Question 1 1.33 / -0
In a triangle, two vertices are (2, 3) and (4, 0), and its circumcentre is (2, z) for some real number z. The circumradius is
Question 2 1.33 / -0
The number of solutions to sin x = 6/x with 0 ≤ x ≤ 12
is
Solution
Clearly, curves meet each other twice in 2
- 3
4
- 5
6
- 7
8
- 9
10
- 11
Therefore, total number of solutions from 0 ≤ x ≤ 12
= 10.
Question 3 1.33 / -0
The value of
equals
Solution
I
N =
=
=
=
= 0 (as denominator
)
Question 4 1.33 / -0
The number of relations R from an m-element set A to an n-element set B satisfying the condition (a
1 b
1 ) ε R
1 (a
1 b
2 ) ε R
b
1 = b
2 for a ε A, b
1 , b
2 ε B is
Solution
The number of relations R from an m-element set A to an n-element set B = m C0 + m C1 + m C2 n2 + .......... m Cm nm = (1 + n)m ............ only one element from A
Question 5 1.33 / -0
Let a, b, c, d be the numbers in the set {1, 2, 3, 4, 5, 6} such that the curves y = 2x3 + ax + b and 2x3 + cx + d have no point in common. The maximum possible value of (a – c)2 + b – d is
Solution
If a - c = 0 and
Max. (a - c)
2 + (b - d) = 0 + 5 = 5
Question 6 1.33 / -0
Let n be a natural number and let a be a real number. The number of zeroes of x2n + 1 – (2n + 1)x + a = 0 in the interval [-1, 1] is
Solution
f(x) = x
2n + 1 - (2n + 1)x + a
f'(x) = (2n + 1)(x
2n - 1)
x = -1
point of local maxima
x = -1
point of local minima
Graph of y = f(x), when a = 0.
f(x) is decreasing on [-1, 1].
Thus, f(x) will have at most one root on [-1, 1] that too will happen if f(-1)
0 and f(1)
0, i.e. for a ∈ [-2n, 2n]
Hence, at most one for every value of a.
Question 7 1.33 / -0
Let S
n =
k denote the sum of the first n positive integers. The numbers S
1 , S
2 , S
3 ,
.... S
99 are written on 99 cards. The probability of drawing a card with an even number written on it is
Solution
S
n =
is even (n : 1, 2, ......... 99)
Sn is even only when n is a multiple of 4 or n is a multiple of 4 - 1.
Then, total number of favourable cases = 24 + 25 = 49
So, probability of drawing a card with an even number written on it = P[E] =
Question 8 1.33 / -0
Let S = {1, 2, 3, ...., n} and A = {(a, b) | 1
a, b
n} = S
S. A subset B of A is said to be a good subset if (x, x) ∈ B for every x ∈ S. Then, the number of good subsets of A is
Solution
Number of elements in B for a good set = nn - 1
Question 9 1.33 / -0
A circle touches the parabola y2 = 4x at (1, 2) and also touches its directrix. The y-coordinate of the point of contact of the circle and the directrix is:
Solution
y
2 = 4x
2y
= 4
m
T =
=
= 1
Circle
S +
= 0
(x + 1)
2 + (y -
)
2 +
(x + 1) = 0 ---(i)
Differentiate.
2(x + 1) + 2(y -
)
+
= 0
At x = 1, y = 2
4 + 2(2 -
) m
T +
= 0
= 2
- 8 ---(ii)
From equation of circle,
2
2 + (2 -
)
2 + 2
= 0
- 4
+ 8 + 2(2
- 8) = 0
= 8
=
Question 10 1.33 / -0
Let f: R
R be a differentiable function such that f(a) = 0 = f(b) and f'(a)f'(b) > 0 for some a < b. The minimum number of roots of f'(x) = 0 in the interval (a, b) is:
Solution
Given f is differentiable over R, hence it is continuous and differentiable on [a, b].
f'(a) f'(b) > 0
⇒ Either both are positive or both are negative
If both are positive from MVT:
If f'(a) > 0 ⇒
> 0 such that f(x) > f(a)
x
(a, a +
)
If f'(b) > 0 ⇒
> 0 such that f(x) < f(b)
x
(b -
, b)
Hence, we can picture the graph of f(x) as to cut x - axis atleast once in (a, b) and since it is continuous and odd, it becomes:
Thus f'(x) has atleast two roots.
Similarly, if both are negative:
Question 11 1.33 / -0
Consider the regions A = {(x, y) | x2 + y2 ≤ 100} and B = {(x, y) | sin (x + y) > 0 | in the plane. Then, the area of the region A ∩ B is:
Solution
x
2 + y
2 ≤ 100
inside of a circle
sin (x + y) > 0
⇒
⇒ x + y = c
equation of a line
Required area = Shaded region =
Question 12 1.33 / -0
Let ao = 0 and an = 3an - 1 + 1 for n ≥ 1. Then, the remainder obtained after dividing a2010 by 11 is:
Solution
a
n = 3a
n - 1 + 1
a
2010 = 3a
2009 + 1
= 3(3a
2008 + 1) + 1 = 3
2 a
2008 + 3 + 1
= 3
3 a
2007 + 3 + 3 + 1
.
.
.
= 3
2010 a
0 + (3 + 3 + ... + 3) + 1
(2009 times)
= 0 + 6027 + 1 = 6028
Remainder =
= 0
Question 13 1.33 / -0
All the points (x, y) in the plane satisfying the equation x2 + 2x sin(xy) + 1 = 0 lie on a/an
Solution
x
2 + (2x) sin(xy) + 1 = 0
2 sin(xy) =
L.H.S. = R.H.S. = 2 or sin(xy) = -1
Then, x = -1
sin(-y) = 1
sin y = -1
Hence, the equation represents a pair of straight lines.
Question 14 1.33 / -0
In triangle ABC, we are given that 3 sin A + 4 cos B = 6 and 4 sin B + 3 cos A = 1. Then, the measure of the angle C is
Solution
Squaring and adding both the equations, we get
9 + 16 + 24 sin (A + B) = 37
sin(A + B) =
A + B =
C =
(wrong)
A + B =
C =
= 30 degree because C =
does not follow equation 3 sin A + 4 cos B = 6
Question 15 1.33 / -0
Let [x] denote the largest integer not exceeding x and {x} = x – [x]. Then,
is equal to
Solution
I = 2012
dx
Using King Property on I = 2012
dx
⇒ 2I = 2012
⇒ 2I = 2012
⇒ I = 1006
Question 16 1.33 / -0
Let X be a non-empty set and let P(X) denote the collection of all subsets of X. Define f: X × P(X)
R by
f(x, A ∪ B)
Then, f(x, A ∪ B) equals
Question 17 1.33 / -0
Two line segments AB and CD are constrained to move along the x and y axes, respectively, in such a way that the points A, B, C and D are concyclic. If AB = a and CD = b, then the locus of the centre of the circle passing through A, B, C and D in polar coordinates is:
Solution
and similarly,
Polar coordinates of centre of the circle be
.
⇒
and
⇒ f =
Therefore,
Question 18 1.33 / -0
For x, t ∈ R, let p
t (x) = (sin t)x
2 - (2cos t)x + sin t be a family of quadratic polynomials in x with variable coefficients.
Let A(t) =
Which of the following statements are true?
1. A(t) < 0 for all t
2. A(t) has infinitely many critical points
3. A(t) = 0 for infinitely many t
4. A'(t) < 0 for all t
Solution
⇒
⇒ A'(t) =
At t =
, A(t) =
and A'(t) = 1
So, statements 1 and 4 are false.
Hence, option 2 is the correct answer.
Question 19 1.33 / -0
A box contains coupons labelled 1, 2, 3 ... n. A coupon is picked at random and the number x is noted. The coupon is put back into the box and a new coupon is picked at random. The new number is y. Then, the probability that one of the numbers x and y divides the other is (In the options below, [r] denotes the largest integer less than or equal to r.)
Solution
Let x = 1
Favourable outcomes = (1, 1), (1, 2) ... (1, n)
Number of favourable outcomes when x = 1 =
Therefore, number of favourable outcomes when x = 1 or y = 1 =
Number of favourable outcomes when x = 2 or y = 2 but x
1, y
1 =
Similarly, number of favourable outcomes when x = k or y = k but x, y
{1, 2 ... k - 1} =
So, probability =
=
Question 20 1.33 / -0
Let P be a closed polygon with 10 sides and 10 vertices (assume that the sides do not intersect except at the vertices). Let k be the number of interior angles of P that are greater than 180°. The maximum possible value of k is
Solution
The sum of angles of closed polygon with 10 sides is
.
So, the maximum number of possible obtuse angles is 7.
Question 21 1.33 / -0
Four metallic plates, each of surface area (of one side) A, are placed at a distance d apart from each other. The two outer plates are connected to a point P and the two inner plates to another point Q as shown in the figure.
Then, the capacitance of the system is
Solution
Potential difference between both the capacitors is the same.
So, we can say that these are in parallel combination.
As we find that C
eq = (Effective number of capacitors) C = 2 ×
Question 22 1.33 / -0
Two identical conducting spheres carry identical charges. If the spheres are set at a certain distance apart, they repel each other with a force F. A third conducting sphere, identical to the other two, but initially unchanged, is then touched to the one sphere and then to the other before being removed. The force between the original two spheres is now
Solution
Charge divides sphere B =
and
Then, on touching
sphere to A, division of charge
Now, force between
F' =
F' =
Question 23 1.33 / -0
A uniform non-deformable cylinder of mass m and radius R is rolling without slipping on a horizontal rough surface. The force of friction
Solution
There is no slide. So, V =
at the point of contact. Because of this, there will be no relative motion at this contact point. So, frictional force would be zero.
Question 24 1.33 / -0
A small body is released from a height H of an inclined plane. At the bottom of the plane is a loop of radius R as shown.
Ignoring friction, the minimum H required for the body to just complete the loop (that is, reach the point 0) is
Solution
From energy conservation, potential energy at height H = Kinetic energy at the lowest point of circular path.
mgH =
mv
2 ..............................(1)
To complete the circular motion, minimum velocity at lowest point will be v =
.................(2)
From eq. (1) and (2),
mgH =
m(5gR)
H =
R
Question 25 1.33 / -0
A planet orbits in an elliptical path of eccentricity 'e' around a massive star considered fixed at one of the foci. The point in space where it is closest to the star is denoted by P and the point where it is farthest is denoted by A. Let V
P and V
A be the respective speeds at P and A. Then,
Solution
Using conservation of angular momentum about S,
V
p (a - ae) = V
A (a + ae)
Now, solving,
Question 26 1.33 / -0
The ratio of the speed of sound to the average speed of an air molecule at 300 K and 1 atmospheric pressure is close to
Solution
=
So, the nearest option is 1.
Question 27 1.33 / -0
The potential energy of a point particle is given by the expression
. A dimensionless combination of the constants
,
and
is:
Solution
From the equation,
= L
So,
will be dimensionless.
Question 28 1.33 / -0
An unpolarised beam of light of intensity I0 passes through two linear polarisers making an angle of 30° with respect to each other. The emergent beam will have an intensity of
Solution
First polariser I
0 becomes I
0 /2.
Now, from Malus's law,
I = I
0 /2cos
2 I = I
0 /2(3/4)
⇒ Final intensity will be
.
Question 29 1.33 / -0
The circuit shown consists of a switch (S), a battery (B) of emf E, a resistance R, and an inductor L.
The current in the circuit at the instant the switch is closed, is
Solution
Using the equation,
Put t = 0, we get
I = 0
Just when the battery is closed, inductor provides infinite resistance to the current flow.
Therefore, current is zero initially.
Question 30 1.33 / -0
Three transparent media of refractive indices μ
1 , μ
2 and μ
3 are stacked as shown. A ray of light follows the path shown. No light enters the third medium.
Then,
Solution
At first, the incident light ray is deviated towards the normal.
Therefore,
Also, at second incidence, TIR takes place.
Therefore,
Also,
because for the same angle in medium
, angle in
medium is less.
Therefore,
Question 31 1.33 / -0
A bus driving along at 39.6 kmph is approaching a person who is standing at the bus stop, while honking repeatedly at an interval of 30 seconds. If the speed of the sound is 330 m/s, at what interval will the person hear the horn?
Solution
v = 39.6 km/hr = 11 m/s
t
1 =
and t
2 =
Hence, time interval,
Now,
seconds
Hence, t
2 = 29 seconds
Question 32 1.33 / -0
A piece of hot copper at 100°C is plunged into a pond at 30°C. The copper cools down to 30°C, while the pond, being huge, stays at its initial temperature. Then,
Solution
Using theory of entropy, as the temperature increases, randomness decreases.
Question 33 1.33 / -0
Two bulbs of identical volumes are connected by a small capillary and are initially filled with an ideal gas at temperature T. Bulb 2 is heated to maintain a temperature 2T while bulb 1 remains at temperature T. Assuming throughout that the heat conduction by the capillary is negligible, the ratio of final mass of the gas in bulb 2 to the initial mass of the gas in the same bulb is close to
Solution
Mole conservation n
1 + n
2 = n
Initial number of moles = n
1 = n
2 =
When temp of 1 vessel is T and the temperature of other is 2T.
Now, n
1 + n
2 = n
Mass of gas
n
1 Therefore,
Question 34 1.33 / -0
At 23°C, a pipe open at both ends resonates at a frequency of 450 hertz. At what frequency does the same pipe resonate on a hot day when the speed of sound is 4 percent higher than it would be at 23°C?
Solution
⇒
⇒
⇒ f
2 = 1.04 × 450
= 468 Hz
Question 35 1.33 / -0
In the circuit shown, the switch is closed at time t = 0.
Which of the graphs shown below best represents the voltage across the inductor, as seen on an oscilloscope?
Solution
As potential difference across inductor during current growth is given by |e| =
Question 36 1.33 / -0
Four students measure the height of a tower. Each student uses a different method and each measures the height many times. The data for each are plotted below. The measurement with highest precision is
Solution
Highest precision (sharpness) is maximum in graph (I).
Question 37 1.33 / -0
A blackbox (BB) which may contain a combination of electrical circuit elements (resistor, capacitor or inductor) is connected with other external circuit elements as shown below in the figure (a). After the switch (S) is closed at time t = 0, the current (I) as a function of time (t) is shown in the figure (b).
From this, we can infer that the blackbox contains
Solution
The given I-t graph is for L-R series circuit.
Question 38 1.33 / -0
A wheel of radius R with an axle of radius R/2 is shown in the figure and is free to rotate about a frictionless axis through its centre and perpendicular to the page. Three forces (F, F, 2F) are exerted tangentially to the respective rims as shown in the figure.
The magnitude of the net torque acting on the system is nearly
Solution
From the given figure, we can calculate the torque as follows:
+ FR + 2FR = 3.5FR
Question 39 1.33 / -0
A ray of light incident on a glass sphere (refractive index
) suffers total internal reflection before emerging out exactly parallel to the incident ray. The angle of incidence was
Solution
Now,
2 cos r =
r = 30° and i = 60°
So, incident angle = 60°
Question 40 1.33 / -0
The figure below shows pressure variation in two different sound waves in air with time at a given position. Both the figures are drawn to the same scale.
Which of the following statements is true?
Solution
From the given graph we can say that wave 1 has higher frequency than wave 2. We all know that wavelength is inversely proportional to the frequency. So, wave 1 has shorter wavelength than wave 2.
Question 41 1.33 / -0
Among
, the species that acts as a Bronsted acid as well as a Bronsted base is
Solution
According to Bronsted-Lowery acid-base theory:
(1) A Bronsted-Lowry acid is any species that is capable of donating a proton (H
+ ).
(2) A Bronsted-Lowry base is any species that is capable of accepting a proton (H
+ ).
(3) An amphoteric is any species, that can act as both a Bronsted-Lowry acid and a Bronsted-Lowry base.
Among the given species,
acts as both a Bronsted-Lowry acid and a Bronsted-Lowry base as shown below:
Bronsted base:
Bronsted acid:
Also,
are Bronsted-Lowry bases and
is a Bronsted-Lowry acid.
Question 42 1.33 / -0
The reaction of butanal with n -propylmagnesium bromide followed by hydrolysis gives a/an
Solution
The product formed is heptan-4-ol which is achiral as it does not have any stereo centre.
Question 43 1.33 / -0
The conjugate bases for HCO3 - and NH3 , respectively are
Solution
Conjugate base
Conjugate base
Question 44 1.33 / -0
The hydrogen ion concentration in a mixture of 10 ml of 0.1 M H2 SO4 and 10 ml of 0.1 M KOH solution in water is
Solution
Concentration of H
+ ions in H
2 SO
4 solution = 2 x 0.1 = 0.2 M
So, number of moles of H
+ ions in 10 ml H
2 SO
4 solution =
= 0.002
Concentration of OH
- ions in 0.1 KOH solution = 1 x 0.1 = 0.1 M
So, number of moles of OH
- ions in 10 ml KOH solution =
= 0.001
After mixing, remaining moles of H
+ ions = 0.002 - 0.001 = 0.001
So, concentration of H
+ ions in mixture of solutions =
Question 45 1.33 / -0
The most stable conformation of 2,3-dibromobutane is:
Solution
The conformation represented in option (3) is staggered conformation which is most stable due to least repulsions between the same groups.
Question 46 1.33 / -0
The major product in the following reaction is:
Question 47 1.33 / -0
Among the following graphs showing the variation of rate (K) with temperature (T) for a reaction, the one that exhibits Arrhenius behaviour over the entire temperature range is:
Solution
According to Arrhenius equation:
Taking natural log both sides, we get
In K =
+ ln A
Hence, we plot a graph of ln K vs 1/T, with slop
and intercept ln A.
It is clear from the above observation that the slop is negative and the intercept is positive. Thus, option (4) is correct.
Question 48 1.33 / -0
Doping silicon with boron produces a/an
Solution
As boron is trivalent impurity, it will produce p-type semiconductor.
Question 49 1.33 / -0
The concentration of a substance undergoing a chemical reaction becomes one-half of its original value after time t regardless of the initial concentration. The reaction is an example of a
Solution
For a first order reaction, half life period is independent of initial concentration of the reaction.
Question 50 1.33 / -0
In the nuclear reaction:
+ X,
'X' is:
Solution
In β-decay, a neutron is converted to a proton and the process creates an electron which leads to increase in atomic number by 1, while there is no change in atomic mass.
Question 51 1.33 / -0
For a first order reaction R
P, the rate constant is k. If the initial concentration of R is [R
o ], the concentration of R at any time 't' is given by the expression:
Solution
For a first order reaction:
kt = ln
Or
Or [R] = [R
o ]e
-kt
Question 52 1.33 / -0
Cyclohexene is reacted with bromine in CCl4 in the dark. The product of the reaction will be
Solution
The addition of cyclohexene to bromine in dark in the presence of CCl
4 gives 1,2-dibromocyclohexane. It is an example of electrophilic addition reaction in which bromine is polarised by the approaching π-bond in the cyclohexene.
The process of bromination is an example of ante addition.
Question 53 1.33 / -0
The rate of gas phase chemical reactions generally increases rapidly with the rise in temperature. This is mainly because the
Solution
With the increase in temperature, the molecules of gas move faster and 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. Energy distribution at two different temperatures is shown below.
Question 54 1.33 / -0
The major product of the following reaction is
Solution
The reaction is an example of Pinacole-Pinacolone reaction, with classical ring expansion rearrangement.
Question 55 1.33 / -0
The products formed in the oxidation of NaBH4 by I2 are
Solution
Na BH
4 + I
2 NaI + B
2 H
6 + H
2
Question 56 1.33 / -0
Boiling points in i – iii follows the order
Solution
The correct order of boiling points is
Question 57 1.33 / -0
The variation of solubility of four different gases (G
1 , G
2 , etc.) in a given solvent with pressure at a constant temperature is shown in the plot.
The gas with the highest value of Henry's law constant is:
Solution
According to Henry's law,
P = K
H .x (g) [where 'x' is mole fraction of gas]
x (g) = P ×
Slope =
It is clear from the above equation that the higher the value of K
H at a particular pressure, the lower is the solubility of the gas in the liquid.
Question 58 1.33 / -0
The numbers of lone pairs on Xe in XeF2 and XeF4 respectively are:
Solution
Structure of XeF
2 :
Number of Ione pairs on Xe = 3
Structure of XeF
4 :
Number of lone pairs on Xe = 2
Question 59 1.33 / -0
For isothermal reversible expansion of an ideal gas,
Solution
Since the process is isothermal,
Question 60 1.33 / -0
An ionic compound is formed between a metal M and a non metal Y. If M occupies half the octahedral voids in the cubic close packed arrangement formed by Y, the chemical formula of the ionic compound is
Solution
For CCP (cubic close packed arrangement):2 Y4 or MY2 .
Question 61 1.33 / -0
Plants are attracted to light through the hormonal action of which of the following?
Solution
The plant hormone auxin is responsible for a plant's ability to grow towards the light. The cells on the plant that are the farthest from the light have a chemical called auxin, that reacts when phototropism occurs. This causes the plant to have elongated cells on the farthest side from the light.
Question 62 1.33 / -0
Nucleotides are monomers of DNA. Each nucleotide consists of a
Solution
DNA is a polymer. The monomer units of DNA are nucleotides and the polymer is known as a "polynucleotide". Each nucleotide consists of a 5-carbon sugar (deoxyribose), a nitrogen containing base attached to the sugar and a phosphate group.
Question 63 1.33 / -0
Earthworms are bisexual but still cross-fertilisation is common. This is because
Solution
Earthworms are bisexual but still cross-fertilisation is common. This is because spermatozoa and ova mature at different times in the same earthworm.
Question 64 1.33 / -0
Wooden doors and windows swell up in the rainy season by
Solution
The process of adsorption of water by solid particles of a substance, without forming a solution, is called 'imbibition'. Wooden doors and windows adsorb water in humid rainy season and increase in their volume so that they are hard to open or close. This occurs by the process of imbibition.
Question 65 1.33 / -0
Peptic ulcers are caused by
Solution
Helicobacter pylori (H. pylori) is a bacteria that can cause a stomach infection and inflammation. This is the main organism that causes peptic ulcers.
Question 66 1.33 / -0
By which of the following mechanisms is glucose reabsorbed from the glomerular filtrate by the kidney tubule?
Solution
Glucose is present in blood plasma and glomerular filtrate, but not present in urine (normally). This is because glucose is selectively reabsorbed in the proximal convoluted tubule. It is reabsorbed from the filtrate into the blood by active transport (symport with Na+ ions).
Question 67 1.33 / -0
Plant roots are usually devoid of chlorophyll and cannot perform photosynthesis. However, three are exceptions. Which of the following plant roots can perform photosynthesis?
Solution
Tinospora has long, slender and hanging aerial adventitious roots. These roots are green and photosynthetic.
Question 68 1.33 / -0
In a food chain such as grass
deer
lion, the energy cost of respiration as a proportion of total assimilated energy at each level would be
Solution
With increase in every trophic level, the amount of energy decreases. Hence, the energy cost of respiration will increase with every trophic level.
Question 69 1.33 / -0
If the total number of photons falling per unit area of a leaf per minute is kept constant, then which of the following will result in maximum photosynthesis?
Solution
Rate of photosynthesis is fastest in blue and red lights. Hence, option 3 is correct.
Question 70 1.33 / -0
When a person is suffering from high fever, it is sometimes observed that the skin has a reddish tinge. Why does this happen?
Solution
This happens because of higher blood circulation to the skin. This helps to release more heat from the body.
Question 71 1.33 / -0
Conversion of the Bt protoxin produced by Bacillus thuringiensis to its active form in the gut of the insects is mediated by
Solution
Bt crystal toxins include the Cry proteins (crystal toxins) and Cyt proteins (cytolytic toxins). When these crystal proteins are exposed to the alkaline environment of the gut of susceptible larvae, the crystals are solubilised and proteolytically processed at the N-terminus and/or the C-terminus by mid-gut proteases to yield the active protease-resistant toxin.
Question 72 1.33 / -0
Estimate the order of the speed of propagation of an action potential or nerve impulse.
Solution
The mean conduction velocity of an action potential ranges from 1 metre per second (m/s) to over 100 m/s, and in general, increases with axonal diameter.
Question 73 1.33 / -0
The Na+ /K+ pump is present in the plasma membrane of mammalian cells where it
Solution
The sodium-potassium pump works by pumping two potassium ions into the cell and pumping out three sodium ions using the energy from an ATP molecule.
Question 74 1.33 / -0
Patients who have undergone organ transplants are given anti-rejection medications to
Solution
Immunosuppressive therapy used in transplantation prevents activation and proliferation of alloreactive T-lymphocytes.
Question 75 1.33 / -0
The process of organogenesis starts as soon as the three germ layers are formed in a developing embryo. The human brain is formed from the
Solution
Organogenesis begins with the process of neurulation. This neurulation begins with the formation of primitive streak in epiblast (ectoderm), which leads to the formation of neural tube (nerve cord).
Question 76 1.33 / -0
An electrode is placed in the axoplasm of a mammalian axon and another electrode is placed just outside the axon. The potential difference measured will be
Solution
When a nerve or muscle cell is at "rest", its membrane potential is called the resting membrane potential. In a typical neuron, this is about –70 millivolts (mV).
Question 77 1.33 / -0
Genetic content of a cell reduces to half during
Solution
The process of meiosis involves two divisions of the genetic material. The first division is called the reduction division – or meiosis I – because it reduces the number of chromosomes from 46 chromosomes or 2n to 23 chromosomes or n (n describes a single chromosome set). Hence, option (1) is the correct answer.
Question 78 1.33 / -0
The two enzymatic activities associated with RuBisCo are
Solution
RuBisCo is a bifunctional enzyme. It possesses both carboxylase and oxygenase activities. Both the activities are found on the same active site. The enzyme acts as an oxygenase during photorespiration and as a carboxylase during photosynthesis.
Question 79 1.33 / -0
In orange and lemon, the edible part of the fruit is
Solution
The outer ovary wall in citrus fruits becomes the thick spongy layer of the rind, while the inner ovary wall becomes very juicy with several seeds.
Question 80 1.33 / -0
In vertebrates, 'glycogen' is stored chiefly in
Solution
Glycogen is a readily mobilised storage form of glucose. The two major sites of glycogen storage are the liver and skeletal muscles.
Question 81 1.33 / -0
Define the sequence {a
n },
n > 0, where: a
n =
for
Then,
equals
Question 82 1.33 / -0
The sum of all the absolute values of the differences of the numbers 1, 2, 3,..., n, taken two at a time, i.e.
equals
Solution
S =
|i-j|
= |1 - 1| + |1 - 2| + ..... + |1 - n| =
= |2 - 2| + ...... + |2 - n| =
S =
= n +
1 C
3
Question 83 1.33 / -0
A polynomial P(x) with real coefficients has the property that P''(x) ≠ 0 for all x. Suppose P(0) = 1 and P'(0) = -1. What can you say about P(1)?
Solution
Since P''(x) ≠ 0 ⇒ a ≠ 02 + bx + c2 - x + 1
Question 84 1.33 / -0
Let f(x) =
for all x ≠ 1. Let f
1 (x) = f(x), f
2 (x) = f(f(x)) and generally f
n (x) = f(f
n - 1 (x)) for n > 1. Let P = f
1 (2)f
2 (3)f
3 (4)f
4 (5), then which of the following is a multiple of P?
Solution
P = f(2).f(f(3))f(f(f(4)))f(f(f(5))) =
=
= (3)(3)(f(4))f(f(5))
=
=
= (15)(5) = 75
375 is multiple of 75.
Question 85 1.33 / -0
Among all cyclic quadrilaterals inscribed in a circle of radius R with one of its angles equal to 120°, consider the one with maximum possible area. Its area is
Solution
Using sine formula,
Area = 2 ×
×
R × R =
R
2
Question 86 1.33 / -0
Suppose m, n are positive integers such that 6m + 2m+n .3W + 2n = 332. The value of the expression m2 + mn + n2 is
Solution
6m + 2m+n .3w + 2n = 332m exceeds 332.2 + mn + n2 = 4 + 6 + 9 = 19
Question 87 1.33 / -0
Let A = {θ ∈ R | cos2 (sin θ) + sin2 (cosθ) = 1} and B = {θ ∈ R | cos(sin θ) sin(cos θ) = 0}. Then, A ∩ B
Solution
A = {θ ∈ R | cos
2 (sinθ) = cos
2 (cosθ)}
A = {θ ∈ R | sinθ = nπ ± cosθ}
A = {θ ∈ R | sinθ ± cosθ = nπ}
A = {θ ∈ R | sinθ ± cosθ = 0}
∵ sinθ ± cosθ ∈ [-
,
]
⇒ A = nπ ±
.....(1)
B ≡ {θ ∈ R | cos (sinθ) sin(cosθ) = 0} by A
B ≡ {θ ∈ R | cos (± cosθ) sin(cosθ) = 0}
B ≡ {θ ∈ R |
sin(2cosθ) = 0}
∴ 2cosθ = pπ, p ∈ I
cosθ =
∴ p = 0 only
∴ cosθ = 0
θ = 2kπ ±
.......(2)
From equation (1) and (2)
So A ∩ B =
Question 88 1.33 / -0
The arithmetic mean and the geometric mean of two distinct 2-digit numbers x and y are two integers one of which can be obtained by reversing the digits of the other (in base 10 representation). Then, x + y equals
Solution
Suppose AM and GM be ab and ba respectively
AM = 10a + b
GM = 10b + a
= 10a + b, xy = (10b + a)
2 (x - y)
2 = (x + y)
2 - 4xy
= 4[10b + a]
2 - 4[10b + a]
2 (x - y)
2 = 22 × 18 (a + b) (a - b)
(x - y)
2 = 11 × 9 × 4 (a + b) (a - b)
This should be a perfect square of an integer, only possible case when a + b = 11
a - b = 1
Then, a = 6 and b = 5.
Then, x + y = 2(10a + b) = 2(10 × 6 + 5) = 130
Question 89 1.33 / -0
Let XY be the diameter of a semicircle with centre O. Let A be a variable point on the semicircle and B another point on the semicircle such that AB is parallel to XY. The value of ÐBOY for which the inradius of triangle AOB is maximum, is
Question 90 1.33 / -0
In a tournament with five teams, each team plays against every other team exactly once. Each game is won by one of the playing teams and the winning team scores one point, while the losing team scores zero. Which of the following is not necessarily true?
Solution
Let teams be T
1 , T
2 , T
3 , T
4 and T
5 .
Now, we can have 5 teams with the scores of 2 points each.
Matches are
This score board contradicts option D.
Therefore, D is not necessarily true.
Question 91 1.33 / -0
A solid uniform sphere having a mass M, radius R, and moment of inertia
MR
2 rolls down a plane inclined at an angle θ to the horizontal starting from rest. The coefficient of static friction between the sphere and the plane is μ. Then,
Solution
Torque balance about point A
We get acceleration
Now from balance of forces,
M
= Mgsin
-
So, from above
Question 92 1.33 / -0
Consider 1 kg of liquid water undergoing change in phase to water vapour at 100°C, the vapour pressure is 1.01 × 10
5 N-m
2 and the latent heat of vaporisation is 22.6 × 10
5 J kg
2 . The density of liquid water is 10
3 kg-m
-3 and that of vapour is
kg-m
-3 . The change in internal energy in this phase change is nearly
Solution
From the first law of thermodynamics,
22.6
10
5 =
Question 93 1.33 / -0
Consider three concentric metallic spheres A, B and C of radii a, b, c respectively where a is
Solution
(Given)
⇒
⇒
Now charge on C = -Q
Question 94 1.33 / -0
A simple pendulum is released from rest at the horizontally stretched position. When the string makes an angle θ with the vertical, the angle
which the acceleration vector of the bob makes with the string is given by
Solution
From energy conservation:
a
c =
a
t =
Now,
=
⇒
Question 95 1.33 / -0
A narrow parallel beam of light falls on a glass sphere of radius R and refractive index μ at normal incidence. The distance of the image from the outer edge is given by
Solution
Using
For the first refracting surface, we have
For the second refracting surface, we have
Replacing V
1 by
and solving for V
f ,
we get
First image is real and second is virtual.
Question 96 1.33 / -0
At time t = 0, a container has N0 radioactive atoms with a decay constant λ. In addition, c numbers of atoms of the same type are being added to the container per unit time. How many atoms of this type are there at t = T?
Solution
N
0 = Initial nucleon
At t = 0, N = N
0 Addition is at a constant rate
⇒
Integrating, we get
N =
Therefore,
Question 97 1.33 / -0
A constant amount of an ideal gas undergoes the cyclic process ABCA in the PV diagram shown below.
The path BC is isothermal. The work done by the gas during one complete cycle, beginning and ending at A, is nearly
Solution
⇒
= 0 ..................................................(1)
Process BC is an isothermal process.
So, P
C V
C = P
B V
B 500 × 2 = 200 × V
B V
B = 5m
3 = 200[V
B - V
A ] = 200[5 - 2] = 600 kJ .................................(2)
....................(3)
Therefore, net work done is (from all given equations)
Total work = -300 kJ
Question 98 1.33 / -0
A particle moves in a plane along an elliptic path given by
. At point (0, b), the x-component of velocity is u. The y-component of acceleration at this point is
Solution
⇒
Differentiating w.r.t time
u
x = u at (0, b)
u
y = 0
Again, diff. w.r.t time
Now, acceleration at (0, b) is
a
y =
Question 99 1.33 / -0
The figure below shows a circuit and its input voltage v
i as a function of time t.
Assuming the diodes to be ideal, which of the following graphs depicts the output voltage v
0 as a function of time t?
Solution
V
0 = V
i , when no current flow through 1 k
For negative values of V
i ,
i from D
1 = 0 (always)
i from D
2 = 0 up to 3 V
So, from 0 to -3 V, V
i = V
o From -3 to -4 V, V
o = -3 V
For positive values of V
i ,
i from D
1 = 0 up to 1 V
i from D
2 = 0 (always)
Hence, for 0 to 1 V, V
i = V
0 For 1 to 4 V, V
o = 1 V
Question 100 1.33 / -0
A metallic prong consists of 4 rods made of the same material, cross-section and same lengths as shown. The three forked ends are kept at 100
° C and the handle end is at 0°C. The temperature of the junction is
Solution
Let the temperature of the junction be T.
Then,
3w - 3T = T
T = 75°C
Question 101 1.33 / -0
When Co(II) chloride is dissolved in concentrated HCl, a blue solution is obtained. Upon dilution with water, the colour changes to pink because
Solution
Aqueous cobalt(II) chloride solution has a pink colour because of six water molecules surrounding one cobalt(II) ion which forms octahedral hexaaquacobalt(II) complex, [Co(H
2 O)
6 ]
2+ .
When sufficient amount of concentrated HCl is added to this solution, water molecules from the inner complex sphere are gradually displaced, and blue tetrachlorocobaltate (II) complex is formed which has a regular tetrahedron structure. This can be simplified as:
However, the pink colour is reestablished by diluting the solution with water. This observation can be explained on the basis of Le Chatelier's principle by considering the equilibrium that exists in the above chemical system.
In the case of aqueouscobalt(II) chloride solution, the equilibrium can be disturbed by changing chloride ion concentration. When we add hydrochloric acid, the equilibrium shifts to the right in order to consume extra chloride ions, and blue tetrachlorocobaltate(II) is formed. Contrary to it, adding more water leads to the blue complex's decomposition and formation of pink hexaaquacobalt(II) complex.
Question 102 1.33 / -0
In the following transformation
Reagents 1 and 2, respectively, are
Question 103 1.33 / -0
A metal is irradiated with light of wavelength 660 nm. Given that the work function of the metal is 1.0 eV, the de Broglie wavelength of the ejected electron is close to
Solution
Kinetic energy = h
v - Work function
KE =
KE = 0.878 eV
Question 104 1.33 / -0
The following data is obtained for a reaction, X + Y
Products.
The overall order of the reaction is
Solution
According to rate law:
Rate (R) = K[X]
x [Y]
y Order of reaction = x + y
By experiments (1) and (2):
x = 2
Similarly, by experiments (1) and (3):
y = 3
Thus, order of the reaction is: 2 + 3 = 5.
Question 105 1.33 / -0
The solubility product of Mg(OH)2 is 1.0 × 10-12 . If concentrated NaOH solution is added to a 0.01 M aqueous solution of MgCl2 , then pH at which precipitation of Mg2+ occurs is:
Solution
The dissociation of Mg(OH)
2 is given as:
The solubility product is given as:
K
sp = [Mg
2+ ][OH
- ]
2 Let [OH
- ] = x
x =
x =
x = 10
-5 pOH = −log([OH
- ])
pOH = −log(10
−5 ) = 5
At room temperature,
pOH + pH = 14
Or pH = 14 − 5 = 9
Question 106 1.33 / -0
A solution containing 8.0 g of nicotine in 92 g of water freezes 0.925 degrees below the normal freezing point of water. If the molal freezing depression constant Kf = 1.85°C kg mol-1 , then the molar mass of nicotine is
Solution
Change in freezing point can be calculated as:
M
B = 173.91 g/mol
≈ 174 g/mol
[Note: Actual molar mass of nicotine is 162 g/mol.]
Question 107 1.33 / -0
In the following conversion,
the major products X and Y, respectively are
Question 108 1.33 / -0
Emulsification of 10 ml of oil in water produces 2.4 × 1018 droplets. If the surface tension at the oil – water interface is 0.03 Jm-2 and the area of each droplet is 12.5 × 10 - 16 m2 , then the energy spent in the formation of oil droplets is
Solution
Total droplets = 2.4 × 1018 18 × 12.5 × 1016 2 m2 Energy consumption2
Question 109 1.33 / -0
The standard electrode potential of Zn2+ /Zn is –0.76 V and that of Cu2+ /Cu is 0.34 V. The emf (V) and the free energy change (kJ/mol), respectively for a Daniel cell will be
Question 110 1.33 / -0
In aqueous solution, [Co(H2 O)6 ]2+ (X) reacts with molecular oxygen in the presence of excess liquor NH3 to give a new complex Y. The number of unpaired electrons in X and Y, respectively are
Question 111 1.33 / -0
E. coli has an optimal temperature of 37°C. Which of the following is an INCORRECT explanation for this?
Solution
Cell membrane becomes more permeable at high temperature. E. coli has an optimal temperature of 37°C. At this temperature, all the processes required for synthesising new cells occur at maximal rate. Hence, option 1 is incorrect.
Question 112 1.33 / -0
Energetically unfavourable reactions occur in human cells through
Solution
Cells use a strategy called reaction coupling, in which an energetically favourable reaction (like ATP hydrolysis) is directly linked with an energetically unfavourable (endergonic) reaction. The linking often happens through a shared intermediate, meaning that a product of one reaction is "picked up" and used as a reactant in the second reaction.
Question 113 1.33 / -0
The restriction endonuclease EcoR-I recognises and cleaves DNA sequence as shown below:
Solution
Let's start with 10 kb fragment, and go through from one end to the other, and mark every time a 'G' occurs. Theoretically those could all be the beginning of EcoRI sites. There will be approximately 2500 such 'G's. Now to match the pattern, we need an 'A' in the second position, so mark all those sites that contain the dinucleotide 'GA', there will be about 2500 x 0.25 = 625 of them. The next match we need is position 3, another 'A', so in a similar fashion, let's go through the positions of all those GA dinucleotides, and just mark the ones that have an 'A' at the next position. So that will be about 625 x 0.25, or approximately 156 'GAA's distrubuted randomly along the 10 kb fragment.
Question 114 1.33 / -0
Insects constitute the largest animal group on earth. About 25-30% of the insect species are known to be herbivores. In spite of such huge herbivore pressure, globally, green plants have persisted. One possible reason for this persistence is that
Solution
Change in preference of herbivores for plants has enabled the survival of plants for long.
Question 115 1.33 / -0
The following DNA sequence (5 '
3') specifies part of a protein coding sequence, starting from position I. Which of the following mutations will give rise to a protein that is shorter than the full-length protein?
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A T G C A A G A T A T A G C T
Solution
An open reading frame (ORF) is the part of a reading frame that has the potential to be translated. An ORF starts with ATG and ends with TAA or TGA. So, insertion of A between 10 and 11 will lead to a stop codon and premature termination of transcription.
Question 116 1.33 / -0
You mark two ink-spots along the height at the base of a coconut tree and also at the top of the tree. When you examine the spots next year when the tree has grown taller, you will see that
Solution
Tree trunk growth is coordinated with the increase of tree height and width. When buds begin opening in the early Spring, cells in the trunk and limbs get the signal to increase in girth by dividing and in height by elongating.
Question 117 1.33 / -0
E.coli was pulsed with tritiated thymidine for 5 min and then transferred to normal medium. Which of the following observations would be correct after one cell division?
Solution
DNA replication is semi conservative. Hence, when DNA is transferred to the normal medium, only one strand will be radioactive.
Question 118 1.33 / -0
The length of one complete turn of a DNA double helix is
Solution
One complete 360 degree turn of the DNA helix covers 10 base in length and a physical distance of 34 angstrom.
Question 119 1.33 / -0
Presence of plastids in plasmodium suggests it is a
Solution
An apicoplast is a derived non-photosynthetic plastid found in most Apicomplexa, including malaria parasites such as Plasmodium falciparum . It originated from an alga through secondary endosymbiosis.
Question 120 1.33 / -0
Lions in India are currently restricted to Gir, Gujarat. Efforts are being made to move them to other parts of the country. This is because they are most susceptible to extinction due to infectious diseases under the following conditions when present as
Solution
Animals present as one large population are most susceptible to extinction due to infectious diseases.
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