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What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes 3/5 of the presentweight at the equator. Equatorial radius of the earth is 6400 km.
True weight at equator, W = mg
Observed weight at equator,
W ' = mg ' = 3/5 mg
At equator, latitude λ = 0
Using the formula,
mg ' = mg  mRω^{2} cos^{2}λ
3/5 mg = mg  mRω^{2} cos^{2}0 = mg  mRω^{2 }
⇒ mRω^{2 } = mg  3/5 mg = 2/5 mg
Block A of mass m and block B of mass 2m are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless pulley as shown in figure.
The wedge is inclined at 45° to the horizontal on both the sides. If the coefficient of friction between the block A and the wedge is 2/3 and that between the block B and the wedge is 1/3 and both the blocks A and B are released from rest, the acceleration of A will be
The surface charge density of a thin charged disc of radius σ is s. The value of the electric field at the centre of the disc is σ/2∈_{0}. With respect to the field at the centre, the electric field along the axis at a distance R from the centre of the disc
The molecules of a given mass of a gas have r.m.s. velocity of 200 ms^{–1} at 27°C and 1.0 × 10^{5} Nm^{–2} pressure. When the temperature and pressure of the gas are respectively, 127°C and 0.05 × 10^{5} Nm^{–2}, the r.m.s. velocity of its molecules in ms^{–1 }is :
Here v_{1} = 200 m/s;
temperature T_{1 }= 27°C = 27 + 273 = 300 k
temperature T_{2} = 127° C = 127 + 273 = 400 k, V = ?
R.M.S. Velocity, V ∝ √T
An inductor of inductance L = 400 mH and resistors of resistance R_{1} = 2Ω and R_{2} = 2Ω are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is
Growth in current in LR_{2} branch when switch is closed is given by
Hence, potential drop across
Two wires are made of the same material and have the same volume. However wire 1 has crosssectional area A and wire 2 has crosssectional area 3A. If the length of wire 1 increases by Δx on applying force F, how much force is needed to stretch wire 2 by the same amount?
As shown in the figure, the wires will have the same Young’s modulus (same material) and the length of the wire of area of crosssection 3A will be ℓ/3 (same volume as wire 1).
For wire 1,
For wire 2,
From (i) and (ii),
⇒ F ' = 9F
Two spheres of different materials one with double the radius and onefourth wall thickness of the other are filled with ice. If the time taken for complete melting of ice in the larger sphere is 25 minute and for smaller one is 16 minute, the ratio of thermal conductivities of the materials of larger spheres to that of smaller sphere is
Radius of small sphere = r Thickness of small sphere = t Radius of bigger sphere = 2r Thickness of bigger sphere = t/4 Mass of ice melted = (volume of sphere) × (density of ice)
Let K_{1} and K_{2} be the thermal conductivities of larger and smaller sphere.
For bigger sphere,
For smaller sphere,
A biconvex lens h as a radius of curvature of magnitude 20 cm. Which one of the following options best describe the image formed of an object of height 2 cm placed 30 cm from the lens?
In the figure below, what is the potential difference between the point A and B and between B and C respectively in steady state
The equivalent circuit is shown in figure.
V_{1} + V_{2} = 100
and 2V_{1} = 6V_{2}
On solving above equations, we get
V_{1} = 75V, V_{2} = 25V
A radioactive element X converts into another stable element Y. Half life of X is 2 hrs. Initially only X is present. After time t, the ratio of atoms of X and Y is found to be 1 : 4, then t in hours is
Let N_{0} be the number of atoms of X at time t = 0.
Then at t = 4 hrs (two half lives)
∴ N_{x}/N_{y} = 1/3
and at t = 6 hrs (three half lives)
The given ratio 1/4 lies between 1/3 and 1/7.
Therefore, t lies between 4 hrs and 6 hrs.
The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 × 10^{–11} Pa^{–1} and density of water is 10^{3} kg/m^{3}.What fractional compression of water will be obtained at the bottom of the ocean?
Compressibility of water,
K = 45.4 × 10^{–11} Pa^{–1}
density of water P = 10^{3} kg/m^{3}
depth of ocean, h = 2700 m
We have to find ΔV/V = ?
As we know, compressibility,
So,(ΔV/V) = KPgh
= 45.4 × 10^{–11} × 10^{3} × 10 × 2700
= 1.2258 × 10^{–2}
A frictionless wire AB is fixed on a sphere of radius R. A very small spherical ball slips on this wire. The time taken by this ball to slip from A to B is
Acceleration of body along AB is g cos q Distance travelled in time t sec =
From ΔABC, AB = 2R cos θ
A string of length ℓ is fixed at both ends. It is vibrating in its 3^{rd} overtone with maximum amplitude 'a'. The amplitude at a distance ℓ/3 from one end is
For a string vibrating in its n^{th} overtone (n + 1)^{th} harmonic)
A deuteron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to the magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same B is
In the circuit shown in the figure, find the current in 45 Ω.
Kepler's third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance r between sun and planet i.e. T^{2} = Kr^{3} here K is constant. If the masses of sun and planet are M and m respectively then as per Newton's law of gravitation force of attraction between them is here G is gravitational constant. The relation between G and K is described as
As we know, orbital speed,
Time period
Squarring both sides,
Find the number of photon emitted per second by a 25 watt source of monochromatic light of wavelength 6600 Å. What is the photoelectric current assuming 3% efficiency for photoelectric effect ?
Pin = 25W, λ = 6600 Å = 6600 × 10^{–10} m nhv = P
⇒ Number of photons emitted/sec,
3% of emitted photons are producing current
A ray of light of intensity I is incident on a parallel glass slab at point A as shown in diagram. It undergoes partial reflection and refraction. At each reflection, 25% of incident energy is reflected. The rays AB and A'B' undergo interference. The ratio of I_{max} and I_{min} is :
A capillary tube of radius r is immersed vertically in a liquid such that liquid rises in it to height h (less than the length of the tube). Mass of liquid in the capillary tube is m. If radius of the capillary tube is increased by 50%, then mass of liquid that will rise in the tube, is
New mass
The drift velocity of electrons in silver wire with crosssectional area 3.14 × 10^{–6} m^{2} carrying a current of 20 A is. Given atomic weight of Ag = 108, density of silver = 10.5 × 10^{3} kg/m^{3}.
Number of electrons per kg of silver
Number of electrons per unit volume of silver
A parallel plate capacitor of area ‘A’ plate separation ‘d’ is filled with two dielectrics as shown. What is the capacitance of the arrangement?
(∵ C_{1} and C_{2} are in series and resultant of these two in parallel with C_{3})
In the Young’s doubleslit experiment, the intensity of light at a point on the screen where the path difference is λ is K, (λ being the wave length of light used). The intensity at a point where the path difference is λ/4, will be :
For path differ ence λ, phase difference = 2π rad.
For path difference λ/4, phase difference
= π/2 rad.
As K = 4I_{0} so intensity at given point where path difference is λ/4
The mass of _{7}N^{15} is 15.00011 amu, mass of _{8}O^{16} is 15.99492 amu and m_{p} = 1.00783 amu. Determine binding energy of last proton of _{8}O^{16}.
M(_{8}O^{16}) = M (_{7}N^{15}) + 1m_{P}
binding energy of last proton
= M (N^{15}) + m_{P} – M (_{1}O^{16})
= 15.00011 + 1.00783 – 15.99492
= 0.01302 amu = 12.13 MeV
A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to Xaxis while semicircular portion of radius R is lying in YZ plane. Magnetic field at point O is:
A stone projected with a velocity u at an angle θ with the horizontal reaches maximum height H_{1}. When it is projected with velocity u at an angle with the horizontal, it reaches maximum height H_{2}. The relation between the horizontal range R of the projectile, heights H_{1} and H_{2} is
and
If the series limit wavelength of Lyman series for the hydrogen atom is 912 Å then the series limit wavelength for Balmer series of hydrogen atoms is
For limiting wavelength of Lyman series
For limiting wavelength of Balmer series
n_{1} = 2, n_{2} = ∞
∴ λ_{B} = 4λ_{L} = 4 × 912 Å.
In the shown arrangement of the experiment of the meter bridge if AC corresponding to null deflection of galvanometer is x, what would be its value if the radius of the wire AB is doubled?
At null point
If radius of the wire is doubled, then the resistance of AC will change and also the resistance of CB will change. But since R_{1}/R_{2} does not change so, R_{3}/R_{4} should also not change at null point. Therefore the point C does not change.
A 1 kg mass is attached to a spring of force constant 600 N/m and rests on a smooth horizontal surface with other end of the spring tied to wall as shown in figure. A second mass of 0.5 kg slides along the surface towards the first at 3m/s. If the masses make a perfectly inelastic collision, then find amplitude and time period of oscillation of combined mass.
The frequency of vibration of string is given by
Here p is number of segments in the string and / is the length. The dimensional formula for m will be
Now, dimensional formula of R.H.S.
[p will have no dimension as it is an integer] = ML^{–1}T^{0}.
So, dimensions of m will be ML^{–1}T^{0}.
For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index:
The angle of minimum deviation is given as
δ_{min} = i + e–A
for minimum deviation
δ_{min} = A then
2A = i + e
in case of δ_{min} i = e
2A = 2i r_{1} = r_{2} = A/2
i = A = 90°
from smell’s law
1 sin i = n sin r_{1}
sin A = n sin A/2
when A = 90° = _{imin}
then n_{min} = √2
i = A = 0 n_{max} = 2
Consider elastic collision of a particle of mass m moving with a velocity u with another particle of the same mass at rest. After the collision, the projectile and the struck particle move in directions making angles θ_{1} and θ_{2} respectively with the initial direction of motion. The sum of the angles θ_{1} + θ_{2}, is:
A conducting circular loop is placed in a uniform magnetic field of 0.04 T with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at 2 mm/s. The induced emf in the loop when the radius is 2 cm is
Induced emf in th e loop is gi ven by
e = B. dA/dt where A is the area of the loop.
r = 2cm = 2 × 10^{–2} m
dr = 2 mm = 2 × 10^{–3} m
dt = 1s
= 0.32 π × 10^{–5}V
= 3.2 π ×10^{–6}V
= 3.2 π μV
Figure below shows two paths that may be taken by a gas to go from a state A to a state (c) In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be
In cyclic process ABCA
Q_{cycle} = W_{cycle}
Q_{AB} + Q_{BC} + Q_{CA} = ar. of ΔABC
+ 400 + 100 + Q_{C→A} = 1/2 (2 × 10^{–3}) (4 × 10^{4})
⇒ Q_{C→A }= – 460 J
⇒ Q_{A→C }= + 460 J
Two resistances at 0°C with temperature coefficient of resistance α_{1} and α_{2} joined in series act as a single resistance in a circuit. The temperature coefficient of their single resistance will be
R_{1} = R_{0}(1 + α_{1}t) + R_{0}(1 + α_{2}t)
Comparing with R = R_{0} (1+ αt)
Two identical charged spheres suspended from a common point by two massless strings of lengths l, are initially at a distance d (d << l ) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as :
From figure tan θ = F_{e}/mg ≅ θ
Differentiating eq. (1) w.r.t. time
so x^{2}(v) ∝ q Replace q from eq. (2)
x2(v) ∝ x^{3/2} or v ∝ x^{–1/2}
A point particle of mass 0.1 kg is executing S.H.M. of amplitude of 0.1 m. When the particle passes through the mean position, its kinetic energy is 8 x 10^{3} Joule. Obtain the equation of motion of this particle if this initial phase of oscillation is 45°.
A source of sound S emitting waves of frequency 100 Hz and an observor O are located at some distance from each other. The source is moving with a speed of 19.4 ms^{1} at an angle of 60° with the source observer line as shown in the figure. The observor is at rest. The apparent frequency observed by the observer is (velocity of sound in air 330 ms^{1})
Here, original frequency of sound, f_{0} = 100 Hz
Speed of source V_{s} = 19.4 cos 60° = 9.7
19.4 cos 60° = 9.7
From Doppler's formula
Apparent frequency f^{1} = 103 Hz
A resistor of resistance R, capacitor of capacitance C and inductor of inductance L are connected in parallel to AC power source of voltage ε_{0} sin ωt. The maximum current through the resistance is half of the maximum current through the power source. Then value of R is
A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d/2 in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:
By covering aperture, focal length does not change. But intensity is reduced by 1/4 times, as aperture diameter d/2 is covered.
∴ New focal length = f and intensity = 3I/4.
A circular disc of radius R and thickness R/6 has moment inertia I about an axis passing through its centre perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter is
According to problem disc is melted and recasted into a solid sphere so their volume will be same.
Moment of inertia of disc
Moment of inertia of sphere
In PO^{3}_{4}, the formal charge on each oxygen atom and the P  O bond order respectively are
Bond order
Three unit negative charge is being shared by four O atoms. Formal charge = –3/4
The decreasing order of the ionization potential of the following elements is
Closed shell (Ne), half filled (P) and completely filled configuration (Mg) are the cause of higher value of I.E.
Knowing that the chemistry of lanthanoids(Ln) is dominated by its + 3 oxidation state, which of the following statements is incorrect?
Most of the Ln^{3+} compounds except La^{3+} and Lu^{3+} are coloured due to the presence of felectrons.
Which of the following arrangements does not represent the correct order of the property stated against it?
Fe^{2+} electr onic configura tion is [Ar] 3d^{6}. Since CN is strong field ligand d electrons are paired . In Ni(CO)_{4} O. S. of Ni is zero electron ic configuration is [Ar}3d^{8} 4s^{2}. In presence of CO it is [Ar] 3d^{10} 4s^{0}, electrons are paired. Electronic configuration of Ni^{2+} [Ar]3d^{8 }4s^{0}, due to CN^{–} ligand all electrons are paired. Co^{3+} is [Ar] 3d^{6} since F is weak ligand hence paramagnetic.
The hypothetical complex chlorodiaquatriamminecobalt (III) chloride can be represented as
The complex chlorodiaquatriammine cobalt (III) chloride can have the structure [CoCl(NH_{3})_{3}(H_{2}O)_{2}]Cl_{2}
The normality of 26% (wt/vol) solution of ammonia (density = 0.855 ) is approximately :
Wt. of NH_{3} = 26g = 26/17 g eg = 1.53 g eg
Vol. of soln. = 100 mc= 0.1 L
∴ Normality = 1.53/0.1 = 15.3N
1.25 g of a sample of Na_{2}CO_{3} and Na_{2}SO_{4} is dissolved in 250 ml solution. 25 ml of this solution neutralises 20 ml of 0.1N H_{2}SO_{4}.The % of Na_{2}CO_{3} in this sample is
Let the amount of Na_{2}CO_{3} present in the mixture be xg . Na_{2}SO_{4} will not react with H_{2}SO_{4}. Then
Which of the following compound has all the four types (1°, 2°, 3° and 4°) of carbon atoms?
C_{1} and C_{5} are 1°, C_{3} is 2°, C_{4} is 3°, and C_{2} is 4°.
A rapid umbrella type inversion rapidly converts the structure III to its enantiomer; hence the two enantiomers are not separable.
Catalytic hydrogen ation of alkyn es gives cisalkene which in turn adds deuterium atoms in presence of H_{2} again in cismanner forming meso2, 3dideuterobutane.
Give the possible structure of X in the following reaction :
D_{2}SO_{4} transfers D^{+}, an electrophile, to form hexadeuterobenzene.
An aromatic compound has molecular formula C_{7}H_{7}Br. Give the possible isomers and the appropriate method to distinguish them.
Which of the following method gives better yield of pnitrophenol?
Thus here, oxidation of phenol is minimised by forming pnitrosophenol.
Formation of polyethylene from calcium carbide takes place as follows
CaC_{2} + 2H_{2}O → Ca (OH)_{2} + C_{2}H_{2}
C_{2}H_{2} + H_{2} → C_{2}H_{4}
nC_{2}H_{4} → (CH_{2}  CH_{2} )n
The amount of polyethylene obtained from 64.1 kg of CaC_{2} is
The concerned chemical reactions are
Thus 64 kg of CaC_{2 }gives 26 kg of acetylene which in turn gives 28 kg of ethylene whose 28 kg gives 28 kg of the polymer, polythene.
The most likely acidcatalysed aldol condensation products of each of the two aldehydes I and II will respectively be
Sometimes, the colour observed in Lassaigne’s test for nitrogen is green. It is because
Although blue coloured ferric ferrocyanide is formed but due to the presence of yellow coloured Fe^{3+} salts, the blue colour gives the shade of green.
Fructose on reduction gives a mixture of two alcohols which are related as
Ketoses on reduction produce a new chiral carbon leading to the formation of two isomeric alcohols which are diastereomeric as well as C2 epimers.
What will happen when D(+)glucose is treated with methanolic —HCl followed by Tollens’ reagent?
Reaction of D(+)glucose with methanolic —HCl leads to formation of methyl glucoside (C_{1}—OH group is methylated) which, being acetal, is not hydrolysable by base, so it will not respond Tollens’ reagent.
Which of the followings forms the base of talcum powder?
Magnesium hydrosilicate forms base of Talcum powder.
BHT is the important antioxidant used in food.
The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at
For Balmer n_{1} = 2 and n_{2} = 3;
An e^{–} has magnetic quantum number as –3, what is its principal quantum number?
When m = – 3, l = 3, ∴ n = 4 .
At what temperature, the rate of effusion of N_{2 }would be 1.625 times than that of SO_{2} at 50^{o}C?
The average kinetic energy of an ideal gas per molecule in SI unit at 25°C will be
The degree of dissociation of PCl_{5} (α) obeying the equilibrium PCl5 ⇌ PCl_{3} + Cl_{2} is related to the equilibrium pressure by
In a closed system, A(s) ⇌ 2B(g) + 3C(g), if partial pressure of C is doubled, then partial pressure of B will be
For a particular reversible reaction at temperature T, ΔH and ΔS were found to be both +ve. If T_{e} is the temperature at equilibrium, the reaction would be spontaneous when
At equilibrium ΔG = 0
Hence, ΔG = ΔH – T_{e}ΔS = 0
For a spontaneous reaction
ΔG must be negative which is possible only
if ΔH – TΔS < 0
∴ ΔH < TΔS
Given
Based on data provided, the value of electron gain enthalpy of fluorine would be:
Applying Hess’s Law
617= 161+ 520 + 77 + E.A. + (–1047)
E.A. = –617 + 289 = –328 kJ mol^{–1}
∴ electron affinity of fluorine = –328 kJ mol^{–1}
The percentage hydrolysis of 0.15 M solution of ammonium acetate, K_{a} for CH_{3}COOH is 1.8 × 10^{–5} and K_{b} for NH_{3} is 1.8 × 10^{–5}
For a sparingly soluble salt A_{p}B_{q}, the relationship of its solubility product Ls → K_{sp} with its solubility (S) is
Let the solubility be S mol/liter Thus,
K_{sp} = [A ^{+}]_{p }[B^{–}]^{q}
= [Sp]^{p} [Sq]^{q }= p ^{p}.q^{q} (S)^{p+q} .
Consider the reaction :
Cl_{2}(aq) + H_{2}S(aq) → S(s) + 2H^{+} (aq) + 2Cl^{–} (aq)
The rate equation for this reaction is rate = k[Cl_{2}][H_{2}S]
Which of these mechanisms is/are consistent with this rate equation?
A. Cl_{2} + H_{2}S → H^{+} + Cl^{–} + Cl^{+} + HS^{–} (slow)
Cl^{+} + HS^{–} → H^{+} + Cl^{–} + S (fast)
B. H_{2}S ⇌ H^{+} + HS^{} (fast equilibrium)
Cl_{2} + HS^{} → 2Cl^{} + H^{+ }+ S (Slow )
In the reaction, P + Q → ? R + S
The time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is
For P, if t_{50%} = x
then t_{75%} = 2x
This is true only for first order reaction. So, order with respect to P is 1.
Further the graph shows that concentration of Q decreases with time. So rate, with respect to Q, remains constant. Hence, it is zero order wrt Q.
So, overall order is 1+ 0 = 1
The EMF of th e cell Tl/Tl^{+} (0.001M)  Cu^{2+} (0.01M) /Cu is 0.83. The cell EMF can be increased by
The oxidation potential
and reduction potential ∝ concentration of ions. The cell voltage can be increased by decreasing the concentration of ions around anode or by increasing the concentration of ions around cathode
Electrolysis is carried out in three cells
(A) 1.0 M CuSO_{4} Pt electrode
(B) 1.0 M CuSO_{4} copper electrodes
(C) 1.0 M KCl Pt electrodes
If volume of electrolytic solution is maintained constant in each of the cell, which is correct set of pH changes in (A), (B) and (C) cell respectively?
The equilibrium constant for the disproportionation reaction 2Cu^{+} (aq) → Cu(s) + Cu^{2+} (ag) at 25°C
(E^{o} Cu^{+} / Cu = 0.52V, E^{o}Cu^{2+} / Cu= 0.16V) is
The reaction
At equilibrium E_{cell} = 0
∴ K_{c} = 1.2 x 10^{6}
The non stoichiometric compound Fe_{0.94}O is formed when x % of Fe^{2+} ions are replaced by as many 2/3 Fe^{3+} ions, x is
The number of Fe^{3+ }ions replacing x Fe^{2+ }ions = 2x/3 vacancies of cations
But x/3 = 1 – 0.94 = 0.06, x = 0.06 × 3
= 0.18 = 18%
Al (at. wt 27) crystallizes in the cubic system with a cell edge of 4.05 Å. Its density is 2.7 g per cm^{3}.
Determine the unit cell type calculate the radius of the Al atom
Hence it is face centred cubic unit lattice.
Again 4r = a√2 = 5.727 Å
∴ r = 1.432 Å
A compound of Xe and F is found to have 53.5% of Xe. What is oxidation number of Xe in this compound ?
Xe = 53.5 % ∴ F = 46.5%
Relative number of atoms Xe
Simple ratio Xe = 1 and F = 6 ; Molecular formula is XeF_{6}
O.N.of Xe is +6
Directions: Choose the word which best expresses the meaning of the given word.
CORPULENT
Directions: Choose the word which best expresses the meaning of the given word.
EMBEZZLE
Directions: Choose the word which is the exact OPPOSITE of the given words.
ARROGANT
Directions: Choose the word which is the exact OPPOSITE of the given words.
EXODUS
Directions: Read the following passage and answer the questions that follows.
At this stage of civilisation, when many nations are brought in to close and vital contact for good and evil, it is essential, as never before, that their gross ignorance of one another should be diminished, that they should begin to understand a little of one another's historical experience and resulting mentality. It is the fault of the English to expect the people of other countries to react as they do, to political and international situations. Our genuine goodwill and good intentions are often brought to nothing, because we expect other people to be like us. This would be corrected if we knew the history, not necessarily in detail but in broad outlines, of the social and political conditions which have given to each nation its present character.
Q. According to the author of 'Mentality' of a nation is mainly product of its
Directions: Read the following passage and answer the questions that follows.
At this stage of civilisation, when many nations are brought in to close and vital contact for good and evil, it is essential, as never before, that their gross ignorance of one another should be diminished, that they should begin to understand a little of one another's historical experience and resulting mentality. It is the fault of the English to expect the people of other countries to react as they do, to political and international situations. Our genuine goodwill and good intentions are often brought to nothing, because we expect other people to be like us. This would be corrected if we knew the history, not necessarily in detail but in broad outlines, of the social and political conditions which have given to each nation its present character.
Q. The need for a greater understanding between nations
Directions: Read the following passage and answer the questions that follows.
At this stage of civilisation, when many nations are brought in to close and vital contact for good and evil, it is essential, as never before, that their gross ignorance of one another should be diminished, that they should begin to understand a little of one another's historical experience and resulting mentality. It is the fault of the English to expect the people of other countries to react as they do, to political and international situations. Our genuine goodwill and good intentions are often brought to nothing, because we expect other people to be like us. This would be corrected if we knew the history, not necessarily in detail but in broad outlines, of the social and political conditions which have given to each nation its present character.
Q. The character of a nation is the result of its
Directions: Read the following passage and answer the questions that follows.
At this stage of civilisation, when many nations are brought in to close and vital contact for good and evil, it is essential, as never before, that their gross ignorance of one another should be diminished, that they should begin to understand a little of one another's historical experience and resulting mentality. It is the fault of the English to expect the people of other countries to react as they do, to political and international situations. Our genuine goodwill and good intentions are often brought to nothing, because we expect other people to be like us. This would be corrected if we knew the history, not necessarily in detail but in broad outlines, of the social and political conditions which have given to each nation its present character.
Q. According to the author his countrymen should
Directions: In questions below, each passage consist of six sentences. The first and sixth sentence are given in the beginning. The middle four sentences in each have been removed and jumbled up.
These are labelled as P, Q, R and S. Find out the proper order for the four sentences.
S1: A force of exists between everybody in the universe.
P: Normally it is very small but when the one of the bodies is a planet, like earth, the force is considerable.
Q: It has been investigated by many scientists including Galileo and Newton.
R: Everything on or near the surface of the earth is attracted by the mass of earth.
S: This gravitational force depends on the mass of the bodies involved.
S6: The greater the ma ss, the greater is the earth's force of attraction on it. We can call this force of attraction gravity.
Q. The Proper sequence should be:
Directions: In questions below, each passage consist of six sentences. The first and sixth sentence are given in the beginning. The middle four sentences in each have been removed and jumbled up.
These are labelled as P, Q, R and S. Find out the proper order for the four sentences.
S1: Calcutta unlike other cities kepts its trams.
P: As a result there horrendous congestion.
Q: It was going to be the first in South Asia.
R: They run down the centre of the road
S: To ease in the city decided to build an underground railway line.
S6: The foundation stone was laid in 1972.
Q. The Proper sequence should be:
Directions: Pick out the most effective word from the given words to fill in the blank to make the sentence meaningfully complete.
Q. The miser gazed ...... at the pile of gold coins in front of him.
Directions: Pick out the most effective word from the given words to fill in the blank to make the sentence meaningfully complete.
Q. I saw a ...... of cows in the field.
Directions: Read the each sentence to find out whether there is any grammatical error in it. The error, if any will be in one part of the sentence. The letter of that part is the answer. If there is no error, the answer is 'd'. (Ignore the errors of punctuation, if any).
Directions: Read the each sentence to find out whether there is any grammatical error in it. The error, if any will be in one part of the sentence. The letter of that part is the answer. If there is no error, the answer is 'd'. (Ignore the errors of punctuation, if any).
Directions: Read the each sentence to find out whether there is any grammatical error in it. The error, if any will be in one part of the sentence. The letter of that part is the answer. If there is no error, the answer is 'd'. (Ignore the errors of punctuation, if any).
Select a suitable figure from the four alternatives that would complete the figure matrix.
In each row, the third figure comprises of a black circle and only those line segments which are not common to the first and the second figures.
Select a suitable figure from the four alternatives that would complete the figure matrix.
In each row (as well as each column), the third figure is a combination of all the, elements of the first and the second figures.
Select a suitable figure from the four alternatives that would complete the figure matrix.
In each row, the third figure is a collection of the common elements (line segments) of the first and the second figures.
Direction: Choose the correct alternative that will continue the same pattern and replace the question mark in the given series.
3, 4, 7, 7, 13, 13, 21, 22, 31, 34, ?
The given sequence is a combination of two series :
I. 3, 7, 13, 21, 31, ? and II. 4, 7, 13, 22, 34
The pattern in I is + 4, + 6, + 8, + 10,.....
The pattern in II is + 3, + 6, + 9, + 12,.....
So, missing term = 31 + 12 = 43.
Introducing a boy, a girl said, "He is the son of the daughter of the father of my uncle." How is the boy related to the girl?
The father of the boy's uncle is the grandfather of the boy and daughter of the grandfather is sister of father.
Direction: In these series, you will be looking at both the letter pattern and the number pattern. Fill the blank in the middle of the series or end of the series.
QAR, RAS, SAT, TAU, _____
In this series, the third letter is repeated as the first letter of the next segment. The middle letter, A, remains static. The third letters are in alphabetical order, beginning with R.
Direction: In these series, you will be looking at both the letter pattern and the number pattern. Fill the blank in the middle of the series or end of the series.
DEF, DEF_{2}, DE_{2}F_{2}, _____, D_{2}E_{2}F_{3}
In this series, the letters remain the same: DEF.
The subscript numbers follow this series: 111, 112, 122, 222, 223, 233, 333, ...
Direction: In each question below are given two statements followed by two conclusions numbered I and II. You have to take the given two statements to be true even if they seem to be at variance from commonly known facts. Read the conclusion and then decide which of the given conclusions logically follows from the two given statements, disregarding commonly known facts.
Statements: Raman is always successful. No fool is always successful.
Conclusions:
I. Raman is a fool.
II. Raman is not a fool.
Since both the premises are universal and one premise is negative, the conclusion must be universal negative and should not contain the middle term. So, only II follows.
Direction: In each question below are given two statements followed by two conclusions numbered I and II. You have to take the given two statements to be true even if they seem to be at variance from commonly known facts. Read the conclusion and then decide which of the given conclusions logically follows from the two given statements, disregarding commonly known facts.
Statements: Some desks are caps. No cap is red.
Conclusions:
I. Some caps are desks.
II. No desk is red.
Since one premise is particular and the other premise is negative, the conclusion must be particular negative and should not contain the middle term. So, it follows that 'Some desks are not red'. However, I is the converse of the first premise and thus it holds.
Choose the set of figures which follows the given rule.
Rule: Closed figures losing their sides and open figures gaining their sides.
Let f (x) = , then fof (x) = x, provided that:
⇒ (ac + dc)x^{2} + (bc + d^{2}  bc  a^{2}) x – ab – bd = 0, ∀ x ∈ R
⇒ (a + d)c = 0, d^{2}  a^{2} = 0 and (a + d)b = 0
⇒ a + d = 0
Two finite sets have m and n elements. The number of subsets of the first set is 112 more than that of the second set. The values of m and n respectively are,
2^{m}  2^{n} = 112 ⇒ 2^{n} (2^{mn}  1)= 16.7
∴ 2^{n} (2^{mn}  1) = 2^{4} (2^{3} 1)
Comparing we get n = 4 and m – n = 3
⇒ n = 4 and m = 7
If A and B are positive acute angles satisfying 3 cos ^{2} A + 2 cos ^{2} B = 4 and , Then the value of A + 2B is equal to :
Given, 3 cos^{2}A +2 cos^{2} B = 4
⇒ 2 cos^{2} B 1 = 4  3cos^{2} A 1
⇒ cos 2B = 3 (1  cos ^{2} A) = 3 sin^{2}A...(1)
and 2 cos B sin B = 3 sin A cos A
sin 2B = 3 sin A cos A ...(2)
Now, cos (A + 2B) = cos A cos 2B – sin A sin 2B
= cos A (3 sin^{2} A) – sin A (3 sin A cos A) = 0 [using eqs. (1) and (2)]
⇒ A +2B = π/2
If sin θ_{1} + sin θ_{2} + sin θ_{3} = 3, then cos θ_{1} + cos θ_{2} + cos θ_{3} =
Since, sin θ_{1} + sin θ_{2} + sin θ_{3} = 3
∴ sin θ_{1} = sin θ_{2} = sin θ_{3} = 1 ⇒ θ_{1} = θ_{2} = θ_{3} = π/2
∴ cos θ_{1} + cos θ_{2} + cos θ_{3} = 0
If tan (cot x) = cot (tan x), then sin 2x is equal to :
Given,
The general solution of the equation sin 2x + 2sin x +2 cos x+ 1 = 0 is
Given, sin 2x + 2 sin x + 2 cos x + 1 = 0
⇒ 1 + sin 2x + 2(sin x + cos x ) = 0
⇒ (sin x + cos x ) 2 + 2(sin x + cos x) = 0
⇒ (sin x + cos x ) (sin x + cos x + 2) = 0
∴ sin x + cos x = 0 or sin x + cos x = 2
But, sin x + cos x = 2 is inadmissible.
Since,  sin x  ≤ 1,  cos x  ≤ 1
In a ΔABC, if and the side a = 2, then area of the triangle is
⇒ cot A = cot B = cot C
⇒ A = B = C = 60° ⇒ ΔABC is equilateral
Hence, area of
Given,
∴ 2tan^{–1} a – 2tan^{–1}b = 2 tan^{–1} x
⇒ tan^{–1} a – tan^{–1} b = tan^{–1} x
The arithmetic mean of numbers a, b, c, d, e is M. What is the value of (a – M) + (b – M) + (c – M) + (d – M) + (e – M) ?
Given
⇒ a + b + c + d + e = 5 M
⇒ a + b + c + d + e – 5 M = 0
⇒ (a – M) + (b – M) + (c – M) + (d – M) + (e – M) = 0
Hence, required value = 0
The fourth term of an A.P. is three times of the first term and the seventh term exceeds the twice of the third term by one, then the common difference of the progression is
Let the progression be a, a + d, a + 2d,
Then x 4 = 3x_{1} ⇒ a + 3d = 3a ⇒ 3d = 2a ...(i)
Again x_{7 }= 2x_{3} + 1
⇒ a + 6d = 2(a + 2d) + 1 ⇒ 2d = a + 1 ...(ii)
Solving (i) and (ii), we get
a = 3, d = 2
If log a, log b, and log c are in A.P. and also log a – log 2b, log 2b – log 3c, log 3c – log a are in A.P., then
The series is
If z_{1} = √3 + i√3 an d z_{2} = √3 + i, then the complex number lies in the:
Hence, lies in the first quadrant asboth real and imaginary parts of this number are positive.
 A  = 0 as the matrix A is singular
Apply R_{2} → R_{2} – 2R_{1} and R_{3} → R_{3} – 3R_{1} and expand.
–2(4 – 3λ) + 4(4 – 2λ) = 0
⇒ 8 – 2λ = 0 ⇒ λ = 4
For λ = 4, the second and the third column are proportional.
Let α_{1}, α_{2 }and β_{1}, β_{2} be the roots of ax^{2} + bx + c = 0 and px^{2} + qx + r = 0 respectively. If the system of equations α_{1}y + α_{2}z = 0 and β_{1}y + β_{2}z = 0 has a nontrivial solution, then
Since α_{1}, α_{2} and β_{1}, β_{2} are the roots of ax^{2} + bx + c = 0 and px^{2} + qx + r = 0 respectively, therefore
Since the given system of equation has a nontrivial solution
If [ ] denotes the greatest integer less than or equal to the real number under consideration and –1 ≤ x < 0; 0 ≤ y < 1; 1 ≤ z < 2 , then the value of the determinant is
Since, –1 ≤ x < 0
∴ [x] = –1
0 ≤ y < 1 ∴ [y] = 0
1 ≤ z < 2 ∴ [z] = 1
∴ Given determinant =
If α, β are the roots of the equations x^{2} – 2x– 1 = 0, then what is the value of α^{2} β^{–2}+ α^{ –2} β^{2}
If a, b and c are real numbers then the roots of the equation (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are always
Given equation is (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0
⇒ 3x^{2} – 2(b + a + c) x + ab + bc + ca = 0
Now, here A = 3, B = – 2 (a + b + c)
C = ab + bc + ca
The number of points at which the function is discontinuous is:
The function logx is not defined at x = 0, so, x = 0 is a point of discontinuity. Also for f(x) to be defined, log x ≠ 0 ⇒ x ≠ ± 1.
Hence, 0, 1, –1 are three points of discontinuity.
We have,
Since Lf'(0) = Rf'(0), therefore f(x) is differentiable at x = 0
Since differentiability ⇒ continuity, therefore f(x) is continuous at x = 0.
The set of all values of a for which the function f(x) = (a^{2} – 3a + 2) (cos^{2}x/4 –sin^{2}x/4) + (a –1) x + sin 1 does not possess critical points is
Match List I with List II and select the correct answer using the code given below the lists:
Codes:
(A) Graph of f(x) = cos x cuts xaxis at infinite number of points. (5 of list II)
(B) Graph of f(x) = In x cuts xaxis in only one point. (4 of list II)
(C) Graph of f(x) = x^{2} – 5x + 4 cuts x axis in two points (2 of list II)
(D) Graph of f(x) = e^{x} cuts yaxis in only one point. (3 of list II)
What is the xcoordinate of the point on the curve f (x) = √x (7x – 6), where the tangent is parallel to xaxis?
f(x) = √x (7x – 6) = 7x^{3/2} – 6x^{1/2}
When tangent is parallel to x axis f '(x) = 0
A wire 34 cm long is to be bent in the form of a quadrilateral of which each angle is 90°. What is the maximum area which can be enclosed inside the quadrilateral?
Let one side of quadrilateral be x an d another side be y
so, 2(x + y) = 34
or, (x + y) = 17 ...(i)
We know from the basic principle that for a given perimeter square has the maximum area, so, x = y and putting this value in equation (i)
x = y = 17/2
Area = x . y =
Consider the following statements in respect of the function f (x) = x^{3} – 1, x ∈ [–1, 1]
I. f (x) is increasing in [– 1, 1]
II. f (x) has no root in (– 1, 1).
Which of the statements given above is/are correct?
Since f (x) is an increasin g function in [– 1, 1] and it has a root in (– 1, 1).
∴ Only statement I is correct.
At an extreme point of a function f (x), the tangent to the curve is
At an extreme point of a function f (x), slope is always zero.
Thus, At an extreme point of a function f (x), the tangent to the curve is parallel to the xaxis.
The curve y = xe^{x} has minimum value equal to
Let y = xe^{x}.
Differentiate both side w.r.t. ‘x’.
Hence, y = xe^{x} is minimum function and y_{min} =
A ray of light coming from the point (1, 2) is reflected at a point A on the xaxis and then passes through the point (5, 3). The coordinates of the point A is
The equation x^{2}  2√3xy + 3y^{2}  3x + 3√3y  4 = 0 represents
We have
Thus abc + 2fgh  af^{2}  bg^{2}  ch^{2} = 0
Hence the equation represents a pair of straight lines.
∴ the lines are parallel. The distance between them
The line joining (5, 0) to ( (10 cos θ, 10 sin θ) is divided internally in the ratio 2 : 3 at P. If θ varies, then the locus of P is
Let P(x, y) be the point dividing the join of A and B in the ratio 2 : 3 internally, then
Squaring and adding (i) and (ii), we get the required locus ( x  3)^{2 }+ y^{2} = 16, which is a circle.
The number of integral values of λ for which x^{2} + y^{2} + λx + (1 – λ)y + 5 = 0 is the equation of a circle whose radius cannot exceed 5, is
Radius ≤ 5
∴ λ = 7,  6,  5, .......,7, 8 , in all 16 values
The lengths of the tangent drawn from any point on the circle 15x^{2} + 15y^{2}  48x + 64y = 0 to the two circles 5x^{2} + 5y^{2} – 24x + 32y + 75 = 0 and 5x^{2} + 5y^{2} – 48x + 64y + 300 = 0 are in the ratio of
The length of the chord x + y = 3 intercepted by the circle x^{2} + y^{2}  2x  2y  2 = 0 is
The locus of the point of intersection of two tangents to the parabola y^{2} = 4ax, which are at right angle to one another is
The parabola having its focus at (3, 2) and directrix along the yaxis has its vertex at
Vertex of the parabola is a point which lies on the axis of the parabola, which is a line ⊥ to the directrix through the focus, i.e., y = 2 and equidistant from the focus and directrix x = 0, so that the vertex is
The number of values of r satisfying the equation
^{39}C_{3r1}  ^{39}C_{r}^{2} = ^{39}C_{r21}  ^{39}C_{3r }is
^{39}C_{3r1}  ^{39}C_{r}^{2} = ^{39}C_{r21}  ^{39}C_{3r}
⇒ ^{39}C_{3r1} + ^{39}C_{3r }= ^{39}C_{r21} + ^{39}C_{r2}
⇒ ^{40}C_{3r} = ^{40}C_{r}^{2 }
⇒ r^{2} = 3r or r^{2} = 40  3r
⇒ r = 0, 3 or – 8, 5
3 and 5 are the values as the given equation is not defined by r = 0 and r = –8. Hence, the number of values of r is 2.
Comparing we get n + 1 = 6 ⇒ n = 5
All the words that can be formed using alphabets A, H, L, U and R are written as in a dictionary (no alphabet is repeated). Rank of the word RAHUL is
No. of words starting with A are 4 ! = 24
No. of words starting with H are 4 ! = 24
No. of words starting with L are 4 ! = 24
These account for 72 words
Next word is RAHLU and the 74th word RAHUL.
If the sum of odd numbered terms and the sum of even numbered terms in the expansion of (x + a)^{n} are A and B respectively, then the value of (x^{2} – a^{2})^{n} is
= A + B ....(1)
Similarly, (x – a)^{n} = A – B .....(2)
Multiplying eqns. (1) and (2), we get
(x^{2}  a^{2} )^{n} = A^{2}  B^{2}
If the third term in the expansion of [ x + x^{ log10 x} ]5 is 10^{6}, then x may be
Put log _{10} x = y, th e given expr ession becomes (x + x^{y})^{5}.
T_{3} = ^{5}C_{2} . x^{3} ( x^{y})^{2} = 10x ^{3+2y }= 10^{6} (given)
⇒ (3 + 2y) log_{10} x = 5 log_{10} 10 = 5
⇒ (3 + 2y)y = 5
∴ x = 10 or x = (10)^{–5/2}
If three vertices of a regular hexagon are chosen at random, then the chance that they form an equilateral triangle is:
Three vertices can be selected in ^{6}C_{3} ways.
The only equilateral triangles possible are A_{1}A_{3}A_{5} and A_{2}A_{4}A_{6 }
A man takes a step forward with probability 0.4 and backward with probability 0.6. The probability that at the end of eleven steps he is one step away from the starting point is
As 0.4 + 0.6 = 1, the man either takes a step forward or a step backward. Let a step forward be a success and a step backward be a failure.
Then, the probability of success in one step
The probability of failure in one step
In 11 steps he will be one step away from the starting point if the numbers of successes and failures differ by 1.
So, the number of successes = 6
The number of failures = 5 or the number of successes = 5,
The number of failures = 6
∴ the required probability
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