Sample Test Paper (STP) For ResoFAST-2017 · PDF file · 2016-08-30Toll Free : 1800...

STP1718

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005

Website : www.resonance.ac.in | E-mail : [email protected]

Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 STPRESOFASTADV17 - 1

Index

The sample test papers are only for reference and guidance. The sample papers given in the booklet are actually the papers of previousyear's ResoFAST conducted by Resonance for its various courses.Note : Resonance reserves the right to change the pattern of selection test (ResoFAST). Pervious year papers do not guarantee thatthe papers for this year selection test will be on the same pattern. However, the syllabus of the test paper will be equivalent to the syllabusof qualifying school/board examination and as given on page no. 4.

Copyright reserved 2017-18.All rights reserved. Any photocopying, publishing or reproduction of full or any part of this material is strictly prohibited. This material belongs to only theapplicants of RESONANCE for its various Selection Tests (ResoFAST) to be conducted for admission in Academic Session 2017-18. Any sale/resale ofthis material is punishable under law. Subject to Kota Jurisdiction only.

Sample Test Paper (STP) For ResoFAST-2017

S.No. Contents Target Page No.

1 How to Prepare for the Resonance's Forward Admission & Scholarship Test (ResoFAST) - 2017 ResoFAST 2017 2

2 General Instructions for the Examination Hall ResoFAST 2017 3

3 Syllabus for ResoFAST-2017 ResoFAST 2017 4

4Sample Test Paper- I : For Class-X Appearing students (Moving from Class-X to Class-XI ) For the students applying for VIKAAS (JA*) Courses

JEE(Main + Advanced) 2019 9

5Sample Test Paper-I Answer key & Hints & Solution : For Class-X Appearing students (Moving from Class-X to Class-XI ) For the students applying for VIKAAS (JA*) Courses


6Sample Test Paper-II : For Class-XI Appearing students (Moving from Class-XI to Class-XII).For the students applying for VISHWAAS (JF) Course


7Sample Test Paper-II Answer key & Hints & Solution : For Class-XI Appearing students (Moving from Class-XI to Class-XII).For the students applying for VISHWAAS (JF) Course


8Sample Test Paper-III : For Class-XII Appearing students (Moving from Class-XII to Class-XIII) For the students applying for VISHESH (JD*) Courses


9Sample Test Paper-III Answer key & Hints & Solution : For Class-XII Appearing students (Moving from Class-XII to Class-XIII) For the students applying for VISHESH (JD*) Courses


10 Sample ORS Answer Sheet for Resonance's Forward Admission & Scholarship Test (ResoFAST) - 2017 ResoFAST 2017 84

STP1718




For Class-X appearing students (Class-X to Class-XI Moving) :

Study thoroughly the books of Science (Physics & Chemistry) and Maths of Classes

IX & X. (NCERT & Respective Board)

For Class-XI appearing students (Class-XI to Class-XII Moving):

1. Study thoroughly the books of Physics, Chemistry and Maths of Class XI (Respective

Board).

2. Refer to the following books (only Class-XI syllabus) to increase the level of competence:

For Physics : Concepts of Physics by H.C. Verma Vol. I & II, NCERT Books

For Chemistry : NCERT Books(XI & XII), A text book of Physical Chemistry

(8th Edition), Shishir Mittal, Disha Publications, Concise Inorganic

Chemistry, J.D. Lee, Wiley-India Edition, Vogel’s Qualitative Analysis for

the JEE (7th Edition), G. Svehla & Shishir Mittal, Pearson Education,Organic

Chemistry : Clayden, Greeves, Warren and Wothers, Oxford University,

A guide book to Mechanism In Organic Chemistry (6th Edition), Peter Sykes,

Pearson Education

For Maths : Higher Algebra By Hall & Knight; Co-ordinate Geometry By

S.L. Loney ; Plane Trigonometry By S.L. Loney, Problem book in high school

by A.I.Prilepko

For Class-XII appearing students (Class-XII to Class-XIII Moving):

1. Study thoroughly the books of Physics, Chemistry and Maths of Classes XI & XII

(Respective Board).

2. Refer to the following books (Class-XI & Class-XII syllabus) to increase the level of

competence :

For Physics : Concepts of Physics by H.C. Verma Vol-I & II

For Chemistry : Physical Chemistry By R.K. Gupta, Organic Chemistry By

Morrison & Boyd, Organic Chemistry By I. L. Finar, Inorganic Chemistry By J.D.

Lee, Objective Chemistry By Dr. P. Bahadur

For Maths : Higher Algebra By Hall & Knight; Co-ordinate Geometry By S.L.

Loney; Plane Trigonometry By S.L. Loney, Differential Calculus By G.N. Berman;

Integral Calculus By Shanti Narayan; Vector Algebra By Shanti Narayan ;

A Das Gupta (subjective).

How to prepair for the Resonance’s Forward Admission & Scholarship Test (ResoFAST) - 2017

STP1718




GENERAL INSTRUCTIONS IN THE EXAMINATION HALL(ijh{kk Hkou ds fy, lkekU; funsZ'k)

1. This booklet is your Question Paper. ¼;g iqfLrdk vkidk iz'u&i=k gS½

2. The Question Paper Code is printed on the top right corner of this sheet. ¼iz'u&i=k dksM bl i`"B

ds Åij nk;sa dksus esa Nik gqvk gS½

3. Blank papers, clip boards, log tables, slide rule, calculators, mobile or any other electronic

gadgets in any form are not allowed to be used. ¼[kkyh dkxt] fDyi cksMZ] y?kqx.kd lkj.kh] LykbM :y]

dSYdqysVj] eksckby ;k vU; fdlh bySDVªWkfud midj.k ds fdlh Hkh :i esa mi;ksx dh vkKk ugha gS½

4. Write your Name & Application Form Number in the space provided in the bottom of this

booklet. (bl i`"B ds uhps fn;s x;s fjDr LFkku esa viuk uke o vkosnu QkWeZ la[;k vo'; Hkjs a½

5. Before answering the paper, fill up the required details in the blank space provided in the Objective

Response Sheet (ORS). (iz'u&i=k gy djus ls igys] ORS&'khV esa fn;s x;s fjDr LFkkuks a esa iwNs x;s

fooj.kksa dks Hkjsa½

6. Do not forget to mention your paper code and Application Form Number neatly and clearly in

the blank space provided in the Objective Response Sheet (ORS) / Answer Sheet. ¼mÙkj&iqfLrdk

esa fn;s x;s fjDr LFkku esa vius iz'u&i=k dk dksM o viuk vkosnu QkWeZ la[;k Li"V :i ls Hkjuk uk Hkwysa½

7. No rough sheets will be provided by the invigilators. All the rough work is to be done in the blank

space provided in the question paper. ¼fujh{kd ds }kjk dksbZ jQ 'khV ugha nh tk;sxhA jQ dk;Z iz'u&i=k

esa fn;s x;s [kkyh LFkku esa gh djuk gS½

8. No query related to question paper of any type is to be put to the invigilator.

¼fujh{kd ls iz'u&i=k ls lEcfU/kr fdlh izdkj dk dksbZ iz'u uk djsas½

Question Paper ¼i z’ u&i =k½

9. Marks distribution of questions is as follows. ¼iz'uks a ds izkIrkadks dk fooj.k fuEu izdkj ls gSA½

Correct Wrong Blank

1 to 20 39 to 46 52 to 59Only one correct

(dsoy , d fodYi l gh)3 –1 0

21 to 28 47 to 50 60 to 63One or more than one correct Answer

(, d ; k , d l s v f/kd fodYi l gh)4 0 0

29 to 36 Comprehensions (v uqPNsn) 4 –1 0

37 to 38 51 64Matrix Match Type

(eSfVªDl l qesy i zdkj )6 [1, 2, 3, 6] 0 0

Marks to be awardedPart - I (Mathematics)

Part - II (Physics)

Part - III (Chemistry)

Type

Name : _________________________________ Application Form Number : _______________

STP1718




CLASS - X (CHEMISTRY)Basic : Cooling by evaporation. Absorption of heat. All things accupyspace, possess mass. Definition of matter ; Elementary idea aboutbonding.

Solid, liquid and gas : characteristics-shape, volume, density;change of state - melting, freezing, evaporation, condensation,sublimation.

Elements, compounds and mixtures :Heterogeneous andhomogeneous mixtures; Colloids and suspension.

Mole concept : Equivalence - that x grams of A is chemically notequal to x grams of B ; Partical nature, basic units : atoms andmolecules ; Law of constant proportions ; Atomic and molecularmasses;Relationship of mole to mass of the particles and numbers ;Valency ; Chemical formulae of common compounds.

Atomic structure : Atoms are made up of smaller particles :electrons, protons, and neutrons. These smaller particles are presentin all the atoms but their numbers vary in different atoms.Isotopes and isobars.

Gradations in properties : Mendeleev periodic table.

Acids, bases and salts : General properties, examples and uses.Types of chemical reactions : Combination, decomposition,displacement, double displacement, precipitation, neutralisation,oxidation and reduction in terms of gain and loss of oxygen andhydrogen.

Extractive metallurgy : Properties of common metals ; Briefdiscussion of basic metallurgical processes.

Compounds of Carbon : Carbon compounds ; Elementary ideaabout bonding ; Saturated hydrocarbons, alcohols, carboxylic acids(no preparation, only properties).Soap - cleansing action of soap.

CLASS - X (MATHEMATICS)Number Systems :

Natural Numbers, Integers, Rational number on the number line. Even- odd integers, prime number, composite numbers, twin primes,divisibility tests, Co-prime numbers, LCM and HCF of numbers.Representation of terminating/non-terminating recurring decimals, onthe number line through successive magnification. Rational numbersas recurring/terminating decimals. Ratio and proportions.

Polynomials :Polynomial in one variable and its Degree. Constant, Linear, quadratic,cubic polynomials; monomials, binomials, trinomials, Factors andmultiplex. Zeros/roots of a polynomial/equation.Remainder theorem, Factor Theorem. Factorisation of quadratic andcubic polynomialsStandard form of a quadratic equation ax2 + bx + c = 0, (a 0).Relation between roots and coefficient of quadratic and relationbetween discriminant and nature of roots.

Linear Equation :Linear equation in one variable and two variable and their graphs.Pair of linear equations in two variables and their solution andinconsistency

Arithmetic Progressions (AP) :Finding the nth term and sum of first n terms.

Trigonometry :

Trigonometric ratios of an acute angle of a right-angled triangle,Relationships between the ratios.Trigonometric ratios of complementary angles and trigonometricidentities. Problems based on heights and distances.

Coordinate Geometry :

The cartesian plane, coordinates of a point, plotting points in theplane, distance between two points and section formula (internal).Area of triangle. Properties of triangle and quadrilateral. (Square,Rectangle rhombus, parallelogram).

Geometry :

Lines :Properties of parallel and perpendicular lines.Triangle :Area of a triangle, Properties of triangle, similarity and congruencyof triangles.Medians, Altitudes, Angle bisectors and related centres.Geometrical representation of quadratic polynomials.Circle :Properties of circle, Tangent, Normal and chords.

Mensuration :Area of triangle using Heron’s formula and its application in findingthe area of a quadrilateral.Area of circle ; Surface areas and volumes of cubes, cuboids,spheres (including hemispheres) and right circular cylinders/conesand their combinations.

Statistics :Mean, median, mode of ungrouped and grouped data.

Probability :Classical definition of probability, problems on single events.

Logarithm & exponents :Logarithms and exponents and their properties.

Interest :Problem based on simple interest, compound interest and discounts.

Mental Ability :Problem based on data interpretation, family relations, Logicalreasoning.

Direct & Indirect variations :Ratios & proportions, Unitary method, Work and time problems.

CLASS - X (PHYSICS)Mechanics : Uniform and non-uniform motion along a straight line ;Concept of distance and displacement, Speed and velocity,accelaration and relation ship between these ; Distance-time andvelcocity - time graphs.Newton’s Law of motion ; Relationship between mass, momentum,force and accelaration ; work done by a force ; Law of conservationof energy.Law of gravitation ; acceleration due to gravity.Electricity and magnetism : Ohm’s law ; Series and parallelcombination of resistances ; Heating effect of current.

Magnetic field near a current carrying straight wire, along the axis ofa circular coil and inside a solenoid ; Force on current carryingconductor ; Fleming’s left hand rule ; Working of electric motor ;Induced potential difference and current

Electric generator : Principle and working ; Comparision of AC andDC ; Domestic electric circuits.

Optics : Rectilinear propagation of light ; Basic idea of concavemirror and convex lens ; Laws of refraction ; Dispersion.

CLASS - XI (CHEMISTRY)Some Basic Concepts of Chemistry : Particulate nature of matter,laws of chemical combination, Dalton’s atomic theory : concept ofelements, atoms and molecules.

Syllabus of ResoFAST-2017

STP1718




Atomic and molecular masses. Mole concept and molar mass ;percentage composition and empirical and molecular formula ;chemical reactions, stoichiometry and calculations based onstoichiometry.Structure of Atom : Discovery of electron, proton and neutron ;atomic number, isotopes and isobars.Thompson’s model and its limitations, Rutherford’s model and itslimitations, concept of shells and sub-shells, dual nature of matterand light, de Broglie’s relationship, Heisenberg uncertainty principle,concept of orbitals, quantum numbers, shapes of s, p, and d orbitals,rules for filling electrons in orbitals - Aufbau principle, Pauli exclusionprinciple and Hund’s rule, electronic configuration of atoms, stabilityof half filled and completely filleld orbitals.Classification of Elements and Periodicity in Properties :Significance of classification, brief history of the development ofperiodic table, trends in properties of elements - atomic radii, ionicradii, inert gas radii, ionization enthalpy, electron gain enthalpy,electronegativity, valence.

Chemical Bonding and Molecular Structure :Valence electrons, ionic bond, covalent bond, bond parameters, Lewisstructure, polar character of covalent bond, covalent character ofionic bond, valence bond theory, resonance, geometry of covalentmolecules, VSEPR theory, concept of hybridization involving s, p andd orbitals and shapes of some simple molecules,molecular orbital theory of homonuclear diatomic molecules (qualitativeidea only), hydrogen bond.

States of Matter : Gases and Liquids :Three states of matter, intermolecular interactions, type of bonding,melting and boiling points, role of gas laws in elucidating the conceptof the molecule, Boyle’s law, Charles’ law, Gay Lussac’s law,Avogadro’s law, ideal behavior, empirical derivation of gas equation,Avogadro’s number ideal gas equation, deviation from ideal behaviour,Liquefaction of gases, critical temperature.Liquid State - Vapour pressure, viscosity and surface tension(qualitative idea only, no mathematical derivations)Thermodynamics :Concepts of system, types of systems, surroundings, work, heat,energy, extensive and intensive properties, state functions.First law of thermodynamics - internal energy and enthalpy, heatcapacity and specific heat, measurement of U and H, Hess’s lawof constant heat summation, enthalpy of bond dissociation,combustion, formation, atomization sublimation, phase transition,ionization, and dilution.Introduction of entropy as a state function, free energy change forspontaneous and non-spontaneous process, equilibrium.

Equilibrium : Equilibrium in physical and chemical processes,dynamic nature of equilibrium, law of mass action, equilibriumconstant, factors affecting equilibrium - Le Chatelier’s principle ;ionic equilibrium - ionization of acids and bases, strong and weakelectrolytes, degree of ionization concept of pH. Hydrolysis of Salts(elementary idea), buffer solutions, solubility product, common ioneffect (with illustrative examples).

Redox Reactions : Concept of oxidation and reduction, redoxreactions,oxidation number, balancing redox reactions, applications of redoxreaction.

Hydrogen : Position of hydrogen in periodic table, occurrence,isotopes, preparation, properties and uses of hydrogen ; hydrides -ionic, covalent and interstitial ; physical and chemical properties ofwater, heavy water ; hydrogen peroxide - preparation, reactionsand structure ; hydrogen as a fuel.

s-Block Elements (Alkali and Alkaline Earth Metals) :Group 1 and Group 2 elements :General introduction, electronic configuration, occurrence, anomalousproperties of the first element of each group, diagonal relationship,trends in the variation of properties (such as ionization enthalpy,atomic and ionic radii), trends in chemical reactivity with oxygen,water, hydrogen and halogens ; uses.

Preparation and properties of some important compoundsSodium carbonate, sodium chloride, sodium hydroxide and sodiumhydrogen carbonateCaO, CaCO3, and industrial use of lime and limestone, Ca.

General Introduction to p-Block Elements :Group 13 elements : General introduction, electronic configuration,occurrence, variation of properties, oxidation states, trends inchemical reactivity, anomalous properties of first element of thegroup ;Boron - physical and chemical properties, some importantcompounds ; borax, boric acids, boron hydrides. Aluminium : uses,reactions with acids and alkalies.Group 14 elements ; General introduction, electronic configuration,occurrence, variation of properties, oxidation states, trends inchemical reactivity, anomalous behaviour of first element. Carbon -catenation, allotropic forms, physical and chemical propeties ; usesof some important compounds : oxides.Important compounds of silicon and a few uses : silicon tetrachloride,silicones, silicates and zeolites.Principles of qualitative analysis : Determinantion of one anionand one cation in a given saltCations - Pb2 + , Cu2+, As3+, Al3+, Fe3+, Mn2+, Ni2 +, Zn2+, Co2+, Ca2+, Sr2+,

Ba2+, Mg2+, 4NH

Anions - ,NO,SO,SO,S,CO –2

–24

–23

–2–23

–3

–242

–34

––––3

–3 COOCHOC,PO,,Br,Cl,NO,NO

(Note : Insoluble salts excluded)

Organic chemistry - Some Basic Principles and TechniquesGeneral introduction, methods of purification, qualitative andquantitative analysis, classification and IUPAC nomenclature oforganic compounds.Electronic displacements in a covalent bond : free radicals,carbocations, carbanions ; electrophiles and nucleophiles, types oforganic reactions

Classification of Hydrocarbons : Alkanes : Nomenclature,isomerism, conformations (ethane only), physical propeties,chemical reactions including free radical mechanism ofhalogenation, combustion and pyrolysis.

Alkenes : Nomenclatures, structure of double bond (ethene),geometrical isomerism, physical properties, methods of preparation; chemical reactions : addition of hydrogen, halogen, water, hydrogenhalides (Markovnikov’s addition and peroxide ef fect),ozonolysis, oxidation, mechanism of electrophilic addition.Alkynes : Nomenclature, structure of triple bond (ethyne), physicalproperties, methods of preparation, chemical reactions : acidiccharacter of alkynes, addition reaction of - hydrogen, halogens,hydrogen halides and water.

Aromatic hydrocarbons : Introduction, IUPAC nomenclature ;Benzene : resonance, aromaticity ; chemical properties : mechanismof electrophilic substitution - nitration sulphonation, halogenation,Friedel Craft’s alkylation and acylation ; directive influence offunctional group in mono-substituted benzene ; carcinogenicity andtoxicity.

CLASS - XI (MATHEMATICS)Functions :Sets and their representations. Empty, finite and infinite sets, Subsets,Union and intersection of sets, Venn diagrams.Pictorial representation of a function domain, co-domain and rangeof a function domain and range of constant, identity, polynomial,rational, modulus, signum and greatest integer functions with theirgraphs. Sum, difference, product and quotients of functions.Trigonometric Functions :Measuring angles in radians and in degrees and conversion fromone measure to another. Signs of trigonometric functions and sketchof their graphs. Addition and subtraction formulae, formulae involvingmultiple and sub-mult iple angles. General solution oftrigonometric equations.

Complex NumberAlgebra of complex numbers, addition, multiplication, conjugation,polar representation, properties of modulus and principal argument,triangle inequality, cube roots of unity, geometric interpretations.

Quadratic equations :Quadratic equations with real coefficients, formation of quadraticequations with given roots, symmetric functions of roots.

STP1718




Sequence & Series :Arithmetic, geometric and harmonic progressions, arithmetic,geometric and harmonic means, sums of f inite arithmetic andgeometric progressions, infinite geometric series, sums of squaresand cubes of the first n natural numbers.

Logarithm & exponents :Logarithms and exponents and their properties. Exponential andlogarithmic series.

Binomial Theorem :Binomial theorem for a positive integral index, properties of binomialcoefficients. Binomial theorem for any index.

Permutations and combinations : Problem based on fundamental counting principle, Arrangement ofalike and different objects, Circular permutation, Combination,formation of groups.Straight Line :Cartesian coordinates, distance between two points, sectionformulae, shift of origin. Equation of a straight line in various forms,angle between two lines, distance of a point from a line; Lines throughthe point of intersection of two given lines equation of the bisector ofthe angle between two lines, concurrency of lines; Centroid,orthocentre, incentre and circumcentre of a triangle.

Conic Sections :Equation of a circle in various forms, equations of tangent, normaland chord. Parametric equations of a circle, intersection of a circlewith a straight line or a circle, equation of a through the points ofintersection of two circles and those of a circle and a straight line.Equations of a parabola, ellipse and hyperbola in standard form, theirfoci, directrices and eccentricity, parametric equations, equations oftangent and normal locus problems.Mental Ability :Problem based on data interpretation, family relations & Logicalreasoning.

CLASS - XI (PHYSICS)

General : Units and dimensions, dimensional analysis; least count,significant figures; Methods of measurement and error analysis forphysical quantities pertaining to the following experiments:Experiments based on using Vernier calipers and screw gauge(micrometer), Determination of g using simple pendulum, Young’smodulus by Searle’s method.

Mechanics : Kinematics in one and two dimensions (Cartesiancoordinates only), projectiles; Uniform Circular motion; Relativevelocity.

Newton’s laws of motion; Inertial and uniformly accelerated framesof reference; Static and dynamic friction; Kinetic and potential energy;Work and power; Conservation of linear momentum and mechanicalenergy.

Systems of particles; Centre of mass and its motion; Impulse; Elasticand inelastic collisions.

Law of gravitation; Gravitational potential and field; Acceleration dueto gravity; Motion of planets and satellites in circular orbits; Escapevelocity.

Rigid body, moment of inertia, parallel and perpendicular axestheorems, moment of inertia of uniform bodies with simplegeometrical shapes; Angular momentum; Torque; Conservation ofangular momentum; Dynamics of rigid bodies with fixed axis ofrotation; Rolling without slipping of rings, cylinders and spheres;Equilibrium of rigid bodies; Collision of point masses with rigid bodies.

Linear and angular simple harmonic motions.

Hooke’s law, Young’s modulus.

Pressure in a fluid; Pascal’s law; Buoyancy; Surface energy andsurface tension, capillary rise; Viscosity (Poiseuille’s equationexcluded), Stoke’s law; Terminal velocity, Streamline flow, equationof continuity, Bernoulli’s theorem and its applications.

Waves : W ave motion (plane waves only), longitudinal andtransverse waves, superposition of waves; Progressive andstationary waves; Vibration of strings and air columns;Resonance;Beats; Speed of sound in gases; Doppler effect (in sound).

Thermal physics : Thermal expansion of solids, liquids and gases;Calorimetry, latent heat; Heat conduction in one dimension; Elementaryconcepts of convection and radiation; Newton’s law of cooling; Idealgas laws; Specific heats (Cv and Cp for monoatomic and diatomicgases); Isothermal and adiabatic processes, bulk modulus of gases;Equivalence of heat and work; First law of thermodynamics and itsapplications (only for ideal gases); Blackbody radiation: absorptiveand emissive powers; Kirchhoff ’s law; Wien’s displacement law,Stefan’s law.

CLASS - XII (CHEMISTRY)Physical ChemistryGeneral topics : Concept of atoms and molecules; Dalton’s atomictheory; Mole concept; Chemical formulae; Balanced chemicalequations; Calculations (based on mole concept) involving commonoxidation-reduction, neutralisation, and displacement reactions;Concentration in terms of mole fraction, molarity, molality and normality.

Gaseous and liquid states : Absolute scale of temperature, idealgas equation; Deviation from ideality, van der Waals equation; Kinetictheory of gases, average, root mean square and most probablevelocit ies and their relation with temperature; Law of partialpressures; Vapour pressure; Diffusion of gases.Atomic structure and chemical bonding : Bohr model, spectrumof hydrogen atom, quantum numbers; Wave-particle duality, de Brogliehypothesis; Uncertainty principle; Qualitative quantum mechanicalpicture of hydrogen atom, shapes of s, p and d orbitals; Electronicconfigurations of elements (up to atomic number 36); Aufbau principle;Pauli’s exclusion principle and Hund’s rule; Orbital overlap and covalentbond; Hybridisation involving s, p and d orbitals only; Orbital energydiagrams for homonuclear diatomic species; Hydrogen bond; Polarityin molecules, dipole moment (qualitative aspects only); VSEPR modeland shapes of molecules (linear, angular, triangular, square planar,pyramidal, square pyramidal, trigonal bipyramidal, tetrahedral andoctahedral).

Energetics : First law of thermodynamics; Internal energy, workand heat, pressure-volume work; Enthalpy, Hess’s law; Heat ofreaction, fusion and vapourization; Second law of thermodynamics;Entropy; Free energy; Criterion of spontaneity.

Chemical equilibrium : Law of mass action; Equilibrium constant,Le Chatelier’s principle(effect of concentration, temperature and pressure); Significanceof G and Go in chemical equilibrium; Solubility product, commonion effect, pH and buffer solutions; Acids and bases (Bronsted andLewis concepts); Hydrolysis of salts.

Electrochemistry : Electrochemical cells and cell reactions;Standard electrode potentials; Nernst equation and its relation to DG;Electrochemical series, emf of galvanic cells; Faraday’s laws ofelectrolysis; Electrolytic conductance, specific, equivalent and molarconductivity, Kohlrausch’s law; Concentration cells.

Chemical kinetics : Rates of chemical reactions; Order ofreactions; Rate constant; First order reactions; Temperaturedependence of rate constant (Arrhenius equation).

Solid state : Classification of solids, crystalline state, seven crystalsystems (cell parameters a, b, c, ), close packed structure of solids(cubic), packing in fcc, bcc and hcp lattices; Nearest neighbours,ionic radii, simple ionic compounds, point defects.

Solutions : Raoult’s law; Molecular weight determination fromlowering of vapour pressure, elevation of boiling point and depressionof freezing point.

Surface chemistry : Elementary concepts of adsorption (excludingadsorption isotherms); Colloids: types, methods of preparation andgeneral properties; Elementary ideas of emulsions, surfactants andmicelles (only definitions and examples).

STP1718




Nuclear chemistry : Radioactivity: isotopes and isobars; Propertiesof rays; Kinetics of radioactive decay (decay series excluded),carbon dating; Stability of nuclei with respect to proton-neutron ratio;Brief discussion on fission and fusion reactions.

Inorganic Chemistry

Isolation/preparation and properties of the following non-metals : Boron, silicon, nitrogen, phosphorus, oxygen, sulphur andhalogens; Properties of allotropes of carbon(only diamond and graphite), phosphorus and sulphur.

Preparation and properties of the following compounds :Oxides, peroxides, hydroxides, carbonates, bicarbonates, chloridesand sulphates of sodium, potassium, magnesium and calcium; Boron:diborane, boric acid and borax; Aluminium: alumina, aluminium chlorideand alums; Carbon: oxides and oxyacid (carbonic acid); Silicon:silicones, silicates and silicon carbide; Nitrogen: oxides, oxyacidsand ammonia; Phosphorus: oxides, oxyacids (phosphorus acid,phosphoric acid) and phosphine; Oxygen: ozone and hydrogenperoxide; Sulphur: hydrogen sulphide, oxides, sulphurous acid,sulphuric acid and sodium thiosulphate; Halogens: hydrohalic acids,oxides and oxyacids of chlorine, bleaching powder; Xenon fluorides.

Transition elements (3d series) : Definition, generalcharacteristics, oxidation states and their stabilities, colour (excludingthe details of electronic transitions) and calculation of spin (onlymagnetic moment), Coordination compounds: nomenclature ofmononuclear coordination compounds, cis-trans and ionisationisomerisms, hybridization and geometries of mononuclearcoordination compounds (linear, tetrahedral, square planar andoctahedral).

Preparation and properties of the following compounds :Oxides and chlorides of tin and lead; Oxides, chlorides and sulphatesof Fe2+, Cu2+ and Zn2+; Potassium permanganate, potassiumdichromate, silver oxide, silver nitrate, silver thiosulphate.

Ores and minerals : Commonly occurring ores and minerals ofiron, copper, tin, lead, magnesium, aluminium, zinc and silver.

Extractive metallurgy : Chemical principles and reactions only(industrial details excluded); Carbon reduction method (iron and tin);Self reduction method (copper and lead); Electrolytic reduction method(magnesium and aluminium); Cyanide process (silver and gold).

Principles of qualitative analysis : Groups I to V (only Ag+, Hg2+,Cu2+, Pb2+, Bi3+, Fe3+, Cr3+, Al3+, Ca2+, Ba2+, Zn2+, Mn2+ and Mg2+);Nitrate, halides (excluding fluoride), sulphate and sulphide.

Organic Chemistry

Concepts : Hybridisation of carbon; Sigma and pi-bonds; Shapesof simple organic molecules; Structural and geometrical isomerism;Optical isomerism of compounds containing up to two asymmetriccentres, (R,S and E,Z nomenclature excluded); IUPAC nomenclatureof simple organic compounds (only hydrocarbons, mono-functionaland bi-functional compounds); Conformations of ethane and butane(Newman projections); Resonance and hyperconjugation; Keto-enoltautomerism; Determination of empirical and molecular formulae ofsimple compounds (only combustion method); Hydrogen bonds:definition and their effects on physical properties of alcohols andcarboxylic acids; Inductive and resonance effects on acidity andbasicity of organic acids and bases; Polarity and inductive effects inalkyl halides; Reactive intermediates produced during homolytic andheterolytic bond cleavage; Formation, structure and stability ofcarbocations, carbanions and free radicals.Preparation, properties and reactions of alkanes : Homologousseries, physical properties of alkanes (melting points, boiling pointsand density); Combustion and halogenation of alkanes; Preparationof alkanes by Wurtz reaction and decarboxylation reactions.Preparation, properties and reactions of alkenes and alkynes: Physical properties of alkenes and alkynes (boiling points, densityand dipole moments); Acidity of alkynes; Acid catalysed hydration ofalkenes and alkynes (excluding the stereochemistry of addition andelimination); Reactions of alkenes with KMnO4 and ozone; Reductionof alkenes and alkynes; Preparation of alkenes and alkynes byelimination reactions; Electrophilic addition reactions of alkenes withX2, HX, HOX and H2O (X=halogen); Addition reactions of alkynes;Metal acetylides.

Reactions of Benzene : Structure and aromaticity; Electrophilicsubstitution reactions: halogenation, nitration, sulphonation, Friedel-Crafts alkylation and acylation; Effect of ortho, meta and para directinggroups in monosubstituted benzenes.

Phenols : Acidity, electrophilic substitution reactions (halogenation,nitration and sulphonation); Reimer-Tieman reaction, Kolbe reaction.

Characteristic reactions of the following (including thosementioned above): Alkyl halides: rearrangement reactions of alkyl carbocation, Grignardreactions, nucleophilic substitution reactions; Alcohols:esterif ication, dehydration and oxidation, reaction with sodium,phosphorus halides, ZnCl2/concentrated HCl, conversion of alcoholsinto aldehydes and ketones; Ethers:Preparation by Williamson’sSynthesis; Aldehydes and Ketones: oxidation, reduction, oximeand hydrazone formation; aldol condensation, Perkin reaction;Cannizzaro reaction; haloform reaction and nucleophilic additionreactions (Grignard addition); Carboxylic acids: formation of esters,acid chlorides and amides, ester hydrolysis; Amines: basicity ofsubstituted anilines and aliphatic amines, preparation from nitrocompounds, reaction with nitrous acid, azo coupling reaction ofdiazonium salts of aromatic amines, Sandmeyer and related reactionsof diazonium salts; carbylamine reaction; Haloarenes: nucleophilicaromatic substitution in haloarenes and substituted haloarenes(excluding Benzyne mechanism and Cine substitution).

Carbohydrates: Classification; mono- and di-saccharides (glucoseand sucrose); Oxidation, reduction, glycoside formation andhydrolysis of sucrose.

Amino acids and peptides : General structure (only primarystructure for peptides) and physical properties.

Properties and uses of some important polymers : Naturalrubber, cellulose, nylon, teflon and PVC.

Practical organic chemistry : Detection of elements (N, S,halogens); Detection and identification of the following functionalgroups: hydroxyl (alcoholic and phenolic), carbonyl (aldehyde andketone), carboxyl, amino and nitro; Chemical methods of separationof mono-functional organic compounds from binary mixtures.

CLASS - XII (MATHEMATICS)Complex Number and Quadratic equations :

Algebra of complex numbers, addition, multiplication, conjugation,polar representation, properties of modulus and principal argument,triangle inequality, cube roots of unity, geometric interpretations.Quadratic equations with real coefficients, formation of quadraticequations with given roots, symmetric functions of roots.

Sequence & Series :Arithmetic, geometric and harmonic progressions, arithmetic,geometric and harmonic means, sums of finite arithmetic andgeometric progressions, infinite geometric series, sums of squaresand cubes of the first n natural numbers.

Logarithms and their properties. Permutations and combinations,Binomial theorem for a positive integral index, properties of binomialcoeff icients.Binomial theorem for any index, exponential and logarithmic series.

Matrices & Determinants :Matrices as a rectangular array of real numbers, equality of matrices,addition, multiplication by a scalar and product of matrices, transposeof a matrix, determinant of a square matrix of order up to three,inverse of a square matrix of order up to three, properties of thesematrix operations, diagonal, symmetric and skew- symmetric matricesand their properties, solutions of simultaneous linear equation in twoor three variables.

Probability :Addition and multiplication rules of probability, conditional probability,baye’s theorem, independence of events, computation of probabilityof events using permutations and combinations.

Straight Line :Cartesian coordinates, distance between two points, sectionformulae, shift of origin. Equation of a straight line in various forms,

STP1718




angle between two lines, distance of a point from a line; Lines throughthe point of intersection of two given lines equation of the bisector ofthe angle between two lines, concurrency of lines; Centroid,orthocentre, incentre and circumcentre of a triangle.

Conic Section :Equation of a circle in various forms, equations of tangent, normaland chord. Parametric equations of a circle, intersection of a circlewith a straight line or a circle, equation of a through the points ofintersection of two circles and those of a circle and a straight line.Equations of a parabola, ellipse and hyperbola in standard form, theirfoci, directrices and eccentricity, parametric equations, equations oftangent and normal locus problems.

Three dimensions :Direction cosines and direction ratios, equation of a straight line inspace, equation of a plane, distance of a point from a plane

Vectors :Addition of vectors, scalar multiplication, dot and cross products,scalar triple products and their geometrical interpretations. Positionvector of a point dividing a line segment in a given ratio. Projection ofa vector on a line.

Function :Real valued functions of a real variable, into, onto and one-to-onefunctions, sum, difference, product and quotient of two functions,composite functions, absolute value, polynomial, rational,trigonometric, exponential and logarithmic functions. Even and oddfunctions, inverse of a function, composite function.

Limit, Continuity & Derivability :Limit and continuity of a function, limit and continuity of the sum,difference, product and quotient of two functions, L’Hospital rule ofevaluation of limits of functions even and odd functions, inverse of afunction, continuity of composite function. intermediate value propertyof continuous functions.

Differentiation :Derivative of a function, derivative of the sum, difference, productand quotient of two functions, chain rule, derivatives of polynomial,rational, trigonometric, inverse trigonometric, exponential andlogarithmic functions. Derivatives of implicit functions, derivatives upto order two.

Tangent & Normal :Geometrical interpretation of the derivative, tangents and normal.

Maxima & Minima :Increasing and decreasing functions, maximum and minimum valuesof a function, rolle’s theorem and Lagrange’s Mean value theorem.

Integral calculus :Integration as the inverse process of differentiation, indefinite integralsof standard functions, integration by parts, integration by the methodsof substitution and partial fractions.Definite integrals and their properties, fundamental theorem of integralcalculus. Application of definite integrals to the determination of areasinvolving simple curves.Formation of ordinary differential equations, solution of homogeneousdifferential equations, separation of variables method, linear firstorder differential equations.

Trigonometry :Trigonometric functions, their periodicity and graphs addition andsubtraction formulae, formulae involving multiple and sub-multipleangles, general solution of trigonometric equations.Relations between sides and angles of a triangle, sine rule, cosinerule, half-angle formula and the area of a triangle, inverse trigonometricfunctions (principal value only).

CLASS - XII (PHYSICS)General : Units and dimensions, dimensional analysis; least count,significant figures; Methods of measurement and error analysis forphysical quantities pertaining to the following experiments:Experiments based on using Vernier calipers and screw gauge(micrometer), Determination of g using simple pendulum, Young’smodulus by Searle’s method, Specific heat of a liquid using calorimeter,focal length of a concave mirror and a convex lens using u-v method,Speed of sound using resonance column, Verification of Ohm’s lawusing voltmeter and ammeter, and specific resistance of the materialof a wire using meter bridge and post office box.

Mechanics : Kinematics in one and two dimensions (Cartesiancoordinates only), Projectile Motion; Uniform Circular Motion; RelativeVelocity.

Newton’s laws of motion; Inertial and uniformly accelerated framesof reference; Static and dynamic friction; Kinetic and potential energy;Work and power; Conservation of linear momentum and mechanicalenergy.

Systems of particles; Centre of mass and its motion; Impulse; Elasticand inelastic collisions.Law of gravitation; Gravitational potential and field; Acceleration dueto gravity; Motion of planets and satellites in circular orbits; Escapevelocity.

Rigid body, moment of inertia, parallel and perpendicular axestheorems, moment of inertia of uniform bodies with simple geometricalshapes; Angular momentum; Torque; Conservation of angularmomentum; Dynamics of rigid bodies with fixed axis of rotation;Rolling without slipping of rings, cylinders and spheres; Equilibriumof rigid bodies; Collision of point masses with rigid bodies.

Linear and angular simple harmonic motions.

Hooke’s law, Young’s modulus.

Pressure in a fluid; Pascal’s law; Buoyancy; Surface energy andsurface tension, capillary rise; Viscosity (Poiseuille’s equationexcluded), Stoke’s law; Terminal velocity, Streamline flow, equationof continuity, Bernoulli’s theorem and its applications.

Waves : W ave motion (plane waves only), longitudinal andtransverse waves, superposition of waves; Progressive andstationary waves; Vibration of strings and air columns;Resonance;Beats; Speed of sound in gases; Doppler effect (in sound).

Thermal physics : Thermal expansion of solids, liquids and gases;Calorimetry, latent heat; Heat conduction in one dimension; Elementaryconcepts of convection and radiation; Newton’s law of cooling; Idealgas laws; Specific heats (Cv and Cp for monoatomic and diatomicgases); Isothermal and adiabatic processes, bulk modulus of gases;Equivalence of heat and work; First law of thermodynamics and itsapplications (only for ideal gases); Blackbody radiation: absorptiveand emissive powers; Kirchhoff ’s law; Wien’s displacement law,Stefan’s law.

Electricity and magnetism : Coulomb’s law; Electric f ield andpotential; Electrical potential energy of a system of point chargesand of electrical dipoles in a uniform electrostatic field; Electric fieldlines; Flux of electric field; Gauss’s law and its application in simplecases, such as, to f ind field due to infinitely long straight wire,uniformly charged infinite plane sheet and uniformly charged thinspherical shell.

Capacitance; Parallel plate capacitor with and without dielectrics;Capacitors in series and parallel; Energy stored in a capacitor.

Electric current; Ohm’s law; Series and parallel arrangements ofresistances and cells; Kirchhoff ’s laws and simple applications;Heating effect of current.

Biot–Savart’s law and Ampere’s law; Magnetic field near a current-carrying straight wire, along the axis of a circular coil and inside along straight solenoid; Force on a moving charge and on a current-carrying wire in a uniform magnetic field.Magnetic moment of a current loop; Effect of a uniform magnetic fieldon a current loop; Moving coil galvano- meter, voltmeter, ammeterand their conversions.Electromagnetic induction: Faraday’s law, Lenz’s law; Self and mutualinductance; RC, LR and LC circuits with d.c. and a.c. sources.

Optics: Rectilinear propagation of light; Reflection and refraction atplane and spherical surfaces; Total internal reflection; Deviation anddispersion of light by a prism; Thin lenses; Combinations of mirrorsand thin lenses; Magnification.

Wave nature of light: Huygen’s principle, interference limited to Young’sdouble-slit experiment.

Modern physics : Atomic nucleus; Alpha, beta and gammaradiations; Law of radioactive decay; Decay constant; Half-life andmean life; Binding energy and its calculation; Fission and fusionprocesses; Energy calculation in these processes.Photoelectric effect; Bohr’s theory of hydrogen-like atoms;Characteristic and continuous X-rays, Moseley’s law; de Brogliewavelength of matter waves.

STP1718




SAMPLE TEST PAPER - I(For Class-X Appearing / Passed Students)

Course : VIKAAS (JA*)

Correct Wrong Blank


(dsoy , d fodYi l gh)3 –1 0



29 to 36 Comprehensions (v uqPNsn) 4 –1 0

37 to 38 51 64Matrix Match Type



Part - II (Physics)


Type

PART - I (Hkkx- I)

SECTION - I ([k.M- I)

Straight Objective Type (lh/ks oLrqfu"B izdkj )

This section contains 20 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of

which ONLY ONE is correct.

bl [k.M esa 20 iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA

1. ‘P + Q’ means ‘P is the brother of Q’. ‘P – Q means P is the mother of Q and ‘P × Q’ means ‘P is the sisterof Q’. Which of the following means that M is the maternal uncle of R ?

‘P + Q’ dk vFkZ ‘Q dk HkkbZ P gSA’ ‘P – Q dk vFkZ gS Q dh ek¡ P gS rFkk ‘P × Q’ dk vFkZ gS Q dh cgu ‘P gSA fuEu esa

ls fdldk lk vFkZ R dk ekek M gSA(A) M – R + K (B) M + K – R (C) M + K × Q (D) M – K + R

2. 8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys can finish it in 14 days, thenthe ratio of the time taken by one man alone to that by one boy alone to finish the work is

,d dk;Z dks 8 O;fDr rFkk 12 yMds 10 fnuksa esa djrs gSa tcfd blh dk;Z dks 6 O;fDr rFkk 8 yMds 14 fnuksa esa djrs gS]

rks ,d vdsys O;fDr }kjk dk;Z dks iwjk djus esa fy;k x;k le; rFkk ,d vdsys yMds }kjk dk;Z dks iwjk djus esa fy, x;s

le; dk vuqikr gS &

(A) 1 : 2 (B) 2 : 1 (C) 1 : 1 (D) 1 : 3

3. If A =

3

4

3

, B =

2

5

3

, C = (0.4)2, D = (–1.3)2, then

;fn A =

3

4

3

, B =

2

5

3

, C = (0.4)2, D = (–1.3)2, rks

(A) A > B > C > D (B) D > A > B > C (C) D > B > C > A (D) D > C > A > B

4. Water runs into a cylindrical tank , of diameter 4 m and height 5 m , through a pipe of radius 2 cm ,at the rate of 1/10 m per second . Find the time taken by the tank to fill up .

,d csyukdkj VSad ftldk O;kl 4 m o Å¡pkbZ 5 m

gS dks ,d 2 cm f=kT;k okyh uyh ls 1/10 m izfr lSd.M dh

nj ls Hkjk tkrk gSA VSad dks Hkjus esa dqy le; yxsxk :

(A) 150 hr (B) 150 hr 50 min 50 sec

(C) 138 hr 53 min 20 sec (D) 100 hr 50 min

STP1718




5. In the figure, PQ is parallel to ST, then QRS is equal to :

fn;s x;s fp=k esa] PQ rFkk ST lekUrj gS] rks QRS cjkcj gS

(A) 30° (B) 40° (C) 20° (D) 60°

6. The remainder when x4 – y4 is divided by x + y is :

tc x4 – y4 dks x + y ls Hkkx fn;k tkrk gS rks 'ks"kQy gS -

(A) 0 (B) x + y (C) x2 – y2 (D) 2y4

7. The number 10n – 1 is divisible by 11 for :

(A) odd values of n (B) even values of n

(C) all values of n (D) n, a multiple of 11

la[;k 10n – 1 ,11 ls HkkT; gS -

(A) n ds fo"ke ekuksa ds fy;s (B) n ds le ekuksa ds fy;s

(C) n ds lHkh ekuksa ds fy;s (D) n, 11 ds xq.kd ds fy;s

8. The perimeter of one face of a cube is 20m, then its volume is :

?ku ds ,d Qyd dk ifjeki 20m gks rks mldk vk;ru gS -

(A) 800 m3 (B) 1000 m3 (C) 125 m3 (D) 400 m3

9. The distance between the points A( – 7, 7) and B( – 2, – 5) is :

(A) 13 units (B) 14 units

(C) 15 units (D) none of these

fcUnq A( – 7, 7) rFkk B( – 2, – 5) ds chp nwjh gS -

(A) 13 bdkbZ (B) 14 bdkbZ

(C) 15 bdkbZ (D) buesa ls dksbZ ugha

10. The age of father is twice that of the elder son. Ten years hence the age of the father will be three times that

of the younger son. If the difference of ages of the two sons is 15 years, then the age of father is

(A) 50 years (B) 55 years (C) 60 years (D) 70 years

firk dh mez cM+s iq=k dh mez ls nksxquh gSA nl o"kZ ckn firk dh mez NksVs iq=k dh mez ls rhu xquk gksxhA ;fn nksuksa iq=kksa

dh mez dk vUrj 15 o"kZ gks] rks firk dh mez gksxh&

(A) 50 o"kZ (B) 55 o"kZ (C) 60 o"kZ (D) 70 o"kZ

11. In a railway compartment, 6 seats are vacant on a bench. In how many ways can 3 passenger sit on them

,d jsyos dksp esa] ,d csap ij 6 lhV [kkyh gS bu lhVksa ij rhu ;k=kh fdrus rjhds ls cSB ldrs gS&

(A) 100 (B) 109 (C) 107 (D) 120

STP1718




12. Let R1 and R

2 are the remainders when the polynomials x3 + 2x2 – 5ax – 7 and x3 + ax2 – 12x + 6 are divided

by x – 1 and x + 2 respectively. If 2R1 + R

2 = 2, then the value of a is

cgqin x3 + 2x2 – 5ax – 7 dks x – 1 ls foHkkftr djrs gS] rks 'ks"kQy R1 rFkk cgqin x3 + ax2 – 12x + 6 dks x + 2 ls

foHkkftr djrs gSa] rks 'ks"kQy R2 gSA ;fn 2R

1 + R

2 = 2 gS] rks a =

(A) 2 (B) 4 (C) 6 (D) 0

13. A cone is cut half way through its axis and parallel to the base, the volumes of two portions are in the ratio :

,d 'kadq mlds v{k ds fcYdqy e/; ls o vk/kkj ds lekUrj dkVk tkrk gS rks nksuksa Hkkxksa ds vk;ru dk vuqikr gS&

(A) 1 : 1 (B) 1 : 3 (C) 1 : 7 (D) 1 : 4

14. The condition for equation , x2 b

x + c = 0 (b and c are integers) to have two consecutive integers as

its roots is :

;fn nks Øekxr iw.kkZ ad lehdj.k x2 b x + c = 0 (b o c iw.kkZ ad) ds ewy gS rks :

(A) b2 c2 = 1 (B) b2 4 c = 1 (C) c2 4

b2 = 1 (D) b = c

15. The sum of all the roots of 4x3 – 8x2 – 63x – 9 = 0 is

4x3 – 8x2 – 63x – 9 = 0 ds lHkh ewyksa dk ;ksxQy gS&

(A) 8 (B) 2 (C) – 8 (D) – 2

16. Given : x > 0, y > 0, x > y and z 0. The inequality which is not always correct is ?

fn;k gS : x > 0, y > 0, x > y rFkk z 0, dkSulh vlfedk lnSo lgh ugha gksxh\

(A) x + z > y + z (B) x – z > y – z (C) xz > yz (D) 2z

x > 2z

y

17. Which of the following is/are correct(1) factors of x3 + 1 are (x + 1), (x2 + x + 1) (2) Factors of x3 + 1 are (x + 1), (x2 – x + 1)(3) Factors of x3 – 1 are (x – 1), (x2 – x + 1) (4) Factors of x3 – 1 are (x – 1), (x2 + x + 1)(A) 1 (B) 2, 3 (C) 3 (D) 2, 4

fuEu esa dkSulk/ls lgh gS ?

(1) x3 + 1 ds xq.ku[k.M (x + 1), (x2 + x + 1) gS (2) x3 + 1 ds xq.ku[k.M (x + 1), (x2 – x + 1) gS

(3) x3 – 1 ds xq.ku[k.M (x – 1), (x2 – x + 1) gS (4) x3 – 1 ds xq.ku[k.M (x – 1), (x2 + x + 1) gS(A) 1 (B) 2,3 (C) 3 (D) 2, 4

18. If points (8, 1), (k, – 4) and (2, – 5) are collinear, then the value of k is

;fn fcUnq (8, 1), (k, – 4) rFkk (2, – 5) lejsa[kh; gS] rks k =(A) 2 (B) 3 (C) 4 (D) 5

19. A body strikes the floor vertically with a velocity u and rebounds at the same speed. The change in velocitywould be(A) u (B) 3u (C) 2u (D) zero

,d oLrq u osx ls {kSfrt ry ij m/okZ/kj Vdjkrh gS rFkk leku pky ls okil m/okZ/kj fn'kk esa tkrh gSA osx esa ifjorZu

gksxk -

(A) u (B) 3u (C) 2u (D) 'kwU;

STP1718




20. The value of x satisfying the equation (considering only positive root)

625625/625625 = 2/x is

x ds eku tcfd lehdj.k

625625/625625 = 2/x dks larq"V djrs gS&

(dsoy /kukRed ewy ysrs gq,)

(A) 6 (B) 3 (C) 3 (D) 2/9

SECTION - II ([k.M- II)

Multiple Correct Answer Type (cgqy lgh fodYi izdkj)

This section contains 8 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONE OR MORE is/are correct.

bl [k.M esa 8 iz'u gSaA izR;sd iz'u ds mÙkj ds fy, 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls

vf/kd lgh gSA

21. The difference between areas of outside and inside surfaces of a cylindrical metallic pipe 14 cm long is44 cm2. If the pipe is made of 99 cubic centrimeters of metal and outer radii and inner radii are R and rrespectively, then

14 cm yEcs ,d /kkrq ls cus csyukdkj ikbZi dh ckÐ; rFkk vkUrfjd lrgksa ds {ks=kQyksa ds e/; varj 44 cm2 gSA ;fn ikbZi

99 /ku lsVhehVj /kkrq ls cuk;k tkrk gS rFkk ikbZi dh ckÐ; f=kT;k o vkUrfjd f=kT;k Øe'k% R rFkk r gS] rks &

(A) R = 2.5 (B) R = 2 (C) r = 1.5 (D) r = 2

22. If the following system of linear equations 2x + 3y = 7 , 2x + ( + ) y = 28 has infinite number of

solutions, then

;fn js[kh; lehdj.kk s a ds fuEu fudk; 2x + 3y = 7 , 2x + ( + ) y = 28 ds vuUr gy gS a ]

rks &

(A) = 4 (B) = 8 (C) = 4 (D) = 8

23. Which of the following is/are correct :

fuEu eas ls dkSu lk lR; gS &

(A) –6 2 > – 4 6 (B) –7 2 < – 4 6

(C) log0.4

3 < log4 3 (D) 347 = 2 + 3

24. The points which trisect the line segment joining the points (0, 0) and (9, 12) are

fcUnq tks fcUnq (0, 0) rFkk (9, 12) dks feykus okys js[kk dks lef=kHkkftr djrs gSa] gS&

(A) (3, 4) (B) (8, 6) (C) (6, 8) (D) (4, 0)

25. Which of the following when simplified reduces to unity ?

fuEu esa ls dksu lk ljy djus ij eku 1 ds cjkcj nsrk gS ?

(A)2

3log log4 81log

3(B) log

2 6 + log

2

3

2

(C)

27

64log

6

1–

2

3 (D) 2

7log (1 + 2 – 3 6)

STP1718




26. A fast train takes 3 hours less than a slow train for a journey of 900 km. If the speed of slow train is 15 km/

hr. less than that the fast train, then the speeds of the two trains can be

,d nzqr xfr ls pyus okyh Vsªu /kheh xfr ls pyus okyh Vsªu ls 900 fd-eh- nwjh r; djus esa 3 ?k.Vs dk le; de ysrh

gSA ;fn /kheh xfr ls pyus okyh Vsªu dh xfr nzqr xfr ls pyus okyh Vsªu dh xfr ls 15 fd-eh-@?k.Vk de gS] rc nksuksa

Vsªuksa dh xfr gks ldrh gS&

(A) 75 km/h fd-eh-@?k.Vk (B) 80 km/h fd-eh-@?k.Vk

(C) 60 km/h fd-eh-@?k.Vk (D) 65 km/h fd-eh-@?k.Vk

27. If Bsin

Asin= p,

Bcos

Acos= q, then ;fn

Bsin

Asin= p,

Bcos

Acos= q, rks

(A) tan2A = 22

22

q)p–1(

p)1–q((B) tan2A = 22

22

p)q–1(

q)1–p(

(C) tan2B = 2

2

p–1

1–q(D) p2tan2A = q2 tan2B

28. A man on the top of vertical tower observes a car moving at a uniform speed coming directly towards the footof the tower. If it takes 12 minutes for the angle of depression to change from 30º to 45º, After this it will reachthe tower in

,d O;fDr m/okZ/kj ehukj ds 'kh"kZ ls ,d leku xfr ls ehukj ds ikn dh vksj lh/ks vkrh gqbZ dkj dks ns[krk gSA ;fn ;g

voueu dks.k (angle of depression) 30º ls 45º gksus esa 12 feuV ysrk gS] rks og Vkoj rd igqpus esa fdruk le; ysxk

(A) 13

12

minutes feuV (B) 6( 3 + 1) minutes feuV

(C) 13

12

minutes feuV (D) 6( 3 – 1) minutes feuV

SECTION - III ([k.M - III)

Comprehension Type (cks/ku çdkj)This section contains 2 paragraphs. Based upon each paragraph, there are 4 questions. Each question has4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.bl [k.M esa 2 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 4 ç'u gSA çR;sd ç'u ds 4 fodYi

(A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA

Paragraph for Question Nos. 29 to 32iz'u 29 ls 32 ds fy, vuqPNsn

If (x–) is the highest common factor (HCF) of x2+px+q and x2+qx+p, then answer the following questions;fn x2 + px + q rFkk x2 + qx + p dk egÙke lekiorZd (HCF), (x – ) gks] rks fuEu iz'uksa ds mÙkj nhft;sA

29. The value of p + q + 1 is an(A) Even number (B) odd number (C) Irrational number (D) None of thesep + q + 1 dk eku gksxk&

(A) le la[;k (B) fo"ke la[;k (C) vifjes; la[;k (D) buesa ls dksbZ ugha

30. The value of p3 + q3 + 1 isp3 + q3 + 1 dk eku gS&

(A) 3pq (B) (p2 + q2 + 1 – pq – p – q)(C) (pq + p + q – p2 – q2 – 1) (D) None of these (buesa ls dksbZ ugha)

STP1718




31. The value of (p – q)2 + 3

4 (p3 + q3 + 1) is k , then k is

(A) a prime number (B) a composite number(C) neither prime nor composite (D) none of these

(p – q)2 + 3

4 (p3 + q3 + 1) dk eku k gS] rks k gS&

(A) ,d vHkkT; la[;k (B) ,d la;qDr la[;k

(C) u rks vHkkT; rFkk u gh la;qDr (D) buesa ls dksbZ ugha

32. The value of k + q + p isk + q + p dk eku gS&

(A) 2 (B) 4 (C) 1 (D) 0

Paragraph for Question Nos. 33 to 36

iz'u 33 ls 36 ds fy, vuqPNsn

The following questions are based on the given data for an examination.A. Candidates appeared 10500

B. Passed in all the five subjects 5685C. Passed in three subject only 1498D. Passed in two subject only 1250

E. Passed in one subject only 835F. Failed in English only 78G. Failed in maths only 275

H. Failed in Physics only 149I. Failed in chemistry only 147J. Failded in Biology only 221

,d ijh{kk ds fn;s x;s vkadMks+ ds vk/kkj ij fuEu iz'uksa dk mÙkj nhft,&

A. ijh{kk esa 'kkfey dqy ijh{kkFkhZ 10500

B. lHkh ik¡p fo"k;ks esa mÙkh.kZ 5685

C. dsoy rhu fo"k;ks esa mÙkh.kZ 1498

D. dsoy nks fo"k;ks esa mÙkh.kZ 1250

E. dsoy ,d fo"k; esa mÙkh.kZ 835

F. vaxszth fo"k; esa vuqÙkh.kZ 78

G. xf.kr fo"k; esa vuqÙkh.kZ 275

H. HkkSfrdh fo"k; esa vuqÙkh.kZ 149

I. jlk;u fo"k; esa vuqÙkh.kZ 147

J. tho foKku fo"k; esa vuqÙkh.kZ 221

33. How many candidates failed in all the subjects ?

lHkh fo"k;ksa esa vuqÙkh.kZ ijh{kkFkhZ dh la[;k gS&

(A) 4815 (B) 3317

(C) 2867 (D) 362

34. How many candidates passed at least in four subjects ?

dsoy pkj fo"k;ksa esa mÙkh.kZ ijh{kkFkhZ dh la[;k gS&

(A) 6555 (B) 5685

(C) 1705 (D) 870

STP1718




35. How many candidates failed because of having failed in four or less subjects ?

pkj ;k pkj ls de fo"k;ks esa vuqÙkh.kZ ijh{kkFkhZ;ksa dh la[;k gS&

(A) 4815 (B) 4453 (C) 3618 (D) 2368

36. How many candidates failed in only two subjects ?(A) 4815 (B) 3618 (C) 1498 (D) none of these

dsoy nks fo"k;ksa esa vuqÙkh.kZ ijh{kkFkhZ;ksa dh la[;k gS&

(A) 4815 (B) 3618 (C) 1498 (D) buesa ls dksbZ ugha

SECTION - IV ([k.M - IV)

Matrix - Match Type (eSfVªDl&lqesy izdkj)

This section contains 2 questions. Each question contains statements given in two columns whichhave to be matched. Statements in Column are labelled as A,B,C and D whereas statements inColumn are labelled as p,q,r and s. The answers to these questions

have to be appropriately bubbled as illustrated in the following example.

If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q ,

p q r sp q r sp q r sp q r s

ABCD

p q r s

then the correctly bubbled matrix will look like the following :

bl [k.M esa 2 iz'u gSaA izR;sd iz'u esa nks dkWye esa oDRO; (statements) fn;s gq, gSa ftudk lqesy (match) djuk gSA dkWye

(Column-I) esa fn;s x;s oDrO;ksa (A, B, C, D) dks dkWye (Column-II) esa fn;s x;s oDrO;ksa (p, q, r, s) ls lqesy djuk gSA

bu iz'uksa ds mÙkj fn;s x;s mnkgj.k ds vuqlkj mfpr cqYyksa dks dkyk djds n'kkZuk gSA p q r sp q r sp q r sp q r s

ABCD

p q r s

;fn lgh lqesy A-p, A-r, B-p, B-s, C-r, C-s rFkk D-q gSa] rks lgh fof/k ls dkys fd,

x;s cqYyksa dk 4 x 4 esafVªDl (matrix) uhps n'kkZ;k x;k gSA

37. Let A, B and C be three points whose coordinates are (2, 3), (3, b) and (5, 7) respectively.

ekuk fd A, B rFkk C rhu fcUnq gS] ftuds funsZ'kkad Øe'k% (2, 3), (3, b) rFkk (5, 7) gSaA

Column – Column –

(A) If area of ABC is 2

1 unit, then positive value of b is (p) 6

;fn ABC dk {ks=kQy 2

1 bdkbZ gS] rks b dk /kukRed eku gS&

(B) If A, B and C are collinear, then value of 3b is (q) 5

;fn A, B rFkk C lajs[kh; gS] rks 3b dk eku gS&

(C) If A, B and C are collinear and B divides AC in the ratio : 1, (r) 4then value of 4( + 1) is

;fn A, B rFkk C lajs[kh; gS rFkk B ] AC dks : 1 ds vuqikr esa

foHkkftr djrk gS rks 4( + 1) dk eku gS&

(D) If b = 4, then the length of the longest side of PQR is (s) 13

;fn b = 4 gS] rks PQR dh lcls yEch Hkqtk dh yEckbZ gS&

STP1718




38. Match the following : fuEu dks lqesfyr dhft,&

Column – (LrEHk – ) Column – (LrEHk – )(A) If f(x) = – x2 + 2x + 5, then maximum value of f(x) is (p) 3

;fn f(x) = – x2 + 2x + 5, rc f(x) dk vf/kdre eku gksxk&

(B) If a, b N and a2 – b2 = 11, then a = (q) 6;fn a, b N vkSj a2 – b2 = 11 gS] rks a =

(C) Number of tangents common to two circles (r) 8touching each other externally isnks o`Ùk tks ,d nwljs dks ckg~; Li'kZ djrs gS]

dh mHk;fu"B Li'kZ js[kkvksa dh la[;k gS&

(D) If one root of the equation 3x2 – 9 = k2x – k is 2, then (s) – 4a value of 4k is;fn lehdj.k 3x2 – 9 = k2x – k dk ,d ewy 2 gS] rks4k dk ,d eku gS &

PART - II (Hkkx - II)

SECTION - I ([k.M- I)Straight Objective Type (lh/ks oLrqfu"B izdkj)

This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), outof which ONLY ONE is correct.bl [k.M esa 8 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA

39. In the figure shown a convex mirror of radius of curvature 20 cm is shown. An object O is placed in frontof this mirror. Its ray diagram is shown. How many mistakes are there in the ray diagram (AB is itsprincipal axis):fp=k esa 20 cm oØrk f=kT;k okyk mÙky niZ.k n'kkZ;k x;k gSA ,d oLrq O niZ.k ds lkeus j[kh tkrh gSA bldk fdj.k

fp=k n'kkZ;k x;k gSA blds fdj.k fp=k esa fdruh xyfr;ka (AB bldh eq[; v{k gS) gS \

(A) 3 (B) 2 (C) 1 (D) 0

40. A particle moves in a straight line obeying the V–t graph as shown in figure. The average velocity of theparticle from t = 0 to t = T is :

(A) Zero (B) 0V

2(C)

02V

3(D) 03V

4

V0

V

Tt0

ljy js[kk ij xfr'khy d.k fp=k esa iznf'kZr osx≤ xzkQ dk ikyu djrk gSA t = 0 ls t = T le;kUrjky ds e/; d.k

dk vkSlr osx gksxk :

(A) 'kwU; (B) 0V

2(C)

02V

3(D) 03V

4

STP1718




41. A lift is moving upwards with a constant speed of 5 m/s. The speed of one of the pulleys is 5 m/s as shown.Then the speed of second pulley is :(A) 5m/s (B) zero (C) 7.5 m/s (D) 10 m/s

,d fy¶V 5 m/s dh fu;r pky ls Åij dh vksj xfreku gSA fp=k esa n'kkZ;s x, vuqlkj ,d vkn'kZ f?kjuh dk osx 5 m/s gSrks nwljh f?kjuh dh pky gksxh %

(A) 5m/s (B) 'kwU; (C) 7.5 m/s (D) 10 m/s

42. The frequency of a man’s voice is 300 Hz and its wavelength is 1 meter. If the wavelength of a child’s voice is1.5 m, then the frequency of the child’s voice is:,d vkneh ds vkokt dh vko`fÙk 300 Hz rFkk rjaxnS/;Z 1 ehVj gSA vxj cPps dh vkokt dh rjaxnS/;Z 1.5 ehVj gSrks cPps ds vkokt dh vko`fÙk gksxh :(A) 200 Hz (B) 150 Hz (C) 400 Hz (D) 350 Hz.

43. A circuit has fuse of 5A. What is the maximum number of 100W (220 V) bulb that can be safely used in the circuit?,d ifjiFk esa 5 ,fEi;j dk ¶;wt gSA 100 okV (220 oksYV) ds cYcksa dh vf/kdre~ la[;k D;k gksxh tks ifjiFk esa lqjf{kr

:i ls iz;qä fd;s tk ldrs gS&

(A) 5 (B) 7 (C) 9 (D) 11

44. The work done by a satellite of mass m in going once a round the earth in an orbit of radius r is :m nzO;eku ds mixzg }kjk r f=kT;k ds d{k esa i`Foh dk ,d pDdj yxkus esa fd;k x;k dk;Z gS :(A) r2 mg (B) m r G (C) Zero 'kwU; (D) Infinite vuUr

45. A light ray incident on a surface of quartz from air with angle of incidence 45º . What is the angle of refraction

and velocity of light in quartz ? ( Refractive index of quartz = 2 )DokV~Zt dh lrg ij ok;q ls 45º ds dks.k ij izdk'k fdj.k vkifrr gSA viorZu dks.k rFkk izdk'k dk osx gksxk : (DokV~Zt

dk viorZukad = 2 )

(A) = 30º , V = 3

2 2 10+8 m/s (B) = 60º , V = 3 2 108 m/s

(C) = 30º , V = 3 2

2 108 m/s (D) = 45º , V =

32

108 m/s

46. Electrostatics force between two point charges is given by :

nks fcUnq vkos'kksa ds e/; fLFkj fuEu fu;e ls fn;k tkrk gS %

(A) Newton law U;wVu fu;e (B) Maxwell law esDlosy fu;e

(C) Lenz Law ySat fu;e (D) Coloumbs Law dqykWe fu;e

SECTION - II ([k.M - II)

Multiple Correct Answers Type (cgqy lgh mÙkj izdkj)

This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B),(C) and (D), out of which ONE OR MORE THAN ONE is/are correct.

bl [k.M esa 4 cgq lgh mÙkj izdkj ds iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls ,d ;k ,d

ls vf/kd fodYi lgh gS ¼gSa½A

STP1718




47. A particle is moving along straight line which of the following option are correct :(A) If its acceleration is increasing its speed may be decreasing

(B) If its acceleration is zero at certain instant its speed may be non–zero at that instant

(C) If its acceleration is zero for some interval of time its speed remains constant during that interval

(D) If acceleration of particle is constant its speed may be zero at some instant

,d d.k ljy js[kk ds vuqfn'k xfr'khy gS fuEu esa ls dksulk fodYi lgh gS

(A) ;fn bldk Roj.k c<rk gS bldh pky ?kV ldrh gS

(B) ;fn bldk Roj.k fdlh {k.k 'kwU; gS bldh pky bl {k.k ij v'kwU; gks ldrh gS

(C) ;fn bldk Roj.k dqN le;kUrjky ds fy, 'kwU; gS bldh pky bl le;kUrjky esa vpj jgrh gS

(D) ;fn d.k dk Roj.k vpj gS bldh pky fdlh {k.k 'kwU; gks ldrh gS

48. There are two massless springs A and B of spring constant KA and K

B respectively and K

A > K

B . If

WA and W

B be denoted as work done on A and work done on B respectively, then

(A) If they are compressed to same distance, WA > W

B

(B) If they are compressed by same force (upto equilibrium state) WA < W

B

(C) If they are compressed by same distance, WA = W

B

(D) If they are compressed by same force (upto equilibrium state) WA > W

B

nks nzO;eku jfgr fLiazx A rFkk B ftuds fLiazx fu;rkad Øe'k% KA rFkk K

B gS] tgkW K

A > K

B gSA] ;fn A rFkk B ij

fd;s tkus okys dk;Z Øe'k% WA rFkk W

B gks a] rks

(A) ;fn nksuks dks leku nwjh rd laihfM+r fd;k tk;s] WA > W

B

(B) ;fn nksuks a dks leku cy ls laihfMr fd;k tk;s] (lkE;oLFkk rd), WA < W

B

(C) ;fn nksuks a dks leku nwjh rd laihfM+r fd;k tk;s, WA = W

B

(D) ;fn nksuks a dks leku cy ls laihfMr fd;k tk;s (lkE;oLFkk rd ) WA > W

B

49. Which of the following statement is true ?

(A) Like poles repel

(B) Unlike poles attract

(C) Magnet may have only one pole

(D) On breaking the magnet, the magnetism of its pieces vanish.

fuEufyf[kr esa ls dkSulk dFku lR; gaS \

(A) pqEcd ds leku /kqzoksa esa izfrd"kZ.k gksrk gSA

(B) pqEcd ds vleku /kqzoksa esa vkd"kZ.k gksrk gSA

(C) pqEcd dk dsoy ,d gh /kzqo gks ldrk gSA

(D) pqEcd ds VqdM+s djus ij] VqdM+ksa dk pqEdRo lekIr gks tkrk gSA

50. Which of the following are non-renewable sources of energy :

(A) Solar energy (B) Petroleum

(C) Natural gas (D) Uranium

fuEu es ls dkSu mTkkZ ds vuohuhdj.k L=kksr gS

(A) lkSj mTkkZ (B) isVªksfy;e

(C) izkdfrd xSl (D) ;wjsfu;e

STP1718




SECTION - III ([k.M- III)

Matrix - Match Type (eSfVªDl&lqesy izdkj)

This section contains 1 question. The question contains statements given in twocolumns which have to be matched. Statements (A, B, C, D) in Column-I have to bematched with statements (p, q, r, s) in Column-II. The answers to these questions haveto be appropriately bubbled as illustrated in the following example.


ABCD

p q r s

If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q, then the correctly matched answer will looklike in bubbled matrix. One option in Column–I may have one or more than one correct options in Column–II.Marks will be awarded only if an option in Column–I is matched with all correct options in Column-II. Eachcorrect matching will be awarded 1 mark and if all the matchings are correct then 6 marks will be awarded.

bl [k.M esa 1 iz'u gSaA izR;sd iz'u esa nks dkWye esa oDRO; (statements) fn;s gq, gSa ftudk lqesy

(match) djuk gSA dkWye (Column-I) esa fn;s x;s oDrO;ksa (A, B, C, D) dks dkWye (Column-II) esa fn;s

x;s oDrO;ksa (p, q, r, s) ls lqesy djuk gSA bu iz'uksa ds mÙkj fn;s x;s mnkgj.k ds vuqlkj mfpr cqYyksa

dks dkyk djds n'kkZuk gSA


ABCD

p q r s

;fn lgh lqesy A-p, A-r, B-p, B-s, C-r, C-s rFkk D-q gSa] rks lgh fof/k ls dkys fd, x;s cqYyksa dk 4 x 4 esafVªDl (matrix)

esa n'kkZ;k x;k gSA dkWye–I ds izR;sd fodYi dk dkWye–II ds ,d ;k ,d ls T;knk fodYiksa ls lqesy djk;k tk ldrk gSA vad

mlh fLFkfr esa iznku fd;s tk;sxsaA tcfd dkWye–I ds fdlh fodYi dk dkWye–II ds lHkh lgh fodYiksa ls lqesy djk;k x;kA izR;sd

lgh lqesy ds fy, 1 vad iznku fd;s tk;sxsa rFkk tc lHkh lqesy lgh gksxsa rc 6 vad iznku fd;s tk;sxsaA

51. Match the items given in Column I with those in Column II.Column-I Column-II

(A) Electric fuse (p) Magnetic effect of current(B) Electric generator (q) Heating effect of current(C) Electro motor (r) Chemical effect of current(D) Dry cell (s) Electromagnetic inductiondkWye I ds enksa dks dkWye II ds enksa ls lqesy dhft;s %

dkWye-I dkWye-II(A) fo|qr ¶;wt (p) fo|qr /kkjk dk pqEcdh; izHkko

(B) fo|qr tfu=k (q) fo|qr /kkjk dk rkih; izHkko

(C) fo|qr&eksVj (r) fo|qr /kkjk dk jklk;fud izHkko

(D) 'kq"d lsy (s) fo|qr&pqEcdh; izsj.k

PART - III


Straight Objective Type (lh/ks oLrqfu"B izdkj)This section contains 8 multiple choice questions. Each question has choices (A), (B), (C) and (D), outof which ONLY ONE is correct.

bl [k.M esa 8 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA

52. Consider the processes izØeksa ij fopkj dhft,

(i) 2

1H

2 (g) + aq. H+ (aq.) (ii) 2O(g) O

2(g)

(iii) NH4+ (g) + Cl– (g) NH

4Cl(s) (iv) P

4 (black) + 5O

2(g) P

4O

10(s)

Which of the above does not represent Hformation

of the product.

mijksDr esa ls dkSulk izØe mRikn ds HfuekZ.k

dks iznf'kZr ugh djrk gS\

(A) I,IV (B) II, IV (C) I, II, III (D) II , III, IV

STP1718




53. Maximum oxidation state of oxygen is:

vkWDlhtu dh vf/kdre vkWDlhdj.k voLFkk gS&

(A) +6 (B) +4 (C) +2 (D) –2

54. In the reaction 4P + 3KOH + 3H2O 3KH2PO2 + PH3fuEu vfHkfØ;k] 4P + 3KOH + 3H2O 3KH2PO2 + PH3 esa](A) P undergoes reduction only (P dk dsoy vi;pu gksrk gSA)(B) P undergoes oxidation only (P dk dsoy vkWDlhdj.k gksrk gSA)(C) P undergoes both oxidation and reduction (P dk vkWDlhdj.k o vip;u nksuksa gksrk gSA)(D) neither undergoes oxidation nor reduction (u rks vkWDlhdj.k gksrk gS vkSj u gh vip;u gksrk gSA )

55. The value of Planck’s constant is 6.63 × 10–34 Js. The velocity of light is 3 × 108 m/sec. Which value isclosest to the wavelength of light with frequency of 8 × 1015 sec–1 ?Iykad fLFkjkad dk eku 6.63 × 10–34 Js gSA izdk'k dk osx 3 × 108 [email protected] gSA dkSulk rjaxnS/;Z eku izdk'k dh

vko`fr 8 × 1015 lSd.M–1 ds lehi gS ?(A) 5 × 10–18 m (B) 4 × 10–8 m (C) 3 × 107 m (D) 2 × 10–25 m

56. The maximum & minimum oxidation state of Bromine is:

czksehu dh vf/kdre rFkk U;wure vkWDlhdj.k voLFkk dkSulh gS\

(A) +6, –2 (B) +7, –1 (C) +1, –1 (D) +5, –1

57. Which of the following is the IUPAC nomenclature of ethyl methyl ketone?

(A) Propan-1-one (B) Butanone (C) Butan-1-one (D) Propan-2-onefuEu esa ls dkSulk uke ,fFky esfFky dhVksu dk IUPAC ukedj.k gS\

(A) izksisu-1-vkWu (B) C;wVsukWu (C) C;wVsu-1-vkWu (D) izksisu-2-vkWu

58. Which of the following reaction is Finkelstein reaction?fuEu esa ls dkSulh vfHkfØ;k fQadyLVhu vfHkfØ;k gS\

(A) R–I + NaCl Acetone R–Cl + NaI (B) R–Cl + AgF RF + AgCl

(C) R–Cl + NaI Acetone R–I + NaCl (D) R–Cl + NaI OH2 R–I + NaCl

59. What is the relation between 3-Ethylpentane and 3-Methylhexane ?(A) Chain isomers (B) Position isomers (C) Functional isomers (D) Relation3-,fFkyisUVsu vkSj 3-esfFkygsDlsu ds e/; D;k lEcU/k gS \

(A) Ja[kyk leko;oh (B) fLFkfr leko;oh (C) fØ;kRed leko;oh (D) lEcU/k

SECTION - II ([k.M - II)Multiple Correct Answers Type (cgqy lgh mÙkj izdkj)

This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A),(B), (C) and (D), out of which ONE OR MORE THAN ONE is/are correct.bl [k.M esa 4 cgq lgh mÙkj izdkj ds iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls ,d ;k ,d


60. When SO3 is taken in 10 L vessel an equilibrium is established between SO

3, SO

2 and O

2

SO3(g) SO

2(g) +

1

2O

2(g)

The density of the equilibrium mixture is found to be 1.6 gm/l at equilibrium temperature 300 K. Which of

the following conclusions can be made using the data given above ?

(A) Number of moles of SO3 initially is 0.2

(B) Total number of moles is 0.5 at equilibrium.

(C) Number of moles of SO3 at equilibrium can’t be calculated as additional data is required.

(D) Partial pressure of SO3 is 0.492 atm at equilibrium.

STP1718




tc 10 L ik=k esa SO3 dks fy;k tkrk gS rks SO3, SO2 o O2 ds e/; ,d lkE; LFkkfir gksrk gS &

SO3(g) SO2(g) + 1

2O

2(g)

lkE; feJ.k dk ?kuRo lkE; rkieku 300 K ij 1.6 gm/ gS] fn;s x;s mijksDr vk¡dM+ks dk iz;ksx dj fuEu esa ls dkSuls

fu"d"”kZ cuk;s tk ldrs gS \

(A) izkjEHk esa SO3 ds eksyks dh la[;k 0.2 gSA

(B) lkE; ij eksyks dh dqy la[;k 0.5 gSA

(C) lkE; ij SO3 ds eksyks dh la[;k dh x.kuk ugh dj ldrs gS D;ksfd vfrfjDr vk¡dM+ks dh vko';drk gksrh gSA

(D) lkE; ij SO3 dk vkaf'kd nkc 0.492 atm gSA

61. Which of the following can act as oxidant as well as reductant?

fuEu esa ls dkSulk@dkSuls vkWDlhdkjh ds lkFk&lkFk vipk;ddkjh dh Hkkafr dk;Z dj ldrk gS@ldrs gSa\

(A) H2O

2(B) ClO

2(C) SO

2(D) F

2

62. Which of the following is/are electron deficient compounds ?

fuEu esa ls dkSulk@dkSuls bysDVªkWu U;wu ;kSfxd gS@gSa \

(A) NaBH4

(B) B2H

6(C) AlCl

3(D) LiAlH

4

63. Which is / are the correct IUPAC name of the corresponding structures

(A)

OH

CH3

NHCH CH2 3

3-(N-Ethylamino)-2-methylcyclohexan-1-ol

(B) 3-Bromo-2–chlorobutanal

(C) 3-Cyanopentanoic acid

(D) CH3–CH–CH–CH3

CH3 CH2–CH3 2, 3-Dimethyl Pentane

mijksDr esa ls dkSulh lajpukvksa ds lgh IUPAC uke n'kkZ;s x;s gS\

(A)

OH

CH3

NHCH CH2 3

3-(N-,fFky,ehuks)-2-esfFkylkbDyksgsDlsu-1-vkWy

(B) 3-czkseks-2-DyksjksC;wVsusy

(C) 3-lk;uksisUVsukWbd vEy

(D) CH3–CH–CH–CH3

CH3 CH2–CH3 2, 3-MkbesfFky isUVsu

STP1718




SECTION - III ([k.M- III)Matrix - Match Type (eSfVªDl&lqesy izdkj)

This section contains 1 question. Each question contains statements given in two columns which have tobe matched. Statements (A, B, C, D) in Column-I have to be matched with statements (p, q, r, s) in Column-II. The answers to these questions have to be appropriately bubbled as illustrated in the following example.If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q, then thecorrectly will look like in bubbled matrix. Each statement in Column–I mayhave one or more than one correct options in Column–II. One Mark will beawarded for each statement in Column–I is matched with all correct optionsin Column-II. If all the statements in column-I are correctly matched to their


ABCD

p q r s

correct options in column-II , then 6 marks will be awarded.bl [k.M esa 1 iz'u gSaA izR;sd iz'u esa nks dkWye esa oDRO; (statements) fn;s gq, gSa ftudk lqesy (match) djuk gSA dkWye

(Column-I) esa fn;s x;s oDrO;ksa (A, B, C, D) dks dkWye (Column-II) esa fn;s x;s oDrO;ksa (p, q, r, s) ls lqesy djuk gSA bu

iz'uksa ds mÙkj fn;s x;s mnkgj.k ds vuqlkj mfpr cqYyksa dks dkyk djds n'kkZuk gSA

;fn lgh lqesy A-p, A-r, B-p, B-s, C-r, C-s rFkk D-q gSa] rks lgh fof/k ls dkys fd, x;s

cqYyksa dk 4 x 4 esafVªDl (matrix) esa n'kkZ;k x;k gSA dkWye–I ds izR;sd oDÙkO; dk dkWye–II ds ,d ;k ,d ls T;knk fodYiksa ls lqesy djk;k tk ldrk gSA dkWye–I ds izR;sd oDÙkO;


ABCD

p q r s

dk dkWye–II ds lHkh lgh fodYiksa ls lqesy djkus ds fy, ,d vad iznku fd;k tk;sxkA ;fn

dkWye–I ds lHkh OkDÙkO;ksa dk dkWye–II ds lHkh lgh fodYiksa ls lqesy fd;k x;k rks 6 vadiznku fd;s tk;sxsaA

64. Column – I Column – II

(A) At low pressure (p) Z = 1 + RT

pb

(B) At higher pressure (q) Z = 1 – RTV

a

m

(C) At low density of gas (r) gas is more compressible than ideal gas

(D) For H2 and He at 0°C (s) gas is less compressible than ideal gas

dkWye– I dkWye – II

(A) U;wu nkc ij (p) Z = 1 + RT

pb

(B) mPp nkc ij (q) Z = 1 – RTV

a

m

(C) xSl ds fuEu ?kuRo ij (r) xSl vkn'kZ xSl dh rqyuk esa vf/kd laihM~; gSA

(D) 0°C ij H2 rFkk He ds fy, (s) xSl vkn'kZ xSl dh rqyuk esa de laihM~; gSA

ANSWER KEY1. (B) 2. (A) 3. (C) 4. (C) 5. (C) 6. (A) 7. (B)

8. (C) 9. (A) 10. (A) 11. (D) 12. (A) 13. (C) 14. (B)

15. (B) 16. (C) 17. (D) 18. (B) 19. (C) 20. (B) 21. (AD)

22. (AD) 23. (ABCD) 24. (AC) 25. (AC) 26. (AC) 27. (AC) 28. (AB)

29. (A) 30. (A) 31. (C) 32. (D) 33. (D) 34. (A) 35. (B)

36. (C) 37. (A) (r),(B) (s),(C) (p),(D) (q)

38. (A) (q); (B) (q); (C) (p); (D) (q, s)39. (B) 40. (B) 41. (A) 42. (A) 43. (D) 44. (C) 45. (C)46. (D) 47. (ABCD) 48. (AB) 49. (AB) 50. (BCD)51. (A) – q ; (B) – s ; (C) – p ; (D) – r

52. (D) 53. (C) 54. (C) 55. (B) 56. (B) 57. (B) 58. (C)

59. (A) 60. (AC) 61. (ABC) 62. (BC) 63. (ABCD)

64. (A) - q, r ; (B) - p,s ; (C) - q,r ; (D) - p,r

STP1718




HINTS & SOLUTION TO SAMPLE TEST PAPER-I

1. M + K – R

2. 1 : 2

3. D > B > C > A

4. 138 hr 53 min 20 sec

5. 20°

6. 0

7. even values of n (n ds le ekuksa ds fy;s)

8. 125 m3

9. 13 units (bdkbZ)

10. 50 years

11. Total ways dqy rjhds = 6C3.3! = !3!3

!66. =

6

4.5.6.6 = 120

12. 2

13. volume of cone ABCF

v1 =

3

1 (r

2)2 h.

vol. of frustum

v2 =

3

1h [r

12 + r

22 + r

1 r

2]

2

1

r

r =

h

h2

r1

= 2r2

So, v2 =

3

1 h [7r

22)

v2 =

3

1 (r

2)2h

hr73

1

hr3

1

V

V

22

22

2

1

7

1

v

v

2

1

14. b2 4 c = 1

15. 2

16. Let ekuk z < 0 zx < zyso xz > zy is not correct alwaysvr% xz > zy lnSo lgh ugha gksxhA

STP1718




17. x3 + 1 = (x + 1)(x2 + x + 1)x3 – 1 = (x – 1)(x2 + x + 1)so (2)&(4) are correct vr% (2)vkSj(4) lgh gSA

18. 3

19. 2u

20. 3

21. R = 2.5

22.28

73

2

2

= 4, + = 12 = 8

23. –6 2 > – 4 6 ; –7 2 < – 4 6 ; log0.4

3 < log4 3 ; 347 = 2 + 3

24.

(0, 0) (9, 12)

A C D B

C divides AB in the ratio 1 : 2.

Co-ordinates of C

12

012,

21

09 = C (3, 4) D divides AB in the ratio of 2 : 1.

Co-ordinates of D

12

024,

12

018 = D (6, 8) Trisection points are (3, 4) and (6, 8).

25. (A)2

3log log4 3log 8)3( =

2

3log 8log 22 =

2

3log

2

3 = 1

(B) log2 6 + log

2 3

2= log

2 6 ·

3

2 1

(C) – 6

1

27

64log

2/3 =

6

1

6

3

23

2log

= 1

(D)2

7log

6

3–21 =

2

7log

2

5 1

26.15–x

900 –

x

900 = 3

x

300–

15–x

3003 = 3 4500 = x2 – 15x x2 – 15x – 4500 = 0 x =

2

1800022515

x =2

13515 = 75, – 60

x – 15 = 60Ans. 75 km/h & 60 km/h

27. sinB = p

Asin, cos B =

q

Acos 2

2

p

Asin+ 2

2

q

Acos = 1

STP1718




2

2

p

Atan + 2q

1 = 1 + tan2 AA

tan2A = 22

22

q)p–1(

p)1–q(

sin A = psinB, cosA = q cosB 1 = p2 sin2B + q2 cos2 B

1 + tan2 B = p2 tan2 B + q2 tan2B = 2

2

p–1

1–q

28.13

12

minutes (feuV) ; 6( 3 + 1) minutes (feuV)

29. Since x – is HCF of x2 + px + q and x2 + qx + pp q and (x – ) is a factor of x2 + px + q as well as x2 + qx + p

2 + p + q = 0 .... (i)2 + qa + p = 0 .... (ii)

Subtracting (p – q) = (p – q) = 1 p + q + 1 = 0 by (i)

Hindi pw¡fd x2 + px + q o x2 + qx + p dk egÙke lekiorZd (x – gSp q rFkk (x – ) O;atd x2 + px + q rFkk x2 + qx + p nksuksa dk ,d xq.ku[k.M gksxkA

2 + p + q = 0 .... (i)2 + qa + p = 0 .... (ii)

?kVkus ij (p – q) = (p – q) = 1 p + q + 1 = 0 by (i)

30. If p + q + 1 = 0then p3 + q3 + 13 = 3pq

Hindi ;fn p + q + 1 = 0

rks p3 + q3 + 13 = 3pq

31. p2 + q2 – 2pq + 3

4 (3pq) = p2 + q2 + 2pq

= (p + q)2 = (– 1)2 = 1

32. 0

33. 362

34. 6555

35. 4453

36. 1498

37. Area of ABC is D = 2

1

175

1b3

132

= 2

1 (– 3b + 13)

(A) given D = 2

1

2

1(– 3b + 13) =

2

1 – 3b + 13 = 1

b = 4 &3

14

(B) If A, B, C are Collinear = 0 (– 3b + 13) = 0 3b = 13

STP1718




(C) A, B, C are collinear b = 3

13

also B divider AC in the ratio : 1

(3, b) =

1

37,

1

25

1

25

= 3 &1

37

= 3

13

3

13b

= 2

1

Hence 4( + 1) = 6

(D) If P(2, 3), Q(3, 4) & R(5, 7) PQ = 211 PR > QR > PQ

QR = 1394

PR = 5169

38. (A) (q), (B) (q), (C) (p), (D) (q, s)

39. Error – Error –

40. Vave

= Total displacement

Total time l e;dqy

foLFkki udqy =

00

1TV V2T 2

41. In the frame of the lift first pulley will be stationary so velocity of second pulley will be zero & so in the frameof ground it moves with 5 m/sfy¶V ds funsZ'k rU=k esa iqyh fojke esa gksxhA vr% f}rh; iqyh dk osx 'kwU; gksxkA vr% tehu funsZ'k rU=k esa ;g 5 m/s ls

xfreku gksxhA

42. (A) f1

1 = f

2

2

(300) (1) = (f2) (1.5)

200 Hz = f2

43. Current rating of fuse = 5APower = 100 W, Potential difference = 220V

Current through each bulb = A11

5

220

100

differencePotential

Power

Number of bulbs = 1111/5

5

bulbeachthroughCurrent

fuseofratingCurrent

44. Displacment is zero for a complete cycles work done is also zero

45. = 30º , V = 3 2

2 108 m/s

46. Coloumbs Law dqykWe fu;e

47. (A) If acceleration and velocity in opposite direction then speed of particle may decrease.(B) In SHM, acceleration is zero at mean position but speed is non-zero.(C) If acceleration is zero then speed remains constant.(D) In vertical motion of particle at heighest point of motion speed is zero but acceleration is constant.(A) ;fn osx o Roj.k foifjr fn'kk esa gS rc d.k dh pky ?kVrh gSA

STP1718




(B) SHM esa, ek/; fLFkfr ij Roj.k 'kwU; gksrk gS ysfdu pky v'kwU; gksrh gSA

(C) ;fn Roj.k 'kwU; gS rc pky vpj gksrh gSA

(D) d.k dh m/okZ/kj xfr esa mPpre fcUnq ij pky 'kwU; gksrh gS ysfdu Roj.k vpj gksrk gSA

48. If the springs are compressed to same amount :

WA =

2

1 K

A x2 ; W

B =

2

1 K

B x2 K

A > K

B W

A > W

B

If the springs are compressed by same force.

F = KA x

A = K

Bx

B ; x

A =

AK

F ; x

B =

BK

F ;

B

A

W

W =

2B

2

B

2A

2

A

K

FK

2

1

K

F.K

2

1

= A

B

K

K

Hence, WA < W

B

49. Like poles repel. (pqEcd ds leku /kqzoksa esa izfrd"kZ.k gksrk gSA)Unlike poles attract (pqEcd ds vleku /kqzoksa esa vkd"kZ.k gksrk gSA)

50. Solar energy is renewable source of energy and the others are non-renewable.lkSj-mtkZ mtkZ dk uohuhdj.k L=kksr gS rFkk vU; vuohuhdj.k L=kksr gS

51. (A) – q ; (B) – s ; (C) – p ; (D) – r

52. Hformation

respresent the enthalpy of formation of a substance from it’s elements in their standard state.H

fuekZ.k ekud voLFkk esa] buds rRoksa ls inkFkZ ds fuekZ.k dh ,UFkSYih dks iznf'kZr djrh gSA

53. Oxygen only shows +2 to –2 oxidation state.

vkWDlhtu dsoy +2 ls –2 vkWDlhdj.k voLFkk n’kkZrk gSA

54. 0 +1 –34P + 3KOH + 3H2O 3KH2PO2 + PH3Hence P is undergoing oxidation as well as reduction. 0 +1 –34P + 3KOH + 3H2O 3KH2PO2 + PH3vr% P dk vkWDlhdj.k rFkk vip;u nksuksa gksrk gSA

55. 15

8

108

103c

= 3.75 × 10–8 m.

56. Br show oxidation state varying from –1 to +7.

57.

CH3–C–CH2–CH3

O – Ethyl methyl ketone

Butanone – IUPAC name.

58. In Finkelstein reaction Iodine is prepared by exchange of chlorine in the presence of non-polar solvent acetone.

59. and (rFkk)

Carbon skeleton is different in both compounds.nksuksa ;kSfxdksa esa dkcZu <+kapk fHkUu&fHkUu gSA

60. Equilibrium mass = 16 g = Initial mass (mass conservation)

Initial moles of SO3 =

16

80= 0.2

All other conclusions require additional data i.e. equilibrium constant or degree of dissociation.

STP1718




gy- lkE; nzO;eku = 16 g = izkjfEHkd nzO;eku ¼nzO;eku laj{k.k½

SO3 ds izkjfEHkd eksy =

16

80= 0.2

lHkh vU; fu”d”kksZ ds fy, vfrfjDr vk¡dM+ks dh vko’;drk gksrh gS vFkkZr~ lkE; fu;rkad ;k fo;kstu dh ek=kk dh

vko’;drk gksrh gSA

61. Substance in maximum or minimum oxidation state never acts as both oxidant and reductant.

inkFkZ tks vf/kdre ;k U;wure vkWDlhdj.k vad esa gks og dHkh Hkh vkWDlhdkjh o vipk;d nksuksa dh rjg O;ogkj ugha dj

ldrk gSA

62. (B) B2H

6 is a electron deficient compound as boron has only six electron in its valence shell.

(C) AlCl3 is a electron deficient compound as aluminium has only six electron in its valence shell.

(B) B2H

6 ,d bysDVªkWu U;wu ;kSfxd gSa D;ksafd cksjkWu ds la;ksth dks'k esa dsoy N% bysDVªkWu ik;s tkrs gSA

(C) AlCl3 ,d bysDVªkWu U;wu ;kSfxd gSa D;ksafd ,Y;qfefu;e ds la;ksth dks'k esa dsoy N% bysDVªkWu gksrs gSA

63. The correct IUPAC name of is Cyclohexanecarbaldehyde

dk lgh IUPAC uke lkbDyksgsDlsudkcZ,fYMgkbM gSA

64. (A) At low pressure, b is negligible in comparison to Vm.

2mV

aP (Vm) = RT

RT

PVm = Z = 1 – RTV

a

m < 1

So, gas is more compressible than ideal gas.

(B) At high pressure, 2mV

a is negligible in comparison to P..

P (Vm – b) = RT

RT

PVm = Z = 1 + RT

Rb < 1.

So, gas is less compressible than ideal gas.

(C) Low density of gas means pressure is low so, at low pressure Z = 1 – RTV

a

m < 1 and gas is

more compressible than ideal gas.(D) At 0ºC H2 and He have a 0.

So, Z = 1 + RT

Pb and gas is less compressible than ideal gas.

Sol. (A) fuEu nkc ij b, Vm dh rqyuk esa ux.; gSA

2mV

aP (Vm) = RT

RT

PVm = Z = 1 – RTV

a

m < 1

vr%] xSl] vkn'kZ xSl ls vf/kd lEihM~; gSaA

(B) mPp rki ij 2mV

a, P dh rqyuk esa ux.; gSA

P (Vm – b) = RT RT

PVm = Z = 1 + RT

Rb < 1.

vr%] xSl] vkn'kZ xSl ls de lEihM~; gSaA

(C) xSl dk fuEu ?kuRo vFkkZr~ nkc de gS vr% fuEu nkc ij Z = 1 – RTV

a

m < 1 rFkk xSl] vkn'kZ xSl ls vfèkd

lEihM~; gSaA

(D) 0ºC ij H2 rFkk He ds fy, a 0 gksrk gSA

vr%, Z = 1 + RT

Pb rFkk xSl] vkn'kZ xSl ls de lEihM~; gSaA

STP1718




SAMPLE TEST PAPER-II(For Class-XI Appearing / Passed Students)

Course : VISHWAAS (JF)

Correct W rong Blank


(d soy , d fod Yi l gh)3 -1 0

16 to 21 42 to 45 62 to 63One or m ore than one correct Answ er

(, d ; k , d l s v f/kd fod Yi l gh)4 0 0

22 to 30 46 to 51 64 to 66 Com prehensions (v uqPNsn) 4 -1 0

31 52 67Matrix Match Type

(eSfVªDl l qesy i zd kj )6 [1, 2, 3, 6] 0 0

Marks to be aw ardedPart - I (Mathem atics)

Part - II (Physics)

Part - III (Chem istry)

Type

PART - I (Hkkx- I)


Straight Objective Type (lh/ks oLrqfu"B izdkj)

This section contains 15 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONLY ONE is correct.

bl [k.M esa 15 iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA

1. ‘P + Q’ means ‘P is the brother of Q’. ‘P – Q means P is the mother of Q and ‘P × Q’ means ‘P is the sister

of Q’. Which of the following means that M is the maternal uncle of R ?

‘P + Q’ dk vFkZ ‘Q dk HkkbZ P gSA’ ‘P – Q dk vFkZ gS Q dh ek¡ P gS rFkk ‘P × Q’ dk vFkZ gS Q dh cgu ‘P gSA fuEu esa

ls fdldk lk vFkZ R dk ekek M gSA(A) M – R + K (B) M + K – R (C) M + K × Q (D) M – K + R




le; dk vuqikr gS &

(A) 1 : 2 (B) 2 : 1 (C) 1 : 1 (D) 1 : 3

3. If cos – sin , 2

1, cos + sin are in G.P. then possible value of is

;fn cos – sin , 2

1, cos + sin xq.kksÙkj Js.kh esa gS rks dk laHkkfor gS -

(A) 3

(B)

6

(C)

4

(D)

9

4. If the coefficients of (2r + 4)th, (r – 2)th terms in the expansion of (1 + x)18 are equal, then r is

;fn (1 + x)18 esa (2r + 4)osa in dk xq.kkad (r – 2)osa in ds xq.kkad ds cjkcj gS] rks r dk eku gS&

(A) 4 (B) 6 (C) 8 (D) 7

5. If z = )i43)(i–1)(i3–2(

)i4–3)(i32)(i1(

= a + ib then a2 + b2 =

;fn z = )i43)(i–1)(i3–2(

)i4–3)(i32)(i1(

= a + ib rc a2 + b2 =

(A) 132 (B) 144 (C) 25 (D) 1

STP1718




6. The equation of the normal to the circle x2 + y2 = 2 x which is parallel to the line x + 2

y = 3 is

o`Ùk x2 + y2 = 2 x ds vfHkyEc dk lehdj.k] tks js[kk x + 2

y = 3 ds lekUrj gS] gS &

(A) x 2 y 1 = 0 (B) x + 2 y 1 = 0

(C) 4 x 3 y 4 = 0 (D) none of these buesa ls dksbZ ugha

7. A segment of a line x + y = 2 of length 44 cm is bent to form a circle keeping the point (1,1) as fixed. If thecentre of the circle so formed do not lie in first quadrant, then the centre of the circle is:

ljy js[kk x + y = 2 ds ,d 44 cm dh yEckbZ ds [k.M dks eksM+dj ,d o`Ùk bl rjg cuk;k tkrk gS fd ml fLFkr fcUnq

(1, 1) ;Fkkor jgsA ;fn bl rjg cus o`Ùk dk dsUnz izFke prqFkk±'k esa ugha gks] rks o`Ùk dk dsUnz gS -

(A)

2

71,

2

71 (B)

2

71,

2

71

(C)

2

71,

2

71 (D)

2

71,

2

71

8. If a = 4/3 cot2 30º + 3 sin2 60º – 3 cosec2 60º – 3/4 tan2 30º and b = 3 tan2 45º + cos0º – cot 90º then logb(a)

is

(A) equal to 2 (B) equal to –1 (C) equal to 1/2 (D) equal to 1/2

;fn a = 4/3 cot2 30º + 3 sin2 60º – 3 cosec2 60º – 3/4 tan2 30º rFkk b = 3 tan2 45º + cos0º – cot 90º rc logb(a)

dk eku gS&

(A) 2 ds cjkcj (B) –1 ds cjkcj (C) 1/2 ds cjkcj (D) 1/2 ds cjkcj

9. If x1 < x

2 and x

1, x

2 are the roots of x2 – 26x + 120 = 0 then 21 xx =

;fn x1 < x

2 vkSj x

1, x

2 lehdj.k x2 – 26x + 120 = 0 ds ewy gS rc 21 xx =

(A) 5 + 1 (B) 5 + 2 (C) 3 + 2 (D) 5 + 3

10. The smallest value of n satisfying n+1C6 + nC

4 > n+2C

5 – nC

5 is :

n+1C6 + nC

4 > n+2C

5 – nC

5 dks larq"B djus okyk x dk lcls NksVk eku gS -

(A) 11 (B) 10 (C) 9 (D) 8

11. The number of ways in which candidates A1, A

2, ...... A

10 can be ranked if A

1 is always above A

2 is

Øep;ksa dh la[;k gksxh ftlesa ijh{kkFkhZ A1, A

2, ...... A

10 dks ojh;rk Øe esa j[kk tk ldrk gS ;fn A

1 ls Åij gS A

2 ls

Åij gS -

(A) (10)! (B) 9! (C) 2

19! (D)

2

1(10)!

12. Coefficient of xn in !1

x1 +

!2

)x1( 2+

!3

)x1( 3 + .........

!1

x1 +

!2

)x1( 2+

!3

)x1( 3 + .........esa xn dk xq.kkad gS

(A) e (B) n! (C) e(n!) (D) !n

e

13. If x2 – 6x + 5 < 0, x2 – 5x + 6 > 0 then x

;fn x2 – 6x + 5 < 0, x2 – 5x + 6 > 0 rc x

(A) (1, 2) (3, 5) (B) (1, 3) (4, 5) (C) (2, 3) (4, 5) (D) (1, 4) (5, 6)

STP1718




14. Three pieces of cakes of weights 2

14 lbs,

4

36 Ibs and

5

17 Ibs respectively are to be divided into parts of

equal weights. Further, each must be as heavy as possible. If one such part is served to each guest, thenwhat is the maximum number of guests that could be entertained ?

rhu fVfd;k ftuds Hkkj Øe'k% 2

14 ikm.M,

4

36 ikm.M vkSj

5

17 ikm.M gS] dks vf/kdre laHko leku Hkkj ds VqdMksa esa foHkkftr

fd;k tkrk gSA ;fn bl izdkj ds ,d dks izR;sd esgeku dks ijkslk tkrk gS] rc esgekuksa dh vf/kdre la[;k gksxh ?(A) 54 (B) 72 (C) 20 (D) 41

15. If a, b, c are real and distinct numbers, then the value of )ac(.)cb(.)ba(

)ac()cb()ba( 333

is

;fn a, b, c okLrfod vkSj fofHkUu la[;k,a gS] rks )ac(.)cb(.)ba(

)ac()cb()ba( 333

dk eku gS&

(A) 1 (B) a b

c

(C) 2 (D) none of these (buesa ls dksbZ ugha)


Multiple Correct Answer Type (cgqy lgh fodYi izdkj)This section contains 6 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONE OR MORE is/are correct.bl [k.M esa 6 iz'u gSaA izR;sd iz'u ds mÙkj ds fy, 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls

vf/kd lgh gSA

16. For the ellipse 12x2 + 4y2 + 24x – 16y + 25 = 0.

(A) the centre is (–1, 2) (B) the length of axes are and 3

(C) ecentricity of the ellipse is 3

2(D) the area of the ellipse is

2

3

nh?kZ o`Ùk 12x2 + 4y2 + 24x – 16y + 25 = 0 ds fy, -

(A) dsUnz (–1, 2) gSA (B) v{kksa dh yEckbZ o 3 gSA

(C) nh?kZo`Ùk dh mRdsUnzrk 3

2gSA (D) nh?kZ o`Ùk dk {ks=kQy

2

3 gSA

17. The lines L1 : x 2y + 6 = 0 & L

2 : x 2y 9 = 0 are tangents to the same circle . If the point of

contact of L1 with the circle is ( 2, 2), then :

(A) the centre of the circle is ( 7/2, 5) (B) the centre of the circle is ( 1/2, 1)(C) area of the circle is 45/4 sq. units(D) the point of contact of L

2 with the circle has the coordinates (1 , 4)

js[kk,¡ L1 : x 2y + 6 = 0 ,oa L

2 : x 2y 9 = 0 ,d gh o`Ùk dh Li'kZ js[kk,¡ gSA ;fn L

1 dk o`Ùk ds lkFk Li'kZ

fcUnq ( 2, 2) gks] rks&(A) o`Ùk dk dsUnz ( 7/2, 5) gS(B) o`Ùk dk dsUnz ( 1/2, 1) gS(C) o`Ùk dk {ks=kQy 45/4 oxZ bdkbZ gS

(D) L2 dk o`Ùk ds lkFk Li'kZ fcUnq (1 , 4) gSA

18. If z = 20 21i + 21 20 i then the principle value of arg(z) can be:

;fn z = 20 21i + 21 20 i rks arg(z) ds eq[; eku gks ldrs gS&

(A) 4

(B) 34

(C) 4

(D) 34

STP1718




19. Solution(s) of the equation x2log

xlog

4

2 = x8log3

x4log4

8

8 is/are

lehdj.k x2log

xlog

4

2 = x8log3

x4log4

8

8 ds gy gS@gSa &

(A) 2–2 (B) 2 (C) 2–4 (D) 2

1

20. The normal y = mx – 2am – am3 to the parabola y2 = 4ax subtends a right angle at the vertex ifijoy; y2 = 4ax dk vfHkyEc y = mx – 2am – am3, 'kh"kZ ij ledks.k vUrfjr djrk gS] ;fn

(A) m = 3 (B) m = – 3 (C) m = 2– (D) m = 2

21. If a , b , c are positive numbers in G.P. and log

ac5

, log

c5b3

, log

b3a

are in A.P, then which

of the following is/are correct ?

(A) b = 35

c (B) a , b , c cannot be the sides of a triangle

(C) a = 9c25

(D) a = 9

5 c

;fn /kukRed la[;k,a a , b , c xq.kksÙkj Js<h esa gS vkSj log

ac5

, log

c5b3

, log

b3a

lekUrj Js<h esa gS] rc

fuEu esa ls dkSulk@dkSuls dFku lgh gS ?

(A) b = 35

c (B) a , b , c f=kHkqt dh Hkqtk,a ugha gks ldrh

(C) a = 9c25

(D) a = 9

5 c

SECTION - III ([k.M - III)Comprehension Type (cks/ku çdkj)

This section contains 3 paragraphs. Based upon each paragraph, there are 3 questions. Each question has4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.bl [k.M esa 3 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 3 ç'u gSA çR;sd ç'u ds 4 fodYi


Paragraph for Question Nos. 22 to 24 (ç'u 22 ls 24 ds fy, vuqPNsn)

Dictionary of different words is being formed by arranging the letter of word "ARRANGE".

'kCn "ARRANGE" ds v{kjksa dks O;ofLFkr djds fofHkUu 'kCnksa dk 'kCn&dks"k cuk;k cuk;k x;k gSA

22. The number of words in which the two 'R' are not together is'kCnks a dh la[;k ftuesa nksuks a R ,d lkFk ugha vk,] gS &

(A) 1260 (B) 660 (C) 900 (D) 240

23. The number of words in which neither two 'R' nor two 'A' come together is'kCnks a dh la[;k ftuesa u rks nksuks a A ,d lkFk vk, vkSj u gh nksuks a R ,d lkFk vk,] gS &

(A) 1260 (B) 900 (C) 240 (D) 660

24. The rank of word 'ARRANGE' in the dictionary is

'kCndks"k esa 'ARRANGE' 'kCn dk Øe gS &

(A) 340 (B) 341 (C) 342 (D) 343

STP1718





ç'u 25 ls 27 ds fy, vuqPNsn

Let ||x – a| – b| = k. Then(i) k = 0, b > 0 equation has two solutions(ii) b > k > 0 equation has four solutions(iii) b = k > 0 equation has three solutions(iv) 0 < b < k equation has two solutionsekukfd ||x – a| – b| = k gS] rks

(i) k = 0, b > 0 lehdj.k ds nks gy gSaA

(ii) b > k > 0 lehdj.k ds pkj gy gSaA

(iii) b = k > 0 lehdj.k ds rhu gy gSaA

(iv) 0 < b < k lehdj.k ds nks gy gSaA

25. If number of solutions of ||x + 1| – 2| = 1 is m, then m =;fn ||x + 1| – 2| = 1 ds gyksa dh la[;k m gS] rks m =(A) 1 (B) 2 (C) 3 (D) 4

26. If number of solutions of ||x – 2| – 3| = m is , then =(where m is obtained in Q.No. 25);fn ||x – 2| – 3| = m ds gyksa dh la[;k gS] rks =

(tgk¡ m iz'u la[;k 25 ls izkIr gksrk gSA)(A) 1 (B) 2 (C) 3 (D) 4

27. Number of solutions of ||x – 2| – 5| = + 3 is(where is obtained in Q.No. 26)||x – 2| – 5| = + 3 ds gyksa dh la[;k gS &

(tgk¡ iz'u la[;k 26 ls izkIr gksrk gSA)(A) 1 (B) 2 (C) 3 (D) 4



Consider the letters of the word MATHEMATICS. There are eleven letters some of them are identical.Letters are classified as repeating and non-repeating letters. Set of repeating letters = {M, A, T}. Set ofnon-repeating letters = {H, E, I, C, S}

;fn MATHEMATICS 'kCn ds v{kj fy, tk, rks X;kjg v{kjksa esa ls dqN v{kj ,d ckj vkrs gS rFkk dqN dh iqujko`fr

gksrh gSA iqujkorhZ v{kjksa dk leqPp; = {M, A, T} gSA tcfd viqujkorhZ = {H, E, I, C, S} gSaA

28. Possible number of words taking all letters at a time such that atleast one repeating letter is at oddposition in each word, is

lHkh v{kjksa dks ,d ysdj dqy fdrus 'kCn cuk, tk ldrs gS ftuesa de ls de ,d iqujko`rhZ v{kj fo"ke LFkku ij jgsA

(A) !2!2!2

!9(B) !2!2!2

!11(C) !2!2!2

!11 – !2!2

!9(D) !2!2

!9

29. Possible number of words taking all letters at a time such that in each word both M's are together andboth T's are together but both A's are not together, is

lHkh v{kjksa dks ,d lkFk fy, tk, rkfd izR;sd v{kj esa nksuksa M lkFk vk, rFkk nksuksa T Hkh lkFk&lkFk vk, ijUrq nksuksa A

lkFk&lkFk ugha vk, rc cuus okys laHkkfor 'kCnksa dh la[;k gS &

(A) 7 ! . 8C2

(B) !2!2!2

!11 – !2!2

!10(C) !2!2

!4!6(D) !2!2!2

!9

30. Possible number of words in which no two vowels are together, is

mu 'kCnksa dh la[;k Kkr dhft, ftuesa dksbZ Loj lkFk&lkFk u vk,A

(A) !4)!2(

!7!83 (B) !2

!7 . 8C

4 . !2

!4(C) 2)!4()!2(

!7!8(D) )!4()!2(

!7!84

STP1718





Matrix - Match Type (eSfVªDl&lqesy izdkj)This section contains 1 questions. Each question contains statementsgiven in two columns which have to be matched. Statements in Column are labelled as A,B,C and D whereas statements in Column are labelledas p,q,r and s. The answers to these questions have to be appropriatelybubbled as illustrated in the following example.If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q ,


ABCD

p q r s


bl [k.M esa 1 iz'u gSaA izR;sd iz'u esa nks dkWye esa oDRO; (statements) fn;s gq, gSa ftudk

lqe sy (match) djuk g SA dk Wye (Column-I) es a fn; s x; s oDrO; k s a

(A, B, C, D) dks dkWye (Column-II) esa fn;s x;s oDrO;ksa (p, q, r, s) ls lqesy djuk

gSA bu iz'uksa ds mÙkj fn;s x;s mnkgj.k ds vuqlkj mfpr cqYyksa dks dkyk djds n'kkZuk gSA



31. Column – ¼LrEHk –½ Column – ¼LrEHk –½

(A) Maximum value of for which solution of (p) 13 sin x + 4 cos x = exists is

dk vf/kdre eku ftlds fy, 3 sin x + 4 cos x =

ds gy fo|eku gS] gS &

(B) Number of solutions of sin2 + 3 cos = 3 in [–, ] is (q) 3

[–, ] esa sin2 + 3 cos = 3 ds gyks dh la[;k gS &

(C) Number of common tangent of x2 + y2 = 4 and (r) 5(x – 3)2 + (y – 4)2 = 9 is

o`Ùkks x2 + y2 = 4 rFkk (x – 3)2 + (y – 4)2 = 9 dh mHk;fu"V

Li'kZ js[kkvksa dh la[;k gS &

(D) Length of chord intercepted on the line x + y = 2 by (s) 0the circle (x – 1)2 + (y – 1)2 = 9/4 is

o`Ùk (x – 1)2 + (y – 1)2 = 9/4 }kjk js[kk x + y = 2 ij dkVh xbZ

vUr%[kf.Mr thok dh yEckbZ gS &




This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), outof which ONLY ONE is correct.


32. Kerosene oil is flowing through a tube of length & radius r . The pressure difference between two ends

of the tube is P , then the viscosity of oil is given by = v4

)xr(P 22 where v is velocity of oil at a

distance x from the axis of tube . From this relation the dimensions of viscosity are : yEckbZ vkSj r f=kT;k dh fdlh V~;wc (uyh) esa dsjksflu rsy cg jgk gSA ;fn uyh esa nksauks fljksa ij nkckUrj P gS , rc rsy

dh ';kurk ;fn = v4

)xr(P 22 ls nh tkrh gSA tgk¡ v V~;wc dh v{k ls x nwjh ij rsy dk osx gSA bl lw=k ls ';kurk

dh foek gskxh :(A) M L

2 T

2 (B) M L 1

T 1 (C) M0 L0 T0 (D) M L T

3

STP1718




33. A 1m long wire having tension of 100 N and of linear mass density 0.01 kg/m is fixed at end A and free at endB. The point C which is 20 cm from end B is constrained to be stationary. To create resonance in this wire,the minimum frequency of the tuning fork will be :1m yEcs rkj esa ruko 100 N rFkk js[kh; nzO;eku ?kuRo 0.01 kg/m gS dks fljs A ij fdyfdr rFkk fljs B ij eqDr j[kk x;k

gSA fcUnq C tks fljs B ls 20 cm ij gS] fLFkj jgus ds fy, ckf/kr gSA rkj esa vuqukn ds fy, Lofj=k dh U;wure vkofÙk gksxh

(A) 125 Hz (B) 150 Hz (C) 300 Hz (D) 275 Hz

34. A block of mass M placed on an horizontal turn table at distance'r' from its axis of rotation, Table rotets with constant angular velocity

0, it is known that block not slide on the table, fricition coefficient

between block and table is '' then friction force on the block.(A) Mg along radius(B) Mg along tangent(C) M

02r along tangent

(D) M02r along radius.

,d M nzO;eku dk CykWd {kSfrt esa ?kweus okyh Vscy (turn table) ij ?kw.kZu

v{k dh 'r' nwjh ij j[kk gSA Vscy ?kw.kZu v{k ds ikfjr fu;r dks.k osx 0 ls

?kw.kZu dj jgh gSA ;g Kkr gS fd CykWd Vscy ij fQlyrk ugh gS, rFkk CykWd

o Vscy ds e/; ?k"kZ.k xq.kkad '' gS rks CykWd ij yxus okyk ?k"kZ.k cy gksxkA

(A) Mg f=kT;k ds vuqfn'k

(B) Mg Li'kZ js[kk ds vuqfn'k

(C) M02r Li'kZ js[kk ds vuqfn'k

(D) M02r f=kT;k ds vuqfn'k

35. Three blocks and a uniform rod of mass 1.5 kg are connectedwith light strings. The coefficient of friction between 2.5 kg blockand ground is

k = 1 and the pulleys are smooth and fixed. Now

the system is released. The tension at a point, which is at a

height of 1 cm from the lowest point of the rod is : (g = 10 m/s2)

rhu CykWd rFkk 1.5 kg dh le:i NM+ gYdh Mksjh }kjk ;qfXer gSA

2.5 kg CykWd o Hkwfe ds e/; ?k"kZ.k xq.kkad k = 1 rFkk f?kjfu;k¡ fpduh o

tM+or~ gSA fudk; dks eqDr fd;k tkrk gSA NM+ ds fuEure fcUnq ls 1

cm Å¡pkbZ ij fLFkr fcUnq ij ruko gksxk (g = 10 m/s2)

(A) 20 N (B) 22.5 N(C) 40 N (D) 45 N

36. Two rods are moving with constant velocity v in perpendicular direction to their length as shown in figure. Thevelocity of point of intersection of two rods will be :nks NM+ viuh yEckbZ ds yEcor~ fn'kk esa fp=kkuqlkj fu;r osx v ls xfr dj jgh gSA nks auks NM+ks a ds çfrPNsnu fcUnq

dk osx gksxk &

(A) v cosec (B) v cos (C) v sin (D) 2vsin

STP1718




37. A planet is revolving in an elliptical orbit around the sun as shown in figure. The areal velocity (area swappedby the radius vector with respect to sun in unit time) is :

(A) r1 v1 (B) 2

1 r2 v2

(C) 2

211

r

r v

2

1(D) dependent on the position of planet from sun

fp=kkuqlkj ,d xzg lw;Z ds pkjksa vksj nh?kZo`Ùkkdkj d{kk esa ifjØe.k dj jgk gSA {ks=kh; osx ¼bdkbZ le; esa lw;Z ds lkis{k

f=kT;h; lfn'k }kjk ?ksjk x;k {ks=kQy½ gSa &

(A) r1 v1 (B) 2

1 r2 v2

(C) 2

211

r

r v

2

1(D) lw;Z ls xzg dh fLFkfr ij fuHkZj

38. A massless conical flask filled with a liquid is kept on a table in a vacuum. the force exerted by the liquid onthe base of the flask is W

1. The force exerted by the flask on the table is W

2.

(A) W1 = W

2(B) W

1 > W

2(C) W

1 < W

2

(D) the force exerted by the liquid on the walls of the flask is (W1 W

2).

nzo ls Hkjs gq, nzO;eku ghu 'kaDokdkj ik=k dks fuokZr esa est ij j[kk gSA nzo }kjk ik=k ds vk/kkj ij vjksfir cy W1 gSA

ik=k }kjk esat ij vjksfir cy W2 gSa rks

(A) W1 = W

2(B) W

1 > W

2(C) W

1 < W

2

(D) ik=k dh nhokjksa ij nzo }kjk vkjksfir cy W1 W

2 gSA

39. The moment of inertia of a system of four rods, each of length and mass m, about the axis shown is :,d fudk; tks pkj NM+ks ls feydj cuk gS] çR;sd dk nzO;eku m rFkk yEckbZ gS rks bl fudk; dk fp=k esa n'kkZbZ xbZ

v{k ds lkis{k tM+Ro vk?kw.kZ gksxkA

(A) 3

2m2 (B) 2m2 (C) 3m2 (D)

3

8m2

40. Two particles moving towards each other with same speed meet in time (t1). When speed of one becomes 2

times and other’s half they meet in time t2 moving is same direction then

2

1

t

t is :

nks d.k ,d nwljs dh rjQ leku pky ls pyrs gq, (t1) le; esa feyrs gSA ;fn ,d d.k dh pky nqxquh rFkk nwljs d.k

dh pky vk/kh dj nh tk, rks leku fn'kk esa pyrs gq, ,d nwljs ls t2 le; esa feyrs gSA

2

1

t

t Kkr djks :

(A) 3

4(B)

2

1(C)

4

5(D)

4

3

STP1718




41. Block A of mass m is placed on a plank B. A light support S is fixed on plank B and is attached with the blockA with a spring of spring constant K. Consider that initially spring is in its natural length, and the system isat rest. Find the maximum compression in the spring, if the plank B is moved with a constant acceleration‘a’. (All the surfaces are smooth) :

m nzO;eku dk ,d CykWd A, Iykad B ij fLFkr gSA ,d gYdk n`<+ vk/kkj S, Iykad B ij fLFkr gS rFkk bl n`<+ vk/kkj ls]

fLçax fu;rkad K okyh fLçax ds }kjk CykWd A, tqM+k gqvk gSA ekuk fLçax çkjEHk esa çkÑfrd yEckbZ esa gS rFkk fudk; izkjEHk

esa fojke esa gSA ;fn Iykad B dks fu;r Roj.k a fn;k tkrk gS rks fLçax esa vf/kdre ladqpu gksxk (lHkh lrg fpduh gS) :

(A) ma

2K(B)

2ma

K(C)

ma

K(D)

4ma

K



This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B),(C) and (D), out of which ONE OR MORE THAN ONE is/are correct.



42. In the figure shown, coefficient of restitution between A and B is e = 2

1, then :

(A) velocity of B after collision is 2

v(B) impulse between two during collision is

4

3mv

(C) loss of K.E. during collision is 16

3mv2 (D) loss of K.E. during collision is

4

1mv2

fp=kkuqlkj A vkSj B esa e/; çR;koLFkku xq.kkad e = 2

1, gksxk :

(A) VDdj ds ckn B dk osx 2

vgksxkA (B) nksauks ds e/; VDdj ds nkSjku vkosx

4

3mv gksxkA

(C) VDdj ds nkSjku xfrt ÅtkZ esa gkfu16

3mv2 gksxhA (D) VDdj ds nkSjku xfrt ÅtkZ esa gkfu

4

1mv2 gksxhA

43. In a standing transverse wave on a string :(A) In one time period all the particles are simultaneously at rest twice.(B) All the particles must be at their positive extremes simultaneously once in one time period.(C) All the particles may be at their positive extremes simultaneously once in a time period.(D) All the particles are never at rest simultaneously.Mksjh esa vuqçLFk vçxkeh rjax esa &

(A) ,d vkorZdky esa lHkh d.k ,d lkFk nks ckj fojkekoLFkk esa vkrs gSA

(B) lHkh d.k muds /kukRed pje fLFkfr ij ,d vkorZdky esa ,d ckj ,d lkFk vo'; vkrs gSA

(C) lHkh d.k muds /kukRed pje fLFkfr ij ,d vkorZdky esa ,d ckj ,d lkFk vk ldrs gSA

(D) lHkh d.k ,d lkFk fojkekoLFkk ij ugha gksrs gSA

STP1718




44. For a particle moving along a straight line velocity-time graph is shown.Which of the following is/are incorrect for motion from t = 0 to t = t3.(A) Particle has changed it’s direction 2 time in it’s motion.(B) Through-out the motion displacement is increasing.(C) Acceleration is always either positive or zero.(D) Total displacement is 3/4th of the distance travelled.ljy js[kk esa xfr djrs gq, ,d d.k dk osx≤ xzkQ fp=k esa iznf'kZr gSA t = 0 ls

t = t3 rd xfr ds fy, fuEu esa ls dkSulk@dkSuls vlR; gSA

(A) d.k viuh xfr esa 2 ckj xfr dh fn'kk dks ifjofrZr djsxkA

(B) lEiw.kZ xfr esa foLFkkiu c<+ jgk gSA

(C) Roj.k lnSo ;k rks /kukRed gksxk ;k 'kwU; gksxkA

(D) dqy foLFkkiu r; dh xbZ nqjh dk 3/4ok¡ gksxkA

45. The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circleand CDA is half of an ellipse. Then,

(A) the process during the path A B is isothermal

(B) heat flows out of the gas during the path B C D

(C) work done during the path A B C is zero

(D) positive work is done by the gas in the cycle ABCDA

,d vkn'kZ xSl dks pØ ABCDA ls xqtkjk tkrk gS vkSj bldk P-V xzkQ fp=k esa fn[kk;k x;k gSA ABC ,d v/kZ&o`Ùk gS

vkSj CDA vk/kk nh?kZo`Ùk gSA rc

(A) çfØ;k A B lerkih; gSA

(B) iFk B C D ds nkSjku xSl ls Å"ek ckgj fudyrh gSA

(C) iFk A B C ds nkSjku fd;k x;k dk;Z 'kwU; gSA

(D) pØ ABCDA esa xSl }kjk /kukRed dk;Z fd;k x;k gSA


Comprehension Type (c) cks/ku çdkj)This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have tobe answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.bl [k.M esa 2 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 3 cgq&fodYih ç'u ds mÙkj nsus gSA

çR;sd ç'u ds 4 fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA


ç'u 46 ls 48 ds fy, vuqPNsnA block of mass 10 kg oscillates simple harmonically when it is attached with a spring. The graph ofoscillation between the displacement verses time is given then answer the following questions.10 kg nzO;eku dk ,d CykWd A ljy vkorhZ :i esa nksyu djrk gS tc ;g fLizax ls tqM+k gqvk gSA foLFkkiu le; ds e/

; nksyu dk xzkQ fp=k esa n'kkZ;k x;k gSA rks fuEu iz'uksa ds mÙkj nhft;sA

xA

A/2

2s4.4s

time

46. Time period of oscillation of block will be :CykWd ds nksyu dk vkorZdky gksxkA

(A) 4 sec. (B) 5 sec. (C) 4.8 sec. (D) 4.4 sec

47. Equation of SHM of the block :CykWd dh ljy vkorZ xfr dh lehdj.k gksxhA

(A) x = A sin (t + 30°) (B) x = A sin

(t + 45°) (C) x = A sin

(t + 150°) (D) x = A sin

(t – 30°)

STP1718




48. The graph for velocity vs time will be :osx≤ xzkQ gksxkA

(A)

A/2

time (B) time

(C) time

–

(D) time

–



A particle undergoes uniform circular motion. The velocity and angular velocity of the particle at an

instant of time is j4i3v

m/s and j6ix

rad/sec.

,d d.k dk ,dleku o`Ùkh; xfr djrk gSA fdlh {k.k ij d.k dk osx rFkk dks.kh; osx Øe'k% j4i3v

m/s rFkk

j6ix

rad/sec gSA

49. The value of x in rad/s is(A) 8 (B) – 8 (C) 6 (D) can't be calculated

x dk eku jsfM;[email protected] esa gS &

(A) 8 (B) – 8 (C) 6 (D) ifjdyu ugh fd;k tk ldrk

50. The radius of circle in metres is(A) 1/2 m (B) 1 m (C) 2 m (D) can't be calculatedo`Ùk dh f=kT;k ehVj esa gS &

(A) 1/2 m (B) 1 m (C) 2 m (D) ifjdyu ugh fd;k tk ldrk

51. The acceleration of particle at the given instant isfn;s x;s {k.k ij d.k dk Roj.k gS &

(A) 25 (B) 42 m (C) 30 m (D) 50 m

SECTION - IV ([k.M- IV)Matrix - Match Type (eSfVªDl&lqesy izdkj)

This section contains 1 question. The question contains statements given in twocolumns which have to be matched. Statements (A, B, C, D) in Column-I have to bematched with statements (p, q, r, s) in Column-II. The answers to these questionshave to be appropriately bubbled as illustrated in the following example.


ABCD

p q r s

If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q, then the correctly matched answer will looklike in bubbled matrix. One option in Column–I may have one or more than one correct options in Column–II. Markswill be awarded only if an option in Column–I is matched with all correct options in Column-II. Each correct matchingwill be awarded 1 mark and if all the matchings are correct then 6 marks will be awarded.bl [k.M esa 1 iz'u gSaA izR;sd iz'u esa nks dkWye esa oDRO; (statements) fn;s gq, gSa ftudk lqesy

(match) djuk gSA dkWye (Column-I) esa fn;s x;s oDrO;ksa (A, B, C, D) dks dkWye (Column-II)esa fn;s x;s oDrO;ksa (p, q, r, s) ls lqesy djuk gSA bu iz'uksa ds mÙkj fn;s x;s mnkgj.k ds vuqlkj

mfpr cqYyksa dks dkyk djds n'kkZuk gSA


ABCD

p q r s

;fn lgh lqesy A-p, A-r, B-p, B-s, C-r, C-s rFkk D-q gSa] rks lgh fof/k ls dkys fd, x;s cqYyksa dk 4 x 4 esafVªDl (matrix)esa n'kkZ;k x;k gSA dkWye–I ds izR;sd fodYi dk dkWye–II ds ,d ;k ,d ls T;knk fodYiksa ls lqesy djk;k tk ldrk gSA vad



STP1718




52. A particle is projected from level ground. Assuming projection point as origin, x-axis along horizontaland y-axis along vertically upwards. If particle moves in x-y plane and its path is given by y = ax – bx2

where a, b are positive constants. Then match the physical quantities given in column-I with the valuesgiven in column-II. (g in column II is acceleration due to gravity in negative y-direction.) Column I Column II

(A) Horizontal component of velocity (p) b

a

(B) Time of flight (q) b4

a2

(C) Maximum height (r) b2

g

(D) Horizontal range (s) gb

a2 2

,d d.k dks tehu ls iz{ksfir fd;k tkrk gSA iz{ksi.k fcUnq dks ewyfcUnq, x-v{k dks {kSfrt ds vuqfn'k rFkk y-v{k dks

Å/okZ/kj Åij dh vksj ekfu;sA ;fn d.k x-y ry esa xfr djrk gS rFkk bldk iFk y = ax – bx2 }kjk fn;k tkrk gS] tgk¡

a, b /kukRed fu;rkad gSaA rks dkWye-I esa nh xbZ HkkSfrd jkf'k;ks a dks dkWye-II esa fn;s x;s ekuksa ls lqesfyr dhft, &

(dkWye-II esa g xq:Roh; Roj.k _.kkRed y fn'kk esa gS)dkWye-I dkWye-II

(A) osx dk {kSfrt osx (p) b

a

(B) mM~M;u dky (q) b4

a2

(C) vf/kdre Å¡pkbZ (r) b2

g

(D) {kSfrt ijkl (s) gb

a2 2

PART - III (Hkkx - III)

SECTION - I ([k.M- I)Straight Objective Type (lh/ks oLrqfu"B izdkj)

This section contains 9 multiple choice questions. Each question has choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.bl [k.M esa 9 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA

53. Consider the processesizØeksa ij fopkj dhft,

(i) 2

1H

2 (g) + aq. H+ (aq.) (ii) 2O(g) O

2(g)

(iii) NH4+ (g) + Cl– (g) NH

4Cl(s) (iv) P

4 (black) + 5O

2(g) P

4O

10(s)

Which of the above does not represent Hformation

of the product.mijksDr esa ls dkSulk izØe mRikn ds H

fuekZ.k dks iznf'kZr ugh djrk gS\

(A) I,IV (B) II, IV (C) I, II, III (D) II , III, IV

54. Identify the element whose most stable oxidation state is different from other three?

,sls rRo dks igpkfu, ftldh lokZf/kd LFkk;h vkWDlhdj.k voLFkk vU; rhu ls fHkUu gksrh gS\

(A) S (B) Se (C) Te (D) O

55. In the reaction 4P + 3KOH + 3H2O 3KH2PO2 + PH3fuEu vfHkfØ;k] 4P + 3KOH + 3H2O 3KH2PO2 + PH3 esa](A) P undergoes reduction only (P dk dsoy vi;pu gksrk gSA)(B) P undergoes oxidation only (P dk dsoy vkWDlhdj.k gksrk gSA)(C) P undergoes both oxidation and reduction (P dk vkWDlhdj.k o vip;u nksuksa gksrk gSA)(D) neither undergoes oxidation nor reduction (u rks vkWDlhdj.k gksrk gS vkSj u gh vip;u gksrk gSA )

STP1718




56. The value of Planck’s constant is 6.63 × 10–34 Js. The velocity of light is 3 × 108 m/sec. Which value isclosest to the wavelength of light with frequency of 8 × 1015 sec–1 ?Iykad fLFkjkad dk eku 6.63 × 10–34 Js gSA izdk'k dk osx 3 × 108 [email protected] gSA dkSulk rjaxnS/;Z eku izdk'k dh

vko`fr 8 × 1015 lSd.M–1 ds lehi gS ?(A) 5 × 10–18 m (B) 4 × 10–8 m (C) 3 × 107 m (D) 2 × 10–25 m

57. In the thiocyanate ion, SCN– three resonating structure are possible with the electron-dot method as shownin figure :

S = C = N:: ::

1–

(x)

The decreasing order of % contribution in resonance hybrid is :(A) x > y > z (B) y > z > x (C) z > x > y (D) cannot predicted.bysDVªkWu&MkWV fof/k }kjk Fkk;kslk;usV vk;u SCN– dh rhu vuqukfn lajpukvksa dks fuEu fp=kksa ds }kjk çnf'kZr fd;k tkrk

gS %

S = C = N:: ::

1–

(x)

vuqukfn ladj esa buds izfr'kr ;ksxnku dk ?kVrk gqvk Øe gksxk %

(A) x > y > z (B) y > z > x (C) z > x > y (D) Kkr ugha fd;k tk ldrk gSaA

58. In which of the following molecules number of lone paris and bond pairs on central atom are not equal ?fuEu esa ls fdl v.kq esa dsfUnz; ijek.kq ij ,dkdh ;qXeksa dh la[;k rFkk ca/kh ;qXeksa dh la[;k leku ugha gS \

(A) H2O (B) I3– (C) O2F2 (D) SCl2

59. What is the IUPAC name of iso-octane.

(A) 2,3,4 trimethyl pentane (B) 2,2,4 trimethyl pentane

(C) 2,2,5 trimethyl pentane (D) 1,2,3 trimethyl pentane

vkblks&vkWDVsu dk IUPAC uke D;k gS\

(A) 2,3,4 VªkbZesfFky isUVsu (B) 2,2,4 VªkbZesfFky isUVsu

(C) 2,2,5 VªkbZesfFky isUVsu (D) 1,2,3 VªkbZesfFky isUVsu

60. The order of electron density in the benzene ring in the following compounds :fuEufyf[kr ;kSfxdksa esa csUthu oy; ds bysDVªkWu ?kuRo dk lgh Øe dkSulk gS :

OH OC H2 5

(A) I > II > III > IV (B) II > IV > I > III (C) IV > III > II > I (D) I > III > IV > II

61. The major product of the reaction

vfHkfØ;k dk eq[; mRikn

CH2CH3 + Cl

2 h is gS&

(A) CH2–CH2–Cl

(B) CH–CH3

Cl

(C)

CH2CH3

Cl (D)

CH2CH3

Cl

STP1718






This section contains 2 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C)and (D), out of which ONE OR MORE THAN ONE is/are correct.



62. When SO3 is taken in 10 L vessel an equilibrium is established between SO

3, SO

2 and O

2

SO3(g) SO

2(g) +

1

2O

2(g)

The density of the equilibrium mixture is found to be 1.6 gm/l at equilibrium temperature 300 K. Which of

the following conclusions can be made using the data given above ?

(A) Number of moles of SO3 initially is 0.2

(B) Total number of moles is 0.5 at equilibrium.

(C) Number of moles of SO3 at equilibrium can’t be calculated as additional data is required.

(D) Partial pressure of SO3 is 0.492 atm at equilibrium.

tc 10 L ik=k esa SO3 dks fy;k tkrk gS rks SO3, SO2 o O2 ds e/; ,d lkE; LFkkfir gksrk gS &

SO3(g) SO2(g) + 1

2O

2(g)

lkE; feJ.k dk ?kuRo lkE; rkieku 300 K ij 1.6 gm/ gS] fn;s x;s mijksDr vk¡dM+ks dk iz;ksx dj fuEu esa ls dkSuls

fu"d"”kZ cuk;s tk ldrs gS \

(A) izkjEHk esa SO3 ds eksyks dh la[;k 0.2 gSA

(B) lkE; ij eksyks dh dqy la[;k 0.5 gSA

(C) lkE; ij SO3 ds eksyks dh la[;k dh x.kuk ugh dj ldrs gS D;ksfd vfrfjDr vk¡dM+ks dh vko';drk gksrh gSA

(D) lkE; ij SO3 dk vkaf'kd nkc 0.492 atm gSA

63. Which of the following will not give iodoform test?

fuEu esa ls dkSu vk;ksMksQkeZ ijh{k.k ughs nsrs gSa\

(A)

CH3–C–CH3

O(B)

CHO

(C) CH3–CH

2–CH

2–OH (D) CH

3–CH

2–OH

SECTION - III ([k.M - III)Comprehension Type (c) cks/ku çdkj)

This section contains 1 paragraphs. Based upon each paragraph, 3 multiple choice questions have tobe answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.bl [k.M esa 1 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 3 cgq&fodYih ç'u ds mÙkj nsus gSA çR;sd ç'u

ds 4 fodYi (A), (B), (C) rFkk (D) gS, ftuesa ls flQZ ,d lgh gSA

Paragraph for Question Nos. 64 to 66ç'u 64 ls 66 ds fy, vuqPNsn

The addition of hydrogen to an alkene is an exothermicreaction & the enthalpy change is called the heat of hydrogenation. More branched alkenes are more stableand have less potential energy. The alkene having higher potential energy releases more heat on hydrogenation.In the figure the enthalpy changes of three isomeric alkenes has been shown.

STP1718




,Ydhu ij gkbMªkstu dk ;ksx ,d m"ek{ksih vfHkfØ;k gS vkSj bleas

gksus okyk ,UFksYih esa ifjorZu gkbMªkstuhdj.k dh Å"ek dgykrh

gSA vf/kd 'kkf[kr ,Ydhu vf/kd LFkk;h gksrh gS vkSj foHko ÅtkZ

de gksrh gSA ,Ydhu ftldh foHko ÅtkZ vf/kd gksrh gS oks vf/kd

ek=kk esa gkbMªkstuhdj.k dh Å"ek fudkyrh gS fn;s x;s fp=k esa rhu

leko;oh ,Ydhuksa dk ,UFksYih ifjorZu fn[kk;k x;k gSA

64. From the above statement & graph identify P if P, Q & R are the isomers of butene.;fn P, Q vkSj R C;wVhu ds leko;oh gS] rks fn;s x;s dFku dks vkSj xzkQ dks ns[krs gq;s P dks igpkfu;sA

(A) (B) CH3 – CH

2 – CH = CH

2 (C) (D)

65. Which of the following can be R if P, Q and R are isomeric pentenes.fn;s P, Q vkSj R isUVhu ds leko;oh gS rks fuEu esa ls R gksxkA

(A) 23

3

CHCHCHCH|

HC

(B) 33

3

CHCHCCH|

HC

(C) 322

3

CHCHCCH|

HC

(D) Both A and C (A rFkk C nksuksa)

66. Which of the following alkene is most stable?fuEu esa ls dkSulk ,fYdu lokZf/kd LFkk;h gS\

(A) (B) (C) (D)

SECTION - IV ([k.M - IV)Matrix - Match Type ¼eSfVªDl&lqesy izdkj½

This section contains 1 questions. Each question contains statementsgiven in two columns which have to be matched. Statements in Column are labelled as A,B,C and D whereas statements in Column are labelledas p,q,r and s. The answers to these questions have to be appropriatelybubbled as illustrated in the following example.If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q ,


ABCD

p q r s

then the correctly bubbled matrix will look like the following :bl [k.M esa 1 iz'u gSaA izR;sd iz'u esa nks dkWye esa oDRO; (statements) fn;s gq, gSaftudk lqesy (match) djuk gSA dkWye (Column-I) esa fn;s x;s oDrO;ks a

(A, B, C, D) dks dkWye (Column-II) esa fn;s x;s oDrO;ksa (p, q, r, s) ls lqesy djuk gSA bu

iz'uksa ds mÙkj fn;s x;s mnkgj.k ds vuqlkj mfpr cqYyksa dks dkyk djds n'kkZuk gSA



ABCD

p q r s


STP1718




67. Column-I Column-II

Unbalanced Reactions Characteristics

(A) H2(g) + O2(g) H2O (l) (p) Limiting reagent is O2.

1 mole H2 is mixed with 64 g of O2

(B) CS2(l) + O2(g) SO2(g) + CO2(g) (q) 50% of excess reagent is left after

152 g CS2 is mixed with 22.4 of O2 gas at STP completion of reaction.

(C) C(s) + O2(g) CO2(g) (r) No reactants are left behind after

24 g Carbon is mixed with 1 mole O2 completion of reaction.

(D) NO(g) + O2(g) NO2(g) (s) The products formed contain a total

2 NA molecules of NO are mixed with 1 mole O2 of 3 NA atoms.

dkWye-I dkWye-II

vlarqfyr vfHkfØ;k,¡ vfHky{k.k

(A) H2 (g) + O2(g) H2O () (p) lhekar vfHkdeZd O2 gSA

1 eksy H2 dks 64 g O2 ds lkFk fefJr fd;k tkrk gSA

(B) CS2() + O2(g) SO2(g) + CO2(g) (q) vfHkfØ;k iw.kZ gksus ds i'pkr~

152 g CS2 dks STP ij 22.4 L O2 xSl ds lkFk 50% vkf/kD; vfHkdeZd 'ks"”k cp tkrk gSA

fefJr fd;k tkrk gSA

(C) C(s) + O2(g) CO2(g) (r) vfHkfØ;k ds iw.kZ gksus ds i'pkr~ dksbZ

24 g dkcZu dks O2 ds 1 eksy v.kq ds lkFk fefJr fd;k tkrk gSA dksbZ Hkh vfHkdkjd 'ks"”k ugha cprk gSA

(D) NO(g) + O2(g) NO2(g) (s) cuk;s x;s mRikn] dqy 3 NA ijek.kq

NO ds 2NA v.kq dks 1 eksy O2 ds lkFk fefJr fd;k tkrk gSA j[krs gSaA

ANSWER KEY

1. (B) 2. (A) 3. (B) 4. (B) 5. (D) 6. (B) 7. (A)

8. (C) 9. (A) 10. (B) 11. (D) 12. (D) 13. (A) 14. (D)

15. (D) 16. (ABC) 17. (BCD) 18. (ABCD) 19. (BC) 20. (CD) 21. (ABC)

22. (C) 23. (D) 24. (C) 25. (D) 26. (B) 27. (C) 28. (B)

29. (A) 30. (A) 31. (A) (r),(B) (p),(C) (q),(D) (q)

32. (B) 33. (A) 34. (D) 35. (D) 36. (A) 37. (B) 38. (B)

39. (D) 40. (D) 41. (B) 42. (BC) 43. (AC) 44. (ACD) 45. (BD)46. (C) 47. (A) 48. (C) 49. (B) 50. (A) 51. (D)

52. (A) - r ; (B) - s ; (C) - q ; (D) - p

53. (D) 54. (D) 55. (C) 56. (B) 57. (A) 58. (B) 59. (B)

60. (D) 61. (B) 62. (AC) 63. (BC) 64. (B) 65. (B) 66. (D)

67. (A - S) ; (B - P, S) ; (C - P,Q,S) ; (D - R)

STP1718




HINTS & SOLUTION TO SAMPLE TEST PAPER-II

1. M + K – R

2. 1 : 2

3. (cos – sin )(cos + sin ) = 2

1 cos 2 – sin 2 =

2

1 cos 2 =

2

1 =

6

4. 6

5. a2 + b2 = |z|2 = )169)(11)(94(

)169)(94)(11(

= 1

6. Centre of circle (x – 1)2 + y2 = 1 is (1, 0)Equation of normal is x + 2y = passing through centre of circle, 1 + 0 = = 1 x + 2y – 1 = 0o`Ùk (x – 1)2 + y2 = 1 dk dsUnz (1, 0) gSA

vfHkyEc dk lehdj.k x + 2y = gS tks o`Ùk ds dsUnz ls xqtjrk gS rc 1 + 0 = = 1 x + 2y – 1 = 0

7.

2

71,

2

71

8. a = 3

4cot2 30º + 3 sin2 60º – 3cosec2 60º – 3/4 tan2 30º =

3

4×3 + 3×

4

3–3 ×

3

4–

4

3×

3

1

= 4 + 4

9 – 4 –

4

1 = 2 b = 3 tan2 45 + cos 0º– cot 90º

b = 3 × 1 + 1 – 0

log42 = 0 2

2 2log =2

1 log22 =

2

1

9. x2 – 26x + 120 = 0 x

1 = 6, x

2 = 20 2xx = 206 = 526 = 5 + 1

10. n+1C6 + nC

4 > n+2C

5 – nC

5 n+1C

6 + nC

4 + nC

5 > n+2C

5 n+1C

6 + n+1C

5 > n+2C

5 n+2C

6 > n+2C

5 n + 2 > 6 + 5

n > 9 Least value of n is 10. n dk U;wure eku 10.

11. In any arrangement either A1 as above A

2 or A

2 is above A

1.

i.e, in half of the arrangements A1 is above A2

Required arrangements = 2

1(10)!

fdlh Øep; esa A1, A

2 ls Åij ;k A

2, A

1 ls Åij gSA

vFkkZr~ A1, A

2 ls Åij] dqy Øep;ksa ds vk/ks Øep;ksa esa gS vfHk"B Øep; =

2

1(10)!

12. G.E = e(1+x) – 1 = e . ex – 1

coeff. of xn is e. !n

1

STP1718




13. x2 – 6x + 5 < 0 1 < x < 5

x2 – 5x + 6 > 0 x < 2 or x > 3 x (1, 2) (3, 5)

14. HCF

5

36,

4

27,

2

9 = )5,4,2(LCM

)36,27,9(HCF =

20

9 Ibs = weight of each piece.

Total weight dqy Hkkj= 5

36

4

27

2

9 = 18.45

Maximum no. of guests = 9

2045.18 = 41.

15. none of these (buesa ls dksbZ ugha)

16. the centre is (–1, 2) (dsUnz (–1, 2) gSA )

the length of axes are and 3 (v{kksa dh yEckbZ o 3 gSA)

ecentricity of the ellipse is 3

2 (nh?kZo`Ùk dh mRdsUnzrk

3

2gSA )

17. the centre of the circle is ( 1/2, 1) (o`Ùk dk dsUnz ( 1/2, 1) gS)

area of the circle is 45/4 sq. units (o`Ùk dk {ks=kQy 45/4 oxZ bdkbZ gS)

the point of contact of L2 with the circle has the coordinates (1 , 4)

(L2 dk o`Ùk ds lkFk Li'kZ fcUnq (1 , 4) gSA)

18.4

; 34

; 4

; 34

19. Put log2x = t

log42x =

2

1 (log

2 2 + log

2x) =

2

1 (1 + t)

Hence the given equation is

t1

t

= )t3(3

)t2(2

or t2 + 3t – 4 = 0 or t = 1, – 4 = log2x

x = 2 or 2–4

izfrLFkkihr djus ij log2x = t

log42x =

2

1 (log

2 2 + log

2x) =

2

1 (1 + t)

vr% fn;k x;k lehdj.k fuEu gksxkA

t1

t

= )t3(3

)t2(2

;k t2 + 3t – 4 = 0 ;k t = 1, – 4 = log2x x = 2 ;k 2–4

20. Making y2 = 4ax homogeneous with the help of y = mx – 2am – am3 ........(i)

y = mx – 2am – am3 ........(i) dh lgk;rk ls y2 = 4ax dk le?kkrh; cukrs gSA

or 3amam2

y–mx

= 1 then y2 = 4ax

3amam2

y–mx

y2 =

3mm2

y–mxx4

(2m + m3) y2 – 4mx2 + 4xy = 0

Angle between the lines represented by equation (i) is /2. Coefficient of x2 + coefficient of y2 = 0

lehdj.k (i) js[kkvks ds e?; dksa.k /2 iznf'kZr djrh gSA

STP1718




x2 dk xq.kkad + y2 dk xq.kkad = 0 2m + m3 – 4m = 0 m3 – 2m = 0

m2 = 2, m 0 m = ± 2

21. b = 35

c ; a , b , c cannot be the sides of a triangle a , b , c f=kHkqt dh Hkqtk,a ugha gks ldrh ; a = 9c25

22. Total words = !2!2

!7 = 1260

No. of words when two R's are not together= 1260 – No. of ways when two R's are together

= 1260 – !2

!6 = 900

dqy 'kCn = !2!2

!7 = 1260

'kCnks a dh la[;k ftuesa nksuks a R ,d lkFk ugha vk,

= 1260 – 'kCnks a dh la[;k ftuesa nksuks a R ,d lkFk vk,

= 1260 – !2

!6 = 900

23. No. of words when 2R's are together = !2

!6 = 360

No. of words when 2A's are together = 360No. of words when 2R's & 2A's are together = 5! = 120Hence number of words when neither 2R's nor 2A's are together

= Total – (360 – 360) + 120= 1260 – 720 + 120 = 660

'kCnks a dh la[;k tc nksuks a R ,d lkFk vk, = !2

!6 = 360

'kCnks a dh la[;k tc nksuks a A ,d lkFk vk, = !2

!6 = 360

'kCnks a dh la[;k tc nksuks a A vkSj nksuks a R ,d lkFk vk, = 5! = 120

vr% 'kCnks a dh la[;k ftuesa u rks nksuks a A ,d lkFk vk, vkSj u gh nksuks a R ,d lkFk vk,

= dqy – (360 – 360) + 120= 1260 – 720 + 120 = 660

24. AAEGNRR

Number of words starting with 'AA' = !2

!5 = 60

Number of words starting with 'AE' = !2

!5 = 60

Number of words starting with 'AG' = !2

!5 = 60

Number of words starting with 'AN' = !2

!5 = 60

Number of words starting with ARA = 4! = 24Number of words starting with ARE = 4! = 24Number of words starting with ARG = 4! = 24Number of words starting with ARN = 4! = 24Number of words starting with ARRAE = 2! = 2Number of words starting with ARRAG = 2! = 2Number of words starting with ARRANE = 1Next word is ARRANGEHence Rank of ARRANGE is = 342

STP1718




25. here 2 > 1 > 0 4 solutions;gk¡ 2 > 1 > 0 4 gy gSaA

26. here 0 < 3 < m = 4 2 solutions;gk¡ 0 < 3 < m = 4 2 gy gSaA

27. here + 3 = 5 and 5 = 5 > 0 3 solutions;gk¡ + 3 = 5 rFkk 5 = 5 > 0 3 gy gSaA

28. Since there are 5 even places and 3 pairs of repeated letters therefore at least one of these must be atan odd place.

the number of ways = !2!2!2

!11

pw¡fd ;gk¡ ikap leLFkku gS rFkk iqujkorhZ v{kjksa ds rhu ;qXe gS ftuesa ls de ls de fo"ke LFkku ij jgsaA

dqy rjhdks dh la[;k = !2!2!2

!11

29. Make a bundle of both M's and another bundle of T's. Then except A's we have 5 letters remaining soM's, T's and the letters except A's can be arranged in 7 ! ways total number of arrangements = 7 ! × 8C

2

nksuksa M rFkk nksuksa T dk lewg cukb,A A dks NksM+us ij 5 v{kj 'ks"k jg tkrs gS rkfd M, T ,oa A dks NksM+dj 7 rjhdksa ls

O;ofLFkr dj ldrs gSA

O;oLFkkvksa dh dqy la[;k = 7 ! × 8C2

30. Consonants can be placed in !2!2

!7 ways

Then there are 8 places and 4 vowels

Number of ways = !2!2

!7. 8C

4 !2

!4 = !4)!2(

!7!83

O;atuksa LFkkuksa ds rjhds !2!2

!7 gks ldrs gSA

vc 8 LFkkuksa ij 4 Loj O;ofLFkr djus gSaA

rjhdksa dh la[;k = !2!2

!7. 8C

4 !2

!4 = !4)!2(

!7!83

31. (A) 3 sin x + 4 cos x 5 maximum value of is 5

(B) sin2 + 3 cos = 3i.e. cos2 – 3 cos + 2 = 0 cos = 1, 2but cos = 2 is not possible = 0 is the only solution

(C) Centre and radius of the circle x2 + y2 = 4 are (0, 0) and 2 respectivelycentre and radius of the circle (x – 3)2 + (y – 4)2 = 9 are (3, 4) and 3 respectivelydistance between the centres = 5Sum of the radii = 5 the two circles touch each other externally number of common tangents = 3

(D) Centre and radius of the circle 4(x – 1)2 + 4 (y – 1)2 = 9 are (1, 1) and 2

3 respectively

equation of the chord is x + y = 2since point (1, 1) lies on the chord the chord is a diameter of the circle length of chord is 3

STP1718




(A) 3 sin x + 4 cos x 5 dk vf/kdre eku 5 gS

(B) sin2 + 3 cos = 3vFkkZr~ cos2 – 3 cos + 2 = 0 cos = 1, 2ysfdu cos = 2 tks laHko ugha gSA

dsoy = 0 gy gSA

(C) o`Ùk x2 + y2 = 4 ds dsUnz rFkk f=kT;k Øe'k% (0, 0) rFkk 2 gSAo`Ùk (x – 3)2 + (y – 4)2 = 9 ds dsUnz rFkk f=kT;k Øe'k% (3, 4) rFkk 3 gSAdsUnzks ds e/; nwjh = 5f=kT;kvksa dk ;ksx = 5 nksuks o`Ùk ,d&nwljs dks ckÐ; Li'kZ djrs gSA

mHk;fu"B Li'kZ js[kkvksa dh la[;k = 3

(D) o`Ùk 4(x – 1)2 + 4 (y – 1)2 = 9 dk dsUnz rFkk f=kT;k Øe'k% (1, 1) rFkk 2

3 gSA

thok dh lehdj.k x + y = 2 gSApw¡fd fcUnq (1, 1) thok ij fLFkr gSA

thok o`Ùk dk ,d O;kl gSA

thok dh yEckbZ 3 gSA

32. [] = ]LLT[

]L][TML[1

221

= ML–1T–1 .

33. Length of BC to AC is 80

20 =

4

1yEckbZ BC dk AC ls vuqikr

80

20 =

4

1

So, value of loops in BC to AC will also be 1 : 4 vr% BC rFkk AC esa ik'kksa dh la[;k dk vuqikr 1 : 4 gksxkA

So, of vibration will be vr% dEiUu dh vko`fÙk gksxhA

f1 =

T

)2.0(4

1 = 125 Hz f

2 =

T

)2.0(4

3 = 275 Hz.

34. 0 = constant So a

t = 0

anet

= 2r

2t aa = a

ralong radius. f=kT;k ds vuqfn'k

Which is provided by friction f = M02r

tks fd ?k"kZ.k }kjk izkIr gksxkA

35. a = 5.225.15

)g5.2)1(g2()g5.6(

a = 11

g2 m/sec2

to find the tension at the given point, draw the FBD of the part :fufnZ"V fcUnq ij ruko Kkr djus ds fy, Hkkx dk FBD cuk,

5.5 g – T = (5.5) (a)Solving gy djus ij T = 4.5 g

T = 45 N.

36.

STP1718




dt

dx = v sin – v´

dt

dy = – v cos

y = x tan (always) (lnSo)

dt

dy =

dt

dx tan

v´ = v cosec

37. As per Kepler's law, areal velocity remain constantdsiyj ds fu;e ls {ks=kh; pky fu;r jgrh gS

dt

dA =

m2

J =

m2

rvm 11 = 2

1r1 v1 =

2

1r2 v2 .

38. Net force on massless flask must be zero.ik=k ij dqy cy 'kwU; gksuk pkfg,A

39. Moment of inertia about 2 : (2 ds lkis{k cyk?kw.kZ)

2 = 4

º45sin

3

m 22

= 3

m2 2

Apply perpendicular axis theorem, yEcor~ v{k çes; ds mi;ksx ls

1 =

2 + mh2 =

3

m2 2 + 4m

2

2

1

=

3

8 m2

40. d

t1 =

2

d

/2 2

t2 =

3

d2

4

3

t

t

2

1

41.

In the frame of B (B ds funsZ'k rU=k esa)W

pseudo + W

spring = K = 0

STP1718




+ ma xmax

+ (– 2max

1Kx

2) = 0

xmax

= 2ma

K

42.

(After collision)v = v

A + v

B.......(1)

2

1 =

v

v–v AB

vB – v

A =

2

v.......(2)

vB =

4

v3

vA =

4

v

IB =

0–4

v3m =

4

mv3

loss in K.E. = 2

1mv2 =

2

1m

22

4

v3

4

v =

16

3mv2.

43. In one time period all the particles are simultaneously at rest twice.,d vkorZdky esa lHkh d.k ,d lkFk nks ckj fojkekoLFkk esa vkrs gSA ; ‘ All the particles may be at their positive extremes simultaneously once in a time period.lHkh d.k muds /kukRed pje fLFkfr ij ,d vkorZdky esa ,d ckj ,d lkFk vk ldrs gSA

44. As area of v-t graph gives displacement and velocity is always positive, so particle is moving only in adirection and since velocity is decreasing, so acceleration is negative also and displacement = distance.v-t xzkQ dk {ks=kQy foLFkkiu nsrk gS rFkk osx lnSo /kukRed gksxk vr% d.k dsoy mlh fn'kk esa xfr dj jgk gS rFkk osx

?kV jgk gS vr% Roj.k _.kkRed gksxk rFkk foLFkkiu = nwjh

45. (A) process is not isothermal

(B) volume decreases and temperature decreases

U = negative

W = negative

so Q = negative

(C) Work done in process A B C is positive

(D) Cycle is clockwise, so work done by the gas is positive.

(A) çØe lerkih; ugha gSA

(B) vk;ru ?kV jgk gS o rkieku Hkh ?kV jgk gS

U = _.kkRed

W = _.kkRed

vr% Q = _.kkRed

(C) çØe A B C eas fd;k x;k dk;Z /kukRed gSA

(D) pØh; çØe nf{k.kko`Ùk gS, blfy, xSl }kjk fd;k x;k dk;Z /kukRed gSA

STP1718




46. 4.8 sec.

47. x = A sin (t + 30°)

48.time

–

Sol. 49 to 51. (Moderate)

The angular velocity and linear velocity are mutually perpendicular

v = 3x + 24 = 0 or x = – 8

The radius of circle r = v

= 10

5 =

2

1 meter

The acceleration of particle undergoing uniform circular motion is

va

= )j4i3()j6i8( = k50

dks.kh; osx rFkk js[kh; osx ijLij yEcor~ gSA

v = 3x + 24 = 0 or x = – 8

o`Ùk dh f=kT;k r = v

= 10

5 =

2

1 ehVj

,d leku o`Ùkh; xfr esa d.k dk Roj.k

va

= )j4i3()j6i8( = k50

52. Equation of path is given asy = ax – bx2

Comparing it with standard equation of projectile;

y = x tan – 22

2

cosu2

xg

tan = a, 22 cosu2

g = b

Horizontal component of velocity = u cos = b2

g

Time of flight T = g

sinu2 = g

tan)cosu(2 =

gb

a2

g

ab2

g2

2

Maximum height H = g2

sinu 22 =

g2

]tan.cosu[ 2 =

b4

a

g2

a.b2

g2

2

Horizontal range R = g

2sinu2 = g

)cosu()sinu(2 =

b

a

g

b2

ga.

b2

g2

iFk dk lehdj.k fn;k x;k gS &

y = ax – bx2

STP1718




bldh iz{ksI; ds ekud lehdj.k ls rqyuk djus ij

y = x tan – 22

2

cosu2

xg

tan = a, 22 cosu2

g = b

osx dk {kSfrt ?kVd = u cos = b2

g

mM~M;u dky T = g

sinu2 = g

tan)cosu(2 =

gb

a2

g

ab2

g2

2

vf/kdre Å¡pkbZ H = g2

sinu 22 = g2

]tan.cosu[ 2 =

b4

a

g2

a.b2

g2

2

{kSfrt ijkl R = g

2sinu2 = g

)cosu()sinu(2 = b

a

g

b2

ga.

b2

g2

53. Hformation

respresent the enthalpy of formation of a substance from it’s elements in their standard state.

HfuekZ.k

ekud voLFkk esa] buds rRoksa ls inkFkZ ds fuekZ.k dh ,UFkSYih dks iznf'kZr djrh gSA

54. S has +6 as most stable oxidation state.

Se has +6 as most stable oxidation state.

Te has +6 as most stable oxidation state.

O has –2 as most stable oxidation state.

55. 0 +1 –34P + 3KOH + 3H2O 3KH2PO2 + PH3

Hence P is undergoing oxidation as well as reduction.

0 +1 –34P + 3KOH + 3H2O 3KH2PO2 + PH3

vr% P dk vkWDlhdj.k rFkk vip;u nksuksa gksrk gSA

56. 15

8

108

103c

= 3.75 × 10–8 m.

57. (i) Negative charge should be at more electronegative atom.(ii) Less formal charge provides more stability.

(i) _.kkos'k vf/kd fo|qr_.kh ijek.kq ij gksuk pkfg,A

(ii) de vkSipkfjd vkos'k vf/kd LFkkf;Ro iznku djrk gSA

58. (A)

(B) Number of bond pairs around I = 2. Number of lone pairs around I = 3.

I ds pkjksa vkSj ca/k ;qXeksa dh la[;k = 2 ; I ds pkjksa vkSj ,dkdh ;qXeksa dh la[;k = 3

STP1718




(C) (D)

59.

CH3–C–CH2–CH–CH3

CH3 CH3

CH3

60. +m group increases electron density in the benzene ring.

61. It is a type of free radical substitution reaction, so here most stable intermediate is

CH–CH3

So •Cl will attack on it & form CH–CH3

Cl

62. Equilibrium mass = 16 g = Initial mass (mass conservation)

Initial moles of SO3 =

16

80= 0.2

All other conclusions require additional data i.e. equilibrium constant or degree of dissociation.

lkE; nzO;eku = 16 g = izkjfEHkd nzO;eku ¼nzO;eku laj{k.k½

SO3 ds izkjfEHkd eksy =

16

80= 0.2

lHkh vU; fu”d”kksZ ds fy, vfrfjDr vk¡dM+ks dh vko’;drk gksrh gS vFkkZr~ lkE; fu;rkad ;k fo;kstu dh ek=kk dh

vko’;drk gksrh gSA

63.

CHO

; CH3–CH

2–CH

2–OH

Sol. (64 to 66)

Heat of hydrogenation alkeneofstability

1

gkbMªkstuhdj.k dh m"ek Ro dk LFkkf;fYdukssa,

1

(D) is conjugated alkene with more number of -hydrogen atoms.

Therefore it is more stabilised by conjugation and hyper conjugation.

(D) ;g v.kq vfrla;qXeu rFkk la;qXeu ds dkj.k fn;s x;s v.kqvksa esa lokZf/kd LFkk;h gSA

STP1718




67. (A) 2H2(g) + O2(g) 2H2O (l)

1 mole 64

2 mole32

1 mole

(LR)

% of excess reagent left =2 0.5

1002

= 75%

Total number of atoms in the product formed = 1 × NA × 3 = 3 N

A (S)

(B) CS2(l) + 3O

2(g) 2SO2(g) + CO2(g)

152

76= 2 mole mole 1

4.22

4.22

2

3mole

3

1 mole

(LR) (p)

% of excess reagent left = 1002

31

2

= 83.33 %

Total number of atoms in the product formed = 3

2 × NA × 3 + 3

1 × NA × 3 = 3 N

A (S)

(C) C(s) + O2(g) CO2(g)

12

24= 2 mole 1 mole 1 mole

(LR) (p)

% of excess reagent left = 1002

12

= 50% (Q)

Total number of atoms in the product formed = 1 × NA × 3 = 3 N

A (S)

(D) 2NO(g) + O2(g) 2NO2(g)

A

A

N

N 2= 2 mole 1 mole 2 mole

No reactants are left behind after completion of reaction. (R)Total number of atoms in the product formed = 2 × NA × 3 = 6 NA

gy (A) 2H2(g) + O2(g) 2H2O ()

1 eksy mole 232

64 1 eksy

(LR)

vkf/kD; esa fy;s x;s vfHkdeZd ds ‘ks”k cps Hkkx dk % = 1002

5.02

= 75%

cuk;s x;s mRikn esa ijek.kqvksa dh dqy la[;k = 1 × NA × 3 = 3 NA (S)

(B) CS2() + 3O2(g) 2SO2(g) + CO2(g)

76

152= 2 eksy mole 1

4.22

4.22

3

2eksy

3

1eksy

(LR) (p)


31

2

= 83.33 %

cuk;s x;s mRikn esa ijek.kqvksa dh dqy la[;k = × NA × 3 + × NA × 3 = 3 NA (S)(C) C(s) + O2(g) CO2(g) = 2 eksy 1 eksy 1 eksy

(LR) (p)


31

2

= 50% (Q)

cuk;s x;s mRikn esa ijek.kqvksa dh dqy la[;k = 3

2 × NA × 3 + 3

1 × NA × 3 = 3 NA (S)

(D) 2NO(g) + O2(g) 2NO2(g)

A

A

N

N 2= 2 eksy 1 eksy 2 eksy

vfHkfØ;k ds iw.kZ gksus ds i’pkr,dksbZ vfHkdkjd‘ks”k ugha cprk gSA (R)cuk;s x;s mRikn esa ijek.kqvksa dh dqy la[;k = 2 × NA × 3 = 6 NA

STP1718




SAMPLE TEST PAPER-III(For Class-XII Appearing / Passed Students)

Course : VISHESH (JD*)

Correct Wrong Blank


(dsoy , d fodYi l gh)3 -1 0



22 to 30 46 to 51 64 to 66 Comprehensions (v uqPNsn) 4 -1 0

31 52 67Matrix Match Type



Part - II (Physics)


Type

PART - I (Hkkx- I)



This section contains 15 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONLY ONE is correct.bl [k.M esa 15 iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA

1. ‘P + Q’ means ‘P is the brother of Q’. ‘P – Q means P is the mother of Q and ‘P × Q’ means ‘P is the sisterof Q’. Which of the following means that M is the maternal uncle of R ?‘P + Q’ dk vFkZ ‘Q dk HkkbZ P gSA’ ‘P – Q dk vFkZ gS Q dh ek¡ P gS rFkk ‘P × Q’ dk vFkZ gS Q dh cgu ‘P gSA fuEu esals fdldk lk vFkZ R dk ekek M gSA(A) M – R + K (B) M + K – R (C) M + K × Q (D) M – K + R




le; dk vuqikr gS &

(A) 1 : 2 (B) 2 : 1 (C) 1 : 1 (D) 1 : 3

3. If cos – sin , 2

1, cos + sin are in G.P. then possible value of is

;fn cos – sin , 2

1, cos + sin xq.kksÙkj Js.kh esa gS rks dk laHkkfor gS -

(A) 3

(B)

6

(C)

4

(D)

9

4. The fundamental period of cos(sin x) + cos(cos x) is :

cos(sin x) + cos(cos x) dk eq[; vkorZ gS -

(A) 2

(B) (C)

2

3(D) 2

5. If z = )i43)(i–1)(i3–2(

)i4–3)(i32)(i1(

= a + ib then a2 + b2 =

;fn z = )i43)(i–1)(i3–2(

)i4–3)(i32)(i1(

= a + ib rc a2 + b2 =

(A) 132 (B) 144 (C) 25 (D) 1

STP1718




6. If f(x) = 2x

x

)1–t( dt, then |f ‘()| = ( is cube root of unity)

;fn f(x) = 2x

x

)1–t( dt, gks rks |f ‘()| = (tgk¡ bdkbZ dk ?kuewy gSA)

(A) 3 (B) 2 3 (C) 3 3 (D) 4 3

7. The vectors x i – 2 j + 5 k and i + y j – z k are collinear then z

xy2

=

lfn'k x i – 2 j + 5 k vkSj i + y j – z k lajs[kh; gS rc z

xy2

=

(A) 5

4(B)

5

3(C) –

5

3(D) –

5

4

8. If f(x) = 2x + 3, g(x) = x2 + 7 then the value(s) of x for which (fog)(x) = 25 is

;fn f(x) = 2x + 3, g(x) = x2 + 7 gks rks x dk eku gksxk ftlds fy, (fog)(x) = 25 gS -

(A) 0 (B) ±1 (C) ±2 (D) ±2

1

9. If x1 < x

2 and x

1, x

2 are the roots of x2 – 26x + 120 = 0 then 21 xx =

;fn x1 < x

2 vkSj x

1, x

2 lehdj.k x2 – 26x + 120 = 0 ds ewy gS rc 21 xx =

(A) 5 + 1 (B) 5 + 2 (C) 3 + 2 (D) 5 + 3

10. The smallest value of n satisfying n+1C6 + nC

4 > n+2C

5 – nC

5 is :

n+1C6 + nC

4 > n+2C

5 – nC

5 dks larq"B djus okyk x dk lcls NksVk eku gS -

(A) 11 (B) 10 (C) 9 (D) 8

11. The number of ways in which candidates A1, A

2, ...... A

10 can be ranked if A

1 is always above A

2 is

Øep;ksa dh la[;k gksxh ftlesa ijh{kkFkhZ A1, A

2, ...... A

10 dks ojh;rk Øe esa j[kk tk ldrk gS ;fn A

1 ls Åij gS A

2 ls Åij gS -

(A) (10)! (B) 9! (C) 2

19! (D)

2

1(10)!

12. Coefficient of xn in !1

x1 +

!2

)x1( 2+

!3

)x1( 3 + .........

!1

x1 +

!2

)x1( 2+

!3

)x1( 3 + .........esa xn dk xq.kkad gS

(A) e (B) n! (C) e(n!) (D) !n

e

13. If x2 – 6x + 5 < 0, x2 – 5x + 6 > 0 then x

;fn x2 – 6x + 5 < 0, x2 – 5x + 6 > 0 rc x

(A) (1, 2) (3, 5) (B) (1, 3) (4, 5) (C) (2, 3) (4, 5) (D) (1, 4) (5, 6)

STP1718




14. If det A = 7, A =

cbg

fed

cba

then det (2A)–1 is equal to :

;fn det A = 7, A =

cbg

fed

cba

rc det (2A)–1 cjkcj gS -

(A) 14

1(B)

49

1(C)

56

1(D)

2

7

15. The system of equations kx + y + z = 1, x + ky + z = k and x + y + kz = k2 does not have unique solution if

k is equal to :

lehdj.kksa kx + y + z = 1, x + ky + z = k vkSj x + y + kz = k2 dk fudk; vf}rh; gy ugh j[krk gS ;fn k cjkcj gS -

(A) 0 (B) 1 (C) –1 (D) –14


Multiple Correct Answer Type (cgqy lgh fodYi izdkj)

This section contains 6 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONE OR MORE is/are correct.

bl [k.M esa 6 iz'u gSaA izR;sd iz'u ds mÙkj ds fy, 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls

vf/kd lgh gSA

16. Let f be differentiable for all x. If f (1) = 2 and f (x) 2 for x [ 1, 6] then;fn f izR;sd x ds fy, vodyuh; gSA ;fn f (1) = 2 ,oa f (x) 2 x [ 1, 6] rc(A) f (6) < 8 (B) f (6) 8 (C) f (6) > 5 (D) f (6) 5

17. The lines L1 : x 2y + 6 = 0 & L

2 : x 2y 9 = 0 are tangents to the same circle . If the point of contact

of L1 with the circle is ( 2, 2), then :

(A) the centre of the circle is ( 7/2, 5) (B) the centre of the circle is ( 1/2, 1)

(C) area of the circle is 45/4 sq. units

(D) the point of contact of L2 with the circle has the coordinates (1 , 4)

js[kk,¡ L1 : x 2y + 6 = 0 ,oa L

2 : x 2y 9 = 0 ,d gh o`Ùk dh Li'kZ js[kk,¡ gSA ;fn L

1 dk o`Ùk ds lkFk Li'kZ fcUnq

( 2, 2) gks] rks&

(A) o`Ùk dk dsUnz ( 7/2, 5) gSA (B) o`Ùk dk dsUnz ( 1/2, 1) gSA

(C) o`Ùk dk {ks=kQy 45/4 oxZ bdkbZ gSA (D) L2 dk o`Ùk ds lkFk Li'kZ fcUnq (1 , 4) gSA

18. If z = 20 21i + 21 20 i then the principle value of arg(z) can be:

;fn z = 20 21i + 21 20 i rks arg(z) ds eq[; eku gks ldrs gS&

(A) 4

(B) 34

(C) 4

(D) 34

19. Let f(x) = min {1, cosx, 1 - sinx}; x , then

(A) f is not differentiable at ‘0’ (B) f is differentiable at /2(C) f has a local maxima at ‘0’ (D) none of these;fn f(x) = min {1, cosx, 1 - sinx}; x , rks -

(A) f, ‘0’ ij vodyuh; ugha gSA (B) f, /2 ij vodyuh; gSA

(C) f, ‘0’ ij mPpre fcUnq j[krk gSA (D) buesa ls dksbZ ugha

STP1718




20. a

0

dx)xa(f)x(f is equal to

a

0

dx)xa(f)x(f =

(A) a

0

dx)x(f2 (B) a

0

dx)xa(f2 (C) a

0

dx)xa(f (D) a

0

dx)x(f

21. Solution of the differential equation

3

dx

dy

– xdx

dy + y = 0 is/are :

vody lehdj.k

3

dx

dy

– xdx

dy + y = 0 ds laHko gy gS -

(A) y = 2x – 8 (B) y = x – 1 (C) y = 2/3x

33

2(D) y =

2/3x33

2


Comprehension Type (cks/ku çdkj)

This section contains 3 paragraphs. Based upon each paragraph, there are 3 questions. Each question has4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.bl [k.M esa 3 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 3 ç'u gSA çR;sd ç'u ds 4 fodYi




consider a function f(x) = min {tan x, cot x}, x R.Qyu f(x) = min {tan x, cot x}, x R ysus ij

22. Range of f(x) isf(x) dk ifjlj gS&

(A) [– 1, 0] (B) (– , – 1) (0, 1) (C) (– , ) (D) (– , – 1] [0, 1]

23. Points of discontinuity of f(x) aref(x) ds vlrr~rk okys fcUnq gS&

(A)

n,2

n(B)

n,

4

n(C)

n,

8

n(D)

n,

12

n

24. Points of non-differentiability of f(x) aref(x) ds os fcUnq tgk¡ Qyu vodyuh; ugha gS] gS&

(A)

n,4

n(B)

n,

6

n(C)

n,

3

n(D)

n,

8

n.



Let ||x – a| – b| = k. Then(i) k = 0, b > 0 equation has two solutions(ii) b > k > 0 equation has four solutions(iii) b = k > 0 equation has three solutions(iv) 0 < b < k equation has two solutions

STP1718




ekukfd ||x – a| – b| = k gS] rks

(i) k = 0, b > 0 lehdj.k ds nks gy gSaA

(ii) b > k > 0 lehdj.k ds pkj gy gSaA

(iii) b = k > 0 lehdj.k ds rhu gy gSaA

(iv) 0 < b < k lehdj.k ds nks gy gSaA

25. If number of solutions of ||x + 1| – 2| = 1 is m, then m =;fn ||x + 1| – 2| = 1 ds gyksa dh la[;k m gS] rks m =

(A) 1 (B) 2 (C) 3 (D) 4

26. If number of solutions of ||x – 2| – 3| = m is , then =(where m is obtained in Q.No. 25);fn ||x – 2| – 3| = m ds gyksa dh la[;k gS] rks =

(tgk¡ m iz'u la[;k 25 ls izkIr gksrk gSA)(A) 1 (B) 2 (C) 3 (D) 4

27. Number of solutions of ||x – 2| – 5| = + 3 is(where is obtained in Q.No. 26)||x – 2| – 5| = + 3 ds gyksa dh la[;k gS &

(tgk¡ iz'u la[;k 26 ls izkIr gksrk gSA)(A) 1 (B) 2 (C) 3 (D) 4



Consider the letters of the word MATHEMATICS. There are eleven letters some of them are identical.Letters are classified as repeating and non-repeating letters. Set of repeating letters = {M, A, T}. Set ofnon-repeating letters = {H, E, I, C, S}

;fn MATHEMATICS 'kCn ds v{kj fy, tk, rks X;kjg v{kjksa esa ls dqN v{kj ,d ckj vkrs gS rFkk dqN dh iqujko`fr

gksrh gSA iqujkorhZ v{kjksa dk leqPp; = {M, A, T} gSA tcfd viqujkorhZ = {H, E, I, C, S} gSaA

28. Possible number of words taking all letters at a time such that atleast one repeating letter is at oddposition in each word, is

lHkh v{kjksa dks ,d ysdj dqy fdrus 'kCn cuk, tk ldrs gS ftuesa de ls de ,d iqujko`rhZ v{kj fo"ke LFkku ij jgsA

(A) !2!2!2

!9(B) !2!2!2

!11(C) !2!2!2

!11 – !2!2

!9(D) !2!2

!9

29. Possible number of words taking all letters at a time such that in each word both M's are together andboth T's are together but both A's are not together, is

lHkh v{kjksa dks ,d lkFk fy, tk, rkfd izR;sd v{kj esa nksuksa M lkFk vk, rFkk nksuksa T Hkh lkFk&lkFk vk, ijUrq nksuksa A

lkFk&lkFk ugha vk, rc cuus okys laHkkfor 'kCnksa dh la[;k gS &

(A) 7 ! . 8C2

(B) !2!2!2

!11 – !2!2

!10(C) !2!2

!4!6(D) !2!2!2

!9

30. Possible number of words in which no two vowels are together, is

mu 'kCnksa dh la[;k Kkr dhft, ftuesa dksbZ Loj lkFk&lkFk u vk,A

(A) !4)!2(

!7!83 (B) !2

!7 . 8C

4 . !2

!4(C) 2)!4()!2(

!7!8(D) )!4()!2(

!7!84

STP1718





Matrix - Match Type (eSfVªDl&lqesy izdkj)This section contains 1 questions. Each question contains statementsgiven in two columns which have to be matched. Statements in Column are labelled as A,B,C and D whereas statements in Column are labelledas p,q,r and s. The answers to these questions have to be appropriatelybubbled as illustrated in the following example.If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q ,


ABCD

p q r s


bl [k.M esa 1 iz'u gSaA izR;sd iz'u esa nks dkWye esa oDRO; (statements) fn;s gq, gSa ftudk

lqe sy (match) djuk g SA dk Wye (Column-I) es a fn; s x; s oDrO; k s a


gSA bu iz'uksa ds mÙkj fn;s x;s mnkgj.k ds vuqlkj mfpr cqYyksa dks dkyk djds n'kkZuk gSA



31. Column– Column–

(A) If 1 2

1 0 01

A 0 1 1 and A A cA d6

0 2 4

, (p) 5

then (c + d) is

(B) If a b c 5,

then the value of a b b c c a

is (q) 2

(C) – tan2 – cot2 + sec2 + cosec2 is equal to (r) 3

(D) Let 'a' is degree and 'b' is order of the differential (s) 9/2equation of all parabolas, whose axis is the x-axis,then (a + b) is equal to

LrEHk - I LrEHk- II

(A) ;fn A =

42–0

110

001

rFkk 1 21A 0 1 1 and A A cA d

6 , (p) 5

gS] rks (c + d) gS&

(B) ;fn a b c 5,

gS] rks a b b c c a

dk eku gS& (q) 2

(C) – tan2 – cot2 + sec2 + cosec2 cjkcj gS& (r) 3

(D) ekukfd a ?kkr gS rFkk lHkh ijoy;ksa ftldh v{k x-v{k gS] ds (s) 9/2

vody lehdj.k dh dksfV b gS] rks (a + b) =

STP1718







This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.


32. Kerosene oil is flowing through a tube of length & radius r . The pressure difference between two ends

of the tube is P, then the viscosity of oil is given by = v4

)xr(P 22 where v is velocity of oil at a

distance x from the axis of tube . From this relation the dimensions of viscosity are : yEckbZ vkSj r f=kT;k dh fdlh V~;wc (uyh) esa dsjksflu rsy cg jgk gSA ;fn uyh esa nksauks fljksa ij nkckUrj P gS , rc rsy

dh ';kurk ;fn = v4

)xr(P 22 ls nh tkrh gSA tgk¡ v V~;wc dh v{k ls x nwjh ij rsy dk osx gSA bl lw=k ls ';kurk

dh foek gskxh :(A) M L

2 T

2 (B) M L 1

T 1 (C) M0 L0 T0 (D) M L T

3

33. Pressure versus temperature graph of an ideal gas of equal number ofmoles of different volumes are plotted as shown in figure. Choose thecorrect alternative :(A) V

1 = V

2,V

3 = V

4 and V

2 > V

3

(B) V1 = V

2,V

3 = V

4 and V

2 < V

3

(C) V1 = V

2= V

3 = V

4

(D) V4 > V

3 > V

2 > V

1 nkc ds laxr rki dk xzkQ fdlh vkn'kZ xSl ds leku eksyks a dh la[;k ds fy,]

vyx&vyx vk;ru ij] fp=k esa iznf'kZr gS] rks lgh rF; gksxk :

(A) V1 = V

2,V

3 = V

4 rFkk V

2 > V

3

(B) V1 = V

2,V

3 = V

4 rFkk V

2 < V

3

(C) V1 = V

2= V

3 = V

4

(D) V4 > V

3 > V

2 > V

1

34. A wall is moving with constant velocity u towards a fixed source of sound of frequency 'f'. The velocityof sound is 'v'. The wavelength of the sound reflected by the wall is,d nhokj ,d fu;r osx u ds lkFk xfr dj jgh gS ;g xfr f vko`fÙk ds fLFkj /ofu L=kksr dh rjQ gSA /ofu dk osx

v gSA nhokj }kjk ijkofrZr /ofu dh rjaxnS/;Z gksxh &

(A) f

v(B)

f

uv (C)

f

uv (D)

f

v.

uv

uv

35. An object is moving parallel to principal axis of a concave mirror of focal length 100 cm. Distance between

object and principal axis is 1 cm. The tangent of angle between velocity of image and principal axis is nearly

,d fcUnq oLrq 100 cm Qksdl nwjh ds vory niZ.k dh eq[; v{k ds lekUrj xfr'khy gSA eq[; v{k o oLrq ds e/; nwjh

1 cm gSA eq[; v{k ,oa izfrfcEc ds osx ds e/; dks.k dk Li'kZT;k (tan ) dk eku yxHkx gksxk %

(A) 0.01 (B) 0.02 (C) 0.03 (D) 0.04

STP1718




36. A 1m long wire having tension of 100 N and of linear mass density 0.01 kg/m is fixed at end A and free at endB. The point C which is 20 cm from end B is constrained to be stationary. To create resonance in this wire,the minimum frequency of the tuning fork will be :1m yEcs rkj esa ruko 100 N rFkk js[kh; nzO;eku ?kuRo 0.01 kg/m gS dks fljs A ij fdyfdr rFkk fljs B ij eqDr j[kk x;k gSA

fcUnq C tks fljs B ls 20 cm ij gS] fLFkj jgus ds fy, ckf/kr gSA rkj esa vuqukn ds fy, Lofj=k dh U;wure vkofÙk gksxh :

(A) 125 Hz (B) 150 Hz (C) 300 Hz (D) 275 Hz37. An ideal gas undergoes a cyclic process shown in the graph, in which one process is isochoric, one process

is isothermal and one process is adiabatic. During the isothermal process, 40 J heat is released by the gas,and during the isochoric process, 80 J heat is absorbed by the gas. If work done by the gas during adiabatic

process is W1 and during isothermal process is W

2 then

1

2

W

W will be equal to:

,d vkn'kZ xSl dks pØh; çØe ls xqtkjk tkrk gS ftlesa ,d çØe le&vk;rfu;] ,d çØe lerkih; rFkk ,d çØe

:)ks"e gSA lerkih; çØe ds nkSjku xSl 40 J Å"ek fu"dkf'kr djrh gS] rFkk le&vk;rfu; çØe ds nkSjku xSl 80 J

Å"ek vo'kksf"kr djrh gSA ;fn xSl }kjk fd;k x;k dk;Z :)ks"e çØe ds nkSjku W1 rFkk lerkih; çØe ds nkSjku W2

gks rks 1

2

W

W dk vuqikr fdlds cjkcj gksxk :

(A) –2 (B) 2 (C) 4 (D) –1/2

38. The least count of a stop watch is 1/5 sec. The time of 20 oscillations of a pendulum is measured to be

25 sec. The maximum percentage error in the measurement of time will be :

,d fojke ?kM+h dk vYirekad 1/5 lSd.M gSA ,d yksyd ds 20 nksyu dk le; 25 lSd.M ekik tkrk gSA le; ds ekiu

esa vf/kdre~ izfr'kr~ =kqfV gksxhA

(A) 0.1% (B) 0.8% (C) 1.8% (D) 8%

39. In a Young’s double-slit experiment, a thin sheet of mica is placed over one of the two slits. As a result, the

center of the fringe pattern (on the screen) shifts by an amount corresponding to 30 dark bands. The wavelength

of the light in this experiment is 480nm and the refractive index of the mica is 1.60. The mica thickness is

(approximately) :

,d ;ax f}fNnz iz;ksx esa] nks fNnzksa esa ls ,d ds lkeus ,d iryh vHkzd dh ijr j[kh tkrh gSA ftlds ifj.kke Lo:i ijns

ij dsfUnz; fÝat izk:i 30 vfnIr fÝat ds cjkcj foLFkkfir gksrk gSA bl iz;ksx esa iz;qDr izdk'k dk rjaxnS/;Z 480nm gS]

,oa vHkzd dk ijkorZukad 1.60 gSA vHkzd ijr dh eksVkbZ yxHkx gksxhA

(A) 0.090mm (B) 0.012mm (C) 0.014mm (D) 0.024mm

STP1718




40. Two identical bulbs are connected to an ac source by using an inductor and a capacitor in series, with thebulbs as shown then the brightness of B1 and B2 will be -(A) Same for both cases. (B) More for B1 than B2

(C) Depends on the frequency of source (D) More for B2 than B1

~

V

C

L

B1

B2

nks ,d tSls cYo ,d izR;korhZ /kkjk L=kksr ls fp=kkuqlkj tksMs+ x;s gSA ,d izsjd dq.Myh rFkk la/kkfj=k] cYcksa ds Js.kh Øe

esa mi;ksx djrs gq, yxs gS rks B1 rFkk B2 dh ped gksxh %

(A) nksuksa esa leku gksxh (B) B1 esa B2 ls vf/kd

(C) L=kksr dh vko`fÙk ij fuHkZj djsxhA (D) B2 eas B1 dh vis{kk vf/kd

41. By splitting a total mass M in two masses M1 and M2, the maximum value of tension that can be producedin the string 1 as shown in figure is :,d nzO;eku M dks nks Hkkxksa M1 rFkk M2 esa rksM+k tkrk gSA fp=k esa n'kkZ;h xbZ jLlh 1 esa vf/kdre~ ruko D;k gksxkA

(A) Mg (B) 4 Mg (C) 2

Mg(D)

4

Mg






42. In the ac circuit shown, XL=7 ,R=4 and X

C=4 .The reading of the

ideal voltmeter V2 is 8 2 V..

(A) The reading of the ideal ammeter will be 2A:(B) The reading of the ideal ammeter will be 1A(C) The reading of ideal voltmeter V1 will be 7 V :(D) The reading of ideal voltmeter V1 will be 10 V : iznf'kZr izR;korhZ ifjiFk esa, XL = 7 ,R = 4 rFkk XC = 4 vkn'kZ oksYVehVj

dk ikB~;kad V2 = 8 2 V gSA

(A) vkn'kZ vehVj dk ikB~;kad 2A gksxkA (B) vkn'kZ vehVj dk ikB~;kad 1A gksxkA

(C) vkn'kZ oksYVehVj V1 dk ikB~;kad 7V gksxkA (D) vkn'kZ oksYVehVj V1 dk ikB~;kad 10 V gksxkA

STP1718




43. A non–viscous fluid of density is flowing in a tube as shown in figure. Area of section–(1) is double that ofsection–(2). Centre of mass of section–(2) is h height above the Centre of mass of section–(1) and level ofwater in tube–1 is ‘h’ height above that in tube–2. Then :

(A) Velocity of fluid of section–(1) is gh

3

(B) Velocity of fluid at section–(1) is 2gh

3

(C) Work done by gravitational force per unit volume from section–(1) to section–(2) is = gh(D) Pressure difference at point 1 & 2 is 2gh

h

h

1

2 tube-1

tube-2

h

h

1

2 ufydk-1

ufydk-2

fp=k esa n'kkZ,uqlkj ?kuRo dk v';ku nzo ufydk eas çokfgr gSA [k.M–(1) dk {kS=kQy [k.M–(2) ds {kS=kQy dk nqxquk gSA

[k.M–(2) dk nzO;eku dsUnz] [k.M–1 ds nzO;eku dsUnz ls ‘h’ Å¡pkbZ ij gS rFkk ufydk–1 esa ry] ufydk–2 esa ry ls ‘h’

Å¡pkbZ Åij gS rks :

(A) [k.M–(1) esa nzo dk osx gh

3 gSA

(B) [k.M–(1) esa nzo dk osx 2gh

3 gSA

(C) [k.M–(1) ls [k.M–(2) esa xq:Roh; cy }kjk çfr bdkbZ vk;ru fd;k x;k dk;Z = gh gSA

(D) fcUnq 1 o fcUnq 2 esa nkckUrj 2gh gSA44. A particle is moving in a straight line according to graph given below : (a is acceleration, v is velocity, x is

position, t is time) :uhps fn;s x;s xzkQ ds vuqlkj ,d d.k ljy js[kk esa xfr'khy gS rks (a Roj.k gS, v osx gS, x fLFkfr gS, t le; gSA) :

(A)

x

a

(B) (C)

x

t

(D)

45. In the figure shown, coefficient of restitution between A and B is e = 2

1, then :

(A) velocity of B after collision is 2

v(B) impulse between two during collision is

4

3mv

(C) loss of K.E. during collision is 16

3mv2 (D) loss of K.E. during collision is

4

1mv2

STP1718




fp=kkuqlkj A vkSj B esa e/; çR;koLFkku xq.kkad e = 2

1, gksxk :

(A) VDdj ds ckn B dk osx 2

vgksxkA (B) nksauks ds e/; VDdj ds nkSjku vkosx

4

3mv gksxkA

(C) VDdj ds nkSjku xfrt ÅtkZ esa gkfu16

3mv2 gksxhA (D) VDdj ds nkSjku xfrt ÅtkZ esa gkfu

4

1mv2 gksxhA


Comprehension Type (c) cks/ku çdkj)This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have tobe answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.bl [k.M esa 2 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 3 cgq&fodYih ç'u ds mÙkj nsus gSA

çR;sd ç'u ds 4 fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA



In the shown figure, there are two long fixed parallel conducting rails (having negligible resistance) andare separated by distance L. A uniform rod of resistance R and mass M is placed at rest on frictionlessrails. Now at time t = 0, a capacitor having charge Q

0 and capacitance C is connected across rails at

ends a and b such that current in rod(cd) is from c towards d and the rod is released. A uniform andconstant magnetic field having magnitude B exists normal to plane of paper as shown. (Neglectacceleration due to gravity)fp=k esa ,d nwljs ls L nwjh ij ux.; izfrjks/k okyh nks lekUrj yEch pkyd iVfj;k

iznf'kZr dh xbZ gSA M nzO;eku ,oa R izfrjks/k dh ,d ,dleku NM+ ?k"kZ.k jfgr

iVfj;ksa ij fLFkj voLFkk esa j[kh gqbZ gSA vc t = 0 le; ij vkos'k Q0 ,oa /kkfjrk

C dk ,d la/kkfj=k iVfj;ksa ds a ,oa b fcUnqvksa ds e/; bl izdkj tksM+k tkrk gS

fd NM+ (cd) esa fo|qr /kkjk c ls d dh vksj izokfgr gksrh gS rFkk NM+ eqä gks tkrh

gS ,d leku fu;r pqEcdh; {ks=k B dkxt ds ry ds yEcor~ fp=kkuqlkj gS

(xq:Roh; Roj.k dks ux.; ekfu;s)

Ba

b

L

c

d

46. When the speed of rod is v, the charge on capacitor is : tc NM+ dh pky v gS rks la/kkfj=k ij vkos'k gS &

(A)

vLBC

RMQ0 (B)

vBRC

MQ

20 (C)

v

LB

MQ0 (D)

vRCB

MQ0

47. When the acceleration of rod is zero, the speed of rod is :tc NM+ dk Roj.k 'kwU; gS rks NM+ dh pky gS &

(A) 222

20

CLBM

BCRQ

(B)

CLBM

LBQ22

0

(C) 442

0

CRBM

LBQ

(D) 222

0

CLBM

LBQ

48. When the acceleration of rod is zero, the charge on capacitor is :tc NM+ dk Roj.k 'kwU; gS rks la/kkfj=k ij vkos'k gS &

(A) CLBM

QCLB22

022

(B) 222

0342

CLBM

QCRB

(C) 442

0342

CRBM

QCRB

(D) 222

022

CLBM

QCLB

STP1718






A block of mass 10 kg oscillates simple harmonically when it is attached with a spring. The graph ofoscillation between the displacement verses time is given then answer the following questions.10 kg nzO;eku dk ,d CykWd A ljy vkorhZ :i esa nksyu djrk gS tc ;g fLizax ls tqM+k gqvk gSA foLFkkiu le; ds e/

; nksyu dk xzkQ fp=k esa n'kkZ;k x;k gSA rks fuEu iz'uksa ds mÙkj nhft;sA

xA

A/2

2s4.4s

time

49. Time period of oscillation of block will be :

CykWd ds nksyu dk vkorZdky gksxkA

(A) 4 sec. (B) 5 sec. (C) 4.8 sec. (D) 4.4 sec

50. Equation of SHM of the block :

CykWd dh ljy vkorZ xfr dh lehdj.k gksxhA

(A) x = A sin (t + 30°) (B) x = A sin

(t + 45°) (C) x = A sin

(t + 150°) (D) x = A sin

(t – 30°)

51. The graph for velocity vs time will be :

osx≤ xzkQ gksxkA

(A)

A/2

time (B) time

(C) time

–

(D) time

–

SECTION - IV ([k.M- IV)Matrix - Match Type (eSfVªDl&lqesy izdkj)

This section contains 1 question. The question contains statements given in two columnswhich have to be matched. Statements (A, B, C, D) in Column-I have to be matched withstatements (p, q, r, s) in Column-II. The answers to these questions have to be appropriatelybubbled as illustrated in the following example.


ABCD

p q r s

If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q, then the correctly matched answer will looklike in bubbled matrix. One option in Column–I may have one or more than one correct options in Column–II.Marks will be awarded only if an option in Column–I is matched with all correct options in Column-II. Eachcorrect matching will be awarded 1 mark and if all the matchings are correct then 6 marks will be awarded.bl [k.M esa 1 iz'u gSaA izR;sd iz'u esa nks dkWye esa oDRO; (statements) fn;s gq, gSa ftudk lqesy

(match) djuk gSA dkWye (Column-I) esa fn;s x;s oDrO;ksa (A, B, C, D) dks dkWye (Column-II) esa fn;s

x;s oDrO;ksa (p, q, r, s) ls lqesy djuk gSA bu iz'uksa ds mÙkj fn;s x;s mnkgj.k ds vuqlkj mfpr cqYyksa

dks dkyk djds n'kkZuk gSA


ABCD

p q r s

;fn lgh lqesy A-p, A-r, B-p, B-s, C-r, C-s rFkk D-q gSa] rks lgh fof/k ls dkys fd, x;s cqYyksa dk 4 x 4 esafVªDl (matrix)

esa n'kkZ;k x;k gSA dkWye–I ds izR;sd fodYi dk dkWye–II ds ,d ;k ,d ls T;knk fodYiksa ls lqesy djk;k tk ldrk gSA vad



STP1718




52. Match the column-I with Column-II. Consider a solid neutral conducting sphereof radius 2R having a concentric cavity of radius R. Points A , D and B are at

distances 3 R , 2

R3 &

2

R from centre C respectively. S

1 & S

2 are inner and

outer surface of the hollow sphere. q0 is a point charge which can be placed

either at point A or B or C or D. ekfu;s dh 2R f=kT;k dk ,d Bksl mnklhu pkyd xksyk gS] tks R f=kT;k dh ldsUnzh;

xqgk (cavity) j[krk gSA dsUnz C ls fcUnq A , D o B Øe'k% 3R , 2

R3 o

2

R nwjh ij

gSA S1 o S

2 [kks[kys xksys dh vkUrfjd o ckgjh lrg gSA ,d fcUnq vkos'k q

0 dks fcUnq

A ;k B ;k C ;k D esa ls fdlh ,d LFkku ij j[k ldrs gSA

Match the description in column-I with description in column-IIdkWye-I esa nh xbZ O;oLFkk dk dkWye-II esa nh xbZ O;k[;k ls feyku dhft,A

Column-I Column-I(A) point charge q

0 is at C (p) Induced charge on S

1 is –q

0

fcUnq vkos'k q0 fcUnq C ij gSA S

1 ij izsfjr vkos'k –q

0 gSA

(B) point charge q0 is at B (q) Induced charge on S

2 will be distributed uniformly

fcUnq vkos'k q0 fcUnq B ij gSA S

2 ij izsfjr vkos'k le:i forfjr gksxkA

(C) point charge q0 is at A (r) V

D = R3

kq0

fcUnq vkos'k q0 fcUnq A ij gSA

(D) point charge q0 is at D (s) E

D = 0

fcUnq vkos'k q0 fcUnq D ij gSA

PART - III (Hkkx - III)


Straight Objective Type (lh/ks oLrqfu"B izdkj)This section contains 9 multiple choice questions. Each question has choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.bl [k.M esa 9 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA

53. Which of the following will give carylamine reaction?

fuEu esa ls dkSu dkfcZy ,ehu vfHkfØ;k nsxk\

(A) CH3NH

2(B)

NH–CH3 (C) (CH

3)2NH (D) (CH

3)3N

54. 1 mol CH3COOH is added in 250 g benzene. Acetic acid dimerises in benzene due to hydrogen bond.

Kb of benzene is 2 K kgmol–1. The boiling point has increased by 6.4K. % dimerisation of acetic acid is :

(A) 50 (B) 40 (C) 30 (D) 20

1 eksy CH3COOH dks 250 g csathu esa feykrs gSA gkbMªkstu ca/k ds dkj.k ,flfVd vEy csathu esa f}ydhd`r gks tkrk gSA

csathu ds Kb dk eku 2 K kgmol–1 gSA DoFkukad esa 6.4K dh o`f) gksrh gSA ,flfVd vEy dk % f}ydhdj.k (dimerisation)

gS %

(A) 50 (B) 40 (C) 30 (D) 20

55. The substances, P, Q and R have coagulation values 3, 0.6, 0.08 for a metal sol respectively. Their flocculat-ing powers are in the ratio :

fdlh /kkrq lkWy ds fy, inkFkZ P, Q o R ds LdUnu eku Øe'k% 3, 0.6, 0.08 gSA budh Å.kZu {kerk (flocculating powers)

dk vuqikr gS %

(A) 0.0267 : 5 : 1 (B) 1 : 5 : 37.5 (C) 0.08 : 0.6 : 3 (D) 1 : 0.2 : 0.0267

STP1718




56. The following electrochemical cell is taken.Cu | Cu2+(aq) || Ag+ (aq) | Ag, and has emf E

1 > 0. By which of the following actions, E

1 increases?

(A) Adding NH3 to the cathodic chamber. (B) Adding HCl to the cathodic chamber.

(C) Adding AgNO3 to the anodic chamber. (D) Adding NH

3 to the anodic chamber.

fo|qrjklk;fud lsy Cu | Cu2+(aq) || Ag+ (aq) | Ag fy;k x;k gS rFkk bldk emf E1 > 0 gSA fuEu esa ls fdl izØe

}kjk E1 c<+rk gS \

(A) dSFkksM Hkkx esa NH3 feykus ij (B) dSFkksM Hkkx esa HCl feykus ij

(C) ,uksM Hkkx esa AgNO3 feykus ij (D) ,uksM Hkkx esa NH

3 feykus ij

57. Which of the following complexes is a paramagnetic inner orbital complex?fuEUk esa ls dkSulk ladqy ,d vuqpqEcdh; vkUrfjd d{kd ladqy gS \

(A) [Fe(CN)6]4– (B) [CoCl4]2– (C) [Cu(NH3)4]2+ (D) [Ni(CN)4]2–

58. Which of the following is the major product of the given reaction is

nh x;h vfHkfØ;k esa fuEu esa ls eq[; mRikn dkSulk gS\

OH 3(i) CHCl NaOH(aq.)

(ii) H

(A) OH

CHO (B)

CHO

OH

(C)

CHO

OH

(D)

CHO

OH

59. Chlorine gas is obtained by various reactions but not by–

(A) KMnO4 + Conc HCl (B) KCl + F

2

(C) KCl + K2Cr

2O

7 + Conc H

2SO

4 (D) MnO2 + Conc HCl

dbZ vfHkfØ;kvks }kjk Dyksjhu xSl izkIr gksrh gS ysfdu bues ls dkSulh vfHkfØ;k ds }kjk izkIr ugh gksrh gS&

(A) KMnO4 + lkUnz HCl (B) KCl + F

2

(C) KCl + K2Cr

2O

7 + lkUnz H

2SO

4 (D) MnO2 + lkUnz HCl

60. Which one of the following benzene ring has maximum electron density ?fuEufyf[kr esa ls dkSulh csUthu oy; lokZf/kd bysDVªkWUk ?kuRo j[krh gS \

(A)

CH3

OCH3

(B)

CH3

OH

(C)

CH3

NHCOCH3

(D)

CH3

CH O2 H

61. product. The major product obtained is

mRiknA vfHkfØ;k ls izkIr eq[; mRikn gS &

(A) (B) (C) (C)

STP1718









62. Which of the following show oxidation state of +8 ?

fuEu esa ls dkSu +8 vkWDlhdj.k voLFkk n’'kkZrk gS\

(A) Os (B) Xe (C) He (D) Ru

63. Correct statement about peptide linkage in a protein molecule is/are correct ?(A) It is amide linkage(B) It has partial double bond character.(C) It is hydrophilic in nature(D) It connects protein molecules through H-bonds.

izksVhu v.kq esa ik;s tkus okys isIVkbM ca/k ds fo"k; esa fuEu esa ls dkSulk@dkSuls dFku lgh gS \

(A) ;g ,d ,ekbM ca/k gSA

(B) blesa vkaf'kd f}ca/k ik;k tkrk gSA

(C) ;g tyLusgh izd`fr dk gksrk gSA

(D) ;g H-ca/kksa }kjk izksVhu v.kqvksa ls tqM+k gksrk gSA


Comprehension Type (c) cks/ku çdkj)This section contains 1 paragraphs. Based upon each paragraph, 3 multiple choice questions have tobe answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.bl [k.M esa 1 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 3 cgq&fodYih ç'u ds mÙkj nsus gSA çR;sd ç'u

ds 4 fodYi (A), (B), (C) rFkk (D) gS, ftuesa ls flQZ ,d lgh gSA



Carbonyl compound which contains –H gives aldol condenation reaction in presence of alkaline medium.The reaction between two molecules of acetaldehyde take place as follows in presence of base.

dkcksZfuy ;kSfxd tks –H j[krs gS {kkjh; ek/;e esa ,YMksy la?kuu vfHkfØ;k nsrs gS] {kkj dh mifLFkfr esa ,flVsfYMgkbM ds

nks v.kqvksa ds e/; vfHkfØ;k fuEu izdkj lEiUu gksrh gSA

STP1718




64. Aldol condensation reaction is given by,YMkWy la?kuu vfHkfØ;k nsrk gS

(A) C6H5–CHO (B) CX3 – CHO (C) O2N CHO (D) C6H5–CH2–CHO

65. + Product (mRikn)

(A) Ph – CH = CH – (CH2)

5 – CHO (B) Ph – (CH

2)

5 – CH = CH – CHO

(C) (D)

66. Intramolecular aldol condensation reaction is given by(A) 2,5-diketone (B) 2,7-diketone (C) 2,6 and 2,8-diketone (D) All of thesevUr% vkf.od ,YMksy la?kuu vfHkfØ;k fuEu }kjk nh tkrh gS

(A) 2,5-MkbdhVksu (B) 2,7-MkbdhVksu (C) 2,6 rFkk 2,8-MkbdhVksu (D) mijksDr lHkh


Matrix - Match Type ¼eSfVªDl&lqesy izdkj½

This section contains 1 questions. Each question contains statementsgiven in two columns which have to be matched. Statements in Column are labelled as A,B,C and D whereas statements in Column are labelledas p,q,r and s. The answers to these questions have to be appropriatelybubbled as illustrated in the following example.If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q ,


ABCD

p q r s

then the correctly bubbled matrix will look like the following :bl [k.M esa 1 iz'u gSaA izR;sd iz'u esa nks dkWye esa oDRO; (statements) fn;s gq, gSa

ftudk lqesy (match) djuk gSA dkWye (Column-I) esa fn;s x;s oDrO;ks a


gSA bu iz'uksa ds mÙkj fn;s x;s mnkgj.k ds vuqlkj mfpr cqYyksa dks dkyk djds n'kkZuk

gSA



ABCD

p q r s


STP1718




67. Match the following for a 1st order reaction A products with time on the x axis.

,d izFke dksfV dh vfHkfØ;k A mRiknksa ds fy, x v{k ij le; ds lkFk fuEu dks lqesfyr dhft,A

Column I Column II

dkWye I dkWye II

(A) [A] v/s time (p)

[A] rFkk le;

(B) dt

]A[d v/s [A] (q)

dt

]A[d rFkk [A]

(C) dt

]A[d v/s time (r)

dt

]A[d rFkk le;

(D) log [A] v/s time (s)

log [A] rFkk le;

ANSWER KEY

1. (B) 2. (A) 3. (B) 4. (A) 5. (D) 6. (C) 7. (D)8. (C) 9. (A) 10. (B) 11. (D) 12. (D) 13. (A) 14. (C)15. (B) 16. (BC) 17. (BCD) 18. (ABCD) 19. (AC) 20. (AB) 21. (ABCD)

22. (D) 23. (A) 24. (A) 25. (D) 26. (B) 27. (C) 28. (B)29. (A) 30. (A) 31. (A) (p), (B) (p), (C) (q), (D) (r)

32. (B) 33. (A) 34. (D) 35. (A) 36. (A) 37. (A) 38. (B)

39. (D) 40. (C) 41. (C) 42. (AD) 43. (BD) 44. (ABD) 45. (BC)

46. (C) 47. (B) 48. (A) 49. (C) 50. (A) 51. (C)

52. (A) – p,q,s ; (B) – p,q,s ; (C) – r,s ; (D) – q,s

53. (A) 54. (B) 55. (B) 56. (D) 57. (C) 58. (B) 59. (C)

60. (B) 61. (B) 62. (ABD) 63. (ABCD) 64. (D) 65. (D) 66. (D)

67. [A – r] ; [B – q] ; [C – r] ; [D – p].

STP1718




HINTS & SOLUTION TO SAMPLE TEST PAPER-III

1. M + K – R

2. 1 : 2

3. (cos – sin )(cos + sin ) = 2

1 cos 2 – sin 2 =

2

1 cos 2 =

2

1 =

6

4. cos

x2

sin + cos

x2

cos = cos(cos x) + cos(– sin x)

= cos(cos x) + cos(sin x) Period is 2

5. a2 + b2 = |z|2 = )169)(11)(94(

)169)(94)(11(

= 1

6. f ‘(x) = (x2 – 1)2x – (x – 1) = 2x3 – 3x + 1

f ‘() = 23 – 3 + 1 = 3 – 3 = 3

i

2

3

2

1––1 = 3

i

2

3–

2

3

|f ‘()| = 3 4

3

4

9 = 3 3

7. Since the vectors are collinear therefore 1

x = y

2– =

z–

5

pwfd lfn'k lajs[kh; gS blfy, 1

x = y

2– =

z–

5 xy = – 2,

z

y =

5

2

z

xy2

= xy, , z

y = –2

5

2 = –

5

4

8. f(g(x)) = 25 f(x2 + 7) = 25 2(x2 + 7) + 3 = 25 x2 = 4 x = ± 2

9. x2 – 26x + 120 = 0 x

1 = 6, x

2 = 20

2xx = 206 = 526 = 5 + 1

10. n+1C6 + nC

4 > n+2C

5 – nC

5 n+1C

6 + nC

4 + nC

5 > n+2C

5 n+1C

6 + n+1C

5 > n+2C

5

n+2C6 > n+2C

5 n + 2 > 6 + 5 n > 9

Least value of n is 10.

n dk U;wure eku 10.

11. In any arrangement either A1 as above A

2 or A

2 is above A

1.

i.e, in half of the arrangements A1 is above A2

Required arrangements = 2

1(10)!

STP1718




12. G.E = e(1+x) – 1 = e . ex – 1

coeff. of xn is e. !n

1

13. x2 – 6x + 5 < 0 1 < x < 5

x2 – 5x + 6 > 0 x < 2 or x > 3

x (1, 2) (3, 5)

14. |2A|–1 = |A2|

1 = |A|2

13 =

7.8

1 =

56

1

15.k11

1k1

11k

= 0 k(k2 – 1) – 1(k – 1) + 1(1 – k) = 0 k3 – 3k + 2 = 0 k = 1

16. f (6) > 5

17. the centre of the circle is ( 1/2, 1) (o`Ùk dk dsUnz ( 1/2, 1) gS) ;

area of the circle is 45/4 sq. units (o`Ùk dk {ks=kQy 45/4 oxZ bdkbZ gS) ;

the point of contact of L2 with the circle has the coordinates (1 , 4)

(L2 dk o`Ùk ds lkFk Li'kZ fcUnq (1 , 4) gSA)

18.4

19. f is not differentiable at ‘0’ (f, ‘0’ ij vodyuh; ugha gS) ;

f has a local maxima at ‘0’ (f, ‘0’ ij mPpre fcUnq j[krk gS)

20. I = a

0

a

0

a

0

a

0

a

0

a

0

dx)xa(f2dx)x(f2dx)x(fdx)x(fdx)xa(fdx)x(f

21. y = 2x – 8 ; y = x – 1 ; y = 2/3x

33

2; y =

2/3x33

2

Sol.22 to 24

22. Range of f(x) = (– , – 1] [0, 1]f(x) dk ifjlj (– , – 1] [0, 1] gS

STP1718




23. Points of discontinuity are , 2

, 0 ........ which can be put in the form of

n,

2

n

f(x) ds vlr~rk okys fcUnq , 2

, 0 ........ gS] tks fd

n,

2

nds :i esa fu:fir fd;s tk ldrs gSA

24. The points of non-differentiability are , ± ,2

,4

3

± 4

, 0...... which can be put in form of .n,

4

n

fcUnqvksa ± ,2

,4

3

± 4

, 0...... ij Qyu vodyuh; ugha gS tks fd .n,

4

n

ds :i esa fu:fir

fd;s tk ldrs gSA

25. here 2 > 1 > 0 4 solutions;gk¡ 2 > 1 > 0 4 gy gSaA

26. here 0 < 3 < m = 4 2 solutions;gk¡ 0 < 3 < m = 4 2 gy gSaA

27. here + 3 = 5 and 5 = 5 > 0 3 solutions;gk¡ + 3 = 5 rFkk 5 = 5 > 0 3 gy gSaA

28. Since there are 5 even places and 3 pairs of repeated letters therefore at least one of these must be atan odd place.

the number of ways = !2!2!2

!11

pw¡fd ;gk¡ ikap leLFkku gS rFkk iqujkorhZ v{kjksa ds rhu ;qXe gS ftuesa ls de ls de fo"ke LFkku ij jgsaA

dqy rjhdks dh la[;k = !2!2!2

!11

29. Make a bundle of both M's and another bundle of T's. Then except A's we have 5 letters remaining soM's, T's and the letters except A's can be arranged in 7 ! ways

total number of arrangements = 7 ! × 8C2

nksuksa M rFkk nksuksa T dk lewg cukb,A A dks NksM+us ij 5 v{kj 'ks"k jg tkrs gS rkfd M, T ,oa A dks NksM+dj 7 rjhdksa ls

O;ofLFkr dj ldrs gSA

O;oLFkkvksa dh dqy la[;k = 7 ! × 8C2

30. Consonants can be placed in !2!2

!7 ways

Then there are 8 places and 4 vowels

Number of ways = !2!2

!7. 8C

4 !2

!4 = !4)!2(

!7!83

O;atuksa LFkkuksa ds rjhds !2!2

!7 gks ldrs gSA

vc 8 LFkkuksa ij 4 Loj O;ofLFkr djus gSaA

rjhdksa dh la[;k = !2!2

!7. 8C

4 !2

!4 = !4)!2(

!7!83

STP1718




31. (A)

1 0 0

A 0 1 1 and A A cA d

0 2 4

1 21A 0 1 1 and A A cA d

6

6A–1 = A2 + cA + dI6I = A3 + cA2 + dAA3 + cA2 + dA – 6I = 0 ..............(i)

Now characteristics equationA – I = 0

–42–0

1–10

00–1

= 0

(1 – ) { (1 – ) (4 – ) + 2 } = 0 3 –62 + 11 – 6 = 0Putting = AA3 – 6A2 + 11A – 6I = 0 ..............(ii)Comparing (i) & (ii), we get c = – 6, d = 11 c + d = 5

(B) a b c 5,

accbba

= 2cba

= 5

(D) Parabola will be of the form y2 = 4(x – )

Differentiating equation 2y dx

dy = 4

2

dx

dy

+ y 2

2

dx

yd = 0

b = 2a = 1

a + b = 3

(A)

1 0 0

A 0 1 1 and A A cA d

0 2 4

1 21A 0 1 1 and A A cA d

6

6A–1 = A2 + cA + dI6I = A3 + cA2 + dAA3 + cA2 + dA – 6I = 0 ..............(i)

vfHkyk{kf.kd lehdj.k

A – I = 0

STP1718




–42–0

1–10

00–1

= 0

(1 – ) { (1 – ) (4 – ) + 2 } = 0 3 –62 + 11 – 6 = 0

= A j[kus ijA3 – 6A2 + 11A – 6I = 0 ..............(ii)

(i) ,oa (ii) ls rqyuk djus ij c = – 6, d = 11 c + d = 5

(B) a b c 5,

accbba

= 2cba

= 5

(D) ijoy; y2 = 4(x – ) :i dk gksxkA

vodfyr lehdj.k 2y dx

dy = 4

2

dx

dy

+ y 2

2

dx

yd = 0

b = 2a = 1

a + b = 3

32. [] = ]LLT[

]L][TML[1

221

= ML–1T–1 .

33.

tan' > tan

2143 V

nR

V

nR

V

nR

V

nR

V1 = V

2 > V

3 = V

4

34. Frequency of sound reflected by wall f' = fv

uv

Wavelength of sound reflected by wall = f

v

uv

uv

f

uv

nhokj }kjk ijkofrZr /ofu dh vko`fÙk f' = fv

uv

nhokj }kjk ijkofrZr /ofu dh rjaxnS/;Z = f

v

uv

uv

f

uv

STP1718




35.

h

tan = h

f = 0. 01

36. Length of BC to AC is 80

20 =

4

1

yEckbZ BC dk AC ls vuqikr 80

20 =

4

1

So, value of loops in BC to AC will also be 1 : 4

vr% BC rFkk AC esa ik'kksa dh la[;k dk vuqikr 1 : 4 gksxkA

So, of vibration will be vr% dEiUu dh vko`fÙk gksxhA

f1 =

T

)2.0(4

1 = 125 Hz

f2 =

T

)2.0(4

3 = 275 Hz.

37. For isothermal process lerkih; çØe ds fy,

W2 = QBC = –40 J

For complete cycle pØh; çØe ds fy,

Qcycle = Wcycle

–40 + 80 + 0 = –40 + 0 + W

W1 = 80 J

38. Maximum permissible % error in time

le; esa vf/kdre~ laHko % =kqfV

t100

t

=

0.2100

25

= 0.8 %

39. t ( – 1) = n = 30

t = 930 480 10

0.6

= 0.024 mm

40. The brightness of B1 and B2 depend upon the currents in them which depends upon the freqeuncy of the

source. (Since XL = L and XC = C

1

)

STP1718




AlternativeSince due to frequency, resistance will be difference, so the voltage drop would be different, so the brighnessdepends on frequency.

B1 rFkk B2 /kkjk ij fuHkZj djrh gS] tks L=kksr dh vko‘fr ij fuHkZj djrh gSA (pwafd XL = L rFkk XC = C

1

)

osdfYid

pawfd vko‘fr ds dkj.k izfrjks/k esa vUrj gksxkA vr% foHko iru fHkUu&fHkUu gksxkA vr% ped vko‘fr ij fuHkZj djrh gSA

41. Tension in the system is

fudk; esa ruko

M

)xM(x2T

g

for T to the maximum ;

vf/kdre~ T ds fy, ;

dx

dT = 0

M – 2x = 0

x = 2

M Tmax =

M2

M

2

M2

g = 2

Mg

42. Reading ikB~;kad V2 = 2 2R CV V = 2 2

CR X

2 28 2 4 4 = 4 2 = 2 Amp

V1 = 22L CR X X

= 222 4 7 4 = 2 × 5 = 10 volt

43. Applying Bernoulli is equation from section–(1) and (2)[k.M–(1) ls [k.M–(2) esa cjukWyh lehdj.k yxkus ij

P1 +

1

2V1

2 + gh1 = P2 + 1

2v2

2 + gh2

P1 + 1

2V2 + 0 = P2 +

1

2(2v)2 + gh

and rFkk P1 – P2 = g(2h)

Solving we get, gy djus ij çkIr gksxk

V = 2gh

3

(C) Work done by gravitation force per unit value Wgr = decrease in gravitational PTvalue = gh1 – gh2

xq:Roh; cy }kjk çfr bdkbZ vk;ru fd;k x;k dk;Z Wgr = xq:Roh; fLFkfrt ÅtkZ esa deh PTvalue = gh1 – gh2

Wgr = 0 – gh(D) Work done by elastic force volume, çR;kLFk cyksa }kjk çfr bdkbZ vk;ru ij fd;k x;k dk;Z

We = decrease in elastic P.E.vol = decrease in pressureWe = çR;kLFk fLFkfrt ÅtkZ esa deh P.E.vol = nkc esa deh

= P1 – P2 = g(2h).

STP1718




44. Given fn;k x;k gS v = x & t = 0v = 1, x = 1

dv dxadt dt

& dvv. xdx

a = v = x

v v

1 1

v.dv x.dx v v2 2v x1 1 v = x

dx xdt

v v

1 0

dx dtx nx t

tx e

45.

(After collision)v = v

A + v

B.......(1)

2

1 =

v

v–v AB

vB – v

A =

2

v.......(2)

vB =

4

v3

vA =

4

v

IB =

0–4

v3m =

4

mv3

loss in K.E. = 2

1mv2 =

2

1m

22

4

v3

4

v =

16

3mv2.

Sol. 46 to 48At any instant t, the charge on capacitor q, velocity of rod v and the current I through rod are as shown.fdlh Hkh {k.k t ij la/kkfj=k ij vkos'k q, NM+ dk osx v ,oa NM+ ls /kkjk I fuEu izdkj iznf'kZr fd;s tk ldrs gSA

mdt

dv = B

L =

dt

dqB L .... (1)

q

Q

v

0 0

dqBLdvm

a

b

L

c

d

v+q I

solving we get (gy djus ij) q = Q0 –

BL

Mv.... (2)

Also ,oa C

q = BLV + R = BLv – R

dt

dq.... (3)

from equation (1) and (3) [leh- (1) ,oa (3) ls]

STP1718




C

q = BLv +

dt

dv

BL

mR.... (4)

from (4) when [(4) ls tc] dt

dv = 0

C

q = BLv .... (5)

From (2) and (5). At instant acceleration is zero(2) ,oa (5) ls tc Roj.k 'kwU; gks rc

v = CLBM

LBQ22

0

and q =

CLBM

QCLB22

022

49. 4.8 sec.

50. x = A sin (t + 30°)

51. time

–

52. (A) p holds by Gauss theorem if we take a Gaussian surface of radius 2

R3

p xkWml izes; ds vuqlkj gS ;fn ge 2

R3 f=kT;k dh xkWmlh;u lrg ysrs gSA

(B) q holds by Gauss theorem if we take a Gaussian surface of radius 2

R3

q xkWml izes; ds vuqlkj gS ;fn ge 2

R3 f=kT;k dh xkWmlh;u lrg ysrs gSA

ED = 0. Induced charge on S

2 will be distributed uniformly, V

D = R2

kq0

ED = 0. S

2 ij izsfjr vkos'k le:i forfjr gksrk gS, V

D = R2

kq0

(C) Induced charge on S2 will not be distributed uniformly, E

D = 0 (conductor)

S2 ij izsfjr vkos'k vle:i forfjr gksrk gS, E

D = 0 (pkyd)

VD = V

C =

R3

kq0

(D) Charge given to conductor resides on the outer surface.pkyd dks fn;k x;k vkos'k mldh ckgjh lrg ij fLFkr gksrk gSA

53. Only 1º Amine give this reaction

54. Tb = i K

b .m

Given molality (eksyyrk) = 250

10001 = 4m, 6.4 = i 2 4 or i = 0.8

For dimerisation (f}ydhdj.k ds fy,)

i = 1 – 2

0.8 – 1 =

2

–or = 0.4 40%

STP1718




55. Flocculation power valuenCoagulatio

1

Å.kZu {kerk ekuLdanu

1

P : Q : R

3

1:

6.0

1:

08.0

1

or 1 : 5 : 37.5

56. Cu | Cu2+(aq) || Ag+ (aq) | Ag. E1 > 0.

(A) Due to NH3 addition, concentration of Ag+ decreases due to complex formation. Hence, E decrease.

(B) Due to addition of HCl, Ag+ is consumed to form AgCl solid. Concentration of Ag+ decreases. So, Edecreases.(C) Adding AgNO

3 to anodic chamber result in direct reaction between Cu and Ag+

Cu + 2Ag+ Cu2+ + 2AgMore Cu2+ produced, hence E decreases.(D) Adding NH

3 in anodic chamber results in complex formation, [Cu(NH

3)

4 ]2+. Hence Cu2+ concentration

decreases. So E increases.Sol. Cu | Cu2+(aq) || Ag+ (aq) | Ag. E

1 > 0.

(A) NH3 feykus ij ladqy fuekZ.k ds dkj.k Ag+ dh lkUnzrk ?kVrh gSA vr% E ?kVrk gSA

(B) HCl feykus ij Ag+ Bksl AgCl fuekZ.k esa iz;qDr gksrk gSA ftlls Ag+ dh lkUnzrk ?kV tkrh gSA blfy, E ?kVrk gSA(C) ,uksM Hkkx esa AgNO

3 feykus ds ifj.kkeLo:i Cu o Ag+ ds e/; lh/kh vfHkfØ;k (direct reaction) gksrh gSA

Cu + 2Ag+ Cu2+ + 2Ag

vf/kd Cu2+ mRiUu gksrs gSa vr% E ?kVsxkA(D) ,uksM Hkkx esa NH

3 feykus ij ladqy [Cu(NH

3)

4 ]2+ fuekZ.k gksrk gS vr% Cu2+ dh lkUnzrk ?kVrh gS blfy, E c<+rk gSA

57. [Fe(CN)6]4–

Fe2+, t2g6 eg

0

diamagnetic, inner orbital

[CoCl4]2–

Co2+, t2g5 eg

2

paramagnetic, outer orbital

[Cu(NH3)4]2+

Cu2+, d9 square planar, dsp2

paramagnetic, inner orbital

[Ni(CN)4]2–

Ni2+ d8, square planarinner orbital, diamagnetic

gy % [Fe(CN)6]4–

Fe2+, t2g6 eg

0

diamagnetic, inner orbitalizfrpqEcdh; vkUrfjd d{kd

[CoCl4]2–

Co2+, t2g5 eg

2

paramagnetic, outer orbitalvuqpqEcdh; cká d{kd

[Cu(NH3)4]2+

Cu2+, d9 oxZ leryh;, dsp2

vuqpqEcdh;, vkUrfjd d{kd

[Ni(CN)4]2–

Ni2+ d8, oxZ leryh;

vkUrfjd d{kd] izfrpqEcdh;

STP1718




58. (B)

CHO

OH

(This is Reimer Tiemann reaction)

59. (C) Here orange red vapours of CrO2Cl

2 [Chromyl chloride] is formed

(C) ;gk¡ CrO2Cl

2 ¼Øksekby dkyksjkbM½ [Chromyl chloride] fd yky ukjaxh jax dh ok"i curh gSA

60. Due to + M effect of – OH group and hyperconjugation of – CH3 group

–OH lewg ds +M izHkko vkSj –CH3 lewg ds vfrla;qXeu izHkko ds dkj.kA

61. If bromine in acetic acid is used, bromination takes place without decarboxylation.,lhfVd vEy esa czksehu iz;qDr djus ij fodkcksZfDlyhdj.k ds fcuk czksehuhdj.k gksrk gSA

OH

COOHOH

2

3

Br

CH COOH

OH

COOHOH

Br

62. Os, Ru, Xe only gives +8 O.S.

63. It is amide linkage ;g ,d ,ekbM ca/k gSA ;It has partial double bond character. blesa vkaf'kd f}ca/k ik;k tkrk gSA;It is hydrophilic in nature ;g tyLusgh izd`fr dk gksrk gSA ;It connects protein molecules through H-bonds. ;g H-ca/kksa }kjk izksVhu v.kqvksa ls tqM+k gksrk gSA

64. – H hydrogen is present. – H gkbMªkstu mifLFkr gSA

65.

66. Polycarbonyl compound with hydrogen gives intramolecular aldol condensation reaction in presence ofalkaline medium, to give entrophically favoured rings.ikWyhdkcksZfuy ;kSfxd ftuesa gkbMªkstu mifLFkr gksrs gS] {kkjh; ek/;e esa vUrjkv.kqd ,YMksy la?kuu (intramolecularaldol condensation) vfHkfØ;k nsrs gSA

67. [A – r] ; [B – q] ; [C – r] ; [D – p].For Ist order reaction

STP1718




01

05

03

15

25

07

27

11

21

09

19

29

13

23

A B C D

02

06

04

26

08

28

12

22

10

20

30

14

24

Objective Response Sheet (ORS)

ResoFAST-2016

Application Form No.

D D M M Y Y Y Y

Signature of Invigilator

Invigilator Signature:____________________

Invigilator Name: _______________________

______________________________________

English Hindi

Signature of Candidate:

E H

0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3

4 4 4 4 4 4 4 4

5 5 5 5 5 5 5 5

6 6 6 6 6 6 6 6

7 7 7 7 7 7 7 7

8 8 8 8 8 8 8 8

9 9 9 9 9 9 9 9

A AB BC CD DRight Wrong

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

16 A B C D

17 A B C D

18 A B C D

P Q R S31P Q R S

P Q R S

P Q R S

39

43

41

45

47

A B C D

40

44

42

46

48

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

49 A B C D

50 A B C D

CBSE

ICSE

STATE

OTHER

In case of other name of board

_____________________________

TEST DATE :

37 A B C D

36 A B C D

35 A B C D

34 A B C D

33 A B C D

32 A B C D

38 A B C D

51 A B C D

P Q R S

P Q R S

52

67

P Q R S

P Q R S

P Q R S

P Q R S

P Q R S

P Q R S

60

64

62

66

A B C D

61

65

63

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

58 A B C D

57 A B C D

56 A B C D

55 A B C D

54 A B C D

53 A B C D

59 A B C D

General

SC

OBC

ST

DS

PH

Sample Test Paper (STP) For ResoFAST-2017 · PDF file · 2016-08-30Toll Free : 1800...

Documents

Transcript of Sample Test Paper (STP) For ResoFAST-2017 · PDF file · 2016-08-30Toll Free : 1800...