Test: BITSAT Past Year Paper- 2015
150 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers | Test: BITSAT Past Year Paper- 2015
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An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth. The height (h) of the satellite above the earth’s surface is (Take radius of earth as Re)
The escape velocity from earth is given by
ve =
The orbital velocity of a satellite revolving around earth is given by
where, Me = mass of earth, Re = radius of earth, h = height of satellite from surface of earth.
By the relation GMe = gR2s
So, 
Dividing equation (i) by (ii), we get
Given, 
Squaring on both side, we get
or Re + h = 2Re i.e., h = Re
In figure, two blocks are separated by a uniform strut attached to each block with frictionless pins. Block A weighs 400N, block B weighs 300N, and the strut AB weigh 200N. If µ = 0.25 under B, determine the minimum coefficient of friction under A to prevent motion.
Consider FBD of structure.
Applying equilibrium equations,
Av + Bv = 200 N … (i)
AH = BH … (ii)
From FBD of block B,
BH + FB cos 60° – NB sin 60° = 0
NB cos 60° – BV – 300 + FB sin 60° = 0
FB = 0.25 NB
BH – 0.74 NB = 0 … (iii)
– BV + 0.71 NB = 300 … (iv)
FBD of block A
FA – AH = 0
NA – AV = 400 … (v)
FA = µA NA
∴ µANA – AH = 0 … (vi)
On solving above equations, we get NA = 650 N, FA = 260N, FA = µANA
Two tuning forks with n atur al fr equen cies 340 Hz each move relative to a stationary observer. One fork moves away from the observer, while the other moves towards the observer at the same speed. The observer hears beats of frequency 3 Hz. Find the speed of the tuning forks.
Let v = speed of sound and vS = speed of tuning forks. Apparent frequency of fork moving towards the observer is 
Apparent frequency of the fork moving away from the observer is 
If f is the number of beats heard per second. th en f = n1 – n2
putting v = 340 m/s, f = 3, n = 340 Hz we get,
The displacement of a particle is given at time t, by: x = A sin (-2ωt ) + B sin 2ωt Then,
The displacement of the particle is given by:
This motion represents SHM with an amplitude: 
and mean position B/2.
A ray parallel to principal axis is incident at 30° from normal on concave mirror having radius of curvature R. The point on principal axis where rays are focussed is Q such that PQ is
From similar triangles,
or 
Thus
PQ = PC – QC = 
A solid sph ere of r adius R has a charge Q distributed in its volume with a charge density ρ = kra, where k and a are constants and r is the distance from its centre. If the electric field at r = R/2 is 1/8 times that at r = R, the value of a is
Using Gauss's law, we have
For 
Given, E2 = E1/8
or 
∴ 
or a = 2.
A charged particle moving in a uniform magnetic field and losses 4% of its kinetic energy. The radius of curvature of its path changes by
As we know F = qvB = mv2/r
And KE = k = 1/2 mv2
Therefore percentage changes in radius of path,
Calculate th e wavelength of ligh t used in an interference experiment from the following data : Fringe width = 0.03 cm. Distance between the slits and eyepiece through which the interference pattern is observed is 1m. Distance between the images of the virtual source when a convex lens of focal length 16 cm is used at a distance of 80 cm from the eyepiece is 0.8 cm.
Given: fringe with β = 0.03 cm, D = 1 m = 100 cm
Distance between images of the source = 0.8 cm.
Image distance v = 80 cm
Object distance = u
Using mirror formula, 
⇒ u = 20 cm
Magnification, 
Magnification 
Fringe width 
or, 
Therefore, wavelength of light used λ = 6000 Å
The masses of blocks A and B are m and M respectively. Between A and B, there is a constant frictional force F and B can slide on a smooth horizontal surface. A is set in motion with velocity while B is at rest. What is the distance moved by A relative to B before they move with the same velocity?
For th e blocks A and B FBD as sh own below
Equations of motion
Relative acceleration, of A w.r.t. B,
Initial relative velocity of A w.r.t. B, uAB = v0
using equation v2 = u2 + 2as
i.e., Distance moved by A relative to B
An elastic string of un stretched length L and force constant k is stretched by a small length x. It is further stretched by another small length y. The work done in the second stretching is
In the string elastic force is conservative in nature.
∴ W = – ΔU
Work done by elastic force of string,
W = – (UF – Ui) = Ui – UF.
Therefore, the work done against elastic force 
A body is thrown vertically upwards from A, the top of the tower, reaches the ground in time t1. If it is thrown vertically downwards from A with the same speed, it reaches the ground in time t2. If it is allowed to fall freely from A, then the time it takes to reach the ground is given by
Let the body is projected ver tically upwards from A with a speed u0.
Using equation , s = ut + (1/2)at2
For case (1) – h = 
For case (2) – h = 
Subtracting eq (2) from (1), we get
Putting the value of u0 in eq (2), we get
For case 3, u0 = 0, t = ?
Comparing eq. (4) and (5), we get
0.5 mole of an ideal gas at constant temperature 27°C kept inside a cylinder of length L and crosssection area A closed by a massless piston.
The cylinder is attached with a conducting rod of length L , cross-section ar ea (1/9) m2 and thermal conductivity k, whose other end is maintained at 0°C. If piston is moved such that rate of heat flow through the conducing rod is constant then velocity of piston when it is at height L/2 from the bottom of cylinder is : [Neglect any kind of heat loss from system ]
work done per unit time = 
⇒ 
⇒ 
⇒ 
A conducting square loop is placed in a magnetic field B with its plane perpendicular to the field. The sides of the loop start shrinking at a constant rate α. The induced emf in the loop at an instant when its side is ‘a’ is
At any time t, the side of the square a = (a0 – α t), where a0 = side at t = 0.
At this instant, flux through the square :
⇒ E = – B.2 (a0 – α t) (0 – α) = +2αaB
Th e beam of light h as th r ee wavelength s 4144Å, 4972Å and 6216 Å with a total intensity of 3.6 × 10–3 Wm2 equally distributed amongst the three wavelengths. The beam falls normally on the area 1 cm2 of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in 2s.
As we know, thresh old wavelength 
⇒ λ0 = 5404 Å
Hence, wavelength 4144 Å and 4972 Å will emit electron from the metal surface.
For each wavelength energy incident on the surface per unit time = intensity of each × area of the surface wavelength 
Therefore, energy incident on the surface for each wavelength in 2s
E = (1.2 × 10–7) × 2 = 2.4 × 10–7 J
Number of photons n1 due to wavelength 4144 Å
Number of photons n2 due to the wavelength 4972 Å
Therefore total number of photoelectrons liberated in 2s,
N = n1 + n2
= 0.5 x 1012 + 0.575 x 1012
= 1.075 x 1012
A squar e gate of size 1 m × 1m is h inged at its mid-point. A fluid of density ρ fills the space to the left of the gate. The force F required to hold the gate stationary is
The net force acting on the gate element of width dy at a depth y from the surface of the fluid, is
dy = p0 + ρgy - p0) x 1dy
= ρgydy
Torque about the hinge is
Net torque experienced by the gate is
⇒ F = ρg/6
i.e., The force F required to hold the gate stationary is pg/6
When 0.50 Å X-r ays strike a material, the photoelectrons from the k shell are observed to move in a circle of radius 23 mm in a magnetic field of 2 × 10–2 tesla acting perpendicularly to the direction of emission of photoelectrons. What is the binding energy of k-shell electrons?
As we know,
F = qvB = m
The kinetic energy of the photoelectron 
= 
= 2.97 × 10–15 J
Energy of the incident photon = hc/λ = 12.4/0.50 = 24.8keV
Therefore, Binding energy = 24.8 – 18.6 = 6.2 keV
In CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. If the base resistance is 1kΩ and the current amplification of the transistor is 100, the input signal voltage is
Given : Voltage across the collector V0 = 2 V; collector resistance, Rc = 2 × 103 Ω; Base resistance RB = 1 × 103 Ω; Input signal voltage, Vi = ?
V0 = ICRC = 2
Current gain 
At the corners of an equilateral tri angle of side a (1 metre), three point charges are placed (each of 0.1 C). If this system is supplied energy at the rate of 1 kw, then calculate the time required to move one of the mid-point of the line joining the other two.
Initial potential energy of the system 
Let charge at A is moved to mid-point O, Then final potential energy of thhe system 
Work done = Uf – Ui = 18 × 107 J
Also, energy supplied per sec = 1000 J (given)
Time required to move one of the mid-point of the line joining the other two
A vessel of volume 20L contains a mixture of hydrogen and helium at temperature of 27°C and pressure 2 atm. The mass of mixture is 5g. Assuming the gases to be ideal, the ratio of mass of hydrogen to that of helium in the given mixture will be
Let ther e are n1 moles of hydrogen and n2 moles of helium in the given mixture. As Pv = nRT
Then the pressure of the mixture 
⇒ 
or, 
or, n1 + n2 = 1.62 ... (1)
The mass of the mixture is (in grams)
n1 × 2 + n2 × 4 = 5
⇒ (n1 + 2n2) = 2.5 ... (2)
Solving the eqns. (1) and (2), we get n1 = 0.74 and n2 = 0.88
Hence, 
The resistance of a wire is R. It is bent at the middle by 180° and both the ends are twisted together to make a shorter wire. The resistance of the new wire is
Resistan ce of wire (R) = ρ(l/A)
If wire is bent in the middle then 
∴ New resistance, 
In a YDSE, the light of wavelength λ = 5000 Å is used, which emerges in phase from two slits a distance d = 3 × 10–7m apart. A transparent sheet of thickness t = 1.5 × 10–7m refractive index μ = 1.17 is placed over one of the slits. what is the new angular position of the central maxima of the interference pattern, from the centre of the screen? Find the value of y.
The path difference when transparent sheet is introduced Δx = (μ – 1)t
If the central maxima occupies position of nth fringe, then (μ – 1)t = n λ = d sin θ
Therefore, angular position of central maxima θ = sin-1 0.085 = 4.88°≈ 4.9
For small angles, sin θ ≈ θ ≈ tan θ
⇒ 
The position of a projectile launched from the origin at t = 0 is given by 
m at t = 2s. If the projectile was launched at an angle θ from the horizontal, then θ is (take g = 10 ms–2)
From question,
Horizontal velocity (initial), ux = 40/2 = 20m/s
Vertical velocity (initial), 50 
⇒ 
or, 50 = 2uy – 20
or, 
Water is flowing on a horizontal fixed surface, such that its flow velocity varies with y (vertical direction) as 
If coefficient of viscosity for water is h, what will be shear stress between layers of water at y = a.
Newton’s law of viscosity, 
Stress = 
At y = a, stress = 
A load of mass m falls from a h eight h on to the scale pan hung from the spring as shown in the figure. If the spring constant is k and mass of the scale pan is zero and the mass m does not bounce relative to the pan, then the amplitude of vibration is
According to energy conservation principle,
If, x1 is maximum elongation in the spring when the particle is in its lowest extreme position. Then,
Amplitude A = X1 – X0 (elongation in spring for equilibrium position)
In an ore containing uranium, the ratio of U238 to Pb206 is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of U238. Take the half-life of U238 to be 4.5 × 109 yr.
Let the initial mass of uranium be M0
Final mass of uranium after time t, M = 3/4(M0)
According to the law of radioactive disintegration.
⇒ t = 1.867 × 109 yr.
A direct current of 5A is superposed on an alternating current I = 10 sin ωt flowing through the wire. The effective value of the resulting current will be
Total carrent, 1 = (5 + 10 sin ωt)
But, 
and 
So, 
A planoconvex lens fits exactly in to a planoconcave lens. Their plane surface are parallel to each other. If the lenses are made of different materials of refractive indices µ1 & µ2 and R is the radius of curvature of the curved surface of the lenses, then focal length of combination is
If F be the equivalent focal length, then
A thin rod of length 4l and mass 4m is bent at the points as shown in figure. What is the moment of inertia of the rod about the axis passes through point O and perpendicular to the plane of paper?
Total moment of inertia
= I1 + I2 + I3 + I4 = 2I1 + 2I2
= 2(l1 + l2) [I3 = I1, I1 = I4]
Now, I2 = I3 = MI2/3
Using parallel axes theorem, we have
Putting all values we get
Moment of inertia, 
One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.
For 2nd line of Balmer series in hydrogen spectrum 
which is satisfied by n = 12 → n = 6.
Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1 and R2, respectively. The ratio of masses of X and Y is
When a charge particle is allowed to move in a uniform magnetic field, then it describes spiral or circular path
Centripetal force, mv2/R = qvB ∴ 
Hence, 
or, m ∝ R2
[∵ V, q and B are constant]
or, 
A glass capillary tube of internal radius r = 0.25 mm is immersed in water. The top end of the tube projected by 2 cm above the surface of the water. At what angle does the liquid meet the tube? Surface tension of water = 0.7 N/m.
Water wets glass and so the angle of contact is zero.
For full rise, neglecting the small mass in the meniscus
As the tube is only 2 cm above the water and so, water will rise by 2 cm and meet the tube at an angle such that,
The liquid will meet the tube at an angle, θ ≌ 70°
A particle of mass 2 m is projected at an angle of 45° with the horizontal with a velocity of 20√2 m/s. After 1s, explosion takes place and the particle is broken into two equal pieces. As a result of explosion, one part comes to rest. The maximum height from the ground attained by the other part is
Given : Initial velocity u0 = 20√2 m/s; angle of projection θ = 45°
Therefore horizontal and vertical components of initial velocity are ux = 20√2 cos 45°= 20m/s and uy = 20√2 sin 45°= 20m/s
After 1s, horizontal component remains unchanged while the vertical component becomes vy = uy – gt
Due to explosion, one part comes to rest.
Hence, from the conservation of linear momentum, vertical component of second part will become v'y = 20m/s.
Therefore, maximum height attained by the second part will be H = h1 + h2
Using, h = ut + (1/2)at2
a = g = 10 m/s2
H = 20 + 15 = 35 m
A 2 m wide truck is moving with a uniform speed v0 = 8 m/s along a straight horizontal road. A pedestrain starts to cross the road with a uniform speed v when the truck is 4 m away from him. The minimum value of v so that he can cross the road safely is 
Let the man starts crossing the road at an angle q as shown in figure. For safe crossing the condition is that the man must cross the road by the time the truck describes the distance 4 + AC or 4 + 2cot θ.
...(i)
For minimum v, 
or 
or 2 cos θ – sin θ = 0
or tan θ = 2
From equation (i), 
A neutron moving with speed v makes a head on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of the neutron for which inelastic collision takes place is
Let speed of neutr on before collision = V Speed of neutron after collision = V1
Speed of proton or hydrogen atom after collision = V2
Energy of excitation = DE From the law of conservation of linear momentum, mv = mv1 + mv2 ...(1)
And for law of conservation of energy, 
...(2)
From squaring eq. (i), we get
...(3)
From squaring eq. (ii), we get
...(4)
From eqn (3) & (4)
As, v1 – v2 must be real, 
⇒ 
The minimum energy that can be absorbed by the hydrogen atom in the ground state to go into the excited state is 10.2 eV. Therefore, the maximum kinetic energy needed is 
2 x 10.2 = 20.4 eV
Vertical displacement of a Planck with a body of mass m on it is varying according to law y = sin ωt + √3 cos ωt. The minimum value of w for which the mass just breaks off the Planck and the moment it occurs first after t = 0, are given by
From, figure, 
∴ y = 2 sin 
amax = -2ω2 = g
For which mass just breaks off the plank 
This will be happen for the first time when 
∴ 
A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charge to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is
From the figure.
The net charge shared between the two capacitors
Q' = Q2 – Q2 = 4CV – CV = 3CV
The two capacitors will have some potential, say V'.
The net capacitance of the parallel combination of the two capacitors C' = C1 + C2 = C + 2C+ 3C
The potential of the capacitors 
The electrostatic energy of the capacitors 
In the circuit shown below, the ac source has voltage V = 20 cos(ωt) volt with ω = 2000 rad/s. The amplitude of the current will be nearest to
= 10Ω
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if
The heat produced is given by 
Thus heat (H) is doubled if both length (l) and radius (r) are doubled.
The frequency of a son ometer wir e is 100 Hz. When the weights producing the tensions are completely immersed in water, the frequency becomes 80 Hz and on immersing the weights in a certain liquid, the frequency becomes 60 Hz. The specific gravity of the liquid is
As we know, frequency
In water, fw = 0.8fair
In liquid, g'/g = (0.6)2 = 0.36
...(2)
From eq. (1) and (2)
A long straight wire along the Z-axis carries a current I in the negative Z-direction. The magnetic vector field 
at a point having coordinates (x, y) in the Z = 0 plane is
The wire carries a current I in the negative z-direction. We have to consider the magnetic vector field
at (x, y) in the z = 0
plane.
Magnetic field 
is perpendicular to OP.
Which of the following pollutants is main product of automobiles exhaust?
NO pollutant is the main product of automobiles exhaust.
The disease caused the high concentr ation of hydrocarbon pollutants in atmosphere is/are
The high concentration of hydrocarbon pollutants in atmosphere causes cancer.
Electronic configuration of element with atomic number 118 will be [Rn]5f146d10 7s27p6. Since its elctronic configuration in the outer most orbit (ns2np6) resemble with that of inert or noble gases, therefore it will be noble gas element.
Which law of the thermodynamics helps in calculating the absolute entropies of various substances at different temperatures?
The third law helps to calculate the absolute entropies of pure substances at different temperature.
The entropy of the substance at different temperature. T may be calculated by the measurement of heat capacity change 
Where ST = Entropy at T K
S0 = Entropy at 0K = Cp . logeT
= 2.303Cp.logT
CoCl3.5NH3.H2O is pink in colour
Cobalt is present in vitamin B12.
Cobalt (60) isotope is used in the treatment of :
Cobalt (60) isotope is used in the treatment of cancer.
PMMA is used in bullet proof glass
What is the correct increasing order of Bronsted bases?
ClO4- < ClO3-< ClO2-< ClO- is the correct increasing order of Bronsted base. With increase in the number of oxygen atoms in the conjugate bases, the delocalisation of the π bond becomes more and more extended. This results in decrease in the electron density. Consequently basicity also decreases.
The boiling point of alkyl halide are higher than those of corresponding alkanes because of
Due to dipole-dipole interaction the boiling point of alkyl halide is higher as compared to corresponding alkanes.
Some salts containing two different metallic elements give test for only one of them in solution, such salts are
Complex compounds contains two different metallic elements but give test only for one of them. Because complex ions such as [Fe (CN)6]4– of K4 [Fe (CN)6], do not dissociate into Fe2+ and CN– ions.
Primary amines (aromatic or aliphatic) on warming with chloroform and alcoholic KOH, gives carbylamine having offensive smell. This reaction is called carbylamine reaction.
Nitrous oxide (i.e., N2O) is the laughing gas.
Anthracene is purified by sublimation. In sublimation, a solid is converted directly into gaseous state on heating without passing through liquid phase.
Zeise’s salt is common name of K[Pt Cl3(η2 = C2H4)]
CaCl2 is produced as a by product in solvay ammonia process.
(i) NaCl + CO2 + NH3 + H2O → NaHCO3 + NH4Cl
(ii) CaCO3 → CO2 + CaO
(iii) 2 NH2Cl + CaO → 2 NH3 + 
Semiconductor materials like Si and Ge are usually purified by
Semiconductor materials like Si and Ge are usually purified by zone refining. Zone refining is based on the principle of fractional crystallisation i.e. difference in solubilities of impurities in solid and molten states of metal, so that the zones of impurities are formed and finally removed.
Order of basic character is NH3 > PH3 > AsH3 > SbH3. Basic-character decreases down the group from N to Bi due to increase in atomic size.
Normal glass is calcium alkali silicate glass made by fusing the alkali metal carbonate, CaCO3 and SiO2.
Exa = 1018
Which of the following is the most basic oxide?
More the oxidation state of the central atom (metal) more is its acidity. Hence SeO2 (O. S. of Se = +4) is acidic. Further for a given O.S., the basic character of the oxides increases with the increasing size of the central atom. Thus Al2O3 and Sb2O3 are amphoteric and Bi2O3 is basic.
Wh ich on e of th e following does not follow octate rule?
BF3 does not follow octate rule because central atom, boron lacks an electron pair.
Thus, it also acts as Lewis acid.
Whch of the following according to LeChatelier’s principle is correct?
According to Le-Chatelier’s principle increase in temperature favours the endothermic reaction while decrease in temperature favour the exothermic reaction. Increase in pressure shifts the equilibrium in that side in which number of gaseous moles decreases.
The efficiency of fuel cell is given by the expression, η is
Efficiency of fuel cell is:
The mass of the substance deposited when one Faraday of charge is passed through its solution is equal to
The mass of the substance deposited when one Faraday of charge is passed through its solution is equal to gram equivalent weight.
The unit of rate constant for reactions of second order is
Unit of rate constant for second order reaction is L mol–1 sec–1.
In a first order reaction with time the concentration of the reactant decreases
For first order reaction [A] = [A0]e–kt
∴ The concentration of reactants will exponentially decreases with time.
The P—P—P angle in P4 molecule and S—S—S angle in S8 molecule is(in degree) respectively
In P4 molecule, the four sp3-hybr idised phosphorous atoms lie at the corners of a regular tetrahedron with ÐPPP = 60°.
In S8 molecule S-S-S angle is 107° rings.
The number of elements present in the d-block of the periodic table is
40 elements are pr esent in d-block.
Which of the following represents hexadentate ligand?
EDTA is hexadentate ligand
Which one of given elements shows maximum number of different oxidation states in its compounds?
Am shows maximum n umber of oxidation states, + 3, + 4, + 5, + 6
Fe3+ ion can be detected by K4[Fe(CN)6]
4Fe3+ + 3K4 [Fe(CN)6] → Fe4 [Fe(CN)6]3+ 12K+
ΔG = ΔH – TΔS; ΔG is positive for a reaction to be non-spontaneous when ΔH is positive and ΔS is negative.
Which one the following removes temporary hardness of water ?
This method is known as Clark's process.
In this method temporary hardness is removed by adding lime water or milk of lime.
Graphite is covalent solid.
Which of the following ionic substances will be most effective in precipitating the sulphur sol?
Which of the following fluorides of xenon is impossible?
Structures of Xenon fluorides
XeF2 : Hybridization sp3d2
Linear
Xe F4 : Hybridization sp3d2
Square planar
XeF6 : Hybribisation sp3d3
Pentagonal pyramidal or distorted octahedral
Calcium silicophosp h ate (a mixture of Ca3(PO4)2 & Ca2SiO4) is called Thomas slag.
A sequence of how many nucleotides in messenger RNA makes a codon for an amino acid?
The sequence of bases in mRNA are read in a serial order in groups of three at a time.
Each triplet of nucleotides (having a specific sequence of bases) in known as codon.
Each codon specifies one amino acid.
Further since, there are four bases. therefore, 43 = 64 triplets or codons are possible.
Which of the following molecule/ion has all the three types of bonds, electrovalent, covalent and co-ordinate :
Bond structure of molecules are :
HCl = H+ – Cl–
hence, clearly NH+4 ion contains all three types of bonds.
DIRECTIONS : Choose the word which best expreses the meaning of the underlined word in the sentence.
Q. Decay is an immutable factor of human life.
‘Immutable’ means ‘unchangeable’. So, option (c) is correct choice.
DIRECTIONS : Choose the word which best expreses the meaning of the underlined word in the sentence.
Q. It was an ignominious defect for the team.
‘ignominious’ means ‘shameful’. So, option (a) is correct choice.
DIRECTIONS : Choose the word which best expreses the meaning of the underlined word in the sentence.
Q. The attitude of western countries towards the third world countries is rather callous to say the least.
‘callous’ means ‘showing or having an insensitive and cruel disregard for others’.
So, option (c) is correct choice.
DIRECTIONS : Fill in the blank.
Q. Freedom and equality are the ______ rights of every human.
Option (d) institutional as the word means relating to principles esp. of law, so legally also every human has rights of freedom and equality.
DIRECTIONS : Fill in the blank.
Q. The team was well trained and strong, but some how their ______ was low.
DIRECTIONS : Fill in the blank.
Q. His speech was disappointing: it ______ all the major issues.
DIRECTIONS : Choose the word which is closest to the opposite in meaning of the underlined word in the sentence.
Q. Hydra is biologically believed to be immortal.
Immortal means living forever, never dying or decaying. So, perishable is the correct opposite to it.
DIRECTIONS : Choose the word which is closest to the opposite in meaning of the underlined word in the sentence.
Q. The Gupta rulers patronised all cultural activities and thus Gupta period was called the golden era in Indian History.
Opposed is the correct answer of this. To patronise means favour or pat on the back.
DIRECTIONS : Choose the word which is closest to the opposite in meaning of the underlined word in the sentence.
Q. The General Manager is quete tactful and handles the workers union very effectively.
Tactful means having or showing skill and senstivity in dealing with others or with defficult essues.
So, in cautious is the correct opposite of factful.
DIRECTIONS : In each ot the following questions, out of the four alternatives, choose the one which can be substituted for the given words/ sentence.
Q. A person who does not believe in any religion
Atheist is the best alternative.
DIRECTIONS : In each ot the following questions, out of the four alternatives, choose the one which can be substituted for the given words/ sentence.
Q. A person who believes that pleasure is the chief good
‘Epicure’ is the best alternative.
DIRECTIONS : In each ot the following questions, out of the four alternatives, choose the one which can be substituted for the given words/ sentence.
Q. A person who is in charge of museum.
DIRECTIONS : Choose the order of the sentences marked A, B, C, D and E to form a logical paragraph.
Q.
A. Tasty and healthy food can help you bring out their best.
B. One minute they are toddlers and next you see them in their next adventure.
C. Your young ones seem to be growing so fast.
D. Being their loving custodians, you always want to see them doing well.
E. Their eye sparkle with curiosity and endless questions on their tongues.
DIRECTIONS : Choose the order of the sentences marked A, B, C, D and E to form a logical paragraph.
Q.
A. It is hoping that overseas friends will bring in big money and lift the morale of the people.
B. But a lot needs to be done to kick start industrial revival.
C. People had big hopes from the new government.
D. So far government has only given an incremental push to existing policies and programmes.
E. Government is to go for big time reforms, which it promised.
DIRECTIONS : Choose the order of the sentences marked A, B, C, D and E to form a logical paragraph.
Q.
A : Forecasting the weather has always been a defficult business.
B : During a period of drought, steams and rivers dried up, the cattle died from thirst and were ruined.
C : Many different things affect the weather and we have to study them carefully to make accurate forecast.
D : Ancient egyptians had no need of weather in the Nille valley hardly ever changes.
E: In early times, when there were no instruments, such as their mometer or the barometer, a man looked for tell tale signs in the sky.
Choose the correct answer figure which will make a complete square on joining with the problem figure
Problem figure
In the following question, five figures are given. Out of them, find the three figures that can be joined to form square.
Choose the answer figure which completes the problem figure matrix.
Problem Figures
The contents of the third figure in each row (and column) are determined by the contents of the first two figures. Lines are carried forward from the first two figures to the third one, except where two lines appear in the same position, in which they are cancelled out.
What is the opposite of 3, if four different positions of dice are as shown below :
From figure, (i), (iii) and (iv), we have concluded that 2, 6, 1 and 5 appear adjacent so 3. clearly, 4 will appear opposite to 3.
In the following questions, one or m ore dots are placed in the figure marked as (A). The figure is followed by four alternatives marked as (a), (b), (c) and (d). One out of these four options contains region(s) common to the circle, square, triangle, similar to that marked by the dot in figure (A).
Problem Figure
In figure (A), the dot is placed in the region which is common to the circle and triangle.
Now, we have to find similar common region in all the four options. Only in figure (c), we find such a region which is common to the circle and triangle.
Complete the series by replaing ‘? mark
G4T, J9R, M20P, P43N, S90L
Neeraj starts walking towards South. After walking 15 m, he turns towards North. After walking 20 m, he turns towards East and walks 10 m. He then turns towards South and walks 5 m. How far is he from his original position and in which direction?
According to the given information, the direction of Neeraj is as following.
The average age of 8 men is increased by 2 yr when one of them whose age is 20 yr is replaced by a new man. What is the age of the new man
Let the average age of 8 men = x yr
Total age of 8 men = 8 x yr
Now, new average age = (x + 2)yr
Total age = 8(x + 2) yr
Difference of ages = 8(x + 2) – 8x
= 8x + 16 – 8x = 16 yr
∴ Age of new man = 20 + 16 = 36 yr
So, the new man is 16yr older to the man by whom the new man is replaced.
Shikha is mother-in-law of Ekta who is sister-inlaw of Ankit. Pankaj is father of Sanjay, the only brother of Ankit. How is Shikha related to Ankit?
The relation is as following:
It is clearly shown that Shikha is the mother of Ankit.
In a queue of children, Arun is fifth from the left and Suresh is sixth from the right. When they interchange their places among themselves, Arun becomes thirteenth from the left. Then, what will be Suresh's position from the right?
Since Arun and Suresh interchange places, so Arun's new position (13th from left) is the same as Suresh's earlier position (6th from right).
So, number of children in the queue = (12 + 1 + 5) = 18.
Now, Suresh's new position is the same as Arun's earlier position fifth from left.
Therefore Suresh's position from the right = (18 – 4) = 14th.
equalsConsider 
If ω is the complex cube r oot of unity, then the value of 
is
Consider 
Which can be written as
Since 
is a G. P.. therfore by sum of infinite G.P, we have
∴ Given expression = -1
The root of the equation 2(1+ i)x2 - 4(2 - i)x - 5 - 3i = 0 which has greater modulus is
Roots = 
upto n terms isConsider 
upto n terms
upto n terms
upto n terms
= (1 + 1 + 1+ ....upto n terms)
tanθ is of period π so that tan 3θ is of period π/3.
If a function f(x) is given by 
then at x = 0, f(x)
Let 
For x = 0, we have f(x) = 0
Thus, we have 
Clearly, 
So, f(x) is not continuous at x = 0.
If g is the inverse of function f and f'(x) = sin x, then g'(x) is equal to
Since, g is the inverse of function f.
Therefore, g(x) = f–1(x)
⇒ f[g(x)] = x
⇒ fog (x) = x, for all x
Differentiate both side, w.r.tx
(By defn of f'(x))
A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides, whereas the remaining (n + 1) coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is 31/42, then n is equal to
Total number of coins= 2n+1
Consider the following events:
E1 = Getting a coin having head on both sides from the bag.
E2 = Getting a fair coin from the bag
A = Toss results in a head
Given: 
and P 
Then,
P(A) = P(E1) P(A /E1) + P(E2) P(A/E2)
n = 10
If φ(x) is a differential function, then the solution of the differential equation dy + {y φ(x) – φ(x) φ'(x)}dx = 0, is
Given differential equation is 
which is a linear differential equation with
∴ Solution is 
The area of the region R = {(x, y):|x| ≤ |y| and x2 + y2 ≤ 1} is
Required area = 4 (Area of the shaded region in first quadrant)
Universal set,
U = {x | x5 – 6x4 + 11x3 – 6x2 = 0}
A = {x | x2 – 5x + 6 = 0}
B = {x | x2 – 3x + 2 = 0}
What is (A ∩ B)' equal to ?
U = {x : x5 – 6x4 + 11x3 – 6x2 = 0}
Solving for values of x, we get
U = {0, 1, 2, 3}
A = {x : x2 – 5x + 6 = 0}
Solving for values of x, we get
A = {2, 3}
and B = {x : x2 – 3x + 2 = 0}
Solving for values of x, we get B = {2, 1}
A ∩ B = {2}
∴ (A ∩ B)' = U – (A ∩ B)
= {0, 1, 2, 3} – {2} = {0, 1, 3}
then 4x2 - 4 xy cosα + y2 is equal to
⇒ 4 - y2 -4x2 + x2y2 = 4 cos2 α + x2 y2 - 4 xy cosα
⇒ 4x2 + y2 - 4xycosα = 4sin2α.
If 
then the value of 2a1 + 23a3 + 25a5+ .... is
Let
By using
= a1 = a3 = a5 = ... = 0
Hence, 2a1 + 23a3 + 25a5 + ....= 0
Let a, b and c be three vectors satisfying a × b = (a ×c), |a| = |c| = 1, |b| = 4 and |b × c| = √15 . If b – 2c = λa, then λ equals
Let θ be the angle between b and c.
Now given, 
The total number of 4-digit numbers in which the digits are in descending order, is
Total number of arrangements of 10 digits 0, 1, 2, ...., 9 by taking 4 at a time = 10C4 × 4!
We observe that in every arrangement of 4 selected digits there is just one arrangement in which the digits are in descending order.
∴ Required number of 4-digit numbers.
The line which is parallel to X-axis and crosses the curve y = √x at an angle of 45°, is
Given equation of a line parallel to X-axis is y = k.
Given equation of the curve is y = √x ,
On solving equation of line with the equation of curve, we get x = k2
Thus the intersecting point is (k2, k)
It is given that the line y = k intersect the curve y = √x at an an gle of π/4. This means that the slope of the tangent to 
In a ΔABC, the lengths of th e two larger sides are 10 and 9 units, respectively. If the angles are in AP, then the length of the third side can be
Let A, B and C be the three angles of ΔABC and
Let a = 10 and b = 9
It is given that the angles are in AP.
∴ 2B = A + C on adding B both the sides, we get 3B = A + B + C
⇒ 3B = 180° ⇒ B = 60°
Now, we know 
The arithmetic mean of the data 0, 1, 2, ...... , n with frequencies 1, nC1, nC2,....., nCn is
Since, 
where xi are observations with frequencies fi, i = 1, 2, ........n The required mean is given by
The mean square deviation of a set of n observation x1, x2, .... xn about a point c is defined as 
The mean square deviations about – 2 and 2 are 18 and 10 respectively, the standard deviation of this set of observations is
We have 
and 
and
Let S be the focus of the parabola y2 = 8x and PQ be the common chord of the circle x2 + y2 – 2x – 4y = 0 and the given parabola. The area of DPQS is
The parametr ic equation s of the parabola y2 = 8x are x = 2t2 and y = 4t.
and the given equation of circle is x2 + y2 – 2x – 4y = 0
On putting x = 2t2 and y = 4t in circle we get
4t4 + 16t2 – 4t2 – 16t = 0
⇒ 4t2 + 12t2 – 16t = 0
⇒ 4t (t3 + 3t – 4) = 0
⇒ t(t – 1) (t2 + t + 4) = 0
⇒ t = 0, t = 1
Thus the coordinates of points of intersection of the circle and the parabola are Q (0, 0) and P(2, 4). Clearly these are diametrically opposite points on the circle.
The coordinates of the focus S of the parabola are (2, 0) which lies on the circle.
= 4 sq. units.
The number of real roots of the equation ex–1 + x – 2 = 0 is
Let f(x) = ex–1 + x – 2
check for x = 1
Then, f (1) = e0 + 1 – 2 = 0
So, x = 1 is a real root of the equation f(x) = 0 Let x = α be the other root such that α > 1 or α < 1. Consider the interval [1, α ] or [α,1].
Clearly f(1) = f(α) = 0
By Rolle’s theorem f'(x) = 0 has a root in (1, α) or in (α, 1).
But f'(x) = ex–1 + 1 > 0, for all x. Thus, f'(x)≠ 0 , for any x ∈ (1, α) or x ∈ (α,1) , which is a contradiction.
Hence, f(x) = 0 has no real root other than 1.
Minimise 
Subject to 
is a LPP with number of constraints
Constraints will be
So, total number of constraints = m + n
A bag contains 3 red and 3 white balls. Two balls are drawn one by one. The probability that they are of different colours is.
Let A ≡ event that drawn ball is red B ≡ event that drawn ball is white Then AB and BA are two disjoint cases of the given event.
∴ P (AB + BA) = P(AB) + P (BA)
Let M be a 3 × 3 non-singular matrix with det (M) = α. If [M–1 adj (adj (M)] = KI, then the value of K is
We know that, M (adj M) = |M| I
Replacing M by adj M, we get adj M [adj (adj M) = det (adj M) I = det (M) M–1 [adj (adj M) = α2l
Hence, K = α
Tangents are drawn from the origin to the curve y = cos x. Their points of contact lie on
Let (x1, y1) be one of the points of contact.
Given curve is y = cos x
Now the equation of the tangent at (x1,y1) is
⇒y -y1 = - sin x1(0- x1)
Since, it is given that equation of tangent passes through origin.
∴ 0 - y1 = - sin x1 (0-x1)
⇒ y1 = – x1 sin x1 ...(i)
Also, point (x1, y1) lies on y = cos x.
∴ y1 = cos x1
From Eqs. (i), (ii) , we get
⇒ 
Hence, the locus of (x1, y1) is x2 = y2 + y2x2 ⇒ x2y2 = x2 – y2
The slope of the tangent to the curve y = ex cos x is minimum at x = α, 0 ≤ a ≤ 2π, then the value of α is
Let m be the slope of the tangent to the curve
y = ex cos x.
Then, 
Diff. w.r.t ‘x’
⇒ 
=-2ex sinx
Clearly, 
Thus, y is minimum at x = π.
Hence the value of α = π.
Two lines 
are coplanar. Then, α can take value (s)
The equations of given lines can be written as
Since, these lines are coplanar.
Therefore, 
The eccentricity of an ellipse, with its centre at the origin, is 1/2. If one of the directrices is x = 4 , then the equation of the ellipse is:
Equation of ellispe is
has a local minimum at
is of the form 
and
equality holds for x = 1
Squaring, 
Differentiating,
If 
and 
then which one of the following is correct?
As given,
and 
⇒ B=0
∴ A = 1 and B = 0 is correct
If a and b are non-zero roots of x2 + ax + b = 0 then the least value of x2 + ax + b is
As given a and b are the roots of the equation x2 + ax + b = 0
⇒ sum of roots, a + b = – a
⇒ b = – 2a ...(1)
and product of roots, ab = b
⇒ ab – b = 0
⇒ b (a – 1) = 0
if b = 0 then a = 0
if b ≠ 0 then a = 1 and b = – 2
so, the expression will be, f (x) = x2 + x – 2
⇒ 
So, f (x) will be minimum, if 
i.e. when x = -(1/2)
⇒ minimum value of function = -(9/4)
Let us assume the functions f(x) and g(x) given by
f(x) = tanx – x and g(x) = x – sinx, for 0<x<π/2
Now, f'(x) = sec2x – 1 and
g'(x) = 1 – cos x
The degree of the differential equation satisfying 
Put x = sin θ and y = sin φ
⇒ cos θ + cos φ = a (sin θ – sin φ)
⇒ 2 cos 
Differentiate 
so the degree is one
Let f(x) be a polynomial of degree three satisfying f(0) = – 1 and f(1) = 0. Also, 0 is a stationary point of f(x). If f(x) does not have an extremum at x = 0, then the value of 
is
Let f(x) = ax3 + bx2 + cx + d
Put x = 0 and x = 1
Then, we get f(0) = –1 and f(1) = 0
⇒ d = – 1 and a + b + c + d = 0
⇒ a + b + c = 1 ...(i)
It is given that x = 0 is a stationary point of f(x), but it is not a point of extremum.
Therefore, f"(0) = 0 = f " (0) and f"(0) =0
Now, f(x) = ax3 + bx2 + cx + d
⇒ f'(x) = 3ax2 + 2bx + c,
f " (x) = 6ax + 2b and f"' (x) = 6a
f' = 0, f "' (0) =0 and f "' (0) = 0≠0
⇒ c = 0, b = 0 and a ≠ 0
From Eqs. (i) and (ii), we get a = 1, b = c = 0 and d = – 1
Put these values in f(x)
we get f(x) = x3 – 1
Hence, 
if (i)-1 ≤ x -3 ≤ 1 ⇒ 2 ≤ x ≤ 4 and
(ii) 9-x2 > 0 ⇒ -3 < x < 3
Taking common solution of (i) and (ii), we get 2 ≤ x < 3
∴ Domain = [2, 3)
If the lines p1x + q1y = 1, p2x + q2y = 1 and p3x + q3y = 1 be concurrent, then the points (p1, q1), (p2, q2) and (p3, q3)
The equations of the lines are
p1x + q1y - 1 = 0 ...(i)
p2x + q2y - 1 = 0 ...(ii)
and p3x + q3y - 1 = 0 ...(iii)
As they are concurrent,
This is also the condition for the points (p1, q1), (p2, q2) and (p3, q3) to be collinear.
Area of the circle in which a chord of length √2 makes an angle π/2 at the centre, is
Let AB be the chord of length √2. Let O be the centre of the circle and let OC be the perpendicular from O on AB.
Then, AC = BC = 
In ΔOBC, we have
OB= BC cosec 45°
∴ Area of the circle = π(OB)2 = π sq units
then the value of (m2 – n2) sin2 B iscos A = n cos B and sin A = m sin B
Squaring and adding, we get
1 = n2 cos2B + m2sin2B
⇒ 1 = n2 (1 – sin2B) + m2sin2B
∴ (m2 – n2) sin2B = 1 – n2
If complex number z1, z2 and 0 are vertices of equilateral triangle, then 
is equal to
z1, z2, 0 are vertices of an equilateral triangle, so we have 
⇒ 
If ρ = {(x, y) |x2 + y2 = 1; x, y ∈ R}. Then, ρ is
Obviously, the relation is not reflexive and transitive, but it is symmetric, because
x2 + x2 = 2x2 ≠ 1
and x2 + y2 = 1, y2 + z2 = 1
⇒ x2 + z2 = 1
But x2 + y2 = 1 ⇒ y2 + x2 = 1
A line makes the same angle θ with each of the X and Z-axes. If the angle β, which it makes with Y-axis, is such that sin2 β = 3sin2 θ, then cos2 θ equals
Let l, m and n be the direction cosines.
Then, l = cos θ, m = cos β, n = cos θ
we have l2 + m2 + n2 = 1
⇒ tan2θ = 2/3
If in a binomial distribution n = 4, P(X = 0) = 16/81, then P(X = 4) equals
Given n = 4 and P(X = 0) = 16/81
Let p be the probability of success and q that of failure in a trial.
Then, P(X = 0) = 4C0p0q4 = 16/81
⇒ 
∴ P(X = 4) = 4C4p4q0 = p4 = (1/3)4 = 1/81
Let f : R → R be a function such that
f (x + y) = f (x) + f (y), 
If f (x) is differentiable at x = 0, then which one of the following is incorrect?
Let f(x + y) = f(x) + f(y), 
Put x = 0 = y
⇒ f(0) = f(0) + f(0)
⇒ f(0) = 0
Now, 
Now, 
⇒ f (x) = x f'(0) + C
But f(0) = 0
∴ C = 0
Hence, f(x) = x f'(0), 
Clearly, f(x) is everywhere continuous and differentiable and f'(x) is constant.
If binomial coefficients of three consecutive terms of (1 + x)n are in HP, then the maximum value of n is
Let the coefficients of rth, (r + 1)th, and (r + 2)th terms be in HP.
Then, 
⇒ n2 – 4nr + 4r2 + n = 0
⇒ (n – 2r)2 + n = 0
which is not possible for any value for n.
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