SAMPLE TEST PAPERS - Resonance - Admission Session : 2016-17 SAMPLE TEST PAPERS (Resonance National...

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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 ResoNET conducted by Resonance for its various courses.

Note : Resonance reserves the right to change the pattern of selection test (ResoNET). Pervious year papers do not guarantee that thepapers 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.

For More Practice of RESONANCE NATIONAL ENTRANCE TEST (ResoNET) - 2016Resonance selection test papers of last few years with answer key, hints & solutions are available on demand. Following sets ofPractice Test Papers (PTPs), in hard copy, are available with us :

S.No. Sample Pa per Code

De tails of PTPs Course (Code ) Target Rema rk

1 Set-A 10 Papers Set for Class-X Appearing/Passed s tudents VIKAAS (JA) &

VIPUL (JB) JEE(Main + Advanced) 2018Answer key,

Hints & Solutions

2 Set-B 10 Papers Set for Class-XI Appearing/Passed students VISHW AAS (JF) JEE(Main + Advanced) 2017 Only Answer key

3 Set-C 10 Papers Set for Class-XII Appearing Passed students VISHESH (JD) & VIJAY (JR)

JEE(Main + Advanced) 2017 Answer key, Hints & Solutions

Interested students may collect the same from Resonance Study Centres or Corporate Office at Kota (at Plot No. A-46, A-52, Near CityMall, Jhalawar Road, Reception) by paying an additional fees of Rs.300/- only per set. Any of the above Practice Test Papers (PTPs)sets may be procured through post / courier from 'Resonance Eduventures Ltd' by sending a Bank Demand Draft (DD) of Rs. 300/- infavour of 'Resonance' and payable at Kota. A student may send the request application on plain paper along with prerequisite fees tothe institute to collect any of the sets of Practice Test Papers (PTPs). Please, mention clearly your name and roll number (ApplicationForm No.) on the back of the DD and which set of Practice Test Papers (Set A, B or C) is required by you in the request application.Resonance Selection Test Papers of last few years with Answer key, Hints & Solutions are available on demand. ResoNET Papers areavailable Online too.ResoNET Online Practice Test Papers (OPTPs) :

S. No. Details of OPTPs Course Code Target Fee(Taxes included)

1 3 Tests for Class-X Appearing/Passed students VIKAAS (JA) & VIPUL (JB) JEE(Main + Advanced) 2018 Rs. 300/-

2 6 Tests for Class-X Appearing/Passed students VIKAAS (JA) & VIPUL (JB) JEE(Main + Advanced) 2018 Rs. 500/-

3 3 Tests for Class-XI Appearing/Passed students VISHWAAS (JF) JEE(Main + Advanced) 2017 Rs. 300/-

4 6 Tests for Class-XI Appearing/Passed students VISHWAAS (JF) JEE(Main + Advanced) 2017 Rs. 500/-

5 3 Tests for Class-XII Appearing/Passed students VISHESH (JD) & VIJAY (JR) JEE(Main + Advanced) 2017 Rs. 300/-

6 6 Tests for Class-XII Appearing/Passed students VISHESH (JD) & VIJAY (JR) JEE(Main + Advanced) 2017 Rs. 500/-

Students can buy these Online Test papers at http://elpd.resonance.ac.in

Sample Test Paper (STP) For ResoNET-2016

© Copyright reserved 2016-17.All rights reserved. Any photocopying, publishing or reproduction of full or any part of this material is strictly prohibited. This material belongs to only the applicants of RESONANCE for itsvarious Selection Tests (ResoNET) to be conducted for admission in Academic Session 2016-17. Any sale/resale of this material is punishable under law. Subject to Kota Jurisdiction only.

S.No. Contents Target Page No.

1 How to Prepare for the Resonance National Entrance Test (ResoNET)-2016 ResoNET 2016 2

2 General Instructions for the Examination Hall ResoNET 2016 3

3 Syllabus for ResoNET-2016 ResoNET 2016 4

4Sample Test Paper- I : For Class-X Appearing/Passed students (Moving from Class-X to Class-XI ) For the students applying for VIKAAS (JA) & VIPUL (JB) Courses JEE(Main + Advanced) 2018 9

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

JEE(Main + Advanced) 2018 18

6Sample Test Paper-II : For Class-XI Appearing / Passed students (Moving from Class-XI to Class-XII).For the students applying for VISHWAAS (JF) Course JEE(Main + Advanced) 2017 25

7Sample Test Paper-II Answer key & Hints & Solution : For Class-XI Appearing / Passed students (Moving from Class-XI to Class-XII).For the students applying for VISHWAAS (JF) Course JEE(Main + Advanced) 2017 42

8Sample Test Paper-III : For Class-XII Appearing / Passed students (Moving from Class-XII to Class-XIII) For the students applying for VISHESH (JD) & VIJAY (JR) Courses JEE(Main + Advanced) 2017 52

9Sample Test Paper-III Answer key & Hints & Solution : For Class-XII Appearing / Passed students (Moving from Class-XII to Class-XIII) For the students applying for VISHESH (JD) & VIJAY (JR) Courses JEE(Main + Advanced) 2017 68

10 Sample ORS Answer Sheet for Resonance National Entrance Test (ResoNET)-2016 ResoNET 2016 79

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HOW TO PREPARE FOR THE RESONANCE NATIONAL ENTRANCE TEST (ResoNET) - 2016

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

Study thoroughly the books of Science (Physics & Chemistry) and Maths of ClassesIX & 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 (RespectiveBoard).

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 InorganicChemistry, J.D. Lee, Wiley-India Edition, Vogel’s Qualitative Analysis forthe JEE (7th Edition), G. Svehla & Shishir Mittal, Pearson Education,OrganicChemistry : 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 ByS.L. Loney ; Plane Trigonometry By S.L. Loney, Problem book in high schoolby 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 ofcompetence :

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

For Chemistry : Physical Chemistry By R.K. Gupta, Organic Chemistry ByMorrison & 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).

Note : For further practice, a set of several Practice Test Papers (PTPs) of Resonance may beprocured from the institute. For this, the details are available on Page No.1.

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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 electronicgadgets 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 thisbooklet. (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 ObjectiveResponse 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.kks a dks Hkjs a½

6. Do not forget to mention your paper code and Application Form Number neatly and clearly inthe 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 blankspace 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 ¼iz'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 35 to 42 47 to 54Only one correct

(dsoy , d fodYi l gh)3 -1 0

21 to 24 43 to 46 55 to 58One or more than one correct Answer

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

25 to 32 Comprehensions (vuqPNsn) 4 0 0

33 to 34Matrix 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 : _______________

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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 conserva-tion of energy.Law of gravitation ; acceleration due to gravity.Electricity and magnetism : Ohm’s law ; Series and parallel com-bination of resistances ; Heating effect of current.

Magnetic field near a current carrying straight wire, along the axisof a 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 ResoNET-2016

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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, ion-ization, 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, redox re-actions,oxidation number, balancing redox reactions, applications of redoxreaction.

Hydrogen : Position of hydrogen in periodic table, occurrence, iso-topes, 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 in chemi-cal reactivity, anomalous properties of first element of the group ;Boron - physical and chemical properties, some important compounds; borax, boric acids, boron hydrides. Aluminium : uses, reactionswith 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 tetrachlo-ride, 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 purif ication, 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, hydro-gen halides (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 of func-tional group in mono-substituted benzene ; carcinogenicity and tox-icity.

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.

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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.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 en-ergy; Work and power; Conservation of linear momentum and me-chanical energy.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; Elemen-tary concepts of convection and radiation; Newton’s law of cooling;Ideal gas laws; Specific heats (Cv and Cp for monoatomic and di-atomic gases); Isothermal and adiabatic processes, bulk modulus ofgases; Equivalence of heat and work; First law of thermodynamicsand its applications (only for ideal gases); Blackbody radiation:absorptive and emissive powers; Kirchhoff’s law; Wien’s displace-ment 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).

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.

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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 ChemistryConcepts : 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,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 l ines; Centroid,orthocentre, incentre and circumcentre of a triangle.

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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 (Cartesian co-ordinates 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 en-ergy; Work and power; Conservation of linear momentum and me-chanical energy.

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 : Wave motion (plane waves only), longitudinal and trans-verse waves, superposition of waves; Progressive and stationarywaves; 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; Elemen-tary concepts of convection and radiation; Newton’s law of cooling;Ideal gas laws; Specific heats (Cv and Cp for monoatomic and di-atomic gases); Isothermal and adiabatic processes, bulk modulus ofgases; Equivalence of heat and work; First law of thermodynamicsand its applications (only for ideal gases); Blackbody radiation:absorptive and emissive powers; Kirchhoff’s law; Wien’s displace-ment 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 find field due to infinitely long straight wire, uni-formly charged infinite plane sheet and uniformly charged thin spheri-cal 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 mu-tual inductance; 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; Charac-teristic and continuous X-rays, Moseley’s law; de Broglie wave-length of matter waves.

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Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 9

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

Course : VIKAAS (JA) & VIPUL (JB)

S.No. Subject Sections No. of questions Marks Negative marks Total

1 to 24 Sections-I 24 3 -1 72

25 to 28 Sections-II 4 4 0 16

29 to 36 Sections-III Comprehension (vuqPNsn) (SCQ)

2 x 4 = 8 8 4 -1 32

37 to 44 Sections-I 8 3 -1 24


49 to 56 Sections-I 8 3 -1 24


60 200

Multiple choice questions (MCQ)(dsoy ,d ;k ,d ls vf/kd fodYi lgh)



PART-II(PHYSICS)

PART-III(CHEMISTRY)

Single choice questions (SCQ)(dsoy ,d fodYi lgh)


Nature of Questions


PART-I(MATHS)

Total

PART - I (Hkkx - I)

SECTION - I ([k.M- I)

Straight Objective Type (lh/ks oLrqfu"B izdkj )This section contains 24 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONLY ONE is correct.

bl [k.M esa 24 iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA

1. The median AD of ABC meets BC at D. The internal bisectors of ADB and ADC meet AB and AC at Eand F respectively. Then EF(A) is perpendicular to AD (B) is parallel to BC(C) divides AD in the ratio of AB : AC (D) none of theseABC dh ekf/;dk,a AD Hkqtk BC dks D ij feyrh gSA ADB vkSj ADC ds vkUrfjd dks.k v/kZd AB vkSj AC dks

Øe'k% E vkSj F ij feyrs gS] rc EF(A) AD ds yEcor~ gS (B) BC ds lekUrj gS

(C) AD dks vuqikr AB : AC esa foHkkftr djrk gS (D) buesa ls dksbZ ugha

2. If A is an acute angle such that sin A = 53

, then the value of AcosAsin

AcosAsin133

is

;fn U;wudks.k A bl çdkj gS] fd sin A = 53

gS] rks AcosAsin

AcosAsin133

dk eku gS&

(A) 57

(B) – 51

(C) 1235

(D) None of these (buesaa ls dksbZ ugha)

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3. If the diagonal and the area of a rectange are 25 m and 168 m2 respectively, what is the length of therectangle ?fdlh vk;r ds fod.kZ vkSj {ks=kQy Øe'k% 25 m rFkk 168 m2 gS] rks vk;r dh yEckbZ gksxh ?(A) 17 m (B) 31 m (C) 12 m (D) 24 m

4. If x = 7 + 4 3 then the value of x

1x is

;fn x = 7 + 4 3 gks] rc x

1x dk eku gS&

(A) 8 (B) 6 (C) 5 (D) 4

5. In an A.P., the sum of first n term is 2n5

2n3 2

. Its 25th term is

fdlh lekUrj Js<h ds izFke n inksa dk ;ksx 2n5

2n3 2

gS] rks bldk 25ok¡ in gksxk&

(A) 73 (B) 76 (C) 79 (D) 81

6. In Figure, D and E are the mid-points of sides AB and AC respectively of ABC, Find EDB.fn;s x;s fp=k esa ABC ds D rFkk E Øe'k% AB rFkk AC ds e/;fcUnq gS EDB Kkr djksA]

A

B C

D E5cm

60°

50°

(A) 110° (B) 120° (C) 70° (D) 80°

7. Two circles with centres A and B intersect at P and Q. Which of the following is false?(A) AB is the perpendicular bisector of PQ (B) PQ is the perpendicular bisector of AB(C) APQ = AQP (D) PBA = QBAnks o`Ùk ftuds dsUnz A vkSj B gS] P vkSj Q ij izfrPNsn djrs gSA fuEu esa ls dkSulk xyr gS?(A) AB, PQ dk yEclef}Hkktd gS (B) PQ, AB dk yEclef}Hkktd gS

(C) APQ = AQP (D) PBA = QBA

8. 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

9. In the figure given below, TBP and TCQ are tangents to the circle whose centre is ‘O’, touching the circle at B and C respectively. Also, PBA = 60ºand ACQ = 70° then BTC isfn;s x;s fp=k esa TBP vkSj TCQ o`Ùk ftldk dsUnz O gS] dh Li'kZ js[kk,a gS o`Ùk dks

B vkSj C ij Øe'k% Li'kZ djrh gSA PBA = 60º vkSj ACQ = 70° rc BTCgksxk&

(A) 50º (B) 70º (C) 80º (D) 60º

10. If a selling price of Rs. 24 results in a 20% discount of the list price, the selling price that would result in a30% discount of the list price is(A) Rs. 9 (B) Rs. 27 (C) Rs. 14 (D) Rs. 21;fn 24 :i;s foØ; ewY; gksus ij cktkj ewY; ij 20% dh NqV feyrh gS] rc foØ; ewY; D;k gksxk tc cktkj ewY;

ij 30% dh NqV feyrh gS&

(A) 9 : (B) 27 : (C) 14 : (D) 21 :

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11. The value of (a1/8 + a–1/8) (a1/8 – a–1/8) (a1/4 + a–1/4)(a1/2 + a–1/2) is(a1/8 + a–1/8) (a1/8 – a–1/8) (a1/4 + a–1/4)(a1/2 + a–1/2) dk eku gS&

(A) (a + a–1) (B) (a – a–1) (C) (a2 – a–2) (D) (a1/2 – a–1/2)

12. If a, b, c are real and distinct numbers, then the value of )ac(.)cb(.)ba()ac()cb()ba( 333

is

(A) 1 (B) a b c (C) 2 (D) none of these

;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) buesa ls dksbZ ugha

13. In figure, CDEF is a cyclic quadrilateral, DE and CF are produced to Aand B respectively such that AB || CD. If FED = 80°, find FBA.fn;s x;s fp=k esa CDEF ,d pØh; prqHkZqt gS ftlesa DE vkSj CF dks Øe'k% AvkSj B rd bl izdkj c<+k;k tkrk gS tcfd AB || CD ;fn FED = 80° rcFBA Kkr dhft, -(A) 30° (B) 60°(C) 80° (D) None of these (buesa ls dksbZ ugha)

14. A father’s age is equal to the sum of the ages of his 3 children. In 9 years his age will be equal to the sumof the two eldest son’s ages and 3 years after that his age will be equal to the sum of the ages of his eldestand youngest children. Again 3 years after that his age will be equal to the sum of the ages of his twoyoungest children. find the father’s present age ?(A) 32 years (B) 36 years (C) 40 years (D) 44 yearsfirk dh vk;q mlds rhu iq=kksa dh vk;q ds ;ksx ds cjkcj gSA 9 o"kZ ckn mldh vk;q] nks cM+s iq=kkaas dh vk;q ds ;ksx ds

cjkcj gksxh rFkk mlds 3 o"kZ ckn mldh vk;q lcls cM+s vkSj lcls NksVs iq=k dh vk;q ds ;ksxQy ds cjkcj gksxhA iqu%

3 o"kZ ckn mldh vk;q nks NksVs iq=kksa dh vk;q ds ;ksx ds cjkcj gksxhA firk dh orZeku vk;q crkb;saA

(A) 32 o"kZ (B) 36 o"kZ (C) 40 o"kZ (D) 44 o"kZ

15. If 313–1

sincossin–cos

, then acute angle =

;fn 313–1

sincossin–cos

, rks U;wudks.k =

(A) 60º (B) 30º (C) 45º (D) 90º

16. In the given figure, the value of ‘a’ is :fn;s x;s fp=k esa] ‘a’ dk eku crkvks :(A) 170°

P

B

O

A a b

85°

(B) 20°(C) 10°(D) 5°

17. ABCD is a parallelogram, P is a point on AB such that AP : PB = 3 : 2. Q is a point on CD such thatCQ : QD = 7 : 3. If PQ meets AC at R, then AR : AC isABCD ,d lekUrj prqHkqZt gS] AB ij fcUnq P bl izdkj gS fd AP : PB = 3 : 2 gSA CD ij fcUnq Q bl izdkj gS fd

CQ : QD = 7 : 3. ;fn PQ , AC dks R ij feyrk gS] rc AR : AC gS&(A) 5 : 11 (B) 6 : 13 (C) 4 : 7 (D) 2 : 5

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18. Recurring decimal 352.0 , when expressed in the form qp

, is (where p and q are integers and q 0).

vkorhZ n'keyo 352.0 dks qp

ds :i esa O;Dr djus ij cjkcj gS (tgk¡ p rFkk q iw.kkZad gSa rFkk q 0)

(A) 1000235

(B) 990243

(C) 990233

(D) 990223

.

19. The sum of all natural numbers less than 400 which are not divisible by 6, is :400 ls de lHkh izkÑr la[;kvksa dk ;ksx tks fd 6 ls foHkkftr ugh gS] gksxk :(A) 13266 (B) 66534 (C) 79800 (D) 93066

20. Ram sets his watch at 6 : 10 am on Sunday, which gains 12 minutes in a day. On Wednesday if this watchis showing 2 : 50 pm. What is the correct time ?jke viuh ?kMh jfookj dks 6 : 10 am ij feykrk gS] tks fd ,d fnu esa 12 feuV vkxs gks tkrh gSA cq/kokj dks ;fn ;g

?kMh 2 : 50 pm fn[kk jgh gS] rks lgh le; D;k gksxk ?(A) 1 : 50 pm (B) 2 : 10 pm (C) 2 : 30 pm (D) 3 : 30 pm

21. If , be the roots of x2 – a(x –1) + b = 0, then value of a–

12 + a–

12 + ba

2

is

;fn x2 – a(x –1) + b = 0 ds ewy , gks] rks a–

12 + a–

12 + ba

2 dk eku gS &

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

22. A square sheet of paper is converted into a cylinder by rolling it along its length. What is the ratio of thebase radius to the side of the square ?,d oxkZdkj dkxt dks mldh yEckbZ ds vuqfn'k ?kqekdj ,d flysUMj esa :ikUrfjr fd;k tk;sA rks mldh vk/kkj

f=kT;k dk oxZ dh Hkqtk ls vuqikr D;k gksxk ?

(A) 1

2 (B) 2

(C) 12

(D) 1

23. In which of the following case(s) , the equation , a x2 + b x + c = 0 always has imaginary rootsfuEu esa ls dkSuls fodYi lehdj.k a x2 + b x + c = 0 ds lnSo dkYifud ewy gS&

(A) a > 0 , b > 0 , c > 0 (B) a > 0 , b < 0 , c = 0(C) a > 0 , b = 0 , c > 0 (D) a < 0 , b = 0 , c > 0

24. 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 4 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 4 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

25. The expression (5a – 3b)3 + (3b – 7c)3 – (5a – 7c)3 is divisible by :O;atd (5a – 3b)3 + (3b – 7c)3 – (5a – 7c)3 foHkkftr gksxk

(A) (5a + 3b + 7c) (B) (5a – 3b – 7c) (C) (3b – 7c) (D) (7c – 5a)

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26. (2x2 + 3x + 5)1/2 + (2x2 + 3x + 20)1/2 = 15, therefore x is :(2x2 + 3x + 5)1/2 + (2x2 + 3x + 20)1/2 = 15, rc x gS :(A) (–8/3) (B) (14/5) (C) (–11/2) (D) 4

27. 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, then14 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

28. For any real number a, b, c & d which of the following statement(s) is/are true ?(A) If a > b then ac > bc (B) If a > b then a – c > b – c

(C) If a > b > 0 and c > d > 0, then da

> cb

(D) If a > b and c > d then a – c > b – d.

fdUgha okLrfod la[;kvksa a, b, c ,oa d ds fy, fuEufyf[kr esa ls dkSu ls dFku lR; gS \

(A) ;fn a > b gks] rks ac > bc (B) ;fn a > b gks] rks a – c > b – c

(C) ;fn a > b > 0 ,oa c > d > 0 gks] rks da

> cb

(D) ;fn a > b ,oa c > d gks] rks a – c > b – d

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

This section contains 2 paragraphs. Based upon each paragraph, there are 4 questions. Each questionhas 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 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 32In the following figure, the smaller triangle represents the teachers; the big triangle, the politicians; the circle, the graduates and therectangle, the members of parliament. Different regions are beingrepresented by the letters of english alphabet

On the basis of the about diagram, answer the following questions :29. Who among the following are graduates or teachers but not politicians ?

(A) B, G (B) G, H (C) A, E (D) E, F

30. Who among the following politicians are graduates but not the members of parliament ?(A) B, C (B) L, B (C) D, L (D) A, H, L

31. Who among the following politicians are neither teachers nor graduates ?(A) E, F (B) D, E (C) C, D (D) L, H

32. Who among the following members of parliament is a graduate as well as a teacher?(A) G (B) F (C) C (D) H

iz'u 29 ls 32 ds fy, vuqPNsn

fuEu vkys[k esa] NksVk f=kHkqt] v/;kidksa dks çnf'kZr djrk gS rFkk cM+k f=kHkqt]

usrkvksa dks çnf'kZr djrk gSA o`Ùk] LukÙkd dks çnf'kZr djrk gS rFkk vk;r] laln

ds lnL;ksa dks çnf'kZr djrk gSA çR;sd fofHké {ks=kksa dks vaxzsth o.kZekyk ds v{kjksa

}kjk vyx&vyx çnf'kZr fd;k x;k gS&

Åij fn;s x;s fp=k ds vk/kkj ij fuEu ç'uksa dk mÙkj nhft;s&

29. fuEufyf[kr esa ls dkSuls @lk Lukrd ;k v?;kid gS ijUrq usrk ugha gks ?(A) B, G (B) G, H (C) A, E (D) E, F

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30. fuEufyf[kr esa ls dkSuls @lk usrk Lukrd gS ijUrq laln dk lnL; ugha gS ?(A) B, C (B) L, B (C) D, L (D) A, H, L

31. fuEufyf[kr esa ls dkSuls @lk usrk] u rks v/;kid gS vkSj u gh Lukrd gS ?(A) E, F (B) D, E (C) C, D (D) L, H

32. fuEufyf[kr esa ls dkSuls @lk laln dk lnL;] Lukrd gS rFkk v/;kid gS ?(A) G (B) F (C) C (D) H

Paragraph for Question Nos. 33 to 36Four different integers a, b, c and d form an increasing A.P. One of these numbers is equal to thesum of the squares of the other three numbers. Then

33. The smallest number is :(A) – 2 (B) 0 (C) – 1 (D) 2

34. The common difference of A.P. is(A) 2 (B) 1 (C) 3 (D) 4

35. c, d and d2 are in(A) A.P. (B) G.P. (C) H.P. (D) None of these

36. The value of (a + c)2 + (d – 2c)2 is equal to(A) 0 (B) 1 (C) 2 (D) 4


pkj fHkUu iw.kkZ ad a, b, c rFkk d ,d c<+rh gqbZ lekUrj Js.kh cukrs gSA ftuesa ls ,d la[;k vU; rhu la[;kvks a

ds oxks Z a ds ;ksx ds cjkcj gS] rc

33. lcls NksVh la[;k gS :(A) – 2 (B) 0 (C) – 1 (D) 2

34. lekUrj Js.kh dk lkoZvUrj gS

(A) 2 (B) 1 (C) 3 (D) 4

35. c, d rFkk d2 gks axs(A) lekUrj Js.kh (B) xq.kksÙkj Js.kh (C) gjkRed Js.kh (D) buesa ls dksbZ ugha

36. (a + c)2 + (d – 2c)2 dk eku cjkcj gS

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

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 choices (A), (B), (C) and (D), out ofwhich 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

37. A particle is thrown upward at t = 0 sec from ground with a vertical velocity of 100 m/sec.Distance travelledby the particle in 12 sec.: (Take g = 10 m/s2),d d.k Åij dh rjQ t = 0 sec ij tehu ls 100 m/sec.ds Å/okZ/kj osx ls Qsadk tkrk gS rks 12 sec esa d.k }kjk

r; nwjh gksxh: (g = 10 m/s2)(A) 0 m (B) 500 m (C) 480 m (D) 520 m

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38. An electron enters a magnetic field at right angles to it, as shown in figure. The direction of force acting on the electron will be :(A) to the right (B) to the left(C) out of the page (D) into the page,d bysDVªkWu fp=kkuqlkj pqEcdh; {ks=k eas yEcor~ :i ls izos'k djrk gS rks bysDVªkWu

ij vkjksfir cy dh fn'kk gksxh A

(A) nka;h rjQ (B) cka;h rjQ(C) dkxt ds ry ls ckgj dh rjQ (D) dkxt ds ry esa vUnj dh rjQ

39. If the distance between two masses is doubled, the gravitational attraction between them.(A) Is doubled (B) Becomes four times (B) Is reduced to half (D) Is reduced to a quarternks nzO;ekuks ds e/; nwjh nqxquh dj nh xbZ gS rks buds e/; xq:Rokd"kZ.k cy gksxkA

(A) nqxquk (B) pkj xquk (B) vk/kk jg tk;sxk (D) ,d pkSFkkbZ jg tk;sxk

40. A force vector applied on a mass is represented as F i j k 6 8 10 and accelerates with 1 m/s2. Whatwill be the mass of the body-

fdlh fi.M ij yxk;k x;k cy F i j k 6 8 10 gS] rFkk ;g 1 eh@ls-2 ls Rofjr gksrk gS rks fi.M dk nzO;eku D;k gksxk&

(A) 210 kg (B) 102 kg (C) 10 kg (D) 20 kg41. A particle of mass m at rest is acted upon by a force F for a time t. Its kinetic energy after an

interval t is :m nzO;eku dk ,d d.k fojke ij gS bl ij t le; ds fy, F cy vkjksfir jgrk gSA t le;kUrjky ds ckn bldh xfrt

ÅtkZ gS %

(A) mtF 22

(B) m2tF 22

(C) m3tF 22

(D) m2tF

42. The displacement-time relationship for a particle is given by 20 1 2x a a t a t . The acceleration of the

particle is;fn ,d d.k ds foLFkkiu≤ dk lEcU/k 2

0 1 2x a a t a t gks rks mldk Roj.k gksxk&

(A) 0a (B) 1a (C) 2a (D) 22a43. Light ray AB incidents on a plane mirror XY at an angle of 50º from normal. The second plane mirror is

placed is such a way that the reflected ray BC from the mirror XY retraces its path. Angle of inclination oftwo mirrors will be:izdk'k dh fdj.k AB lery niZ.k XY ij vfHkyEc ls 50º ds dks.k ij vkifrr gksrh gSA nwljk lery niZ.k bl izdkj

j[kk tkrk gS fd lery niZ.k XY ls ijkofrZr fdj.k BC nwljs niZ.k ls ijkofrZr gksdj vius iFk ij okil tkrh gSaA

nksuksa niZ.kksa ds e/; ikjLifjd >qdko gSa :(A) 25º (B) 50º (C) 75º (D) 90º

44. A body moves on three quarters of a circle of radius r. The displacement and distance travelled by it are :,d oLrq r f=kT;k ds o`Ùk ij rhu pkSFkkbZ Hkkx pyrh gS] blds }kjk r; fd;k x;k foLFkkiu o nwjh gS :

(A) r, 3r (B) 2r , 2

r3(C) 2r,

2r3

(D) 0, 2

r3

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

45. A beam of green light is incident from air and after refraction enters water. In comparison to that in air :(A) Speed of light is less in water (B) Frequency of light is less in water(C) Wavelength of light is less in water (D) Speed of light is more in waterok;q ls vkifrr gjs izdk'k dh fdj.k] viorZu ds ckn ty esa izos'k djrh gSA ok;q dh rqyuk esa &

(A) ty esa izdk'k dh pky de gksxh (B) ty esa izdk'k dh vko`fÙk de gksxh

(C) ty esa izdk'k dh rjaxnS/;Z de gksxh (D) ty esa izdk'k dh pky T;knk gksxh

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46. A household electric power outlet (assume 220 V constant voltage) is fused to cut at if the current equals orexceeds 20 Ampere. A 2 kW heater, 1kW Air conditioner and three 100 W bulbs are already running atrated power. If now somebody wants to run a computer then computer can run without causing fuse to burnif power requirement of computer is (neglect losses in current carrying wire),d ?kjsyw fo|qr 'kfDr ifjiFk (220 V vpj ekusa) esa ;fn /kkjk 20 A ;k mlls vf/kd gks tk;s rks ¶;wt VwV tkrk gSA ,d

2 kW dk ghVj, 1kW dk ,;j df.M'kuj rFkk rhu 100 W ds cYc mudh vafdr 'kfDr ij dk;Zjr gSA vc ;fn dksbZ

,d dEI;wVj pykuk pkgrk gS rFkk dEI;wVj ¶;wt tyk;s fcuk gh dk;Z dj ldrk gS ;fn dEI;wVj dh vko';d 'kfDr

gksxh& (/kkjkokgh rkj esa gkfu ux.; ysaos)(A) 1000 W (B) 1100 W (C) 100 W (D) 1200 W

47. Which of the following statements is/are true :(A) An electric motor converts mechanical energy into electrical energy.(B) An electric generator works on the principle of electromagnetic induction.

(C) This symbol is used for variable register..

(D) A wire with a green insulation is usually the live wire of an electric supply.fuEu eas ls dkSulk@ls dFku lR; gSA

(A) fo|qr eksVj ;kaf=kd ÅtkZ dks fo|qr ÅtkZ esa ifjofrZr djrh gSA

(B) fo|r tfu=k fo|qr pqEcdh; izsj.k fl)kUr ij dk;Z djrk gSA

(C) bl ladsr dk iz;ksx ifjofrZ izfrjks/k ds fy;s fd;k tkrk gSA

(D) ?kjks esa fo|qr vkiwfrZ ds fy, gjk fo|qr jks/kh rkj ,d /kkjkokgh rkj ds :i esa iz;qDr fd;k tkrk gSA

48. A cuboid block of mass 12 kg is lying on the ground (Assume airis absent). Take g = 10 m/sec.2 :(A) Pressing force applied by the block on the ground is 120 N.(B) If the surface ABCD is lying on the ground, then pressure(stress) exerted by the block on the ground will be 20 Pa.(C) If surface ABEF is lying on the ground, then the pressure(stress) exerted by the block on the ground will be 60 Pa.(D) If we place the block on the ground such that different planesurfaces lie on the ground, pressure (stress) on the ground willbe maximum when surface BCFG lies on the ground.

12 kg nzO;eku dk ?kukHk ds vkdkj dk ,d Bksl CykWd tehu ij j[kk

gqvk gSA g dk eku = 10 m/sec.2 ysaA

(A) CykWd }kjk tehu ij yxk;k x;k ncko cy 120 N gSA(B) ;fn CykWd dks lrg ABCD ds lgkjs tehu ij j[kk gS] rks CykWd }kjk tehu ij yxk;k x;k nkc (izfrcy) 20 Pa gksxkA(C) ;fn CykWd dks lrg ABEF ds lgkjs tehu ij j[kk gS] rks CykWd }kjk tehu ij yxus okyk nkc (izfrcy) 60 Pa gksxkA(D) ;fn ge CykWd dks tehu ij vyx-vyx lery lrgks ds lgkjs j[ks rks tehu ij nkc (izfrcy) vf/kdre rc gksxk

tc CykWd dks lrg BCFG ds lgkjs tehu ij j[ksA

PART - III (Hkkx - III)SECTION - I ([k.M- I)

Straight Objective Type (lh/ks oLrqfu"B izdkj)This section contains 8 multiple choice questions. Each question has choices (A), (B), (C) and (D),out of 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

49. Which has highest e/m ratio ?fuEu esa ls fdldk e/m vuqikr vf/kdre gS \

(A) He2+ (B) H+ (C) He+ (D) H

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50. For a chemical reaction 3X(g) + Y(g) X3 Y(g), the amount of X3 Y at equilibrium is affected by(A) temperature and pressure (B) temperature only(C) pressure only (D) temperature, pressure and catalystfdlh jklk;fud vfHkfØ;k ds fy, 3X(g) + Y(g) X3 Y(g), lkE; ij X3 Y dh ek=kk fuEu }kjk izHkkfor gksxh&

(A) rki vkSj nkc (B) dsoy rki (C) dsoy nkc (D) rki] nkc vkSj mRizsjd

51. 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, –152. The value of Planck’s constant is 6.63 × 10–34 Js. The velocity of light is 3 × 108 m/sec. Which value is

closest 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

53. 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

54. Lucas reagent is used to convert alcohol into alkyl halide. In this rate of reaction with respect to 1º, 2º, 3ºalcohol will be–(A) 1º > 2º > 3º (B) 3º > 2º > 1º (C) 3º = 2º = 1º (D) None of theseY;wdkl vfHkdeZd dk mi;ksx ,Ydksgy dks ,fYdy gSykbM esa cnyus esa gksrk gSA 1º, 2º o 3º ,Ydksgy ds lUnHkZ esa

vfHkfØ;k dh nj gksxhA

(A) 1º > 2º > 3º (B) 3º > 2º > 1º (C) 3º = 2º = 1º (D) buesa ls dksbZ ugha

55. What is the formula of sodium periodate?fuEu esa ls lksfM;e ijvk;ksMsV dk lw=k dkSulk gS\

(A) NaIO4 (B) NaIO3 (C) NaIO2 (D) NaIO

56. IUPAC name of the compound

33

323

CHCH||

CHCHCHCHNHCH is

(A) 2-(N-methylamino)-4-methylpentane (B) N-Methyl-4-methylpentan-2-amine(C) 2-(N-methylamino)-3-isopropylpropane (D) 2-(N-methylamino)-1, 4, 4-trimethylbutane

;kSfxd

33

323

CHCH||

CHCHCHCHNHCH dk IUPAC uke gSA

(A) 2-(N-esfFky,feuks)-4-esfFkyisUVsu (B) N-esfFky-4-esfFkyisUVsu-2-,sehu(C) 2-(N-esfFky,feuks)-3-vkblksizksfiyizksisu (D) 2-(N-esfFky,ehuks)-1, 4, 4-VªkbZesfFkyC;wVsu


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

57. Which of the given particles are -vely charged?(A) proton (B) electron (C) antiproton (D) positronfuEu esa ls dkSuls d.k _.kvkosf'kr gSa\

(A) izksVksu (B) bysDVªkWu (C) ,UVhizksVksu (D) iksftVªkWu

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58. The pair of compounds having the same general formula.fuEu esa ls dkSulk ;qXe leku lkekU; lw=k j[krk gSaA

(A) and (rFkk) (B) and (rFkk)

(C) and (rFkk) (D) and (rFkk) H—CC—CC—H

59. A hydrogen - like atom has ground state binding energy 122.4 eV. Then :(A) its atomic number is 3(B) a photon of 90 eV can excite it to a higher state(C) a 80 eV photon cannot excite it to a higher state(D) None

,d gkbMªkstu leku ijek.kq dh vk| voLFkk dh cU/ku ÅtkZ 122.4 eV gSa rks &(A) bldk ijek.kq Øekad 3 gSA(B) 90 eV okyk QksVkWu bls mPp voLFkk esa mÙksftr dj ldrk gSA

(C) 80 eV okyk QksVkWu bls mPp voLFkk esa mÙksftr ugha dj ldrk gSA

(D) dksbZ ugah

60. Which compound is/are the isomer of 3-Ethyl-2-methylpentane ?dkSulk@dkSuls ;kSfxd 3-,fFky-2-esfFkyisUVsu ds leko;oh gSa \

(A) (B)

(C) (D)

ANSWER KEY TO SAMPLE TEST PAPER-I1. (B) 2. (D) 3. (D) 4. (D) 5. (B) 6. (A) 7. (B)8. (C) 9. (C) 10. (D) 11. (B) 12. (D) 13. (C) 14. (B)15. (A) 16. (D) 17. (B) 18. (C) 19. (B) 20. (B) 21. (B)22. (A) 23. (C) 24. (B) 25. (CD) 26. (CD) 27. (AD) 28. (BC)29. (C) 30. (A) 31. (D) 32. (B) 33. (C) 34. (B) 35. (B)36. (A) 37. (D) 38. (D) 39. (D) 40. (A) 41. (B) 42. (D)43. (B) 44. (B) 45. (AC) 46. (ABC) 47. (BC) 48. (ABD) 49. (B)50. (A) 51. (B) 52. (B) 53. (B) 54. (B) 55. (A) 56. (B)57. (BC) 58. (ABD) 59. (AC) 60. (ACD)

HINTS & SOLUTION TO SAMPLE TEST PAPER-I1. Since DE is angle bisector

AE : EB = DA : DB .... (i)Since DF is angle bisector AF : FC = DA : DC = DA : DB .... (ii)from (i) and (ii) we getAE : EB = AF : FC EF // BC

3. The diagonal d = 25m. and area A = 168m2. Let ‘I’ be the length and ‘b’ be the width of the rectangle.Therefore, I2+b2 = d2. and Ib = A We can therefore write (I + b)2 = d2 + 2A and (I b)2 = d2 2A.Substituting and solving we get, I + b = 31 and I b = 17. Hence I = 24 andb = 7.

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4. x = 7 + 34

x = 32 x1

= 32 So x + x

1 = 2 + 3 + 32 = 4

5. Sn = 2n5

2n3 2

Sn–1 = 23

(n – 1)2 + 25

(n – 1) an = sn – sn–1 = 23

(2n – 1) + 25

a25 = 76

6. In ABC, we have A+ B + C = 180° 60° + B + 50° = 180° 110° + B = 180° B = 180° – 110° B = 70°Since D and E are the mid-points of AB and AC respectively. Therefore,

DE || BC and DE = 12

BC

Now, DE || BC and the transversal AB cuts them at D and B respectively. EDB + B = 180°

Interior angles on thesameside ofthe transversalaresupplementary

EDB + 70° = 180° EDB = 180° – 70° = 110°

7. AB is perp. bisector of PQAPQ = AQPand PBA = QBABut PQ is perpendicular bisector of AB is not always true(It is true only when the circles are of equal radius)

9. ACB = ABP = 60°ABC = ACQ = 70° BAC = 180° – (60° + 70°) = 50° BOC = 100° BTC = 180° – 100° = 80°

10. Selling price = Rs. 24 = 80% of marked price.Marked price = 30

Now selling price = 70% of marked price = 10070

× 30 = 21

11. (a1/8 + a–1/8)(a1/8 – a–1/8)(a1/4 + a–1/4)(a1/2 + a–1/2)= (a1/4 – a–1/4)(a1/4 + a–1/4)(a1/2 + a–1/2)= (a1/2 – a–1/2)(a1/2 + a–1/2)= (a – a–1).

13. FCD = 180° – 80° = 100°[Opposite angles of a cyclic quadrilateral are supplementary]Now AB || CDABC + DCB = 180° [Co-interior angles]FBA + 100° = 180°So. FBA = 80°

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15. We have (gesa fn;k gS )

313–1

sincossin–cos

)sin(cos–)sin–(cos)sin(cos)sin–(cos

= )31(–)3–1()31()3–1(

[Applying componendo and dividendo] [;ksxkUrjkuqikr

dk mi;ksx djus ij]

sin2–cos2

= 32–2

cot = 31

tan= 3 tan= tan60º = 60º

16. AOB = 2 (APB)b = AOB = 2 (85°) = 170°In AOB, b + 2a = 180° (OA = OB)170° + 2a = 180° a = 5°

17.

QRC and PRA are similar

AR : RC = 10/75/3

= 6 : 7

AR : AC = 6 : 13

18. Let x = 352.0

10x = 35.2

1000x = 235.35. + 9935

990x = 233

x = 990233

x = – 23

, y = 23

19. Sum of all natural numbers less than 400

S1 = 1 + 2 + 3 + 4 +......399 =399 400 79800

2

Sum of all natural number divisible by six.S2 = 6 + 12 + 18 +.......396Number of terms 396 = 6 + (n–1)6n = 66

66sum [2 6 (66 1)6]2

S2 = 13266 S1 – S2 = 79800 – 13266 = 66534.

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20. Incorrect watch covers 1452 min in 1440 min. So, it will cover 1 min in 14521440

min. Therefore it will cover

4840 min in 14521440

× 4840 = 4800 min = 80 h. Therefore 80 h = 3 days and 8 h.

21. 2 – a = –a – b2 – a = –a – b

a–

12 + a–

1 + ba

2 = ba

2ba

1–ba

1–

= 0

22. Surface area of the cylinder = surface area of the square 2rh = a2

Here height ‘h’ of the cylinder = side of the square since it is rolled along its length. 2ra = a2.

base radius ar

2

Ratio of base radius to side of square a 1:a

2 2

24. 2x

625625

625625

625 = 23

625 = 23

So, 2x

23232323

2x

2232

23

2x x = 3

25. (5a – 3b)3 + (3b – 7c)3 – (5a – 7c)3

Let 5a – 3b = x, 3b – 7c = y and 7c – 5a = zx3 + y3 + z3 = 3xyz, if x + y + z = 0.x + y + z = 5a – 3b + 3b – 7c + 7c – 5a = 0(5a – 3b)3 + (3b – 7c)3 + (7c – 5a)3

= 3 (5a – 3b) (3b – 7c) (7c – 5a) It is divisible by 5a – 3b, 3b – 7c, 7c – 5a.

26. (2x2 + 3x + 5)1/2 + (2x2 + 3x + 20)1/2 = 15Let 2x2 + 3x + 5 = y2

y + (y2 + 15)1/2 = 15y2 + 15 = (15 – y)2

y2 + 15 = 225 + y2 – 30y

30y = 210210y 7.30

2x2 + 3x + 5 = 492x2 + 3x – 44 = 02x2 – 8x + 11x – 44 = 02x(x – 4) + 11 (x – 4) = 0x = –11/2 & x = 4.

Hence, two solutions x = –11/2 & x = 4 are possible.

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28. a > b if ;fn c < 0 ac < bc so (A) is incorrect. xyr gSA

a > b a – c > b – c so (B) is correct.lgh gS

If ;fn a > b > 0 , c > d > 0 da

> cb

so (C) is correct. lgh gSA

If ;fn a > b , c > d c = 50, d = 2, a = 10, b = 1 a – c = – 40, b – d = – 1

= – 1 1

29 to 32Sol.

29. A,E

30. B,C

31. L,H

32. F

(33-36)Sol. Let four integers be a–d, a , a + d and a + 2d

where a and d are integers and d > 0. a + 2d = (a – d)2 + a2 + (a + d)2

2d2 – 2d + 3a2 – a = 0 ...............(i)

d = 21

2a6a211 ...............(ii)

Since d is positive integer 1 + 2a – 6a2 > 0

6a2 – 2a – 1 < 0

6

71 < a <

671

a is an integer

a = 0 Put in (ii) d = 1 or 0 but d > 0 d = 1 The four numbers are : –1, 0, 1, 2

35. c, d, d2 1, 2, 4 they are in G.P.

36. (– 1 + 1)2 + (2 – 2)2 = 0

37. F 2r1

. If r becomes double then F reduces to 4F

38. mass nzO;eku m = |a||F|

= 1

200 = 10 2

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39. a = mF

, S = 21

2t

mF

, WF = FS = F

m2Ft2

40. x = a0 + a1 t + a2 t2

dtdx

= 0 + a1 + 2a2 t a = 2

2

dtxd

= 0 + 2a2

41. 50º

N

X40

A

B

C

50º

Y

Ray BC retraces its path when A fall on second mirror perpendicularly, so as per figure L BYC will be euqal to50º.

42. S Q

P

R

r f=kT;k ds o`rh; iFk ij 3/4 Hkkx r; djus ij

(i) Displacement (foLFkkiu) = 22 rr = r 2

(ii) Distance (nwjh) = r243 =

2r3

43. We know that v = f f = constant(i) speed of light

v 1

w > air

so vw < air(ii) vso w < air

ge tkurs gS fd v = f f = fu;r(i) izdk'k dh pky

v 1

w > air

vr% vw < air(ii) vvr% w < air

44. Total power supplied dqy fn xbZ 'kfDr = 20 × 220 = 4400 WattAlready existing load tks yksM ij igys ls mifLFkr gS = 2000 + 1000 + 300 = 3300 WattWe can increase load upto 4400 Watt so A, B & C are correct options.ge yksM 4400 Watt rd c<+k ldrs gSA vr% A, B rFkk C lgh fodYi gSA

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45. Based on informationlwpuk ij vk/kkfjr

46. (A) Pressing force ncko cy = mg = 12 g = 120 N

(B) P = AF

= 23

120

= 20 Pa

(C) P = AF

= 13

120

= 40 Pa

(D) Area of surface BCFG is minimum So, pressure will be maximum. lrg BCFG dk {kS=kQy U;qure gS vr% nkc vf/kdre gksxkA

49. For He2+ PM4

e2 ; For H+

PMe

; For He+ PM4

e ; For H

0

Me

gy% He2+ ds fy, PM4

e2 ; H+ ds fy,

PMe

; He+ ds fy,PM4

e ; H ds fy,

0

Me

50. Equilibrium is affected by pressure and temperature but not by catalyst.gy- lkE; ij nkc o rki dk izHkko iMrk gS ysfdu mRizsjd dk ughaA

51. Br show oxidation state varying from –1 to +7.Br, –1 ls +7 rd ifjofrZr gksus okyh vkWDlhdj.k voLFkk n'kkZrk gSA

52. 15

8

108103c

= 3.75 × 10–8 m.

53.

CH3–C–CH2–CH3

O

– Ethyl methyl ketone

Butanone – IUPAC name.

CH3–C–CH2–CH3

O

– ,fFkyesfFkydhVksu

C;wVsukWu – IUPAC uke.

59. Ground state binding energy = 13.6 Z2 = 122.4 eV. Z = 3.1st excitation energy = 10.2 Z2 = 91.8 eV. an 80 eV electron cannot excite it to a higher state.

gy % vk| voLFkk esa cU/ku ÅtkZ = 13.6 Z2 = 122.4 eV. Z = 3.1st mÙkstu ÅtkZ = 10.2 Z2 = 91.8 eV. ,d 80 eV dk QksVkWu bls mPp voLFkk esa mÙksftr ugha dj ldrk gSA

60. ACD are 8 carbon alkanes but B has only 6 carbons atoms.

vU; lHkh esa 8 dkcZu gSa] tcfd B esa dsoy 6 dkcZu gSaA

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Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 25

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

Course : VISHWAAS (JF)

No. of Questions

Correct Wrong Total

1 to 14 Section-IStraight Objective Type Questions(dsoy ,d fodYi lgh) 14 3 -1 42

15 to 19 Section-IIMultiple Choice Questions(,d ;k ,d ls vf/kd fodYi lgh) 5 4 0 20

20 to 23 Section-III Comprehension (vuqPNsn) (2 Comp. x 2 Q.) 4 3 -1 12







Total 69 222

PART-I(Maths)

PART-II(Physics)

PART-III(Chemistry)

SubjectS.No.Marks to be awarded

Nature of QuestionsSection

PART - I (Hkkx - I)

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

This section contains 14 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONLY ONE is correct.bl [k.M esa 14 iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA

1. If |z – i| 2 and z0 = 5 + 3i then the maximum value of |iz + z0| is

(A) 2 + 31 (B) 7

(C) 31 – 2 (D) None of these

;fn |z – i| 2 rFkk z0 = 5 + 3i gS] rks |iz + z0| dk vf/kdre eku gS

(A) 2 + 31 (B) 7

(C) 31 – 2 (D) buesa ls dksbZ ugha

2. Ram sets his watch at 6 : 10 am on Sunday, which gains 12 minutes in a day. On Wednesday if this watchis showing 2 : 50 pm. What is the correct time ?

jke viuh ?kMh jfookj dks 6 : 10 am ij feykrk gS] tks fd ,d fnu esa 12 feuV vkxs gks tkrh gSA cq/kokj dks ;fn ;g

?kMh 2 : 50 pm fn[kk jgh gS] rks lgh le; D;k gksxk ?(A) 1 : 50 pm (B) 2 : 10 pm(C) 2 : 30 pm (D) 3 : 30 pm

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3. If 2576a 456b is divisible by 15 thenS1 : a may take value 5S2 : a may take value 3S3 : a may take value 9S4 : a may take value 8(A) S1 and S2 are true (B) S1 and S3 are true(C) S1 , S2 and S4 are true (D) S1 and S4 are true

;fn 2576a 456b , 15 ls foHkkT; gS rc -S1 : a dk eku 5 gks ldrk gS

S2 : a dk eku 3 gks ldrk gS



(A) S1 vkSj S2 lR; gS (B) S1 vkSj S3 lR; gS

(C) S1, S2 vkSj S4 lR; gS (D) S1 vkSj S4 lR; gS

4. The value of 12 – 22 + 32 – 42 + 52 – 62 + ....+ 992 – 1002 is :12 – 22 + 32 – 42 + 52 – 62 + ....+ 992 – 1002 dk eku gS&

(A) – 100 (B) – 5050 (C) – 2500 (D) – 2520

5. If ax1 = bc, by1 = ca, cz1 = ab, then xy + yz + zx =(A) xyz (B) 3 xyz (C) x+y+z (D) none of these;fn ax1 = bc, by1 = ca, cz1 = ab, rc xy + yz + zx =(A) xyz (B) 3 xyz (C) x+y+z (D) buesa ls dksbZ ugha

6. If cot = 31

, then the value of

2

2

sin2cos1

is

;fn cot = 31

gS] rks

2

2

sin2cos1

dk eku gS &

(A) 51

(B) 52

(C) 53

(D) None dksbZ ugah

7. Total number of solutions of the equation 2)x1( + 2)x1( = 2 is

(A) 2 (B) 3 (C) 4 (D) infinite

lehdj.k 2)x1( + 2)x1( = 2 ds gyksa dh la[;k gS&

(A) 2 (B) 3 (C) 4 (D) vuUr

8. The arithmetic mean of greatest and least integral values of so that (2 + 1, – 1) and origin lie on the sameside of x – 10y – 2 = 0 is(A) 0 (B) 4 (C) 5 (D) none of these;fn ewy fcUnq rFkk fcUnq (2 + 1, – 1) ds js[kk x – 10y – 2 = 0 ds gh vksj fLFkr gks] rks bl izdkj izkIr ds vf/kdre

rFkk U;wure iw.kk±d ekuksa dk lekUrj ek/; gS &

(A) 0 (B) 4 (C) 5 (D) buesa ls dksbZ ugha

9. The equation of circumcircle of an equilateral triangle is x2 + y2 + 2gx + 2fy + c = 0 and one vertex of thetriangle is (1, 1). The equation of incircle of the triangle is(A) 4(x2 + y2) = g2 + f2 (B) 4(x2 + y2) + 8gx + 8fy = (1 g) (1 + 3g) + (1 f) (1 + 3f)(C) 4(x2 + y2) + 8gx + 8fy = g2 + f2 (D) None of theseleckgq f=kHkqt ds ifjxr o`Ùk dk lehdj.k x2 + y2 + 2gx + 2fy + c = 0 gS rFkk f=kHkqt dk ,d 'kh"kZ (1, 1) gSA rcf=kHkqt ds vUro`Ùk dk lehdj.k gS -(A) 4(x2 + y2) = g2 + f2 (B) 4(x2 + y2) + 8gx + 8fy = (1 g) (1 + 3g) + (1 f) (1 + 3f)(C) 4(x2 + y2) + 8gx + 8fy = g2 + f2 (D) buesa ls dksbZ ugha

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10. The exhaustive range of values of ‘a’ such that the angle between the pair of tangents drawn from (a, a) to

the circle x2 + y2 2x 2y 6 = 0 lies in the range ,3

, is

‘a’ ds ekuksa dk fu''ks"kh ifjlj gksxk tcfd fcUnq (a, a) ls o`Ùk x2 + y2 2x 2y 6 = 0 ij [khph xbZ Li'kZ js[kkvksa

ds ;qXeksa ds e/; dks.k dk ifjlj ,3

esa fLFkr gS -

(A) (1, ) (B) (5, 3) (3, 5)

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

11. The variance of 20 observations is 5. If each observations is multiplied by 2, then the new variance of theresulting observations is20 izs{k.kks a dk izlj.k 5 gSA ;fn izR;sd izs{k.k dks 2 ls xq.kk fd;k tkrk gS rc ifj.kkeh izs{k.k dk u;k izlj.k gS -(A) 10 (B) 20 (C) 30 (D) 40

12. Sita has 4 different and simmi has 7 different toys. Number of ways in which they can exchange their toys sothat each keeps her inital number of toys, islhrk 4 rFkk xhrk 7 fHkUu&fHkUu f[kykSus j[krh gS] rks mu rjhdksa dh la[;k D;k gksxhA tcfd nksuksa viuh izkjfHkd f[kykSuksa

dh la[;k dks leku j[krs gq, vius f[kykSuksa dks ,d&nwljs ls cny ysaA

(A) 329 (B) 334 (C) 332 (D) 345

13. In a series of 5 matches in football (A team has equal chances to win, loose or draw a match). Theprobability that a forecast selected at random has exactly 2 correct predictions, is5 QqVcky espks a dh ,d Ja[kyk esa Bhd 2 espks a ds iwokZuqeku lgh gksus dh izkf;drk (tcfd izR;sd esp esa fdlh ny

ds thrus] gkjus o esp Mªk gksus dh leku laHkkouk gS] gksxh&

(A) 243163

(B) 24340

(C) 24360

(D) 24380

14. If , & are the roots of the equation x3 2x 1 = 0 then,

11 +

11

+

11 has the value equal to

(A) zero (B) 1 (C) 4 (D) 1

;fn lehdj.k x3 2x 1 = 0 ds ewy , ,oa gks] rks

11

+

11

+

11

dk eku gS&

(A) 'kwU; (B) 1(C) 4 (D) 1



vf/kd lgh gSA

15. The last term in the binomial expansion of n

3

212

is

8log

3

3

9.31

, then the fifth term is

f}in O;atd

n3

212

dk vfUre in

8log

3

3

9.31

gS, rks ik¡pok in gS&

(A) 10C4 (B) 2.10C4

(C) 10C6 (D) 21

10C4

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16. If 1z 5 12i and 2z 4 then

(A) maximum 1 2z iz 17 (B) minimum 1 2z 1 i z 13 4 2

(C) minimum 1

22

z 134 4zz

(D) maximum

1

22

z 134 3zz

;fn 1z 5 12i vkSj 2z 4 gks rc

(A) vf/kdre 1 2z iz 17 (B) U;wure 1 2z 1 i z 13 4 2

(C) U;wure1

22

z 134 4zz

(D) vf/kdre

1

22

z 134 3zz

17. If nC4 , nC5 ,

nC6 are in A.P, then a value of n is;fn nC4,

nC5 , nC6 lekUrj Js<h esa gks] rks n dk eku gS&

(A) 14 (B) 11 (C) 7 (D) 8

18. If (t2 , 2t) is one end of a focal chord of the parabola y2 = 4x, then which is (are) true

(A) length of the focal chord is 2

t1t

(B) slope of the focal chord is 1–t

t22

(C) mid-point of focal chord is focus(D) focal-chord is angle bisector of tangent and the normal at (t2 , 2t) to the parabola

;fn (t2 , 2t) ijoy; y2 = 4x dh ukHkh; thok dk ,d fljk gks] rks fuEu esa ls dkSulk@ls lR; gSa&

(A) ukHkh; thok dh yEckbZ

2

t1t

gSA

(B) ukHkh; thok dh izo.krk 1–t

t22 gSA

(C) ukHkh; thok dk e/; fcUnq] ukfHk gSA

(D) ijoy; ds (t2 , 2t) ij Li'kZ js[kk vkSj vfHkyEc dk v)Zd ukHkh;thok gSA

19. The vertices of a ABC are A( 5, 2) ; B(7, 6) and C(5, 4) . Then :(A) measure of angle B is /4(B) equation of the altitude drawn from the vertex C has the equation, 3x + 2y 7 = 0(C) orthocentre of the triangle does not lie inside the ABC

(D) distance between centroid and circumcentre of the ABC is 4 13

3f=kHkqt ABC ds 'kh"kZ A( 5, 2) ; B(7, 6) vkSj C(5, 4) gks] rks(A) dks.k B dk eki /4 gSA(B) 'kh"kZ C ls [khaps x;s 'kh"kZ yEc dk lehdj.k 3x + 2y 7 = 0 gSA(C) f=kHkqt dk yEcdsUnz f=kHkqt ABC ds vUnj fLFkr ugha gSA

(D) f=kHkqt ABC ds dsUnz vkSj ifjdsUnz ds e/; dh nwjh 4 13

3 gSA

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This section contains 2 paragraphs. Based upon each paragraph, there are 2 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 2 ç'u gSA çR;sd ç'u ds 4 fodYi


Paragraph for Question Nos. 20 to 21iz'u 20 ls 21 ds fy, vuqPNsn

Consider the function y = 2 |x + 2| – |3 – x| – 3x + 3.Qyu y = 2 |x + 2| – |3 – x| – 3x + 3 fn;k x;k gSA

20. Number of real solutions of the equation y = 0 is(A) 1 (B) 2 (C) 3 (D) infinitelehdj.k y = 0 ds okLrfod gyksa dh la[;k gSa&

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

21. Number of real solutions of the equation y = 4 is(A) 1 (B) 2 (C) 3 (D) infinitelehdj.k y = 4 ds okLrfod gyksa dh la[;k gSa&

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


The binomial coefficients of the 2nd, 3rd and 4th terms in the expansion ofm

5 3log)2x()310(log 10x

10 22

represent respectively the first, third and fifth terms of an A.P..

m5 3log)2x()310(log 10

x10 22

ds izlkj esa nwljs, rhljs rFkk pkSFks inksa ds f}in xq.kkad Øe'k% ,d lekUrj Js<+h ds

izFke] r`rh; rFkk ik¡posa inksa dks iznf'kZr djrk gSA

22. The value of m is equal tom dk eku gS &

(A) 2 (B) 7 (C) 5 (D) 9

23. The number of values of x for which 6th term in the expansion is 21, arex ds ekuksa dh la[;k ftuds fy, izlkj esa 6ok¡ in 21 gks] gS &(A) 1 (B) 3 (C) 0 (D) 2

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This section contains 14 multiple choice questions. Each question has choices (A), (B), (C) and (D), outof which ONLY ONE is correct.bl [k.M esa 14 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA

24. The current i in the circuit of figure is -fp=k ds ifjiFk esa /kkjk i gS &

(A) .amp451

(B) .amp151

(C) .amp101

(D) .amp51

25. An electron enters a magnetic field at right angles to it, as shown in figure.The direction of force acting on the electron will be :

(A) to the right (nka;h rjQ)

(B) to the left (cka;h rjQ)

(C) out of the page (dkxt ds ry ls ckgj dh rjQ)

(D) into the page (dkxt ds ry esa vUnj dh rjQ)

26. Which of the following represents the displacement-time graph of two objects A and B moving with equal velocity?fuEu esa ls dkSulk foLFkkiu le;&oØ leku osx ls xfreku nks oLrqvks a A rFkk B dks O;Dr djrk gS ?

(A)

Dis

plac

emen

t

Time

A

B

(B)

Dis

plac

emen

t

Time

A

B(C)

Dis

plac

emen

t

Time

(D)

Dis

plac

emen

t

Time

A

B

27. A particle moves from position k6j2i3r1

to position k9j13i14r2

under the action of force

Nk3ji4 . The work done by this force will be

,d d.k] cy Nk3ji4 ds v/khu fLFkfr k6j2i3r1

ls fLFkfr k9j13i14r2

rd xfr djrk gSA bl

cy }kjk fd;k x;k dk;Z gksxk &

(A) 100 J (B) 50 J (C) 200 J (D) 75 J

28. Two particles of combined mass M, placed in space with certain separation, are released. Interaction betweenthe particles is only of gravitational nature and there is no external force present. Acceleration of one particlewith respect to the other when separation between them is R, has a magnitude :nks d.k ftudk la;qDr nzO;eku M gS] vkdk'k esa fuf'pr nwjh ij j[ks gSa ] mudks NksM+k tkrk gSA d.kksa ds e/; dsoy

xq:Rokd"kZ.k cy yxrk gS rFkk vU; dksbZ ckà; cy ugh yxrk gSA fdlh ,d d.k dk nwljs d.k ds lkis{k Roj.k dk ifjek.k]

tc muds e/; R nwjh gS] gksxk &

(A) 2R2GM

(B) 2RGM

(C) 2RGM2

(D) not possible to calculate due to lack of information (lwpukvksa dh deh ds dkj.k x.kuk lEHko ugh gSA)

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29. In the figure shown ABCD is a rectangular smooth tube kept fixed in a vertical plane. A particle isprojected from point A to reach point C with some speed. At the corners B and D velocity changes itsdirection by 90º without any change of its magnitude at that corner. If time taken on paths ABC andADC are t1 and t2 respectively, then: (given > b)fp=kkuqlkj ,d vk;rkdkj fpduh ufydk ABCD m/okZ/kj ry es tM+or gSA ,d d.k dks A ls C ds fy, fdlh pky ls ç{ksfir

fd;k tkrk gSA dksus B o D ij dk osx fn'kk esa 90º ls cny tkrk gS] ijUrq eku leku jgrk gSA ;fn iFk ABC rFkk ADCes fy;k x;k le; Øe'k% t1 o t2 gS rks : (fn;k gS > b)

(A) t1 = t2 (B) t1 > t2 (C) t1 < t2 (D) none of these (buesa ls dksbZ ugha)

30. Two point masses of mass m1 and m2 are placed at point A and Brespectively as shown in figure. Point A is the centre of hollow sphereof uniformly distributed total mass m3. Consider only gravitationalinteraction between all masses and neglect other gravitational forces.Select the incorrect alternative.nks fcUnq nzO;eku ftuds nzO;eku m1 rFkk m2 gS] dks Øe'k% fcUnqvksa A rFkk Bij fp=kkuqlkj j[kk x;k gSA fcUnq A ,d [kks[kys xksys ftldk nzO;eku m3¼le:i forfjr½ gS] dk dsUnz gSA dsoy fn;s x;s nzO;eku ds e/; gh xq:okd"kZ.k

gS vU; xq:Rokd"kZ.k cyksa dks ux.; ekfu;sA vlR; dFku pqfu;sA

m1 m2

A Ba

r

m3

(A) Hollow sphere and point mass m1 moves with same acceleration.(B) m1 and m2 moves with same acceleration.(C) Net force on m1 is non-zero(D) Net force on hollow sphere and point mass m1 as a system is equal to force experienced by point mass m2 in magnitude.(A) [kks[kyk xksyk rFkk fcUnq nzO;eku m1 leku Roj.k ls xfr djsxsa

(B) m1 rFkk m2 leku Roj.k ls xfreku gksxsa

(C) m1 ij dqy cy v'kqU; gksxk

(D) [kks[kys xksys rFkk m1 dks ,d fudk; ekurs gq;s bl ij yxus okys dqy cy dk ifjek.k m2 nzO;eku ij yxus

okys dqy cy ds ifjek.k ds cjkcj gksxk&

31. A bullet of mass 10 g is fired with a rifle. The bullet takes 0.003 s to move through its barrel and leaves witha velocity of 300 ms–1. The force exerted on the bullet by the rifle :(A) 103 N (B) 104 N (C) 105 N (D) zero,d cUnwd ls 10 xzke dh xksyh NksM+h tkrh gSA xksyh uyh ls xqtjus esa 0.003 lsd.M ysrh gS rFkk uyh ls 300 eh-/ls- ds

osx ls fudyrh gSA cUnwd }kjk xksyh ij yxk;k x;k cy gS %

(A) 103 U;wVu (B) 104 U;wVu (C) 105 U;wVu (D) 'kwU;

32. A machine gun fires n bullets per second and the mass of each bullet is m. If the speed of bullet is v, then themagnitude of force exerted on the machine gun is :,d e'khuxu çfr lSd.M n xksfy;k¡ nkxrh gS vkSj çR;sd xksyh dk æO;eku m gSA ;fn xksyh dh pky v gS] rks e'khu xuij vkjksfir cy gksxk&

(A) mng (B) gmnv

(C) mnv (D) mnvg

33. The number of electrons contained in the nucleus of 92U235.

(A) 92 (B) 143 (C) 235 (D) ZeroukfHkd 92U

235 esa bySDVªkWuksa dh la[;k gS &

(A) 92 (B) 143 (C) 235 (D) 'kwU;

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34. An unnumbered clock shows time as 4 : 35 in its mirror image. The real time is :(A) 8 : 25 (B) 6 : 25 (C) 7 : 25 (D) None of these,d fcuk vad fy[kh gqbZ ?kM+h dk niZ.k izfrfcEc 4 : 35 le; crkrk gSA rks lgh le; gS %

(A) 8 : 25 (B) 6 : 25 (C) 7 : 25 (D) buesa ls dksbZ ugha

35. A transformer is used to(A) convert DC into AC (B) convert AC into DC(C) obtain the required DC voltage (D) obtain the required AC voltage

VªkalQkeZj dk mi;ksx fd;k tkrk gS

(A) DC dks AC es cnyus esa (B) AC dks DC es cnyus esa

(C) vfHk"B DC oksYVst izkIr djus ds fy, (D) vfHk"B AC oksYVst izkIr djus ds fy,

36. A particle is projected from point 'A' with velocity 2u at an angle of 45º with the horizontal as shownin the figure. It strikes the inclined plane BC at right angle. The velocity of the particle just before thecollision with the inclined is :,d d.k fp=kkuqlkj {kSfrt ds lkFk 45º ds dks.k ij 2u osx ls fcUnq A ls ç{ksfir fd;k tkrk gSA ;g ry BC ij yEcor~

Vdjkrk gSA urry ls VDdj ds Bhd igys d.k dk osx gksxkA

(A) 2

u 3(B)

2u

(C) 32u

(D) u

37. A homogeneous plate PQRST is as shown in figure. The centre of mass of plate lies at midpoint A of

segment QT. Then the ratio of ab

is (PQ = PT = b; QR = RS = ST = a)

,d lekax IysV PQRST fp=kkuqlkj crk;h xbZ gSA IysV dk nzO;eku dsUnz QT Hkkx ds e/; fcUnq A ij gSA rc vuqikr ab

gS &

(PQ = PT = b; QR = RS = ST = a)

(A) 4

13(B)

213

(C) 2

13(D)

413


This section contains 5 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 5 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


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38. An ideal battery of 60 volt is connected with the resistors as shown above. A1 and A2 are two ideal ammetersand V is an ideal volt meter. Then :(A) Reading of (A1) is 2 Amp (B) Reading of (A2) is 1 Amp(C) Reading of (V) is 60 volt (D) Total power consumed by all the resistors is 120 watt

60 oksYV dh vkn'kZ cSVjh çnf'kZr fp=kkuqlkj çfrjks/k ls tqM+h gSA ;fn A1 rFkk A2 nks vkn'kZ vehVj rFkk V vkn'kZ

oksYVehVj gks rks :(A) (A1) dk ikB;kad 2 Amp gSA (B) (A2) dk ikB;kad 1 Amp gSA(C) (V) dk ikB;kad 60 volt gSA (D) lHkh çfrjks/kks a esa dqy 'kfDr O;; 120 watt gSA

39. A household electric power outlet (assume 220 V constant voltage) is fused to cut at if the current equals orexceeds 20 Ampere. A 2 kW heater, 1kW Air conditioner and three 100 W bulbs are already running at ratedpower. If now somebody wants to run a computer then computer can run without causing fuse to burn if powerrequirement of computer is (neglect losses in current carrying wire),d ?kjsyw fo|qr 'kfDr ifjiFk (220 V vpj ekusa) esa ;fn /kkjk 20 A ;k mlls vf/kd gks tk;s rks ¶;wt VwV tkrk gSA ,d

2 kW dk ghVj, 1kW dk ,;j df.M'kuj rFkk rhu 100 W ds cYc muds vafdr 'kfDr ij dk;Zjr gSA vc ;fn dksbZ ,d

dEI;wVj pykuk pkgrk gS rFkk dEI;wVj ¶;wt tyk;s fcuk gh dk;Z dj jgk gS rks dEI;wVj dks vko';d 'kfDr D;k gksxh

(/kkjkokgh rkj esa gkfu ux.; ysaos)(A) 1000 W (B) 1100 W (C) 100 W (D) 1200 W

40. Choose the correct statement(s) among the following :(A) The magnetic force on a stationary charge is always zero.(B) The magnetic line of force around a straight conductor is circular.(C) The magnetic force on a moving charge is responsible for change in its KE.(D) Magnetic force is a central force.fuEu esa ls lgh dFkuksa dk p;u dhft,&

(A) fLFkj vkos'k ij pqEcdh; cy lnSo 'kwU; gksxkA

(B) lh/ks pkyd ds pkjksa vksj pqEcdh; cy js[kk,a o`Ùkkdkj gksxh

(C) xfreku vkos'k ij pqEcd cy bldh xfrt ÅtkZ esa ifjorZu ds fy, mÙkjnk;h gSA

(D) pqEcdh; cy dsfUæ; cy gSA

41. A cuboid block of mass 12 kg is lying on the ground(Assume air is absent). Take g = 10 m/sec2.(A) Pressing force applied by the block on the ground is 120 N.(B) If the surface ABCD is lying on the ground, then pressure (stress)exerted by the block on the ground will be 20 Pa.(C) If surface ABEF is lying on the ground, then the pressure (stress)exerted by the block on the ground will be 60 Pa.(D) If we place the block on the ground such that different planesurfaces lie on the ground, pressure (stress) on the ground will bemaximum when surface BCFG lies on the ground.

12 kg nzO;eku dk ?kukHk ds vkdkj dk ,d Bksl CykWd tehu ij j[kk gqvk gSA g dk eku = 10 m/sec2 ysaA

(A) CykWd }kjk tehu ij yxk;k x;k ncko cy 120 N gSA(B) ;fn CykWd dks lrg ABCD ds lgkjs tehu ij j[kk gS] rks CykWd }kjk tehu ij yxk;k x;k nkc (izfrcy) 20 Pa gksxkA(C) ;fn CykWd dks lrg ABEF ds lgkjs tehu ij j[kk gS] rks CykWd }kjk tehu ij yxus okyk nkc (izfrcy) 60 Pa gksxkA(D) ;fn ge CykWd dks tehu ij vyx-vyx lery lrgks ds lgkjs j[ks rks tehu ij nkc (izfrcy) vf/kdre rc gksxk


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42. Two identical balls P and Q moving in the xy plane collide at the origin (x = 0, y = 0) of the coordinatesystem. Their velocity components just before the moment of impact were, for ball P, v x = 6 m/s,vy = 0 ; for ball Q, vx = 5 m/s, vy = 2 m/sAs a result of the collision, the ball P comes to rest. The velocity components of the ball Q just aftercollision will be:nks le:i xsans P rFkk Q, xy ry esa xfr djrs gq, ewy fcUnq (x = 0, y = 0) ij Vdjkrh gSA VDdj ds rqjUr igys buds

osx ds ?kVd Øe'k% P ds fy, vx = 6 m/s, vy = 0 ,oa Q ds fy, vx = 5 m/s, vy = 2 m/s gSAVDdj ds dkj.k xsan P :d tkrh gSA VDdj ds rqjUr i'pkr~ xsan Q ds osx ds ?kVd gksxsaA

(A) vy = 1 m/s (B) vy = 2 m/s (C) vx = 2 m/s (D) vx = 1 m/s

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

This section contains 2 paragraphs. Based upon each paragraph, 2 multiple choice questions haveto be 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 2 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


Impending state of motion is a critical border line between static and dynamic states of a body. A blockof mass m is supported on a rough vertical wall by applying a force F as shown in figure. Coefficient ofstatic friction between block and wall is S. The block under the influence of F sin may have a tendencyto move upward or it may be assumed that F sin just prevents downward fall of the block. Read theabove passage carefully and answer the following questions.

xfr dh lhekUr voLFkk ,d fi.M dh LFkSfrd rFkk xR;kRed fLFkfr;ks a ds chp ,d ØkfUrd lhek js[kk gksrh gSA mnzO;eku ds ,d CykWd dks ,d ?k"kZ.k;qä Å/okZ/kj nhokj ij fn[kk;s fp=kkuqlkj ,d cy F vkjksfir dj lgkjk fn;k

x;k gSA CykWd rFkk nhokj ds chp LFkSfrd ?k"kZ.k xq.kkad S gSA F sinds vUrZxr CykWd dh Åij dh vksj xfr dh

izo`fÙk gks ldrh gS ;k ;g ekuk tk ldrk gS fd F sin CykWd dks uhps fxjus ls Bhd jksdrk gSA mijksä vuqPNsn

dks lko/kkuhiwoZd if<+;s rFkk fuEu iz'uks a ds mÙkj nhft,A

43. The minimum value of force F required to keep the block stationary is :CykWd dks fLFkj j[kus ds fy, vko';d cy dk U;wure eku gksxk &

(A) cosmg

(B) cossinmg

(C) cossinmg

(D) tanmg

44. The value of F for which friction force between the block and the wall is zero.F dk eku] ftlds fy, CykWd rFkk nhokj ds chp ?k"kZ.k cy 'kwU; gS] gksxk &

(A) mg (B) sin

mg(C)

cosmg

(D) tan

mg

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An ideal gas initially at pressure p0 undergoes a free expansion (expansion against vacuum underadiabatic conditions) until its volume is 3 times its initial volume. The gas is next adiabatically compressedback to its original volume. The pressure after compression is 32/3 p0.,d vkn'kZ xSl izkjfEHkd nkc p0 ls eqä izlkj (fuokZr esa :)ks"e fLFkfr;ks a esa izlkj) djrh gS tc rd fd bldk vk;ru]

izkjfEHkd vk;ru dk 3 xquk gks tkrk gSA vkxs xSl dks :)ks"e :i ls okil ewy vk;ru rd laihfM+r djrs gSaA laihM+u

ds ckn nkc 32/3 p0 gSA

45. The pressure of the gas after the free expansion is :eqä izlkj ds ckn xSl dk nkc gS &

(A) 3p0 (B) 3/1

0p (C) p0 (D) 3p0

46. The gas :(A) is monoatomic. (B) is diatomic.(C) is polyatomic. (D) type is not possible to decide from the given information.

xSl gS &

(A) ,d ijek.kqd (B) f} ijek.kqd(C) cgq ijek.kqd (D) nh xbZ lwpuk ds vk/kkj ij izdkj ugha crk;k tk ldrkA

PART - III (Hkkx - III)


This section contains 14 multiple choice questions. Each question has choices (A), (B), (C) and (D),out of which ONLY ONE is correct.bl [k.M esa 14 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA

47. Which of the following oxidation number is not shown by Xenon (Xe)?fuEu esas ls dkSulk vkWDlhdj.k vad ftukWu (Xe) ugha n’'kkZrk gS\(A) +5 (B) +4 (C) +6 (D) 0

48. The equilibrium constant for the following two reactions are K1 and K2 respectively.fuEu nks vfHkfØ;kvksa ds fy, lkE; fu;rkad Øe'k% K1 rFkk K2 gSaAXe F6 (g) + H2O (g) XeOF4 (g) + 2HF (g)XeO4 (g) + XeF6 (g) XeOF4 (g) + XeO3F2(g)The equilibrium constant for the given reaction is :uhps nh xbZ vfHkfØ;k dk lkE; fu;rkad gS %

XeO4 (g) + 2HF (g) XeO3F4 (g) + H2O (g)(A) K1 K2

2 (B) K1 –K2 (C) K2 / K1 (D) K2 / K2

49. Which of the following elements does not show +4 as most stable oxidation state?fuEu esa ls dkSulk@dkSuls rRo lokZf/kd LFkk;h vkWDlhdj.k voLFkk +4 n'kkZrk gS\(A) Si (B) Ge (C) Sn (D) Pb

50. Bonds present in CuSO4. 5H2O(s) is(A) Electrovalent and covalent (B) Electrovalent and coordinate(C) Electrovalent, covalent and coordinate (D) Covalent and coordinateCuSO4. 5H2O(s) esa mifLFkr ca/k gS(A) fo|qr la;kstd o lgal;kstd (B) fo|qrla;kstd o milgl;kstd

(C) fo|qrl;kstd ] lgl;kstd o milgl;kstd (D) lgal;kstd o milgla;kstd

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51. Which one of the following statements is correct in relation to ionization enthalpy ?(A) Electron-electron interaction in the outer orbitals of the elements increases the ionization enthalpy.(B) Removal of electron from orbitals bearing higher ‘n’ value is easier than from orbitals having lower ‘n’ value.(C) The ionization enthalpies of iso-electronic species, F–, Ne and Na+ are same because their size andelectron configuration are same.(D) End of valence electrons is indicated by a small jump in ionization enthalpy.vk;uu ,UFkSYih ds lUnHkZ esa fuEu esa ls dkSulk dFku lgh gS\

(A) rRoksa ds ckáre d{kksa esa bysDVªkWu&bysDVªkWu vUr% fØ;k] vk;uu ,UFkSYih c<+krh gSA

(B) mPp ‘n’ eku okys d{kdksa ls bysDVªkWu dks fudkyuk fuEu ‘n’ eku okys d{kdksa dh vis{kk ljy gksrk gSA

(C) lebysDVªkWfud Lih'kht F–, Ne rFkk Na+ dh vk;uu ,UFkSYih dk eku leku gksrk gS D;ksafd budk vkdkj o bysDVªkWfud

vfHkfoU;kl leku gksrk gSA

(D) vfUre la;ksth bysDVªkWu vk;uu ÅtkZ esa NksVs vUrj dks crkrk gSA

52. For the reaction in acidic medium between KMnO4 and H2O2 the number of electrons transferred per mol ofH2O2 are :(A) one (B) two (C) three (D) fourKMnO4 o H2O2 ds chp vEyh; ek/;e esa vfHkfØ;k ds fy, H2O2 ds çfr eksy LFkkukUrfjr bysDVªkWuksa dh la[;k fuEu gS%

(A) ,d (B) nks (C) rhu (D) pkj

53. The correct IUPAC name of the given compound is :(A) 5-Amino-3-bromo-4-chloro-2-hydroxycyclopentane-1-carbonitrile(B) 1-Amino-3-bromo-2-chloro-4-hydroxycyclopentane-5-carbonitirle(C) 2-Amino-4-bromo-3-chloro-5-hydroxycyclopentane-1-carbonitirle(D) 2-Amino-3-chloro-3-bromo-5-hydroxycyclopentane-1-nitrile

fn, x;s ;kSfxd dk lgh IUPAC uke gS %(A) 5-,ehuks-3-czkseks-4-Dyksjks-2-gkbMªkWDlhlkbDyksisUVsu-1-dkcksZukbVªkby

(B) 1-,ehuks-3-czkseks-2-Dyksjks-4-gkbMªkWDlhlkbDyksisUVsu-5-dkcksZukbVªkby

(C) 2-,ehuks-4-czkseks-3-Dyksjks-5-gkbMªkWDlhlkbDyksisUVsu-1-dkcksZukbVªkby

(D) 2-,ehuks-3-Dyksjks-3-czkseks-5-gkbMªkWDlhlkbDyksisUVsu-1-ukbVªkby

54. Name of some compounds are given. Which one is not as per IUPAC system ?

(A) CH – CH – CH – CH – CH – CH – CH3 2 2 2 3

|

|CH CH2 3

CH3

3–Methyl–4–ethyl heptane

(B) 3–Methyl–2–butanol

(C) 2–Ethyl–3–methyl–but–1–ene

(D) 4–Methyl–2–pentyne

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uhps dqN ;kSfxdksa ds uke fn;s x;s gSaA dkSulk uke IUPAC i)fr ds vuqlkj ugha gS\

(A) CH – CH – CH – CH – CH – CH – CH3 2 2 2 3

|

|CH CH2 3

CH3

3–esfFky–4–,fFky gsIVsu

(B) 3–esfFky–2–C;wVsukWy

(C) 2–,fFky–3–esfFky–C;wV–1–bZu

(D) 4–esfFky–2–isUVkbu

55. In photochemical reaction

R–H + X2 h R–X + HX

Where X = (F, Cl, Br, I)Rate of reaction of which halide is fastest(A) F2 (B) Cl2 (C) Br2 (D) I2

izdk'k jklk;fud vfHkfØ;k esas

R–H + X2 h R–X + HX

tgk¡ X = (F, Cl, Br, I)dkSuls gSykbM dh vfHkfØ;k rhoz gksxh\

(A) F2 (B) Cl2 (C) Br2 (D) I2

56. Which of the following atom(s) does not have maximum oxidation state as +7?fuEu esa ls dkSulk@dkSuls ijek.kq +7 vf/kdre vkWDlhdj.k voLFkk ugha j[krs gS\

(A) Cl (B) F (C) I (D) Br

57. Which of the following equations is a balanced one-fuEu esa ls ,d larqfyr lehdj.k dkSulh gS \

(A) 5BiO3– + 22H+ + Mn2+ 5Bi3+ + 7H2O + MnO4

–

(B) 5BiO3– + 14H+ + 2Mn2+ 5Bi3+ + 7H2O + 2MnO4

–

(C) 2BiO3– + 4H+ + Mn2+ 2Bi3+ + 2H2O + MnO4

–

(D) 6BiO3– + 12H+ + 3Mn2+ 6Bi3+ + 6H2O + 3MnO4

–

58. Which of the following are polar ?fuEu esa ls dkSu /kzqoh; v.kq gS \

(A) XeF4 (B) SF6 (C) XeOF4 (D) XeF5–

59. Match Column-I with Column-II and select the correct answer using the codes given below :Column-I (Metals) Column-II (Ores)(A) Tin (p) Calamine(B) Zinc (q) Cassiterite(C) Iron (r) Cerrusite(D) Lead (s) SideriteCodes :

(A) (B) (C) (D) (A) (B) (C) (D)(A) p q r s (B) q p s r(C) s r q p (D) q p r s

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dkWye-I dks dkWye-II ds lkFk lqesfyr dhft;s rFkk fn;s x;s dksM ds vuqlkj lgh mÙkj nhft;sA

dkWye-I (/kkrq) dkWye-II (v;Ld)(A) fVu (p) dSykekbu

(B) ftad (q) dSlhVsjkbV

(C) vk;ju (r) ls:lkbV

(D) ysM (s) flMsjkbV

dksM :(A) (B) (C) (D) (A) (B) (C) (D)

(A) p q r s (B) q p s r(C) s r q p (D) q p r s

60. C6H12 (P) has two types of alkenes that can be reduced to one type of C6H14 (Q). Q is:C6H12 (P) nks izdkj dh ,Ydhu j[krk gS tks fd ,d izdkj ds C6H14 (Q) esa vipf;r gksrk gSA vr% Q gksxk %

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


This section contains 5 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 5 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


61. Which is/are correctly matched with its common name ?

(A)

OH

NO2

O2N NO2 : Picric acid (B)

COOH

COOCH3 : Aspirin

(C)

COOH

OH : Salicylic acid (D)

COOH

COOH : Phthalic acid

fuEu esa ls dkSu&dkSu buds lkekU; uke ds lkFk lgh lqesfyr gSa \

(A)

OH

NO2

O2N NO2 : fifØd vEy (B)

COOH

COOCH3 : ,Lizhu

(C)

COOH

OH : lsfylhfyd vEy (D)

COOH

COOH : FkSfyd vEy

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62. The pair of compounds having the same general formula.

(A) and (B) and

(C) and (D) and H—CC—CC—H

fuEu esa ls dkSulk ;qXe leku lkekU; lw=k j[krk gSaA

(A) rFkk (B) rFkk

(C) rFkk (D) rFkk H—CC—CC—H

63. A gas cylinder containing cooking gas can withstand a pressure of 14.9 atmosphere. The pressure guaze ofcylinder indicates 12 atmosphere at 27 ºC. Due to sudden fire in the building temperature starts rising. Thetemperature at which cylinder will explode is :[kkuk idkus okyh xSl dk flys.Mj 14.9 ok;qe.Myh; nkc lgu dj ldrk gSA 27ºC ij flys.Mj dk nkc xSt

¼nkcekih½ 12 ok;qe.My bafxr djrk gS, fcfYMax esa vpkud vkx yxus ls rkieku c<+ tkrk gS, og rki Kkr djsa ftl ij

flys.Mj QV tk;sxk&

(A) 372.5 K (B) 99.5 ºC (C) 199 ºC (D) 472.5 k

64. Which of the following is/are aromatic compounds :

fuEu esas ls dkSulk@dkSuls ;kSfxd ,jkseSfVd ;kSfxd gS\

(A)

N(B)

N| H

(C)

O (D)

65. A hydrogen - like atom has ground state binding energy 122.4 eV. Then :(A) its atomic number is 3(B) a photon of 90 eV can excite it to a higher state(C) a 80 eV photon cannot excite it to a higher state(D) None,d gkbMªkstu leku ijek.kq dh vk| voLFkk dh cU/ku ÅtkZ 122.4 eV gSa rks &(A) bldk ijek.kq Øekad 3 gSA(B) 90 eV okyk QksVkWu bls mPp voLFkk esa mÙksftr dj ldrk gSA

(C) 80 eV okyk QksVkWu bls mPp voLFkk esa mÙksftr ugha dj ldrk gSA

(D) dksbZ ugah


This section contains 2 paragraphs. Based upon each paragraph, 2 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 2 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

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Paragraph for Question Nos. 66 to 67

If hydrogen atoms (in the ground state) are passed through an homogeneous magnetic field, the beamis split into two parts. This interaction with the magnetic field shows that the atoms must have magneticmoment. However, the moment cannot be due to the orbital angular momentum since = 0. Hence onemust assume existence of intrinsic angular momentum, which as the experiment shows, has only twopermitted orientations.

Spin of the electron produces angular momentum equal to s =

2h)1s(s where s = +

21

.

Total spin of an atom = 2nor

2n

where n is the number of unpaired electron.The substance which contains species with unpaired electrons in their orbitals behave as paramagneticsubstances. The paramagnetism is expressed in terms of magnetic moment.

The magnetic moment of an atom

s = mc2

eh)1s(s

= mc2eh1

2n

2n

s = 2n

s = )2n(n B.M.n – number of unpaired electrons

1. B.M. (Bohr magneton) = mc4eh

If magnetic moment is zero the substance is di-magnetic.


;fn gkbMªkstu ijek.kqvksa ¼vk| voLFkk esa½ dks leku pqEcdh; {ks=k ls xqtkjk tkrk gS rks iqUt nks Hkkxksa esa foHkkftr gks tkrk

gSA pqEcdh; {ks=k ds lkFk vUrj fØ;k ;g crkrh gS fd ijek.kqvksa dk pqEcdh; vk?kw.kZ gksuk pkfg,A fQj Hkh vk?kw.kZ d{kh;

pqEcdh; vk?kw.kZ ds dkj.k ugha gks ldrk D;ksafd = 0 gSA bl izdkj ,d vkUrfjd dks.kh; vk?kw.kZ dh mifLFkfr ekuh tkrh

gSA ;g iz;ksxkRed :i ls iznf'kZr gksrk gS fd ;g nks rjg dk pØ.k j[krk gSA

bysDVªkWu ds ?kw.kZu ls mRiUu dks.kh; laosx s =

2h)1s(s tgka s = +

21

gksrk gSA

,d ijek.kq dk dqy pØ.k = 2n

vFkok 2n

tgka n v;qfXer bysDVªkWu dh la[;k gSA

rRo ftlesa d{kd esa v;qfXer bysDVªkWu ik;s tkrs gSa og vuqpqEcdh; inkFkZ dh rjg O;ogkj djrk gSA pqEcdRo dks pqEcdh;

vk?kw.kZ ds inksa esa O;Dr fd;k tkrk gSA

,d ijek.kq dk pqEcdh; vk?kw.kZ

s = mc2

eh)1s(s

= mc2eh1

2n

2n

s = 2n

s = )2n(n B.M.

n – v;qfXer bysDVªkWu dh la[;k

1. B.M. (cksjesXusVku) = mc4eh

;fn inkFkZ dk pqEcdh; vk?kw.kZ 'kwU; gS rc rRo izfrpqEcdh; gSA

66. If an ion of 25Mn has a magnetic moment of 3.873 B.M. Then Mn is in which state.;fn vk;u 25Mn dk pqEcdh; vk?kw.kZ 3.873 B.M. gS] rks Mn fdl vkWDlhdj.k voLFkk esa gksxk

(A) + 2 (B) + 3 (C) + 4 (D) + 5

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67. Which of the following is a paramagnetic substance.fuEu esaa ls dkSulk ,d vuqpqEcdh; inkFkZ gS \

(A) Mg2+ (B) Cu+ (C) Mn+7 (D) Ti+2


The key concepts of resonance are :Resonance occurs because of the overlaping of orbitals. Double bonds are made up of pi bonds, formed fromthe overlap of 2p orbitals. The electrons in these pi orbitals will be spread over more than two atoms, andhence are delocalized. Both paired and unshared electrons may be delocalized, but all the electrons mustbe conjugated in a pi system. If the orbitals do not overlap (such as in orthogonal orbitals) the structures arenot true resonance structures and do not mix. Molecules or species with resonance structures are generallyconsidered to be more stable than those without them. The delocalization of the electrons lower the orbitalenergies, imparting this stability. The resonance in benzene gives rise to the property of aromaticity. The gainin stability is called the resonance energy. All resonance structures for the same molecule must have thesame sigma framework (sigma bond form from the "head on" overlap of hybridized orbitals). Furthermore,they must be correct Lewis structures with the same number of electrons (and consequent charge) as wellas the same number of unpaired electrons. Resonance structures with arbitrary separation of charge areunimportant, as are those with fewer covalent bonds. These unimportant resonance structures only contributeminimum (or not at all) to the overall. From the above theory of resonance answer the followings.


vuqukn dh eq[; vo/kkj.kk fuEu gSa :vuqukn d{kdksa ds vfrO;kiu ds dkj.k gksrk gSA f}cU/k] cU/kksa ls cus gksrs gaSA tks fd 2p d{kdksa ds vfrO;kiu ls curs gSaA

bu d{kdksa ds bysDVªkWu nks ;k vf/kd ijek.kqvksa ds e/; foLrkfjr gks tkrs gSa] bl izdkj foLFkkuhd`r gksrs gaSA

;qfXer rFkk v;qfXer nksuksa izdkj ds bysDVªkWu foLFkkuhdr gks ldrs gSa fdUrq lHkh bysDVªkWu iz.kkyh esa la;qfXer gksuk pkfg,A

;fn d{kd vfrO;kiu ugha djrs gSa] (tSls fd vkFkksZxksuy d{kd½ rks lajpuk;sa lgh vuquknh lajpuk;sa ugh gksrh vkSj ijLij

fefJr ugha gksrhA v.kq ,oa ,slh Lih'kht (species), ftudh vuquknh lajpuk;sa gksrh gSa lkekU;r% fcuk vuquknh lajpuk okys

Lih'kht dh vis{kk vf/kd LFkk;h ekuh tkrh gSaA bysDVªkWuksa dk foLFkkuhdj.k d{kdksa dh ÅtkZ dks de dj nsrk gS tks iz.kkyh

dks LFkkf;Ro iznku djrh gSSA cSathu esa vuqukn bls ,jkseSfVdrk dk xq.k iznku djrh gSaA LFkkf;Ro dh miyfC/k dks vuquknh

ÅtkZ (resonance energy) dgrs gSaA ,d v.kq dh lHkh vuquknh lajpuk;sa leku flXek ÝseodZ (frame work) j[krh gSvkxs Hkh ;g dg ldrs gSa fd leku bysDVªkWu la[;k rFkk leku v;qfXer bysDVªkWu ds lkFk yqbZl lajpuk dk fuekZ.k gksrk

gSA vuquknh lajpuk vkSj vkos'k dk izFkDdj.k egRoghu gSa] tSlk fd cgqr FkksM+s lgla;ksth ca/k esa gksrk gSA ;s egRoghu

vuquknh lajpuk;as cgqr de ;ksxnku nsrh gaS ¼;k fcYdqy ugh½ mi;qZDr vuquknh fl)kUr ds vk/kkj ij fuEu iz'uksa ds mÙkj

nhft,A

68. The correct resonating structure of 1, 3-butadiene is -1, 3-C;wVkMkbbZu dh lgh vuquknh lajpuk gS -

(A) 2

–

2 CHCH—HC—HC

(B) 22

–CHCH—HC—HC

(C) 22

–HC—CHCH—HC

(D) None of these (buesa ls dksbZ ugha)

69. Which resonating structure is NOT CORRECT?dkSulh vuquknh lajpuk lgh ugh gS\

(A) 22 CHCH—HC

(B)

2

22

NH|

HNC—NH

(C) (D)

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ANSWER KEY TO SAMPLE TEST PAPER-II1. (B) 2. (B) 3. (D) 4. (B) 5. (A) 6. (C) 7. (D)8. (C) 9. (B) 10. (D) 11. (B) 12. (A) 13. (D) 14. (C)15. (AC) 16. (ABD) 17. (AC) 18. (AB) 19. (ABC) 20. (A) 21. (D)22. (B) 23. (D) 24. (C) 25. (D) 26. (C) 27. (A) 28. (B)29. (C) 30. (B) 31. (A) 32. (C) 33. (D) 34. (C) 35. (D)36. (C) 37. (D) 38. (ABD) 39. (ABC) 40. (AB) 41. (ABD) 42. (BD)43. (B) 44. (B) 45. (A) 46. (A) 47. (A) 48. (C) 49. (D)50. (C) 51. (B) 52. (B) 53. (C) 54. (A) 55. (A) 56. (B)57. (B) 58. (C) 59. (B) 60. (B) 61. (ACD) 62. (ABD) 63. (AB)

64. (AD) 65. (AC) 66. (C) 67. (D) 68. (C) 69. (D)

HINTS & SOLUTION TO SAMPLE TEST PAPER-II

1. |iz + z0| = |i(z – i) – 1 +5 + 3i| = |i(z – i) + 4 + 3i| |i| | z – i| + | 4 + 3i | 1 . 2 + 5 = 7


min. Therefore it will cover 4840

min in 14521440


Hindi. xyr pyrh ?kMh 1440 feuV dk le; 1452 feuV esa r; djrh gSA vr% ;g 1 feuV esa 14521440

feuV dk le; r; djsxh

vr% ;g 4840 feuV dk le; 14521440

× 4840 = 4800 feuV = 80 h. Therefore 80 h = 3 days and 8 h.

3. 2576a456b is divisible by 5 b = 0 or 5(2576a456b, 5 ls HkkT; gS b = 0 ;k 5)2576a456b is divisible by 3 35 + a + b = 3, I(2576a456b, 3 ls HkkT; gS 35 + a + b = 3, I) a = 1, 4, 7, 2, 5, 8.

4. (12 – 22) + (32 – 42) + (52 – 62) + ....+ (992 – 1002)= (1 + 2) (1 – 2) + (3 + 4)(3 – 4) + (5 + 6) (5 – 6) +.....+ (99 + 100)(99 – 100)= – 3 – 7 – 11..........199a = – 3, d = – 4 and = – 199 = a + (n – 1) d– 199 = – 3 + (n – 1)(– 4)– 199 + 3 = (n – 1)(– 4)

n – 1 = 4

196

= 49

n = 50

S50 = 2

50[– 3– 199]

= 25 [– 202] = – 5050.

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5. ax1 = bc, by1 = ca, cz1 = abThis ax = abc = by = cz = K .........(1)Taking log, x log a = log K,

or, log K log K log Kx , y , zlog a log b log c

xy + yz + zx

2 1 1 1(log K)(loga)(l og b) (log b)(logc) (log c)(log a)

= 2(log K) [log a log b log c]

(log a)(log b)(log c)

zlog K log K log(abc) (log c )xylog a log b log c log c

logcxy. z xyzlogc

6. cot = 31

= PB

H = 22 BP H = 22 )1()3( = 4 = 2.

So, cos = HB

= 21

and sin = 23

HP So,

2

2

sin2cos1

=

432

411

45

43

= 53

.

7. |1 + x| + | 1– x| = 2 x [– 1, 1]So, infinite solutions vr% vuUr gy gSA

8. Since (0, 0) and (2 + 1, – 1) lie on the same side of x – 10y – 2 = 0 –2 (2 + 1 – 10 ( – 1) – 2) > 02 – 10 + 9 < 01 < < 9least integral value of is 2greatest integral value of is 8

A.M. = 2

82 = 5

9. In an equilateral triangle incentre and circumcentre are the same point. Incentre = (g, f) also, 12 + 12 + 2g + 2f + c = 0 c = 2(g + f + 1)also circumradius = 2 (inradius)

inradius = 2 21 g f c2

The equation of the incircle is

2 2 2 2 2 21 1 1x g y f g f c g f .2 g f 14 4 4

4(x2 + y2) + 8gx + 8fy = (1 g) (1 + 3g) + (1 f) (1 + 3f)

10. The centre of the circle is (1, 1) and radius = 2 2 .Point (a, a) musty lie outside the circle, so 2a2 4a 6 > 0 a < 1 or a > 3.

Now, 2

2 2tan2 2a 4a 6

.

2 2

(a,a)

As 3 6 2 2

2

2

2 2 1 a 2a 3 2 332a 4a 6

a2 2a 15 < 0 3 < a < 5 a (3, 1) (3, 5).

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11. Let the observations be x1, x2, . . . . . . . , x20.

Variance = (S.(D))2 =

20

x–x20

1i

2i

20

x–x20

1i

2i

= 5 or

20

1i

2i x–x = 100 ...(i)

New observations are 2x1, 2x2, . . . . . , 2x20.Let y and x be respectively the A.M.s of new and given observations.

x220

x.....xx2

20x2.....x2x2

y 20212021

New variance =

20

x–x4

20

x2–x2

20

y–x220

1i

2i

20

1i

2i

20

1i

2i

= 4 ×

20

x–x20

1i

2i

= 4

20100

= 20 [Using (i)]

12. Required number of ways are = Exactly one toy is exchanged + exactly two toys exchanged + exactly 3toys exchanged + 4 toys exchanged= 4C1

7C1 + 4C2 7C2 + 4C3

7C3 + 4C4 7C4 = 329

dqy rjhdks dh la[;k = ,d f[kykSuk cnyus ds rjhdksa dh la[;k + nks f[kykSus cnyus ds rjhdksa dh la[;k + rhu f[kykSus

cnyus ds rjhdksa dh la[;k + pkj f[kykSus cnyus ds rjhdksa dh la[;kA

= 4C1 7C1 + 4C2

7C2 + 4C3 7C3 + 4C4

7C4 = 329

13. P = 5C2 2

31

3

32

= 24380

14. x3 – 2x – 1 = 0then (rks) 3 – 2 – 1 = 0 .............(1)

Let (ekukfd) y11

1y1y

from equation (1), we get (lehdj.k (1) ls) 3

1y1y

– 2

1y1y

– 1 = 0

y3 + 4y2 – y = 0 is the equation whose roots are

11

,

11

,

11

(lehdj.k y3 + 4y2 – y = 0 ds ewy

11

,

11

,

11

gSaA)

then (rks)

11

11

11

= – 4

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15. Tr+1 = r

rn3r

n

21)2(C

r = n

Tn+1 = nn

03n

n

21

21)2(C

n

21

=

8log

321

8log

3

33

39.3

1

= 8log

35– 3

3 = 8–5/3 = 2–5 = 521

= 10

21

=

10

21

n

21

=

10

21

n = 10

T4+1 =10C4. 4103 )2(

4

21

=

22

410

21.2.C = 10C4

16. 1 2z 5 12i, z 4

1 2 1 2z iz z z 13 4 17

1 2 1 2z 1 i z z 1 i z

13 4 2

1 2min z 1 i z 13 4 2

2 22 2

4 4z z 4 1 5z z

2 22 2

4 4z z 4 1 3z z

1

22

z 13max4 3zz

and min

1

22

z 134 5zz

17. 2 . nC5 = nC4 + nC6

!)5–n(!5)!n(2

= !)4–n(!4!n

+ !)6–n(!6!n

)5–n(52

= )5–n)(4–n(1

+ 561

12(n – 4) = n2 – 9n + 50n2 – 21n + 98 = 0(n – 7) (n – 14) = 0n = 7 or ;k n = 14

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19. (A) tan B =

32.51

325

=

3133

13

= 1

B = 45º

(B) Slope of DC = – 23

DC line y + 4 = – 23

(x – 5)

3x + 2y – 7 = 0(C) Since D is obtuse angled triangle so orthocentre lies out side the triangle

20 to 21.Sol. For Q.No. 20, 21

at x = – 3, y = 8at x = – 2, y = 4at x = 3, y = 4at x = 4, y = 2

20. Number of real solution of equation y = 0 is one21. Number of real solution of equation y = 4 is infinite

22. mC1 ; mC2 ,

mC3 are in A.P. m = 7

HIndi mC1 ; mC2 ,

mC3 lekUrj Js<+h esa gSA

m = 7

23. T6 = 7C5 )310log( x

2 . 3log)2x( 102 = 21 log10(10 – 3x) + log103

x – 2 = 0 log (10 – 3x) (3x – 2) = 0 (10 – 3x) 3x – 2 = 1

910

. 3x – 91

32x = 1 32x – 10 . 3x + 9 = 0

3x = 1, 9 x = 0, 2

24. This simplified circuit is shown in the figure.bldk ljyhÑr ifjiFk fp=k esa fufnZ"V gSA

3030

30

2V

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= 30 602V

= 202V

i

i

Therefore, current vr% /kkjk i = 202

= 101

A

27. W = F . )r–r( 12 = 100 J

28.

a1 = 221

RMGM

/ M1 a2 = 221

RMGM

/ M2

acceleation of M1 w.r.t. M2arel. = a1 + a2

= 221

R)MM(G

= 2RGM

.

30.1ma

=

3ma

0

1ma

= – 2ma

0

231 mmm FF

= 0

31. Given (fn;k gqvk gS) : m = 10 × 10–3 kg (fdxzk-)v = 300 ms–1 (eh-/ls-), t = .003 s(ls-)

F = t

mv

F = 003.

3001010 3

F = 1000 = 103 N

36. Let the particle strikes the inclined palne BC perpendicularly with a velocity v. Thenekuk d.k urry BC ls v osx ls yEcor~ Vdjkrk gS rc

v cos30º = 2u cos45º v = 32u

.

37. If centre of mass is at AnzO;eku dsUnz A ij gSA

a2 2a

= 21

ab sin 31

b sin

or4

13ab

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39. Total power supplied dqy fn xbZ 'kfDr

= 20 × 220 = 4400 WattAlready existing load tks yksM ij igys ls mifLFkr gS

= 2000 + 1000 + 300 = 3300 WattWe can increase load upto 4400 Watt so A, B & C are correct options.ge yksM 4400 Watt rd c<+k ldrs gSA vr% A, B rFkk C lgh fodYi gSA

40. BVqF

If V = 0 then BF

= 0

By right hand rule, the field line is circular.Magnetic force cannot do work It cannot change kinetic energyMagnetic force is perpendicular to the line joining current elements not a central force.

BVqF

;fn V = 0 rc BF

= 0

nka;s gkFk ds fu;e ls {ks=k js[kk o`Ùkkdkj gksxhA

pqEcdh; cy] dk;Z ugh dj ldrk gSA

;g xfrt ÅtkZ ifjofrZr ugh dj ldrkA

pqEcdh; cy /kkjk vo;o dks feykus okyh js[kk ds yEcor~ gSA

dsfUnz; cy ugha gSA


(B) P = AF

= 23

120

= 20 Pa

(C) P = AF

= 13

120

= 40 Pa


43. F sin + f = mgand rFkk Fcos = Nfor minimum U;wure ds fy, ; f = N = Fcos

Fmin. = cossinmg

44. As pwafd f = 0 F sin = mg

F = sinmg

45. In free expansion, temperature of the gas remains constant, thereforeeqDr çlkj esa xSl dk rki fLFkj jgrk gSA vr%

p0 v0 = p. 3v0 where tgkW v0 = initial volume. izkjfEHkd vk;ru

p = 3p0

46. For adiabatic compression, initial conditions are 3p0 and 3v0. Final volume and pressure arev0 and 32/3

p0.

:}ks"e laihMu ds fy, çkjfEHkd fLFkfr;k¡ 3p0 rFkk 3v0 gSaA vfUre vk;ru rFkk nkc v0 rFkk 32/3 p0. gSA

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3p0 .(3v0) = 32/3 p0(v0) 3–1 = 32/3

or – 1 = 32

= 35

i.e. gas is monoatomicxSl ,d ijek.kqd gSA

47. +5 is not shown by Xenon.ftukWu +5 vkWDlhdj.k vad ugha n'’kkZrk gSA

48. Substracting 1st reaction from second, we will get desired reaction3rd. So K = K2/K1

f}rh; ls 1st dks dks ?kVkus ij ge bfPNr vfHkfØ;k çkIr djsxsaA

3rd. So K = K2/K1

51. As value of ‘n’ i.e. principal quantum number increases, the size of atom increases and the distance betweenvalence electrons and nucleus increases. As a result, the attraction between the valence electrons andnucleus decreases. This fascilitates the easier removal of valence electrons.‘n’ dk eku vFkkZr~ eq[; DokUVe la[;k esa o`f) ds lkFk ijek.kq ds vkdkj esa o`f) gksrh gS rFkk la;ksth bysDVªkWu o ukfHkd

ds e/; nwjh c<+rh gSA ifj.kkeLo:i la;ksth bysDVªkWuksa o ukfHkd ds e/; vkd"kZ.k ?kVrk gS rFkk la;ksth bysDVªkWuks dks fudkyuk

ljy gksrk gSA

52. Valency factor of H2O2 in both the medium is two.nksuksa ek/;e esa H2O2 dk la;kstdrk xq.kkad nks gSA

53.

2-Amino-4-bromo-3-chloro-5-hydroxycyclopentane-1-carbonitirle2-,ehuks-4-czkseks-3-Dyksjks-5-gkbMªkWDlhlkbDyksisUVsu-1-dkcksZukbVªkby

54. CH – CH – CH – CH – CH – CH – CH3 2 2 2 3

|

|CH CH2 3

CH3

4–Ethyl –3–methyl–heptane

CH – CH – CH – CH – CH – CH – CH3 2 2 2 3

|

|CH CH2 3

CH3

4–,fFky –3–esfFkygsIVsu

57.

Reduction

BiO3– + Mn2+ Bi3+ + MnO4

–

Oxidation

(i) 2e + 6H+ + BiO3– Bi3+ + 3H2O

(ii) 4H2O + Mn2+ MnO4– + 8H+ + 5e

–––––––––––––––––––––––––––––––––––––––––(i) × 5 + (ii) × 2 we get14 H+ + 5 BiO3

– + 5Mn2+ 5Bi3+ + 2MnO4– + 7 H2O

is the correct balanced reaction

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Sol.

vi p; u

BiO3– + Mn2+ Bi3+ + MnO4

–

vkWDl hdj .k

(i) 2e + 6H+ + BiO3– Bi3+ + 3H2O

(ii) 4H2O + Mn2+ MnO4– + 8H+ + 5e

–––––––––––––––––––––––––––––––––––––––––(i) × 5 + (ii) × 2 ge izkIr djrs gS

14 H+ + 5 BiO3– + 5Mn2+ 5Bi3+ + 2MnO4

– + 7 H2Olgh larqfyr vfHkfØ;k gSA

58. Polarity depend on net dipole moment. If diple moment = 0 it is nonpolar.

(A) XeF4 = 0 non polar

(B) SF6 is regular octahedral SF6 =0 non-polar

(C) XeOF4 0 polar

(D)

F

FXe

F

F

F

XeF5– 0 non polar

gy& /kqzoh;rk dqy f}/kqzo vk?kw.kZ ij fuHkZj djrh gSA ;fn f}/kqzo vk?kw.kZ = 0 gS rks ;g v/kqzoh; gSA

(A) XeF4 = 0 v/kqzoh;

(B) SF6 v"VQydh; lajpuk gSA SF6 =0 v/kqzoh;

(C) XeOF4 0 /kqzoh;

(D)

F

FXe

F

F

F

XeF5– 0 v/kqzoh;

59. (A) Tin-cassiterite (SnO2) (B) Zinc - calamine (ZnCO3)(C) Iron - siderite (FeCO3) (D) Lead - Cerrusite (PbCO3)Therefore, (B) option is correct.

(A) fVu&dSlhVsjkbV (SnO2) (B) ftad&dSykekbu (ZnCO3)(C) vk;ju&flMsjkbV (FeCO3) (D) ysM&ls:lkbV (PbCO3)blfy;s (B) lgh fodYi gSA

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60.

63. Suppose the cylinder will burst at T2Kekuk fd flys.Mj T2K rki ij QV tkrk gSA

1

122 P

TPT (V1 = V2) = 123009.14

= 372.5 K

65. Ground state binding energy = 13.6 Z2 = 122.4 eV. Z = 3.1st excitation energy = 10.2 Z2 = 91.8 eV. an 80 eV electron cannot excite it to a higher state.vk| voLFkk esa cU/ku ÅtkZ = 13.6 Z2 = 122.4 eV. Z = 3.1st mÙkstu ÅtkZ = 10.2 Z2 = 91.8 eV. ,d 80 eV dk QksVkWu bls mPp voLFkk esa mÙksftr ugha dj ldrk gSA

66. Magnetic moment = )2n(n = 3.873 number of unpaired electron n = 3

25Mn – [Ar] 3d5 4s2 therefore Mn should be in + 4

pqEcdh; vk?kw.kZ = )2n(n = 3.873 v;qfXer bysDVªkWuksa dh la[;k n = 3

25Mn – [Ar] 3d5 4s2 bl izdkj Mn, + 4 esa gksxkA

67. There is two unpaired electron in Ti2+

;gk¡ Ti2+ esa nks v;qfXer bysDVªkWu gSaA

68.

69.

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Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 52

SAMPLE TEST PAPER-III(For Class-XII Appearing / Passed Students)

Course : VISHESH (JD) & VIJAY (JR)

No. of Questions

Correct Wrong Total










Total 69 222

PART-I(Maths)

PART-II(Physics)

PART-III(Chemistry)

SubjectS.No.Marks to be awarded

Nature of QuestionsSection

PART - I (Hkkx - I)

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

This section contains 14 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONLY ONE is correct.bl [k.M esa 14 iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA

1. Let x y1 – y x1 = 0, where –1 < x < 0, for this relation find the sum of squares of all values of dxdy

at

x = –21

.

ekukfd x y1 – y x1 = 0, tgk¡ –1 < x < 0, bl lEcU/k ds fy, x = – 21

ij dxdy

ds lHkh ekuksa ds oxksZa dk ;ksx gS&

(A) – 8 (B) – 4 (C) 1 (D) – 1

2. The area bounded by the curve y = x3 – 4x and x-axis isoØ y = x3 – 4x rFkk x-v{k ls ifjc) {ks=kQy gS&

(A) 4 (B) 8 (C) 16 (D) none of these (bueas ls dksbZ ugha)

3. In a series of 5 matches in football (A team has equal chances to win, loose or draw a match). Theprobability that a forecast selected at random has exactly 2 correct predictions, is5 QqVcky espks a dh ,d Ja[kyk esa Bhd 2 espks a ds iwokZuqeku lgh gksus dh izkf;drk (tcfd izR;sd esp esa fdlh ny

ds thrus] gkjus o esp Mªk gksus dh leku laHkkouk gS] gksxh&

(A) 243163

(B) 24340

(C) 24360

(D) 24380

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4. If a1 b1 c1, a2b2c2 and a3b3c3 are 3-digit even natural numbers and = 333

222

111

bacbacbac

, then is

;fn a1 b1 c1, a2b2c2 ,oa a3b3c3 , 3-vadh; le izkd`r la[;k,¡ gS rFkk = 333

222

111

bacbacbac

, rks

(A) divisible by 2 but not necessarily by 4 (B) divisible by 4 but not necessarily by 8(C) divisible by 8 (D) none of these(A)2 ls HkkT; gS fdUrq 4 ls HkkT; gksuk vko';d ugha gSA (B) 4 ls HkkT; gS fdUrq 8 ls HkkT; gksuk vko';d ugha gSA

(C) 8 ls HkkT; gSA (D) buesa ls dksbZ ugha

5. Domain of definition of the function f(x) = 2x–1–x is

Qyu f(x) = 2x–1–x dk izkUr gSA

(A)

1 ,

21

(B)

21,1

1,

21

(C)

21,

21

(D)

21,

,

21

6. If , & are the roots of the equation x32x 1= 0 then,

11 +

11

+

11 has the value equal to

;fn lehdj.k x3 2x 1 = 0 ds ewy , ,oa gks] rks

11

+

11

+

11

dk eku gS&

(A) zero ('kwU;) (B) 1 (C) 4 (D) 1

7. If x, y > 0, x + y = 60, then greatest value of x(y – 30)2 is;fn x, y > 0, x + y = 60 gS] rks x(y – 30)2 dk vf/kdre eku gS &

(A) 4000 (B) 54000 (C) does not exist (D) none of these(A) 4000 (B) 54000 (C) fo|eku ugha gSA (D) buesa ls dksbZ ugha

8. Sita has 4 different and simmi has 7 different toys. Number of ways in which they can exchange their toysso that each keeps her inital number of toys, islhrk 4 rFkk xhrk 7 fHkUu&fHkUu f[kykSus j[krh gS] rks mu rjhdksa dh la[;k D;k gksxhA tcfd nksuksa viuh izkjfHkd f[kykSuksa

dh la[;k dks leku j[krs gq, vius f[kykSuksa dks ,d&nwljs ls cny ysaA

(A) 329 (B) 334 (C) 332 (D) 345

9. The exhaustive range of values of ‘a’ such that the angle between the pair of tangents drawn from (a, a) to the

circle x2 + y2 2x 2y 6 = 0 lies in the range ,3

, is

‘a’ ds ekuksa dk fu''ks"kh ifjlj gksxk tcfd fcUnq (a, a) ls o`Ùk x2 + y2 2x 2y 6 = 0 ij [khph xbZ Li'kZ js[kkvksa

ds ;qXeksa ds e/; dks.k dk ifjlj ,3

esa fLFkr gS -

(A) (1, ) (B) (5, 3) (3, 5)

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

10. The equation of circumcircle of an equilateral triangle is x2 + y2 + 2gx + 2fy + c = 0 and one vertex of thetriangle is (1, 1). The equation of incircle of the triangle is(A) 4(x2 + y2) = g2 + f2

(B) 4(x2 + y2) + 8gx + 8fy = (1 g) (1 + 3g) + (1 f) (1 + 3f)(C) 4(x2 + y2) + 8gx + 8fy = g2 + f2

(D) None of these

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leckgq f=kHkqt ds ifjxr o`Ùk dk lehdj.k x2 + y2 + 2gx + 2fy + c = 0 gS rFkk f=kHkqt dk ,d 'kh"kZ (1, 1) gSA rcf=kHkqt ds vUro`Ùk dk lehdj.k gS -(A) 4(x2 + y2) = g2 + f2

(B) 4(x2 + y2) + 8gx + 8fy = (1 g) (1 + 3g) + (1 f) (1 + 3f)(C) 4(x2 + y2) + 8gx + 8fy = g2 + f2

(D) buesa ls dksbZ ugha

11. The tangent lines to the circle x² + y² + 6x – 4y – 12 = 0 which are perpendicular to the line4x + 3y + 5 = 0 are given by :o`Ùk x² + y² + 6x – 4y – 12 = 0 dh Li'kZ js[kkvks a tks js[kk 4x + 3y + 5 = 0 ds yEcor~ gks] ds lehdj.k gS&

(A) 4x + 3y 7 = 0 , 4x + 3y + 15 = 0 (B) 4x + 3y 31 = 0 , 4x + 3y + 19 = 0(C) 3x – 4y 42 = 0 , 3x – 4y – 8 = 0 (D) 3x – 4y + 8 = 0 , 3x + 4y + 21 = 0

12.

2/1242

4

)1xx(x1x

dx is equal to :

2/1242

4

)1xx(x1x

dx dk eku gS&

(A) x

1xx 24 + C (B) 22

x11x + C (C) 2

24

x1xx

+ C (D) 4

24

x1xx

13. Ram sets his watch at 6 : 10 am on Sunday, which gains 12 minutes in a day. On Wednesday if this watchis showing 2 : 50 pm. What is the correct time ?jke viuh ?kMh jfookj dks 6 : 10 am ij feykrk gS] tks fd ,d fnu esa 12 feuV vkxs gks tkrh gSA cq/kokj dks ;fn ;g

?kMh 2 : 50 pm fn[kk jgh gS] rks lgh le; D;k gksxk ?(A) 1 : 50 pm (B) 2 : 10 pm (C) 2 : 30 pm (D) 3 : 30 pm

14. A pair of coins is tossed a fixed number of times. If probability of getting both head in exactly 3 trails is sameas the probability of getting both heads in exactly 4 trials, then number of trails equal toflDdks dk ,d ;qXe ,d fuf'pr la[;k ckj mNkyk tkrk gSA ;fn Bhd rhu mNkyksa esa nksuksa ij fpÙk vkus dh izkf;drk]

Bhd pkj mNkyksa esa nksuksa ij fpÙk vkus dh izkf;drk cjkcj gks] rks dqN mNkyksa dh la[;k gksxh&

(A) 7 (B) 15 (C) 21 (D) 14



vf/kd lgh gSA

15. 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

16. If 1z 5 12i and 2z 4 then

(A) maximum 1 2z iz 17 (B) minimum 1 2z 1 i z 13 4 2

(C) minimum 1

22

z 134 4zz

(D) maximum

1

22

z 134 3zz

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;fn 1z 5 12i vkSj 2z 4 gks rc

(A) vf/kdre 1 2z iz 17 (B) U;wure 1 2z 1 i z 13 4 2

(C) U;wure1

22

z 134 4zz

(D) vf/kdre

1

22

z 134 3zz

17. The differential equation of the curve for which the ordinate at x = 0 of any tangent is equal to the correspond-ing subnormal(A) is linear (B) is homogeneous of first degree(C) is of the form variable separable (D) is of second order

oØ] ftlds fy, fdlh Li'kZ js[kk dh x = 0 ij dksfV laxr v/kksyEc ds cjkcj gS] dk vody lehdj.k

(A) js[kh; gSA (B) izFke ?kkr dk le?kkrh; gSA

(C) pj i`FkDdj.kh; :i dk gSA (D) f}rh; dksfV dk gSA

18. If a is a positive integer, then the value of 0x

lim 3x

)axsin–ax(6 may be equal to

;fn a ,d /kukRed iw.kk±d gS] rks 0x

lim 3x

)axsin–ax(6 cjkcj gks ldrk gS &

(A) 1 (B) 8 (C) 27 (D) 100

19. A point on the ellipse x2 + 3y2 = 37 where the normal is parallel to the line 6x – 5y = 2 isnh?kZo`Ùk x2 + 3y2 = 37 ij fcUnq ftl ij vfHkyEc] js[kk 6x – 5y = 2 ds lekUrj gS] gksxk&

(A) (5, – 2) (B) (5, 2) (C) (– 5, 2) (D) (– 5, – 2)


This section contains 2 paragraphs. Based upon each paragraph, there are 2 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 2 ç'u gSA çR;sd ç'u ds 4 fodYi



Equation of a curve is defined implicitly by the equation y3 – 3y + x = 0. The curve is passing through the point

22,210P . Then

,d oØ dh lehdj.k y3 – 3y + x = 0 }kjk iznf'kZr dh tkrh gSA ;g oØ fcUnq 22,210P ls xqtjrh gS] rks

20. Number of points where the curve has vertical tangents aremu fcUnqvksa dh la[;k ftu ij oØ dh Li'kZ js[kk,a m/okZ/kj gS&

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

21. 2

2

dxyd

at the point P : fcUnq P ij 2

2

dxyd

:

(A) 233724

(B) 233724

(C) 3724

3 (D) 3.724

2

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Let us say that a real number is divisible by two non–zero real numbers a and b, if it is divisible by the leastamong the common integral multiples of a and b.In case there is no common integral multiple of a and b, then say that a number is divisible by both a andb if it is divisible by the product ab.ekuk fd ,d okLrfod la[;k] nksa v'kwU; okLrfod la[;kvksa a o b ls HkkT; gS] ;fn ;g nksuks a o b ds U;wure

mHk;fu"V iw.kk±d xq.kt ls HkkT; gksxhA

;fn a o b dk mHk;fu"V iw.kk±d xq.kt ugha gks] rks la[;k a o b nksuksa ls HkkT; gksxh ;fn ;g xq.kuQy ab ls HkkT; gksA

22. The number of integers in {1, 2, ....... 100}, which are divisible by 2 and 3 but not by 12, is(A) 8 (B) 9 (C) 10 (D) None of these{1, 2, ....... 100} esa iw.kkZadks dh la[;k tksfd 2 o 3 ls HkkT; gS ysfdu 12 ls ugha] gS&


23. The number of numbers in 200.......,8,6,4,2 , which are divisible by both 2 and 8 , is(A) 5 (B) 10 (C) 25 (D) None of these

200.......,8,6,4,2 esa la[;kvksa dh la[;k tksfd 2 o 8 nksauks ls HkkT; gS] gS&




This section contains 14 multiple choice questions. Each question has choices (A), (B), (C) and (D), outof which ONLY ONE is correct.bl [k.M esa 14 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA

24. The current i in the circuit of figure is -fp=k ds ifjiFk esa /kkjk i gS &

(A) .amp451

(B) .amp151

(C) .amp101

(D) .amp51

25. An electron enters a magnetic field at right angles to it, as shown in figure.The direction of force acting on the electron will be :,d bysDVªkWu fp=kkuqlkj pqEcdh; {ks=k eas yEcor~ :i ls izos'k djrk gS rks bysDVªkWu ij vkjksfir cy dh fn'kk gksxh A

(A) to the right (nka;h rjQ)

(B) to the left (cka;h rjQ)

(C) out of the page (dkxt ds ry ls ckgj dh rjQ)

(D) into the page (dkxt ds ry esa vUnj dh rjQ)

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26. Which of the following represents the displacement-time graph of two objects A and B moving with equal velocity ?fuEu esa ls dkSulk foLFkkiu le;&oØ leku osx ls xfreku nks oLrqvks a A rFkk B dks O;Dr djrk gS ?

(A)

Dis

plac

emen

t

Time

A

B

(B)

Dis

plac

emen

t

Time

A

B(C)

Dis

plac

emen

t

Time

(D)

Dis

plac

emen

t

Time

A

B

27. Which of the following is not possible for an object of mass 5kg which is released from height 15m.(Take g = 10 m/s2) KE = Kinetic Energy of the object, PE = Potential Energy),d oLrq dks 15 m ÅpkbZ ls eqDr fd;k tkrk gS oLrq dk nzO;eku 5kg gS rks fuEu esa ls dkSulk vlR; gSA

(;gk¡ g = 10 m/s2) KE = oLrq dh xfrt ÅtkZ, PE = fLFkfrt ÅtkZ½

(A) KE = 600 J ; PE = 150 J (B) KE = 150 J ; PE = 600 J(C) KE = 0 J ; PE = 750 J (D) KE = 800 J ; PE = – 50 J

28. Two particles of combined mass M, placed in space with certain separation, are released. Interaction betweenthe particles is only of gravitational nature and there is no external force present. Acceleration of one particlewith respect to the other when separation between them is R, has a magnitude :nks d.k ftudk la;qDr nzO;eku M gS] vkdk'k esa fuf'pr nwjh ij j[ks gSa ] mudks NksM+k tkrk gSA d.kksa ds e/; dsoy

xq:Rokd"kZ.k cy yxrk gS rFkk vU; dksbZ ckà; cy ugh yxrk gSA fdlh ,d d.k dk nwljs d.k ds lkis{k Roj.k dk ifjek.k]

tc muds e/; R nwjh gS] gksxk &

(A) 2R2GM

(B) 2RGM

(C) 2RGM2

(D) not possible to calculate due to lack of information (lwpukvksa dh deh ds dkj.k x.kuk lEHko ugh gSA)

29. In the figure shown ABCD is a rectangular smooth tube kept fixed in a vertical plane. A particle isprojected from point A to reach point C with some speed. At the corners B and D velocity changes itsdirection by 90º without any change of its magnitude at that corner. If time taken on paths ABC andADC are t1 and t2 respectively, then: (given > b)

fp=kkuqlkj ,d vk;rkdkj fpduh ufydk ABCD m/okZ/kj ry es tM+or gSA ,d d.k dks A ls C ds fy, fdlh pky ls ç{ksfir

fd;k tkrk gSA dksus B o D ij dk osx fn'kk esa 90º ls cny tkrk gS] ijUrq eku leku jgrk gSA ;fn iFk ABC rFkk ADCes fy;k x;k le; Øe'k% t1 o t2 gS rks : (fn;k gS > b)

(A) t1 = t2(B) t1 > t2(C) t1 < t2

(D) none of these (mijksDr esa lss dksbZ ugha)

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30. Two point masses of mass m1 and m2 are placed at point A and B respectively as shown in figure. Point A isthe centre of hollow sphere of uniformly distributed total mass m3. Consider only gravitational interactionbetween all masses and neglect other gravitational forces. Select the incorrect alternative.(A) Hollow sphere and point mass m1 moves with same acceleration.(B) m1 and m2 moves with same acceleration.(C) Net force on m1 is non-zero(D) Net force on hollow sphere and point mass m1 as a system is equal to force experienced by point mass m2 in magnitude.

m1 m2

A Ba

r

m3

nks fcUnq nzO;eku ftuds nzO;eku m1 rFkk m2 gS] dks Øe'k% fcUnqvksa A rFkk B ij fp=kkuqlkj j[kk x;k gSA fcUnq A ,d [kks[kys

xksys ftldk nzO;eku m3 ¼le:i forfjr½ gS] dk dsUnz gSA dsoy fn;s x;s nzO;eku ds e/; gh xq:okd"kZ.k gS vU;

xq:Rokd"kZ.k cyksa dks ux.; ekfu;sA vlR; dFku pqfu;sA

(A) [kks[kyk xksyk rFkk fcUnq nzO;eku m1 leku Roj.k ls xfr djsxsa

(B) m1 rFkk m2 leku Roj.k ls xfreku gksxsa

(C) m1 ij dqy cy v'kqU; gksxk

(D) [kks[kys xksys rFkk m1 dks ,d fudk; ekurs gq;s bl ij yxus okys dqy cy dk ifjek.k m2 nzO;eku ij yxus

okys dqy cy ds ifjek.k ds cjkcj gksxkA

31. A bullet of mass 10 g is fired with a rifle. The bullet takes 0.003 s to move through its barrel and leaves witha velocity of 300 ms–1. The force exerted on the bullet by the rifle :(A) 103 N (B) 104 N (C) 105 N (D) zero

,d cUnwd ls 10 xzke dh xksyh NksM+h tkrh gSA xksyh uyh ls xqtjus esa 0.003 lsd.M ysrh gS rFkk uyh ls 300 eh-/ls- ds

osx ls fudyrh gSA cUnwd }kjk xksyh ij yxk;k x;k cy gS %

(A) 103 U;wVu (B) 104 N U;wVu (C) 105 U;wVu (D) 'kwU;

32. A machine gun fires n bullets per second and the mass of each bullet is m. If the speed of bullet is v, then themagnitude of force exerted on the machine gun is : ,d e'khuxu çfr lSd.M n xksfy;k¡ nkxrh gS vkSj çR;sd xksyh

dk æO;eku m gSA ;fn xksyh dh pky v gS] rks e'khu xu ij vkjksfir cy gksxk&

(A) mng (B) gmnv

(C) mnv (D) mnvg

33. The number of electrons contained in the nucleus of 92U235.

(A) 92 (B) 143 (C) 235 (D) ZeroukfHkd 92U

235 esa bySDVªkWuksa dh la[;k gS &

(A) 92 (B) 143 (C) 235 (D) 'kwU;

34. An unnumbered clock shows time as 4 : 35 in its mirror image. The real time is :(A) 8 : 25 (B) 6 : 25 (C) 7 : 25 (D) None of these,d fcuk vad fy[kh gqbZ ?kM+h dk niZ.k izfrfcEc 4 : 35 le; crkrk gSA rks lgh le; gS %

(A) 8 : 25 (B) 6 : 25 (C) 7 : 25 (D) buesa ls dksbZ ugha

35. A transformer is used to(A) convert DC into AC (B) convert AC into DC(C) obtain the required DC voltage (D) obtain the required AC voltageVªkalQkeZj dk mi;ksx fd;k tkrk gSA

(A) DC dks AC es cnyus esa (B) AC dks DC es cnyus esa

(C) vfHk"B DC oksYVst izkIr djus ds fy, (D) vfHk"B AC oksYVst izkIr djus ds fy,

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36. Two non-conducting hemispherical surfaces, which are having uniform charge density are placed on smoothhorizontal surface as shown in figure. Assuming springs are ideal, calculate compression in each spring ifboth the hemispherical surface are just touching each other.nks vpkyd v/kZ&xksyh; i`"B] ftu ij le vkos'k ?kuRo gS] dks fpdus {kSfrt ry ij fp=kkuqlkj j[kk gSA fLizaxks dks vkn'kZ

ekurs gq,] izR;sd fLizax es laihMu Kkr dhft, ;fn nksuksa v/kZxksyh; i`"B ijLij ek=k laifdZr gSA

(A) KR

2

2

0

2

(B) R (C) KR

2

2

0

2

(D) K2R

0

22

37. The radius of a coil of wire with N turns is 0.22 m, and 3.5 A current flows clockwise in the coil asshown. A long straight wire carrying a current 54A toward the left is located 0.05 m from the edge of thecoil. The magnetic field at the centre of the coil is zero tesla. The number of turns N in the coil are :N pDdj okyh ,d dq.Myh dh f=kT;k 0.22 m gS vkSj blesa fp=kkuqlkj nf{k.kkoÙkZ (clockwise) fn'kk esa /kkjk çokfgrgks jgh gSA ,d yEcs lh/ks rkj tks dq.Myh ds fdukjs ls 0.05 m nwjh ij dq.Myh ds ry esa gS] esa ck;ha rjQ 54 A/kkjk çokfgr gks jgh gSA dq.Myh ds dsUnz ij pqEcdh; {ks=k 'kwU; gSA dq.Myh esa ?ksjks a dh la[;k N gS &

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


This section contains 5 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 5 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


38. An ideal battery of 60 volt is connected with the resistors as shown above. A1 and A2 are two ideal ammetersand V is an ideal volt meter. Then :(A) Reading of (A1) is 2 Amp (B) Reading of (A2) is 1 Amp(C) Reading of (V) is 60 volt (D) Total power consumed by all the resistors is 120 watt

60 oksYV dh vkn'kZ cSVjh çnf'kZr fp=kkuqlkj çfrjks/k ls tqM+h gSA ;fn A1 rFkk A2 nks vkn'kZ vehVj rFkk V vkn'kZ

oksYVehVj gks rks :(A) (A1) dk ikB;kad 2 Amp gSA (B) (A2) dk ikB;kad 1 Amp gSA(C) (V) dk ikB;kad 60 volt gSA (D) lHkh çfrjks/kks a esa dqy 'kfDr O;; 120 watt gSA

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39. A household electric power outlet (assume 220 V constant voltage) is fused to cut at if the current equals orexceeds 20 Ampere. A 2 kW heater, 1kW Air conditioner and three 100 W bulbs are already running at ratedpower. If now somebody wants to run a computer then computer can run without causing fuse to burn if powerrequirement of computer is (neglect losses in current carrying wire),d ?kjsyw fo|qr 'kfDr ifjiFk (220 V vpj ekusa) esa ;fn /kkjk 20 A ;k mlls vf/kd gks tk;s rks ¶;wt VwV tkrk gSA ,d

2 kW dk ghVj, 1kW dk ,;j df.M'kuj rFkk rhu 100 W ds cYc muds vafdr 'kfDr ij dk;Zjr gSA vc ;fn dksbZ ,d

dEI;wVj pykuk pkgrk gS rFkk dEI;wVj ¶;wt tyk;s fcuk gh dk;Z dj jgk gS rks dEI;wVj dks vko';d 'kfDr D;k gksxh

(/kkjkokgh rkj esa gkfu ux.; ysaos)(A) 1000 W (B) 1100 W (C) 100 W (D) 1200 W

40. Choose the correct statement(s) among the following :(A) The magnetic force on a stationary charge is always zero.(B) The magnetic line of force around a straight conductor is circular.(C) The magnetic force on a moving charge is responsible for change in its KE.(D) Magnetic force is a central force.fuEu esa ls lgh dFkuksa dk p;u dhft,&

(A) fLFkj vkos'k ij pqEcdh; cy lnSo 'kwU; gksxkA

(B) lh/ks pkyd ds pkjksa vksj pqEcdh; cy js[kk,a o`Ùkkdkj gksxh

(C) xfreku vkos'k ij pqEcd cy bldh xfrt ÅtkZ esa ifjorZu ds fy, mÙkjnk;h gSA

(D) pqEcdh; cy dsfUæ; cy gSA

41. A cuboid block of mass 12 kg is lying on the ground(Assume air is absent). Take g = 10 m/sec2. :12 kg nzO;eku dk ?kukHk ds vkdkj dk ,d Bksl CykWd tehu

ij j[kk gqvk gSA g dk eku = 10 m/sec2. ysaA

(A) Pressing force applied by the block on the ground is 120 N.(B) If the surface ABCD is lying on the ground, then pressure (stress) exerted by the block on the ground will be 20 Pa.(C) If surface ABEF is lying on the ground, then the pressure (stress) exerted by the block on the ground will be 60 Pa.(D) If we place the block on the ground such that different plane surfaces lie on the ground, pressure (stress)on the ground will be maximum when surface BCFG lies on the ground.(A) CykWd }kjk tehu ij yxk;k x;k ncko cy 120 N gSA(B) ;fn CykWd dks lrg ABCD ds lgkjs tehu ij j[kk gS] rks CykWd }kjk tehu ij yxk;k x;k nkc (izfrcy) 20 Pa gksxkA(C) ;fn CykWd dks lrg ABEF ds lgkjs tehu ij j[kk gS] rks CykWd }kjk tehu ij yxus okyk nkc (izfrcy) 60 Pa gksxkA(D) ;fn ge CykWd dks tehu ij vyx-vyx lery lrgks ds lgkjs j[ks rks tehu ij nkc (izfrcy) vf/kdre rc gksxk


42. A conducting rod of length is moved at constant velocity ‘v0’ on two parallel, conducting, smooth, fixedrails, that are placed in a uniform constant magnetic field B perpendicular to the plane of the rails as shown infigure. A resistance R is connected between the two ends of the rail. Then which of the following is/are correct :(A) The thermal power dissipated in the resistor is equal to rate of work done by external person pulling the rod.(B) If applied external force is doubled than a part of external power increases the velocity of rod.(C) Lenz’s Law is not satisfied if the rod is accelerated by external force(D) If resistance R is doubled then power required to maintain the constant velocity v0 becomes half.

X B

xR V0

X

,d pkyd NM+ ftldh yEckbZ gS nks lekUrj ?k"kZ.k fofgu NM+ksa ds chp ‘v0’ vpj osx ls py jgh gSA fp=kuqlkj tks fd

vpj pqEcdh; {ks= B esa tks fd rails ds ry ds vfHkyEcor~ gS] esa j[kk x;k gSA rails dks ,d izfrjks/k R ls tksM+k x;k gSA rc

fuEu esa ls dkSu lgh gS@gSa :(A) izfrjks/k esa Å"eh; 'kfDr cká O;fDr }kjk NM+ dks [khapus esa fd;s x;s dk;Z dh nj ds cjkcj gksxhA

(B) vxj cká cy dks nqxuk fd;k tk, rc] cká 'kfDr dk dqN Hkkx NM+ dh pky dks c<+krk gSA

(C) vxj cká cy }kjk NM+ dks Rofjr fd;k tk, rc ySat fu;e oS| ugha gSA

(D) vxj izfrjks/k R dks nqxuk fd;k tk, rc NM+ dks vpj osx v0 ij cuk;s j[kus dh pkgh xbZ 'kfDr nqxuh gks tk;sxhA

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This section contains 2 paragraphs. Based upon each paragraph, 2 multiple choice questions haveto be 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 2 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 a series L-R circuit, connected with a sinusoidal ac source, the maximum potential difference across Land R are respectively 3 volts and 4 volts.L-R Js.kh ifjiFk tks fd ,d T;koØh; izR;korhZ L=kksr ls tqM+k gS L o R ds fljksa ij vf/kdre foHkokUrj Øe'k% 3 oksYV o4 oksYV gSA

43. At an instant the potential difference across resistor is 2 volts. The potential difference in volt, across theinductor at the same instant will be :tc izfrjks/k ds fljksa ij foHkokUrj 2 oksYV gS rc mlh {k.k izsj.k dq.Myh (inductor) ds fljksa ij foHkokUrj ¼oksYV esa½ D;k

gksxk\

(A) 3 cos 30° (B) 3 cos 60° (C) 6 cos 45° (D) 6

44. At the same instant, the magnitude of the potential difference in volt, across the ac source will bemlh leku {k.k ij izR;korhZ oksYVrk L=kksr ds fljksa ij foHkokUrj dk ifjek.k oksYV esa gksxkA

(A) 3 cos 67° (B) 5 cos 83° (C) 6 cos 97° (D) 0


An ideal gas initially at pressure p0 undergoes a free expansion (expansion against vacuum underadiabatic conditions) until its volume is 3 times its initial volume. The gas is next adiabatically compressedback to its original volume. The pressure after compression is 32/3 p0.,d vkn'kZ xSl izkjfEHkd nkc p0 ls eqä izlkj (fuokZr esa :)ks"e fLFkfr;ks a esa izlkj) djrh gS tc rd fd bldk vk;ru]

izkjfEHkd vk;ru dk 3 xquk gks tkrk gSA vkxs xSl dks :)ks"e :i ls okil ewy vk;ru rd laihfM+r djrs gSaA laihM+u

ds ckn nkc 32/3 p0 gSA

45. The pressure of the gas after the free expansion is :eqä izlkj ds ckn xSl dk nkc gS &

(A) 3p0 (B) 3/1

0p (C) p0 (D) 3p0

46. The gas :(A) is monoatomic. (B) is diatomic.(C) is polyatomic. (D) type is not possible to decide from the given information.xSl gS &

(A) ,d ijek.kqd (B) f} ijek.kqd(C) cgq ijek.kqd (D) nh xbZ lwpuk ds vk/kkj ij izdkj ugha crk;k tk ldrkA

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PART - III (Hkkx - III)


This section contains 14 multiple choice questions. Each question has choices (A), (B), (C) and (D),out of which ONLY ONE is correct.bl [k.M esa 14 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA

47. A mixture of a definite composition of ethanol and water forms :(A) Maximum boiling azeotrope.(B) Minimum boiling azeoptrope.(C) Both maximum and minimum boiling azeotrope.(D) No azeotrope.,FksukWy rFkk ty ds fuf'pr laxBu ls cuk feJ.k gS %

(A) vf/kdre DoFkukad fLFkjDokFkh feJ.kA

(B) U;wure DoFkukad fLFkjDokFkh feJ.kA

(C) vf/kdre DoFkukad rFkk U;wure DoFkukad fLFkjDokFkh feJ.k nksuksaA

(D) fLFkjDokFkh feJ.k ughaA

48. The equilibrium constant for the following two reactions are K1 and K2 respectively.fuEu nks vfHkfØ;kvksa ds fy, lkE; fu;rkad Øe'k% K1 rFkk K2 gSaAXe F6 (g) + H2O (g) XeOF4 (g) + 2HF (g)XeO4 (g) + XeF6 (g) XeOF4 (g) + XeO3F2(g)The equilibrium constant for the given reaction is :uhps nh xbZ vfHkfØ;k dk lkE; fu;rkad gS %

XeO4 (g) + 2HF (g) XeO3F4 (g) + H2O (g)(A) K1 K2

2 (B) K1 –K2 (C) K2 / K1 (D) K2 / K2

49. Structures of some common polymers are given. Which is not correctly presented ?

(A) Teflon n22 –)CFCF( (B) Neoprene

n

222

Cl|

CHCHCHCCH

(C) Terylene (D) Nylon-66 [ NH(CH2)6NHCO(CH2)4–CO–]n

uhps dqN lkekU; cgqydksa dh lajpuk,sa nh xbZ gSaA fuEu esa ls dkSulk lgh lqesfyr ugh gS\

(A) Vs¶ykWu n22 –)CFCF( (B) fu;ksizhu

n

222

Cl|

CHCHCHCCH

(C) Vsjhyhu (D) uk;ykWu-66 [ NH(CH2)6NHCO(CH2)4–CO–]n

50. 106.2 g 1 molal aqueous solution of ethylene glycol is cooled to –3.72ºC. Mass of ice separated duringcooling is (Kf water = 1.86, freezing point of water = 0ºC)106.2 g ,Fkkbyhu Xykbdksy dk 1 eksyy tyh; foy;u –3.72ºC rd B.Mk fd;k tkrk gSA B.Mk djus ds nkSjku cQZ dk

fdruk nzO;eku i`Fkd gksxkA (ty dk Kf = 1.86, ty dk fgekad = 0ºC)(A) 25 g (B) 50 g (C) 60 g (D) 40 g

51. On heating calcium propionate, the product formed is(A) 3-Pentanone (B) 2-Pentanone(C) 3-Methyl-2-butanone (D) Propanone

dSfY'k;e izksfivksusV dks xeZ djus ij] fufeZr mRikn gS %

(A) 3-isUVsukWu (B) 2-isUVsukWu (C) 3-esfFky-2-C;wVsukWu (D) izksisukWu

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52. Nitrogen and oxygen exist as diatomic but their congeners are P4 and S8 respectively because :(A) phosphorus and sulphur are solids.(B) phosphorus and sulphur catenate due to the existence of d-orbitals and form strainless structures.(C) phosphorus and sulphur polymerise as soon as they are formed.(D) catenation tendency of P and S is stronger because of the high P – P and S – S bond energies ascompared to N – N and O – O bond energies.ukbVªkstu rFkk vkWDlhtu] f}ijek.kqd :i esa vfLrRo j[krs gSa] ijUrq leku oxZ esa vxys rRo ¼congeners½ Øe'k% P4 rFkkS8 gSa] D;ksafd :(A) QkLQksjl rFkk lYQj Bksl gSaA

(B) QkLQksjl rFkk lYQj] d-d{kdksa ds vfLrRo ds dkj.k J¡[kfyr ¼Ja[kykc)½ gks tkrs gSa rFkk ruko jfgr lajpuk cukrs gSA

(C) QkLQksjl rFkk lYQj fufeZr gksrs gh ¼tYnh ls tYnh½ cgqyhd`r gks tkrs gSaA

(D) P rFkk S dh J`a[kyu izo`fÙk izcy gksrh gS] D;ksafd P – P rFkk S – S ca/k ÅtkZ;as] N – N rFkk O – O ca/k ÅtkZvksa dh

rqyuk esa mPp gksrh gSA

53. In which of the following pair of elements +1 oxidation state is most stable?rRoksa ds fuEu esa ls dkSulk@dkSuls ;qXe esa +1 vkWDlhdj.k voLFkk lokZf/kd LFkk;h gS\

(A) Na, Ag (B) Na, Tl (C) K, Sc (D) K, Ca

54. Solution having osmotic pressure nearer to that of an equimolar solution of K4[Fe(CN)6] is:og foy;u ftldk ijklj.k nkc K4[Fe(CN)6] ds leeksyj foy;u ds yxHkx cjkcj gksxk %&

(A) Na2SO4 (B) BaCl2 (C) Al2(SO4)3 (D) C12 H22O11

55. What is the most stable oxidation state lead (Pb)?ysM (Pb) ds fy, lokZf/kd LFkk;h vkWDlhdj.k voLFkk D;k gS\

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

56. In which of the following pair of elements +1 oxidation state is most stable?rRoksa ds fuEu esa ls dkSulk@dkSuls ;qXe esa +1 vkWDlhdj.k voLFkk lokZf/kd LFkk;h gS\

(A) Na, Ag (B) Na, Tl (C) K, Sc (D) K, Ca

57. Which of the following is correct IUPAC name of Acrolein?(A) But-2-enal (B) Propenal (C) Propanal (D) Prop-2-yn-1-alfuEu esa ls dkSulk ,Øksyhu dk lgh IUPAC uke gS\(A) C;wV-2-busy (B) izksihusy (C) izksisusy (D) izksi-2-vkbZu-1-,y

58. For the reaction in acidic medium between KMnO4 and H2O2 the number of electrons transferred per mol ofH2O2 are :(A) one (B) two (C) three (D) fourKMnO4 o H2O2 ds chp vEyh; ek/;e esa vfHkfØ;k ds fy, H2O2 ds çfr eksy LFkkukUrfjr bysDVªkWuksa dh la[;k fuEu gS%

(A) ,d (B) nks (C) rhu (D) pkj

59. Which of the following is the least stable carbanion ?fuEu esa ls U;wure LFkk;h dkcZ_.kk;u gS ?(A) HC C– (B) (C6H5)3C

– (C) (CH3)3C– (D) CH3

–

60. Identify the incorrect statement / statements :(i) Alkynes are more reactive than alkenes towards electrophilic addition reaction(ii) Alkynes are less reactive than alkenes towards electrophilic addition reaction(iii) Alkynes decolourise Br2 water(iv) Addition of HBr to alkynes in presence of peroxide proceeds via Markownikoff’s rule(A) (i) & (ii) (B) (ii) & (iii) (C) (i) & (iv) (D) (ii) & (iv)xyr dFkuksa dk p;u dhft,sA

(i) bysDVªkWuLusgh ;ksxkRed vfHkfØ;k ds izfr ,Ydhu dh rqyuk esa ,Ydkbu vf/kd lfØ; gksrs gSA

(ii) bysDVªkWuLusgh ;ksxkRed vfHkfØ;k ds izfr ,Ydhu dh rqyuk esa ,Ydkbu de lfØ; gksrs gSA

(iii) ,Ydkbu czksehu ty dks jaxghu dj nsrs gSA

(iv) ijkWDlkbM dh mifLFkfr esa ,Ydkbu ij HBr dk ;ksx, ekdksZfudkWQ fu;e ls gksrk gSA

(A) (i) ,oa (ii) (B) (ii) ,oa (iii) (C) (i) ,oa (iv) (D) (ii) ,oa (iv)

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This section contains 5 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 5 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


61. Which is/are correct among the following ?

Given : the half cell emf’s are V52.0E,V34.0E 0Cu|Cu

0Cu|Cu 2 .

(A) Cu+1 disproportionates (B) Cu and Cu2+ comproportionates

(C) 0Cu|Cu

0Cu|Cu EE 2 is positive (D) All of these

fuEu esa ls dkSulk@dkSuls lgh gS@gSa \

fn;k x;k gS % v)Z lsy ds fo|qr okgd cy V52.0E,V34.0E 0Cu|Cu

0Cu|Cu 2 gSaA

(A) Cu+1 dk fo"kekuqikrhdj.k gksrk gSA (B) Cu vkSj Cu2+ lekuqikrhr ¼comproportionate½ gks tkrs gSaA

(C) 0Cu|Cu

0Cu|Cu EE 2 /kukRed gSA (D) mijksDr lHkh

62. Phenol is obtained by :fQukWy fuEu esa ls fdlds }kjk izkIr fd;k tkrk gS\

(A) (B)

(C) (D)

63. In a AB unit cell (Rock salt type) assumig cations forming fcc lattice point(A) The nearest neighbours of A are 6B ion.

(B) The nearest neighbours of B are 6A ion.(C) The second nearest neighbours of A are 12A

(D) The packing fraction of AB crystal 38

.

,d AB ,dd dksf"Bdk (jkWd lkYV izdkj) esa] ekukfd /kuk;u, fcc tkyd fcUnq cukrs gSA

(A) A vk;u ds fudVre lehiorhZ 6B vk;u gSA

(B) B vk;u ds fudVre lehiorhZ 6A vk;u gSA

(C) A vk;u ds f}rh; fudVre lehiorhZ 12A gSA

(D) AB fØLVy dh ladqyu n{krk 38

gSA

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64. Which is/are correctly matched with its common name ?

(A)

OH

NO2

O2N NO2 : Picric acid (B)

COOH

COOCH3 : Aspirin

(C)

COOH

OH : Salicylic acid (D)

COOH

COOH : Phthalic acid

fuEu esa ls dkSulk@dkSuls buds lkekU; ukeks ds lkFk lgh lqesfyr gS@gSa \

(A)

OH

NO2

O2N NO2 : fifØd vEy (B)

COOH

COOCH3 : ,Lizhu

(C)

COOH

OH : lsfyflfyd vEy (D)

COOH

COOH : FkSfyd vEy

65. Which of the following is/are aromatic compounds :

fuEu esas ls dkSulk@dkSuls ;kSfxd ,jkseSfVd ;kSfxd gS\

(A)

N(B)

N| H

(C)

O (D)


This section contains 2 paragraphs. Based upon each paragraph, 2 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 2 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


Alcohols undergo acid catalysed elimination reactions to produce alkenes. Because water is lost in theelimination , this reaction is called dehydration reaction. Secondary and tertiary alcohols always give E1reaction in dehydration. Primary alcohols whose -carbon is branched also give E1 reaction. The reactivityof alcohol for elimination reaction is tertiary alcohol > Secondary alcohol > Primary alcohol.,YdksgkWy vEy mRizsfjr foyksiu vfHkfØ;k }kjk ,Ydhu cukrs gSa] D;ksafd foyksiu vfHkfØ;k esa H2O ckgj fudyrk gS vr%

;g vfHkfØ;k futZyhdj.k dgykrh gSA f}rh;d vkSj r`rh;d ,YdksgkWy futZyhdj.k esa lnSo E1 vfHkfØ;k nsrs gSaA izkFkfed

,YdksgkWy ftldk -dkcZu 'kkf[kr gks oks Hkh E1 vfHkfØ;k nsrk gSA ,YdksgkWyksa dh foyksiu dh fØ;k'khyrk dk Øe r`rh;d

,YdksgkWy > f}rh;d ,YdksgkWy > izkFkfed ,YdksgkWy gSA

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66. In the given reaction : 42SOH.conc Alkenes

Total number of alkenes (Including stereo isomers) formed will be(A) Two (B) Six (C) Four (D) Five

nh xbZ vfHkfØ;k esa] 42SOHl kUnz

,Ydhu

cuus okyh dqy ,Ydhuksa ¼f=kfoe leko;oh dks lEefyr djrs gq,½ dh la[;k gksxh &

(A) nks (B) N% (C) pkj (D) ik¡p

67. Which of the following dehydration product (major) is incorrect ?

(A) 42SOH.conc

(B)

42SOH.conc

(C) CH3 CH2 CH2CH2OH

42SOH.conc CH3–CH=CH–CH3

(D) 42SOH.conc

fuEu esa ls dkSulk futZyhdj.k mRikn ¼eq[;½ xyr gSa \

(A)

42SOHl kUnz

(B)

42SOHl kUnz

(C) CH3 CH2 CH2CH2OH

42SOHl kUnz CH3–CH=CH–CH3

(D)

42SOHl kUnz

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For any polyprotic acid, we always consider successive dissociation. The value of equilibrium constant ofsuccessive dissociation decreases due to common ion effect.For example :

H2A is a dibasic acid.

H2A H+ + HA– K1 = ]AH[]HA][H[

2

HA– H+ + AA– – K2 = ]HA[]A][H[

K1 is greater than K2.


fdlh cgqizksfVd vEy esa ge ges'kk Øekxr fo;kstu dk voyksdu djrs gSaA levk;u izHkko ds dkj.k Øekxr fo;kstu ds

lkE; fu;rkad dk eku ?kVrk tkrk gSA

mnkgj.k ds fy, :H2A ,d f}{kkjh; vEy gSA

H2A H+ + HA– K1 = ]AH[]HA][H[

2

HA– H+ + AA– – K2 = ]HA[]A][H[

K1dk eku K2 ls vf/kd gksrk gSA

68. Find the pH of 0.1 M Na2HAsO4.

Use data : (For H3AsO4, pK1 = 5.6, pK2 = 9.6, pK3 = 13.6)

(A) 7.6 (B) 9.6 (C) 11.6 (D) None of these

0.1 M Na2HAsO4 dk pH Kkr dhft;sA

vk¡dM+s iz;qDr dhft;s % (H3AsO4 ds fy,, pK1 = 5.6, pK2 = 9.6, pK3 = 13.6) :

(A) 7.6 (B) 9.6 (C) 11.6 (D) buesa ls dksbZ ugha

69. Find the concentration of H+ ions in an aqueous solution which is saturated with H2S (0.1 M) as well as H2CO3(0.2 M).

Use data : [K1 = 10–7, K2 = 10–14 for H2S, K1 = 4 × 10–7, K2 = 4 × 10–11 for H2CO3]

(A) 3 × 10–4 M (B) 3.83 × 10–4 M (C) 2.83 × 10–4 M (D) None of these

H2S (0.1 M) rFkk (0.2 M) H2CO3 ds lkFk ,d lar`Ir tyh; foy;u esa H+ dh lkUnzrk Kkr dhft;sA

vk¡dM+s iz;qDr dhft;s % [H2S ds fy;s K1 = 10–7, K2 = 10–14, H2CO3 ds fy;s K1 = 4 × 10–7, K2 = 4 × 10–11 ] :

(A) 3 × 10–4 M (B) 3.83 × 10–4 M (C) 2.83 × 10–4 M (D) buesa ls dksbZ ugh

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ANSWER KEY TO SAMPLE TEST PAPER-III1. (C) 2. (B) 3. (D) 4. (A) 5. (A) 6. (C) 7. (C)

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

15. (CD) 16. (ABD) 17. (AB) 18. (ABC) 19. (BD) 20. (C) 21. (B)

22. (A) 23. (A) 24. (C) 25. (D) 26. (C) 27. (D) 28. (B)

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

36. (C) 37. (A) 38. (ABD) 39. (ABC) 40. (AB) 41. (ABD) 42. (ABD)

43. (A) 44. (B) 45. (A) 46. (A) 47. (A) 48. (C) 49. (B)

50. (B) 51. (A) 52. (D) 53. (B) 54. (C) 55. (D) 56. (B)

57. (B) 58. (B) 59. (C) 60. (C) 61. (AC) 62. (ACD) 63. (ABC)

68. (C) 69. (A)

HINTS & SOLUTION TO SAMPLE TEST PAPER-III

1. x y1 – y x1 = 0 x2 (1 + y) = y2(1 + x) and xy > 0 x2 – y2 + xy (x – y) = 0 and xy > 0 (x – y) (x + y + xy) = 0 and xy > 0

either x = y or y = x1

x

x = y

0xy0y0x1andx1xy

dxdy

= 1, for all x (–1, 0)

Hindi. x y1 – y x1 = 0

x2 (1 + y) = y2(1 + x) rFkk xy > 0 x2 – y2 + xy (x – y) = 0 rFkk xy > 0 (x – y) (x + y + xy) = 0 rFkk xy > 0

;k rks x = y ;k y = x1

x

x = y

0xy0y0x1andx1xy

dxdy

= 1, lHkh x (–1, 0) ds fy,

2. The required area vHkh"V {ks=kQy

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= 0

2–

3 x4–x dx – 2

0

3 x4–x dx

=

0

2–

24

x2–4x

–

2

0

24

x2–4x

= – (4 – 8) – (4 – 8 ) = 8.

3. P = 5C2 2

31

3

32

= 24380

4. As a1b1c1, a2b2c2 and a3b3c3 are even natural numbers, each of c1,c2,c3 is divisible by 2.Let C i = 2k i for i = 1,2, 3 thus

= 2 333

222

111

bakbakbak

= 2m, where m is some natural number..

Thus, is divisible by 2. m may not be divisible by 2 that we can see by taking the three numberas 112, 122, 134

= 314212112

= 2

Hindi pwafd a1b1c1, a2b2c2 ,oa a3b3c3 le izkd`r la[;k,¡ gSA c1,c2,c3 esa ls izR;sd 2 ls HkkT; gSA

ekukfd i = 1,2, 3 ds fy, C i = 2k i

= 2 333

222

111

bakbakbak

= 2m, tgk¡ m dksbZ izkd`r la[;k gSA

vr% , 2 ls HkkT; gSA m, 2 ls HkkT; ugha gks ldrk gSA ge rhu la[;k,¡ tSls 112, 122, 134 ysdj ns[k ldrs gSA

= 314212112

= 2

5. 0x1x 2 (1 – x2 0 –1 x 1 )

2x1x squaring both the sides (nksuks a rjQ oxZ djus ij)x2 1 – x2 (x 0 )

2x2 1 x2 21

x 21

or 21x

1 ,

21x

6. x3 – 2x – 1 = 0then (rks) 3 – 2 – 1 = 0 .............(1)

Let (ekukfd) y11

1y1y

from equation (1), we get (lehdj.k (1) ls) 3

1y1y

– 2

1y1y

– 1 = 0

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y3 + 4y2 – y = 0 is the equation whose roots are

11

,

11

,

11

(lehdj.k y3 + 4y2 – y = 0 ds ewy

11

,

11

,

11

gSaA)

then (rks)

11

11

11

= – 4

7. y = 60 – x x(y – 30)2 = x(60 – x – 30)2 = x(x – 30)2

Let f(x) = x(x – 30)2

f (x) = (x – 30)2 + 2x(x – 30) = (x – 30) (3x – 30) f (x) = 0 x = 10, 30f(10) = 4000, f(30) = 0

0xlim f(x) = 0

60xlim f(x) = 54000

greatest value of x(y – 30)2 does not existHindi. y = 60 – x

x(y – 30)2 = x(60 – x – 30)2 = x(x – 30)2

ekukfd f(x) = x(x – 30)2

f (x) = (x – 30)2 + 2x(x – 30) = (x – 30) (3x – 30) f (x) = 0 x = 10, 30f(10) = 4000, f(30) = 0

0xlim f(x) = 0

60xlim f(x) = 54000

x(y – 30)2 dk vf/kdre eku fo|eku ugha gSA

8. Required number of ways are = Exactly one toy is exchanged + exactly two toys exchanged + exactly 3toys exchanged + 4 toys exchanged= 4C1

7C1 + 4C2 7C2 + 4C3

7C3 + 4C4 7C4 = 329

Hindi. dqy rjhdks dh la[;k = ,d f[kykSuk cnyus ds rjhdksa dh la[;k + nks f[kykSus cnyus ds rjhdksa dh la[;k + rhu f[kykSus

cnyus ds rjhdksa dh la[;k + pkj f[kykSus cnyus ds rjhdksa dh la[;kA

= 4C1 7C1 + 4C2

7C2 + 4C3 7C3 + 4C4

7C4 = 329

9. The centre of the circle is (1, 1) and radius = 2 2 .Point (a, a) musty lie outside the circle, so 2a2 4a 6 > 0 a < 1 or a > 3.

Now, 2

2 2tan2 2a 4a 6

.

2 2

(a,a)

As 3 6 2 2

2

2

2 2 1 a 2a 3 2 332a 4a 6

a2 2a 15 < 0 3 < a < 5 a (3, 1) (3, 5).

10. In an equilateral triangle incentre and circumcentre are the same point. Incentre = (g, f) also, 12 + 12 + 2g + 2f + c = 0 c = 2(g + f + 1)also circumradius = 2 (inradius)

inradius = 2 21 g f c2

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The equation of the incircle is

2 2 2 2 2 21 1 1x g y f g f c g f .2 g f 14 4 4

4(x2 + y2) + 8gx + 8fy = (1 g) (1 + 3g) + (1 f) (1 + 3f)

12. dx

x11x

x1x

21

22

3

subtitute x2 + 1 + 2x

1 = t2


min. Therefore it will cover 4840

min in 14521440


Hindi. xyr pyrh ?kMh 1440 feuV dk le; 1452 feuV esa r; djrh gSA vr% ;g 1 feuV esa 14521440

feuV dk le; r; djsxh

vr% ;g 4840 feuV dk le; 14521440

× 4840 = 4800 feuV = 80 h. Therefore 80 h = 3 days and 8 h.

14. )trialoneinsuccess(P41)HH(P

)success4(P)success3(P

4n4

4n

3n3

3n

43.

41C

43.

41C

n = 15

Hindi. P41)HH(P (,d iz;kl esa lQyrk)

P(3 lQyrk) = P(4 lQyrk)4n4

4n

3n3

3n

43.

41C

43.

41C

n = 15

15. 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 3amam2y–mx

= 1

then y2 = 4ax

3amam2y–mx

y2 =

3mm2y–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

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x2 dk xq.kkad + y2 dk xq.kkad = 0 2m + m3 – 4m = 0 m3 – 2m = 0 m2 = 2, m 0 m = ± 2

16. 1 2z 5 12i, z 4

1 2 1 2z iz z z 13 4 17

1 2 1 2z 1 i z z 1 i z 13 4 2

1 2min z 1 i z 13 4 2

2 22 2

4 4z z 4 1 5z z

2 22 2

4 4z z 4 1 3z z

1

22

z 13max4 3zz

and min 1

22

z 134 5zz

17. If y = f(x) is the curve, Y – y = f (x) (X – x) is the equation of the tangent at (x, y) with f (x) = dxdy

. Putting X

= 0, the ordinate of the tangent is y – x f (x). The subnormal at this point is given by y dxdy

, so we have y

dxdy

= y – x dxdy

dxdy

= yxy

This is a homogeneous equation and, by rewriting it as

dydx

= yyx

= yx

+ 1 dydx

– yx

= 1

we see that it is also a linear equation.

Hindi ;fn y = f(x) oØ gSA Y – y = f (x) (X – x) fcUnq (x, y) ij izo.krk f (x) = dxdy

okyh Li'kZ js[kk dk lehdj.k gSA

X = 0 j[kus ij Li'kZ js[kk dh dksfV y – x f (x) gSA bl fcUnq ij v/kksyEc y dxdy

}kjk fn;k tkrk gSA

vr% y dxdy

= y – x dxdy

dxdy

= yxy

;g ,d le?kkrh; lehdj.k gS rFkk bls fuEu rjg Hkh fy[k ldrs gS&

dydx

= yyx

= yx

+ 1 dydx

– yx

= 1

ge ns[krs gS fd ;g ,d js[kkh; lehdj.k Hkh gSA

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18.0x

lim

2x3

)axcosa–a(6 by L'Hospital's Rule.

0xlim

x23axsina6

.2

= a3

Now a3 = 1 i.e. a = 1a3 = 8 i.e. a = 2a3 = 27 i.e. a = 3a3 = 100 i.e. a = (100)1/3 (not a positive integer)

Hindi :0x

lim

2x3

)axcosa–a(6 by L'Hospital's Rule.

0xlim

x23axsina6

.2

= a3

vc a3 = 1 vFkkZr~ a = 1a3 = 8 vFkkZr~ a = 2a3 = 27 vFkkZr~ a = 3a3 = 100 vFkkZr~ a = (100)1/3 (,d /kukRed iw.kk±d ugha gSA)

19. At (x1, y1) slope of tangent dxdy

= 1

1

y3x

((x1, y1) ij Li'kZ js[kk dk >qdko dxdy

= 1

1

y3x

)

Slope of normal (vfHkyEc dk >qdko ) = 1

1

xy3

= 56

x1 = 2y5 1

4y25 2

1 + 3y12 = 37 y1

2 = 4

y = 2x = 5

(5,2), (–5, –2)

20. x = 3y – y3 1y0y330dydx 2

21. 322222

2

2 )y1(3y2

dxdy.

)y1(y

32

dxyd

)y1(31

dxdy

22. LCM of 2 and 3 6 number of numbers divisible by 6 is 16{6, 12, 18, ........96} out of which 8 are not divisible by 12

Hindi 2 o 3 dk LCM 6 6 ls HkkT; la[;kvksa dh la[;k 16 gSA{6, 12, 18, ........96} esa ls 8 la[;k,sa 12 ls HkkT; ugha gSA

23. LCM of 2 & 8 is 8

number of numbers divisible by both 8&2 is 5.

i.e. 85,84,83,82,8

200,128,72,32,8

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Hindi: 2 o 8 dk LCM 8 gS

8&2 nksauks ls HkkT; la[;kvksa dh la[;k 5 gSA

i.e. 85,84,83,82,8

200,128,72,32,8

24. This simplified circuit is shown in the figure. bldk ljyhÑr ifjiFk fp=k esa fufnZ"V gSA

3030

30

2V = 30 602V

= 202V

i

i

Therefore, current vr% /kkjk i = 202

= 101

A

28.

a1 = 221

RMGM

/ M1 a2 = 221

RMGM

/ M2

acceleation of M1 w.r.t. M2arel. = a1 + a2

= 221

R)MM(G

= 2RGM

.

30.1ma

=

3ma

0

1ma

= – 2ma

0

231 mmm FF

= 0

31. Given (fn;k gqvk gS) : m = 10 × 10–3 kg (fdxzk-)v = 300 ms–1 (eh-/ls-), t = .003 s(ls-)

F = t

mv

F = 003.

3001010 3

F = 1000 = 103 N

35. Obtain the required AC voltagevfHk"B AC oksYVst izkIr djus ds fy,

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36.

0

2

2

. R2 = kx

37. )rd(2i.10

= r2

iN 20

N = 2

1

ii

. )rd(r = 5.3

54 . )27.0(22

722.0 = 4

38. Reading of (A1) is 2 AmpReading of (A2) is 1 AmpTotal power consumed by all the resistors is 120 watt(A1) dk ikB;kad 2 Amp gSA(A2) dk ikB;kad 1 Amp gSAlHkh çfrjks/kks a esa dqy 'kfDr O;; 120 watt gSA

39. Total power supplied dqy fn xbZ 'kfDr = 20 × 220 = 4400 WattAlready existing load tks yksM ij igys ls mifLFkr gS = 2000 + 1000 + 300 = 3300 WattWe can increase load upto 4400 Watt so A, B & C are correct options.ge yksM 4400 Watt rd c<+k ldrs gSA vr% A, B rFkk C lgh fodYi gSA

40. BVqF

If V = 0 then BF

= 0

By right hand rule, the field line is circular.Magnetic force cannot do work It cannot change kinetic energyMagnetic force is perpendicular to the line joining current elements not a central force.

BVqF

;fn V = 0 rc BF

= 0

nka;s gkFk ds fu;e ls {ks=k js[kk o`Ùkkdkj gksxhA

pqEcdh; cy] dk;Z ugh dj ldrk gSA ;g xfrt ÅtkZ ifjofrZr ugh dj ldrkA

pqEcdh; cy /kkjk vo;o dks feykus okyh js[kk ds yEcor~ gSA dsfUnz; cy ugha gSA


(B) P = AF

= 23

120

= 20 Pa

(C) P = AF

= 13

120

= 40 Pa


42. Rate of work done by external agent is :

dtdw

= dtdx.LB

= BLv & thermal power dissipated in the resistor = e = (BvL)

clearly both are equal, hence (A) .

STP1617




If applied external force is doubled, the rod will experience a net force and hence acceleration. As aresult velocity increases, hence (B)

Since ; = Re

On doubling ‘R’, current and hence required power becomes half.Since, P = BLvHence (D).cká dkjd }kjk dk;Z fd;s tkus dh nj gS %

dtdw

= dtdx.LB

= BLv & Å"eh; ÅtkZ tks izfrjks/k esa {k; gqbZ gS = e = (BvL)

nksuksa cjkcj gS blfy, (A)vxj yxk;k x;k cy nqxuk fd;k tk, rc NM+ ij ,d ifj.kkeh cy o Roj.k yxsxkA ifj.kkekuqlkj osx c<+rk gS] blfy,

(B)

pw¡fd ; = Re

‘R’, dks nqxuk djus ij] /kkjk vkSj pkgh xbZ 'kfDr vk/kh jgh tk,xhA

blfy, , P = BLvblfy, (D).

43. Let at an instant vR = (VR)mcostekuk fdlh {k.k vR = (VR)m cost

2 = 4 costcost = ½

cost = 60°.Since D;ksafd VL is 90° ahead of VR

vL = (VL)m cos (t + 90 ) = – 3 sint = – 3 sin 60° = – 3 cos 30° | (VL)m| = 3 cos 30°

44. From phasor diagram (VS)m = 2mL

2mR )V()V( = 5 volt.

dyk&dks.k fp=k ls(VS)m = 2mL

2mR )V()V( = 5 volt.

tan = 43

)V)V

mR

mL = 37°

| vS | = | (VS)m cos (t + 37°)|= 5 | cos (60° + 37°)| = 5 | cos 97°| = 5 cos 83°

45. In free expansion, temperature of the gas remains constant, thereforeeqDr çlkj esa xSl dk rki fLFkj jgrk gSA vr%

p0 v0 = p. 3v0 where tgkW v0 = initial volume. izkjfEHkd vk;ru

p = 3p0

46. For adiabatic compression, initial conditions are 3p0 and 3v0 . Final volume and pressure are

v0 and 32/3 p0.

:}ks"e laihMu ds fy, çkjfEHkd fLFkfr;k¡ 3p0 rFkk 3v0 gSaA vfUre vk;ru rFkk nkc v0 rFkk 32/3 p0. gSA

3p0 .(3v0) = 32/3 p0(v0) 3–1 = 32/3

or – 1 = 32

= 35

i.e. gas is monoatomic xSl ,d ijek.kqd gSA

STP1617




47. Maximum boiling azeotrope is formed when Hmix is negative and attractive force molecules of liquids gettingmixed is more.vf/kdre DoFkukad fLFkj DokFkh curk gS tc H

feJ.k _.kkRed gks rFkk nzo dk vkd"kZ.k cy v.kq vf/kd fefJr gksrk gSaA

48. Substracting 1st reaction from second, we will get desired reaction3rd. So K = K2/K1

f}rh; ls 1st dks dks ?kVkus ij ge bfPNr vfHkfØ;k çkIr djsxsaA

3rd. So K = K2/K1

50. Mass of solute in 1 kg water = 62 g

Mass of solute in 106.2 g solution = 106262

x 106.2 = 6.2 g

3.72 = 1.86 x wx621000x2.6

w = 50 g Mass of ice separated = 50 g

gy- 1 kg ty esa foys; dk nzO;eku = 62 g

106.2 g foy;u esa foys; dk nzO;eku = 106262

x 106.2 = 6.2 g

3.72 = 1.86 x wx621000x2.6

w = 50 g izFk`d cQZ dk nzO;eku = 50 g

51. 223 O–C–CHCH||O

Ca 3223 CH–CH–C–CH–CH||O

+ CaCO3

52. N and O have ability to form p-p multiple bonds with it self on account of smaller size of atoms. N – N andO – O bond energies are less on account of repulsion between non-bonded pairs of electrons due to smallersize of atoms. S – S bond energy (265 kJ mol–1) is next to C – C.N rFkk O esa] buds NksVs vkdkj ds dkj.k] Lo;a ds ijek.kq ds lkFk p-p ca/k fuekZ.k dh {kerk gksrh gSA NksVs ijek.kq vkdkj

ds dkj.k] vcaf/kr bysDVªksu ds ;qXeksa ds e/; izfrd"kZ.k ds dkj.k] N – N rFkk O – O ca/k ÅtkZ;sa de gksrh gSA S–S ca/k ÅtkZ

(265 kJ mol–1) C – C ds ckn vxyh ca/k ÅtkZ gSA

53. Most stable oxidation state of Na is +1 and most stable oxidation state of Tl is +1 because of inert pair effect.Na dh lokZf/kd LFkk;h vkWDlhdj.k voLFkk +1 gS rFkk Tl dh lokZf/kd LFkk;h vkWDlhdj.k voLFkk vfØ; ;qXe izHkko ds

dkj.k +1 gksrh gSA

54. Osmotic pressure will be same for equimolar solutions if Van't Hoff factor is same.;fn okWUVgkWQ xq.kkad leku gS rks leeksyj foy;uksa dh fy;s ijklj.k nkc Hkh leku gksxk

K4[Fe(CN)6] i = 1 + (n–1) = 1 + 4 = 5Al2(SO4)3 i = 1 + (n–1) = 1 + 4 = 5

56. Most stable oxidation state of Na is +1 and most stable oxidation state of Tl is +1 because of inert pair effect.vfØ; ;qXe izHkko ds dkj.k Na dh lokZf/kd LFkk;h vkWDlhdj.k voLFkk +1 rFkk Tl lokZf/kd LFkk;h vkWDlhdj.k voLFkk +1gksrh gSA

58. Valency factor of H2O2 in both the medium is two.nksuksa ek/;e esa H2O2 dk la;kstdrk xq.kkad nks gSA

60. Conceptual (lS)kfUrd)

STP1617




61. 2Cu+1 Cu + Cu+2

2Cu+1 + 2e 2CuCu – 2e Cu+2

––––––––––––––––––––––2Cu+1 Cu+2 + Cu––––––––––––––––––––––

E° 2

)337.0(2521.02 = 0.184

62. (A) + Na2SO3 (B) ationdecarboxylCaONaOH

(l yhdj .kfodkcksZfD

CaONaOH)

(C) + Mg (OH)Br (D) HClNaNO2 OH2

63. Based on FCC crystal, like NaCl lattice

Packing fraction = 3

3B

3A

a

r344r

344 –

=

3 3 3

3 3

3

4 (0.414) 44 43 3(2 2) (2 2)

a a

a

? ?? ? ? ?

? ? ?? ? ? ?? ? ? ? = 0.79

FCC tkyd ij vk/kkfjr] NaCl tkyd dh rjg

ladqyu n{krk = 3

3B

3A

a

r344r

344 –

=

3 3 3

3 3

3

4 (0.414) 44 43 3(2 2) (2 2)

a a

a

? ?? ? ? ?

? ? ?? ? ? ?? ? ? ? = 0.79

66. OH–

H

2

+ +

+

67. Most stable alkene (saytzeff alkene) is major product.vf/kd LFkkbZ ,Ydhu (lSRtSQ ,Ydhu) eq[; mRikn gksxkA

68. pH = 2

pKpK 32 = 2

6.136.9 = 11.6

69. [H+] = 2211 CKCK = 2.01041.010 77 = 3 × 10–4 M

Spec

imen

Cop

y

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