ResoNET 2012 Sample Test Paper

Academic Session : 2012-13

Practice, Persistence and Performance

Sample Test Papers(Resonance National Entrance Test : ResoNET - 2012)

For Yearlong Classroom Contact Programmes (YCCPs) : IIT-JEE Division

For Class - X, XI, XII

IIT-JEE AIEEEAIPMT NTSEKVPYOLYMPIADS SAT

STPFXII1213

Page - 1

Sample Test Paper (STP)For ResoNET-2012I N D E X

Copyright reserved 2012-13.

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 its various Selection Tests (ResoNET) to be conducted for admission in Academic Session 2012-13. Any sale/resale of this material is

punishable under law. Subject to Kota Jurisdiction only.

S.No.

Content EligibilityCourse Applied

Target Page No.

1 How to Prepare for the Resonance National Entrance Test (ResoNET)-2012: List of reference books

N.A. N.A. ResoNET 2012 1

2Resoonance Support Material to Prepare for the Resonance National Entrance Test (ResoNET)-2012


3General Instructions for the Examination Hall


4 Syllabus for ResoNET-2012 N.A. N.A. ResoNET 2012 4

5Sample Test Paper- I (STP-I): Test Paper Class-X Appearing/ Passed

students (Moving from Class-X to Class-XI)

VIKAAS (JA) & VIPUL (JB)

IIT�JEE 2014 9

6Sample Test Paper-I (STP-I) : Answer Key, Hints & Solution

Class-X Appearing/ Passed students (Moving from Class-X to Class-XI)

VIKAAS (JA) & VIPUL (JB)

IIT�JEE 2014 20

7Sample Test Paper-II (STP-II): Test Paper Class-XI Appearing /

Passed students (Moving from Class-XI to Class-XII)

VISHWAAS (JF) IIT�JEE 2013 27

8Sample Test Paper-II (STP-II) : Answer Key, Hints & Solution

Class-XI Appearing / Passed students (Moving from Class-XI to Class-XII)

VISHWAAS (JF) IIT�JEE 2013 42

9Sample Test Paper-III (STP-III) : Test Paper

Class-XII Appearing / Passed students (Moving from Class-XII to Class-XIII)

VISHESH (JD) & VIJAY (JR)

IIT�JEE 2013 50

10Sample Test Paper-III (STP-III) : Answer Key, Hints & Solution

Class-XII Appearing / Passed students (Moving from Class-XII to Class-XIII)

VISHESH (JD) & VIJAY (JR)

IIT�JEE 2013 64

11Sample ORS Answer Sheet for Resonance National Entrance Test (ResoNET)-2012 N.A. N.A. ResoNET 2012 75

The Sample Test Papers (STPs) are only for reference and guidance. The sample papers given in the booklet are actually the papers of previous year'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 the papers for this year

selection test will be on the same pattern. However, the syllabus of the test paper will be equivalent to the syllabus of qualifying school/board examination

and as given on page no. 4.

HOW TO PREPARE FOR THE RESONANCE NATIONAL ENTRANCE TEST (ResoNET) - 2012

For Class-X appearing / passed students (Class-X to Class-XI Moving) :Study thoroughly the books of Science (Physics & Chemistry) and Maths of Classes IX & X. (NCERT & Respective Board)

For Class-XI appearing / passed students (Class-XI to Class-XII Moving):1. Study thoroughly the books of Physics, Chemistry and Maths of Class XI (Respective Board).2. Refer to the following books (only Class-XI syllabus) to increase the level of competence:

Physics : Concepts of Physics by H.C. Verma Vol. I & II Chemistry : NCERT Books Maths : Higher Algebra By Hall & Knight; Co-ordinate Geometry By S.L. Loney ; Plane Trigonometry By S.L. Loney

For Class-XII appearing / passed students (Class-XII to Class-XIII Moving):1. Study thoroughly the books of Physics, Chemistry and Maths of Classes XI & XII (Respective Board).2. Refer to the following books (Class-XI & Class-XII syllabus) to increase the level of competence :

Physics : Concepts of Physics by H.C. Verma Vol-I & II Chemistry : Physical Chemistry By R.K. Gupta, Organic Chemistry By Morrison & Boyd, Organic Chemistry By I. L. Finar,

Inorganic Chemistry By J.D. Lee, Objective Chemistry By Dr. P. Bahadur 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 ; MCQ By A Das Gupta.

STPFXII1213

Page - 2

For More Practice of RESONANCE NATIONAL ENTRANCE TEST (ResoNET) - 2012

Resonance Practice Test Papers (PTPs)

To support the preparation of ResoNET; Resonance Practice Test Papers (PTPs) of last few years with Answer Key, Hints & Solutions

are available on demand. Following Sets of Practice Test Papers (PTPs), in hard copy, are available with us :

S.No.

PTPSet

Content Eligibility Course (Code) Target Remark

1 Set-A 10 Papers Set (PTPs) Class-X Appearing/Passed studentsVIKAAS (JA) &

VIPUL (JB)IIT-JEE 2014

Answer Key, Hints & Solutions

2 Set-B 10 Papers Set (PTPs) Class-XI Appearing/Passed students VISHWAAS (JF) IIT-JEE 2013Answer Key,

Hints & Solutions

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

VIJAY (JR)IIT-JEE 2013

Answer Key, Hints & Solutions

Interested students may collect the PTPs from Resonance Study Centres or Corporate Office at Kota (at Plot No. A-46, A-52, Near

City Mall, 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 Pvt Ltd' by sending a Bank Demand Draft (DD)

of Rs. 300/- in favour of 'Resonance' and payable at Kota. A student may send the request application on plain paper along with

prerequisite fees to the institute to collect any of the sets of Practice Test Papers (PTPs). Please, mention clearly your name and

Roll Number (Application Form 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.

ResoNET Online Practice Test Papers (OPTPs)

To support the preparation of ResoNET; interested students may practice the ResoNET Test Papers online through ResoNET Online

Practice Test Papers (OPTPs) on Resonance Website: http://elpd.resonance.ac.in

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

S. No. Content Eligibility Course Code Target Fee(Taxes included)

1 3 Tests (OPTPs) Class-X Appearing/Passed students VIKAAS (JA) & VIPUL (JB) IIT-JEE 2014 Rs. 300/-

2 6 Tests (OPTPs) Class-X Appearing/Passed students VIKAAS (JA) & VIPUL (JB) IIT-JEE 2014 Rs. 500/-

3 3 Tests (OPTPs) Class-XI Appearing/Passed students VISHWAAS (JF) IIT-JEE 2013 Rs. 300/-

4 6 Tests (OPTPs) Class-XI Appearing/Passed students VISHWAAS (JF) IIT-JEE 2013 Rs. 500/-

5 3 Tests (OPTPs) Class-XII Appearing/Passed students VISHESH (JD) & VIJAY (JR) IIT-JEE 2013 Rs. 300/-

6 6 Tests (OPTPs) Class-XII Appearing/Passed students VISHESH (JD) & VIJAY (JR) IIT-JEE 2013 Rs. 500/-

ResoNET Prepratory Guide (RPG)(Only for Class X to XI moving students)

To support the preparation of ResoNET; interested students (only for Class X to XI moving students) may study, learn, understand

and strengthen the fundamentals of Class IX & X through ResoNET Prepratory Guide (RPG). The salient features of RPG are as

follows :

1. Brief theory of ResoNET syllabus (for Class X to XI moving students).

2. Solved and Unsolved illustrations / examples.

3. Selected questions for practice.

Interested students may collect the RPG from Resonance Study Centres or Corporate Office at Kota (at Plot No. A-46, A-52, Near

City Mall, Jhalawar Road, Reception) by paying an additional fees of Rs.300/- only. The RPG may also be procured through post

/ courier from 'Resonance Eduventures Pvt Ltd' by sending a Bank Demand Draft (DD) of Rs. 300/- in favour of 'Resonance' and

payable at Kota. A student may send the request application on plain paper along with prerequisite fees to the institute to collect

the RPG. Please, mention clearly your name and Roll Number (Application Form No.) and ResNET Prepratory Guide (RPG) on

the back of the DD.

http://elpd.resonance.ac.in

http://elpd.resonance.ac.in

STPFXII1213

Page - 3

Note: The student should consider the following instructions to be written on the front page of the testpaper of ResoNET to be made available in the examination hall.

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

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

d s Å ij nk;sa d ksus esa Nik gqvk gS½

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

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

: y] d SYd qysVj] eksckby ;k vU; fd lh bySDVªWkfud mid j.k d s fd lh Hkh : i esa mi;ksx d h vkKk ugha gS½

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

booklet. (bl i"B d s uhps fn;s x;s fjDr LFkku esa viuk uke o vkosnu Q kWeZ la[;k vo'; Hkjsa½

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

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

fooj.kksa dks Hkjsa½

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

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

esa fn;s x;s fjDr LFkku esa vius iz'u&i=k d k d ksM o viuk vkosnu Q kWeZ la[;k Li"V : i ls Hkjuk uk Hkwysa½7. No rough sheets will be provided by the invigilators. All the rough work is to be done in the blank

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

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

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

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

Question Paper ¼iz'u&i=k½

9. Marks distribution of questions is as follows. ¼iz'uksa d s izkIrkad ks d k fooj.k fuEu izd kj ls gSA½

Correct Wrong Blank

1 to 15 30 to 39 51 to 59Only one correct

(d soy ,d fod Yi lgh)3 -1 0

16 to 19 40 to 43 60 to 61One or more than one correct Answer

(,d ;k ,d ls vf/kd fod Yi lgh)4 0 0

20 to 28 44 to 49 62 to 64 Comprehensions (vuqPNsn) 4 0 0

29 50 65Matrix Match Type(eSfVªDl lqesy izd kj)

6 [1, 2, 3, 6] 0 0

Marks to be awardedPart - I (Mathematics)

Part - II (Physics)

Part - III (Chemistry)

Type

Name : _________________________________ Application Form Number : _______________

Note: The above instructions are for reference purpose only. The actual instructions in the Test Paper ofResoNET in the examination hall may vary.

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

STPFXII1213

Page - 4

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; changeof 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 not equalto x grams of B; Partical nature, basic units: atoms and molecules; Lawof constant proportions; Atomic and molecular masses; Relationship ofmole to mass of the particles and numbers; Valency; Chemical formulaeof common compounds.

Atomic structure: Atoms are made up of smaller particles: electrons,protons, and neutrons. These smaller particles are present in all theatoms 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, displ-acement, double displacement, precipitation, neutralisation, oxidationand reduction in terms of gain and loss of oxygen and hydrogen.

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

Compounds of Carbon: Carbon compounds; Elementary idea aboutbonding; Saturated hydrocarbons, alcohols, carboxylic acids (nopreparation, only properties). Soap - cleansing action of soap._________________________

_________________________

CLASS - X (MATHEMATICS)Number Systems : Natural Numbers, Integers, Rational number onthe number line. Even - odd integers, prime number, composite numbers,twin primes, divisibility tests, Co-prime numbers, LCM and HCF ofnumbers.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 and multiplex. 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). Relationbetween roots and coefficient of quadratic and relation betweendiscriminant and nature of roots.

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

Arithmetic Progressions (AP) : Finding the nth term and sum of firstn 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 the plane, distance between two points and sectionformula (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 andcongruency of 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 finding the 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 singleevents.

Logarithm & exponents : Logarithms and exponents and theirproperties.

Interest : Problem based on simple interest, compound interest anddiscounts.

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

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, accelarationand relation ship between these ; Distance-time and velcocity - timegraphs.Newton�s Law of motion ; Relationship between mass, momentum,

force and accelaration ; work done by a force ; Law of conservationof energy.Law of gravitation ; acceleration due to gravity.

Electricity and magnetism : Ohm�s law ; Series and parallel combi-

nation of resistances ; Heating effect of current.

Magnetic field near a current carrying straight wire, along the axis ofa circular coil and inside a solenoid ; Force on current carrying con-ductor ; 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 concave mir-ror 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 of

elements, atoms and molecules.Atomic and molecular masses. Mole concept and molar mass ;percentage composition and empirical and molecular formula ; chemicalreactions, stoichiometry and calculations based on stoichiometry.

Structure of Atom : Discovery of electron, proton and neutron ;atomic number, isotopes and isobars.Thompson�s model and its limitations, Rutherford�s model and its

limitations, concept of shells and sub-shells, dual nature of matter andlight, de Broglie�s relationship, Heisenberg uncertainty principle, concept

of orbitals, quantum numbers, shapes of s, p, and d orbitals, rules forfilling electrons in orbitals - Aufbau principle, Pauli exclusion principle

Syllabus of ResoNET-2012

STPFXII1213

Page - 5

and Hund�s rule, electronic configuration of atoms, stability of half filled

and completely filleld orbitals.

Classification of Elements and Periodicity in Properties :Significance of classification, brief history of the development of periodictable, trends in properties of elements - atomic radii, ionic radii, inertgas radii, ionization enthalpy, electron gain enthalpy, electronegativity,valence.

Chemical Bonding and Molecular Structure : Valence electrons,ionic bond, covalent bond, bond parameters, Lewis structure, polarcharacter of covalent bond, covalent character of ionic bond, valencebond theory, resonance, geometry of covalent molecules, VSEPR theory,concept of hybridization involving s, p and d orbitals and shapes ofsome 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 concept of the molecule, Boyle�slaw, 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, criticaltemperature.Liquid State - Vapour pressure, viscosity and surface tension(qualitative idea only, no mathematical derivations)

Thermodynamics : Concepts of system, types of systems, sur-roundings, work, heat, energy, extensive and intensive properties,state functions.First law of thermodynamics - internal energy and enthalpy, heat ca-pacity and specific heat, measurement of U and H, Hess�s law of

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 weak electro-lytes, degree of ionization concept of pH. Hydrolysis of Salts (elemen-tary idea), buffer solutions, solubility product, common ion effect (withillustrative examples).

Redox Reactions : Concept of oxidation and reduction, redox reac-tions,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 of water,heavy water ; hydrogen peroxide - preparation, reactions and struc-ture ; hydrogen as a fuel.

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

Preparation and properties of some important compounds:Sodium 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, variationof properties, oxidation states, trends in chemical reactivity, anoma-lous properties of first element of the group ;Boron - physical and chemical properties, some important compoundsborax, boric acids, boron hydrides. Aluminium : uses, reactions withacids and alkalies.Group 14 elements ; General introduction, electronic configuration,occurrence, variation of properties, oxidation states, trends in chemi-

cal reactivity, anomalous behaviour of first element. Carbon - catena-tion, allotropic forms, physical and chemical propeties ; uses of someimportant compounds : oxides.Important compounds of silicon and a few uses : silicon tetrachloride,silicones, silicates and zeolites.

Principles of qualitative analysis : Determinantion of one anion andone cation in a given saltCations - Pb2 + , Cu2+, As3+, Al3+, Fe3+, Mn2+, Ni2 +, Zn2+, Co2+, Ca2+, Sr2+,

Ba2+, Mg2+, 4NH

Anions - ,NO,SO,SO,S,CO �2

�24

�23

�2�23

�3

�242

�34

��3

�3 COOCHOC,PO,,Br,Cl,NO,NO

(Note : Insoluble salts excluded)

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

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

Alkenes : Nomenclatures, structure of double bond (ethene),geometrical isomerism, physical properties, methods of preparation ;chemical reactions : addition of hydrogen, halogen, water, hydrogenhalides (Markovnikov �s addition and peroxide effect),

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, hy-drogen halides and water.

Aromatic hydrocarbons : Introduction, IUPAC nomenclature ;Benzene : resonance, aromaticity ; chemical properties : mechanismof electrophilic substitution - nitration sulphonation, halogenation, FriedelCraft�s alkylation and acylation ; directive influence of functional group

in mono-substituted benzene; carcinogenicity and toxicity._________________________

_________________________

CLASS - XI (MATHEMATICS)Functions: Sets and their representations. Empty, finite and infinitesets, Subsets, Union and intersection of sets, Venn diagrams.Pictorial representation of a function domain, co-domain and range ofa function domain and range of constant, identity, polynomial, rational,modulus, signum and greatest integer functions with their graphs.Sum, difference, product and quotients of functions.

Trigonometric Functions: Measuring angles in radians and indegrees and conversion from one measure to another. Signs oftrigonometric functions and sketch of their graphs. Addition andsubtraction formulae, formulae involving multiple and sub-multiple angles.General solution of trigonometric equations.

Complex Number: Algebra of complex numbers, addition,multiplication, conjugation, polar representation, properties of modulusand principal argument, triangle inequality, cube roots of unity, geometricinterpretations.

Quadratic equations : Quadratic equations with real coefficients,formation of quadratic equations with given roots, symmetric functionsof roots.

Sequence & Series : Arithmetic, geometric and harmonicprogressions, arithmetic, geometric and harmonic means, sums of finitearithmetic and geometric progressions, infinite geometric series, sumsof squares and cubes of the first n natural numbers.

Logarithm & exponents: Logarithms and exponents and theirproperties. Exponential and logarithmic series.

STPFXII1213

Page - 6

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

Permutations and combinations: Problem based on fundamentalcounting principle, Arrangement of alike and different objects, Circularpermutation, Combination, formation of groups.

Straight Line : Cartesian coordinates, distance between two points,section formulae, shift of origin. Equation of a straight line in variousforms, angle between two lines, distance of a point from a line; Linesthrough the point of intersection of two given lines equation of thebisector of the 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 oftangent, normal and chord. Parametric equations of a circle, intersectionof a circle with a straight line or a circle, equation of a through thepoints of intersection of two circles and those of a circle and a straightline.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& Logical reasoning._________________________

_________________________

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: Experi-ments based on using Vernier calipers and screw gauge (microme-ter), Determination of g using simple pendulum, Young�s modulus 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 frames of

reference; Static and dynamic friction; Kinetic and potential energy;Work and power; Conservation of linear momentum and mechanicalenergy.Systems of particles; Centre of mass and its motion; Impulse; Elasticand inelastic collisions.Law of gravitation; Gravitational potential and field; Acceleration due togravity; Motion of planets and satellites in circular orbits; Escape ve-locity.Rigid body, moment of inertia, parallel and perpendicular axestheorems, moment of inertia of uniform bodies with simplegeometrical shapes; Angular momentum; Torque; Conservation of an-gular momentum; Dynamics of rigid bodies with fixed axis ofrotation; Rolling without slipping of rings, cylinders and spheres; Equi-librium 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 and sur-

face tension, capillary rise; Viscosity (Poiseuille�s equation

excluded), Stoke�s law; Terminal velocity, Streamline flow, equation of

continuity, Bernoulli�s theorem and its applications.

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

Thermal physics : Thermal expansion of solids, liquids and gases;Calorimetry, latent heat; Heat conduction in one dimension; Elementaryconcepts of convection and radiation; Newton�s law of cooling; Ideal

gas laws; Specific heats (Cv and Cp for monoatomic and diatomicgases); Isothermal and adiabatic processes, bulk modulus of gases;Equivalence of heat and work; First law of thermodynamics and itsapplications (only for ideal gases); Blackbody radiation: absorptiveand emissive powers; Kirchhoff�s law; Wien�s displacement law,

Stefan�s law.

_________________________

_________________________

CLASS - XII (CHEMISTRY)PHYSICAL CHEMISTRY

General topics : Concept of atoms and molecules; Dalton�s atomic

theory; Mole concept; Chemical formulae; Balanced chemical equations;Calculations (based on mole concept) involving common oxidation-reduction, neutralisation, and displacement reactions; Concentration interms 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 probablevelocities and their relation with temperature; Law of partial pressures;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 covalent

bond; 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, work andheat, pressure-volume work; Enthalpy, Hess�s law; Heat of reaction,

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); Significance ofG and Go in chemical equilibrium; Solubility product, common ioneffect, pH and buffer solutions; Acids and bases (Bronsted and Lewisconcepts); Hydrolysis of salts.

Electrochemistry : Electrochemical cells and cell reactions; Standardelectrode potentials; Nernst equation and its relation to DG;Electrochemical series, emf of galvanic cells; Faraday�s laws of

electrolysis; Electrolytic conductance, specific, equivalent and molarconductivity, Kohlrausch�s law; Concentration cells.

Chemical kinetics : Rates of chemical reactions; Order of reactions;Rate constant; First order reactions; Temperature dependence of rateconstant (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, ionicradii, simple ionic compounds, point defects.

Solutions : Raoult�s law; Molecular weight determination from lowering

of vapour pressure, elevation of boiling point and depression of freezingpoint.

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), carbondating; Stability of nuclei with respect to proton-neutron ratio; Briefdiscussion on fission and fusion reactions.

INORGANIC CHEMISTRYIsolation/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 chloride

STPFXII1213

Page - 7

and alums; Carbon: oxides and oxyacid (carbonic acid); Silicon:silicones, silicates and silicon carbide; Nitrogen: oxides, oxyacids andammonia; Phosphorus: oxides, oxyacids (phosphorus acid, phosphoricacid) and phosphine; Oxygen: ozone and hydrogen peroxide; Sulphur:hydrogen sulphide, oxides, sulphurous acid, sulphuric acid and sodiumthiosulphate; Halogens: hydrohalic acids, oxides and oxyacids ofchlorine, bleaching powder; Xenon fluorides.

Transition elements (3d series) : Definition, general characteristics,oxidation states and their stabilities, colour (excluding the details ofelectronic transitions) and calculation of spin (only magnetic moment),Coordination compounds: nomenclature of mononuclear coordinationcompounds, cis-trans and ionisation isomerisms, hybridization andgeometries of mononuclear coordination compounds (linear, tetrahedral,square planar and octahedral).

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

Ores and minerals : Commonly occurring ores and minerals of iron,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; Shapes ofsimple organic molecules; Structural and geometrical isomerism; Opticalisomerism of compounds containing up to two asymmetric centres,(R,S and E,Z nomenclature excluded); IUPAC nomenclature of simpleorganic compounds (only hydrocarbons, mono-functional and bi-functional compounds); Conformations of ethane and butane (Newmanprojections); Resonance and hyperconjugation; Keto-enol tautomerism;Determination of empirical and molecular formulae of simple compounds(only combustion method); Hydrogen bonds: definition and their effectson physical properties of alcohols and carboxylic acids; Inductive andresonance effects on acidity and basicity of organic acids and bases;Polarity and inductive effects in alkyl halides; Reactive intermediatesproduced during homolytic and heterolytic bond cleavage; Formation,structure and stability of carbocations, 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; Preparation ofalkanes by Wurtz reaction and decarboxylation reactions.

Preparation, properties and reactions of alkenes and alkynes:Physical properties of alkenes and alkynes (boiling points, density anddipole moments); Acidity of alkynes; Acid catalysed hydration of alkenesand alkynes (excluding the stereochemistry of addition and elimination);Reactions of alkenes with KMnO4 and ozone; Reduction of alkenesand alkynes; Preparation of alkenes and alkynes by elimination reactions;Electrophilic addition reactions of alkenes with X2, 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 alkylcarbocation, Grignard reactions, nucleophilic substitution reactions;Alcohols: esterification, dehydration and oxidation, reaction withsodium, phosphorus halides, ZnCl2/concentrated HCl, conversion ofalcohols into aldehydes and ketones; Ethers:Preparation byWilliamson�s Synthesis; Aldehydes and Ketones: oxidation,reduction, oxime and hydrazone formation; aldol condensation, Perkin

reaction; Cannizzaro reaction; haloform reaction and nucleophilicaddition reactions (Grignard addition); Carboxylic acids: formationof esters, acid chlorides and amides, ester hydrolysis; Amines:basicity of substituted anilines and aliphatic amines, preparation fromnitro compounds, 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 and hydrolysisof sucrose.

Amino acids and peptides : General structure (only primary structurefor 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 functional groups: hydroxyl(alcoholic and phenolic), carbonyl (aldehyde and ketone), carboxyl,amino and nitro; Chemical methods of separation of mono-functionalorganic compounds from binary mixtures._________________________

_________________________

CLASS - XII (MATHEMATICS)Complex Number and Quadratic equations: Algebra of complexnumbers, addition, multiplication, conjugation, polar representation,properties of modulus and principal argument, triangle inequality, cuberoots 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 arithmeticand geometric 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 binomialcoefficients.Binomial theorem for any index, exponential and logarithmic series.

Matrices & Determinants: Matrices as a rectangular array of realnumbers, equality of matrices, addition, multiplication by a scalar andproduct of matrices, transpose of a matrix, determinant of a squarematrix of order up to three, inverse of a square matrix of order up tothree, properties of these matrix operations, diagonal, symmetric andskew- symmetric matrices and their properties, solutions ofsimultaneous linear equation in two or three variables.

Probability : Addition and multiplication rules of probability, conditionalprobability, baye�s theorem, independence of events, computation of

probability of events using permutations and combinations.

Straight Line : Cartesian coordinates, distance between two points,section formulae, shift of origin. Equation of a straight line in variousforms, angle between two lines, distance of a point from a line; Linesthrough the point of intersection of two given lines equation of thebisector of the angle between two lines, concurrency of lines; Centroid,orthocentre, incentre and circumcentre of a triangle.

Conic Section: Equation of a circle in various forms, equations oftangent, normal and chord. Parametric equations of a circle, intersectionof a circle with a straight line or a circle, equation of a through thepoints of intersection of two circles and those of a circle and a straightline.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, equationof a straight line in space, equation of a plane, distance of a point froma plane

Vectors: Addition of vectors, scalar multiplication, dot and crossproducts, scalar triple products and their geometrical interpretations.

STPFXII1213

Page - 8

Position vector of a point dividing a line segment in a given ratio. Projectionof a vector on a line.

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

Limit, Continuity & Derivability: Limit and continuity of a function,limit and continuity of the sum, difference, product and quotient of twofunctions, L�Hospital rule of evaluation of limits of functions even and

odd functions, inverse of a function, continuity of composite function.intermediate value property of continuous functions.

Differentiation: Derivative of a function, derivative of the sum,difference, product and quotient of two functions, chain rule, derivativesof polynomial, rational, trigonometric, inverse trigonometric, exponentialand logarithmic functions. Derivatives of implicit functions, derivativesup to order two.

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

Maxima & Minima: Increasing and decreasing functions, maximumand minimum values of a function, rolle�s theorem and Lagrange�s

Mean value theorem.

Integral calculus: Integration as the inverse process of differentiation,indefinite integrals of standard functions, integration by parts, integrationby the methods of 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 first orderdifferential equations.

Trigonometry: Trigonometric functions, their periodicity and graphsaddition and subtraction formulae, formulae involving multiple and sub-multiple angles, 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: Experi-ments based on using Vernier calipers and screw gauge (microme-ter), Determination of g using simple pendulum, Young�s modulus 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 ofsound using resonance column, Verification of Ohm�s law using volt-

meter and ammeter, and specific resistance of the material of a wireusing meter bridge and post office box.

Mechanics : Kinematics in one and two dimensions (Cartesian coor-dinates only), Projectile Motion; Uniform Circular Motion; Relative Ve-locity.Newton�s laws of motion; Inertial and uniformly accelerated frames of

reference; Static and dynamic friction; Kinetic and potential energy;Work and power; Conservation of linear momentum and mechanicalenergy.Systems of particles; Centre of mass and its motion; Impulse; Elasticand inelastic collisions.

Law of gravitation; Gravitational potential and field; Acceleration due togravity; 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 angular momen-tum; Dynamics of rigid bodies with fixed axis of rotation; Rolling with-out 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 and sur-

face tension, capillary rise; Viscosity (Poiseuille�s equation excluded),

Stoke�s law; Terminal velocity, Streamline flow, equation of continuity,

Bernoulli�s theorem and its applications.

Waves : Wave motion (plane waves only), longitudinal and transversewaves, superposition of waves; Progressive and stationary waves;Vibration of strings and air columns;Resonance; Beats; Speed of soundin gases; Doppler effect (in sound).

Thermal physics : Thermal expansion of solids, liquids and gases;Calorimetry, latent heat; Heat conduction in one dimension; Elementaryconcepts of convection and radiation; Newton�s law of cooling; Ideal

gas laws; Specific heats (Cv and Cp for monoatomic and diatomicgases); Isothermal and adiabatic processes, bulk modulus of gases;Equivalence of heat and work; First law of thermodynamics and itsapplications (only for ideal gases); Blackbody radiation: absorptiveand emissive powers; Kirchhoff�s law; Wien�s displacement law,

Stefan�s law.

Electricity and magnetism : Coulomb�s law; Electric field and poten-

tial; Electrical potential energy of a system of point charges and ofelectrical dipoles in a uniform electrostatic field; Electric field lines; Fluxof electric field; Gauss�s law and its application in simple cases, such

as, to find field due to infinitely long straight wire, uniformly chargedinfinite plane sheet and uniformly charged thin spherical shell.

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

Electric current; Ohm�s law; Series and parallel arrangements of resis-

tances 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 a longstraight solenoid; Force on a moving charge and on a current-carryingwire in a uniform magnetic field.Magnetic moment of a current loop; Effect of a uniform magnetic fieldon a current loop; Moving coil galvanometer, voltmeter, ammeter andtheir conversions.Electromagnetic induction: Faraday�s law, Lenz�s law; Self and mutual

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 mirrors andthin lenses; Magnification.Wave nature of light: Huygen�s principle, interference limited to Young�s

double-slit experiment.

Modern Physics : Atomic nucleus; Alpha, beta and gamma radia-tions; Law of radioactive decay; Decay constant; Half-life and meanlife; Binding energy and its calculation; Fission and fusion processes;Energy calculation in these processes.Photoelectric effect; Bohr�s theory of hydrogen-like atoms; Character-

istic and continuous X-rays, Moseley�s law; de Broglie wavelength of

matter waves.

STPFXII1213

Page - 9

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

Course : VIKAAS (JA) & VIPUL (JB)

Correct Wrong Blank


(d soy ,d fod Yi lgh)3 -1 0


(,d ;k ,d ls vf/kd fod Yi lgh)4 0 0

29 to 36 Comprehensions (vuqPNsn) 4 0 0

37 to 38 51 64Matrix Match Type(eSfVªDl lqesy izd kj)

6 [1, 2, 3, 6] 0 0


Part - II (Physics)


Type

PART - I

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

This section contains 20 multiple choice questions. Each question has choices (A), (B), (C) and (D), outof which ONLY ONE is correct.bl [k.M esa 20 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA

1. The HCF of 20x5 � 38x4 + 43x4 � 25x2 and 15x5 � 19x4 + 4x3 is20x5 � 38x4 + 43x4 � 25x2 vkSj 15x5 � 19x4 + 4x3 dk e-l-i- gS&(A) x2(x2 + 1) (B) x2(x2 � 1) (C) x2(x + 1) (D) x2(x � 1)

2. A valid reason for concluding that 635 is not a perfect square of an integer is that635 fdlh iw.kk±d dk oxZ ugha gS bldk ,d lgh dkj.k gS fd&(A) It is an odd integer (;g ,d fo"ke iw.kk±d gS) (B) it ends with 5 (;g 5 ij lekIr gS )(C) it ends with 35 (;g 35 ij lekIr gS ) (D) sum of its odd digits is 8 (blds fo"ke vadksa dk ;ksx 8 gS)

3. The missing number in the following sequence is :1, 3, 3, 6, 7, 9, ?, 12, 21vuqØe 1, 3, 3, 6, 7, 9, ?, 12, 21 dk vKkr in gS&(A) 10 (B) 11 (C) 12 (D) 13

4. Given : x > 0, y > 0, x > y and z 0. The inequality which is not always correct is ?fn;k gS : x > 0, y > 0, x > y rFkk z 0, dkSulh vlfedk lnSo lgh ugha gksxh\

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

x > 2z

y

5. A bag contains '5 red balls and some blue balls' only. If the probability of drawing a blue ball is double theprobability of a red ball, the number of blue balls in the bag is,d FkSys esa dsoy '5 yky xsans rFkk dqN uhyh xsans' gSaA ;fn ,d uhyh xsan dks fudkyus dh izkf;drk ,d yky xsan dks fudkyusdh izkf;drk dh nqxquh gS] rks FkSys esa uhyh xsanksa dh la[;k gS &(A) 5 (B) 10 (C) 15 (D) 20

6. 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

.

STPFXII1213

Page - 10

7. The radii of the bases of two right circular solid cones of same height are 3 cm and 4 cm, respectively. The conesare melted and recast into a solid sphere of radius 5 cm. Then the heights of the given cones isnks leku Å ¡pkbZ okys ledks.kh; oÙkkdkj Bksl 'kadqvksa ds vk/kkj dh f=kT;k,¡ Øe'k% 3 cm rFkk 4 cm gSA 'kadqvksa dks fi?kykdj5 cm f=kT;k dk ,d Bksl xksyk cuk;k tkrk gS] rks fn;s x;s 'kadqvksa dh Å ¡pkbZ;k¡ gS &(A) 10 (B) 30 (C) 20 (D) 90

8. How many numbers between 200 and 600 are divisible by 4, 5, and 6 ?la[;k 200 rFkk 600 ds chp esa fdrus iw.kk±d 4, 5 vkSj 6 ls HkkT; gSa \(A) 6 (B) 7 (C) 8 (D) 9

9. There are four prime numbers written in ascending order. The product of the first three is 385 and that of the lastthree is 1001. The last number is :pkj vHkkT; la[;kvksa dks c<+rs Øe esa fy[kk tkrk gSA izFke rhu dk xq.kuQy 385 gS rFkk vafre rhu dk xq.kuQy 1001 gSArks vafre la[;k gS :(A) 11 (B) 13 (C) 17 (D) 19

10. A is thrice as efficient as B, and B is twice as efficient as C. If A, B and C work together, how long will they taketo complete a job which B completes in 10 days ?A, B ls rhu xquk dk;Z dq'ky gS rFkk B, C ls nks xquk dk;Z dq'ky gSA ;fn A, B o C lkFk esa dk;Z djrs gS rks os fdrus fnuesa lkFk feydj dk;Z dks lekIr djasxs ;fn B vdsyk ml dk;Z dks 10 fnu esa iwjk djrk gS ?

(A) 9

20 days (fnu) (B)

911

days (fnu)

(C) 3 days (fnu) (D) None of these (buesa ls dksbZ ugha)

11. In the given figure, EF || AD and ED || AC. If BF = 4 cm, FD = 6 cm and BE = 8 cm, then BC = _______.nh xbZ vkd fr esa EF || AD vkSj ED || AC. ;fn BF = 4 cm, FD = 6 cm rFkk BE = 8 cm, rc BC =_______.

(A) 12 cm (B) 15 cm(C) 25 cm (D) none of these (buesa ls dksbZ ugha)

12. If ABC is a quarter circle and a circle is inscribed in it and if AB = 1 cm, findradius of smaller circle.;fn ABC ,d oÙk dk pkSFkkbZ gS] ftlds vanj ,d oÙk cuk;k tkrk gS vkSj ;fnAB = 1 cm NksVs oÙk dh f=kT;k Kkr djksA

(A) 12 (B) 2/12 B C

A

1cm

1cm

(C) 2/12 (D) 221

13. If the polynomial ax3 � 4x2 + 3x + 3 when divided by x � 3 leaves the remainder 2 more than the remainder left on

dividing x3 � 4x + a in divided by x �2, then the value of a is

;fn cgqin ax3 � 4x2 + 3x + 3 dks tc x � 3 ls foHkkftr fd;k tkrk gS] rks izkIr 'ks"kQy] x3 � 4x + a dks x �2 ls foHkkftrdjus ij izkIr 'ks"kQy ls 2 vf/kd gS] rks a dk eku gksxk �(A) 0 (B) 1 (C) 2 (D) 3

14. A man, whose eyes are at a height of 10 m above water level, is standing on the deck of a ship. He observes theangle of elevation of the top of a vertical tower as 45° and the angle of depression of the image of the top of the

tower in water as 60°. The distance of the tower from the man is

,d O;fDr ftlds vk¡[ks ty Lrj ls 10 m Å ¡pkbZ ij gS ,d tgkt ds Msd (deck) ij [kMk gSA og ns[krk gS fd ,d Vkojds 'kh"kZ dk mUu;u dks.k 45° gS rFkk Vkoj ds 'kh"kZ ds ty esa izfrfcEc dk voueu dks.k 60° gSA O;fDr ls Vkoj dh nwjh gksxh%

(A) 10 )26( (B) 10 )13( (C) 20 )13( (D) 20 )26(

STPFXII1213

Page - 11

15. If m, n are natural numbers, m > n sum of mth and nth term of an increasing AP is 2m and their product ism2 � n2 , then (m + n)th term of the AP is;fn m, n izkd r la[;k;s gks o m > n gks] ;fn c<+rh gqbZ lekUrj Js<+h ds m osa rFkk n osa in dk ;ksx 2m gks rFkk mudk xq.kuQym2 � n2 gks] rks lekUrj Js<h dk (m + n) oka in gksxk &

(A) nmnm 22

(B)

nm)nm( 22

(C) (m � 2)2 (D) m2 + n2 + mn

16. If two regular polygon are such that ratio between their number of sides is 1 : 2 and ratio of measures of theirinterior angle is 4 : 5, then the number of sides of polygons are;fn nks lecgqHkqt bl izdkj gS fd mudh Hkqtkvksa dk vuqikr 1 : 2 gS rFkk muds vUr%dks.kksa dk vuqikr 4 : 5 gks] rks cgqHkqtksaesa Hkqtkvksa dh la[;k gS �(A) 4, 8 (B) 3, 6 (C) 5, 10 (D) 6, 12

17. p,q and r are three positive numbers and Q = 2

rqp . If (Q�p) : (Q�q) : (Q�r) = 2 : 5 : 7, then find the ratio of

p,q and r ?

;fn p,q o r rhu /kukRed la[;k,sa gks rFkk Q = 2

rqp ;fn (Q�p) : (Q�q) : (Q�r) = 2 : 5 : 7 rks p,q o r dk vuqikr gksxk?

(A) 9 : 7 : 12 (B) 12 : 9 : 7 (C) 12 : 7 : 9 (D) 7 : 9 : 12

18. 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 sum

of the two eldest son�s ages and 3 years after that his age will be equal to the sum of the ages of his eldest

and 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 ?

firk dh mez mlds rhu cPpksa dh mez ds ;ksx ds cjkcj gSA 9 o"kksZ i'pkr mldh mez mlds nks cM+s iq=kksa dh mez ds ;ksx dscjkcj gksxh rFkk mlds 3 o"kZ i'pkr mldh mez mlds lcls cM+s o lcls NksVs iq=k dh mez ds ;ksx ds cjkcj gksxhA iqu% bldsrhu o"kZ i'pkr~ mldh mez mlds nks NksVs iq=kksa dh mez ds ;ksx ds cjkcj gksxha firk dh orZeku mez gksxh ?(A) 32 years (o"kZ) (B) 36 years (o"kZ) (C) 40 years (o"kZ) (D) 44 years (o"kZ)

19. If cos + cos2 = 1, then sin12 + 3 sin10 + 3 sin8 + sin6 + 2 sin4 + 2 sin2 � 2 =

¼;fn cos + cos2 = 1 gS] rks sin12 + 3 sin10 + 3 sin8 + sin6 + 2 sin4 + 2 sin2 � 2 =½(A) 0 (B) 1 (C) 2 (D) 3

20. If , are the roots of x2 + x + 1 = 0 and , are the roots of x2 + 3x + 1 = 0, then ( � ) ( + ) ( + )( � ) =;fn x2 + x + 1 = 0 ds ewy gks rFkk x2 + 3x + 1 = 0 ds ewy , gks] rks ( � ) ( + ) ( + ) ( � ) =(A) 2 (B) 4 (C) 6 (D) 8

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

This section contains 8 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 8 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 lsvf/kd fodYi lgh gS ¼gSa½A

21. Which statement(s) is/are correct? (fuEu dFkuksa esa dkSulk / ls dFku lgh gS&)(A) Quadrilateral with opposite sides parallel is a parallelogram.(foijhr lekUrj Hkqtkvksa okyk prqHkZqt] lekUrj prqHkqZt gksrk gS)(B) A concyclic parallelogram is a rectangle. (,d pØh; lekUrj prqHkqZt] vk;r gksrk gSA)(C) A parallelogram, the diagonals of which intersect at right angles is a square.(,d lekUrj prqHkqZt] ftlds fod.kZ ,d nqljs dks yEcor~ çfrPNsn djrs gS] oxZ gksrk gS)(D) A quadrilateral, the diagonal of which bisect each other at right angles is a rhombus.(,d prqHkqZt ftlds fod.kZ ,d nqljs dks yEcf}Hkkftr djrs gS] leprqHkqZt gksrk gS)

STPFXII1213

Page - 12

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

;fn js[kh; lehd j.kksa d s fuEu fud k; 2x + 3y = 7 , 2x + ( + ) y = 28 d s vuUr gy gSa] rks &(A) = 4 (B) = 8 (C) = 4 (D) = 8

23. In figure ABCD is a square of side 14 cm. With centres A, B, C and D four circlesare drawn such that each circle touch externally two of the remaining three circles.Let x be the area and y be the perimeter of the shadded region, then

fp=k esa] ABCD ,d oxZ gS ftldh Hkqtk 14 cm gSA dsUnzksa A, B, C rFkk D okys pkj oÙkksa dks blizdkj [khapk tkrk gS fd izR;sd oÙk 'ks"k rhuksa oÙkksa esa ls nks dks ckâ; Li'kZ djsaA ;fn Nk;kafdr{ks=k dk {ks=kQy x rFkk ifjeki y gS] rks %

(A) x = 154 (B) x = 42 (C) y = 44 (D) y = 88

24. A sequence is defined by an = n3 � 6n2 + 11n � 6, (n N), then

,d vuqØ e an = n3 � 6n2 + 11n � 6, (n N) }kjk ifjHkkf"kr fd ;k t krk gS] rks&

(A) first three terms are negative ¼igys rhu in _ .kkRed gSA½(B) first three terms are zeroes ¼igys rhu in 'kwU; gSaA½(C) all terms except first three are negative ¼igys rhu d ks NksM+ d j 'ks"k lHkh in _ .kkRed gSA½(D) all terms except first three are positive ¼igys rhu d ks NksM+ d j 'ks"k lHkh in /kukRed gSA½

25. If 1931

= 1 +

b/a1

1

11

11

11

1

then ( ;fn 1931

= 1 +

b/a1

1

11

11

11

1

rks)

(A) ab = 6 (B) a2 + b2 = 13 (C) a � b = �1 (D) a + b = 5

26. If the two sides of a triangle is 14 cm and 22 cm then which can be the third side of that triangle ?;fn ,d f=kHkqt dh nks Hkqtk,¡ 14 cm o 22 cm gS] rks fuEu esa ls dkSulh ml f=kHkqt dh rhljh Hkqtk gks ldrh gS&(A) 7 cm (B) 9 cm (C) 37 cm (D) 34 cm

27. Vijay has been invited for dinner in a club. While walking through the garden path towards the building, heobserve that there is an electric rod on the top of the building. From the point where he is standing, the anglesof elevation of the top of the electric rod and the top of the building are and respectively. If the heights of theelectric rod and the building are p and q respectively, mark all the correct statements.fot; dks ikVhZ esa jkf=k&Hkkst ds fy, cqyk;k tkrk gSA xkMZu esa Vgyrs gq, og ehukj ds 'kh"kZ ij ,d fo|qr NM+ ns[krk gSA ftlfcUnq ij og [kM+k gS ogk¡ ls fo|qr NM+ ds 'kh"kZ vkSj ehukj ds 'kh"kZ dk mUu;u dks.k Øe'k% vkSj gSA ;fn fo|qr NM+ vkSjehukj dh Å ¡pkb;k¡ Øe'k% p rFkk q gSA lgh dFkuksa dk p;u dhft,A

(A) The height of the tower is

tantantanp (ehukj dh Å ¡pkbZ

tantantanp

gS)

(B) The height of the electric rod is )tan(tan

tanq

(fo|qr NM+ dh Å ¡pkbZ

)tan(tantanq

gS)

(C) The height of the tower is

tantantanp

(ehukj dh Å ¡pkbZ

tantantanp

gS)

(D) The height of the electric rod is

tan)tan(tanq

(fo|qr NM+ dh Å ¡pkbZ

tan)tan(tanq gS)

28. If the point (2, 0), (7, 0),

235

,29 , (x, y) are the vertices of a rhombus, then a value of x is

;fn fcUnq (2, 0), (7, 0),

235

,29 , (x, y) leprqHkqZt ds 'kh"kZ gks] rks x dk eku gS &

STPFXII1213

Page - 13

(A) 2

19(B) �

21

(C) 29

(D) 21

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

This section contains 2 paragraphs. Based upon each paragraph, 4 multiple choice questions have to beanswered. 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 4 cgq&fodYih ç'uksa ds mÙkj nsus gSA çR;sd ç'uds 4 fodYi (A), (B), (C) rFkk (D) gS , ft uesa ls flQZ ,d lgh gSA

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

In the following figure, the smaller triangle represents the teachers; thebig triangle, the politicians; the circle, the graduates and the rectangle,the members of parliament. Different regions are being represented by theletters of english alphabetfuEu vkys[k esa] NksVk f=kHkqt] v/;kidksa dks çnf'kZr djrk gS rFkk cM+k f=kHkqt] usrkvksa dksçnf'kZr djrk gSA oÙk] LukÙkd dks çnf'kZr djrk gS rFkk vk;r] laln ds lnL;ksa dksçnf'kZr djrk gSA çR;sd fofHké {ks=kksa dks vaxzsth o.kZekyk ds v{kjksa }kjk vyx&vyxçnf'kZr fd;k x;k gS&

On the basis of the about diagram, answer the following questions :Å ij fn;s x;s fp=k ds vk/kkj ij fuEu ç'uksa dk mÙkj nhft;s&

29. Who among the following are graduates or teachers but not politicians ?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

30. Who among the following politicians are graduates but not the members of parliament ?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. Who among the following politicians are neither teachers nor graduates ?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. Who among the following members of parliament is a graduate as well as a teacher?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 36(iz'u 33 ls 36 ds fy, vuqPNsn)

Read the following information carefully and answer the questions given below itSix flats on a floor in two rows facing North and Sourth are alloted to P, Q, R, S, T and U. Q gets a North facingflat and is not next to S. S and U get diagonally opposite flats. R next to U, gets a Sourth facing flat and T getsa North facing flatfn xbZ lwpuk dks /;kuiwoZd if<+;s vkSj uhps fn;s x;s iz'uksa ds mÙkj nhft, &,d ry ij N% edku nks iafDr;ksa esa ftuds eq[; }kj mÙkj vkSj nf{k.k fn'kk esa gS] P, Q, R, S, T vkSj U dks vkoafVr gSA Qds edku dk eq[; }kj mÙkj esa gS vkSj ;g S ds edku ls vxyk ugha gS S rFkk U ds edku fod.kZ dh foijhr fn'kkvksa esa fLFkrgSA R, U ds vkxs rFkk nf{k.k esa eq[; }kj j[krk gS vkSj T ds edku dk eq[; }kj mÙkj esa gS&

33. Which of the following combinations get South facing flats ?fuEu esa ls fdl lewg ds edkuksa dk eq[; }kj nf{k.k esa gS ?(A) QTS (B) UPT (C) URP (D) None of these (buesa ls dksbZ ugha)

STPFXII1213

Page - 14

34. Whose flat is between Q and S ? (Q rFkk S ds chp fLFkr edku dkSulk gS ?)(A) T (B) U (C) R (D) P

35. If the flats of T and P are interchanged, whose flat will be next to that of U ?;fn T rFkk P vius edku ijLij cny ysrs gS rks U ds edku ls vkxs fdldk edku gksxk \(A) P (B) Q (C) R (D) T

36. Flats other than SU, which are diagonally opposite to each other areSU ds vykok os edku tks fod.kZ ds foijhr fljksa ij fLFkr gS&(A) QP (B) QR (C) PT (D) PS

SECTION - IV ([k.M- IV)

Matrix - Match Type (eSfVªDl&lqesy izdkj)

This section contains 2 question. The question contains statements given in two columns which have to bematched. Statements (A, B, C, D) in Column-I have to be matched with statements (p, q, r, s) in Column-II. Theanswers to these questions have to be appropriately bubbled as illustrated in the following example.If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q, then thecorrectly matched answer will look like in bubbled matrix. One option inColumn�I may have one or more than one correct options in Column�II. Marks

will be awarded only if an option in Column�I is matched with all correct options

in Column-II. Each correct matching will be awarded 1 markand if all the matchings are correct then 6 marks will be awarded.

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

ABCD

p q r s

bl [k.M esa 2 iz'u gSaA izR;sd iz'u esa nks dkWye esa oDRO; (statements) fn;s gq, gSa ftudk lqesy (match) djuk gSA dkWye(Column-I) esa fn;s x;s oDrO;ksa (A, B, C, D) dks dkWye (Column-II) esa fn;s x;s oDrO;ksa (p, q, r, s) ls lqesy djuk gSA bu iz'uksads mÙkj fn;s x;s mnkgj.k ds vuqlkj mfpr cqYyksa dks dkyk djds n'kkZuk gSA ;fn lgh lqesy A-p, A-r, B-p, B-s,

C-r, C-s rFkk D-q gSa] rks lgh fof/k ls dkys fd, x;s cqYyksa dk 4 x 4 esafVªDl (matrix) esa n'kkZ;k x;k gSA dkWye�I ds izR;sd fodYidk dkWye�II ds ,d ;k ,d ls T;knk fodYiksa ls lqesy djk;k tk ldrk gSA vad mlh fLFkfr esa iznku fd;s tk;sxsaA tcfd dkWye�I ds fdlh fodYi dk dkWye�II ds lHkh lgh fodYiksa ls lqesy djk;k x;kA izR;sd lgh lqesy ds fy, 1 vad iznku fd;s tk;sxsa rFkktc lHkh lqesy lgh gksxsa rc 6 vad iznku fd;s tk;sxsaA

37. Column (LrEHk) � Column (LrEHk) �

(A) If and are the roots of x2 � 5x + 3 = 0, then 19

22 is equal to (p) 1

;fn x2 � 5x + 3 = 0 d s ewy o gSa] rks 19

22 cjkcj gS&

(B) The value of 5494141)52( in simplified form is (q) 2

5494141)52( dk ljyre :i gS

(C) The value of 2

321210

in simplified form is (r) � 1

2

321210

dk ljyre :i gS

(D) If a + b + c = 0, then 444

2222

cba

)cba(

has value equal to (s) 7

;fn a + b + c = 0 gks] rks 444

2222

cba

)cba(

dk eku gS

STPFXII1213

Page - 15

38. Column � ¼LrEHk �½ Column � ¼LrEHk �½

(A) If the points (1, 2), (�5, 6) and (a, � 2) are collinear, (p) 0then the value of a is

;fn fcUnq (1, 2), (� 5, 6) rFkk (a, � 2) lajs[k gS] rks a d k eku gS&

(B) The value of log2 log

2 log

3 log

3 273 is (q) 2

log2 log

2 log

3 log

3 273 d k eku gS&

(C) If the system of equations 3x + y = 1 (r) 4(2k � 1) x + (k � 1) y = 2k + 1 has no solution,

then the value of k is

;fn lehd j.k fud k; 3x + y = 1

(2k � 1) x + (k � 1) y = 2k + 1 d k d ksbZ gy ugha gS] rksk d k eku gS&

(D) ABC is an isosceles right triangle with AC = BC, (s) 7then AB2 = xAC2 , where the value of x is

ABC ,d lef}ckgq led ks.k f=kHkqt gS] ft lesa AC = BC,

rks AB2 = xAC2, t gk¡ x d k eku gS&

PART - II (Hkkx - II)

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

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

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

(A) 2 amp. (B) .amp21

(C) .amp41

(D) 1 amp.

40. If the focal length of a concave mirror is 10 cm and the object is placed at a distance of 12 cm from the mirrorthen find the distance between the object and image formed. : (10 lseh- Qksdl nwjh ds ,d vory niZ.k ls 12

lseh- nwjh ij ,d fcEc j[kk gSA izfrfcEc rFkk oLrq ds e/; nwjh Kkr djksA)

(A) 60 cm (B) 48 cm (C) 72 cm (D) 1150

cm

41. Soft iron is used to manufacture electromagnets because their : (ueZ yksgk fo|qr pqEcd cukus esa iz;qä gksrk gS D;ksafd )

(A) coercive force is high (mldk fuxzkgh cy vf/kd gSA)(B) retentivity is high (mldh /kkj.k 'khyrk vf/kd gSA)(C) area of hysteresis curve is large (mlds 'kSfFkY; oØ dk {ks=kQy vf/kd gksrk gSA)(D) magnetic saturation limit is high but retentivity and coercive force are small(mldh ykSg larIr lhek vf/kd gksrh gS ysfdu xzg.k 'khyrk rFkk fuxzkgh cy fuEu gSA)

42. A body of mass 10 kg revolves in a circle of diameter 0.40 m, making 1000 revolutions per minute. Calculate it�slinear velocity. (10 kg nzO;eku dh ,d oLrq 0.40 m O;kl okys ,d oÙkkdkj iFk ij 1000 pDdj izfr ehuV ls xfr djjgh gSA oLrq fd js[kh; pky Kkr djksA)

STPFXII1213

Page - 16

(A) 3

200m/s (B)

32

m/s (C) 3

20m/s (D) 2.0 m/s

43. Force of 50 N is applied to stop a body of mass 20 kg moving initially with a speed of 15 m/s, then the distancetravelled by body before it stops : (,d 20 kg nzO;eku okyh xfreku oLrq dks jksdus ds fy, bl ij 50 N cy yxk;k tkrkgSA izkjEHk esa oLrq 15 m/s dh pky ls xfreku gSA rks :dus ls igys oLrq }kjk r; dh nwjh Kkr djksA) :(A) 40 m (B) 45 m (C) 50 m (D) 100 m

44. Two solid spheres of same radius (R) and of same material are placed in such a way that their centres are 2Rapart. The gravitational force between them is directly proportional to :(nks Bksl xksys] izR;sd dh f=kT;k (R) gS rFkk leku inkFkZ ls cus gS] budks bl izdkj j[kk tkrk gS fd nksuksa ds dsUnzks ds e/; dhnwjh 2R gSA buds e/; xq:Rokd"kZ.k cy fuEu ds lekuqikfr gSA)(A) R2 (B) R�2 (C) R4 (D) R�4

45. A balloon of mass 1000 kg is floating at some height. If 100 kg mass is released from the balloon (withoutchanging volume of the ballon). Then the acceleration of the balloon is : [g = 10 ms�2](1000 kg nzO;eku dk xqckjk iFoh ry ls dqN Å pkbZ ij lkE;koLFkk esa mM+ jgk gSA ;fn xqckjs ls 100 kg nzO;eku fudky fn;ktk, ¼xqckjs dk vkdkj cnys fcuk½ rks xqckjs dk Roj.k gksxkA) [g = 10 ms�2]

(A) 1.1 ms�2 upward (Å ij dh vksj) (B) 1.1 ms�2 downward (uhps dh vksj)(C) 10 ms�2 upward (Å ij dh vksj) (D) 10 ms�2 downward (uhps dh vksj)

46. Light ray AB incidents on a plane mirror XY at an angle of incidence 50º. The second plane mirror is placed in

such a way that the reflected ray BC (from the mirror XY) retraces its path after reflection from the second mirror.Angle of inclination of two mirrors will be :¼izdk'k dh fdj.k AB lery niZ.k XY ij 50º ds vkiru dks.k ij vkifrr gksrhgSA 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 viusiFk ij ykSV tkrh gSaA nksuksa niZ.kksa ds e/; ijkLifjd >qdko dks.k gSa)(A) 25º (B) 50º (C) 75º (D) 90º


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 vfèkdfodYi lgh gS ¼gSa½A

47. A beam of green light is incident from air and after refraction enters water. In comparision to that in air :(ok;q ls vkifrr gjs izd k'k d h fd j.k] viorZu d s ckn t y esa izos'k d jrh gSA ok;q d h rqyuk esa) &(A) Speed of light is less in water (t y esa izd k'k d h pky gksxh)(B) Frequency of light is less in water (t y esa izd k'k d h vkofÙk d e gksxh)(C) Wavelength of light is less in water (t y esa izd k'k d h rjaxnS/;Z d e gksxh)(D) Speed of light is more in water (t y esa izd k'k d h pky T;knk gksxh)

48. A tunnel is made across the earth passing through its centre. A particle is released from the one end of thetunnel when the particle is at the centre of the earth then which of the following is non zero.(,d lqajx iFoh ds dsUnz ls xqtjrh gSA ,d d.k] lqjax ds ,d fljs ls NksM+k tkrk gS tcfd d.k iFoh ds dsUnz ij gS rks fuEuesa ls dkSu ls v'kwU; gSA) �(A) mass (nzO;eku) (B) weight (Hkkj) (C) acceleration (Roj.k) (D) velocity (osx)

49. Mark the correct statement(s) . (lgh dFku dks bafxr dhft;s) �(A) if speed of a body is varying, its velocity must be varying .

(;fn ,d oLrq dh pky cnyrh gS rks mldk osx fuf'pr :i ls ifjofrZr gksxk )(B) if velocity of a body is varying, its speed must be varying

(;fn oLrq dk osx ifjofrZr gksrk gS rks mldh pky Hkh fuf'pr :i ls ifjofrZr gksxh)

STPFXII1213

Page - 17

(C) a body moving with varying velocity may have constant speed(oLrq tks ifjofrZr osx ls xfreku gS fd pky fu;r jg ldrh gS)

(D) a body moving with varying speed may have constant velocity(,d oLrq ftldh pky ifjofrZr gksrh gS rks bldk osx fu;r jg ldrk gS S)

50. From the top of a tower of height 200 m, a ball A is projected up with 10 m s 1 and two seconds later

another ball B is projected vertically down with the same speed. Then : (Take g = 10 m/s2)(200 m Å ¡pkbZ dh ehukj ls ,d xsan dks Å ij dh rjQ 10 m/s ds osx ls iz{ksfir djus ds nks lsd.M ckn nqljh xsan B dksleku pky ls uhps dh rjQ iz{ksfir fd;k tkrk gS rks) (g = 10 m/s2 fyft;s)(A) both A and B will reach the ground simultaneously (A rFkk B nksuksa lkFk lkFk tehu ij igq¡prh gS )(B) the ball A will hit the ground 2 seconds later than B hitting the ground

(xsans A, xsan B ds Vdjkus ds nks lsd.M i'pkr~ tehu ij Vdjkrh gS)(C) both the balls will hit the ground with the same velocity (nksuksa xsan leku osx ls tehu dks Vdjkrh gSa)(D) none of these (buesa ls dksbZ ugha)

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

This section contains 1 question. Each question contains statements given in two columns which have to bematched. Statements (A, B, C, D) in Column-I have to be matched with statements (p, q, r, s) in Column-II. Theanswers to these questions have to be appropriately bubbled as illustrated in the following example.If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q, then thecorrectly will look like in bubbled matrix. Each statement in Column�I may have

one or more than one correct options in Column�II. One Mark will be awarded for

each statement in Column�I is matched with all correct options in Column-II. If all

the statements in column-I are correctly matched to theircorrect options in column-II , then 6 marks will be awarded.


ABCD

p q r s

bl [k.M esa 1 iz'u gSaA izR;sd iz'u esa nks dkWye esa oDRO; (statements) fn;s gq, gSa ftudk lqesy (match) djuk gSA dkWye(Column-I) esa fn;s x;s oDrO;ksa (A, B, C, D) dks dkWye (Column-II) esa fn;s x;s oDrO;ksa (p, q, r, s) ls lqesy djuk gSA buiz'uksa d s mÙkj fn;s x;s mnkgj.k d s vuqlkj mfpr cqYyksa d ks d kyk d jd s n'kkZuk gSA ;fn lgh lqesyA-p, A-r, B-p, B-s, C-r, C-s rFkk D-q gSa] rks lgh fof/k ls dkys fd, x;s cqYyksa dk 4 x 4 esafVªDl (matrix) esa n'kkZ;k x;k gSAdkWye�I ds izR;sd oDÙkO; dk dkWye�II ds ,d ;k ,d ls T;knk fodYiksa ls lqesy djk;k tk ldrk gSA dkWye�I ds izR;sd oDÙkO;dk dkWye�II ds lHkh lgh fodYiksa ls lqesy djkus ds fy, ,d vad iznku fd;k tk;sxkA ;fn dkWye�I ds lHkh OkDÙkO;ksa dk dkWye�II ds lHkh lgh fodYiksa ls lqesy fd;k x;k rks 6 vad iznku fd;s tk;sxsaA

51. A particle is moving along a straight line. Its velocity varies with time as v = kt, where k is a positive constant andt is the time. Match the graphs in Column with the statements in Column and indicate your answer bydarkening appropriate bubbles in the 4 × 4 matrix given in the ORS.

,d d.k ,d ljy js[kk ds vuqfn'k xfr dj jgk gSA bldk osx le; ds lkFk bl izdkj cnyrk gS v = kt, tgk¡ k /kukRed fu;rkadgS vkSj t le; gSA dkWye esa xzkQksa dks dkWye esa fn;s x;s oDrO;ksa ls lqesy dhft, vkSj vius mÙkj dks nh xbZ ORS esa mfprcqycqyksa dks dkyk dj 4 × 4 eSfVªDl esa n'kkZb;sA

Column (dkWye-) Column (dkWye-)

(A) Acceleration versus time curve (p) Roj.k vkSj le; oØ

(B) Acceleration versus displacement curve (q) Roj.k vkSj foLFkkiu oØ

(C) Velocity versus time curve (r) osx vkSj le; oØ

STPFXII1213

Page - 18

(D) Displacement versus time curve (s) foLFkkiu vkSj le; oØ

PART - III


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

52. The highest value of e/m (charge / mass ratio) for anode rays has been observed when the discharge tube isfilled with : (,uksM fdj.kksa ds fy, e/m (vkos'k / nzO;eku vuqikr) dk vf/kdre eku izkIr gksrk gS tc foltZu ufydk esa fuEuesa ls dkSulh xSl Hkjh gqbZ gks %)(A) nitrogen (ukbVªkstu) (B) oxygen (vkWDlhtu) (C) hydrogen (gkbMªkstu) (D) helium (ghfy;e)

53. Which of the following are isoelectronic species? (fuEu esa ls lebysDVªkWfud Lih'kht dkSulh gS %)3CH ,

2NH , 4NH , 3NH

(i) (ii) (iii) (iv)(A) I, II, III (B) II, III, IV (C) I, II, IV (D) II, I

54. The time taken for 80 g of uranium (t1/2

= 4.5 × 109 years) to decay to 10 g would be :(80 g ;wjsfu;e (t

1/2 = 4.5 × 109 o"kZ) ds 10 g rd {k; gksus esa fdruk le; yxsxkA)

(A) 4.5 × 109 years (B) 9.0 × 109 years (C) 13.5 × 109 years (D) 18.0 × 109 years

55. Volume of the 10 g of gas X is 5.6 litre at STP. What will be the molecular weight of X ?(STP ij 10 g xSl X dk vk;ru 5.6 yhVj gSA X dk v.kqHkkj D;k gksxk \)(A) 40 (B) 50 (C) 60 (D) 20

56. In which of the following arrangement, the order is not according to the property indicated against it ?(fuEu esa ls dkSulk Øe] n'kkZ;s x;s xq.kksa ds vuqlkj lgh ugha gSA)(A) Al3+ < Mg2+ < Na+ < F� ; ionic size (vk;fud vkdkj)(B) B < C < N < O ; first ionisation enthalpy (izFke vk;uu ,UFkSYih)(C) < Br < F < Cl ; electron gain enthalpy (with negative sign) (bysDVªkWu xzg.k ,UFkSYih (_ .kkRed fpUg lfgr))

(D) Li < Na < K < Rb ; metallic radius (?kkfRod f=kT;k)

57. The simplest formula of a compound containing 50% X (atomic mass 10 u) and 50% Y (atomic mass 20 u) is :(50% X (ijekf.od nzO;eku 10 u) rFkk 50% Y (ijekf.od nzO;eku 20 u) ;qDr ;kSfxd dk ljyre lw=k gS %)(A) XY

2(B) X

2Y (C) X

2Y

3(D) XY

58. What is the pH of 0.001M NaOH aqueous solution ? (0.001M NaOH tyh; foy;u dh pH gksxh \)(A) 3 (B) 11 (C) 6 (D) 14

59. The structure of 4-Methyl- 2-hexyne is : (4-esfFky - 2-gsDlkbu dh lajpuk gS %)

(A)

3

323

CH|

CH�CH�CH�CHCH�CH (B)

3

323

CH|

CH�CC�CH�CH�CH

(C)

3

323

CH|

CH�CH�CH�CC�CH (D)

3

3222

CH|

CH�CC�CH�CH�CH

STPFXII1213

Page - 19


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 lsvf/kd fodYi lgh gS ¼gSa½A

60. Identify the correct with respect to ozone : (vkstksu ds lanHkZ esa lR; dFku dks igpkfu,A)(A) Ozone is formed in the upper atmosphere by a photochemical reaction involving dioxygen.

(vkstksu dk fuekZ.k Å ijh okrkoj.k esa MkbZvkWDlhtu dh izdk'k jklk;fud vfHkfØ;k }kjk gksrk gSA)(B) Ozone is more reactive than dioxygen. (vkstksu MkbZvkWDlhtu ls vf/kd fØ;k'khy gSA)(C) Ozone is diamagnetic whereas dioxygen is paramagnetic.

(vkstksu izfrpqEcdh; gS tcfd MkbZvkWDlhtu vuqpqEcdh; gSA)(D) Ozone protects the earth�s inhabitants by absorbing ultraviolate-radiations.

(vkstksu ijkcSaxuh fofdj.kksa dks vo'kksf"kr dj iFoh ij mifLFkr lthoksa dh j{kk djrk gSA)

61. Which of the following are non-linear structures ? (fuEu esa ls vjs[kh; lajpuk okys ;kSfxd gSa \)(A) C

2H

2(B) H

2O (C) O

3(D) I

3�

62. Constituent metals of the alloy german silver are : ¼feJ/kkrq teZu flYoj dh ?kVd /kkrq;sa gS %½(A) Zn (B) Cu (C) Fe (D) Ni

63. Which of the following is(are) sulphide ore(s) (fuEu esa ls dkSulk@dkSuls lYQkbM v;Ld gS@gSaA)(A) Galena (xSysuk) (B) Zinc blende (ftad CysMa)(C) Cinnabar (flusckj) (D) Argentite (vtsZUVkbV)

SECTION - III ([k.M- III)

Matrix - Match Type (eSfVªDl&lqesy izdkj)This section contains 1 question. Each question contains statements given in two columns which have to bematched. Statements (A, B, C, D) in Column-I have to be matched with statements (p, q, r, s) in Column-II. Theanswers to these questions have to be appropriately bubbled as illustrated in the following example.If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q, then thecorrectly will look like in bubbled matrix. Each statement in Column�I may have

one or more than one correct options in Column�II. One Mark will be awarded foreach statement in Column�I is matched with all correct options in Column-II. If allthe statements in column-I are correctly matched to theircorrect options in column-II , then 6 marks will be awarded.


ABCD

p q r s

bl [k.M esa 1 iz'u gSA izR;sd iz'u esa nks dkWye esa oDRO; (statements) fn;s gq, gSa ftudk lqesy (match) djuk gSA dkWye(Column-I) esa fn;s x;s oDrO;ksa (A, B, C, D) dks dkWye (Column-II) esa fn;s x;s oDrO;ksa (p, q, r, s) ls lqesy djuk gSA bu iz'uksads mÙkj fn;s x;s mnkgj.k ds vuqlkj mfpr cqYyksa dks dkyk djds n'kkZuk gSA ;fn lgh lqesy A-p, A-r, B-p, B-s,

C-r, C-s rFkk D-q gSa] rks lgh fof/k ls dkys fd, x;s cqYyksa dk 4 x 4 esafVªDl (matrix) esa n'kkZ;k x;k gSA dkWye�I ds izR;sd oDÙkO;dk dkWye�II ds ,d ;k ,d ls T;knk fodYiksa ls lqesy djk;k tk ldrk gSA dkWye�I ds izR;sd oDÙkO; dkdkWye�II ds lHkh lgh fodYiksa ls lqesy djkus ds fy, ,d vad iznku fd;k tk;sxkA ;fn dkWye�I ds lHkh OkDÙkO;ksa dkdkWye�II ds lHkh lgh fodYiksa ls lqesy fd;k x;k rks 6 vad iznku fd;s tk;sxsA

64. Match the column : (fuEu LrEHkksa dks lqesfyr dhft,A)Column-I Column-IIdkWye-I dkWye-II

Molecule ¼v.kq½ Number of C�H bonds ¼C�H ca/kksa dh la[;k½(A) Ethane ¼,sFksu½ (p) 2

STPFXII1213

Page - 20

(B) Ethene ¼,sFkhu½ (q) 4

(C) Ethyne ¼,sFkkbu½ (r) 6

(D) Propane ¼çksisu½ (s) 8

ANSWER KEY TO SAMPLE TEST PAPER-I

1. D 2. C 3. D 4. C 5. B 6. C 7. C

8. A 9. B 10. A 11. C 12. A 13. B 14. B

15. A 16. D 17. B 18. B 19. B 20. D 21. AB

22. AD 23. BC 24. BD 25. ABCD 26. BD 27. AD 28. ABC

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

36. A 37. (A - p),(B - r),(C - s),(D - q) 38. (A - s),(B - p),(C - q),(D - q)

52. C 53. B 54. C 55. A 56. B 57. B 58. B

59. C 60. ABCD 61. BC 62. ABD 63. ABC 64. (A - r),(B - q),(C - p),(D - s).

HINTS & SOLUTION TO SAMPLE TEST PAPER-I1. 20x5 � 38x4 + 43x4 � 25x2

= x2 (20x3 � 38x2 + 43x � 25)

= x2 (x � 1)(20x2 � 18x + 25)15x5 � 19x4 + 4x3 = x3(x � 1)(15x � 4)

H.C.F. e-l-i- = x2(x � 1)

2. All terms which ends with 35 is not a perfect square.(35 ij lekIr gksus okyh lHkh la[;k,a fdlh dk Hkh iw.kZ oxZ ugha gksrh gS)

3.

Required number vHkh"V la[;k = 7 + 6 = 13

4. Let ekuk z < 0 zx < zyso xz > zy is not correct always. (vr% xz > zy lnSo lgh ugha gksxhA)

5. Let there be x blue balls in the bag. Total number of balls in the bag = (5 + x)

P1 = Probability of drawing a blue ball =

x5x

P2 = Probability of drawing a red ball =

x55

P1 = 2P

2

x5

x = 2

x5

5 x = 10

6. Let x = 352.0

STPFXII1213

Page - 21

10x = 35.2

1000x = 235.35. + 9935

990x = 233

x = 990233

x = � 23

, y = 23

7. Let h be the height of each cone

then32

221 R

34

hr31

hr31

322 534

h431

h331

h = 16954 3

= 20 cm.

8. Every such number must be divisible by L.C.M. of 4, 5, 6 i.e.60.Such numbers are 240, 300, 360, 420, 480, 540Clearly, there are 6 such numbers.

9. Let the given prime numbers be a, b, c, d. Then, abc = 385 and bcd = 1001.

1001385

bcdabc

135

da .

So, a = 5, d = 13.

10. B takes 10 days to finish the job

A will take 3

10 days to finish it. C will take 20 days to finish it.

Together they will finish the work in

=

10201

101

1

= 9

20 days.

11. We have EF||AD and ED||ACand BF = 4 cm.

FD = 6 cm.BE = 8 cm.

In triangle BAD EF ||AD

A

E

B F DC

EABE

FDBF

64

=EA8

EA = 4

48 = 12 cm, AB = 8+12 = 20 cm

Now from triangle ABC ED || AC

EABE

DCBD

orDCBD

BD

= EABE

BE

STPFXII1213

Page - 22

BC10

= 208

BC = 8

200= 25 cm

12. In right tangle triangle OABAB2 + BO2 = OA2

x2 + x2 = (1 � x)2

2x2 = x2 + 1 � 2x

x2 + 2x � 1 = 0

x = 2

442�

x = �1 ± 2taking positive sign

x = 2 �1

13.x2x

ax4�x)2�x(

2

3

x3 � 2x2 + 2x2 � 4x

remainder is a 'ks"kQy gS

Now remainder ofvc 'ks"kQy

)3�x(3x3x4�ax 23

= a + 2

9�a9x4�ax3ax

3x3x4�ax)3�x(

2

23

ax3 � 3ax2 + 3ax2 � + 9ax � 4x2 + 12x + 9ax � 27a � 9x � 27

so 27 � 27a + a + 2 = 3

26 = 26a a = 1

14. From the diagram fp=k ds vuqlkj

ABCAC = y � 10

and invkSj ACD esa

AC10y

= tan60º = 3 = 10�y10y

y = 10( 23 )

Required distancevHkh"V nwjh = (y � 10) = 10( 23 ) Ans.

15. Tm + T

n = (m + n) + (m � n) = 2m

Tm . T

n = (m + n) (m � n) = m2 � n2

Let a, d be the first term and common difference respectively.ekuk a, d Øe'k% izFke lkoZvUrj in gSAa + (m � 1) d = m + n

a + (n � 1) d = m � n

(m � n) d = 2n

d = n�m

n2

Tm + n

= a + (m + n � 1) d

STPFXII1213

Page - 23

= [a + (m � 1) d ] + nd

= (m + n) + nn�m

n2

Tm + n

= nmnm 22

16. Let the side of first polygon (ekuk izFke cgqHkqt dh Hkqtk) = x

then the side of second polygon (rks f}rh; cgqHkqt dh Hkqtk) = 2x

Now interior angle of a regular polygon is (vc le cgqHkqt dk vkarfjd dks.k)

= n2

�

(where n is the side) (tgk¡ n Hkqtk gS)

Given (fn;k x;k gS ) 2

1

=

54

x22

�

x2

�

= 54

x/1�1x/2�1

= 54

x = 6

17. Givenfn;k x;k gS Q = 2

rqp and vkSj

2p�Q

= 5

q�Q =

7r�Q

Let2

p�Q=

5q�Q

= 7

r�Q = k

Putting value of Q dk eku j[kus ij2

p�rq =

5q�pr

= 7

r�qp = k

q + r � p = 2k ........(i)

r + p � q = 5k ........(ii)

p + q � r = 7k ........(iii)

Equation (i) + equation (ii) equation (ii) + (iii) Equation (iii) + (i)2r = 7k 2p = 12k 2q = 9k

r = 2k7

p = 6k z = 2k9

p : q : r 6k : 2k9

: 2k7

12 : 9 : 7

18. Let the father's present age be F and C1, C

2, C

3 be the present ages of his children in the decreasing order so

(firk dh orZeku vk;q F vkSj C1, C

2, C

3 ?kVrs Øe esa mlds cPpksa dh vk;q )

F = C1 + C

2 + C

3

F + 9 = (C1 + 9) + (C

2 + 9) F = C

1 + C

2 + 9 C

3 = 9

(F + 12) = (C1 + 12) + (C

3 + 12) F = C

1 + C

3 + 12 C

2 = 12

(F + 15) = (C2 + 15) + (C

3 + 15) F = C

2 + C

3 + 15 C

1 = 15

so F = 15 + 12 + 9F = 36 years (o"kZ)

20. + = � 1, = 1 + = � 3, = 1

STPFXII1213

Page - 24

To find Kkr gS ( � ) ( + ) ( + ) ( � )= ( + � � )( � + � )= ( � ) ( � )= 2 � 2� 2+ 2 { = = 1}= 2 � 2� 2+ 2

= [( + )2� (2)] � [( + )2� 2]= 9 � 8 + 1 = 8

21. only A and B is correct (dsoy A o B lR; gSA)

22. 2873

22

= 4, + = 12 = 8

24. an = n3 � 6n2 + 11n � 6 = (n � 1)(n � 2)(n � 3)

26.

|AB � AC| < BC < AB + AC 8 < BC < 36

27. From figure fp=k ls

tan = AB

qp

tan = ABq

qqp

=

tantan

p tan + q tan = q tan

p =

tan)tan�(tanq

q =

tan�tantanp

28. A.

235

,29

Letekuk A

235

,29

, B(2, 0) and vkSj C(7,0)

are fixed points the position of fourth point D(x, y) can be lie thispkSFkk fcUnq D (x, y) ij fLFkj gks ldrk gSA

STPFXII1213

Page - 25

x1 =

29

+ 5, x2 =

29

� 5

x1 =

219

x2 = �

21

x3 =

29

32.

37. (A) + = 5, = 3 2 + 2 = ( + )2 � 2 = 25 � 6 = 19

(B) )5�2( 5494141 {(a + b)2 = a2 + b2 + 2ab

(2 � 5 ) 2254141

(2 � 5 ) 548141

(2 � 5 ) )52(41

(2 � 5 )(2 + 5 ) = 4 � 5 = � 1

(C)2

3�21210

= 10 + 2 21 + 3 � 2 3 21210

= 13 + 2 21 � 2 21630 = 13 + 2 21� 23212

= 13 + 2 21 � 6 � 2 21 = 7

(D) To findKkr gS 444

2222

cba

)cba(

= ?

= 444

222222444

cba

ac2cb2ba2cba

from A ls a4 + b4 + c4 = 2(a2b2 +b2c2 + c2a2)

]accbba[2

]ca2cb2ba2[2222222

222222

= 2

38. (A) 1(6 + 2) + (�5)(�2 � 2) + a(2 � 6) = 0

8 + 20 � 4a = 0

4a = 28 a = 7

(B) log2 log

2 log

3 log

3 273 = log

2 log

2 log

3 log

3 39 = log

2 log

2 log

3 9 = log

2 log

22 = log

21 = 0

(C)1�k2

3 =

1�k1

1k2

1

3k � 3 = 2k � 1

STPFXII1213

Page - 26

k = 2 1�k

1

1k21

(D) AB2 = AC2 + CB2

AB2 = AC2 + AC2

AB2 = 2AC2 (Since AC = CB)k = 2

40. Given : f = �10 cm, u = �12 cm, v = ?by mirror formula

121

v1

101

u1

v1

f1

by solving, we get v = �60 cm

i.e. image will form in same side of object at 60 cm from mirror.

46. 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.

48. At the centre of the earth gravitational field is zero.So, at centre of gravitational force (i.e. weight) of particle and acceleration of particle is zero, whereas its speedis maximum.

50. s = ut + 2

at2

0 = 10t � 2t10 2

t = 2 sHence A & C are correct.vr% A o C lgh gSA

51. (A), (B)

v = kt Acceleration a = dtdv

= k = constant

a = constant (straraight line)i.e. acceleration is constant irrespective of the value of displacement and time.

(C) v = kt (straight line)

(D) v = ktv = 0 + kt .............(i)

From v = u + at .............(ii) (equation of motion)By comparing (i) & (ii)u = 0, a = kFrom v2 = u2 + 2ax v2 = 0 + 2kx

x = k2

v2

(parabola).

58. [OH�] = 10�3

So, [H+] = 10�11

pH = � log [H+]pH = 11

STPFXII1213

Page - 27

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

Course : VISHWAAS (JF)

Correct Wrong Blank


(dsoy ,d fodYi lgh)3 -1 0


(,d ;k ,d ls vf/kd fodYi lgh)4 0 0


31 52 67Matrix Match Type(eSfVªDl lqesy izdkj)

6 [1, 2, 3, 6] 0 0


Part - II (Physics)


Type

PART - I

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


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

0| is

;fn |z � i| 2 rFkk z0 = 5 + 3i gS] rks |iz + z

0| dk vf/kdre eku gS

(A) 2 + 31 (B) 7 (C) 31 � 2 (D) None of these ¼buesa ls dksbZ ugha½

2. 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

3. 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 dhla[;k dks leku j[krs gq, vius f[kykSuksa dks ,d&nwljs ls cny ysaA(A) 329 (B) 334 (C) 332 (D) 345

4. The equations of the perpendicular bisectors of the sides AB and AC of a ABC are x � y = 0 and

x + y = 0 respectively. If the point A is (1, �2) then the equation of the altitude of ABC through A isf=kHkqt ABC dh Hkqtk AB vkSj AC ds yEc lef}Hkktdksa ds lehdj.k x � y = 0 vkSj x + y = 0 gSA ;fn A(1, �2) gS] rc Als tkus ABC ds 'kh"kZ yEc dk lehdj.k gS&(A) 14x + 23y = 0 (B) 14x � 23y = 0 (C) 2x � y � 4 = 0 (D) x � 2y = 0

5. The digits, from 0 to 9 are written on 10 slips of paper (one digit on each slip) and placed in a box. If three of theslips are drawn and arranged, then the number of possible different arrangements isdkxt ds 10 VqdMkssa ij 0 ls 9 vad (izR;sd VqdMs ij ,d vad ) fy[kdj ,d ckWDl esa j[kk tkrk gSA ;fn bu VqdMksa esa ls rhuVqdMksa dks O;ofLFkr fd;k tkrk gS] rc fofHkUu laHkkfor Øep;ksa dh la[;k gS&(A) 1000 (B) 720 (C) 810 (D) None of these ¼buesa ls dksbZ ugha½

6. First term of a G.P. having positive common ratio exceeds the second term by 2, if the sum of first infinite termsof the series is 50, then the third term of the series is : (/kukRed lkoZvuqikr dh xq.kksÙkj Js<h dk izFke in] nwljs in ls2 vf/kd gSA ;fn bl xq.kksÙkj Js<h ds vuUr inksa dk ;ksxQy 50 gS] rc Js<h dk rhljk in gS&)

(A) 5

32(B)

534

(C) 5

36(D) None of these ¼buesa ls dksbZ ugha½

STPFXII1213

Page - 28

7. Vertices of ABC are A (0 , 0) , B (�3 , �1) and C (�1 , �3) . The equation of line parallel to BC and intersecting

the sides AB and AC such that perpendicular distance of the line from origin is 21

, is

ABC ds 'kh"kZ A (0 , 0) , B (�3 , �1) rFkk C (�1 , �3) gSA BC ds lekUrj js[kk tks AB rFkk AC dks izfrPNsn djrh gS rFkk ftldh

ewy fcUnq ls yEcor~ nwjh 21 gS] dk lehdj.k gS &

(A) x + y � 2

1 = 0 (B) x + y +

2

1 = 0

(C) 2 x + 2 y � 2 = 0 (D) 2 x + 2 y + 2 = 0

8. If the solution set of the inequality )x5x(loglog 259.0

> 0, then contains n integral values, then n =

;fn vlfedk )x5x(loglog 259.0

> 0 ds gyksa dk leqPp; n iw.kk±d eku j[krk gks] rks n =

(A) 7 (B) 8 (C) 6 (D) 10

9. A cone of height 24cm and radius of base 6cm is made up of modelling clay. A child reshapes it in the form of asphere, then the radius of the sphere is,d fpduh feVV~h ls cus gq, 'kadq dh Å ¡pkbZ 24 lseh- rFkk vk/kkj dh f=kT;k 6 lseh- gSA ,d cPpk bls ,d xksys ds :i esa iqu%fufeZr (reshapes) djrk gS] rks xksys dh f=kT;k gS&(A) 2 (B) 4 (C) 5 (D) 6

10. Given figure shows a circle with centre at O, AOB = 30º, and OA = 6 cm,

then area of the shaded region isfn;s x;s fp=k esa ,d oÙk dk dsUnz O gS rFkk AOB = 30º, o OA = 6 cm gS] rksNk;kafdr {ks=k dk {ks=kQy gS&(A) 3 � 9 cm2 (B) 3 cm2

(C) 9 � 3 cm2 (D) 3 � 9 3 cm2

11. If AD = 21

BD, then the value of sin is : (;fn AD = 21

BD gS] rc sin dk eku gS&)

(A) 22 x5y9

x

(B) 22 x5y9

x2

B

A

C

Dy

x

(C) 22 x5y3

x

(D) 22 x4y3

x2

12. The value of g x( )

1

1

g (x) + [x] dx where [ . ] is the greatest integer function and g (x) is continous and

differentiable for all x is ( g x( )

1

1

g (x) + [x] dx dk eku gS tgk¡ [ . ] egÙke iw.kkZad Qyu rFkk g (x) lHkh x ds fy;s

lrr~ o vodyuh; gSA)(A) g(1) g(1) 2 (B) g (1) g(1) + 2 (C) 2 (D) 1

13. Domain of f (x) = xsin|xsin|

1

(dk izkUr gS) is

(A) No solution (dksbZ gy ugha) (B) ((2n + 1) , 2 (n + 1)) , n I

(B) [2n, (2n + 1)], n I (D) None of these ¼buesa ls dksbZ ugha½

STPFXII1213

Page - 29

14. If all words formed by the letters of the word � PARKAR� are arranged alphabatically then 99th word is;fn � PARKAR� 'kCn ds v{kjksa ls 'kCn cukdj vaxzsth 'kCn dks"k ds vuqlkj O;ofLFkr fd;k tkrk gS] rc 99osa 'kCn gksxk&(A) PARAKR (B) PARKRA (C) PARKAR (D) PARARK

15. In the given sequence one number is wrong. Identify the wrong number :4, 7, 16, 43, 126, 367, 1096nh xbZ Js<h esa ,d la[;k xyr gSA xyr la[;k dks igpkfu, :4, 7, 16, 43, 126, 367, 1096(A) 367 (B) 7 (C) 43 (D) 126


This section contains 6 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 6 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

16. If A, B, C are angles of ABC and tan A tan C = 3, tan B tan C = 6, then;fn f=kHkqt ABC ds dks.k A, B, C gS vkSj tan A tan C = 3, tan B tan C = 6, rc

(A) A = 4

(B) tan A tan B = 2

(C) CtanAtan

= 23

(D) tan3A + tan3B = � tan3C + 3 tan A tan B tan C = 0

17. Chords of a hyperbola 2

2

a

x � 2

2

b

y = 1 are drawn passing through a fixed point (2h, 2k), then the locus of their

mid-points is :

vfrijoy; 2

2

a

x� 2

2

b

y= 1 dh ,d thok] ,d fLFkj fcUnq (2h, 2k) ls xqtjrh gS] rc muds e/; fcUnq dk fcUnqiFk gS&

(A) a hyperbola similar to the given hyperbola if 2

2

a

h> 2

2

b

k

(,d vfrijoy; tks fn;s x;s vfrijoy; ds le:i gS ;fn 2

2

a

h> 2

2

b

k)

(B) an ellipse if 2

2

a

h= 2

2

b

k (,d nh?kZoÙk ;fn 2

2

a

h= 2

2

b

k)

(C) a conic with centre (h, k) (,d 'kkado ftldk dsUnz (h, k) gSA)(D) a conic with centre (2h, 2k) (,d 'kkado ftldk dsUnz (2h, 2k) gSA)

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

a

c5, log

c5

b3, log

b3

a are in A.P, then which of the

following is/are correct ?

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

a

c5, log

c5

b3, log

b3

a lekUrj Js<h esa gS] rc fuEu esa

ls dkSulk@dkSuls dFku lgh gS ?

(A) b = 3

5c (B) a , b , c cannot be the sides of a triangle

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

(C) a = 9

c25(D) a =

95

c

STPFXII1213

Page - 30

19. If (t2 , 2t) is one end of a focal chord of the parabola y2 = 4x, then which is (are) true;fn (t2 , 2t) ijoy; y2 = 4x dh ukHkh; thok dk ,d fljk gks] rks fuEu esa ls dkSulk@ls lR; gSa&

(A) length of the focal chord is 2

t1

t

(ukHkh; thok dh yEckbZ

2

t1

t

gS)

(B) slope of the focal chord is 1�t

t22 (ukHkh; thok dh izo.krk

1�t

t22 gS)

(C) mid-point of focal chord is focus (ukHkh; thok dk e/; fcUnq] ukfHk gS)(D) focal-chord is angle bisector of tangent and the normal at (t2 , 2t) to the parabola

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

20. In the figure shown, O is centre of the circle and DAB = 50º, then

uhps fn, x, fp=k esa] oÙk dk dsUnz O gS rFkk DAB = 50º ] rks

(A) x = 100º

(B) y = 140º

(C) x = 110º

(D) y = 130º

21. Let f(x) = min {1, cosx, 1 - sinx}; x , then : (;fn f(x) = min {1, cosx, 1 - sinx}; x , rks)(A) f is not differentiable at �0� (f, �0� ij vodyuh; ugha gS)(B) f is differentiable at /2 (f, /2 ij vodyuh; gS)(C) f has a local maxima at �0� (f, �0� ij mPpre fcUnq j[krk gS)(D) none of these (buesa ls dksbZ ugha)


This section contains 3 paragraphs. Based upon each paragraph, 3 multiple choice questions have to beanswered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.bl [k.M esa 3 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 3 cgq&fodYih ç'uksa ds mÙkj nsus gSA çR;sd ç'uds 4 fod Yi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA

Paragraph for Question Nos. 22 to 24iz'u 22 ls 24 ds fy, vuqPNsn

Let P,Q,R,S and T be five sets about the quadratic equation (a � 5) x2 � 2ax + (a � 4) = 0, a 5 such thatP : All values of a for which the product of roots of given quadratic equation is positive.Q : All values of a for which the product of roots of given quadratic equation is negative.R : All values of a for which the product of real roots of given quadratic equation is positive.S : All value of a for which the roots of given quadratic equation are real.T : All values of a for which the given quadratic equation has imaginary roots.On the basis of above information, answer the following questions :ekuk P,Q,R,S vkSj T f}?kkr lehdj.k (a � 5) x2 � 2ax + (a � 4) = 0, a 5 ds lkis{k ik¡p leqPp; gSAP : a ds lHkh eku ftuds fy, f}?kkr lehdj.k ds ewyksa dk xq.kuQy /kukRed gSAQ : a ds lHkh eku ftuds fy, f}?kkr lehdj.k ds ewyksa dk xq.kuQy _ .kkRed gSAR : a ds lHkh eku ftuds fy, f}?kkr lehdj.k ds okLrfod ewyksa dk xq.kuQy /kukRed gSAS : a ds lHkh eku ftuds fy, f}?kkr lehdj.k ds ewy okLrfod gSAT : a ds lHkh eku ftuds fy, f}?kkr lehdj.k ds ewy dkYifud gSAmijksDr tkudkjh ds vk/kkj ij fuEu iz'uksa ds mÙkj nhft,A

22. Which statement is correct regarding sets P, Q and RleqPp; P, Q rFkk R ds lnHkZ esa dkSulk dFku lR; gS&

STPFXII1213

Page - 31

(A) P Q = (B) R P (C) P Q R � {4, 5} (D) All of the abovewhere R is the set of real numbers. (mijksDr lHkh tgk¡ R okLrfod la[;kvksa dk leqPp; gSA )

23. Which statement is correct regarding sets P, R and TleqPp; P, R rFkk T ds lanHkZ esa dkSulk dFku lR; gS&

(A) P R (B) R T (C) T P (D) P R = T

24. Which statement is correctdkSulk dFku lR; gS&(A) Least positive integer for set R is 2 (leqPp; R dk U;wure /kukRed iw.kk±d 2 gS)(B) Least positive integer for set R is 3 (leqPp; R ds fy, U;wure /kukRed iw.kk±d 3 gS)(C) Greatest positive integer for set T is 3 (leqPp; T ds vf/kdre /kukRed iw.kk±d 3 gS)(D) None of these (buesa ls dksbZ ugha)


In a sequence of (4n + 1) terms the first (2n + 1) terms are in AP whose common difference is 2, and the last (2n+ 1) terms are in GP whose common ratio 0.5. If the middle terms of the AP and GP are equal, then(4n + 1) inksa fd ,d vuqØe esa izFke (2n + 1) in l-Js- esa gS ftldk lkoZvUrj 2 gS] rFkk vfUre (2n + 1) in xq-Js- esa gSftldk lkoZvuqikr 0.5 gS] ;fn l-Js- rFkk xq- Js- ds e/; in leku gks] rks

25. Middle term of the sequence is : (vuqØe dk e/; in gS&)

(A) 12

2.nn

1n

(B) 12

2.nn2

1n

(C) n . 2n (D) None of these (buesa ls dksbZ ugha)

26. First term of the sequence is : (vuqØe dk izFke in gS&)

(A) 12

2.n2n4n

n

(B)

12

2.n2n4n

n

(C)

12

2.nn2n

n

(D)

12

2.nn2n

n

27. Middle term of the GP is : (xq.kksÙkj Js<+h dk e/; in gS&)

(A) 12

2n

n

(B)

12

2.nn

n

(C)

12

nn

(D) 12

n2n


Let S1, S

2, S

3 be the circles x2 + y2 + 3x + 2y + 1 = 0, x2 + y2 � x + 6y + 5 = 0

and x2 + y2 + 5x � 8y + 15 = 0, then

ekukfd S1, S

2, S

3 rhu oÙk Øe'k% x2 + y2 + 3x + 2y + 1 = 0, x2 + y2 � x + 6y + 5 = 0

rFkk x2 + y2 + 5x � 8y + 15 = 0 gks] rks

28. Point from which length of tangents to these three circles is same isog fcUnq ftlls rhuksa oÙkksa ij Li'kZ js[kk dh yEckbZ cjkcj gks] gksxk&(A) (1, 0) (B) (3, 2) (C) (10, 5) (D) (� 2, 1)

29. Equation of circle S4 which cut orthogonally to all given circle is

oÙk S4 tks fd fn;s x;s oÙkksa dks yEcdks.kh; izfrPNsn djrk gS] dk lehdj.k gS&

(A) x2 + y2 � 6x + 4y � 14 = 0 (B) x2 + y2

+ 6x + 4y � 14 = 0

(C) x2 + y2 � 6x � 4y + 14 = 0 (D) x2 + y2 � 6x � 4y � 14 = 0

30. Radical centre of circles S1, S

2, & S

4 is

S1, S

2 ,oa S

4 oÙkksa dk ewyk{k dsUnz gS&

STPFXII1213

Page - 32

(A)

5

8�,

5

3� (B) (3, 2) (C) (1, 0) (D)

2

3�,

5

4�


Matrix - Match Type (eSfVªDl&lqesy izdkj)

This section contains 1 question. The question contains statements given in two columns which have to be matched.Statements (A, B, C, D) in Column-I have to be matched with statements (p, q, r, s) in Column-II. The answers tothese questions have to be appropriately bubbled as illustrated in the following example.If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q, then thecorrectly matched answer will look like in bubbled matrix. One option in Column�I may have one or more than one correct options in Column�II. Marks will be

awarded only if an option in Column�I is matched with all correct options in Column-

II. Each correct matching will be awarded 1 markand if all the matchings are correct then 6 marks will be awarded.


ABCD

p q r s



31. Column - I (LrEHk - I) Column-II (LrEHk-II)

(A) The system of equations lx + y + z = 1, x + ly + z = l, (p) 4x + y + lz = l2 have no solution, then |l| is equal tolehdj.k fudk; lx + y + z = 1, x + ly + z = l, x + y + lz = l2

dk dksbZ gy ugha gS] rks |l| =

(B) If f(x) = sin2x + sin2 (x + 2p /3) + sin2 (x + 4p /3) then the value of (q) 2

8f

15 is equal to

;fn f(x) = sin2x + sin2 (x + 2p /3) + sin2 (x + 4p /3) gks] rks

8f

15 dk eku gS &

(C) Number of odd divisors of 210 × 35 × 72 is equal to (r) 12210 × 35 × 72 ds fo"ke Hkktdksa dh la[;k gS &

(D) Least positive integral solution of the equation (s) 18|2x| � |x � 4| = x + 4 is

lehdj.k |2x| � |x � 4| = x + 4 dk U;wure /kukRed iw.kk±d gy gSA

PART - II



32. The position vector of a particle is given as r

= (t2 � 4t + 6) i� + (t2 ) j� . The time after which the velocity vector and

STPFXII1213

Page - 33

acceleration vector becomes perpendicular to each other is equal to: (fdlh d.k dk fLFkfr lfn'k r

= (t2 � 4t + 6) i�

+ (t2 ) j� ls fn;k tkrk gSA og le; Kkr djks tc osx lfn'k]Roj.k lfn'k ijLij yEcor~ gks) &

(A) 1sec (B) 2 sec (C) 1.5 sec (D) not possible laHko ugha gS

33. The velocity - time graph of a particle is as shown in figure : (fdlh d.k dk osx le; oØ çnf'kZr gS) &

(A) It moves with a constant acceleration throughout. (;g ges'kk fu;r Roj.k ls xfreku gSA)(B) It moves with an acceleration of constant magnitude but changing direction at the end of every two second(;g ges'kk leku ifjek.k ds Roj.k ls pyrk gS ijUrq Roj.k dh fn'kk çR;sd nks lsd.M ds ckn cny tkrh gSA)(C) The displacement of the particle is zero (d.k dk foLFkkiu 'kwU; gSA)(D) The velocity becomes zero at t = 4 second (d.k dk osx t = 4 lsd.M ij 'kwU; gSA)

34. In the figure shown, all the surfaces and pulleys are smooth. Also the stringsare inextensible and light. The acceleration of the 6 Kg block will be: [Take g= 10 m/s2] (fn[kk;s x;s fp=k esa] lHkh lrg rFkk f?kjuh ?k"kZ.kjfgr gS rFkkjLlh vforkU; o gYdh gSA 6 Kg CykWd dk Roj.k gksxk )& [g = 10 m/s2]

(A) 5 m/s2 (B) 1 m/s2

(C) 3 m/s2 (D) 2 m/s2

35. A uniform thin rod is bent in the form of closed loop ABCDEFA as shown in the figure. (,dleku ,d iryh NM+ dkseksM+dj fp=kkuqlkj ,d cUn ywi ABCDEFA cuk;k x;k gSA)

The ratio of moment of inertia of the loop about x-axis to that about y-axis is.(mijksä iz'u esa ywi dk x-v{k ds ifjr% tM+Ro vk?kw.kZ dk y-v{k ds ifjr% tM+Ro vk?kw.kZ ls vuqikr gSA)(A) > 1 (B) < 1 (C) = 1 (D) = 1/2

36. Two particles of medium disturbed by the wave propagation are at x1 = 0 and x

2 = 1cm. The respective

displacements (in cm) of the particles can be given by the equations :(x1 = 0 rFkk x

2 = 1cm ij ek/;e ds nks d.k rjax

lapj.k }kjk çHkkfor gksrs gSA d.kksa ds foLFkkiu (lseh eas) Øe'k% fuEu lehdj.kksa }kjk fn;s tk ldrs gSa) :y

1 = 2sin3t

y2 = 2sin(3t � /8)

The wave velocity is : (rjax osx gS) :(A) 16 cm/sec (B) 24 cm/sec (C) 12 cm/sec (D) 8 cm/sec.

37. A sounding body emitting a frequency of 150 Hz is dropped from a height. During its fall under gravity it crossesa very small balloon moving upwards with a constant velocity of 2m/s one second after it started to fall. Thedifference in the frequency observed by the man in balloon just before and just after crossing the body will be:(given that -velocity of sound = 300m/s; g = 10m/s2)

STPFXII1213

Page - 34

¼,d /ofu mRiUu djus okyk fi.M ftldh vkofÙk 150 Hz gS] mls dqN Å ¡pkbZ ls fxjk;k tkrk gS xq:Roh; {ks=k esa fxjkus ds 1lSd.M ckn ;g ,d cgqr NksVs xqCckjs tks fu;r osx 2m/s ls Å ij dh vksj xfreku gS dks ikj djrk gSa rks xqCckjs esa fLFkr O;fDrds }kjk xqCckjs dks ikj djus ds Bhd igys rFkk Bhd ckn esa vkHkklh vkofÙk esa vUrj gksxk(fn;k x;k gS fd /ofu dk osx = 300m/s; g = 10m/s2½(A) 12 (B) 6 (C) 8 (D) 4

38. In the figure shown, the time period and the amplitude respectively when m is released from rest when the springis relaxed is: (the inclined plane is smooth)(fn[kk;s x;s fp=k esa] fLizax ruko jfgr ,oa fojkekoLFkk esa gS rks m dks eqDr NksM+us ij vkorZdky rFkk vk;ke Øe'k% gksaxs (urry fpduk gSA)

(A) 2m

k,mg

k

sin (B) 2

m

k

sin ,

2 mg

k

sin

(C) 2m

k,

mg

k

cos (D) none of these buesa ls dksbZ ugh

39. Figure shows a light, inextensible string attached to a cart that can

x1x2

A

h=3m

(0,0)

x

yslide along a frictionless horizontal rail aligned along an x axis. Theleft end of the string is pulled over a small pulley, of negligible massand friction and fixed at height h = 3m from the ground level. The cart

slides from x1 = 3 3 m to x2 = 4 m and during the move, tension inthe string is kept constant 50 N. Find change in kinetic energy of thecart in joules. (Use 3 = 1.7)(fp=kkuqlkj ,d gYdh] vfoLrkfjr jLlh dk ,d fljk xkM+h ls tks x v{k ds vuqfn'k fLFkr ?k"kZ.k jfgr jsy&Vªsd ij xfr dj ldrhgS] tksM+k tkrk gSA nwljk fljk gYdh rFkk ?k"kZ.k jghr NksVh f?kjuh }kjk tehu ls h = 3m Å ¡pkbZ ij rkuk gqvk gSA xkM+h x1 =

3 3 m ls x2 = 4 m rd fQlyrh gS rFkk xfr ds nkSjku jLlh esa ruko T = 50 N fu;r j[kk tkrk gS rks xkM+h dh xfrt Å tkZesa ifjorZu ¼twy esa½ gksxk) ( 3 = 1.7)

(A) 50 (B) 150 (C) 200 (D) 350

40. For a fluid which is flowing steadily, the level in the vertical tubes is best represented by(LFkk;h :i ls izokfgr nzo ds fy,] Å /okZ/kj ufy;ksa esa ry loZJs"B :i ls iznf'kZr fd;k tkrk gS) &

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

41. A small coin of mass 40 g is placed on the horizontal surface of a rotating disc. The disc starts from rest and isgiven a constant angular acceleration = 2 rad/s2. The coefficient of static friction between the coin and the discis µ

s = 3/4 and coefficient of kinetic friction is µ

k = 0.5. The coin is placed at a distance r = 1 m from the centre

of the disc. The magnitude of the resultant force on the coin exerted by the disc just before it starts slipping onthe disc is : (Take g = 10 m/s2)

,d NksVk flDdk ftldk nzO;eku 40 xzke gS] dks ,d ?kwerh gq, pdrh ds {kSfrt lrgij j[kk tkrk gSA pdrh fLFkjkoLFkk ls izkjEHk gksrh gS rFkk bls = 2 jsfM;u/lS2 dk fu;r

STPFXII1213

Page - 35

dks.kh; Roj.k iznku djrs gSaA pdrh o flDds ds e/; LFkSfrd ?k"kZ.k xq.kkad µs = 3/4 gS

rFkk xfrd ?k"kZ.k xq.kkad µk = 0.5 gSaA flDds dks pdrh ds dsUnz ls r = 1 eh0 dh nwjh ij

j[kk tkrk gSA flDds ds pdrh ij fQlyus ls Bhd igys pdrh }kjk flDds ij yxk;kx;k dqy cy gksxk&(yhft, g = 10 m/s2)

(A) 0.2 N (B) 0.3 N (C) 0.4 N (D) 0.5 N


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

42. Heat is supplied to a certain homogeneous sample of matter at a uniform rate.Its temperature is plotted against time as shown in the figure. Which of thefollowing conclusions can be drawn?(A) its specific heat capacity is greater in the solid state than in the liquid state.(B) its specific heat capacity is greater in the liquid state than in the solid state.(C) its latent heat of vaporization is greater than its latent heat of fusion.(D) its latent heat of vaporization is smaller than its latent heat of fusion.

fdlh lekax inkFkZ ds uewus dks le:i Å "ek iznku dh xbZA uhps fn;s fp=k ds vuq:imlds rki dk le; ds lkFk xzkQ [khapk x;kA uhps fn;s x;s fu"d"kks± esa dkSulk lgh gS\(A) bldh fof'k"V Å "ek/kkfjrk Bksl voLFkk esa nzo voLFkk ds vis{kk vf/kd gSA rki

le;(B) bldh fof'k"V Å "ek/kkfjrk nzo voLFkk esa Bksl dh vis{kk vf/kd gSA(C) bldh ok"iu dh xqIr Å "ek xyu dh xqIr Å "ek ls vf/kd gSA(D) bldh ok"iu dh xqIr Å "ek xyu dh xqIr Å "ek ls de gSA

43. The acceleration of a particle is, a =

100 x + 50. It is released from x = 2. Here '

a

' & '

x

' are in S.I. units: (,d

d.k dk Roj.k a =100 x + 50 gS] bls x = 2 ls eqDr fd;k tkrk gS] ;gk¡ ' a

' rFkk '

x

' S.I.i)fr esa gS)

(A) the particle will perform SHM of amplitude 2 m. (d.k 2 m vk;ke dh SHM djsxkA )(B) the particle will perform SHM of amplitude 1.5 m. (d.k 1.5 m vk;ke dh SHM djsxkA)(C) the particle will perform SHM of time period 0.63 seconds. (d.k dh SHM dk vkordky 0.63 sec. gksxkA)(D) the particle will have a maximum velocity 15 m/s. (d.k dk vf/kdre osx 15 m/s gksxkA)

44. A rigid body is in pure rotation, that is, undergoing fixed axis rotation. Then which of the following statement(s)are true.,d n<+ fi.M 'kq) ?kw.kZu xfr dj jgk gSA ;g ?kw.kZu xfr ,d fLFkj v{k ds ifjr% gS rks rks fuEu esa ls dkSuls dFku lR; gS &(A) You can find two points in the body in a plane perpendicular to the axis of rotation having same velocity. (?kw.kZuv{k ds yEcor~ ry esa fi.M esa nks ,sls fcUnq gks ldrs gS ftudk osx leku gSA )(B) You can find two points in the body in a plane perpendicular to the axis of rotation having same acceleration.(?kw.kZu v{k ds yEcor~ ry esa fi.M esa nks ,sls fcUnq gks ldrs gS ftudk Roj.k leku gSA )(C) Speed of all the particles lying on the curved surface of a cylinder whose axis coincides with the axis ofrotation is same. (,d csyu ftldh v{k ] ?kw.kZu v{k ds lEikrh gS] ds oØh; i"B ij fLFkr lHkh fcUnqvksa dh pky leku gSA)(D) Angular speed of the body is same as seen from any point in the body. (fi.M ij fLFkr fdlh Hkh fcUnq ds lkis{kfi.M dh dks.kh; pky leku gSA)

45. In a tug of war, the team that exerts a larger tangential force on the ground wins, winning team not moving.Consider the period in which a team is dragging the opposite team by applying a larger tangential force on the

STPFXII1213

Page - 36

ground. Which of the following works are negative? (jLlh&[khap [ksy esa tks Vhe /kjkry ij lcls T;knk Li'kZ js[kh; cyyxkrh gS] thr tkrh gS] thrh gqbZ Vhe fLFkj gS] og le;&vUrjky ekusa ftlesa ,d Vhe nwljh Vhe dks T;knk js[kh; cy yxkdj[khap ysrh gSA dkSulk dk;Z _ .kkRed gksxkA)(A) work through string by the losing team on the wining team.

(gkjh gqbZ Vhe ds }kjk jLlh ds ek/;e ls thrh gqbZ Vhe ij fd;k x;k dk;Z )(B) work by the ground on the wining team. (/kjkry ds }kjk thrh gqbZ Vhe ij fd;k x;k dk;Z)(C) work by the ground on the losing team. (/kjkry ds }kjk gkjh gqbZ Vhe ij fd;k x;k dk;Z)(D) total external work on the two teams. (nksuksa Vheksa ds }kjk fd;k x;k dqy cká dk;Z)


This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to beanswered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.bl [k.M esa 2 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 3 cgq&fodYih ç'u ds mÙkj nsus gSA çR;sd ç'uds 4 fod Yi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA


A uniform bar of length 6 a & mass 8 m lies on a smooth horizontal table. Two point masses m & 2 m moving inthe same horizontal plane with speeds 2 v

andv respectively strike the bar as shown & stick to the bar after

collision. (6 a yEckbZ o 8 m nzO;eku dh ,d le:i NM+ ,d fpduh {kSfrt est ij j[kh gSA nks fcUnq nzO;eku m o 2m leku{kSfrt ry esa Øe'k% 2v o v pky ls n'kkZ;s vuqlkj Vdjkrs gSa ,oa VDdj ds ckn NM+ ls fpid tkrs gSA)

46. Velocity of the centre of mass of the system is : (fudk; ds nzO;eku dsUnz dk osx gS) �

(A) 2v

(B) v (C) 32v

(D) Zero 'kwU;

47. Angular velocity of the rod about centre of mass of the system is: (fudk; ds nzO;eku dsUnz ds ifjr% NM+ dk dks.kh;osx gS) &

(A) 5av

(B) 15a

v(C)

3av

(D) 10a

v

48. Total kinetic energy of the system, just after the collision is: (VDdj ds Bhd ckn fudk; dh dqy xfrt Å tkZ gS) %

(A) 53

mv2 (B) 253

mv2 (C) 153

mv2 (D) 3 mv2


A block B is placed over a cart which in turn lies over a smooth horizontal floor. Block A and block C areconnected to block B with light inextensible strings passing over light frictionless pulleys fixed to the cart asshown. Initially the blocks and the cart are at rest. All the three blocks have mass m and the cart has mass M(M= 3m). Now a constant horizontal force of magnitude F is applied to block A towards right. (,d CykWd B dks ,dxkM+h ds Å ij j[k fn;k tkrk gS tks fd ,d fpdus {kSfrt /kjkry ij gSA CykWd A rFkk CykWd C dks CykWd B ls gYdh ?k"kZ.kjfgr

STPFXII1213

Page - 37

f?kjuh ¼tks xkM+h ls fp=kkuqlkj tqM+h gS½ ls xqtj jgh gYdh vforkU; jLlh ls tksM+k tkrk gSA izkjEHk esa CykWd o xkM+h fojke esagSA izR;sd CykWd dk nzO;eku m gS rFkk xkM+h dk nzO;eku M(M = 3m) gSA vc ,d fu;r {kSfrt cy ftldk ifjek.k F gS bldksnka;h rjQ CykWd A ij vkjksfir fd;k tkrk gSA)

49. Assuming friction to be absent everywhere, the magnitude of acceleration of cart at the shown instant is : ( ekuk?k"kZ.k gj txg vuqifLFkr gS] fn;s x;s {k.k ij xkM+h dk Roj.k gS) �

(A) m6F

(B) m4F

(C) m3F

(D) 0

50. Taking friction to be absent everywhere the magnitude of tension in the string connecting block B and block C is:(;g ekfu, fd ?k"kZ.k gj txg vuqifLFkr gS rks CykWd B o CykWd C dks tksM+us okyh jLlh esa ruko dk ifjek.k gS) &

(A) 9F

(B) 6F

(C) 3F

(D) 3F2

51. Let the coefficient of friction between block B and cart is ( > 0) and friction is absent everywhere else. Then themaximum value of force F applied to block A such that there is no relative acceleration between block B and cartis : (ekuk fd CykWd B o xkM+h ds e/; ?k"kZ.k xq.kkad

( > 0) gS rFkk gj txg ?k"kZ.k vuqifLFkr gSA rc cy F dk vf/kdre eku D;k

gksxk tks fd CykWd A ij bl rjg ls yxk;k tkrk gS fd CykWd B o xkM+h ds e/; dksbZ lkisf{kd Roj.k u gks) &(A) mg (B) 2 mg (C) 3 mg (D) 4 mg


Matrix - Match Type (eSfVªDl&lqesy izdkj)This section contains 1 question. The question contains statements given in two columns which have to be matched.Statements (A, B, C, D) in Column-I have to be matched with statements (p, q, r, s) in Column-II. The answers tothese questions have to be appropriately bubbled as illustrated in the following example.If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q, then thecorrectly matched answer will look like in bubbled matrix. One option in Column�I may have one or more than one correct options in Column�II. Marks will be

awarded only if an option in Column�I is matched with all correct options in Column-

II. Each correct matching will be awarded 1 markand if all the matchings are correct then 6 marks will be awarded.


ABCD

p q r s



52. In column-I condition on velocity, force and acceleration of a particle is given. Resultant motion is described in

column-II. u

= initial velocity, F

= resultant force and v

= instantaneous velocity. (. (u

, v

, and F

are nonzero

vectors) LrEHk-I esa ,d d.k ds osx] ml ij cy o mldk Roj.k fn;s x;s gSA ifj.kkeh xfr LrEHk-II esa nh xbZ gSA u

= izkjfEHkd

osx, F

= ifj.kkeh cy rFkk v

= rkR{kf.kd osx (u

, v

, rFkk F

v'kwU; lfn'k gS)&

STPFXII1213

Page - 38

Column-I (LrEHk-I) Column-II (LrEHk-II)

(A) 0Fu

and F

= constant (p) path will be circular path

0Fu

rFkk F

= fu;r iFk oÙkkdkj gksxk

(B) 0Fu

and F

= constant (q) speed will increase

0Fu

rFkk F

= fu;r pky c<+sxhA

(C) 0Fv

all the time and |F

| = constant (r) path will be straight line and the particle always remains in one plane.

0Fv

lHkh le; ij rFkk |F

| = fu;r vkSj iFk ljy js[kk gksxk

d.k ges'kk ,d gh ry esa jgrk gSA

(D) j�3i�2u

and acceleration at all time j�9i�6a

(s) path will be parabolic

j�3i�2u

rFkk lHkh le; ij Roj.k j�9i�6a

iFk ijoy; gksxk

PART - III


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

53. The radius of the first Bohr orbit of hydrogen atom is 0.59 Å. The radius of the third orbit of He+ will be :(gkbMªkstu ijek.kq ds fy, izFke cksj d{kk dh f=kT;k 0.59 Å gSA He+ dh rrh; d{kk dh f=kT;k fuEu gksxhA )(A) 8.46 Å (B) 0.705 Å (C) 1.59 Å (D) 2.38 Å

54. Bond order of N2 molecule is : (N

2 v.kq dk cU/kØe fuEu gS %)

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

55. In the equilibrium reaction 2HI (g) H2 + I

2 which of the following expressions is true ?

(lkE; vfHkfØ;k 2HI (g) H2 + I

2 esa] fuEu esa ls dkSulk O;atd lgh gSA)

(A) Kp = K

c(B) K

c = 2K

p(C) K

p > K

c(D) K

c = K

p (RT)2

56. Which of the following diagram correctly describes the behaviour of a fixed mass of an ideal gas ?(T is measured in K) (,d vkn'kZ xSl ds fuf'pr nzO;eku ds O;ogkj dks fuEu esa ls dkSulk vkjs[k crkrk gS \ (T dks K }kjkekfir fd;k x;k gS))

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

57. Number of H+ ions present in 250 ml of lemon juice of pH = 3 is :(pH = 3 okys 250 ml yseu twl esa mifLFkr H+ vk;uksa dh la[;k fuEu gS :)(A) 1.506 1022 (B) 1.506 1023 (C) 1.506 1020 (D) 13.012 1021

58. Baking powder is : (cSfdx ikmMj fuEu gS %)(A) NaHCO

3(B) NaHCO

3 .6H

2O (C) Na

2CO

3(D) Na

2CO

3.10H

2O

59. Which of the following gives propyne on hydrolysis ? (fuEu esa ls dkSu ty vi?kV~u ij izksikbu nsrk gS \)(A) Al

4C

3(B) Mg

2C

3(C) B

4C (D) La

4C

3

STPFXII1213

Page - 39

60. The correct IUPAC name of the given compound is : (fn, x;s ;kSfxd dk lgh IUPAC uke gS %)

(A) 5-Amino-3-bromo-4-chloro-2-hydroxycyclopentane-1-carbonitrile (5-,ehuks-3-czkseks-4-Dyksjks-2-gkbMªkWDlhlkbDyksisUVsu-1-dkcksZukbVªkby )(B) 1-Amino-3-bromo-2-chloro-4-hydroxycyclopentane-5-carbonitirle (1-,ehuks-3-czkseks-2-Dyksjks-4-gkbMªkWDlhlkbDyksisUVsu-5-dkcksZukbVªkby )(C) 2-Amino-4-bromo-3-chloro-5-hydroxycyclopentane-1-carbonitirle (2-,ehuks-4-czkseks-3-Dyksjks-5-gkbMªkWDlhlkbDyksisUVsu-1-dkcksZukbVªkby )(D) 2-Amino-3-chloro-3-bromo-5-hydroxycyclopentane-1-nitrile (2-,ehuks-3-Dyksjks-3-czkseks-5-gkbMªkWDlhlkbDyksisUVsu-1-ukbVªkby )

61. The number of possible enantiomer pairs that can be produced during monochlorination of Butane isC;qVsu ds eksuksDyksjhuhdj.k ds nkSjku cuus okys lEHko izfrfcEch ;qXeksa dh la[;k gS %

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


This section contains 2 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C)and (D), out of which ONE OR MORE THAN ONE is/are correct.bl [k.M esa 2 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 vfèkdfodYi lgh gS ¼gSa½A

62. Which of the following methods of expressing concentration are independent of temperature ?(lkUnzrk dks O;Dr djus okyh og fof/k tks rkieku ij fuHkZj ugha djrh gaS] fuEu gS ?)

(A) Molarity (eksyjrk) (B) Molality (eksyyrk)(C) % w/w (Hkkj@Hkkj izfr'kr) (D) Mole fraction (eksy&fHkUu)

63. Which amongs the following are correctly matched. (fuEufyf[kr esa ls dkSu lgh :i ls lqesfyr gS :)

(A) & CH2=CH�CH=C=CH�CH=CH2 are ring chain isomers.

( rFkk CH2=CH�CH=C=CH�CH=CH2 oy; Ja[kyk leko;oh gSaA )

(B) & are positional isomers.

( rFkk fLFkfr leko;oh gSaA )

STPFXII1213

Page - 40

(C) & are chain isomers.

( rFkk Ja[kyk leko;oh gSaA)

(D) & are functional group isomers.

( rFkk fØ;kRed lewg leko;oh gSaA)


This section contains 1 paragraphs. Based upon each paragraph, 3 multiple choice questions have to beanswered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.bl [k.M esa 1 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 3 cgq&fodYih ç'u ds mÙkj nsus gSA çR;sd ç'u ds4 fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA

Paragraph for Question Nos. 64 to 66Conditions of geometrical isomerism :

(I) Geometrical isomerism arises due to the presence of a double bond or a ring structure.

(i.e. , � N = N � or ring structure)

Due to the rigidity of double bond or the ring structure to rotate at the room temperature the molecules exist intwo or more orientations. This rigidity to rotation is described as restricted rotation / hindered rotation / norotation.

(a) (Restricted Rotation)

(b) (Restricted Rotation)

(II) Different groups should be attached at each doubly bonded atom. For example

and are identical but not geometrical isomers.

On the other hand, following types of compounds can exist as geometrical isomers :

C = Ca

b

a

b , C = C

a

b

a

dor C = C

a

b

d

e

ç'u 64 ls 66 ds fy, vuqPNsnT;kferh; leko;ork ds fy;s vko';d 'krs± %(I) T;kfefr; leko;ork ;kSfxd esa f}cU/k ;k oy; lajpuk ds mifLFkr gksus ij mRiUu gksrh gSA

(tSls fd , � N = N � ;k oy; lajpuk)

lkekU;r% ;kSfxd esa mifLFkr f}cU/k ;k oy; lajpuk ls tqM+s gq;s ijek.kqvksa ;k lewgksa dk n<+rk ls tqM+s gksus ds dkj.k eqDr ?kw.kZulaHko ugha gksrk gS] ftlls mlds nks ;k vf/kd :i gh izkIr gksrs gSaA ?kw.kZu esa mRiUu ;g dfBukbZ gh izfrcfU/kr ?kw.kZu (restricted

rotation), v?kw.kZu (no rotation) dgykrh gSA

(a) (izfrcfU/kr ?kw.kZu)

STPFXII1213

Page - 41

(b) (izfrcfU/kr ?kw.kZu)

(II) T;kferh; leko;fo;ksa esa f}cU/k }kjk tqM+s dkcZu ijek.kqvksa dh 'ks"k nksuksa la;kstdrk,sa fHkUu&fHkUu ijek.kqvksa ;k lewgksa ls tqM+hgksuh pkfg;sA mnkgj.k ds :i esa

vkSj le:i lajpuk;sa gS u fd T;kfefr leko;ohA

blds foijhr] fuEu izdkj ds ;kSfxd T;kferh; leko;ork iznf'kZr djrs gSaA

C = Ca

b

a

b , C = C

a

b

a

dvFkok C = C

a

b

d

e

64. Which of the following compounds will show geometrical isomerism ?(fuEu esa ls dkSulk ;kSfxd T;kferh; leko;ork n'kkZrk gS \)(A) But-2-ene (C;wV-2-bZu) (B) Propene (izksihu)

(C) But-1-ene (C;wV-1-bZu) (D) Benzene (csUthu)

65. Which is a pair of geometrical isomers ?fuEu esa ls dkSulk ;qXe T;kferh; leko;ork n'kkZrk gS \

(I) (II)

(III) (IV)

(A) (I) and (rFkk) (II) (B) (I) and (rFkk) (III) (C) (II) and (rFkk) (IV) (D) (III) and (rFkk) (IV)

66. Which of the following compound can not show geometrical isomerism ?(fuEu esa ls dkSulk ;kSfxd T;kehrh; leko;ork ugha n'kkZrk gS \)

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


Matrix - Match Type (eSfVªDl&lqesy izdkj)This section contains 1 question. Each question contains statements given in two columns which have to be matched.Statements (A, B, C, D) in Column-I have to be matched with statements (p, q, r, s) in Column-II. The answers tothese questions have to be appropriately bubbled as illustrated in the following example.If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q, then the correctlywill look like in bubbled matrix. Each statement in Column�I may have one or more

than one correct options in Column�II. One Mark will be awarded for each statementin Column�I is matched with all correct options in Column-II. If all the statements incolumn-I are correctly matched to theircorrect options in column-II , then 6 marks will be awarded.


ABCD

p q r s

bl [k.M esa 1 iz'u gSA izR;sd iz'u esa nks dkWye esa oDRO; (statements) fn;s gq, gSa ftudk lqesy (match) djuk gSA dkWye(Column-I) esa fn;s x;s oDrO;ksa (A, B, C, D) dks dkWye (Column-II) esa fn;s x;s oDrO;ksa (p, q, r, s) ls lqesy djuk gSA bu iz'uksads mÙkj fn;s x;s mnkgj.k ds vuqlkj mfpr cqYyksa dks dkyk djds n'kkZuk gSA ;fn lgh lqesy A-p, A-r, B-p, B-s,

STPFXII1213

Page - 42

C-r, C-s rFkk D-q gSa] rks lgh fof/k ls dkys fd, x;s cqYyksa dk 4 x 4 esafVªDl (matrix) esa n'kkZ;k x;k gSA dkWye�I ds izR;sd oDÙkO;dk dkWye�II ds ,d ;k ,d ls T;knk fodYiksa ls lqesy djk;k tk ldrk gSA dkWye�I ds izR;sd oDÙkO; dkdkWye�II ds lHkh lgh fodYiksa ls lqesy djkus ds fy, ,d vad iznku fd;k tk;sxkA ;fn dkWye�I ds lHkh OkDÙkO;ksa dkdkWye�II ds lHkh lgh fodYiksa ls lqesy fd;k x;k rks 6 vad iznku fd;s tk;sxsaA

67. Columm - I Columm - II(dkWye-I) (dkWye-II)

(A) (Gsystem

) T,P

= 0 (p) Process is in equilibrium ((G

fudk;)

T,P = 0) (vfHkfØ;k lkE; esa gSA)

(B) Ssystem

+ Ssurrounding

> 0 (q) Process is nonspontaneous (S

fudk; + S

okrkoj.k > 0) (vfHkfØ;k vLor% gSA)

(C) Ssystem

+ S surrounding

< 0 (r) Process is spontaneous (S

fudk; + S

okrkoj.k < 0) (vfHkfØ;k Lor% gSA)

(D) (Gsystem

) T,P

> 0 (s) System is unable to do useful work ((G

fudk;)

T,P > 0) (fudk; mi;qDr dk;Z ds v;ksX; gSA)

ANSWER KEY TO SAMPLE TEST PAPER-II

1. B 2. B 3. A 4. C 5. B 6. A 7. B

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

15. D 16. AB 17. AC 18. ABC 19. AB 20. AD 21. AC

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

29. D 30. A 31. (A - q), (B - r), (C - s), (D - p)

32. A 33. B 34. C 35. B 36. B 37. A 38. A

39. A 40. A 41. D 42. AC 43. BCD 44. CD 45. ACD

46. D 47. A 48. A 49. D 50. C 51. C

52. (A - r) ; (B - q,s) ; (C - p) ; (D - q,r) 53. D 54. A 55. A 56. A

57. C 58. A 59. B 60. C 61. A 62. BCD 63. ACD

64. A 65. C 66. A 67. (A - p, s); (B - r) ; (C - q, s) ; (D - q, s).

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

2. (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..........199

a = � 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

STPFXII1213

Page - 43

S50 = 2

50[� 3� 199]

= 25 [� 202] = � 5050.

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

1 7C

1 + 4C

2 7C

2 + 4C

3 7C

3 + 4C

4 7C

4 = 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 cnyusds rjhdksa dh la[;k + pkj f[kykSus cnyus ds rjhdksa dh la[;kA= 4C

1 7C

1 + 4C

2 7C

2 + 4C

3 7C

3 + 4C

4 7C

4 = 329

7. Slope of line parallel to line BC is � 1. (js[kk BC ds lekUrj js[kk dh izo.krk � 1 gSA)Equation of line parallel to BC is x + y + = 0. (BC ds lekUrj js[kk dh lehdj.k x + y + = 0 gSA)

21

2

00

= ±

2

1

Since, line x + y + = 0, intersects AB and AC,pw¡fd js[kk x + y + = 0, AB vkSj AC dks dkVrh gS]

= 2

1

x + y + 2

1 = 0

8. )x5x(loglog 259.0

> 0

log5 )x5x( 2

< 1

(x2 + 5 + x)1/2 < 5 and x2 + x + 5 > 0 x2 + 5 + x < 25 x2 + x � 20 < 0

(x + 5) (x � 4) < 0

x (�5, 4)

n = 8

11. AD = 21

BD

Given AB = x

AD + BD = x

21

BD + BD = x

BD = 32

x

STPFXII1213

Page - 44

Now by ABC

BC = 22 xy

And by BCD

CD = 22 BCBD

= 222 xyx

94

CD = 3

x5y9 22

sin = CDBD

=

3

x5y9

x32

22

= 22 x5y9

x2

Sol. (22 to 24) a � 5 0

x2 �

5�aa2

x +

5�a4�a

= 0

If roots are and , then;fn ewy vkSj gks] rks

For P : =

5�a4�a

> 0

or a (�,4) (5, )P = {a : a (�, 4) (5, ) }

For Q : =

5�a4�a

< 0

or a (4, 5) Q = {a : a (4,5)}For R : D 0 and (vkSj) > 0

2

2

)5�a(

a4� )5�a(

)4�a(4 0 and (vkSj)

5�a4�a

> 0

2)5�a(

20�a9 0 and (vkSj)

5�a4�a

> 0

a

,

920

and (vkSj) a (�,4) (5, )

R =

),5(4,

920

a:a

For S : D 0

S =

}5{,

920

a:a

For T : D < 0

a < 9

20

T =

920

,�a:a

STPFXII1213

Page - 45

22. (a) P Q =

(b) R P(c) P Q = {a : a {(�,4) (5,) (4,5)}

= R � {4,5}

23. T P

24. R =

),5(4,

920

a:a

Least positive integer of R is 3.

Sol. (25 to 27)Let 1st term be a . and common difference is 2. (ekuk izFke in a vkSj lkoZvUrj 2 gSA)

T2n + 1

= a + 4n = A (say) r = 21

Middle term of AP = Tn + 1

(lekUrj Js<+h d k e/; in = Tn + 1

)

Middle term of GP = T3n + 1

(xq- Js<+h d k e/; in = T3n + 1

)

Tn + 1

= a + 2n T3n + 1

= A . rn = n2

)n4a(

(a + 2n) = n2

n4a 2n a + 2n2n = a + 4n

a =12

2.n2n4n

n

T

2n + 1= a + 4n =

12

2.n2n4n

n

+ 4n

= 12

2.n2n

n

=

12

n2n

1n

T3n + 1

= n2

n4a =

)12(2

n2nn

1n

= 12

n2n

28. Point from which length of tangents to these circle is same is radical centreS

1 � S

2 = 0 4x � 4y � 4 = 0 x � y � 1 = 0

S2 � S

3 = 0 �6x + 14y � 10 = 0 -3x + 7y � 5 = 0

3x � 3y � 3 = 0

4y � 8 = 0 y = 2 x = 3

29. If circle be drawn taking radical centre as centre and length of tangents from radical centre to any circle asradius will cut all the three circles orthogonally

Length of tangent = 14949 = 1S = 27Equation of circle (x � 3)2 + (y � 2)2 = 27

S4 : x2 + y2 � 6x � 4y � 14 = 0

30. S1 � S

2 = 0 x � y � 1 = 0

S1 � S

4 = 0 9x + 6y + 15 = 0 3x + 2y + 5 = 0

3x � 3y � 3 = 0

5y + 8 = 0 y = �8/5 x = �3/5

31. (A) D =

11

11

11

= 0

3 + 2 � 3 = 0 = 1, = �2, = 1 for = 1, there are infinite number of solutionsfor = �2, there is no solution || = 2

(B) f(x) = 1 � cos (2x + 2/3) cos (2/3) + sin2(x + 4/3)

= 23

� 21

cos 2x � 21

38

x2cos34

x2cos = 23

f 2315

8f

15 = 12

STPFXII1213

Page - 46

(C) Number of odd divisors = (5 + 1) (2 + 1) = 18(D) We do not require negative solutions.

consider 0 x 4 2x + x � 4 = x + 4 2x = 8 x = 4Since x = 4 is a solution, therefore, we do not require solutions greater than 4.Thus the least positive integral is 4.

32. r

= (t2 � 4t + 6) i� + t2 j� ; v

=dtrd

= (2t � 4) i� + 2t j� , a

=dtvd

= 2 i� + 2 j�

if a

and v are perpendicular ;fn a

rFkk v ijLij yEcor~ gS

a

. v

= 0

(2 i� + 2 j� ). ((2t � 4) i� + 2t j� ) = 0

8t � 8 = 0

t = 1 sec.Ans. t = 1 sec.

34. All the blocks will be having the same.

acceleration along the length of the string.

So, Applying Newtons law along the string on A,B & C.

6g � 2g sin300 � 2g = (6 + 2 + 2)a

3g = 10a a = 10

g3

or a = 3 m/s2 Ans. a = 3 m/s2

35. Moment of inertia of semicircular portions about x and y axes are same. But moment of inertia of straightportions about x-axis is zero.v/kZoÙkkdkj Hkkxks ds x rFkk y v{kksa ds ifjr% tM+Ro vk?kw.kZ leku gS ijUrq x-v{k ds ifjr% lh/ks Hkkxksa dk tM+Ro vk?kw.kZ 'kwU; gSA

x <

yor

y

x

36. Given = 3

f =

2 = 1.5,

Also x = 1.0 cm

Given, = x2

8

=

2 × 1

= 16 cm v = f = 16 × 1.5 = 24 cm/sec

37.

svv

vvff 00

when approaching : fa =

10300

2300150

when receding :

103002300

150rf fa � fr 12 Hence (A).

38. At equilibrium :mg sin = kA

A = Ksinmg

and time period for spring block system is :

T = km

2

Hence (A)

STPFXII1213

Page - 47

39. Displacement of the point of 'A' of the string

= 22 )3()33( � 22 34

= 6 � 5 = 1 m

k = Work done by tension = 50 × 1 = 50 Joule.

40. (A) From continuity equation, velocity at cross-section (1) is more than that at cross-section (2).(A) lkarR; lehd j.k ls vuqizLFk d kV (1) ij osx ] vuqizLFk d kV (2) ij osx ls vf/kd gSAHence ; P

1 < P

2 (bl izd kj ; P

1 < P

2 )

Hence (A) . (blfy, (A))

41. The friction force on coin just before coin is to slip will be : f = µs

mg

flDds ds fQlyus ls igys flDds ij ?k"kZ.k cy : f = µs mg

Normal reaction on the coin ; N = mgflDds ij vfHkyEc çfrfØ;k cy ; N = mgThe resultant reaction by disk to the coin isflDds ij fMLd }kjk ifj.kkeh çfrfØ;k gS

= 22 fN = 22s

2 )mg()mg( = mg 21

= 40 × 10�3 × 10 × 169

1 = 0.5 N

42. Slope of graph is greater in the solid state i.e., temperature is rising faster, hence lower heat capacity.The transition from solid to liquid state takes lesser time, hence latent heat is smaller.

xzkQ dk <ky Bksl voLFkk esa T;knk gSA i.e., rki rsth ls c<+rk gS] vr% Å "ek/kkfjrk de gSA Bksl ls nzo ifjorZu esa de le;yxk vr% xqIr Å "ek de gSA

43. a = � 100x + 50.

2 = 100.and mean position is at �100x + 50 = 0 x = 1/2 m.and as the particle is released at x = 2; so the amplitude will be 1.5 m.

T = 1022

= 0.63 sec. V

max = A = 10 ×

23

= 15 m/s.

44. All points in the body, in plane perpendicular to the axis of rotation revolve in concentric circles. All pointslying on circle of same radius have same speed (and also same magnitude of acceleration) but differentdirections of velocity ( also different directions of acceleration)Hence there cannot be two points in the given plane with same velocity or with same acceleration.As mentioned above, points lying on circle of same radius have same speed.Angular speed of body at any instant w.r.t. any point on body is same by definition.oLrq esa ?kw.kZu v{k ds yEcor~ ry ij lHkh fcUnq ladsUnzh; oÙkksa esa ifjHkze.k djrs gSaA leku f=kT;k ds oÙk ij lHkh fcUnqvksad h leku pky ¼rFkk Roj.k d k Hkh leku ifjek.k½ gSa ysfd u osx d h fn'kk ¼Roj.k d h fn'kk Hkh½ fHkUu gSaAblfy;s fn;s x;s ry esa nks ,sls fcUnq laHko ugh gS ft ud k osx ¼rFkk Roj.k Hkh leku½ leku gksAÅ ij ;g crk;k x;k gS fd leku f=kT;k d h oÙk ij fcUnqvksa d h leku pky gSAfd lh {k.k ij oLrq d h d ks.kh; pky oLrq ij fLFkr fd lh fcUnq d s lkis{k ifjHkk"kk ls leku gksrh gSA

Sol. (46 to 48)(i) Cons. linear momentum

STPFXII1213

Page - 48

� 2m.v + 2v.m = 0 = MVcm

Vcm

= 0(ii) As ball sticks to Rod

Conserving angular momentum about C2v.m. 2a + 2mva =

=

222

a4.ma.m212

a36.m8

6mv.a = 30 ma2. = a5v

(iii) KE =21

2 = 21

. 30 ma2 × 2

2

a25

v =

5mv3 2

.

49. The free body diagram of cart isxkM+h d k eqDr oLrq fp=k gSAHence net horizontal force on cart is zero.acceleration of cart is zero.blfy;s xkM+h ij usV {kSfrt cy 'kwU; gSAxkM+h d k Roj.k 'kwU; gSA

50. The acceleration of each block is equal and equal to m3F

.

Tension in required string can be found by applying Newton's second law to block C.

T2 = ma =

3F

.

izR;sd CykWd d k Roj.k leku gS rFkk m3F

d s cjkcj gSA

CykWd C ij U;wVu d k f}rh; fu;e yxkus ij vko';d Mksjh esa ruko Kkr fd ;k t k ld rk gSA

T2 = ma =

3F

.

51. For block B to just slip on the cart, the friction force on cart is mg. The net force on cart is thus mg.

Hence acceleration of cart is a = m3mg

= 3g

.

required force F = (m + m + m + M) a = 2mgxkM+h ij CykWd B d s Bhd fQlyus d s fy,] xkM+h ij ?k"kZ.k cy mg gSA xkM+h ij usV cy bl izd kj mg gSA

blfy;s] xkM+h d k Roj.k a = m3mg

= 3g

.

vko';d cy F = (m + m + m + M) a = 2mg

52. (A) F

= constant and 0Fu

Therefore initial velocity is either in direction of constant force or opposite to it. Hence the particle willmove in straight line and speed may increase or decrease.

(B) 0Fu and F

= constantinitial velocity is perpendicular to constant force, hence the path will be parabolic with speed of particleincreasing.(C) 0Fv

means instantaneous velocity is alway perpendicular to force. Hence the speed will remain

constant. And also |F|

= constant. Since the particle moves in one plane, the resulting motion has to becircular.(D) j�3i�2u

and j�9i�6a . Hence initial velocity is in same direction of constant acceleration, therefore

particle moves in straight line with increasing speed.

(A) F

= fu;r rFkk 0Fu

STPFXII1213

Page - 49

vr% izkjfEHkd osx ;k rks fu;r cy dh fn'kk esa gksxk ;k bld s foifjr gksxkA vr% d.k lh/kh js[kk esa xfr d jsxk vr% pkyc<+ ;k ?kV ld rh gSA

(B) 0Fu

rFkk F

= fu;rizkjfEHkd osx] fu;r cy d s yEcor~ gksxk] vr% iFk ijoy; gksxk ft lesa d .k d h pky c<+sxhA(C) 0Fv

vFkkZr~ rkR{kf.kd osx ges'kk cy d s yEcor~ gksxkA vr% pky fu;r gksxhA rFkk |F|

= fu;r] ifj.kkeh xfr

oÙkkd kj gksxhA pwafd d .k ges'kk ,d gh ry esa jgrk gSA(D) j�3i�2u

rFkk j�9i�6a vr% izkjfEHkd osx fu;r Roj.k dh fn'kk esa gh gksxkA vr% d.k lh/kh js[kk esa c<+rh pky d s

lkFk tk;sxkA

53. For H-like particles (like He+),

rn =

Z

2n0a=

2

2)3(Å59.0 = 2.38 Å (Ans. is 2.655 Å)

55. 2HI (g) H2 + I

2

n = 2 � 2 = 0

Kp = K

c (RT)n K

p = K

c (RT)0

Kp = K

c

56. By Charle's law, at constant TV

,P = const. i.e., V = k.T. Hence, plot of V vs T is a straight line passing through

the origin. (pkYlZ ds fu;ekuqlkj fLFkj nkc ij TV

,P = fLFkjkad i.e., V = k.T vr% V vs T dk xzkQ ,d ljy js[k gS tks fd

ewy fcUnq ls xqtj jgk gSA )

60.

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

61. CH3�CH2�CH2�CH3 h/Cl2 +

(d + )

62. Only Molarity depend on temperature molality, mole -fracction & % w/w do not depend on temperature.

63. & are functional group isomers.

rFkk fØ;kRed lewg leko;oh gSaA

64. Follow conditions of geometrical isomerism. (T;kferh; leko;ork iznf'kZr djus dh 'krsZ ns[kksA)



STPFXII1213

Page - 50

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

Course : VISHESH (JD) & VIJAY (JR)

Correct Wrong Blank


(dsoy ,d fodYi lgh)3 -1 0


(,d ;k ,d ls vf/kd fodYi lgh)4 0 0


31 52 67Matrix Match Type(eSfVªDl lqesy izdkj)

6 [1, 2, 3, 6] 0 0


Part - II (Physics)


Type

PART - I

SECTION - I ([k.M- I)

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

This section contains 15 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for itsanswer, out of which ONLY ONE is correct.bl [k.M esa 15 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA

1. The curve y � exy + x = 0 has a horizontal tangent at the point �oØ y � exy + x = 0 ds fdl fcUnq ij {kSfrt Li'kZ js[kk gS&(A) (1, 1) (B) (0, 1) (C) (1, 0) (D) no point (dksbZ fcUnq ugha)

2. b

and c

are two non-collinear vectors such that a

|| ( cb

) , then ( ba

).( ca

) is equal to

b

,oa c

nks vlajs[kh; lfn'k bl izdkj gS fd a

|| ( cb

) , rks ( ba

).( ca

) =

(A) )cb(|a| .2

(B) )ca(|b| .2

(C) )ba(|c| .2

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

3. If p is length of tangent from the point (1, 2) to the circle 3x2 + 3y2 + 12x + 18y � 1 = 0 then 3p2 =;fn fcUnq (1, 2) ls oÙk 3x2 + 3y2 + 12x + 18y � 1 = 0 ij [khaph xbZ Li'kZ js[kk dh yEckbZ p gks] rks 3p2 =

(A) 60 (B) 62 (C) 64 (D) none of these (buesa ls dksbZ ugha)

4. If Sn =

n

1r42 rr1

r, then S10 is equal to (;fn Sn =

n

1r42 rr1

r, rks S10 cjkcj gS&)

(A)111210

(B)421110

(C) 111110

(D) 11155

5. If (8cos4x � 4cos3x � 8cos2x + 3cosx + 1) is expressed as (cos px � cos qx), then

;fn (8cos4x � 4cos3x � 8cos2x + 3cosx + 1) d ks (cos px � cos qx) d s : i esa O;Dr fd ;k t k;s] rks(A) p = 4, q = 3 (B) p = 8, q = 3 (C) p = 8, q = 2 (D) none of these (buesa ls dksbZ ugh)

6. The sum of all the numbers which can be formed by using the digits 1, 3, 5, 7 all at a time and which haveno digit repeated isvadks 1, 3, 5, 7 ls fufEkZr lHkh vadks dk ;ksx D;k gksxk] tcfd fdlh Hkh vad dh iqujkofÙk ugha dh xbZ gks\(A) 16 × 4! (B) 1111 × 3! (C) 16 × 1111 × 3! (D) 16 × 1111 × 4!.

STPFXII1213

Page - 51

7. If z1 & z1 represent adjacent vertices of a regular polygon of n sides with centre at the origin & if

Im

Re

z

z

1

1

2 1 then the value of n is equal to :

;fn n Hkqt kvksa d s le: i cgqHkqt ft ld k d sUnz ewy fcUnq ij gks] d s nks vklUu 'kh"kZ z1 o z1

ls fu: fir fd;s t krs gS rFkk

Im

Re

z

z

1

1

2 1 gks] rks n d k eku gS&

(A) 8 (B) 12 (C) 16 (D) 24

8. The graph of the function y = f(x) is as shown in the figure. Thenwhich one of the following could represent the graph of the functiony = f(x) ?

Qyu y = f(x) dk vkys[k fp=kkuqlkj gSA rc Qyu y = f(x) dk vkys[k gS&

(A)

1 20�2 x

y

1(B)

1 20�2x

1

yyy

(C)

1 20�2x

1

0

yyy

(D)

1 20�2x

1

yyy

9. An extreme value of 4sin2x + 3cos2x � 24 sin x

2 � 24 cos

x

2, where 0 x

2

, is

4sin2x + 3cos2x � 24 sin x

2� 24 cos

x

2 dk ,d pje eku] tgk¡ 0 x

2

gS] gS &

(A) 4 + 2 (B) 4 (1 � 6 2 ) (C) 21 (D) 4

10. The number of ways to select three numbers in AP from the first 2n natural numbers is.izFke 2n izkd r la[;kvksa esa ls rhu la[;k,¡ tks fd lekUrj Js<+h esa gSa] pquus ds rjhdksa dh dqy la[;k gS &(A) n(n + 1) (B) n(n � 1) (C) n2 � 1 (D) n2 + 1

11. 2xx)x1(

dx = (

2xx)x1(

dx =)

(A) x1

)1x(2

+ C (B)

1x

)1x(2

+ C (C)

1x

)1x(2

+ C (D)

1x

)1x(2

+ C

12. A tangent to the circle x2 + y2 = 1 is perpendicular to a tangent to the circle x2 + y2 = 9. If locus of the point ofintersection of the tangents is x2 + y2 = a, then the value of a isoÙk x2 + y2 = 1 dh ,d Li'kZ js[kk] oÙk x2 + y2 = 9 dh ,d Li'kZ js[kk ds yEcor~ gSA ;fn Li'kZ js[kkvksa ds izfrPNsn fcUnq dkfcUnqiFk x2 + y2 = a gks] rks a dk eku gS &

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

13. An ellipse with foci (3, 1) and (1, 1) passes through the point (1, 3) its eccentricity isfcUnq (1, 3) ls xqtjus okys rFkk (3, 1) vkSj (1, 1) ukfHk okys nh?kZoÙk dh mRdsUnzrk gS&

(A) 1�2 (B) 1�3 (C) 2�3 (D) 3�2

STPFXII1213

Page - 52

14. Three numbers are chosen at random without replacement from {1,2,.......10}. The probability that theminimum of the chosen number is 3 or their maximum is 7 isleqPp; {1,2,.......10} esa ls rhu la[;k,a ;knfPNd (fcuk okil izfrLFkkfir fd;s) pquh tkrh gS] rks pquh xbZ la[;kvksa esa lsU;wure la[;k 3 rFkk vf/kdre la[;k 7 gksus dh izkf;drk gksxh&

(A) 2011

(B) 207

(C) 4011

(D) 407

15. If A =

001�

01�0

1�00

then (;fn A =

001�

01�0

1�00

gS] rks )

(A) A is zero matrix (A 'kwU; vkO;wg gS) (B) A2 = I

(C) A�1 does not exist (A�1 dk vfLrRo ugha gS) (D) A = (�1) I

SECTION - II ([k.M- II)

Multiple Correct Answer Type (cgqy lgh fodYi izdkj)

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for itsanswer, out of which ONE OR MORE is/are correct.bl [k.M esa 6 cgq fodYi iz'u gSaA izR;sd iz'u ds mÙkj ds fy, 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d lsvf/kd lgh gSA

16. If maximum and minimum values of the determinant

x2sin1xcosxsin

x2sinxcos1xsin

x2sinxcosxsin1

22

22

22

are and respectively,,

then :

;fn lkjf.kd

x2sin1xcosxsin

x2sinxcos1xsin

x2sinxcosxsin1

22

22

22

ds vf/kdre ,oa U;wure eku Øe'k% rFkk gS] rks&

(A) + 99 = 4 (B) 3 � 17 = 26 (C)

= 3

(D) a triangle can be construced having its sides as � , + and + 3(Hkqtk,a � , + rFkk + 3ls ,d f=kHkqt cuk;k tk ldrk gSA)

17. g(x) = f(x2) + f(1 � x2) and f(x) < 0 for 0 x 1, theng(x) = f(x2) + f(1 � x2) rFkk 0 x 1 d s fy, f(x) < 0 gS] rks &

(A) g(x) decreases in

2

1,0 ( g(x) ,

2

1,0 esa Ðkleku gSA )

(B) g(x) increases in

2

1,0 ( g(x),

2

1,0 esa o/kZeku gSA )

(C) g(x) decreases in

1,

2

1( g(x),

1,

2

1 esa Ðkleku gSA )

(D) g(x) increases in

1,

2

1( g(x) ,

1,

2

1 esa o/kZeku gSA )

STPFXII1213

Page - 53

18. The differential equation, xdxdy

+

dxdy3

= x2 ( vod y lehd j.k xdxdy

+

dxdy3

= x2 )

(A) is of order 1 (d h Ø e 1 gS) (B) is of degree 2 (d h d ksfV 2 gS)(C) is linear (js[kh; gS) (D) is non linear (vjs[kh; gS)

19. A line passes through (3, 0). The slope of the line, for which its intercept between y = x � 2 and y = �x + 2

subtends a right angle at the origin, may be.fcUnq (3, 0) ls xqtjus okyh ,d ljy js[kk dk js[kkvksa y = x � 2 vkSj y = �x + 2 ds e/; vUr%[k.M ewyfcUnq ij ledks.k cukrkgS] rks bl js[kk dh izo.krk gS &

(A) 2 (B) 2 (C) 2

1(D)

2

1

20. Let f(x) = 2]x[1

])x[(tan

, then which of the following is/are TRUE ?

( [.] represents greatest integer function )

ekukfd f(x) = 2]x[1

])x[(tan

gS] rks fuEu esa ls dkSuls dFku lR; gS &

(tgk¡ [.] egÙke iw.kkZd Qyu dks crkrk gS )

(A) f(x) is a constant function (f(x) ,d vpj Qyu gS)(B) f(x) is continuous at x = 0 (x = 0 ij f(x) lrr~ gS)(C) f(x) is discontinuous at x = 0 (x = 0 ij f(x) vlrr~ gS )(D) f(x) is continous function for all values of x (x ds lHkh ekuksa ds fy, f(x) lrr~ gS)

21. If the equation 4x � a . 2x � a + 3 = 0 has at least one real solution, then a

;fn lehdj.k 4x � a . 2x � a + 3 = 0 de ls de ,d okLrfod gy gS] rks a (A) (0, ) (B) (1, ) (C) [2, ) (D) [�6, 2]

SECTION - III ([k.M- III)

Comprehension Type (cks/ku izdkj)This section contains 3 groups of questions. Each group has 3 multiple choice questions based on a paragraph.Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct.bl [k.M esa 3 iz'u lewg gSaA izR;sd lewg esa ,d vuqPNsn ij vk/kkfjr 3 cgq fodYih gSaA izR;sd iz'u ds mÙkj ds fy, 4 fodYi(A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA


Consider system of equations ekukfd lehdj.k fudk;x � y + z + 1 = 0

2x + 3y + z � 3 = 0

x + y + 3z � 2 = 0

22. System of equations has unique solution, then : (lehdj.k fudk; vf}rh; gy j[krk gS] rks &)(A) �3, 14 (B) �13, 4

(C) �3, 4 (D) no value of (dk dksbZ eku ugha)

23. System of equations has infinitely many solutions, then : (lehdj.k fudk; ds gyksa dh la[;k vuUr gS] rks)(A) = 2 (B) = 4(C) = �13 (D) no value of (dk dksbZ eku ugha)

STPFXII1213

Page - 54

24. System of equations is in-consistent, then : (lehdj.k fudk; laxr ugha gks] rks&)(A) = �14 (B) = 4, �3

(C) = �13 (D) no value of (dk dksbZ eku ugha)


Read the following passage carefully and answer the questions.fuEu vuqPNsn dks /;kuiwoZd i<+dj uhps fn;s x, iz'uksa ds mÙkj nhft,

Let f(x) x3 � ax2 + bx � 1, g(x) x3 � bx2 + ax � 1 be polynomials with complex coefficients and , be rootsof f(x) = x3.ekukfd f(x) = x3 � ax2 + bx � 1, g(x) = x3 � bx2 + ax � 1 nks cgqin gSa] ftuds xq.kkad okLrfod ;k dkYifud gks ldrs gSavkSj , lehdj.k f(x) = x3 ds ewy gSaA

25. If roots of g(x) = 0 are in A.P. then;fn g(x) = 0 ds ewy lekUrj Js<+h esa gks] rks(A) a = b (B) 2a3 + 27 = 9ab (C) 9ab = 2b3 + 27 (D) ab3 + 27 = 9ab

26. If a, b are roots of x2 + x + 2 = 0, then f(x) g(x);fn a, b lehdj.k x2 + x + 2 = 0 ds ewy gks] rks f(x) g(x)(A) x6 + x5 + x4 + x3 + x2 + x + 1 (B) x6 � x5 + x4 � x3 + x2 � x + 1

(C) � x6 + x5 � x4 + x3 � x2 + x � 1 (D) x6 + 2x5 � x4 + 2x3 � x2 + 2x + 1

27. If a + b = �1 and a1, a

2 are values of 'a' for which , are connected by

+

=

34

, then equation with roots

a1, a

2 is

;fn a + b = �1 vkSj 'a' ds nks eku a1, a

2 gSa ftuds fy, , lEcU/k

+

=

34

ls tqM+s gq, gSa] rks a1, a

2 ewyksa okyh

lehdj.k gSa&(A) 4x2 � 3x + 3 = 0 (B) 3x2 � 4x + 3 = 0 (C) 4x2 + 3x + 3 = 0 (D) 3x2 + 4x + 3 = 0


A bag contains 8 black, 3 red and 9 white balls. If 3 balls are drawn at random, the probability that,d cSx esa 8 dkyh , 3 yky rFkk 9 lQsn xsans gSA ;fn 3 xsan ;knfPNd :i ls pquh tkrh gS

28. They are 2 black and 1 white is2 dkyh rFkk 1 lQsn gksus dh izkf;drk gksxh&

(A) 9521

(B) 9517

(C) 9523

(D) 9619

29. The balls are one of each colour isizR;sd jax dh ,d ,d xsan gksus dh izkf;drk gS&

(A) 9521

(B) 9518

(C) 9517

(D) 9623

30. The balls are drawn in the order black, red, white isxasns dkyh] yky] lQsn ds Øe esa fudkyus dh izkf;drk gS&

(A) 9518

(B) 9521

(C) 953

(D) 957

STPFXII1213

Page - 55

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

This section contains 1 questions. Each question contains statements givenin two columns which have to be matched. Statements in Column are labelledas A,B,C and D whereas statements in Column are labelled as p,q,r ands. The answers to these questions have to be appropriately bubbled asillustrated in the following example.If the correct matches are A-p , A-r , B-p , B-s , C-r , C-s and D-q ,


ABCD

p q r s

then the correctly bubbled matrix will look like the following :

bl [k.M esa 1 iz'u gSaA izR;sd iz'u esa nks dkWye esa oDRO; (statements) fn;s gq, gSa ftudkl qesy (match) d juk gSA d kWy e (Column-I) esa fn;s x;s oDrO;ksa(A, B, C, D) dks dkWye (Column-II) esa fn;s x;s oDrO;ksa (p, q, r, s) ls lqesy djuk gSAbu iz'uksa ds mÙkj fn;s x;s mnkgj.k ds vuqlkj mfpr cqYyksa dks dkyk djds n'kkZuk gSA;fn lgh lqesy A-p, A-r, B-p, B-s, C-r, C-s rFkk D-q gSa] rks lgh fof/k ls dkys fd,


ABCD

p q r s

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

31. Column � Column �

¼LrEHk �½ ¼LrEHk �½

(A) Number of ways of distributing 7 books to 2 children is (p) 126

7 iqLrd ksa d ks 2 cPpksa esa ck¡Vus d s d qy rjhd s gSa&

(B) The value of 2C2 + 3C2 + 4C2 + ............ + 11C2 is (q) 1282C2 + 3C2 + 4C2 + ............ + 11C2 d k eku gS&

(C) The number of five digit numbers in which every digit exceeds the (r) 136immediately preceding digit is5 vad ks d h ,slh fd ruh la[;k,¡ gksxh ft ud k izR;sd vad BhdiwoZ vad ls T;knk gS \

(D) The coefficient of x24 in expansion of

17

32

x

1x

is (s) 220

17

32

x

1x

d s izlkj esa x24 d k xq.kkad gS

PART - II


This section contains 10 multiple choice questions. Each question has choices (A), (B), (C) and (D), outof which ONLY ONE is correct.bl [k.M esa 10 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA

32. A man wearing a hat of extended length 12 cm is running in rain falling vertically downwards with speed10 m/s. The maximum speed with which man can run, so that rain drops do not fall on his face (the length of hisface below the extended part of the hat is 16 cm) will be : (,d O;fDr 12 cm mHkjh gqbZ yEckbZ dh ,d Vksih igus gq,10 m/s ds osx ls Å /okZ/kj uhps dh vksj fxj jgh cjlkr esa nkSM+rk gSA vkneh ds nkSM+us dh vf/kdre pky] rkfd cjlkr dhcawnsa mlds psgjs ij ugha fxjs ¼mlds psgjs dh Vksih ds mHkjs gq, Hkkx ds uhps dh yEckbZ 16 cm gS½ gksxh) &(A) 7.5 m/s (B) 13.33 m/s (C) 10 m/s (D) zero 'kwU;

STPFXII1213

Page - 56

33. A particle is projected at an angle of 60º from a horizontal plane. Then the path of projectile cannot appear a

straight line to an observer: (,d d.k dks {kSfrt ry ls 60° ds dks.k ij iz{ksfir fd;k tkrk gS rks fdlh izs{kd dks iz{ksI;dk iFk ,d ljy js[kk ugha izrhr gks ldrk gSA ;fn izs{kd ) &(A) moving with constant acceleration in horizontal direction

({kSfrt fn'kk esa fu;r Roj.k ls xfreku gksA)(B) moving with constant velocity in horizontal direction

({kSfrt fn'kk esa fu;r osx ls xfreku gksA)(C) moving downwards with constant acceleration '

g

'

(fu;r Roj.k g ls uhps dh vksj xfreku gksA)

(D) none of these. (buesa ls dksbZ ugha)

34. A particle is moving in the vertical plane. It is attached at one end of a string of length whose other end

is fixed. The velocity at the lowest point is u. The tension in the string is T

and velocity of the particle is v

at any position. Then, which of the following quantity will remain constant.(,d d.k ,d Å /okZ/kj ry esa xfr'khy gSA bldks ,d yEckbZ dh jLlh ds ,d fljs ij tksM+k x;k gS ftld k nwljk fljk

fLFkj gSA fuEure fcUnq ij osx u gSA jLlh esa ruko T

rFkk fd lh fLFkfr ij d.k d k osx v

gS rks fuEu esa ls dkSulh jkf'kfLFkjkad gksxh) :

(A) vT

(B) kinetic energy (C) Gravitational potential energy (D) vT

(A) vT

(B) xfrt Å t kZ (C) xq:Roh; fLFkfrt Å t kZ (D) vT

35. A solid conducting sphere of radius a is enclosed in a concentric shell of radius b > a. The charge on the innersphere is 3 C and on the conducting shell is � 10 C. The potential difference between the conductors is V. If+ 10 C additional charge is given to the outer shell, the new potential difference is(f=kT;k a d k ,d Bksl pkyd xksyk b > a f=kT;k d s ,d lad sUnzh; d ks'k ls f?kjk gqvk gSA vkUrfjd xksys ij 3 C vkos'kgS vkSj pkyd d ks'k ij � 10 C vkos'k gSA pkydksa ds chp foHkokUrj V gSA ;fn ckgjh dks'k dks + 10 C vfrfjDr vkos'kfn;k t k;] rks u;k foHkokUrj gksxk) &(A) V (B) V/2 (C) 2V (D) Zero ('kwU;)

36. In the circuit shown, the galvanometer shows zero current.The value of resistance R is :(fn[kk, ifjiFk esa xsYosuksehVj esa/kkjk dk eku 'kwU; gS rks çfrjks/k R dk eku gksxk) :(A) 1 (B) 2 (C) 4 (D) 9

37. The gravitational potential energy of a satellite revolving around the earth in circular orbit is � 4 MJ. The additional

energy (in MJ) that should be given to the satellite so that it escapes from the gravitational field of earth is :(Assume earth's gravitational force to be the only gravitational force on the satellite and no atmospheric resis-tance).(iFoh ds pkjks vksj oÙkh; d{kk esa ?kweus okys fdlh mixzg dh xqLRoh; fLFkrht Å tkZ � 4MJ gSA mixzg dks nh xbZ vfrfjDr Å tkZdk eku (MJ esa), ftlls ;g iFoh ds xq:Roh; {ks=k ls iyk;u dj tk;sA)¼ekuuk gS fd mixzg ij dsoy iFoh dk xq:Roh; cy dk;Z djrk gS vkSj dksbZ ok;qe.Myh; izfrjks/k ugha gSA½(A) 2 (B) 3 (C) 4 (D) 1

38. Figure shows a part of network of a capacitor and resistors.The potential indicated at A, B and C are with respect tothe ground. The charge on the capacitor in steady state is:

B

A

C

24

8

4

+8V

+4V

+6V

1 F 10V

(fn;s x;s fp=k esa ,d la/kkfj=k vkSj izfrjks/kksa d s rU=k d k ,d Hkkxfn[kk;k x;k gSA fcUnq A, B vkSj C ij fn[kk;s x;s foHko /kjkry¼Hkwfe½ d s lkis{k gSA LFkk;h voLFkk esa] la/kkfj=k ij vkos'k gSA)

(A) 4 C (B) 6 C (C) 10 C (D) 16 C

STPFXII1213

Page - 57

39. The radius of a coil of wire with N turns is 0.22 m, and 3.5 A current flows clockwise in the coil as shown.A long straight wire carrying a current 54A toward the left is located 0.05 m from the edge of the coil. Themagnetic field at the centre of the coil is zero tesla. The number of turns N in the coil are :(N pDdj okyh ,d d q.Myh d h f=kT;k 0.22 m gS vkSj blesa fp=kkuqlkj nf{k.kkoÙkZ (clockwise) fn'kk esa /kkjk çokfgr gksjgh gSA ,d yEcs lh/ks rkj tks dq.Myh d s fd ukjs ls 0.05 m nwjh ij dq.Myh ds ry esa gS] esa ck;ha rjQ 54 A /kkjk çokfgrgks jgh gSA d q.Myh d s d sUnz ij pqEcd h; {ks=k 'kwU; gSA d q.Myh esa ?ksjksa d h la[;k N gS) &

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

40. A long capillary tube of mass 20 mg, radius

2mm and negligible thickness, is

partially immersed in a liquid of surface tension 0.1 N/m. Take angle of contactzero and neglect buoyant force of liquid. The force required to holdthe tube vertically, will be - (g = 10 m/s2)

(nzO;eku 20 mg] f=kT;k

2mm ,oa ux.; eksVkbZ dh ,d yEch dsf'kdk uyh dks 0.1

N/m i"B ruko okys nzo esa vkaf'kd : i ls Mqcksrs gSaA Li'kZ d ks.k d ks 'kwU; ysrs gq,rFkk nzo d s mRIykou cy d ks ux.; ekurs gq;s V~;wc d ks Å /oZ : i ls id M+s j[kus d sfy, vko';d cy d k eku gksxk \) (g = 10 eh-@ls-2)(A) 2 mN (B) 1 mN (C) 0.5 mN (D) 4 mN

41. Rigidity modulus of steel is and its young�s modulus is Y. A piece of steel of cross�sectional area �A� is

changed into a wire of length L and area A/10 then :LVhy d k n<+rk xq.kkad rFkk ;ax izR;kLFkrk xq.kkad Y gSA �A� vuqizLFk d kV {ks=kQy d s LVhy ds VqdM+s dks L yEckbZ rFkkA / 10 {ks=kQy d s rkj d s : i esa cnyk t krk gSA rc(A) Y increases and decrease (B) Y and remains the same(C) both Y and increase (D) both Y and decrease(A) Y c<+sxk rFkk ?kVsxkA (B) Y rFkk leku jgsaxs A(C) Y rFkk nksuksa c<saxsA (D) Y rFkk nksuksa ?kVsaxsA

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 lsvf/kd fodYi lgh gS ¼gSa½A

42. Figure shows two blocks A and B connected to an ideal pulley string system.In this system when bodies are released then : (neglect friction and take g= 10 m/s2)fp=k esa n'kkZ;s vuqlkj nks fi.M A rFkk B ,d vkn'kZ f?kjuh&jLlh fudk; ls tqM+s gSA blfudk; esa tc fi.Mksa dks eqDr djrs gSa rks % ¼?k"kZ.k dks ux.; ekusa rFkkg = 10 eh-@ls-2 ysa½(A) Acceleration of block A is 1 m/s2

(A fi.M d k Roj.k 1 eh-/ls-2 gSA)(B) Acceleration of block A is 2 m/s2

(A fi.M d k Roj.k 2 eh-/ls-2 gSA)(C) Tension in string connected to block B is 40 N (B ls t qM+h jLlh esa ruko 40 N gSA)(D) Tension in string connected to block B is 80 N (B ls t qM+h jLlh esa ruko 80 N gSA)

STPFXII1213

Page - 58

43. A beam of light having frequency is incident on an initially neutral metal of work function (h > ). Then(vkofÙk dk izdk'k iqUt dk;ZQyu (h > ) okys /kkrq ¼izkjEHk esa vukosf'kr½ ij vkifrr gksrk gS rks)&(A) all emitted photoelectrons have kinetic energy equal to (h � ). (lHkh mRlftZr QksVks bysDVªkWuksa dh xfrt Å tkZ (h � ) ds cjkcj gksxhA)(B) all free electrons in the metal, that absorb photons of energy h completely, may not be ejected out of

the metal. (/kkrq esa mifLFkr lHkh eqä bysDVªkWu tks h Å tkZ ds QksVkWuksa dks iw.kZr;k vko'kksf"kr djrs gS] /kkrq ds ckgjmRlftZr ugh gks ldrs gSA)

(C) after being emitted out of the metal, the kinetic energy of photoelectrons decreases continuously aslong as they are at a finite distance from metal. (/kkrq ds ckgj mRlftZr gksus ds i'pkr~ QksVks bysDVªkWu dh xfrtÅ tkZ fujUrj rc rd ?kVrh gS tc rd os /kkrq ls ,d fu'fpr nwjh ij gksrs gSA)

(D) the emitted photo electrons move with constant speed after escaping from the field of metal. (/kkrq ds {ks=k ls ckgj fudyus ds i'pkr~ mRlftZr QksVks bysDVªkWu fu;r pky ls xfr djrs gSA)

44. In young�s double slit experiment, slits are arranged in such a way that besides central bright fringe, there

is only one bright fringe on either side of it. Slit separation d for the given condition cannot be (if iswavelength of the light used) : (;ax d s f}fNnz iz;ksx esa] fNnzksa d ks bl izd kj ls O;ofLFkr fd ;k t krk gS fd d sUnzh;ped hyh fÝ Ut d s vykok] d soy ,d ped hyh fÝUt bld s fd lh ,d rjQ curh gSA mijksä fLFkfr d s fy, fNnzksa d se/; nwjh ugha gks ld rh (mi;qZä izd k'k d h rjaxnS/;Z gS) :(A) (B) /2 (C) 2 (D) 3/2

45. Both a point source of sound & an observer are at rest. If air starts blowoing then

(/ofu dk ,d fcUnq L=kksr o ,d Jksrk nksukas fLFkj gSA ;fn gok cguk izkjEHk gks tkrh gS rks) �(A) The velocity of sound will change (/ofu dk osx ifjofrZr gksxkA)(B) The frequency detected by the observer will change (Jksrk }kjk lquh x;h vkofÙk ifjofrZr gksxhA)(C) The wavelength of sound will change (/ofu dh rjaxnS/;Z ifjofrZr gksxhA)(D) The intensity of the sound will change. (/ofu dh rhozrk ifjofrZr gksxhA )

SECTION - IIIComprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to beanswered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

[k.M - IIIc) cks/ku çdkj

bl [k.M esa 2 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 3 cgq&fodYih ç'u ds mÙkj nsus gSA çR;sd ç'u ds4 fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA


A monoatomic ideal gas is filled in a nonconducting container. The gas can be compressed by a movablenonconducting piston. The gas is compressed slowly to 12.5% of its initial volume.,d ijek.kqd ,d vkn'kZ xSl d ks vpkyd ik=k esa Hkjk t krk gSA xSl d ks ,d xfr d jus ;ksX; vpkyd fiLVu ls laihfMrfd ;k t k ld rk gSA xSl d ks izkjfEHkd vk;ru d s 12.5% rd /khjs /khjs laihfMr fd ;k t krk gSA

46. The percentage increase in the temperature of the gas is : (xSl d s rki esa izfr'kr of) ) &(A) 400% (B) 300% (C) � 87.5% (D) 0%

47. The ratio of the initial adiabatic bulk modulus of the gas to the final value of adiabatic bulk modulus of the gas is: (xSl ds izkjfEHkd :)ks"e vk;ru izR;kLFkrk xq.kkad dk vfUre :)ks"e vk;ru izR;kLFkrk xq.kkad ds lkFk vuqikr gS)(A) 32 (B) 1 (C) 1/32 (D) 1/4

STPFXII1213

Page - 59

48. The ratio of work done by the gas to the change in internal energy of the gas is :(xSl }kjk fd ;s x;s d k;Z d k xSl d h vkUrfjd Å t kZ esa ifjorZu d s lkFk vuqikr gS) &(A) 1 (B) �1 (C) (D) 0


In the figure shown the loop placed in the plane of the paper has radius 'R', self inductance 'L' and zero resistance.A uniform magnetic field B = B

0sint is applied (B

0 and are positive constants) normal to the plane of the

paper. Initially the loop has zero current : (çnf'kZr fp=k esa dkxt ds ry esa j[ks ywi dh f=kT;k 'R', LoçsjdRo 'L' rFkk 'kwU;çfrjks/k gSA ,dleku pqEcdh; {ks=k B = B

0sint (B

0 rFkk /kukRed fu;rkad gS) dkxt ds ry ds yEcor~ vkjksfir fd;k tkrk

gSA çkjEHk esa ywi esa 'kwU; /kkjk gSA)

49. Direction of current in the loop at t =

3 : (t =

3 ij ywi esa /kkjk dh fn'kk gksxh) :

(A) clockwise (nf{k.kkorZ)(B) anticlockwise (okekorZ)(C) no current will flow at that time (ml le; ij dksbZ /kkjk çokfgr ugha gksxh)(D) cannot be determined (x.kuk ugha dj ldrs)

50. The current in the loop at time t =

65

is : (t =

65

le; ij ywi esa /kkjk gksxh) :

(A) L2

RB 20 clockwise (nf{k.kkorZ) (B)

L2

RB 20 anticlockwise (okekorZ)

(C) L2

RB 3 20 clockwise (nf{k.kkorZ) (D)

L2

RB 3 20 anticlockwise (okekorZ)

51. The magnitude of self induced emf in the loop will : (ywi esa Loçsfjr fo|qr okgd cy dk ifjek.k) &(A) depend on value of L (L ds eku ij fuHkZj gksxkA) (B) not depend on value of R (R ds eku ij fuHkZj ugha gksxkA)(C) not depend on B

0 (B

0 ij fuHkZj ugha gksxkA) (D) depend on B

0 (B

0 ij fuHkZj gksxkA)





ABCD

p q r s


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

STPFXII1213

Page - 60

dkWye�II ds lHkh lgh fodYiksa ls lqesy djkus ds fy, ,d vad iznku fd;k tk;sxkA ;fn dkWye�I ds lHkh OkDÙkO;ksa dkdkWye�II ds lHkh lgh fodYiksa ls lqesy fd;k x;k rks 6 vad iznku fd;s tk;sxsaA

52. Two identical uniform solid smooth spheres each of mass m each approach each other with constant velocitiessuch that net momentum of system of both spheres is zero. The speed of each sphere before collision is u. Boththe spheres then collide. The condition of collision is given for each situation of column-I. In each situation ofcolumn- information regarding speed of sphere(s) is given after the collision is over. Match the condition ofcollision in column- with statements in column-.nks le:i ,dleku fpdus xksykdkj xksys ftudk izR;sd dk nzO;eku m gS ,d nwljs dh rjQ fu;r osxksa ls bl rjg ls utnhdvk jgs gS fd nksuksa xksys ds fudk; dk ifj.kkeh laosx 'kwU; gSA izR;sd xksys dh Vddj iwoZ pky u gSA nksuksa xksys rc Vdjkrs gSAVDdj dh 'krZ LrEHk-I dh izR;sd fLFkfr ds fy, nh xbZ gSA LrEHk- dh izR;sd fLFkfr esa Vddj [kRe gksus ds ckn dh lwpuk nhxbZ gSA VDdj dh 'krZ tks fd LrEHk- esa nh xbZ gS og LrEHk- esa fn;s x;s oDrO;ksa ls lqesfyr djksA

Column- Column-(A) Collision is perfectly elastic and head on (p) speed of both spheres after collision is u(B) Collision is perfectly elastic and oblique (q) velocity of both spheres after

collision is different

(C) Coefficient of restitution is e = 21

and (r) speed of both spheres after collision

collision is head on is same but less than u.

(D) Coefficient of restitution is e = 21

and (s) speed of one sphere may be more than u.

collision is obliqueLrEHk- LrEHk-

(A) VDd j iw.kZr;k izR;kLFk gS o lEeq[k gS (p) VDd j d s ckn nksuksa xksyks d h pky u gSA(B) VDd j iw.kZr;k izR;kLFk gS o fr;Zd gS (q) VDd j d s ckn nksuksa xksyks d s osx

vyx vyx gS

(C) VDd j d k izR;koLFku e = 21

rFkk (r) VDd j d s ckn nksuksa xksyks d h pky leku gS

VDd j lEeq[k gSA ij u ls d e gSA

(D) VDd j d k izR;koLFku xq.kkad e = 21

(s) ,d xksys d h pky u ls vf/kd gks

rFkk VDd j fr;Zd gSA Hkh ld rh gSA

PART - IIISECTION - I ([k.M- I)

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

53. Which has maximum internal energy at 290 K ? (290 K ij vf/kdre vkUrfjd Å tkZ fdldh gksrh gS \)(A) Neon gas ¼fu;kWu xSl½ (B) Nitrogen gas ¼ukbVªkstu xSl½(C) Ozone gas ¼vkstksu xSl½ (D) Equal ¼leku½

54. 2.5 L of a sample of a gas at 27°C and 1 bar pressure is compressed to a volume of 500 mL keeping the

temperature constant, the percentage increase in pressure is :

27°C rki vkSj 1 ckj nkc ij ,d xSl ds 2.5 L uewus dks 500 mL ds vk;ru rd LkEihfM+r djrs gSa] ftlesa rki fu;r jgrkgS] rc nkc esa fdrus % of) gksxhA(A) 100 % (B) 400 % (C) 500% (D) 80%

STPFXII1213

Page - 61

55. Which of the following contains the greatest number of atoms ? (fuEu es ls dkSu ijek.kq dh vf/kdre la[;k j[krk gS\)(A) 1.0 g of butane (C4H10) (1.0 g C;wVsu (C4H10) (B) 1.0 g of nitrogen (N2) (1.0 g ukbVªkstu (N2)

(C) 1.0 g of silver (Ag) (1.0 g flYoj (Ag) (D) 1.0 g of water (H2O) (1.0 g ty (H2O)

56. Select the correct statement : (lgh d Fku pqfu;s )(A) Perxenate ion is [XeO6]4� with octahedral geometry. (ijft usV vk;u [XeO6]4� d h T;kfefr v"BQyd h; gS)(B) XeF2 is bent molecule with 3 lone pairs (l.p) (3 ,d kd h ;qXe (l.p.) d s lkFk XeF2 eqM+h gqbZ vkd fr d k v.kq gS)(C) XeOF4, XeF4, XeO2F2 all contains one lone pair only (XeOF4, XeF4, XeO2F2 lHkh dsoy ,d ,dkdh ;qXe j[krs gS)(D) None of these ¼buesa ls d ksbZ ugha½

57. In electrolysis of Al2O3 by Hall-Heroult process : (gkWy&gsjkYV fof/k ds }kjk Al2O3 ds fo|qr vi?kVu esa %)(A) cryolite Na3[AlF6] lowers the melting point of Al2O3 and increases its electrical conductivity.

Øk;ksykbZV Na3[AlF6], Al2O3 ds xyukad dks de djrk gS rFkk bldh fo|qr pkydrk dks c<+krk gSA(B) Al is obtained at cathode and probably CO2 at anode

Al dSFkksM+ ij izkIr gksrk gS rFkk lEHkor;k CO2 ,suksM ij izkIr gksrh gSA(C) both (A) and (B) are correct ((A) rFkk (B) nksuksa gh lgh gSA)(D) none of the above is correct ( mijksDr esa ls dksbZ ughaA)

58. The correct IUPAC name of the complex:

fuEu ladqy dk lgh IUPAC uke gS %

C = N

OH

OHH3C

C = NH3C

��

��CoCl2 is :

(A) Dichloridodimethylglyoximecobalt(II) ¼MkbZDyksjhMksMkbZesfFkyXykbvkWfDledksckYV(II)

(B) Bis(dimethyglyoxime)dichloridocobalt(II) ¼fcl~¼MkbZesfFkyXykbvkWfDle½MkbZDyksjhMksdksckYV (II)(C) Dimethylglyoximecobalt(II) chloride ¼MkbZesfFkyXykbvkWfDledksckYV(II) DyksjkbM½(D) Dichlorido(dimethylglyoximato)cobalt(II) ¼MkbZDyksjhMksfcl~¼MkbZesfFkyXykbvkWfDle-N, N)-dksckYV(II)

59. Observe each pair of cations. In which case (s) first is more stable than the second :fuEu /kuk;u ds ;qXeksa dk izs{k.k dhft,A buesa ls fdl ;qXe esa igyk /kuk;u nwljs dh vis{kk vf/kd LFkk;h gS :

(W) , (X) ,

(Y) , (Z) ,

(A) Only in W (B) Only in X and Y (C) Only in Z (D) Only in W and Z

STPFXII1213

Page - 62

60. The solvolysis in aqueous acetone is fastest in,tyh; ,lhVksu esa lcls rhoz nj ls foyk;dhdj.k fØ;k nsrk gSA

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

61. Which of the following reaction is not feasible ?dkSulh vfHkfØ;k lEHko ugha gS \

(A) 4LiAlH

(B) 3NaHCO + CO2 + H2O

(C) HCl Cl

(D)

3

3

3

CH|

Br�C�CH|

CH

ether/Na 23

3

CHC�CH|

CH

+

3

3

3

CH|

H�C�CH|

CH

SECTION - II ([k.M - II)

Multiple Correct Answers Type (cgqy lgh mÙkj izdkj)

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

bl [k.M esa 2 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 lsvf/kd fodYi lgh gS ¼gSa½A

62. Given EAg Ag

/ = 0.80V, EMg Mg2

/ = �2.37V, , ECu Cu2

/ = 0.34V, EHg Hg2

/ = 0.79 V..

Which of the following statements is/are correct

fn;k x;k gSA EAg Ag

/ = 0.80V, EMg Mg2

/ = �2.37V, , ECu Cu2

/ = 0.34V, EHg Hg2

/ = 0.79 V gSA

fuEufyf[kr esa ls dkSulk@dkSuls dFku lR; gSaA(A) AgNO3 can be stored in copper vessel (AgNO3 dks dkWij ik=k esa j[k ldrs gSA)(B) Mg(NO3)2 can be stored in copper vessel (Mg(NO3)2 dks dkWij ik=k esa j[k ldrs gS)(C) CuCl2 can be stored in silver vessel (CuCl2 dks flYoj ik=k esa j[k ldrs gSA)(D) HgCl2 can be stored in copper vessel (HgCl2 dks dkWij ik=k esa j[k ldrs gSA)

STPFXII1213

Page - 63

63. Which of the following compounds will have C2 axis of symmetry.fuEu esa ls dkSuls ;kSfxdksa esa C2 leferh dk v{k mifLFkr gSaA

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

SECTION - IIIComprehension Type

This section contains 1 paragraphs. Based upon each paragraph, 3 multiple choice questions have to beanswered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

[k.M - IIIc) cks/ku çdkj

bl [k.M esa 1 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 3 cgq&fodYih ç'u ds mÙkj nsus gSA çR;sd ç'u ds4 fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA


Compound (X) (C8H7N) on treatment with cold SnCl2/dil.HCl (aqueous), produces Y (C8H8O). (Y) on heatingwith sodium acetate in acetic anhydride followed by acidification forms (Z) (C10H10O2). All these compounds(X,Y,Z) on oxidation under severe conditions form a diacid which gives only one mononitro isomer in nitratingmixture.;kSfxd (X) (C8H7N) dh vfHkfØ;k B.Ms SnCl2/ruq HCl (tyh;), ls djkus ij Y (C8H8O) izkIr gksrk gSA (Y) dks ,lhfVd,ugkbMkbM esa lksfM;e ,lhVsV ds lkFk xeZ djus ij rFkk vEyh;dj.k djus ij ;kSfxd (Z) (C10H10O2) izkIr gksrk gSA ;s lHkh;kSfxd (X,Y,Z) dk dfBu ifjfLFkfr;ksa esa vkWDlhdj.k djus ij Mkb,flM cukrs gSa tks fd ukbVªhdj.k feJ.k ls vfHkfØ;k djusij dsoy eksuksukbVªks leko;oh nsrs gSaA

64. The reductive ozonolysis products of (Z) are :

(Z) dk vipk;d vkstksuh vi?kVu djus ij izkIr mRikn gksxk %

(A)

CH3

CHO + CHO

COOH

(B) CHOH C3 +

COOH

COOH

(C)

COOH

CHOCOOH +

CH3

(D) CHO + H C3

COOH

CHO

65. What is not true about (Y) ? ((Y) ds fy, D;k lgh ugha gS \)(A) It can give cannizzaro reaction (;s dSfutkjks vfHkfØ;k ns ldrk gSA)(B) It gives Iodoform test positive (;s vk;MksQkeZ ijh{k.k /kukRed nsrk gSA)(C) It gives Tollen's test positive (;s VkWysu ijh{k.k /kukRed nsrk gSA)(D) It gives 2, 4-DNP test positive (;s 2, 4-DNP ijh{k.k /kukRed nsrk gSA)

STPFXII1213

Page - 64

66. (X) + CO2

The compound W is :

(X) vkaf'kd tyvi?kVu

+ CO2

;kSfxd W gS %

(A) (B)

(C) (D)





ABCD

p q r s


C-r, C-s rFkk D-q gSa] rks lgh fof/k ls dkys fd, x;s cqYyksa dk 4 x 4 esafVªDl (matrix) esa n'kkZ;k x;k gSA dkWye�I ds izR;sd oDÙkO;dk dkWye�II ds ,d ;k ,d ls T;knk fodYiksa ls lqesy djk;k tk ldrk gSA dkWye�I ds izR;sd oDÙkO; dkdkWye�II ds lHkh lgh fodYiksa ls lqesy djkus ds fy, ,d vad iznku fd;k tk;sxkA ;fn dkWye�I ds lHkh OkDÙkO;ksa dkdkWye�II ds lHkh lgh fodYiksa ls lqesy fd;k x;k rks 6 vad iznku fd;s tk;sxsaA

67. Match each of the reactions given in column I with the corresponding product(s) given in column II.dkWye I esa fn;s x;s izR;sd vfHkfØ;k dks dkWye II esa fn;s x;s laxr mRikn@mRiknksa ds lkFk lqesy dhft,A

Column I Column IIdkWye I dkWye II

(A) Cu + dil HNO3 (p) NO

(B) Cu + conc HNO3 (q) NO2

(C) Zn + dil HNO3 (r) N2O

(D) Zn + conc HNO3 (s) Cu(NO3)2 / Zn(NO3)2

STPFXII1213

Page - 65

ANSWER KEY TO SAMPLE TEST PAPER-III1. B 2. A 3. B 4. D 5. A 6. C 7. A8. B 9. B 10. B 11. A 12. B 13. A 14. C15. B 16. ABC 17. BC 18. ABD 19. CD 20. ABD 21. C22. C 23. D 24. B 25. C 26. A 27. B 28. A29. B 30. C31. (A) (q), (B) (s), (C) (p), (D) (r)32. A 33. D 34. A 35. A 36. A 37. A 38. A39. A 40. B 41. B 42. B,D 43. B,.C 44. B,C 45. A,C46. B 47. C 48. B 49. B 50. B 51. D52. (A) � p,q ; (B) � p,q ; (C) � q,r ; (D) � q,r

53. C 54. B 55. A 56. A 57. C58. A 59. D 60. A 61. C 62. BC

63. ABCD 64. D 65. B 66. A67. A� p, s ; B�q, s ; C� r, s ; D� q, s

HINTS & SOLUTION TO SAMPLE TEST PAPER-III

1. y + x = exy dxdy

+ 1 = exy

ydxdy

x

Now for horizontal tangent (vc {kSfrt Li'kZ js[kk ds fy,) dxdy

= 0 1 = exy (y) .........(1)

Hence option (B) is satisfying the relation (i) (vr% fodYi (B) lEcU/k (i) dks larq"V djrk gSA)

2. a

|| ( cb

)

so a

perpendicular to b

, a

perpendicular to c

(vr% a

, b

ds yEcor~ gS , a

, c

ds yEcor~ gSA )

)ca(.)ba(

= 21 n�|c||a|n�|b||a| . b�a�n�1

= 212 n�n�|c||b||a| .

c�a�n� 2

= | a

|2 )cb(

.

3. x2 + y2 + 4x + 6y � 31

= 0

p = 31

2.61.441 = 362

3p2 = 62

5. Given expression = 8 cos2x (cos2x � 1) � (4cos3x � 3cosx) + 1

= � 8 cos2x sin2x � cos 3x + 1

= 1 � 2sin2 2x � cos 3x

= cos 4x � cos 3x

9. 4 sin2x + 3 cos2x � 24 (sin 2x

+ cos 2x

)

= 3 + sin2x � 24 (sin 2x

+ cos 2x

)

STPFXII1213

Page - 66

Let sin 2x

+ cos 2x

= t. Then t2 = 1 + sin x, 1 t 2 and the expression becomes

(ekukfd sin 2x

+ cos 2x

= t gS] rks t2 = 1 + sin x, 1 t 2 rFkk O;atd fuEu eas ifjofrZr gks tkrk gSA)

3 + (t2 � 1)2 � 24t = t4 � 2t2 � 24t + 4

Let (ekukfd ) f(t) = t4 � 2t2 � 24t + 4

f(t) = 4t3 � 4t � 24 = 4(t � 2) (t2 + 2t + 3) < 0 { 1 t 2 }

f(t) is a decreasing function. (f(t) ,d Ðkleku Qyu gSA)

minimum value of f(t) is at t = 2 and minimum value = 4 (1 � 6 2 )

(t = 2 ij f(t) dk U;wure eku = 4 (1 � 6 2 ))

maximum value of f(t) is at t = 1 and maximum value = � 21

(t = 1 ij f(t) dk vf/kdre eku = � 21)

10. Let m be the smallest of the choosen numbers and d be the common difference.then m + 2d 2n

1 d

2

mn2where [ . ] is the greatest integer function.

For a given value of m, the number of choices of d is

2mn2

The total number of ways =

2n2

1m2

mn2

= (n � 1) + (n � 1) + (n � 2) + (n � 2) + .....+ 1 + 1

= 2{(n � 1) + (n � 2) + .... + 1} = n(n � 1)

11. Let = 2xx)x1(

dx

Put 1 + x = t1

x2

1 dx = 2t

1 dt

=

2

1t1

1t1

1 2t

dt2 =

1t2

dt2

= 1t22 + c = �2 1x1

2

+C

= �2 x1

x1

= 2

x1

1x

.

12. Given circles arex2 + y2 = 1 ..... (1)x2 + y2 = 9 ..... (2)

P1

3

C

From the figure, CP = 91 = 10 the locus is x2 + y2 = 10a = 10

STPFXII1213

Page - 67

13. SP + SP = 2aa = 12

2ae = SS = 2 e = 1�2

14. The number of ways choosing 3 numbers from 1,2,.......10 is 10C3 = 120

if 3 < a < b, the number of ways of chosing a, b is 7C2 = 21

if c < d < 7, the number of ways of chosing c, d is 6C2 = 15

if 3 < e < 7, there are 3 numbers for e the required probability = 120

3�1521=

4011

15.

001�

01�0

1�00

001�

01�0

1�00

=

100

010

001

16. Applying C1 C

1 + C

2 + C

3 , we get

x2sin1xcosx2sin2

x2sinxcos1x2sin2

x2sinxcosx2sin2

2

2

2

=

100

010

x2sinxcosx2sin2 2

(Applying R2 R

2 � R

1 , R

3 R

3 � R

1 )

= 2 + sin2x = 3, = 1Now, � = 2, + = 4 and + 3 = 6. Thus ( � ) + ( + ) = 6 = ( + 3)So, � , + , + 3 cannot form a triangle.

C1 C

1 + C

2 + C

3 dk iz;ksx djus ij

x2sin1xcosx2sin2

x2sinxcos1x2sin2

x2sinxcosx2sin2

2

2

2

=

100

010

x2sinxcosx2sin2 2

(R2 R

2 � R

1 , R

3 R

3 � R

1 ds iz;ksx djus ls)

= 2 + sin2x = 3, = 1vc � = 2, + = 4 rFkk + 3 = 6. vr% ( � ) + ( + ) = 6 = ( + 3)

vr% � , + , + 3 ,d f=kHkqt ugha cuk ldrh gSA

17. g(x) = 2x { f(x2) � f(1 � x2) }

Since f(x) is decreasing (pwafd f(x) Ðkleku gSA)

hence for (vr%) x2 < 1 � x2 x < 2

1, g(x) > 0

x2 > 1 � x2 x > 2

1, g(x) < 0

decreasing in

1,

2

1and increasing in

2

1,0

(

1,

2

1 esa Ðkleku gS rFkk

2

1,0 esa o/kZeku gSA)

18. xdxdy

+

dxdy3

= x2 x

2

dxdy

� x2

dxdy

+ 3 = 0

it is of order 1, degree 2 and is non linear

STPFXII1213

Page - 68

19. Equation of straight lines is (ljy js[kk dk lehdj.k)(y � x + 2) ( y + x � 2) = 0 ....(1) y2 � (x � 2)2 = 0 y2 � x2 + 4x � 4 = 0

Let equation of line passing through (3, 0) is (;g (3, 0) ls xqtjrh gSA)y = mx � 3m ....(2)

Homogenising (1) with the help of (2), we get (lehdj.k (1) dks (2) dh lgk;rk ls le?kkr cukus ij)

y2 � x2 + 4x

m3mxy

� 4

2

m3mxy

= 0

Now, coefficient of x2 + coefficient of y2 = 0 (x2 dk xq.kkad + y2 dk xq.kkad = 0)

m3m4

+ 2m9

4 (1 + m2) = 0

34

� 2m9

4 (1 + m2) = 0 m = ±

2

1

21. t = 2x t2 � at � a + 3 = 0

2

2a

�t

=

4a2

+ a � 3 = 4

)2�a)(6a( 0

a (�, �6] [2, )

t = 2

)2�a)(6a(a = 2x > 0

a cannot be negative and a 2

Sol. (22, 23, 24)Augmented matrix is

231

332

111�1

133

122R�RRR2�RR

,

~

3210

5250

1111

233 R

51

RR

)1(3)2(5

1200

52�50

1111

~

25

1200

52�50

1111

2

unique solution if 2 � � 12 0 4, �3

If = � 3 or 4, then

2000

52�50

1111

no solution

there is no value of for which the given system of equations has infinite solutions

STPFXII1213

Page - 69

25. Let A � d, A, A + d be roots

A � d + A + A + d = b A = 3b

must satisfy g(x) = 0

27b3

� 9

b3

+ 3

ab � 1 = 0

9ab = 2b3 + 27

26. a + b = �1, ab = 2, a2 + b2 + 2ab = 1 a2 + b2 = �3

f(x) . g(x) = x6 � (a + b) x5 + (a + b + ab) x4

� (a2 + b2 + 2) x3 + (a + b + ab) x2 � (a + b) x + 1

= x6 + x5 + x4 + x3 + x2 + x + 1

27. f(x) = x3 ax2 � bx + 1 = 0

+ = ab

, = a1

+ = � 1 � a1

, = a1

Given

22

= 34

2)( 2

= 34

a1

a1

2a1

12

= 34

1 + 2 . a1

+ 2a

1 � 2 .

a1

= a34

1 + 2a

1 =

a34

3a2 + 3 = 4a 3x2 � 4x + 3 = 0 is required equation

28. Probability = 3

201

92

8

C

C.C =

18.19.203.9.7.8

= 9521

29. Probability = 3

201

91

31

8

C

C.C.C=

18.19.206.9.3.8

= 9518

30. P(BRW) = 208

. 193

. 189

= 953

31. (A) Required = 27 = 128

(B) 2C2 + 3C2 + 4C2 + ............ + 11C2= 3C3 + 3C2 + ........ + 11C2 = 12C3 = 220

(C) Required = selecting 5 digits out of 9 = 9C5 = 126

(D) 17Cr (x2)17 � r

r

3x

1

r =

524217

= 2

17C2 = 136

STPFXII1213

Page - 70

32. VR/G(x)

= 0 , VR/G(y)

= 10 m/sLet, velocity of man = vekuk fd vkneh dk osx = v

tan = 1216

= 34

then, vR/man

= v (opposite to man)

12 cm

16

cm

For the required condition :

rc, vR/man

= v (vkneh ds foifjr)

nh xbZ fLFkfr ds fy, :

tan = )x(M/R

)y(M/R

V

V =

v10

= 34

V = 4

3x10 = 7.5 Ans.

34. The tension vector T

is always normal to velocity vector. Hence vT

is always zero. A is correct choice.

35. The potential difference between the two conductors (nks pkyd ksa d s chp foHkokUrj)

V = 0

1

4Q

b

1

a

1

is independent of the charge on the outer shell. (t ks fd ckgjh d ks'k ij vkos'k ls LorU=k gSA)

Q1Q2

a

b Potential difference will remain same. (blfy, foHkokUrj leku jgsxkA)

36.

If pot. drop between A and B is also 2V, then no currrent will pass through the gelvanomter.

Pot. drop across R =

5R12

R = 2

12 R = 2R + 10R = 1

37. PE = �4 MJ

TE = �2MJ

The additional energy required to make the satellite escape = +2MJ.mixzg d s iyk;u d s fy, vfrfjDr Å t kZ = +2MJ.

38.

4V 6V 6V

4 2 4

8 1F 10V

6V

18 1F 10V

E =

41

21

41

44

26

48

= 6 V; Q = 4V × 1F = 4C

39. )rd(2i.10

=

r2

iN 20 N =

2

1

ii

. )rd(r

= 5.3

54 . )27.0(22

722.0 = 4

STPFXII1213

Page - 71

40. The free body diagram of the capillary tube is as shown in the figure. Net force F required to hold tube is(d ksf'kd k uyh d k cy js[kk fp=k n'kkZ;k x;k gSA V~;wc d ks id M+s jgus d s fy, d qy cy F gS)F = force due to surface tension at cross-section (vuqizLFk d kV ij i"B ruko d s d kj.k cy )

(S1 + S

2) + weight of tube (V~;wc d k Hkkj)

F

Free body daigramof capillary tube

T×2 R

mg

S1

S2

T×2 R

F

d sf'kd k uyh cy js[kk fp=k

T×2 R

mg

S1

S2

T×2 R

= (2RT + 2RT) + mg = 4RT + mg

= 4 ×

2 × 10�3 × 0.1 + 20 × 10�6 × 10 = 1 mN

41. and Y are properties of material.These coefficients are independent of geometry of body.

42. Applying NLM on 40 kg block40 fdxzk fi.M ij U;wVu ds fu;e yxkus ij

400 � 4T = 40 a

For 10 kg block T = 10.4 aSolving a = 2m/s2

40kg

10kg

4aT

2T2T

4T

a

10kg

4aT

2T2T

4T

a

10kg

4aT

2T2T

4T

a

10 fdxzk- fi.M ds fy, T = 10(4 a)

gy djus ij a = 2m/s2

T = 80 N

43. (A) All emitted photo electrons have kinetic energies ranging from 0 to hn � f0. Hence false.

(B) Some free electron may lose energy within metal due to collision and cannot escape the metal. Hencetrue.(C) Since metal is positively charged it will attracted emitted photoelectrons. Hence true.(A) lHkh mRlftZr QksVks bysDVªkWuksa dh xfrt Å tkZ 0 ls h �

0 ijkl esa gksrh gSA vr% oDrO; xyr gSA

(B) /kkrq esa mifLFkr eqä bysDVªkWuksa dh VDdj ds i'pkr~ Å tkZ esa gkfu gks ldrh gS rFkk /kkrq ls ckgj ugh fudy ldrs gSAvr% oDrO; lR; gSA(C) pwafd /kkrq /kukRed :i ls vkosf'kr gS ;g mRlftZr QksVksa bysDVªkWuksa dks vkdf"kZr djsxk vr% oDrO; lR; gSA

44. d sin = For first bright fringe. (There will be one bright fringe on both sides of central bright fringe and both arecalled first bright fringe.)izFke ped hyh fÝUt d s fy, (d sUnzh; fÝUt d s nksauks vksj ped hyh fÝUt igyh ped hyh fÝUt d gykrh gSA)

sin = 1d

d .

If d = 2, two maxima will be formed.

;fn d = 2 gS rks nks mfPp"B cusaxsA

STPFXII1213

Page - 72

46. Let initial temperature and volume be T0 and V0. Since the process is adiabatic, the final temperature and

volume is TV �1 = T0V0�1 ( =

35

for monoatomic gas)

T = 32

0

00 8/V

VT

= 4T0

percentage increase in temperature of gas is4 × 100 � 100 = 300%

ekuk izkjfEHkd rki o vk;ru T0 o V0 gSA pwafd izØ e : )ks"e gS vr% vfUre rki o vfUre vk;ru esa lEcU/k

TV �1 = T0V0�1 ( =

35

,d ijek.kqd xSl d s fy,)

T = 32

0

00 8/V

VT

= 4T0

xSl d s rki esa izfr'kr of)4 × 100 � 100 = 300%

47. Adiabatic Bulk modulus B = � VdVdP

= P = v

nRT

: )ks"e vk;ru izR;kLFkr xq.kkad B = � VdVdP

= P = v

nRT

TV

V

T

BB

0

0

f

i = 0

0

0

0

T4

8/V

V

T =

321

48. For adiabatic process dQ = 0: )ks"e izØ e esa dQ = 0

dU + dW = 0 or dUdW

= �1.

49. at t =

3, B is increasing in inward direction

t =

3, ij B vUnj dh vksj c<+ jgk gS

50. Li

at t =

65

ij

= BA

= B0 sin

65

. R2

= 2B0

.R2 = Li i = L2RB 2

0

52. In all cases speed of balls after collision will be same. In case of elastic collision speed of both balls aftercollision will be u, otherwise it will be less than u.

STPFXII1213

Page - 73

lHkh fLFkfr;ksa esa VDd j d s ckn xsanksa d h pkys leku gksxhA izR;sd fLFkfr esa izR;kLFk VDd j d s ckn nksuksa xsanks d h pkysu gS vU;Fkk ;g u ls d e gksxhA

53. Internal energy of a gas ¼xSl dh vkfUrfjd Å tkZ½ = 2f

nRT

Internal energy of a gas ¼xSl dh vkfUrfjd Å tkZ½ f

54. Using (mi;ksx) p1V

1 = P

2V

2 1 × 2.5 = 0.5 × P

2 = 5 bar.

% increase in pressure (% nkc esa of) ) = bar1

bar)1�5( × 100% = 400 %.

55. Compound mole of molecule mole of atom No. of atom;kSfxd v.kq ds eksy ijek.kq ds eksy ijek.kq dh la[;k

C4H

10 581

14 × 581

5814

NA

N2 28

12 ×

281

141

NA

Ag108

1108

1108

1N

A

H2O

181

3×181

6NA

Hence C4H

10 has maximum number of atoms. (vr% C

4H

10 ijek.kq dh vf/kdre la[;k j[krk gSA)

56.

Xe

O¯

O¯O¯

O¯

O

OPerxenate ion [XeO ]6

4-

Xe

F

FXeF (Linear)2

Xe

F

FF

F

XeOF (one l.p.)4

O

Xe

F

FF

F

XeOF (one l.p.)4

Xe

F

O

O

F

XeO F (one l.p.)2 2

STPFXII1213

Page - 74

57. Na3[AlF

6] 3NaF + AlF

3

NaF and AlF3 both are ionic compounds and so ionise to give ions. This increases the electrical conductivity

and lowers the melting point of Al2O

3 .

At cathode : Al3+ (melt) + 3e� Al.At anode : C(s) + O2� (melt) � CO (g) + 2e� ; C(s) + 2O2� (melt) � CO

2 (g) + 4e�.

Na3[AlF

6] 3NaF + AlF

3

NaF, AlF3 nksuksa vk;fud ;kSfxd gS tks vk;fur gksdj vk;u nsrs gSA blfy;s fo|qr pkydrk c<+krs gSA rFkk Al

2O

3 ds xyukad

fcUnq dks de djrs gSAdSFkksM : Al3+ (xfyr) + 3e� Al.

,uksM % C(s) + O2� (xfyr) � CO (g) + 2e� ; C(s) + 2O2� (xfyr) � CO2 (g) + 4e�.

58. Correct name is dichloridodimethylglyoximatecobalt (II)

lgh uke MkbZDyksjhMksMkbZesfFkyXykbvkWfDledksckYV(II)

59. In (W) since C�H bond is weaker than C�D bond so hyperconjugation stability is more in I.In (X) only +I effect is present which is more for �C(CD3).In (Y) only +I which is more for �CD3In (Z) �I effect of �CCl3 group will make II cation highly unstable.(W) esa C�H ca/k C�D ca/k ls nqcZy gksrk gS] blfy, vfrla;qXeu LFkk;hRo I esa vf/kd gksxkA(X) esa dsoy +I izHkko mifLFkr gS] blfy, �C(CD3) dk izHkko vf/kd gksxkA(Y) esa dsoy +I izHkko mifLFkr gS] blfy, �CD3 dk izHkko vf/kd gksxkA(Z) esa �CCl3 dk �I izHkko gS] blfy, II /kuk;u vR;f/kd vLFkkbZ gksxkA

60. (Most stable carbocation)

61. (A) is feasible ((A) lEHko gSA)(B) is feasible. ((B) lEHko gSA)

(C) is not feasible. ((C) lEHko ugha gSA)

(D) is feasible by product in Wurtz reaction. ((D) oqV~l vfHkfØ;k esa ;g mRikn lEHko gSA)

STPFXII1213

Page - 75

62. Lower S.R.P. containing ion can displace higher S.R.P. containing ion.U;wure ekud vip;u foHko ;qDr vk;u mPpre ekud vip;u foHko ;qDr vk;u dks foLFkkfir dj ldrk gSA

64.

66.

67. (A) 3Cu + 8HNO3 (dilute HNO

3) 2NO + Cu(NO

3)

2 + 4H

2O

(B) Cu + 4HNO3 (concentrated) 2NO

2 + Cu(NO

3)

2 + 2H

2O

(C) 4Zn + 10HNO3 (dilute) 4Zn(NO

3)

2 + N

2O + 5H

2O

(D) Zn + 4HNO3 (concentrated) Zn(NO

3)

2 + 2NO

2 + 2H

2O

Corporate Office:Tel. No. : Fax : e-mail :

Resonance Eduventures Pvt. Ltd. , A-46, A-52, Near City Mall, Jhalawar Road, Kota (Raj.)-3240050744- 3012222, 3022222, 2437144 I 022-39167222, 0744-2427144 I [email protected],

Study Centres: Jaipur, Bhopal, Lucknow, Nagpur, Udaipur, Bhubaneswar, Patna, Jodhpur, Ajmer, Indore, Sikar: (STD Code) 3192222Delhi, Kolkata, Mumbai, Ahmedabad: (STD Code) 31922222 | Visakhapatnam: 0891-2757575

sms RESO at 56677 | [email protected] | www.resonance.ac.in | Toll Free : 1800 200 2244 | WAPsite: m.resonance.ac.in

ResonanceEdu.blogspot.comfacebook.com/ResonanceEdu twitter.com/ResonanceEdu linkedin.com/in/ResonanceEduwww.youtube.com/resowatch

IIT-JEE AIEEEAIPMT NTSEKVPYOLYMPIADS SAT

mailto:[email protected],

http://www.youtube.com/resowatch

ResoNET 2012 Sample Test Paper

Documents

Transcript of ResoNET 2012 Sample Test Paper