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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 R_{e})
The escape velocity from earth is given by
v_{e }=
The orbital velocity of a satellite revolving around earth is given by
where, M_{e} = mass of earth, R_{e} = radius of earth, h = height of satellite from surface of earth.
By the relation GM_{e} = gR^{2}_{s}
So,
Dividing equation (i) by (ii), we get
Given,
Squaring on both side, we get
or R_{e} + h = 2R_{e }i.e., h = R_{e}
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)
A_{H} = B_{H} … (ii)
From FBD of block B,
B_{H} + F_{B} cos 60° – N_{B} sin 60° = 0
N_{B} cos 60° – B_{V} – 300 + F_{B} sin 60° = 0
F_{B} = 0.25 N_{B}
B_{H} – 0.74 N_{B} = 0 … (iii)
– B_{V} + 0.71 N_{B} = 300 … (iv)
FBD of block A
F_{A} – A_{H} = 0
N_{A} – A_{V} = 400 … (v)
F_{A} = µ_{A} N_{A}
∴ µ_{A}N_{A} – A_{H} = 0 … (vi)
On solving above equations, we get N_{A} = 650 N, F_{A} = 260N, F_{A }= µ_{A}N_{A}
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 v_{S }= 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 = n_{1} – n_{2}
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 ^{2}ωt 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 = kr^{a}, 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, E_{2} = E_{1}/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 = mv^{2}/r
And KE = k = 1/2 mv^{2}
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, u_{AB} = v_{0}
using equation v^{2} = u^{2} + 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 = – (U_{F} – U_{i}) = U_{i }– U_{F}.
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 t_{1}. If it is thrown vertically downwards from A with the same speed, it reaches the ground in time t_{2}. 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 u_{0}.
Using equation , s = ut + (1/2)at^{2}
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, u_{0} = 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 , crosssection ar ea (1/9) m^{2} 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 = (a_{0} – α t), where a_{0} = side at t = 0.
At this instant, flux through the square :
⇒ E = – B.2 (a_{0} – 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} Wm^{2} equally distributed amongst the three wavelengths. The beam falls normally on the area 1 cm^{2} 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
⇒ λ_{0} = 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 n_{2} due to the wavelength 4972 Å
Therefore total number of photoelectrons liberated in 2s,
N = n_{1} + n_{2}
= 0.5 x 10^{12} + 0.575 x 10^{12}
= 1.075 x 10^{12}
A squar e gate of size 1 m × 1m is h inged at its midpoint. 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 = p_{0} + ρ_{g}y  p_{0}) 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 Xr 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 kshell 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 V_{0} = 2 V; collector resistance, R_{c} = 2 × 10^{3} Ω; Base resistance R_{B} = 1 × 10^{3 }Ω; Input signal voltage, V_{i} = ?
V_{0} = I_{C}R_{C} = 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 midpoint of the line joining the other two.
Initial potential energy of the system
Let charge at A is moved to midpoint O, Then final potential energy of thhe system
Work done = U_{f} – U_{i} = 18 × 10^{7} J
Also, energy supplied per sec = 1000 J (given)
Time required to move one of the midpoint 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 n_{2} moles of helium in the given mixture. As Pv = nRT
Then the pressure of the mixture
⇒
or,
or, n_{1 }+ n_{2} = 1.62 ... (1)
The mass of the mixture is (in grams)
n_{1} × 2 + n_{2} × 4 = 5
⇒ (n_{1} + 2n_{2}) = 2.5 ... (2)
Solving the eqns. (1) and (2), we get n_{1} = 0.74 and n_{2} = 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^{–7}m apart. A transparent sheet of thickness t = 1.5 × 10^{–7}m 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), u_{x} = 40/2 = 20m/s
Vertical velocity (initial), 50
⇒
or, 50 = 2u_{y} – 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, x_{1} is maximum elongation in the spring when the particle is in its lowest extreme position. Then,
Amplitude A = X_{1} – X_{0} (elongation in spring for equilibrium position)
In an ore containing uranium, the ratio of U^{238} to Pb^{206} is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of U^{238}. Take the halflife of U^{238} to be 4.5 × 10^{9} yr.
Let the initial mass of uranium be M_{0}
Final mass of uranium after time t, M = 3/4(M_{0})
According to the law of radioactive disintegration.
⇒ t = 1.867 × 10^{9} 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 µ_{1} & µ_{2} 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
= I_{1} + I_{2} + I_{3} + I_{4} = 2I_{1} + 2I_{2}
= 2(l_{1} + l_{2}) [I_{3} = I_{1}, I_{1} = I_{4}]
Now, I_{2} = I_{3} = MI^{2}/3
Using parallel axes theorem, we have
Putting all values we get
Moment of inertia,
One of the lines in the emission spectrum of Li^{2+} has the same wavelength as that of the 2^{nd} 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 2^{nd} 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 R_{1} and R_{2}, 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, mv^{2}/R = qvB ∴
Hence,
or, m ∝ R^{2}
[∵ 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 u_{0 }= 20√2 m/s; angle of projection θ = 45°
Therefore horizontal and vertical components of initial velocity are u_{x} = 20√2 cos 45°= 20m/s and u_{y} = 20√2 sin 45°= 20m/s
After 1s, horizontal component remains unchanged while the vertical component becomes v_{y} = u_{y} – 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 = h_{1} + h_{2}
Using, h = ut + (1/2)at^{2}
a = g = 10 m/s^{2}
H = 20 + 15 = 35 m
A 2 m wide truck is moving with a uniform speed v_{0} = 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 = V_{1}
Speed of proton or hydrogen atom after collision = V_{2}
Energy of excitation = DE From the law of conservation of linear momentum, mv = mv_{1} + mv_{2} ...(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, v_{1 }– v_{2 }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
a_{max} = 2ω^{2} = 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' = Q_{2} – Q_{2} = 4CV – CV = 3CV
The two capacitors will have some potential, say V'.
The net capacitance of the parallel combination of the two capacitors C' = C_{1} + C_{2} = 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, f_{w} = 0.8f_{air}
In liquid, g'/g = (0.6)^{2} = 0.36
...(2)
From eq. (1) and (2)
A long straight wire along the Zaxis carries a current I in the negative Zdirection. 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 zdirection. 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]5f^{14}6d^{10} 7s^{2}7p^{6}. Since its elctronic configuration in the outer most orbit (ns^{2}np^{6}) 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 S_{T} = Entropy at T K
S_{0} = Entropy at 0K = C_{p} . log_{e}T
= 2.303C_{p}.logT
CoCl_{3}.5NH_{3}.H_{2}O is pink in colour
Cobalt is present in vitamin B_{12}.
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?
ClO_{4}^{} < ClO_{3}^{}< ClO_{2}^{}< 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 dipoledipole 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 K_{4} [Fe (CN)_{6}], do not dissociate into Fe^{2+} 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., N_{2}O) 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.
The common name of K[PtCl_{3}(η^{2}.C_{2}H_{4})] is
Zeise’s salt is common name of K[Pt Cl_{3}(η^{2} = C_{2}H_{4})]
CaCl_{2} is produced as a by product in solvay ammonia process.
(i) NaCl + CO_{2 }+ NH_{3} + H_{2}O → NaHCO_{3} + NH_{4}Cl
(ii) CaCO_{3} → CO_{2} + CaO
(iii) 2 NH_{2}Cl + 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 NH_{3} > PH_{3} > AsH_{3} > SbH_{3}. Basiccharacter 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, CaCO_{3} and SiO_{2}.
Exa = 10^{18}
Which of the following is the most basic oxide?
More the oxidation state of the central atom (metal) more is its acidity. Hence SeO_{2} (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 Al_{2}O_{3} and Sb_{2}O_{3} are amphoteric and Bi_{2}O_{3 }is basic.
Wh ich on e of th e following does not follow octate rule?
BF_{3} 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 LeChatelier’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] = [A_{0}]e^{–kt}
∴ The concentration of reactants will exponentially decreases with time.
The P—P—P angle in P_{4 }molecule and S—S—S angle in S_{8 }molecule is(in degree) respectively
In P_{4} molecule, the four sp^{3}hybr idised phosphorous atoms lie at the corners of a regular tetrahedron with ÐPPP = 60°.
In S_{8} molecule SSS angle is 107° rings.
The number of elements present in the dblock of the periodic table is
40 elements are pr esent in dblock.
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
Fe^{3+} ion can be detected by K_{4}[Fe(CN)_{6}]
4Fe^{3+} + 3K_{4} [Fe(CN)_{6}] → Fe_{4} [Fe(CN)_{6}]_{3}+ 12K^{+}
ΔG = ΔH – TΔS; ΔG is positive for a reaction to be nonspontaneous 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
XeF_{2} : Hybridization sp^{3}d^{2}
Linear
Xe F^{4} : Hybridization sp^{3}d^{2}
Square planar
XeF_{6} : Hybribisation sp^{3}d^{3}
Pentagonal pyramidal or distorted octahedral
Calcium silicophosp h ate (a mixture of Ca_{3}(PO_{4})_{2} & Ca_{2}SiO_{4}) 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, 4^{3} = 64 triplets or codons are possible.
Which of the following molecule/ion has all the three types of bonds, electrovalent, covalent and coordinate :
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 motherinlaw of Ekta who is sisterinlaw 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.
Consider
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)x^{2}  4(2  i)x  5  3i = 0 which has greater modulus is
Roots =
Consider 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:
E_{1} = Getting a coin having head on both sides from the bag.
E_{2} = Getting a fair coin from the bag
A = Toss results in a head
Given: and P
Then,
P(A) = P(E_{1}) P(A /E_{1}) + P(E_{2}) P(A/E_{2})
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 x^{2} + y^{2} ≤ 1} is
Required area = 4 (Area of the shaded region in first quadrant)
Universal set,
U = {x  x^{5} – 6x^{4} + 11x^{3} – 6x^{2} = 0}
A = {x  x^{2} – 5x + 6 = 0}
B = {x  x^{2} – 3x + 2 = 0}
What is (A ∩ B)' equal to ?
U = {x : x^{5} – 6x^{4} + 11x^{3} – 6x^{2} = 0}
Solving for values of x, we get
U = {0, 1, 2, 3}
A = {x : x^{2} – 5x + 6 = 0}
Solving for values of x, we get
A = {2, 3}
and B = {x : x^{2} – 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}
If then 4x^{2}  4 xy cosα + y^{2} is equal to
⇒ 4  y^{2} 4x^{2 }+ x^{2}y^{2} = 4 cos^{2} α + x^{2} y^{2}  4 xy cosα
⇒ 4x^{2} + y^{2}  4xycosα = 4sin^{2}α.
If then the value of 2a_{1} + 2^{3}a_{3} + 2^{5}a_{5}+ .... is
Let
By using
= a_{1} = a_{3 }= a_{5 }= ... = 0
Hence, 2a_{1} + 2^{3}a_{3} + 2^{5}a^{5} + ....= 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 4digit 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 = ^{10}C^{4} × 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 4digit numbers.
The line which is parallel to Xaxis and crosses the curve y = √x at an angle of 45°, is
Given equation of a line parallel to Xaxis is y = k.
Given equation of the curve is y = √x ,
On solving equation of line with the equation of curve, we get x = k^{2}
Thus the intersecting point is (k^{2}, 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, ^{n}C_{1}, ^{n}C_{2},....., ^{n}C_{n} is
Since, where x_{i} are observations with frequencies f_{i,} i = 1, 2, ........n The required mean is given by
The mean square deviation of a set of n observation x_{1}, x_{2}, .... x_{n} 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 y^{2} = 8x and PQ be the common chord of the circle x^{2} + y^{2} – 2x – 4y = 0 and the given parabola. The area of DPQS is
The parametr ic equation s of the parabola y^{2} = 8x are x = 2t^{2} and y = 4t.
and the given equation of circle is x^{2} + y^{2 }– 2x – 4y = 0
On putting x = 2t^{2} and y = 4t in circle we get
4t^{4 }+ 16t^{2} – 4t^{2} – 16t = 0
⇒ 4t^{2} + 12t^{2} – 16t = 0
⇒ 4t (t^{3} + 3t – 4) = 0
⇒ t(t – 1) (t^{2} + 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 e^{x–1} + x – 2 = 0 is
Let f(x) = e^{x–1} + x – 2
check for x = 1
Then, f (1) = e^{0} + 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) = e^{x–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 nonsingular 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) = α^{2}l
Hence, K = α
Tangents are drawn from the origin to the curve y = cos x. Their points of contact lie on
Let (x_{1}, y_{1}) be one of the points of contact.
Given curve is y = cos x
Now the equation of the tangent at (x_{1},y_{1}) is
⇒y y_{1} =  sin x_{1}(0 x_{1})
Since, it is given that equation of tangent passes through origin.
∴ 0  y_{1} =  sin x_{1} (0x_{1})
⇒ y_{1} = – x_{1 }sin x_{1} ...(i)
Also, point (x_{1}, y_{1}) lies on y = cos x.
∴ y_{1} = cos x_{1}
From Eqs. (i), (ii) , we get
⇒
Hence, the locus of (x_{1}, y_{1}) is x^{2} = y^{2} + y^{2}x^{2} ⇒ x^{2}y^{2 }= x^{2} – y^{2}
The slope of the tangent to the curve y = e^{x} 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 = e^{x} cos x.
Then,
Diff. w.r.t ‘x’
⇒ =2e^{x} 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
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 nonzero roots of x^{2} + ax + b = 0 then the least value of x^{2} + ax + b is
As given a and b are the roots of the equation x^{2} + 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) = x^{2} + 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) = sec^{2}x – 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) = ax^{3} + bx^{2} + 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) = ax^{3} + bx^{2} + cx + d
⇒ f'(x) = 3ax^{2} + 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) = x^{3} – 1
Hence,
if (i)1 ≤ x 3 ≤ 1 ⇒ 2 ≤ x ≤ 4 and
(ii) 9x^{2} > 0 ⇒ 3 < x < 3
Taking common solution of (i) and (ii), we get 2 ≤ x < 3
∴ Domain = [2, 3)
If the lines p_{1}x + q_{1}y = 1, p_{2}x + q_{2}y = 1 and p_{3}x + q_{3}y = 1 be concurrent, then the points (p_{1}, q_{1}), (p_{2}, q_{2}) and (p_{3}, q_{3})
The equations of the lines are
p_{1}x + q_{1}y  1 = 0 ...(i)
p_{2}x + q_{2}y  1 = 0 ...(ii)
and p_{3}x + q_{3}y  1 = 0 ...(iii)
As they are concurrent,
This is also the condition for the points (p_{1}, q_{1}), (p_{2}, q_{2}) and (p_{3}, q_{3}) 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
If then the value of (m^{2} – n^{2}) sin^{2} B is
cos A = n cos B and sin A = m sin B
Squaring and adding, we get
1 = n^{2} cos^{2}B + m^{2}sin^{2}B
⇒ 1 = n^{2} (1 – sin^{2}B) + m^{2}sin^{2}B
∴ (m^{2} – n^{2}) sin^{2}B = 1 – n^{2}
If complex number z_{1}, z_{2} and 0 are vertices of equilateral triangle, then is equal to
z_{1}, z_{2}, 0 are vertices of an equilateral triangle, so we have
⇒
If ρ = {(x, y) x^{2} + y^{2} = 1; x, y ∈ R}. Then, ρ is
Obviously, the relation is not reflexive and transitive, but it is symmetric, because
x^{2 }+ x^{2} = 2x^{2} ≠ 1
and x^{2} + y^{2 }= 1, y^{2} + z^{2} = 1
⇒ x^{2} + z^{2} = 1
But x^{2} + y^{2} = 1 ⇒ y^{2} + x^{2} = 1
A line makes the same angle θ with each of the X and Zaxes. If the angle β, which it makes with Yaxis, is such that sin^{2} β = 3sin^{2} θ, then cos^{2} θ equals
Let l, m and n be the direction cosines.
Then, l = cos θ, m = cos β, n = cos θ
we have l^{2} + m^{2} + n^{2} = 1
⇒ tan^{2}θ = 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) = ^{4}C_{0}p^{0}q^{4} = 16/81
⇒
∴ P(X = 4) = ^{4}C_{4}p^{4}q^{0} = p^{4} = (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,
⇒ n^{2} – 4nr + 4r^{2 }+ n = 0
⇒ (n – 2r)^{2} + n = 0
which is not possible for any value for n.
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