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A rifle man, who together with his rifle has a mass of 100 kg, stands on a smooth surface and fires 10 shots horizontally. Each bullet has a mass 10 g and a muzzle velocity of 800 ms^{–1}. The velocity which the rifle man attains after firing 10 shots is
According to law of conservation of
momentum,
100v = 1/1000 x 10 x 800
i.e., v = 0.8 ms^{–1}.
A train accelerating uniformly from rest attains a maximum speed of 40 ms^{–1} in 20 s. It travels at the speed for 20 s and is brought to rest with uniform retardation in further 40 s. What is the average velocity during the period ?
(i) v = u + at_{1}
40 = 0 + a × 20
a = 2 m/s^{2}
v^{2 }– u^{2} = 2as
40^{2} – 0 = 2 × 2 s_{1}
s_{1} = 400 m
(ii) s_{2} = v × t_{2} = 40 × 20 = 800 m
(iii) v = u + at
0 = 40 + a × 40
a = –1 m/s^{2}
0^{2} – 402 = 2(–1)s_{3}
s_{3} = 800 m
Total distance travelled = s_{1} + s_{2} + s_{3}
= 400 + 800 + 800 = 2000 m
Total time taken = 20 + 20 + 40 = 80 s
Average velocity 2000/80 = 25m/s
A projectile is fired with a velocity u making an angle q with the horizontal. What is the magnitude of change in velocity when it is at the highest point –
Initially u = cosθ iˆ + usinθ ˆj .
At highest point v = u cosθ iˆ
∴ difference is u sin q.
For the equation F = A^{a}v^{b}d^{c}, where F is the force, A is the area, v is the velocity and d is the density, the values of a, b and c are respectively
[MLT^{–2}] = [L^{2a}] × [L^{b}T^{–b}][M^{c}L^{–3c}]
= [M^{c}L^{2a + b} –3^{c}T^{–b}]
Comparing powers of M, L and T, on both
sides, we get
c = 1, 2a + b –3c = 1, –b = –2 or b = 2
Also, 2a + 2 – 3(1) =1 Þ 2a = 2 or a = 1
∴ This is 1, 2, 1
A person with his hand in his pocket is skating on ice at the rate of 10m/s and describes a circle of radius 50 m. What is his inclination to vertical: (g = 10 m/sec2)
Since surface (ice) is frictionless, so the
centripetal force required for skating will be
provided by inclination of boy with the
vertical and that angle is given as
tan θ = v2/rg
= where v is speed of skating &
r is radius of circle in which he moves.
A small block of mass m is kept on a rough inclined surface of inclination q fixed in a elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be :
An equilateral prism of mass m rests on a rough horizontal surface with coefficient of friction µ.
A horizontal force F is applied on the prism as shown in the figure.
If the coefficient of friction is sufficiently high so that the prism does not slide before toppling, then the minimum force required to topple the prism is –
The tendency of rotation will be about the point C.
For minimum force, the torque of F about C has to be equal to the torque of mg about C.
∴
A spherically symmetric gravitational system of particles has a mass density where r_{0 }is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed V as a function of distance r (0 < r < ∞) from the centre of the system is represented by
The load versus elongation graph for four wires is shown. The thinnest wire is
Th e wor k done in blowin g a soap bubble of surface tension 0.06 × Nm^{–1} from 2 cm radius to 5 cm radius is
= 0.003168 J
The wavelength of radiation emitted by a body depends upon
One mole of O_{2 }gas having a volume equal to 22.4 Litres at 0°C and 1 atmospheric pressure in compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is
In a thermodynamic process, the pressure of a fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8 J of work is done on the gas. If the initial internal energy of the gas was 30 J, then the final internal energy will be
Given ΔQ = – 20 J, W = – 8 J
Using Ist law ΔQ = ΔU + ΔW
⇒ ΔQ = – 20 + 8 = – 12 J
Uf = – 12 + 30 = 18 J
In the kinetic theory of gases, which of these statements is/are true ?
(i) The pressure of a gas is proportional to the mean speed of the molecules.
(ii) The root mean square speed of the molecules is proportional to the pressure.
(iii) The rate of diffusion is proportional to the mean speed of the molecules.
(iv) The mean translational kinetic energy of a gas is proportional to its kelvin temperature.
Two balloons are filled one with pure he gas and other with air respectively. If the pressure and temperature of these balloons are same, then the number of molecules per unit volume is
Assumingthe balloons have the same volume, as pV= nRT. If P, V and T are the same, n the number of moles present will be the same, whether it is He or air.
Hence, number of molecules per unit volume will be same in both the balloons.
Two particles P and Q describe S.H.M. of same amplitude a, same frequency f along the same straight line. The maximum distance between the two particles is a√2 . The initial phase difference between the particle is –
x_{1} = a sin(ωt + ϕ_{1}), x_{2} = a sin(ωt + ϕ_{2} )
⇒
To maximize
⇒
⇒
⇒
A tunnel has been dug through the centre of the earth and a ball is released in it. It executes S.H.M. with time period
=
=
= 84.6 min
A soun d source, em itting sound of con stan t frequency, moves with a constant speed and crosses a stationary observer. The frequency (n) of sound heard by the observer is plotted against time (t). Which of the following graphs represents the correct variation ?
When a string is divided into three segments of length l_{1}, l_{2}, and l_{3} the fundamental frequencies of these three segments are v_{1}, v_{2} and v_{3} respectively. The original fundamental frequency (v) of the string is
Fundamental frequency is given by
Since, P divided into l_{1 }, l_{2 }and l_{3} segments
Herel = l_{1} + l_{2} + l_{3}
So
Two point dipoles pkˆ and p/2 kˆ are located at(0, 0, 0) and (1m, 0, 2m) respectively. The resultant electric field due to the two dipoles at the point (1m, 0, 0) is
The given point is at axis of
dipole and at equatorial line of dipole so that field at given point.
Electric field in the region is given by then the correct expression for the potential in the region is [assume potential at infinity is zero]
Three capacitors C_{1} = 1 µF, C_{2} = 2 µF and C_{3} = 3 µF are connected as shown in figure, then the equivalent capacitance between points A and B is
C_{eq} = (1 + 2 + 3)μF = 6 μF
Two long coaxial and conducting cylinders of radius a and b are separated by a material of conductivity s and a constant potential difference V is maintained between them, by a battery. Then the current, per unit length of the cylinder flowing from one cylinder to the other is –
Now
From (1) :
A wire X is half the diameter and half the length of a wire Y of similar material. The ratio of resistance of X to that of Y is
A narrow beam of protons and deuterons, each having the same momentum, enters a region of uniform magnetic field directed perpendicular to their direction of momentum. The ratio of the radii of the circular paths described by them is
Since, the radius of circular path of a charged particle in magnetic field is
Now, the radius of circular path of charged particle of given momentum r and magnetic field B is given by
For th e cir cuit (figur e), the cur rent is to be measured. The ammeter shown is a galvanometer with a resistance R_{G} = 60.00W converted to an ammeter by a shunt resistance r_{s} = 0.02W. The value of the current is
RG = 60.00W, shunt resistance, rs = 0.02W
Total resistance in the circuit is RG + 3 = 63W
Hence, I = 3/63 = 0.048 A
Resistance of the galvanometer converted to an ammeter is,
Total resistance in the circuit = 0.02 + 3
= 3.02W
Hence, I = 3/3.02 = 0.99 A
The susceptibility of a magnetism at 300 K is 1.2 × 10^{–5}. The temperature at which the susceptibility increases to 1.8 × 10^{–5} is
⇒
⇒
A coil 10 turns and a resistance of 20W is connected in series with B.G. of resistance 30W. The coil is placed with its plane perpendicular to the direction of a uniform magnetic field of induction 10^{–2} T. If it is now turned through an angle of 60° about an axis in its plane. Find the charge induced in the coil. (Area of a coil = 10^{–2} m²)
Given : n = 10 turns, R_{coil} = 20W, R_{G} = 30W, Total resistance in the circuit = 20 + 30 = 50W.
=
=
= 1 × 10^{–5} C (Charge induced in a coil)
Voltage V and current i in AC circuit are given by V = 50 sin (50 t) volt, i = 50 sin. The power dissipated in the circuit is
Given V = 50 sin (50 t) V
Maximum voltage, V0 = 50 V,
Maximum current, i_{0} = 50 mA = 50 × 10^{–3 }A
Power dissipated,
=
Resolving power of the telescope will be more, if the diameter of the objective is
The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be 20 cm. The focal length of lenses are
∴ f_{0} = 9f_{e}
Also f_{0} + f_{e} = 20 (Qfinal image is at infinity)
9 f_{e} + f_{e} = 20, f_{e} = 2 cm, ∴ f0 = 18 cm
The angular size of the central maxima due to a single slit diffraction is (a ⇒ slit width)
Angular size of central maxima is
Find the final intensity of light (I"), if the angle between the axes of two polaroids is 60°.
From the first polaroid
From second polaroid
=
The threshold wavelength of the tungsten is 2300 Å. If ultraviolet light of wavelength 1800 Å is incident on it, then the maximum kinetic energy of photoelectrons would be about –
Gr aph betwen stopping potential for most energetic emitted photoelectrons (V_{s}) with frequency (u) of incident radiation on metal is given below. Value of AB/BC, in graph is [where h = plank’s constant, e = electronic charge]
If hydrogen atom, an electron jumps from bigger orbit to smaller orbit so that radius of smaller orbit is onefourth of radius of bigger orbit. If speed of electron in bigger orbit was v, then speed in smaller orbit is
Radius of the orbit , r_{n} ∝ n^{2}
⇒
⇒
Velocity of electron in n^{th} orbit
⇒ v_{n small} = 2(v_{n big}) = 2v
A nucleus of uranium decays at rest into nuclei of thorium and helium. Then :
In an explosion a body breaks up into two pieces of unequal masses both part will have numerically equal momentum and lighter part will have more velocity.
U → Th + He
sinc m_{He} is less so KE_{He} will be more.
Let binding energy per nucleon of nucleus is denoted by E_{bn} and radius of nucleus is denoted as r. If mass number of nuclei A, B and 64 and 125 respectively then
r increases with increasing A mass number So, r_{A} < r_{B} as mass number of A is smaller E_{bn} decreases with increasing A for A > 56, ^{56}Fe has highest E^{bn} value.
So, E_{bn} for A = 64 is larger as compared to E_{bn} for nucleus with A = 125
For a CE transistor amplifier, the audio signal voltage across the collector resistance of 2.0 kW is 2.0 V. Suppose the current amplification factor of the transistor is 100, What should be the value of R_{B} in series with V_{BB} supply of 2.0V if the dc base current has to be 10 times the signal current?
The output ac voltage is 2.0 V. So, the ac collector current i_{C} = 2.0/2000 = 1.0 mA.
The signal current through the base is,
therefore given by
i_{B} = i_{C} /β = 1.0 mA/100 = 0.010 mA.
The dc base current has to be 10 × 0.010
= 0.10 mA.
R_{B} = (V_{BB} – V_{BE} ) /I_{B}.
Assuming V_{BE} = 0.6 V, R_{B} = (2.0 . 0.6 )/0.10
= 14 kW.
The final boolean expression is,
The formation of CO and CO_{2} illustrates the law of
Formation of CO and CO_{2 }illustrates the law of multiple proportion that is constant mass of C reacts with different masses of oxygen. These masses here bears simple ratio of 1 : 2.
The wave number of the limiting line in Lyman series of hydrogen is 109678 cm^{–1}. The wave number of the limiting line in Balmer series of He^{+} would be :
R_{H} = 109678 cm^{–1}
Wave number of the limiting line in
Balmer series of He^{+}
=
=
= 109678 cm^{–1}
Th e va len cy sh el l of elemen t A con ta in s 3 electrons while the valency shell of element B contains 6 electrons. If A combines with B, the probable formula of the compound formed will be
The element A is ns^{2}p^{1} and B is ns^{2}p^{4}. They can form compound of the type A_{2}B_{3}.
The enthalpy of sublimation of aluminium is 330 kJ/mol. Its I^{st}, II^{nd} and III^{rd} ionization enthalpies are 580, 1820 and 2740 kJ respectively. How much heat has too be supplied (in kJ) to convert 13.5 g of aluminium into Al^{3+} ions and electrons at 298 k
Heat needed too be supplied
per mol = 330 + 580 + 1820 + 2740
= 5470 kJ
Heat required = 0.5 × 5470 kJ = 2735 kJ
Which one of the following pairs is isostructural (i.e., having the same shape and hybridization)?
BF_{4}^{} hybridisation sp^{3}, tetrahedral
structure.
NH_{4}^{+} hybridisation sp^{3}, tetrahedral
structure.
N_{2} and O_{2} are converted into mono anions, N_{2}^{–} and O_{2}^{–} respectively. Which of the following statements is wrong ?
We know that in O_{2} bond, the order is 2
and in O_{2}^{–} bond, the order is 1.5. Therefore the wrong statements is (b).
If the enthalpy of vaporization of water is 186.5 kJmol^{–1}, the entropy if its vaporization will be :
Given enthalpy of vaporization,
ΔH = 186.5kJmol^{1}
Boiling point of water
= 100°C =100 + 273 = 373K
Entropy change,
= 0.5 k JK^{–1}mol^{–1}
Th e heats of neutra lisation of CH_{3}CO OH, HCOOH, HCN and H_{2}S are – 13.2, – 13.4, – 2.9 and – 3.8 kCal per equivalent respectively. Arrange the acids in increasing order of acidic strength.
The greater the (negative value) of heat of neutralisation, the more is the strength of the acid. Hence,
HCOOH > CH_{3}COOH > H_{2}S > HCN
K_{c} for the the reaction, [Ag(CN)_{2}]^{–} Ag+ + 2CN^{–}, the equillibrium constant at 25°C is 4.0 × 10^{–19}, then the silver ion concentration in a solution which was originally 0.1 molar in KCN and 0.03 molar in AgNO_{3} is :
0.04 ≈0.04
∴
a = 7.5 × 10^{–18}
The ratio of oxidation states of Cl in potassium chloride to that in potassium chlorate is
KCl x = –1 KClO_{3}
potassium chloride +1 + x – 6 = 0
x = + 5 potassium chlorate.
∴ Ratio of oxidation state of Cl = 1/5
Which of the following among alkali metal is most reactive ?
Amongst alkali metals, Cs is most reactive because of its lowest IE.
Which of the following compounds has wrong IUPAC name?
The compound which gives th e most stable carbonium ion on dehydration is
The correct order of increasing C  O bond length CO, CO_{3}^{2} , CO_{2} is:
bond order. O = C = O has second lowest bond length due to double bond.
has highest has bond length due to lowest bond order which is due to resonance.
An organic compound A (C_{4}H_{9}Cl) on reaction with Na/diethyl ether gives a hydrocarbon which on monochlorination gives only one chloro derivative, then A is
When rain is accompanied by a t hunderstorm, the collected rain water will have a pH value:
Normal rain water has pH 5.6. Thunderstorm results in the formation NO and HNO_{3} which lowers the pH.
An elemental crystal has a density of 8570 kg/m^{3}. The packing efficiency is 0.68. The closest distance of approach between neighbouring atom is 2.86 Å. What is the mass of one atom approximately?
The packing efficiency = 0.68, means the given lattice is BCC.
The closest distance of approach = 2r
Let atomic weight of the element = a
∴
a = 8.57 × 3 × (3.3)^{3} × 0.1
= 92.39 ≌ 93 amu
Identify the correct or der of solubilty of Na_{2}S. CuS and ZnS in aqueous medium
The correct order of solublity of sulphides is Na_{2}S > ZnS > CuS
In the cell reaction
E^{0}_{cell }= 0.46 V. By doubling the concentration of Cu2+, E^{0}_{cell }is
Cu^{+}_{aq} is un stable in solution a n d un der goes simultaneous oxidation and reduction according to the reaction :
choose correct Eº for above reaction if
Now
Again
= 0.53 – 0.15 = 0.38 V.
The reduction of peroxydisulphate ion by I^{– }ion is expressed by . If rate of disappearance of I^{–} is 9/2 × 10^{–3} mol lit^{–1} s^{–1}, what is the rate of formation of 2SO_{4}^{2} during same time?
∴
= 3 × 10^{–3} mol Lit^{–1} s^{–1}
A gaseous reaction
There is increase in pressure from 100 mm to 120 mm in 5 minutes. The rate of disappearance of X_{2} is
The increase in pressure shows the increase in conc. of Z. Rate of appearance of
Z = 120100/5 = 4 mm min–1
Rate of disappearance of X_{2} = 2 × rate of appearance of Z
= 2 × 4 mm min–1 = 8 mm min^{–1}
Two substances R and S decompose in solution independently, both following first order kinetics.
The rate constant of R is twice that of S. In an experiment, the solution initially contained 0.5 millimoles of R and 0.25 of S. The molarities of R and S will be equal just at the end of time equal to
The isoelectricpoint of a colloidially dispersed material is the pH value at which
At isoelectric point there is no migration of dispersed phase in an electric field.
Which of the following halogens exhibit only one oxidation state in its compounds ?
Fluorine, since it is the most electronegative element.
Star ch can be used as an indicator for the detection of traces of
I_{2} gives blue colour with starch.
Which one of t h e foll owing arr a n g em ents represents the correct order of electron gain enthalpy (with negative sign) of the given atomic species?
The amount of energy released when an electron is added to an isolated gaseous atom to produce a monovalent anion is called electron gain enthalpy.
Electron affinity value generally increase on moving from left to right in a period however there are exceptions of this rule in the case of those atoms which have stable configuration. These atoms resist the addition of extra electron, therefore the low value of electron affinity
On the other hand Cl, because of its
comparatively bigger size than F, allow the addition of an extra electron more easily.
Most of the transition metal compounds (ionic as well as covalent) are coloured both in the solid state and in aqueous so lution in contrast to the compounds of s andp block elements due to the presence of incomplete dsubshell.
The correct order of magnetic moments (spin only values in B.M.) is:
Number of unpaired electrons in central
atom
The greater the number of unpaired electrons, the higher the value of magnetic
moment
Gammexane is C_{6}H_{6}Cl_{6} or (6, 6, 6). It is a saturated compound so no double bond is there in it.
Consider the following phenols :
The decreasing order of acidity of the above phenols is
Electron withdrawing group (–NO_{2})
increases the acidity while electron
releasing group (–CH_{3}, –H) decreases
acidity.
Also effect will be more iffunctional group is present at para position than ortho
and meta position.
The ionization constant of phenol is higher than that of ethanol because :
The acidic nature of phenol is due to the
formation of stable phenoxide ion in
solution
The phenoxide ion is stable due to
resonance.
The negative charge is delocalized in the benzene ring which is a stabilizing factor in the phenoxide ion and increase acidity of phenol. wheras no resonance is possible in alkoxide ions(RO^{–}) derived from alcohol. The negative charge is localized on oxygen atom. Thus, alcohols are not acidic.
Which one of the following monomers gives the polymer neoprene on polymerization ?
Which of the following can possibly be used as analgesic without causing addiction and modification?
Which among the following is not an antibiotic?
Oxytocin is a hormone (nanopetide) which contracts uterus after the child birth and produces lactation in the mammary glands.
Which of the following ions can be separated by aq. NH_{4}OH in presence of NH_{4}Cl
Cu^{2+} is of group II and Al^{3+} is of group III of cation analysis.
3.92 g of fer rous ammon ium sulph ate react completely with 50 ml N/10 KMnO_{4} solution. Thepercentage purity of the sample is
Eq of KMnO_{4} used =
∴ Eq of FAS reacted = 0.005
∴ weight of FAS needed
= 0.005 × 392 = 1.96 g
Thus percentage purity of FAS is 50%
DIRECTIONS: Out of the four alternatives, choose the one which express the correct meaning of the word.
AUGMENT
Augment means make greater, so increase is the correct option.
DIRECTIONS: Out of the four alternatives, choose the one which express the correct meaning of the CONSOLATION
Consolation means ‘comfort received by a person after a loss or disappointment’, so comfort is correct option.
DIRECTIONS: Out of the four alternatives, choose the one which express the correct meaning of the word.
AUXILIARY
Auxiliary means ‘providing additional help’, so supplemental is correct option.
DIRECTIONS: Choose the word opposite is meaning to the given word.
AUSPICIOUS
Auspicious means ‘favourable’, so ‘unfavourable’ is best opposite word for it.
DIRECTIONS: Choose the word opposite is meaning to the given word.
RECOMPENSE
Recompense means ‘reward given for loss, so ‘penalty’ is the correct opposite word for it.
DIRECTIONS: Choose the word opposite is meaning to the given word.
IMPEDE
Impede means ‘hinder’ or ‘obstruct’, so
‘push’ is correct opposite word for it.
DIRECTIONS: A part of sentence is underlined. Belence are given alternatives to the underlined part a, b, c and d which may improve the sentence. Choose the correct alternative.
Q. They requested me to follow them.
Here a sense of command is depicted insentence, sowe should use ‘ordered’ for proper meaning of sentence.
DIRECTIONS: A part of sentence is underlined. Belence are given alternatives to the underlined part a, b, c and d which may improve the sentence. Choose the correct alternative.
Q. She did not believed me.
Sentence is in past tense and V_{1} is used in those sentence which contain ‘did’, so option (c) is correct.
DIRECTIONS: A part of sentence is underlined. Belence are given alternatives to the underlined part a, b, c and d which may improve the sentence. Choose the correct alternative.
Q. I am fine, what about you?
No improvement is needed as sentence is right.
DIRECTIONS: Fill in the blanks.
Q. They were afraid ........... the lion, so they dropped the idea of hunting in jungle.
Afraid agrees with preposition ‘of’,so option (d) is correct.
DIRECTIONS: Fill in the blanks.
Q. Our company signed a profitable ...... last month.
Normally, company signs a contract or deal, so use of ‘deal’ is proper here.
DIRECTIONS: Fill in the blanks.
Q. What is your ......... for tonight?
The question gives a sense of query about normal routine of some special/specific day, so use of ‘plan’ is more proper here.
DIRECTIONS: Arrange the following sentences in correct pattern and mark at the correct combination.
1. Today we live in modern technology era.
P. We have a log of problems now.
Q. We want to get everything in one day.
R. Ancient time was quite pleasant.
S. We has no problems then. 6. Perhaps greed is the main cause for this.
DIRECTIONS: Arrange the following sentences in correct pattern and mark at the correct combination.
1. He is a common man.
P. Yesterday our city saw a brutal crime.
Q. Police is trying to arrest innocent persons.
R. The criminals are well known.
S. Police as well as whole system in corrupt. 6. Police will arrest him as he is an easy target because of being a common man.
DIRECTIONS: Arrange the following sentences in correct pattern and mark at the correct combination.
1. I want to change the room.
P. Last month I got a job.
Q. I had been living there for six months.
R. The office is far from the room.
S. I want to cut expenses of travelling. 6. Hopefully I will do this next week.
In a certain code language, ‘SAFER’ is written as ‘5@3#2’ and ‘RIDE’ is written as ‘2©%#’, how would ‘FEDS’ be written in that code?
Find the missing number from the given response
From the given responses,
4 × 2 × 3 × 3 = 72
9 × 4 × 2 × 10 = 720
Similarly, 6 × 20 × 1 × 6 = 720
If the fir st an d second letters in the word DEPRESSION were interchanged, also the third and fourth letters, the fifth and the sixth letters and so on, then which of the following would be seventh letter from the right.
Since, consecutive two letters are interchanged. Therefore,
7th from Right.
Today is Thrusday. The day after 59 days will be
Every day of week repeats after seven days.
Hence, 59 = 7 × 8 + 3 = 56 + 3
It will be Thursday after 56 days.
∴ 57th day = Thursday ⇒ 58th day = Friday
59th day = Saturday ⇒ 60th day = Sunday
∴ It will be Sunday after 59 days.
Which of the following represents coal mines, factories and fields?
Both coal mines and factories are located in the fields.
Find out the missing term in the series. 1, 8, 27, ? , 125, 216
From the given series,
If ‘+’ means ‘×’, ‘–’ means ‘+’, ‘×’ means ‘÷’ and ‘÷’ means ‘–’, then 6 – 9 + 8 × 3 ÷ 20 = ?
Interchanging the symbols as given in the above question, the above equation
becomes.
= 6 + 24 – 20 = 10
Here are some words translated from an artificial language.
mallon piml means blue light
mallon tifl means blue berry
arpan tifl means rasp berry
Which word could means ‘light house’?
A piece of paper is folded and penched as shown in the figure below
How will it appear when unfolded?
Given set can be written as
(A – B) ∪ (B – A) = (A ∪ B) – (A ∩ B)
(By definition of symmetric difference)
Hence, (A \ B) ∪ (B \ A) = (A ∪ B) \ (A ∩ B)
f (x) is defined if
⇒
⇒
⇒
⇒ 0 < x < 1
=
=
The solution of (2 cos x – 1) (3 + 2 cos x) = 0 in the interval 0 ≤ x ≤ 2π is
We have (2cos x – 1) (3 + 2 cos x) = 0If 2 cos x – 1=0, then cos x = 1/2
∴x = π/3, πp/3
If 3 + 2 cos x = 0, the cos x = –3/2
which is not possible.
23n – 7n – 1 Taking n = 2;
26 – 7 × 2 – 1
= 64 – 15 = 49
The gr eatest positive in teger, wh ich divides n (n + 1)(n + 2)(n + 3) for all n ∈ N , is
The product of r consecutive integers is divisible by r ! . Thus n (n+ 1 ) (n + 2) (n + 3)is divisible by 4 ! = 24.
If z = x + iy, z^{1/3} = a – ib, then
where k is equal to
=
= 0
The complex number z = z + iy which satisfies the equation , lies on
[if  zz_{1} = z+z_{2}  , then it is a perpendicular
bisector of z_{1} and z_{2}]
Hence, perpendicular bisector of (0, 3) and (0, – 3) is X–axis.
The number of all three elements subsets of the set {a_{1}, a_{2}, a_{3} . . . a_{n}} which contain a_{3} is
The number of three elements subsets containing a_{3 }is equal to the number of ways of selecting 2 elements out of n –1 elements. So, the required number of subsets is ^{n –1}C_{2}.
In how many ways can a committee of 5 made out 6 men and 4 women containing atleast one woman?
The coefficient of x^{4} in the expansion of (1 + x + x^{2} + x^{3})^{11}, is
We have coefficient of x4 in (1 + x + x^{2} + x^{3})^{11}
= coefficient of x^{4} in (1 + x^{2})^{11} (1 + x)^{11}
= coefficient of x^{4} in (1 + x)^{11} + coefficient of x^{2}
=
= 990
If T_{0}, T_{1}, T_{2}.....Tn represent the terms in the expansion of (x + a)n, then (T_{0} –T_{2} + T_{4} – .......)^{2} + (T_{1} – T_{3} + T_{5} – .....)^{2 }=
From the given condition, replacing a by ai and – ai respectively, we get
and
Multiplying (ii) and (i) we get required result i.e.,
If the (2p)^{th} term of a H.P. is q and the (2q)^{th} term is p, then the 2(p + q)^{th} term is
If a is the first term and d is the common
difference of the associated A.P.
⇒
=
∴
=
=
The product of n positive numbers is unity, then their sum is :
Since, product of n positive number is unity.
⇒ x_{1} x_{2} x_{3} ....... x_{n} = 1 ..(i)
Using A.M. ≥ GM
⇒
Hence
If P_{1} and P_{2} be the length of perpendiculars from the origin upon the straight lines x secθ + y cosecθ= a and x cosθ – y sinqθ = a cos2θ respectively, then the value of 4P_{1}^{2} + P_{2}^{2}.
We have P_{1} = length of perpendicular from (0, 0) on x secθ + y cosecqθ = a
=
P_{2}= Length of the perpendicular from (0, 0)on x cosθ – y sin q = a cos2θ
Now
The angle of intersection of the two circles x^{2} + y^{2} – 2x – 2y = 0 and x^{2} + y^{2 }= 4, is
An arch of a bridge is semielliptical with major axis horizontal. If the length the base is 9 meter and the highest part of the bridge is 3 meter from the horizontal; the best approximation of the height of the arch. 2 meter from the centre of the base is
The equation of the ellipse is
Where centre is assumed as origin and base as xaxis. Put x = 2, we get
(approximately)
Let
Taking log on both sides, we get
=
=
log y = –1
y = e^{1} = 1/e
We know that
∴
A bag contains 5 brown and 4 white socks. A man pulls out 2 socks. Find the probability that they are of the same colour.
Let A = event of two socks being brown.
B = event of two socks being white.
Then P (A) =
=
Now, since A and B are mutually exclusive events, so required probability
=
Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12}. Then, the relation is
(3, 3), (6, 6), (9, 9), (12, 12), ∈R. R is not symmetric as (6, 12) ∈R but (12, 6) ÎR.
R is transitive as the only pair which
needs verification is (3, 6) and (6, 12) ∈R.
⇒ (3, 12) ∈ R
Let f : R →R be a function defined by , where m ≠ n , then
Let f : R → R be a function defined by
For any (x, y) ∉ R
Let f (x) = f (y)
⇒
∴ f is one – one
Let α ∈ R such that f (x) = α
⇒
⇒
So, f is not onto.
If is square root of identity matrix of order 2 then –
⇒ α^{2 }+ βγ = 1
The value of λ , for which the lines 3x  4y= 13, 8x  11y= 33 and 2 x  3y + λ=0 are concurrent is
For concurrency of 3 lines the determinant of coefficients of equations should be 0.
i.e.
⇒ 3(11λ99)+4(8λ+66) 13(24+22) = 0
⇒ 33λ  297 + 32λ + 264 + 312  286 = 0
⇒ λ  583+ 576 = 0
⇒ λ = 7
Let f(x) =
Then which one of the following is true?
We have;
=
∴ which does not exist.
∴ f is not differentiable at x = 1
Also
= –sin 1 + cos 1
∴ f is differentiable at x = 0
The interval in which the function 2x^{3} + 15 increases less rapidly than the function 9x^{2} – 12x, is –
The fuel charges for running a train are proportional to the square of the speed generated in miles per hour and costs ' 48 per hour at 16 miles per hour. The most economical speed ifthe fixed charges i.e. salaries etc. amount to ' 300 per hour is
Let the speed of the train be v and distance to be covered be s so that total time taken is s/v hours. Cost of fuel per hour = kv^{2} (k is constant)
Also 48 = k. 16^{2} by given condition k = 3/16
∴ Cost to fuel per hour 3/16 v^{2}
Other charges per hour are 300. Total running cost,
Dividing the numerator and denominator by cos^{2}x, we get
Putting tan x = t ⇒ sec^{2}x dx = dt, we get
⇒
Let
Adding (1) and (2), we get
The area bounded by the xaxis, the curve y = f(x) and the lines x =1, x =b, is equal to √b^{2} + 1  2 for all b > 1, then f(x) is
Given
Differentiate with respect to b
⇒
If the middle points of sides BC, CA & AB of triangle ABC are respectively D, E, F then position vector of centre of triangle DEF, when position vector of A, B, C are respectively iˆ + ˆj, ˆj +kˆ, kˆ+iˆ is
The position vector of points D, E, F are
respectively
So, position vector of centre of ΔDEF
=
=
The angle between any two diagonal of a cube is
for a unit cube unit vector along the diagonal
unit vector along the diagonal
∴
Find the angle between the line and the plane 10x + 2y – 11z = 3.
Let q be the angle between the line and the normal to the plane. Converting the given equations into vector form, we have
Here
=
The equation of the right bisector plane of the segment joining (2, 3, 4) and (6, 7, 8) is
If the given points be A (2, 3, 4) and B
(6, 7, 8), then their midpoint N(4, 5, 6) must lie on the plane. The direction ratios of AB are 4, 4, 4, i.e. 1, 1, 1.
The required plane passes through N (4, 5, 6) and is normal to AB. Thus its equation is
1(x  4) +1(y  5) +1(z  6) = 0 ⇒ x + y + z = 15
A bag contains n + 1 coins. It is known that one of these coins shows heads on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is , 7/12 then the value of n is.
A coin is tossed 7 times. Each time a man calls head. Find the probability that he wins the toss on more occasions.
The man has to win at least 4 times.
Reqd. probability =
=
Consider Then number of possible solutions are :
Consider
convert them into equation and solve them and draw the graph of these equations we get y = 1 and x = 3/2
From graph region is finite but numbers of possible solutions are infinite because for different values of x and y we have different or infinite no. of solutions.
⇒ A^{100} = 2^{100}1A \ A^{100} = 2^{99}A
Apply R_{1} → R_{1} – R_{3} and R_{2} → R_{2} – R_{3}, we get
⇒ x[yr + z(q  y)] z[0  y(p  x)] = 0
[Expansion along first row]
⇒
Through the vertex O of a parabola y^{2} = 4x, chords OP and OQ are drawn at right angles to one another. The locus of the middle point of PQ is
Given parabola is y^{2 }= 4x ...(1)
Let
Slope of OP =
and slope of OQ =
Since OP OQ,
Let R (h, k) be the middle point of PQ, then
and k = t_{1} + t_{2} ...(4)
From (4),
[ From(2) and (3)]
Hence locus of R (h, k) is y^{2} = 2x – 8.
Let f (x ) =
If f(x) is continuous at x = π/2 , (p, q) =
=
=
∴
q = 4.
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