n m = 6n will be 9 or 1.n must be 9.
Question 9 1.33 / -0
The number of rectangles that can be obtained by joining four of the twelve vertices of a 12-sided regular polygon is:
Solution
Number of diagonals passing through the centre = 66 C2 = 15
Question 10 1.33 / -0
The number of roots of the equation cos7 θ - sin4 θ = 1 that lie in the interval [0, 2π] is:
Solution
But
So,
,
⇒
Hence, number of the roots of the equation = 2
Question 11 1.33 / -0
The following figure shows the graph of a continuous function y = f(x) on the interval [1, 3]. The points A, B and C have coordinates (1, 1), (3, 2) and (2, 3), respectively, and the lines L
1 and L
2 are parallel, with L
1 being tangent to the curve at C. If the area under the graph of y = f(x) from x = 1 to x = 3 is 4 square units, then the area of the shaded region is:
Solution
Equation of ℓ
2 : y - 1 =
(x - 1)
⇒ 2y - 2 = x -1
⇒ 2y - x = 1
⇒
Slope of ℓ
1 = 1/2
Equation of ℓ
1 = y - 3 =
(x - 2)
⇒ 2y - 6 = x - 2
⇒ 2y - x = 4
Area under f(x) = 4
Shaded area = Area of trapezium DEFG - Area under f(x)
=
× 2 - 4
= 6 - 4 = 2
Question 12 1.33 / -0
How many six-digit numbers are there in which no digit is repeated, even digits appear at even places, odd digits appear at odd places and the number is divisible by 4?
Solution
3 × 3 × 4 × 4 × 5 × 2 = 1440
Question 13 1.33 / -0
Let P be an m × m matrix such that P2 = P. Then, (I + P)n equals
Question 14 1.33 / -0
In the xy-plane, three distinct lines l1 , l2 , l3 concur at a point (λ , 0). Further, the lines l1 , l2 , l3 are normal to the parabola y2 = 6x at the points A = (x1 , y1 ), B = (x2 , y2 ) and C = (x3 , y3 ), respectively. Then, we have
Solution
y = mx - 2am - am
3 Here, a = 3/2 through (λ, 0).
0 = mλ - 2am - am
3 m = 0, λ = 2a + am
2 λ > 2a
λ > 3
Question 15 1.33 / -0
The minimum distance between a point on the curve y = ex and a point on the curve y = loge x is
Solution
Point where slope of tangent is 1 of the curve y = e
x is
y' = e
x = 1
⇒ x = 0
∴ (0, 1)
and for y = Inx point is (1, 0)
∴ required minimum distance is
Question 16 1.33 / -0
Let
be three vectors in the xyz space such that
=
=
≠ 0. If A, B and C are points with position vectors
respectively, then the number of possible positions of the centroid of triangle ABC is
Question 17 1.33 / -0
Let ω be a cube root of unity not equal to 1. Then, the maximum possible value of |a + bω + cω2 |, where a, b, c ∈ {+1, -1}, is
Solution
|a + bω + cω2 |2 | = |2 + 2ω| = 2|ω2 | = 2
Question 18 1.33 / -0
The angles
,
and
of a triangle satisfy the equations 2 sin
+ 3 cos
= 3
and 3 sin
+ 2 cos
= 1. Then, angle
equals?
Solution
...(i)
...(ii)
Sum of squares of equations (i) and (ii),
⇒
If
Putting in equation (i) and (ii), we get
⇒
⇒
= .8 <
cos
as cos
Therefore,
= 150°
= 180° - 150° = 30°
Question 19 1.33 / -0
For a real number x, let [x] denote the largest integer less than or equal to x and {x} = x - [x]. The smallest possible integer value of n for which
exceeds 2013 is
Question 20 1.33 / -0
The number of integers n with 100 ≤ n ≤ 999 and containing at most two distinct digits is
Solution
Total number of integers from "100 to 999" = 999 - 99 = 900
Question 21 1.33 / -0
A parallel plate capacitor is charged fully by using a battery. Without disconnecting the battery, the plates are moved further apart. Then, the
Solution
As we know,
Capacitance
, where A = Area of plate and d = Distance between plates
If d increases, then capacitance C decreases.
Hence, energy stored U =
decreases.
Question 22 1.33 / -0
An ideal diatomic gas is heated at constant pressure. The ratio of the work done to the heat supplied is
Solution
Heat supplied Q = du +
W
PV = RT
PdV = RdT
dT =
Q = C
v dT + PdV
Q = C
v + PdV
Work done at constant pressure, W = PdV
(For diatomic gas, C
V =
)
Question 23 1.33 / -0
A plane electromagnetic wave propagating in the direction of the unit vector
with a speed c is described by electric and magnetic field vectors
and
, respectively. Which of the following relations (in SI units) between
and
can be ruled out on dimensional grounds alone?
Solution
We know that relation between electric field and magnetic field is given by
Where v = Velocity (m/s) and c = Speed of light (m/s)
So, based on dimensional ground only, option (1) satisfies above equation.
Question 24 1.33 / -0
A block of mass m is stationary on a rough plane of mass M inclined at an angle
to the horizontal while the whole set up is accelerating upwards at an acceleration a. If the coefficient of friction between the block and the plane is μ, then the force that the plane exerts on the block is
Solution
Since mass m is stationary on the inclined plane and whole system is accelerated with acceleration a upwards.
So, force on the block would be downward mg and ma.
From newton's action-reaction law, force exerted by plane would be m(g + a) upwards.
Question 25 1.33 / -0
A liquid drop placed on a horizontal plane has a near spherical shape (slightly flattened due to gravity). Let R be the radius of its largest horizontal section. A small disturbance causes the drop to vibrate with frequency v about its equilibrium shape. By dimensional analysis, the ratio
can be: (Here, σ is surface tension and ρ is density.)
Question 26 1.33 / -0
A conducting rod of mass m and length
l is free to move without friction on two parallel long conducting rails, as shown below. There is a resistance R across the the rails. In the entire space around, there is a uniform magnetic field B normal to the plane of the rod and rails. The rod is given an impulsive velocity V
0 .
Finally, the initial energy
Solution
Due to the negative work on rod, K.E. will decrease and finally become zero. But due to energy conservation, K.E. will convert into heat energy of resistor.
Question 27 1.33 / -0
Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is held at a temperature of 100°C, while the other one is kept at 0°C. If the two are brought into contact, then, assuming no heat loss to the environment, the final temperature that they will reach is
Solution
Since heat capacity increases with the temperature, the amount of heat transfer
is more at lower temperature than at higher temperature.
So, the final temperature will be more than 50
o C.
Question 28 1.33 / -0
A cylindrical steel rod of length 0.10 m and thermal conductivity 50 W. m-1 .K-1 is welded end to end to copper rod of thermal conductivity 400 W.m-1 .K-1 and of the same area of cross section but 0.20 m long. The free end of the steel rod is maintained at 100°C and that of the copper rod at 0°C. Assuming that the rods are perfectly insulated from the surroundings, the temperature at the junction of the two rods is
Solution
Resistance of steel,
R
s =
Resistance of copper,
R
c =
Both are connected in series, hence heat current in both will be the same.
So,
T = 20°C
Question 29 1.33 / -0
A bead of mass m is attached to the mid-point of a taut, weightless string of length l and is placed on a frictionless horizontal table.
Under a small transverse displacement x, as shown, if the tension in the string is T, then the frequency of oscillation is
Solution
Let the angle of T with the vertical be
, then F.B.D.
Therefore, 2T cos
= ma
Also,
Given ℓ >> x
Therefore,
+ x
2 Therefore, a =
(negative sign for restoring force)
Or a =
Also, this is similar to the equation of S.H.M, i.e.
a =
Therefore,
and f =
Therefore,
Question 30 1.33 / -0
The following travelling electromagnetic wave Ex = 0, Ey = E0 sin(kx + ω t), Ez = 2E0 sin(kx - ω t) is
Solution
From the equations of Ey and E2 , it is evident that wave is circularly polarised.
Question 31 1.33 / -0
A van der Waals gas obeys the equation of state
(V - nb) = nRT. Its internal energy is given by U = CT -
. The equation of a quasi-static adiabat for this gas is given by:
Solution
For adiabatic process,
dQ = 0 and dU - dW = 0
Here, dU = nC
v dT
Now, given, U = CT-
Therefore,
dU = CdT +
dV
Put this value of dU in dU = dW.
Therefore,
---(1)
Also, P =
Replace it in (1).
Therefore, CdT =
Therefore,
Integrating, we get
ℓn T
C/nR = ℓn (V - nb) + k
constant of integration)
Hence, ℓn(T
C/nR )(V - nb) = k
Or (T
C/nR )(V - nb) = constant
Question 32 1.33 / -0
Two masses m1 and m2 are connected by a massless spring of spring constant k and unstretched length l. The masses are placed on a frictionless straight channel which we consider our x-axis. They are initially at rest at x = 0 and x = l, respectively. At t = 0, a velocity of v0 is suddenly imparted to the first particle. At a later time t0 , the centre of mass of the two masses is at:
Solution
V
COM =
Also, X
COM = X
iCOM + V
COM t
Therefore, X
COM =
Question 33 1.33 / -0
If the speed (v) of the bob in a simple pendulum is plotted against the tangential acceleration (a), then the correct graph will be represented by
Solution
In SHM,
...........(1)
and
.............(2)
From above equations, we get
i.e. ellipse
Question 34 1.33 / -0
A speaker emits a sound wave of frequency f0 . When it moves towards a stationary observer with speed u, the observer measures a frequency f1 . If the speaker is stationary and the observer moves towards it with speed u, the measured frequency is f2 . Then,
Solution
⇒
is always greater than one.
Therefore, f
1 > f
2 .
Question 35 1.33 / -0
In the circuit shown below, the resistances are given in ohms and the battery is assumed ideal with emf equal to 3.0 volts. The resistor that dissipates the most power is
Solution
Power dissipated in R
1 is maximum as its current is maximum.
Question 36 1.33 / -0
The total radiative power emitted by a spherical blackbody with radius R and temperature T is P. If the radius is doubled and the temperature is halved, then the radiative power will be
Question 37 1.33 / -0
A nuclear decay is possible if the mass of the parent nucleus exceeds the total mass of the decay particles.
If M(A, Z) denotes the mass of a single neutral atom of an element with mass number A and atomic number Z, then the minimal condition that the β-decay
will occur is (m
e denotes the mass of the b-particle and the neutrino mass m
v can be neglected):
Solution
Energy equation for the beta decayA Z ) - (m(YA Z+1 ) + me )C2 A Z ) > m(YA Z+1 ) + me .......(1)A Z ) is the mass of the parent nuclei and m(YA Z+1 ) is the mass of the daughter nuclei.A Z ) + Zme A Z+1 ) + (Z + 1)me e > M(A, Z + 1) - (Z + 1)me + me
Question 38 1.33 / -0
An arrangement with a pair of quarter circular coils of radii r and R with a common centre C and carrying a current I is shown.
The permeability of free space is
. The magnetic field at C is
Solution
B due to arc =
Now,
B
total =
B
total =
out of the page
Question 39 1.33 / -0
Photons of energy 7 eV are incident on two metals A and B with work functions 6 eV and 3 eV, respectively. The minimum de Broglie wavelengths of the emitted photoelectrons with maximum energies respectively are
A and
B , where
A /
B is nearly
Solution
De Broglie wavelength is inversely proportional to the momentum P and P is proportional to the square root of kinetic energy.
Now, KE = Energy of photon - work function
So,
Question 40 1.33 / -0
The average distance between molecules of an ideal gas at STP is approximately of the order of
Solution
N = 40 × 6.023 × 10
23 = 24 × 10
24 Average distance =
= 1 nm
Question 41 1.33 / -0
The IUPAC name for the given compound is
Solution
The IUPAC name of the given compound is
Question 42 1.33 / -0
Which of the following has the highest bond angle?
Solution
(1) CH4 bond angle will be 109.5° as it has a tetrahedral molecular geometry with sp3 hybridisation.2 has a bond angle 180°, with sp hybridisation having linear shape.2 O has a bond angle 104.5°. It has sp3 hybridisation. Due to the presence of lone pair - lone pair replusions, bond angle reduces to 104.5°.2 has sp2 hybridisation with bond angle 120°. It has a bent shape.2 has the largest bond angle.
Question 43 1.33 / -0
The equilibrium constant for the reaction N
2 + 3H
2 2NH
3 at 400 K is 41. The equilibrium constant for the reaction
N
2 +
H
2 NH
3 at the same temperature will be close to
Solution
Let the equilibrium constant for the following reaction be K
1 .
Let the equilibrium constant for the following reaction be K
2 .
Squaring both sides of the equation (ii), we get
K
2 2 =
K
2 2 = K
1 K
2 =
K
2 =
K
2 = 6.4
Question 44 1.33 / -0
The newman projection of
is known as the
Solution
It is known as the Gauche conformer because angle between the same groups is 60°.
Question 45 1.33 / -0
For a one-electron atom, the set of allowed quantum numbers is:
Solution
For the given set of quantum numbers:
The value of n >
and
should have values from
to
.
Question 46 1.33 / -0
A zero order reaction, A
Product, with an initial concentration [A]
o has a half life of 0.2 s. If one starts with the concentration 2[A]
o ,
then the half life is
Solution
For a zero order reaction:
Since half life period is directly proportional to initial concentrated of reactant, on doubling the initial concentration, half life period is also doubled.
Hence, First half life period (t
1/2 )
1 = 0.2 s
Final half life period (t
1/2 )
2 = 0.4 s
Question 47 1.33 / -0
The pKa of a weak acid is 5.85. The concentrations of the acid and its conjugate base are equal at a pH of
Solution
∵ [Conjugate base] = [Acid]
pH = pK
a = 5.85
Question 48 1.33 / -0
The equilibrium constant for the following reactions are K
1 and K
2 , respectively.
2P(g) + 3Cl2(g)
2PCl3(g)
PCl3(g) + Cl2(g)
PCl5(g)
Then, the equilibrium constant for the reaction given below is:
2P(g) + 5Cl2(g)
2PCl5(g)
Solution
2P(g) + 3Cl
2 (g)
2PCl
3 (g) ...(i)
PCl
3 (g) + Cl
2 (g)
PCl
5 (g) ...(ii)
2P(g) + 5Cl
2 (g)
2PCl
5 (g) ...(iii)
On multiplying equation (ii) by 2 and adding in (i), we obtain equation (iii).
Therefore, K
3 = K
1 K
2 2
Question 49 1.33 / -0
The bond order of
is
Solution
The molecular electronic configuration of
is:
Bond order =
Question 50 1.33 / -0
The order of acidity of compounds I-IV is:
Solution
(1) All carboxylic acids are more acidic than phenols. The pK
a of all carboxylic acids is of the order 3-7, whereas the pK
a of all phenols is of the order 7-10. The lesser the pKa, the greater is the acidity. This is due to resonance stabilisation of phenoxide ion.
(2) Out of benzene sulphonic acid and benzoic acid, benzene sulphonic acid is more acidic because after de-protonation, it is stabilised by three equally stable resonating structures, whereas in benzoic acid, only 2 equally stable resonating structures are possible. Thus, the more stable the conjugate base, the more is the acidic strength. Hence, benzene sulphonic acid is more acidic.
Therefore, the correct order of acidic strengths is: I < III < II < IV
Question 51 1.33 / -0
In metallic solids, the number of atoms for the face centred and the body centred cubic unit cells, respectively are:
Solution
FCC unit cell:
Number of atoms in the FCC unit cell =
BCC unit cell:
Number of atoms in the BCC unit cell =
Question 52 1.33 / -0
Consider the reaction:
2 NO
2 (g)
2 NO(g) + O
2 (g).
In the given figure, identify the curves X, Y and Z associated with the three species in the reaction:
Solution
Therefore, (Z) is reactant, which is NO
2 , as the concentration of (Z) is decreasing with time.
Rate of disappearance of NO
2 = Rate of formation of NO
So, (X) is NO.
The rate of formation of O
2 is half the rate of formation of NO, hence (Y) is O
2 .
Question 53 1.33 / -0
Assuming ideal behaviour, the enthalpy and volume of mixing of two liquids, respectively are
Solution
For ideal solution,
H
mixing = 0. No heat should be absorbed or evolved during mixing. This is because there is no change in magnitude of the attractive forces in the two components present.
V
mixing = 0. No expansion or contraction on mixing. Volume of solution is the sum of volume of components present in solution.
Question 54 1.33 / -0
For a process to occur spontaneously,
Solution
According to second law of thermodynamics, for spontaneity
Question 55 1.33 / -0
To obtain a diffraction peak, for a crystalline solid with interplane distance equal to the wavelength of incident X-ray radiation, the angle of incidence should be
Solution
From Bragg's equation,
d =
and n = 1
⇒
Question 56 1.33 / -0
The adsorption isotherm for a gas is given by the relation x = ap/(1 + bp), where x is the number of moles of gas adsorbed per gram of the adsorbent, p is the pressure of the gas and a and b are constants. Then, x
Question 57 1.33 / -0
The compound which reacts with excess bromine to produce 2,4,6-tribromophenol is:
Solution
Reaction of salicylic acid with bromine water gives 2,4,6- tribromophenol. Initially, bromination occurs at the carbon bearing the -COOH group, followed by decarboxylation to give
o- bromophenol. But in phenol, the highly activating OH- group undergoes bromination at all three available positions, giving 2,4,6-tribromophenol as shown below:
Question 58 1.33 / -0
The compound that readily tautomerizes is:
Solution
Acetoacetic ester has active methylene group. Thus, it will tautomerize readily.
Question 59 1.33 / -0
In structure of Borax, the number of boron atoms and B-O-B units respectively are
Solution
Structure of Borax:
Number of Boron atoms = 4
Number of B-O-B bonds = 5
Question 60 1.33 / -0
The spin-only magnetic moments of [Fe(NH3 )6 ]3+ and [FeF6 ]3- in BM respectively are
Solution
In both [Fe(NH
3 )
6 ]
3+ and [FeF
6 ]
3- , the oxidation state of Fe is +3.
Electronic configuration of Fe
3+ = [Ar] 3d
5 , 4s
0 In Fe(NH
3 )
6 ]
3+ , NH
3 is a strong field ligand. Thus, pairing of electrons takes place.
Therefore, number of unpaired electrons = 1
Magnetic moment (μ) =
=
In case of [FeF
6 ]
3- , F
- is a weak field ligand. Therefore, no pairing of electrons occurs.
Thus, number of unpaired electrons = 5
Magnetic moment (μ) =
=
Question 61 1.33 / -0
The function of enzymes is to
Solution
Enzymes are macro-molecular biological catalysts. They accelerate chemical reactions by lowering activation energy.
Question 62 1.33 / -0
Which of the following diseases is not sexually transmitted?
Solution
Tuberculosis is not a sexually transmitted disease. Tuberculosis bacillus is transmitted through droplets expelled into the air by an infected person during normal breathing, while talking or sneezing, but especially when coughing.
Question 63 1.33 / -0
Sickle cell anaemia is caused by
Solution
Sickle-cell disease is caused by a single point mutation (a missense mutation) in the beta-haemoglobin gene that converts a GAG codon into GUG, which encodes the amino acid valine, rather than glutamic acid.
Question 64 1.33 / -0
Mitochondria are not associated with which of the following functions?
Solution
Protein glycosylation occurs in the ER and Golgi apparatus.
Question 65 1.33 / -0
Diabetes insipidus is due to
Solution
Diabetes insipidus can be caused by low or absent secretion of the water-balance hormone vasopressin from the pituitary gland of the brain, or by a poor kidney response to this chemical messenger, which is also called antidiuretic hormone.
Question 66 1.33 / -0
A ripe mango, kept with unripe mangoes, causes their ripening. This is due to the release of a gaseous plant hormone called
Solution
Ethylene affects fruit-ripening. Normally, when the seeds are mature, ethylene production increases and builds-up within the fruit, resulting in ripening of the fruits.
Question 67 1.33 / -0
A hormone molecule binds to a specific protein on the plasma membrane inducing a signal. The protein that binds to it is called a/an
Solution
A receptor is a protein molecule that receives chemical signals from outside a cell. These generally bind to ligands which are hormones. When such chemical signals bind to a receptor, they cause some form of cellular/tissue response, e.g. a change in the electrical activity of a cell.
Question 68 1.33 / -0
If the sequence of bases in sense strand of DNA is 5'-GTTCATCG-3', then the sequence of the bases in its RNA transcript would be
Solution
The base pairing follows the complementary base pairing rule. During transcription, thymine is replaced by uracil. Hence, adenine is hydrogen bonded to uracil by two hydrogen bonds. Guanine is hydrogen bonded to cytosine by three hydrogen bonds. The two strands are antiparallel.
Question 69 1.33 / -0
Bacteria can survive by absorbing soluble nutrients via their outer body surface, but animals cannot, because
Solution
Bacteria have a large surface area as compared to volume as against animals which have a small surface area to volume ratio. Hence, bacteria can survive by absorbing soluble nutrients via their outer body surface. Materials must be able to reach all parts of a cell quickly, and when volume is too large relative to surface area, diffusion cannot occur at sufficiently high rates to ensure this.
Question 70 1.33 / -0
If you compare adults of two herbivore species of different sizes, but from the same geographical area, the amount of faeces produced per kg body weight would be
Solution
A large gut size should provide more space, enabling a large microbial diversity to settle. Hence, if per kg body weight is seen, faeces will be more in the herbivore of smaller size.
Question 71 1.33 / -0
A couple went to a doctor and reported that both of them were "carriers" of a particular disorder, their first child was suffering from that disorder and that they were expecting their second child. What is the probability that the new child would be affected by the same disorder?
Solution
This is a case of autosomal recessive inheritance: Two unaffected people, who each carries one copy of the mutated gene for an autosomal recessive disorder (carriers), have a 25 percent chance with each pregnancy of having a child affected by the disorder.
Question 72 1.33 / -0
The fluid part of blood flows in and out of capillaries in tissue to exchange nutrients and waste materials. Under which of the following conditions will fluid flow out from the capillaries into the surrounding tissue?
Solution
There are two forces at work in capillaries that determine whether water stays as part of the blood in a blood vessel or if water leaves to become part of the interstitial fluid. When the arterial blood pressure exceeds the osmotic pressure, the fluid flows out of the capillaries.
Question 73 1.33 / -0
In humans, the composition of a zygote that will develop into a female is
Solution
The genetic composition of a normal female is 44 autosomes and 2 X chromosomes. Hence, 44 A + XX is the correct option.
Question 74 1.33 / -0
In which of the following sets of fractions will you find nucleic acids if you fractionate all the organelles from the cytoplasm of a plant cell?
Solution
Nucleic acids are found in the nucleus. In plant cells, DNA is also present in chloroplasts and mitochondria. Therefore, the nearest option is option (1).
Question 75 1.33 / -0
Circadian rhythm is an endogenously driven cycle for biochemical, physiological and behavioural processes. In humans, the approximate duration of this 'biological clock' is
Solution
Research has shown that human internal clock runs on a daily cycle of 24 hours, 11 minutes.
Question 76 1.33 / -0
As proposed by Watson and Crick in 1953, the original model of DNA is a
Solution
According to Watson and Crick, DNA is a double helical structure. The two strands are antiparallel i.e both have opposite polarity.
Question 77 1.33 / -0
The sequences of four DNA molecules are given below.
Which of these DNA molecules will have the highest melting temperature (Tm)?
Solution
In GC pairing, three H-bonds are present, whereas in AT pairing, two H-bonds are present. The DNA molecule with higher number of GC bases will have a higher melting temperature.
Question 78 1.33 / -0
In green leaves, the light and dark reactions occur in
Solution
In photosynthesis, the light-dependent reaction takes place on the thylakoid membranes. The inside of the thylakoid membrane is called the lumen and outside the thylakoid membrane is the stroma, where the light-independent reactions take place.
Question 79 1.33 / -0
Which of the following organs is/are capable of producing fructose in humans?
Solution
Seminal vesicles produce fructose. It is used to provide energy for swimming sperm cells.
Question 80 1.33 / -0
Species that are most effective at colonising new habitats show
Solution
Dispersal is when individuals or seeds move from one site to a breeding or growing site. Species that show high dispersal ability are most effective at colonising new habitats.
Question 81 1.33 / -0
The area of the region bounded by y = ||x - 3| - 4| - 5 and the x-axis is
Solution
y = ||x - 3| - 4| - 5
From the above graph, it is clear that area of the region y = ||x - 3| - 4| - 5 and x-axis =
BCD +
ACE + Quadrilateral (ABFG)
=
+
+
= 16 + 16 + 17 = 49 square units
Question 82 1.33 / -0
Two points are randomly chosen on the circumference of a circle of radius r. The probability that the distance between the two points is at least r is equal to
Solution
We want the distance between two points to be at least r. Now, when the points A, B are at distance r, then the angle made by arc BA is 60°.
Now as chord AB comes closer to centre, the length of chord AB is increased, that is, it is greater than r and the angle also increases i.e from 60° to 180° and now when chord AB moves way from centre, then the length of chord AB decreases. When chord AB reaches CD, the length of AB is equal to r and the angle changes from 180° to 60°.
So, the angle required for desired conditions = 2(180 - 60) = 240
Total angle made by the circle = 360°
So, required probability =
Question 83 1.33 / -0
Let (x, y) be a variable point on the curve 4x2 + 9y2 - 8x - 36y + 15 = 0. Then, min (x2 - 2x + y2 - 4y + 5) + max (x2 - 2x + y2 - 4y + 5) is
Solution
4x
2 + 9y
2 - 8x - 36y + 15 = 0
4(x
2 - 2x + 1) + 9(y
2 - 4y + 4) - 25 = 0
4(x - 1)
2 + 9(y - 2)
2 = 25
min ((x - 1)
2 + (y - 2)
2 ) + max ((x - 1)
2 + (y - 2)
2 )
=
Question 84 1.33 / -0
Consider
L =
+
+......+
R =
+
+......+
and l =
dx
Question 85 1.33 / -0
Suppose a, b are real numbers such that ab ≠ 0. Which of the following four figures represents the curve (y - ax - b)(bx2 + ay2 - ab) = 0?
Solution
y = ax + b and
Slope = a
For the line, intercept = b
Figure 1:
For line
a < 0, b > 0
Hence, the other figure cannot be an ellipse.
Figure 2:
For line
a > 0, b < 0
Hence, the other figure is a hyperbola.
Similarly, you can check the rest of the two options.
Question 86 1.33 / -0
The value of
min {|x - π|, cos
-1 (cos x)} dx min
is
Question 87 1.33 / -0
The locus of the point P = (a, b), where a and b are real numbers such that the roots of x3 + ax2 + bx + a = 0 are in arithmetic progression is a/an
Solution
Suppose roots of the equations are
Sum of the roots = 3
= -a
= -
...... (1), product of the root =
... (2)
and pair of product b =
3
- d
2 = b
b =
(From equation (2))
(Put value of
from equation (1))
Hence, locus of point P(a, b)
is a parabola whose vertex is on the y-axis.
Question 88 1.33 / -0
For each positive integer n, define f
n (x) = minimum
for 0 ≤ x ≤ 1. Let I
n =
f
n (x)dx, n ≥ 1. Then, I
n =
I
n is equal to
Question 89 1.33 / -0
The vertices of the base of an isosceles triangle lie on a parabola y2 = 4x and the base is a part of the line y = 2x – 4. If the third vertex of the triangle lies on the x-axis, its coordinates are
Solution
(x - 2)
2 = x
x
2 - 5x + 4 = 0
x = 1, 4
C(1, -2)
B(4, 4) as AB = AC
=
On solving, we get
Hence, vertex of the triangle = (
, 0) =
Question 90 1.33 / -0
Let n be a positive integer. For a real number x, let [x] denote the largest integer not exceeding x and {x} = x – [x]. Then,
dx is equal to
Question 91 1.33 / -0
n moles of a van der Waals gas obeying the equation of state (P +
)(V - nb) = nRT, where a and b are gas dependent constants, is made to undergo a cyclic progress that is depicted by a rectangle in the P-V diagram as shown in the figure. What is the hear absorbed by the gas in one cycle?
Solution
As we know:
For cyclic processes,
So,
= Area of the loop in P-V graph
= (P
1 - P
2 )(V
2 - V
1 )
Question 92 1.33 / -0
Two small identical speakers are connected in phase to the same source. The speakers are 3 m apart and at ear level. An observer stands at P, 4 m in front of one speaker as shown. The sound she hears is least intense when the wavelength is
and most intense when the wavelength is
.
Then, possible values of
and
are:
Solution
From figure, path difference = 5 - 4 = 1 m
For constructive interference:
n
= 1 m
For n = 1,
= 1 m
For destructive interference and n = 1
= 2 m
Question 93 1.33 / -0
A radioactive nucleus A has a single decay mode with half-life τ
A . Another radioactive nucleus B has two decay modes 1 and 2. If decay mode 2 were absent, the half-life of B would have been τ
A/2 . If decay mode 1 were absent, the half-life of B would have been 3T
A . If the actual half-life of B is τ
B , then the ratio
is
Question 94 1.33 / -0
A standing wave in a pipe with a length L = 1.2 m is described as
y(x, t) = y
0 sin
sin
Based on the above information, which one of the following statements is incorrect?
(Speed of sound in air is 300 ms
-1 )
Solution
At x = 0 and x = L, y = 0, which means nodes at x = 0, L and anti-node at x = L/2.
Pipe is closed at both ends as node at x = 0, L.
Also
So, wavelength = 1.2 m.
Thus, only option (4) is incorrect.
Question 95 1.33 / -0
Three equal charges +q are placed at the three vertices of an equilateral triangle centred at the origin. They are held in equilibrium by a restoring force of magnitude F(r) = kr directed towards the origin, where k is a constant. What is the distance of the three charges from the origin?
Solution
F(r) = kr
Now, F
net on a particle is 2F
q cos 30° due to the other two charges.
Also, r =
Therefore, a =
Replacing it in F
net , we get
F
net =
This is balanced by F(r).
Therefore,
Therefore, r =
Question 96 1.33 / -0
An ideal gas with constant heat capacity C
v =
nR is made to carry out a cycle that is depicted by a triangle in the figure given below:
The following statement is true about the cycle:
Solution
f = 3
For BA,
For AC,
Now,
Hence,
Question 97 1.33 / -0
Figure given below shows a small mass connected to a string, which is attached to a vertical post. If the ball is released when the string is horizontal as shown, the magnitude of the total acceleration (including radial and tangential) of the mass as a function of the angle θ is
Solution
Radical acceleration =
Tangential acceleration = gcos
Total acceleration =
Question 98 1.33 / -0
A rectangular loop of wire shown below is coplanar with a long wire carrying current I.
The loop is pulled to the right as indicated. What are the directions of the induced current in the loop and the magnetic forces on the left and the right sides of the loop?
Induced current Force on left side F orce on right side (A) Counter clockwise To the left To the right (B) Clockwise To the left To the right (C) Counter clockwise To the right To the left (D) Clockwise To the right To the left
Solution
Flux is inward and it is decreasing, as loop is going away from wire. Decrease of magnetic field induces emf that opposes the decrement of magnetic field.
Therefore, direction of induced current is clockwise.
Force on left side is in left and force on right side is in right.
Question 99 1.33 / -0
In the P-V diagram below, the dashed curved line is an adiabat.
For a process that is described by a straight line joining two points X and Y on the adiabat (solid line in the diagram), heat is
(Hint: Consider the variations in temperature from X to Y along the straight line)
Solution
From the given graph, we can say that up to a point where PV has the maximum value, absorption of heat takes place and after that, heat is released. So, option (3) is correct.
Question 100 1.33 / -0
The three processes in a thermodynamic cycle shown in the figure are : Process 1
2 is isothermal, Process 2
3 is isochoric (volume remains constant) and Process 3
1 is adiabatic. The total work done by the ideal gas in this cycle is 10 J. The internal energy decreases by 20 J in the isochoric process. The work done by the gas in the adiabatic process is –20 J. The heat added to the system in the isothermal process
is
Question 101 1.33 / -0
The addition of 0.643 g of a compound to 50 ml of benzene (density = 0.879 g ml-1 ) lowers the freezing point from 5.51°C to 5.03°C. If the freezing point constant (Kf ) for benzene is 5.12 K kg mol-1 , then the molar mass of the compound is approximately
Solution
W = d × V = 0.879 × 50 = 43.95 g
K
f = 5.12 K kg mol
-1 = 5120 K g mol
-1 As,
5.51 - 5.03 =
= 156 g mol
-1
Question 102 1.33 / -0
In the above sequence of reactions, the major products X and Y are
Solution
NOTE: 'X' is the major product during the bromination of phenol as ortho-bromophenol is also formed.
Question 103 1.33 / -0
(R)-2-bromobutane upon treatment with aq. NaOH gives
Solution
(R)-2-bromobutane upon treatment with aq. NaOH gives (S)-2-butanol with S
N 2 (inversion).
Question 104 1.33 / -0
Among the following, the species that is both tetrahedral and diamagnetic is
Solution
In Ni(CO)
4 , oxidation state of Ni is zero because CO is a neutral ligand.
The electronic configuration of Ni = [Ar] 3d
8 , 4s
0 Since CO is strong field ligand, therefore pairing of electrons takes place as shown below:
As in Ni(CO)
4 , hybridisation of Ni is sp
3 , so it has a tetrahedral structure and it is diamagnetic in nature due to absence of unpaired electrons.
Question 105 1.33 / -0
The numbers of possible enatiomeric pairs from the bromination of I and II, respectively, are
Solution
(1)
(2) During the bromination of 2-methyl but-1-ene, one enatiomeric pair
d- and
l- 1,2-dibromo-2-methyl butane are formed.
Question 106 1.33 / -0
Diborane is formed by the elements as shown in equation (1):
2B(s) + 3H
2 (g)
B
2 H
6 (g) ----------(1)
Given that:
The ΔH° for the reaction (1) is
Question 107 1.33 / -0
If
= -0.440 V and
= 0.770 V, then
is
Solution
Fe
+2 + 2e
- Fe E
o = -0.44
Fe
+3 + e
- Fe
+2 E
o = 0.770
Fe
+3 + 3e
- Fe E
o = ?
E
o =
= -0.037 Volt
Question 108 1.33 / -0
At 300 K, the vapour pressure of two pure liquids, A and B is 100 and 500 mm Hg, respectively. If in a mixture of A and B, the vapour pressure is 300 mm Hg, then the mole fractions of A in the liquid and in the vapour phase, respectively are
Solution
=
as
300 = (-400)X
A + 500
X
A =
∵ X
A + X
B = 1
Therefore, X
B =
Question 109 1.33 / -0
The compound X (C7 H9 N) reacts with benzensulphonyl chloride to give Y (C13 H13 NO2 S), which is insoluble inalkali. The compound X is
Question 110 1.33 / -0
The density and equivalent weight of a metal are 10.5 g cm–3 and 100, respectively. The time required for a current of 3 amp to deposit a 0.005 mm thick layer of the same metal on an area of 80 cm2 is closest to
Solution
As
,
m = d V
w = 10.5 × 80 × 5 × 10
-4 = 42 × 10
-2 42 × 10
-2 =
t = 135 s
Question 111 1.33 / -0
Instead of 3, if 2 bases coded for an amino acid, degeneracy of codons coding for the same amino acid would have
Solution
Instead of 3, if 2 bases coded for an amino acid, degeneracy of codons coding for the same amino acid would have increased.
Question 112 1.33 / -0
The condition varicose veins is swelling of veins, that occurs due to
Solution
The condition varicose veins is swelling of veins, that occurs due to loss of elasticity of the muscular layer. Varicose veins occurs when veins aren't functioning properly. Veins have one-way valves that prevent blood from flowing backward. When these valves fail, the blood begins to collect in the veins rather than continuing towards heart. Hence, the nearest answer is option 1.
Question 113 1.33 / -0
Insectivorous plants that mostly grow on swampy soil use insects as a source of
Solution
Insectivorous plants grow in soils that are deficient in nitrogen. In order to meet their nitrogen requirement, these plants prey on insects.
Question 114 1.33 / -0
When a person begins to fast, after some time, glycogen stored in the liver is mobilised as a source of glucose. Which of the following graphs best represents the change of glucose level (y-axis) in his blood, starting from the time (x-axis) when he begins to fast?
Solution
Glucose level initially starts to decrease but as beta oxidation and gluconeogenesis increase, the levels become constant for some time. When these reserves are also lost, glucose levels decrease again.
Question 115 1.33 / -0
Two students are given two different double-stranded DNA molecules of equal lengths. They are asked to denature the DNA molecule by heating. The DNA given to student A has the following composition of bases (A:G:T:C:35:15:35:15) while that given to student B is (A:G:T:C:12:38:12:38). Which of the following statements is true?
Solution
DNA molecule given to student A would denature faster than that given to student B. This is because the amounts of A and T are higher in DNA sample of student A. A and T are joined by 2 hydrogen bonds and hence, will break faster.
Question 116 1.33 / -0
Consider a locus with two alleles, A and a. If the frequency of AA is 0.25, then what will be the frequency of A under Hardy-Weinberg equilibrium?
Solution
The Hardy-Weinberg Equation: For a population in genetic equilibrium, p + q = 1.0 (The sum of the frequencies of both alleles is 100%.). Hence, the answer is 0.5.
Question 117 1.33 / -0
Rodents can distinguish between many different types of odours. The basis for odour discrimination is that
Solution
Rodents can distinguish between many different types of odours. This is because they have a small number of odorant receptors that bind to many different odorant molecules.
Question 118 1.33 / -0
For a human male, what is the probability that all the maternal chromosomes will end up in the same gamete?
Solution
Each gamete has 23 chromosomes. Hence, the probability that all the maternal chromosomes will end up in the same gamete is 1/23.
Question 119 1.33 / -0
Eco RI and Rsa I restriction endonucleases require 6 bp and 4 bp sequences, respectively for cleavage. In a 10 kb DNA fragment, how many probable cleavage sites are present for these enzymes.
Solution
For Ec
0 R
I × 10000 = 2.44 sites are present.
R
sal × 10000 = 39.06 (Using probability rules)
Question 120 1.33 / -0
Brown fat is specialised adipose tissue with abundant mitochondria and rich blood supply. Brown fat
Solution
BAT is thermogenic tissue. When activated, BAT cells take lipids and run them through the mitochondria to generate heat, rather than synthesise ATP.
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