SAMPLE TEST PAPERS (Resonance National Entrance Test: ResoNET - 2016) For Yearlong Classroom Contact Programmes (YCCPs) : JEE(Main+Advanced) Academic Session : 2016-17 Index STP1617 Sample Test Paper (STP) For ResoNET-2016 S.No. Contents Target Page No. 1 How to Prepare for the Resonance National Entrance Test (ResoNET)-2016 ResoNET 2016 2 2 General Instructions for the Examination Hall ResoNET 2016 3 3 Syllabus for ResoNET-2016 ResoNET 2016 4 4 Sample Test Paper- I : For Class-X Appearing/Passed students (Moving from Class-X to Class-XI ) For the students applying for VIKAAS (JA) & VIPUL (JB) Courses JEE(Main + Advanced) 2018 9 5 Sample Test Paper-I Answer key & Hints & Solution : For Class-X Appearing/Passed students (Moving from Class-X to Class-XI ) For the students applying for VIKAAS (JA) & VIPUL (JB) Courses JEE(Main + Advanced) 2018 18 6 Sample Test Paper-II : For Class-XI Appearing / Passed students (Moving from Class-XI to ClassXII).For the students applying for VISHWAAS (JF) Course JEE(Main + Advanced) 2017 25 JEE(Main + Advanced) 2017 42 JEE(Main + Advanced) 2017 52 Sample Test Paper-III Answer key & Hints & Solution : For Class-XII Appearing / Passed students JEE(Main + Advanced) 2017 (Moving from Class-XII to Class-XIII) For the students applying for VISHESH (JD) & VIJAY (JR) Courses 68 Sample ORS Answer Sheet for Resonance National Entrance Test (ResoNET)-2016 79 7 8 9 10 Sample Test Paper-II Answer key & Hints & Solution : For Class-XI Appearing / Passed students (Moving from Class-XI to Class-XII).For the students applying for VISHWAAS (JF) Course Sample Test Paper-III : For Class-XII Appearing / Passed students (Moving from Class-XII to ClassXIII) For the students applying for VISHESH (JD) & VIJAY (JR) Courses ResoNET 2016 The sample test papers 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. For More Practice of RESONANCE NATIONAL ENTRANCE TEST (ResoNET) - 2016 Resonance selection test papers 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. S a m ple Pa pe r Code De ta ils of P TP s Course (Code ) Ta rge t Re m a rk 1 S e t-A 10 Papers S et for Class-X A ppearing/P assed s tudents VIK A AS (JA ) & V IPUL (JB ) Answer key, JEE (Main + A dvanced) 2018 Hints & S olutions 2 S e t-B 10 Papers S et for Class-XI A ppearing/P assed students VIS HW A AS (JF) JEE (Main + A dvanced) 2017 Only Answer key 3 S e t-C 10 Papers S et for Class-XII A ppearing P assed students VIS HE S H (JD) & VIJAY (JR) JEE (Main + A dvanced) 2017 Answer key, Hints & S olutions Interested students may collect the same 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 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. Resonance Selection Test Papers of last few years with Answer key, Hints & Solutions are available on demand. ResoNET Papers are available Online too. ResoNET Online Practice Test Papers (OPTPs) : S. No. Details of OPTPs 1 3 Tests for Class-X Appearing/Passed students VIKAAS (JA) & VIPUL (JB) JEE(Main + Advanced) 2018 Course Code Target Fee(Taxes included) Rs. 300/- 2 6 Tests for Class-X Appearing/Passed students VIKAAS (JA) & VIPUL (JB) JEE(Main + Advanced) 2018 Rs. 500/- 3 3 Tests for Class-XI Appearing/Passed students VISHWAAS (JF) JEE(Main + Advanced) 2017 Rs. 300/- 4 6 Tests for Class-XI Appearing/Passed students VISHWAAS (JF) JEE(Main + Advanced) 2017 Rs. 500/- 5 3 Tests for Class-XII Appearing/Passed students VISHESH (JD) & VIJAY (JR) JEE(Main + Advanced) 2017 Rs. 300/- 6 6 Tests for Class-XII Appearing/Passed students VISHESH (JD) & VIJAY (JR) JEE(Main + Advanced) 2017 Rs. 500/- Students can buy these Online Test papers at http://elpd.resonance.ac.in © Copyright reserved 2016-17. All rights reserved. Any photocopying, publishing or reproduction of full or any part of this material is strictly prohibited. This material belongs to only the applicants of RESONANCE for its various Selection Tests (ResoNET) to be conducted for admission in Academic Session 2016-17. Any sale/resale of this material is punishable under law. Subject to Kota Jurisdiction only. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 PAGE # 1 STP1617 HOW TO PREPARE FOR THE RESONANCE NATIONAL ENTRANCE TEST (ResoNET) - 2016 For Class-X appearing students (Class-X to Class-XI Moving) : Study thoroughly the books of Science (Physics & Chemistry) and Maths of Classes IX & X. (NCERT & Respective Board) For Class-XI appearing students (Class-XI to Class-XII Moving): 1. Study thoroughly the books of Physics, Chemistry and Maths of Class XI (Respective Board). 2. Refer to the following books (only Class-XI syllabus) to increase the level of competence: For Physics : Concepts of Physics by H.C. Verma Vol. I & II, NCERT Books For Chemistry : NCERT Books(XI & XII), A text book of Physical Chemistry (8th Edition), Shishir Mittal, Disha Publications, Concise Inorganic Chemistry, J.D. Lee, Wiley-India Edition, Vogel’s Qualitative Analysis for the JEE (7th Edition), G. Svehla & Shishir Mittal, Pearson Education,Organic Chemistry : Clayden, Greeves, Warren and Wothers, Oxford University, A guide book to Mechanism In Organic Chemistry (6th Edition), Peter Sykes, Pearson Education For Maths : Higher Algebra By Hall & Knight; Co-ordinate Geometry By S.L. Loney ; Plane Trigonometry By S.L. Loney, Problem book in high school by A.I.Prilepko For Class-XII appearing students (Class-XII to Class-XIII Moving ): 1. Study thoroughly the books of Physics, Chemistry and Maths of Classes XI & XII (Respective Board). 2. Refer to the following books (Class-XI & Class-XII syllabus) to increase the level of competence : For Physics : Concepts of Physics by H.C. Verma Vol-I & II For Chemistry : Physical Chemistry By R.K. Gupta, Organic Chemistry By Morrison & Boyd, Organic Chemistry By I. L. Finar, Inorganic Chemistry By J.D. Lee, Objective Chemistry By Dr. P. Bahadur For Maths : Higher Algebra By Hall & Knight; Co-ordinate Geometry By S.L. Loney; Plane Trigonometry By S.L. Loney, Differential Calculus By G.N. Berman; Integral Calculus By Shanti Narayan; Vector Algebra By Shanti Narayan ; A Das Gupta (subjective). Note : For further practice, a set of several Practice Test Papers (PTPs) of Resonance may be procured from the institute. For this, the details are available on Page No.1. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 PAGE # 2 STP1617 GENERAL INSTRUCTIONS IN THE EXAMINATION HALL (ijh{kk Hkou ds fy, lkekU; funsZ 'k ) 1. This booklet is your Question Paper. ¼;g iq fLrdk vkidk iz ' u&i=k gS ½ 2. The Question Paper Code is printed on the top right corner of this sheet. ¼iz ' u&i=k dksM bl i`"B ds Åij nk;sa dksus esa Nik gqvk gS½ 3. Blank papers, clip boards, log tables, slide rule, calculators, mobile or any other electronic gadgets in any form are not allowed to be used. ¼[kkyh dkxt] fDyi cks M Z ] y?kq x.kd lkj.kh] LykbM :y] dSYdqysVj] eksc kby ;k vU; fdlh bySDVªWkfud midj.k ds fdlh Hkh :i esa mi;ksx dh vkKk ugha gS½ 4. Write your Name & Application Form Number in the space provided in the bottom of this booklet. (bl i` "B ds uhps fn;s x;s fjDr LFkku es a viuk uke o vkos n u QkW e Z la [ ;k vo'; Hkjs a ½ 5. Before answering the paper, fill up the required details in the blank space provided in the Objective Response Sheet (ORS). ( iz ' u&i=k gy djus ls igys ] ORS&'khV es a fn;s x;s fjDr LFkkuks a es a iw Ns 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&iq fLrdk esa fn;s x;s fjDr LFkku esa vius iz'u&i=k dk dksM o viuk vkos n u QkW e Z la [ ;k Li"V :i ls Hkjuk uk Hkwysa½ 7. No rough sheets will be provided by the invigilators. All the rough work is to be done in the blank space provided in the question paper. ¼fujh{kd ds }kjk dksbZ jQ 'khV ugha nh tk;sxhA jQ dk;Z iz'u&i=k esa fn;s x;s [kkyh LFkku esa gh djuk gS½ 8. No query related to question paper of any type is to be put to the invigilator. ¼fujh{kd ls iz'u&i=k ls lEcfU/kr fdlh izdkj dk dksbZ iz'u uk djsas½ 9. QUESTION PAPER ¼iz'u i=k½ Marks distribution of questions is as follows. ¼iz ' uks a ds iz kIrka dks dk fooj.k fuEu iz dkj ls gS A½ Part - I Part - II Part - III (Mathematics) (Physics) (Chemistry) 1 to 20 35 to 42 47 to 54 21 to 24 43 to 46 55 to 58 25 to 32 33 to 34 Marks to be awarded Type Correct Wrong Blank 3 -1 0 4 0 0 4 0 0 6 [1, 2, 3, 6] 0 0 Only one correct (ds oy , d fodYi l gh) One or more than one correct Answer (, d ; k , d l sv f/kd fodYi l gh) Comprehensions (v uq PNsn) Matrix Match Type (eS fVªDl Name : _________________________________ l q esy i zd kj ) Application Form Number : _______________ Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 PAGE # 3 STP1617 Syllabus of ResoNET-2016 CLASS - X (CHEMISTRY) Basic : Cooling by evaporation. Absorption of heat. All things accupy space, possess mass. Definition of matter ; Elementary idea about bonding. Solid, liquid and gas : characteristics-shape, volume, density; change of state - melting, freezing, evaporation, condensation, sublimation. Elements, compounds and mixtures :Heterogeneous and homogeneous mixtures; Colloids and suspension. Mole concept : Equivalence - that x grams of A is chemically not equal to x grams of B ; Partical nature, basic units : atoms and molecules ; Law of constant proportions ; Atomic and molecular masses;Relationship of mole to mass of the particles and numbers ; Valency ; Chemical formulae of common compounds. Atomic structure : Atoms are made up of smaller particles : electrons, protons, and neutrons. These smaller particles are present in all the atoms but their numbers vary in different atoms. Isotopes and isobars. Gradations in properties : Mendeleev periodic table. Acids, bases and salts : General properties, examples and uses. Types of chemical reactions : Combination, decomposition, displacement, double displacement, precipitation, neutralisation, oxidation and reduction in terms of gain and loss of oxygen and hydrogen. Extractive metallurgy : Properties of common metals ; Brief discussion of basic metallurgical processes. Compounds of Carbon : Carbon compounds ; Elementary idea about bonding ; Saturated hydrocarbons, alcohols, carboxylic acids (no preparation, only properties).Soap - cleansing action of soap. CLASS - X (MATHEMATICS) Number Systems : Natural Numbers, Integers, Rational number on the number line. Even - odd integers, prime number, composite numbers, twin primes, divisibility tests, Co-prime numbers, LCM and HCF of numbers. Representation of terminating/non-terminating recurring decimals, on the number line through successive magnification. Rational numbers as 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 and cubic polynomials Standard form of a quadratic equation ax2 + bx + c = 0, (a 0). Relation between roots and coefficient of quadratic and relation between discriminant and nature of roots. Linear Equation : Linear equation in one variable and two variable and their graphs. Pair of linear equations in two variables and their solution and inconsistency Arithmetic Progressions (AP) : Finding the nth term and sum of first n terms. Trigonometry : Trigonometric ratios of an acute angle of a right-angled triangle, Relationships between the ratios. Trigonometric ratios of complementary angles and trigonometric identities. 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 section formula (internal). Area of triangle. Properties of triangle and quadrilateral. (Square, Rectangle rhombus, parallelogram). Geometry : Lines : Properties of parallel and perpendicular lines. Triangle : Area of a triangle, Properties of triangle, similarity and congruency 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/cones and their combinations. Statistics : Mean, median, mode of ungrouped and grouped data. Probability : Classical definition of probability, problems on single events. Logarithm & exponents : Logarithms and exponents and their properties. Interest : Problem based on simple interest, compound interest and discounts. Mental Ability : Problem based on data interpretation, family relations, 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, accelaration and relation ship between these ; Distance-time and velcocity - time graphs. Newton’s Law of motion ; Relationship between mass, momentum, force and accelaration ; work done by a force ; Law of conservation of energy. Law of gravitation ; acceleration due to gravity. Electricity and magnetism : Ohm’s law ; Series and parallel combination of resistances ; Heating effect of current. Magnetic field near a current carrying straight wire, along the axis of a circular coil and inside a solenoid ; Force on current carrying conductor ; Fleming’s left hand rule ; W orking of electric motor ; Induced potential difference and current Electric generator : Principle and working ; Comparision of AC and DC ; Domestic electric circuits. Optics : Rectilinear propagation of light ; Basic idea of concave mirror 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. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 PAGE # 4 STP1617 Atomic and molecular masses. Mole concept and molar mass ; percentage composition and empirical and molecular formula ; chemical reactions, 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 and light, de Broglie’s relationship, Heisenberg uncertainty principle, concept of orbitals, quantum numbers, shapes of s, p, and d orbitals, rules for filling electrons in orbitals - Aufbau principle, Pauli exclusion principle 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 periodic table, trends in properties of elements - atomic radii, ionic radii, inert gas radii, ionization enthalpy, electron gain enthalpy, electronegativity, valence. Chemical Bonding and Molecular Structure : Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character of covalent bond, covalent character of ionic bond, valence bond theory, resonance, geometry of covalent molecules, VSEPR theory, concept of hybridization involving s, p and d orbitals and shapes of some simple molecules, molecular orbital theory of homonuclear diatomic molecules (qualitative idea 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’s law, Charles’ law, Gay Lussac’s law, Avogadro’s law, ideal behavior, empirical derivation of gas equation, Avogadro’s number ideal gas equation, deviation from ideal behaviour, Liquefaction of gases, critical temperature. Liquid State - Vapour pressure, viscosity and surface tension (qualitative idea only, no mathematical derivations) Thermodynamics : Concepts of system, types of systems, surroundings, work, heat, energy, extensive and intensive properties, state functions. First law of thermodynamics - internal energy and enthalpy, heat capacity and specific heat, measurement of U and H, Hess’s law of cons tant heat summation, enthalpy of bond dissoc iation, combustion, formation, atomization sublimation, phase transition, ionization, and dilution. Introduction of entropy as a state function, free energy change for spontaneous and non-spontaneous process, equilibrium. Equilibrium : Equilibrium in physical and chemical processes, dynamic nature of equilibrium, law of mass action, equilibrium constant, factors affecting equilibrium - Le Chatelier’s principle ; ionic equilibrium - ionization of acids and bases, strong and weak electrolytes, degree of ionization concept of pH. Hydrolysis of Salts (elementary idea), buffer solutions, solubility product, common ion effect (with illustrative examples). Redox Reactions : Concept of oxidation and reduction, redox reactions, oxidation number, balancing redox reactions, applications of redox reaction. Hydrogen : Position of hydrogen in periodic table, occurrence, isotopes, preparation, properties and uses of hydrogen ; hydrides ionic, covalent and interstitial ; physical and chemical properties of water, heavy water ; hydrogen peroxide - preparation, reactions and structure ; hydrogen as a fuel. s-Block Elements (Alkali and Alkaline Earth Metals) : Group 1 and Group 2 elements : General introduction, electronic configuration, occurrence, anomalous properties of the first element of each group, diagonal relationship, trends in the variation of properties (such as ionization enthalpy, atomic and ionic radii), trends in chemical reactivity with oxygen, water, hydrogen and halogens ; uses. Preparation and properties of some important compounds Sodium carbonate, sodium chloride, sodium hydroxide and sodium hydrogen carbonate CaO, CaCO3, and industrial use of lime and limestone, Ca. General Introduction to p-Block Elements : Group 13 elements : General introduction, electronic configuration, occurrence, variation of properties, oxidation states, trends in chemical reactivity, anomalous properties of first element of the group ; Boron - physical and chemical properties, some important compounds ; borax, boric acids, boron hydrides. Aluminium : uses, reactions with acids and alkalies. Group 14 elements ; General introduction, electronic configuration, occurrence, variation of properties, oxidation states, trends in chemical reactivity, anomalous behaviour of first element. Carbon catenation, allotropic forms, physical and chemical propeties ; uses of some important compounds : oxides. Important compounds of silicon and a few uses : silicon tetrachloride, silicones, silicates and zeolites. Principles of qualitative analysis : Determinantion of one anion and one cation in a given salt Cations - Pb2 + , Cu2+, As3+, Al3+, Fe3+, Mn2+, Ni2 +, Zn2+, Co2+, Ca2+, Sr2+, Ba2+, Mg2+, NH4 Anions - CO23 – , S2 – , SO32– , SO 24– ,NO2– , NO3– , NO3– , Cl – , Br – , – , PO 34 – , C 2O 24 – CH3 COO – (Note : Insoluble salts excluded) Organic chemistry - Some Basic Principles and Techniques General introduction, methods of purification, qualitative and quantitative analysis, classification and IUPAC nomenclature of organic compounds. Electronic dis plac ements in a covalent bond : free radicals, carbocations, carbanions ; electrophiles and nucleophiles, types of organic reactions Classification of Hydrocarbons : Alkanes : Nomenclature, is omerism, conformations (ethane only), physical propeties, chemical reactions including free radic al mechanism of halogenation, combustion and pyrolysis. Alkenes : Nomenclatures, structure of double bond (ethene), geometrical isomerism, physical properties, methods of preparation ; chemical reactions : addition of hydrogen, halogen, water, hydrogen halides (Markovnikov’s addition and peroxide ef fect), ozonolysis, oxidation, mechanism of electrophilic addition. Alkynes : Nomenclature, structure of triple bond (ethyne), physical properties, methods of preparation, chemical reactions : acidic character of alkynes, addition reaction of - hydrogen, halogens, hydrogen halides and water. Aromatic hydrocarbons : Introduction, IUPAC nomenclature ; Benzene : resonance, aromaticity ; chemical properties : mechanism of electrophilic substitution - nitration sulphonation, halogenation, Friedel Craft’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 infinite sets, Subsets, Union and intersection of sets, Venn diagrams. Pictorial representation of a function domain, co-domain and range of a 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 in degrees and conversion from one measure to another. Signs of trigonometric functions and sketch of their graphs. Addition and subtraction formulae, formulae involving multiple and s ub-multiple angles. General solution of trigonometric equations. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 PAGE # 5 STP1617 Complex Number Algebra of complex numbers, addition, multiplication, conjugation, polar representation, properties of modulus and principal argument, triangle inequality, cube roots of unity, geometric interpretations. Quadratic equations : Quadratic equations with real coefficients, formation of quadratic equations with given roots, symmetric functions of roots. Sequence & Series : Arithmetic, geometric and harmonic progressions, arithmetic, geometric and harmonic means, sums of finite arithmetic and geometric progressions, infinite geometric series, sums of squares and cubes of the first n natural numbers. Logarithm & exponents : Logarithms and exponents and their properties. Exponential and logarithmic series. Binomial Theorem : Binomial theorem for a positive integral index, properties of binomial coefficients. Binomial theorem for any index. Permutations and combinations : Problem based on fundamental counting principle, Arrangement of alike and different objects, Circular permutation, Combination, formation of groups. Straight Line : Cartesian coordinates, distance between two points, section formulae, shift of origin. Equation of a straight line in various forms, angle between two lines, distance of a point from a line; Lines through the point of intersection of two given lines equation of the bisector of the angle between two lines, c oncurrency of lines; Centroid, orthocentre, incentre and circumcentre of a triangle. Conic Sections : Equation of a circle in various forms, equations of tangent, normal and chord. Parametric equations of a circle, intersection of a circle with a straight line or a circle, equation of a through the points of intersection of two circles and those of a circle and a straight line. Equations of a parabola, ellipse and hyperbola in standard form, their foci, directrices and eccentricity, parametric equations, equations of tangent 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 for physic al quantities pertaining to the following experiments: Experiments based on using Vernier calipers and screw gauge (micrometer), Determination of g using simple pendulum, Young’s modulus by Searle’s method. Mechanics : Kinematics in one and two dimensions (Cartesian coordinates only), projectiles; Uniform Circular motion; Relative velocity. 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 mechanical energy. Systems of particles; Centre of mass and its motion; Impulse; Elastic and inelastic collisions. Law of gravitation; Gravitational potential and field; Acceleration due to gravity; Motion of planets and satellites in circular orbits; Escape velocity. Rigid body, moment of inertia, parallel and perpendicular axes theorems, moment of inertia of unif orm bodies with simple geometrical shapes; Angular momentum; Torque; Conservation of angular momentum; Dynamics of rigid bodies with fixed axis of rotation; Rolling without slipping of rings, cylinders and spheres; Equilibrium of rigid bodies; Collision of point masses with rigid bodies. Linear and angular simple harmonic motions. Hooke’s law, Young’s modulus. Pressure in a fluid; Pascal’s law; Buoyancy; Surface energy and surface 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 : W ave motion (plane waves only), longitudinal and transverse waves, s uperposition of waves; Progressive and stationary 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; Elementary concepts of convection and radiation; Newton’s law of cooling; Ideal gas laws; Specific heats (Cv and Cp for monoatomic and diatomic gases); Isothermal and adiabatic processes, bulk modulus of gases; Equivalence of heat and work; First law of thermodynamics and its applications (only for ideal gases); Blackbody radiation: absorptive and 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 in terms of mole fraction, molarity, molality and normality. Gaseous and liquid states : Absolute scale of temperature, ideal gas equation; Deviation from ideality, van der Waals equation; Kinetic theory of gases, average, root mean square and most probable velocities and their relation with temperature; Law of partial pressures; Vapour pressure; Diffusion of gases. Atomic structure and chemical bonding : Bohr model, spectrum of hydrogen atom, quantum numbers; Wave-particle duality, de Broglie hypothesis; Uncertainty principle; Qualitative quantum mechanical picture of hydrogen atom, shapes of s, p and d orbitals; Electronic configurations 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 energy diagrams for homonuclear diatomic species; Hydrogen bond; Polarity in molecules, dipole moment (qualitative aspects only); VSEPR model and shapes of molecules (linear, angular, triangular, square planar, pyramidal, square pyramidal, trigonal bipyramidal, tetrahedral and octahedral). Energetics : First law of thermodynamics; Internal energy, work and heat, 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 of G and Go in chemical equilibrium; Solubility product, common ion effect, pH and buffer solutions; Acids and bases (Bronsted and Lewis concepts); Hydrolysis of salts. Electrochemistry : Electrochemical cells and cell reactions; Standard electrode potentials; Nernst equation and its relation to DG; Electrochemical series, emf of galvanic cells; Faraday’s laws of electrolysis; Electrolytic conductance, specific, equivalent and molar conductivity, Kohlrausch’s law; Concentration cells. Chemical kinetics : Rates of chemical reactions; Order of reactions; Rate constant; First order reactions; Temperature dependence of rate constant (Arrhenius equation). Solid state : Classification of solids, crystalline state, seven crystal systems (cell parameters a, b, c, ), close packed structure of solids (cubic), packing in fcc, bcc and hcp lattices; Nearest neighbours, ionic radii, simple ionic compounds, point defects. Solutions : Raoult’s law; Molecular weight determination from lowering of vapour pressure, elevation of boiling point and depression of freezing point. Surface chemistry : Elementary concepts of adsorption (excluding adsorption isotherms); Colloids: types, methods of preparation and general properties; Elementary ideas of emulsions, surfactants and micelles (only definitions and examples). Nuclear chemistry : Radioactivity: isotopes and isobars; Properties of rays; Kinetics of radioactive decay (decay series excluded), carbon dating; Stability of nuclei with respect to proton-neutron ratio; Brief discussion on fission and fusion reactions. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 PAGE # 6 STP1617 Inorganic Chemistry Isolation/preparation and properties of the following nonmetals : Boron, silicon, nitrogen, phosphorus, oxygen, sulphur and halogens; Properties of allotropes of carbon (only diamond and graphite), phosphorus and sulphur. Preparation and properties of the following compounds : Oxides, peroxides, hydroxides, carbonates, bicarbonates, chlorides and sulphates of sodium, potassium, magnesium and calcium; Boron: diborane, boric acid and borax; Aluminium: alumina, aluminium chloride and alums; Carbon: oxides and oxyacid (carbonic acid); Silicon: silicones, silicates and silicon carbide; Nitrogen: oxides, oxyacids and ammonia; Phosphorus: oxides, oxyacids (phosphorus acid, phosphoric acid) and phosphine; Oxygen: ozone and hydrogen peroxide; Sulphur: hydrogen sulphide, oxides, sulphurous acid, sulphuric acid and sodium thiosulphate; Halogens: hydrohalic acids, oxides and oxyacids of chlorine, bleaching powder; Xenon fluorides. Transition elements (3d series) : Definition, general characteristics, oxidation states and their stabilities, colour (excluding the details of electronic transitions) and calculation of spin (only magnetic moment), Coordination compounds: nomenclature of mononuclear coordination compounds, cis-trans and ionisation is omerisms , hybridization and geometries 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 sulphates of Fe2+ , Cu 2+ and Zn 2+ ; 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. Phenols : Acidity, electrophilic substitution reactions (halogenation, nitration and sulphonation); Reimer-Tieman reaction, Kolbe reaction. Characteristic reactions of the following (including those mentioned above): Alkyl halides: rearrangement reactions of alkyl carbocation, Grignard reactions, nuc leophilic substitution reactions; Alcohols: esterification, dehydration and oxidation, reaction with sodium, phosphorus halides, ZnCl2/concentrated HCl, conversion of alcohols into aldehydes and ketones; Ethers:Preparation by W illiamson’s Synthesis; Aldehydes and Ketones: oxidation, reduction, oxime and hydrazone formation; aldol condensation, Perkin reaction; Cannizzaro reaction; haloform reaction and nucleophilic addition reactions (Grignard addition); Carboxylic acids: formation of esters, acid chlorides and amides, ester hydrolysis; Amines: basicity of substituted anilines and aliphatic amines, preparation from nitro compounds, reaction with nitrous acid, azo coupling reaction of diazonium salts of aromatic amines, Sandmeyer and related reactions of diazonium salts; carbylamine reaction; Haloarenes: nucleophilic aromatic substitution in haloarenes and substituted haloarenes (excluding Benzyne mechanism and Cine substitution). Carbohydrates: Classification; mono- and di-saccharides (glucose and sucros e); Oxidation, reduc tion, glycoside formation and hydrolysis of sucrose. Amino acids and peptides : General structure (only primary structure for peptides) and physical properties. Properties and uses of some important polymers : Natural rubber, 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-functional organic compounds from binary mixtures. CLASS - XII (MATHEMATICS) Complex Number and Quadratic equations : Organic Chemistry Concepts : Hybridisation of carbon; Sigma and pi-bonds; Shapes of simple organic molecules; Structural and geometrical isomerism; Optical isomerism of compounds containing up to two asymmetric centres, (R,S and E,Z nomenclature excluded); IUPAC nomenclature of simple organic compounds (only hydrocarbons, mono-functional and bi-functional compounds); Conformations of ethane and butane (Newman projections); Resonance and hyperconjugation; Keto-enol tautomerism; Determination of empirical and molecular formulae of simple compounds (only combustion method); Hydrogen bonds: definition and their effects on physical properties of alcohols and carboxylic acids; Inductive and resonance effects on acidity and basicity of organic acids and bases; Polarity and inductive effects in alkyl halides; Reactive intermediates produced during homolytic and heterolytic bond cleavage; Formation, structure and stability of carbocations, carbanions and free radicals. Preparation, properties and reactions of alkanes : Homologous series, physical properties of alkanes (melting points, boiling points and density); Combustion and halogenation of alkanes; Preparation of alkanes by Wurtz reaction and decarboxylation reactions. Preparation, properties and reactions of alkenes and alkynes : Physical properties of alkenes and alkynes (boiling points, density and dipole moments); Acidity of alkynes; Acid catalysed hydration of alkenes and alkynes (excluding the stereochemistry of addition and elimination); Reactions of alkenes with KMnO4 and ozone; Reduction of alkenes and 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; Electrophilic substitution reactions: halogenation, nitration, sulphonation, FriedelCrafts alkylation and acylation; Effect of ortho, meta and para directing groups in monosubstituted benzenes. Algebra of complex numbers, addition, multiplication, conjugation, polar representation, properties of modulus and principal argument, triangle inequality, cube roots of unity, geometric interpretations. Quadratic equations with real coefficients, formation of quadratic equations with given roots, symmetric functions of roots. Sequence & Series : Arithmetic, geometric and harmonic progressions, arithmetic, geometric and harmonic means, sums of finite arithmetic and geometric progressions, infinite geometric series, sums of squares and cubes of the first n natural numbers. Logarithms and their properties. Permutations and combinations, Binomial theorem for a positive integral index, properties of binomial coefficients. Binomial theorem for any index, exponential and logarithmic series. Matrices & Determinants : Matrices as a rectangular array of real numbers, equality of matrices, addition, multiplication by a scalar and product of matrices, transpose of a matrix, determinant of a square matrix of order up to three, inverse of a square matrix of order up to three, properties of these matrix operations, diagonal, symmetric and skew- symmetric matrices and their properties, solutions of simultaneous linear equation in two or three variables. Probability : Addition and multiplication rules of probability, conditional probability, baye’s theorem, independence of events, computation of probability of events using permutations and combinations. Straight Line : Cartesian coordinates, distanc e between two points, section formulae, shift of origin. Equation of a straight line in various forms, angle between two lines, distance of a point from a line; Lines through the point of intersection of two given lines equation of the bisector of the angle between two lines, concurrency of lines; Centroid, orthocentre, incentre and circumcentre of a triangle. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 PAGE # 7 STP1617 Conic Section : Equation of a circle in various forms, equations of tangent, normal and chord. Parametric equations of a circle, intersection of a circle with a straight line or a circle, equation of a through the points of intersection of two circles and those of a circle and a straight line. Equations of a parabola, ellipse and hyperbola in standard form, their foci, directrices and eccentricity, parametric equations, equations of tangent and normal locus problems. Three dimensions : Direction cosines and direction ratios, equation of a straight line in space, equation of a plane, distance of a point from a plane Vectors : Addition of vectors, scalar multiplication, dot and cross products, scalar triple products and their geometrical interpretations. Position vector of a point dividing a line segment in a given ratio. Projection of 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 f unctions , absolute value, polynomial, rational, trigonometric, exponential and logarithmic functions. Even and odd functions, inverse of a function, composite function. Limit, Continuity & Derivability : Limit and continuity of a function, limit and continuity of the sum, difference, product and quotient of two functions, L’Hospital rule 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, derivatives of polynomial, rational, trigonometric, inverse trigonometric, exponential and logarithmic functions. Derivatives of implicit functions, derivatives up to order two. Tangent & Normal : Geometrical interpretation of the derivative, tangents and normal. Maxima & Minima : Increasing and decreasing functions, maximum and 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, integration by the methods of substitution and partial fractions. Definite integrals and their properties, fundamental theorem of integral calculus. Application of definite integrals to the determination of areas involving simple curves. Formation of ordinary differential equations, solution of homogeneous differential equations, separation of variables method, linear first order differential equations. Trigonometry : Trigonometric functions, their periodicity and graphs addition 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, cosine rule, half-angle formula and the area of a triangle, inverse trigonometric functions (principal value only). CLASS - XII (PHYSICS) General : Units and dimensions, dimensional analysis; least count, significant figures; Methods of measurement and error analysis for physic al quantities pertaining to the following experiments: Experiments based on using Vernier calipers and screw gauge (micrometer), 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 of sound using resonance column, Verification of Ohm’s law using voltmeter and ammeter, and specific resistance of the material of a wire using meter bridge and post office box. Mechanics : Kinematics in one and two dimensions (Cartesian coordinates only), Projectile Motion; Uniform Circular Motion; Relative Velocity. 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 mechanical energy. Systems of particles; Centre of mass and its motion; Impulse; Elastic and inelastic collisions. Law of gravitation; Gravitational potential and field; Acceleration due to gravity; Motion of planets and satellites in circular orbits; Escape velocity. Rigid body, moment of inertia, parallel and perpendicular axes theorems, moment of inertia of uniform bodies with simple geometrical shapes; Angular momentum; Torque; Conservation of angular momentum; Dynamics of rigid bodies with fixed axis of rotation; Rolling without slipping of rings, cylinders and spheres; Equilibrium of rigid bodies; Collision of point masses with rigid bodies. Linear and angular simple harmonic motions. Hooke’s law, Young’s modulus. Pressure in a fluid; Pascal’s law; Buoyancy; Surface energy and surface 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 transverse waves, superposition of waves; Progressive and stationary 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; Elementary concepts of convection and radiation; Newton’s law of cooling; Ideal gas laws; Specific heats (Cv and Cp for monoatomic and diatomic gases); Isothermal and adiabatic processes, bulk modulus of gases; Equivalence of heat and work; First law of thermodynamics and its applications (only for ideal gases); Blackbody radiation: absorptive and emissive powers; Kirchhoff’s law; Wien’s displacement law, Stefan’s law. Electricity and magnetism : Coulomb’s law; Electric field and potential; Electrical potential energy of a system of point charges and of electrical dipoles in a uniform electrostatic field; Electric field lines; Flux of electric field; Gauss’s law and its application in simple cases, such as, to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell. Capacitance; Parallel plate capacitor with and without dielectrics; Capacitors in series and parallel; Energy stored in a capacitor. Electric current; Ohm’s law; Series and parallel arrangements of resistances and cells; Kirchhoff’s laws and simple applications; Heating effect of current. Biot–Savart’s law and Ampere’s law; Magnetic field near a currentcarrying straight wire, along the axis of a circular coil and inside a long straight solenoid; Force on a moving charge and on a currentcarrying wire in a uniform magnetic field. Magnetic moment of a current loop; Effect of a uniform magnetic field on a current loop; Moving coil galvano- meter, voltmeter, ammeter and their 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 at plane and spherical surfaces; Total internal reflection; Deviation and dispersion of light by a prism; Thin lenses; Combinations of mirrors and thin lenses; Magnification. Wave nature of light: Huygen’s principle, interference limited to Young’s double-slit experiment. M odern physics : Atomic nucleus; Alpha, beta and gamma radiations; Law of radioactive decay; Decay constant; Half-life and mean life; Binding energy and its calculation; Fission and fusion processes; Energy calculation in these processes. Photoelectric effect; Bohr’s theory of hydrogen-like atoms; Characteristic and continuous X-rays, Moseley’s law; de Broglie wavelength of matter waves. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 PAGE # 8 STP1617 SAMPLE TEST PAPER -I (For Class-X Appearing / Passed Students) Course : VIKAAS (JA) & VIPUL (JB) S.No. Subject Sections Nature of Questions No. of questions Marks Negative marks Total Single choice questions (SCQ) 1 to 24 Sections-I PART-I (MATHS) 25 to 28 29 to 36 (dsoy 24 3 -1 72 4 4 0 16 8 4 -1 32 8 3 -1 24 4 4 0 16 8 3 -1 24 4 4 0 16 Multiple choice questions (MCQ) Sections-II Sections-III 37 to 44 ,d fodYi lgh) (dsoy ,d ;k ,d ls vf/kd fodYi lgh) Comprehension (vuqPNsn) 2x4=8 (SCQ) Single choice questions (SCQ) Sections-I (dsoy PART-II (PHYSICS) 45 to 48 Sections-II 49 to 56 Sections-I ,d fodYi lgh) Multiple choice questions (MCQ) (dsoy ,d ;k ,d ls vf/kd fodYi lgh) Single choice questions (SCQ) PART-III (CHEMISTRY) 57 to 60 (dsoy ,d fodYi lgh) Multiple choice questions (MCQ) Sections-II (dsoy ,d ;k ,d ls vf/kd fodYi lgh) 60 Total 200 PART - I (Hkkx - I) SECTION - I ([k.M- I) Straight Objective Type (lh/ks oLrqfu"B izdkj ) This section contains 24 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. bl [k.M esa 24 iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA 1. The median AD of ABC meets BC at D. The internal bisectors of ADB and ADC meet AB and AC at E and F respectively. Then EF (A) is perpendicular to AD (B) is parallel to BC (C) divides AD in the ratio of AB : AC (D) none of these ABC dh ekf/;dk,a AD Hkqtk BC dks D ij feyrh gSA ADB vkSj ADC ds vkUrfjd dks.k v/kZd AB vkSj AC dks Øe'k% E vkSj F ij feyrs gS] rc EF (A) AD ds yEcor~ gS (B) BC ds lekUrj gS (C) AD dks vuqikr AB : AC esa foHkkftr djrk gS (D) buesa ls dksbZ ugha 2. If A is an acute angle such that sin A = 1 sin A cos A 3 , then the value of is 5 sin3 A cos 3 A ;fn U;wudks.k A bl çdkj gS] fd sin A = 3 5 (A) 7 5 (B) – 1 5 1 sin A cos A gS] rks sin3 A cos 3 A (C) 35 12 dk eku gS& (D) None of these (buesa ls dksbZ ugha) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 9 STP1617 3. If the diagonal and the area of a rectange are 25 m and 168 m2 respectively, what is the length of the rectangle ? fdlh vk;r ds fod.kZ vkSj {ks=kQy Øe'k% 25 m rFkk 168 m2 gS] rks vk;r dh yEckbZ gksxh ? (A) 17 m (B) 31 m (C) 12 m (D) 24 m 4. If x = 7 + 4 3 then the value of x ;fn x = 7 + 4 3 gks] rc (A) 8 5. x x 1 (B) 6 x (C) 5 (D) 4 3n 2 5n . Its 25th term is 2 2 fdlh lekUrj Js<h ds izFke n inksa dk ;ksx 6. is dk eku gS& In an A.P., the sum of first n term is (A) 73 1 (B) 76 3n 2 5n gS] rks bldk 25ok¡ in gksxk& 2 2 (C) 79 (D) 81 In Figure, D and E are the mid-points of sides AB and AC respectively of ABC, Find EDB. fn;s x;s fp=k esa ABC ds D rFkk E Øe'k% AB rFkk AC ds e/;fcUnq gS EDB Kkr djksA] A 60° D B (A) 110° (B) 120° 5cm E 50° C (C) 70° (D) 80° 7. Two circles with centres A and B intersect at P and Q. Which of the following is false? (A) AB is the perpendicular bisector of PQ (B) PQ is the perpendicular bisector of AB (C) APQ = AQP (D) PBA = QBA nks o`Ùk ftuds dsUnz A vkSj B gS] P vkSj Q ij izfrPNsn djrs gSA fuEu esa ls dkSulk xyr gS? (A) AB, PQ dk yEclef}Hkktd gS (B) PQ, AB dk yEclef}Hkktd gS (C) APQ = AQP (D) PBA = QBA 8. Water runs into a cylindrical tank , of diameter 4 m and height 5 m , through a pipe of radius 2 cm , at the rate of 1/10 m per second . Find the time taken by the tank to fill up . ,d csyukdkj VSad ftldk O;kl 4 m o Å¡pkbZ 5 m gS dks ,d 2 cm f=kT;k okyh uyh ls 1/10 m izfr lSd.M dh nj ls Hkjk tkrk gSA VSad dks Hkjus esa dqy le; yxsxk : (A) 150 hr (B) 150 hr 50 min 50 sec (C) 138 hr 53 min 20 sec (D) 100 hr 50 min 9. In the figure given below, TBP and TCQ are tangents to the circle whose centre is ‘O’, touching the circle at B and C respectively. Also, PBA = 60º and ACQ = 70° then BTC is fn;s x;s fp=k esa TBP vkSj TCQ o`Ùk ftldk dsUnz O gS] dh Li'kZ js[kk,a gS o`Ùk dks B vkSj C ij Øe'k% Li'kZ djrh gSA PBA = 60º vkSj ACQ = 70° rc BTC gksxk& (A) 50º 10. (B) 70º (C) 80º (D) 60º If a selling price of Rs. 24 results in a 20% discount of the list price, the selling price that would result in a 30% discount of the list price is (A) Rs. 9 (B) Rs. 27 (C) Rs. 14 (D) Rs. 21 ;fn 24 :i;s foØ; ewY; gksus ij cktkj ewY; ij 20% dh NqV feyrh gS] rc foØ; ewY; D;k gksxk tc cktkj ewY; ij 30% dh NqV feyrh gS& (A) 9 : (B) 27 : (C) 14 : (D) 21 : Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 10 STP1617 11. 12. The value of (a1/8 + a–1/8) (a1/8 – a–1/8) (a1/4 + a–1/4)(a1/2 + a–1/2) is (a1/8 + a–1/8) (a1/8 – a–1/8) (a1/4 + a–1/4)(a1/2 + a–1/2) dk eku gS& (A) (a + a–1) (B) (a – a–1) (C) (a2 – a–2) If a, b, c are real and distinct numbers, then the value of (A) 1 (B) a b c (a b)3 (b c )3 (c a )3 is (a b) . (b c ) . (c a) (C) 2 3 ;fn a, b, c okLrfod vkSj fofHkUu la[;k,a gS] rks (A) 1 (D) (a1/2 – a–1/2) (D) none of these 3 3 (a b) (b c ) (c a ) (a b) . (b c ) . (c a) (B) a b c (C) 2 dk eku gS& (D) buesa ls dksbZ ugha 13. In figure, CDEF is a cyclic quadrilateral, DE and CF are produced to A and B respectively such that AB || CD. If FED = 80°, find FBA. fn;s x;s fp=k esa CDEF ,d pØh; prqHkZqt gS ftlesa DE vkSj CF dks Øe'k% A vkSj B rd bl izdkj c<+k;k tkrk gS tcfd AB || CD ;fn FED = 80° rc FBA Kkr dhft, (A) 30° (B) 60° (C) 80° (D) None of these (buesa ls dksbZ ugha) 14. A father’s age is equal to the sum of the ages of his 3 children. In 9 years his age will be equal to the 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 two youngest children. find the father’s present age ? (A) 32 years (B) 36 years (C) 40 years (D) 44 years firk dh vk;q mlds rhu iq=kksa dh vk;q ds ;ksx ds cjkcj gSA 9 o"kZ ckn mldh vk;q] nks cM+s iq=kkaas dh vk;q ds ;ksx ds cjkcj gksxh rFkk mlds 3 o"kZ ckn mldh vk;q lcls cM+s vkSj lcls NksVs iq=k dh vk;q ds ;ksxQy ds cjkcj gksxhA iqu% 3 o"kZ ckn mldh vk;q nks NksVs iq=kksa dh vk;q ds ;ksx ds cjkcj gksxhA firk dh orZeku vk;q crkb;saA (A) 32 o"kZ (B) 36 o"kZ (C) 40 o"kZ (D) 44 o"kZ 15. If cos – sin 1 – 3 , then acute angle = cos sin 1 3 cos – sin 1– 3 ;fn cos sin , rks U;wudks.k = 1 3 (A) 60º 16. 17. (B) 30º In the given figure, the value of ‘a’ is : fn;s x;s fp=k esa] ‘a’ dk eku crkvks : (A) 170° (B) 20° (C) 10° (D) 5° (C) 45º (D) 90º P 85° O A a b B ABCD is a parallelogram, P is a point on AB such that AP : PB = 3 : 2. Q is a point on CD such that CQ : QD = 7 : 3. If PQ meets AC at R, then AR : AC is ABCD ,d lekUrj prqHkqZt gS] AB ij fcUnq P bl izdkj gS fd AP : PB = 3 : 2 gSA CD ij fcUnq Q bl izdkj gS fd CQ : QD = 7 : 3. ;fn PQ , AC dks R ij feyrk gS] rc AR : AC gS& (A) 5 : 11 (B) 6 : 13 (C) 4 : 7 (D) 2 : 5 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 11 STP1617 18. p , is (where p and q are integers and q 0). q Recurring decimal 0.235 , when expressed in the form p vkorhZ n'keyo 0.235 dks q ds :i esa O;Dr djus ij cjkcj gS (tgk¡ p rFkk q iw.kkZad gSa rFkk q 0) (A) 235 1000 (B) 243 990 (C) 233 990 (D) 223 . 990 19. The sum of all natural numbers less than 400 which are not divisible by 6, is : 400 ls de lHkh izkÑr la[;kvksa dk ;ksx tks fd 6 ls foHkkftr ugh gS] gksxk : (A) 13266 (B) 66534 (C) 79800 (D) 93066 20. Ram sets his watch at 6 : 10 am on Sunday, which gains 12 minutes in a day. On Wednesday if this watch is showing 2 : 50 pm. What is the correct time ? jke viuh ?kMh jfookj dks 6 : 10 am ij feykrk gS] tks fd ,d fnu esa 12 feuV vkxs gks tkrh gSA cq/kokj dks ;fn ;g ?kMh 2 : 50 pm fn[kk jgh gS] rks lgh le; D;k gksxk ? (A) 1 : 50 pm (B) 2 : 10 pm (C) 2 : 30 pm (D) 3 : 30 pm 21. If , be the roots of x2 – a(x –1) + b = 0, then value of ;fn x2 – a(x –1) + b = 0 ds ewy , gks] rks (A) 1 22. (B) 0 1 2 1 1 – a 1 + + 2 2 – a – a (C) 2 + 2 – a 2 ab 2 is ab + dk eku gS & (D) 3 A square sheet of paper is converted into a cylinder by rolling it along its length. What is the ratio of the base radius to the side of the square ? ,d oxkZdkj dkxt dks mldh yEckbZ ds vuqfn'k ?kqekdj ,d flysUMj esa :ikUrfjr fd;k tk;sA rks mldh vk/kkj f=kT;k dk oxZ dh Hkqtk ls vuqikr D;k gksxk ? (A) 1 2 (B) 2 (C) 1 2 (D) 1 23. In which of the following case(s) , the equation , a x 2 + b x + c = 0 always has imaginary roots fuEu esa ls dkSuls fodYi lehdj.k a x 2 + b x + c = 0 ds lnSo dkYifud ewy gS& (A) a > 0 , b > 0 , c > 0 (B) a > 0 , b < 0 , c = 0 (C) a > 0 , b = 0 , c > 0 (D) a < 0 , b = 0 , c > 0 24. The value of x satisfying the equation (considering only positive root) 5 2 6 5 2 6 / 5 2 6 5 2 6 = x / 2 is x ds eku tcfd lehdj.k 5 2 6 5 2 6 / 5 2 6 5 2 6 = (dsoy /kukRed ewy ysrs gq,) (A) 6 (B) 3 (C) 3 x/2 dks larq"V djrs gS& (D) 9 / 2 SECTION - II ([k.M- II) Multiple Correct Answer Type (cgqy lgh fodYi izdkj) This section contains 4 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONE OR MORE is/are correct. bl [k.M esa 4 iz'u gSaA izR;sd iz'u ds mÙkj ds fy, 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls vf/kd lgh gSA 25. The expression (5a – 3b)3 + (3b – 7c)3 – (5a – 7c)3 is divisible by : O;atd (5a – 3b)3 + (3b – 7c)3 – (5a – 7c)3 foHkkftr gksxk (A) (5a + 3b + 7c) (B) (5a – 3b – 7c) (C) (3b – 7c) (D) (7c – 5a) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 12 STP1617 26. (2x2 + 3x + 5)1/2 + (2x2 + 3x + 20)1/2 = 15, therefore x is : (2x2 + 3x + 5)1/2 + (2x2 + 3x + 20)1/2 = 15, rc x gS : (A) (–8/3) (B) (14/5) (C) (–11/2) (D) 4 27. The difference between areas of outside and inside surfaces of a cylindrical metallic pipe 14 cm long is 44 cm2. If the pipe is made of 99 cubic centrimeters of metal and outer radii and inner radii are R and r respectively, then 14 cm yEcs ,d /kkrq ls cus csyukdkj ikbZi dh ckÐ; rFkk vkUrfjd lrgksa ds {ks=kQyksa ds e/; varj 44 cm2 gSA ;fn ikbZi 99 /ku lsVhehVj /kkrq ls cuk;k tkrk gS rFkk ikbZi dh ckÐ; f=kT;k o vkUrfjd f=kT;k Øe'k% R rFkk r gS] rks & (A) R = 2.5 (B) R = 2 (C) r = 1.5 (D) r = 2 28. For any real number a, b, c & d which of the following statement(s) is/are true ? (A) If a > b then ac > bc (B) If a > b then a – c > b – c a b (C) If a > b > 0 and c > d > 0, then > (D) If a > b and c > d then a – c > b – d. d c fdUgha okLrfod la[;kvksa a, b, c ,oa d ds fy, fuEufyf[kr esa ls dkSu ls dFku lR; gS \ (A) ;fn a > b gks] rks ac > bc (B) ;fn a > b gks] rks a – c > b – c (C) ;fn a > b > 0 ,oa c > d > 0 gks] rks a b > d c (D) ;fn a > b ,oa c > d gks] rks a – c > b – d SECTION - III ([k.M - III) Comprehension Type (cks/ ku çdkj) This section contains 2 paragraphs. Based upon each paragraph, there are 4 questions. Each 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 ç'u gSA çR;sd ç'u ds 4 fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA Paragraph for Question Nos. 29 to 32 In the following figure, the smaller triangle represents the teachers; the big triangle, the politicians; the circle, the graduates and the rectangle, the members of parliament. Different regions are being represented by the letters of english alphabet 29. On the basis of the about diagram, answer the following questions : Who among the following are graduates or teachers but not politicians ? (A) B, G (B) G, H (C) A, E (D) E, F 30. Who among the following politicians are graduates but not the members of parliament ? (A) B, C (B) L, B (C) D, L (D) A, H, L 31. Who among the following politicians are neither teachers nor graduates ? (A) E, F (B) D, E (C) C, D (D) L, H 32. Who among the following members of parliament is a graduate as well as a teacher? (A) G (B) F (C) C (D) H iz'u 29 ls 32 ds fy, vuqPNsn fuEu vkys[k esa] NksVk f=kHkqt] v/;kidksa dks çnf'kZr djrk gS rFkk cM+k f=kHkqt] usrkvksa dks çnf'kZr djrk gSA o`Ùk] LukÙkd dks çnf'kZr djrk gS rFkk vk;r] laln ds lnL;ksa dks çnf'kZr djrk gSA çR;sd fofHké {ks=kksa dks vaxzsth o.kZekyk ds v{kjksa }kjk vyx&vyx çnf'kZr fd;k x;k gS& 29. Åij fn;s x;s fp=k ds vk/kkj ij fuEu ç'uksa dk mÙkj nhft;s& 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 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 13 STP1617 30. fuEufyf[kr esa ls dkSuls @lk usrk Lukrd gS ijUrq laln dk lnL; ugha gS ? (A) B, C 31. (B) L, B (D) A, H, L fuEufyf[kr esa ls dkSuls @lk usrk] u rks v/;kid gS vkSj u gh Lukrd gS ? (A) E, F 32. (C) D, L (B) D, E (C) C, D (D) L, H 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 Four different integers a, b, c and d form an increasing A.P. One of these numbers is equal to the sum of the squares of the other three numbers. Then 33. 34. 35. 36. The smallest number is : (A) – 2 (B) 0 (C) – 1 (D) 2 The common difference of A.P. is (A) 2 (B) 1 (C) 3 (D) 4 (C) H.P. (D) None of these (C) 2 (D) 4 2 c, d and d are in (A) A.P. (B) G.P. The value of (a + c) 2 + (d – 2c)2 is equal to (A) 0 (B) 1 iz'u 33 ls 36 ds fy, vuqPNsn pkj fHkUu iw. kkZad a, b, c rFkk d ,d c<+rh gqbZ lekUrj Js. kh cukrs gSA ftuesa ls ,d la[ ;k vU; rhu la[ ;kvksa ds oxksZa ds ;ksx ds cjkcj gS] rc 33. lcls NksVh la[ ;k gS : (A) – 2 34. 35. 36. (B) 0 (C) – 1 (D) 2 lekUrj Js. kh dk lkoZvUrj gS (A) 2 (B) 1 (C) 3 (D) 4 c, d rFkk d2 gks a xs (A) lekUrj Js . kh (B) xq . kks Ùkj Js . kh (C) gjkRed Js . kh (D) bues a ls dks bZ ugha (C) 2 (D) 4 (a + c)2 + (d – 2c)2 dk eku cjkcj gS (A) 0 (B) 1 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 which ONLY ONE is correct. bl [k.M esa 8 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA 37. A particle is thrown upward at t = 0 sec from ground with a vertical velocity of 100 m/sec.Distance travelled by the particle in 12 sec.: (Take g = 10 m/s2) ,d d.k Åij dh rjQ t = 0 sec ij tehu ls 100 m/sec.ds Å/okZ/kj osx ls Qsadk tkrk gS rks 12 sec esa d.k }kjk r; nwjh gksxh: (g = 10 m/s2) (A) 0 m (B) 500 m (C) 480 m (D) 520 m Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 14 STP1617 38. An electron enters a magnetic field at right angles to it, as shown in figure. The direction of force acting on the electron will be : (A) to the right (B) to the left (C) out of the page (D) into the page ,d bysDVªkWu fp=kkuqlkj pqEcdh; {ks=k eas yEcor~ :i ls izos'k djrk gS rks bysDVªkWu ij vkjksfir cy dh fn'kk gksxh A (A) nka;h rjQ (B) cka;h rjQ (C) dkxt ds ry ls ckgj dh rjQ (D) dkxt ds ry esa vUnj dh rjQ 39. If the distance between two masses is doubled, the gravitational attraction between them. (A) Is doubled (B) Becomes four times (B) Is reduced to half (D) Is reduced to a quarter nks nzO;ekuks ds e/; nwjh nqxquh dj nh xbZ gS rks buds e/; xq:Rokd"kZ.k cy gksxkA (A) nqxquk (B) pkj xquk (B) vk/kk jg tk;sxk (D) ,d pkSFkkbZ jg tk;sxk 40. A force vector applied on a mass is represented as F 6 i 8 j 10k and accelerates with 1 m/s . What will be the mass of the body2 fdlh fi.M ij yxk;k x;k cy F 6 i 8 j 10k gS] rFkk ;g 1 eh@ls-2 ls Rofjr gksrk gS rks fi.M dk nzO;eku D;k gksxk& (A) 10 2 kg 41. (B) 2 10 kg (C) 10 kg (D) 20 kg A particle of mass m at rest is acted upon by a force F for a time t. Its kinetic energy after an interval t is : m nzO;eku dk ,d d.k fojke ij gS bl ij t le; ds fy, F cy vkjksfir jgrk gSA t le;kUrjky ds ckn bldh xfrt ÅtkZ gS % (A) 42. F2t 2 m (B) F2t 2 2m (C) F2t 2 3m (D) Ft 2m The displacement-time relationship for a particle is given by x a0 a1t a2 t 2 . The acceleration of the particle is ;fn ,d d.k ds foLFkkiu≤ dk lEcU/k x a0 a1t a2 t 2 gks rks mldk Roj.k gksxk& (A) a 0 (B) a1 (C) a 2 (D) 2a2 43. Light ray AB incidents on a plane mirror XY at an angle of 50º from normal. The second plane mirror is placed is such a way that the reflected ray BC from the mirror XY retraces its path. Angle of inclination of two mirrors will be: izdk'k dh fdj.k AB lery niZ.k XY ij vfHkyEc ls 50º ds dks.k ij vkifrr gksrh gSA nwljk lery niZ.k bl izdkj j[kk tkrk gS fd lery niZ.k XY ls ijkofrZr fdj.k BC nwljs niZ.k ls ijkofrZr gksdj vius iFk ij okil tkrh gSaA nksuksa niZ.kksa ds e/; ikjLifjd >qdko gSa : (A) 25º (B) 50º (C) 75º (D) 90º 44. A body moves on three quarters of a circle of radius r. The displacement and distance travelled by it are : ,d oLrq r f=kT;k ds o`Ùk ij rhu pkSFkkbZ Hkkx pyrh gS] blds }kjk r; fd;k x;k foLFkkiu o nwjh gS : (A) r, 3r 3r (B) r 2 , 2 (C) 2r, 3r 2 (D) 0, 3r 2 SECTION - II ([k.M - II) Multiple Correct Answers Type (cgqy lgh mÙkj izdkj) This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE THAN ONE is/are correct. bl [k.M esa 4 cgq lgh mÙkj izdkj ds iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls ,d ;k ,d ls vf/kd fodYi lgh gS ¼gSa½A 45. A beam of green light is incident from air and after refraction enters water. In comparison to that in air : (A) Speed of light is less in water (B) Frequency of light is less in water (C) Wavelength of light is less in water (D) Speed of light is more in water ok;q ls vkifrr gjs izdk'k dh fdj.k] viorZu ds ckn ty esa izo s'k djrh gSA ok;q dh rqyuk esa & (A) ty es a iz dk'k dh pky de gks xh (B) ty es a iz dk'k dh vko` fÙk de gks xh (C) ty es a iz dk'k dh rja xnS /;Z de gks xh (D) ty es a iz dk'k dh pky T;knk gks xh Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 15 STP1617 46. A household electric power outlet (assume 220 V constant voltage) is fused to cut at if the current equals or exceeds 20 Ampere. A 2 kW heater, 1kW Air conditioner and three 100 W bulbs are already running at rated power. If now somebody wants to run a computer then computer can run without causing fuse to burn if power requirement of computer is (neglect losses in current carrying wire) ,d ?kjsyw fo|qr 'kfDr ifjiFk (220 V vpj ekusa) esa ;fn /kkjk 20 A ;k mlls vf/kd gks tk;s rks ¶;wt VwV tkrk gSA ,d 2 kW dk ghVj, 1kW dk ,;j df.M'kuj rFkk rhu 100 W ds cYc mudh vafdr 'kfDr ij dk;Zjr gSA vc ;fn dksbZ ,d dEI;wVj pykuk pkgrk gS rFkk dEI;wVj ¶;wt tyk;s fcuk gh dk;Z dj ldrk gS ;fn dEI;wVj dh vko';d 'kfDr gksxh& (/kkjkokgh rkj esa gkfu ux.; ysaos) (A) 1000 W 47. (B) 1100 W (C) 100 W (D) 1200 W Which of the following statements is/are true : (A) An electric motor converts mechanical energy into electrical energy. (B) An electric generator works on the principle of electromagnetic induction. (C) This symbol is used for variable register.. (D) A wire with a green insulation is usually the live wire of an electric supply. fuEu eas ls dkSulk@ls dFku lR; gSA (A) fo|qr eksVj ;kaf=kd ÅtkZ dks fo|qr ÅtkZ esa ifjofrZr djrh gSA (B) fo|r tfu=k fo|qr pqEcdh; izsj.k fl)kUr ij dk;Z djrk gSA (C) bl ladsr dk iz;ksx ifjofrZ izfrjks/k ds fy;s fd;k tkrk gSA (D) ?kjks esa fo|qr vkiwfrZ ds fy, gjk fo|qr jks/kh rkj ,d /kkjkokgh rkj ds :i esa iz;qDr fd;k tkrk gSA 48. A cuboid block of mass 12 kg is lying on the ground (Assume air is absent). Take g = 10 m/sec.2 : (A) Pressing force applied by the block on the ground is 120 N. (B) If the surface ABCD is lying on the ground, then pressure (stress) exerted by the block on the ground will be 20 Pa. (C) If surface ABEF is lying on the ground, then the pressure (stress) exerted by the block on the ground will be 60 Pa. (D) If we place the block on the ground such that different plane surfaces lie on the ground, pressure (stress) on the ground will be maximum when surface BCFG lies on the ground. 12 kg nzO;eku dk ?kukHk ds vkdkj dk ,d Bksl CykWd tehu ij j[kk gqvk gSA g dk eku = 10 m/sec.2 ysaA (A) CykWd }kjk tehu ij yxk;k x;k ncko cy 120 N gSA (B) ;fn CykWd dks lrg ABCD ds lgkjs tehu ij j[kk gS] rks CykWd }kjk tehu ij yxk;k x;k nkc (izfrcy) 20 Pa gksxkA (C) ;fn CykWd dks lrg ABEF ds lgkjs tehu ij j[kk gS] rks CykWd }kjk tehu ij yxus okyk nkc (izfrcy) 60 Pa gksxkA (D) ;fn ge CykWd dks tehu ij vyx-vyx lery lrgks ds lgkjs j[ks rks tehu ij nkc (izfrcy) vf/kdre rc gksxk tc CykWd dks lrg BCFG ds lgkjs tehu ij j[ksA PART - III (Hkkx - III) SECTION - I ([k.M- I) Straight Objective Type (lh/ks oLrqfu"B izdkj) This section contains 8 multiple choice questions. Each question has choices (A), (B), (C) and (D), out of which ONLY ONE is correct. bl [k.M esa 8 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA 49. Which has highest e/m ratio ? fuEu esa ls fdldk e/m vuqikr vf/kdre gS \ (A) He2+ (B) H+ (C) He+ (D) H Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 16 STP1617 50. For a chemical reaction 3X(g) + Y(g) X3 Y(g), the amount of X3 Y at equilibrium is affected by (A) temperature and pressure (B) temperature only (C) pressure only (D) temperature, pressure and catalyst fdlh jklk;fud vfHkfØ;k ds fy, 3X(g) + Y(g) X3 Y(g), lkE; ij X3 Y dh ek=kk fuEu }kjk izHkkfor gksxh& (A) rki vkSj nkc (B) dsoy rki (C) dsoy nkc (D) rki] nkc vkSj mRizsjd 51. The maximum & minimum oxidation state of Bromine is: czksehu dh vf/kdre rFkk U;wure vkWDlhdj.k voLFkk dkSulh gS\ (A) +6, –2 (B) +7, –1 (C) +1, –1 (D) +5, –1 52. The value of Planck’s constant is 6.63 × 10–34 Js. The velocity of light is 3 × 108 m/sec. Which value is closest to the wavelength of light with frequency of 8 × 1015 sec–1 ? Iykad fLFkjkad dk eku 6.63 × 10–34 Js gSA izdk'k dk osx 3 × 108 [email protected] gSA dkSulk rjaxnS/;Z eku izdk'k dh vko`fr 8 × 1015 lSd.M–1 ds lehi gS ? (A) 5 × 10–18 m (B) 4 × 10–8 m (C) 3 × 107 m (D) 2 × 10–25 m 53. Which of the following is the IUPAC nomenclature of ethyl methyl ketone? (A) Propan-1-one (B) Butanone (C) Butan-1-one fuEu esa ls dkSulk uke ,fFky esfFky dhVksu dk IUPAC ukedj.k gS\ (A) izksisu-1-vkWu (B) C;wVsukWu (C) C;wVsu-1-vkWu 54. (D) Propan-2-one (D) izksisu-2-vkWu Lucas reagent is used to convert alcohol into alkyl halide. In this rate of reaction with respect to 1º, 2º, 3º alcohol will be– (A) 1º > 2º > 3º (B) 3º > 2º > 1º (C) 3º = 2º = 1º (D) None of these Y;wdkl vfHkdeZd dk mi;ksx ,Ydksgy dks ,fYdy gSykbM esa cnyus esa gksrk gSA 1º, 2º o 3º ,Ydksgy ds lUnHkZ esa vfHkfØ;k dh nj gksxhA (A) 1º > 2º > 3º (B) 3º > 2º > 1º (C) 3º = 2º = 1º (D) buesa ls dksbZ ugha 55. What is the formula of sodium periodate? fuEu esa ls lksfM;e ijvk;ksMsV dk lw=k dkSulk gS\ (A) NaIO 4 56. (B) NaIO 3 (C) NaIO 2 (D) NaIO IUPAC name of the compound CH3 NH CH CH2 CH CH3 is | | CH3 CH3 (A) 2-(N-methylamino)-4-methylpentane (C) 2-(N-methylamino)-3-isopropylpropane (B) N-Methyl-4-methylpentan-2-amine (D) 2-(N-methylamino)-1, 4, 4-trimethylbutane ;kSfxd CH3 NH CH CH2 CH CH3 dk IUPAC uke gSA | CH3 | CH3 (A) 2-(N-esfFky,feuks)-4-esfFkyisUVsu (C) 2-(N-esfFky,feuks)-3-vkblksizksfiyizksisu (B) N-esfFky-4-esfFkyisUVsu-2-,sehu (D) 2-(N-esfFky,ehuks)-1, 4, 4-VªkbZesfFkyC;wVsu SECTION - II ([k.M - II) Multiple Correct Answers Type (cgqy lgh mÙkj izdkj) This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE THAN ONE is/are correct. bl [k.M esa 4 cgq lgh mÙkj izdkj ds iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls ,d ;k ,d ls vf/kd fodYi lgh gS ¼gSa½A 57. Which of the given particles are -vely charged? (A) proton (B) electron (C) antiproton (D) positron fuEu esa ls dkSuls d.k _.kvkosf'kr gSa\ (A) izksVksu (B) bysDVªkWu (D) iksftVªkWu (C) ,UVhizksVksu Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 17 STP1617 58. The pair of compounds having the same general formula. fuEu esa ls dkSulk ;qXe leku lkekU; lw=k j[krk gSaA 59. (A) and (rFkk) (B) (C) and (rFkk) (D) and (rFkk) and (rFkk) H—CC—CC—H A hydrogen - like atom has ground state binding energy 122.4 eV. Then : (A) its atomic number is 3 (B) a photon of 90 eV can excite it to a higher state (C) a 80 eV photon cannot excite it to a higher state (D) None ,d gkbMªkstu leku ijek.kq dh vk| voLFkk dh cU/ku ÅtkZ 122.4 eV gSa rks & (A) bldk ijek.kq Øekad 3 gSA (B) 90 eV okyk QksVkWu bls mPp voLFkk esa mÙksftr dj ldrk gSA (C) 80 eV okyk QksVkWu bls mPp voLFkk esa mÙksftr ugha dj ldrk gSA (D) dksbZ ugah 60. Which compound is/are the isomer of 3-Ethyl-2-methylpentane ? dkSulk@dkSuls ;kSfxd 3-,fFky-2-esfFkyisUVsu ds leko;oh gSa \ (A) (B) (C) (D) ANSWER KEY TO SAMPLE TEST PAPER-I 1. 8. 15. 22. 29. 36. 43. 50. 57. (B) (C) (A) (A) (C) (A) (B) (A) (BC) 2. 9. 16. 23. 30. 37. 44. 51. 58. (D) (C) (D) (C) (A) (D) (B) (B) (ABD) 3. 10. 17. 24. 31. 38. 45. 52. 59. (D) (D) (B) (B) (D) (D) (AC) (B) (AC) 4. 11. 18. 25. 32. 39. 46. 53. 60. (D) (B) (C) (CD) (B) (D) (ABC) (B) (ACD) 5. 12. 19. 26. 33. 40. 47. 54. (B) (D) (B) (CD) (C) (A) (BC) (B) 6. 13. 20. 27. 34. 41. 48. 55. (A) (C) (B) (AD) (B) (B) (ABD) (A) 7. 14. 21. 28. 35. 42. 49. 56. (B) (B) (B) (BC) (B) (D) (B) (B) HINTS & SOLUTION TO SAMPLE TEST PAPER-I 1. 3. Since DE is angle bisector AE : EB = DA : DB .... (i) Since DF is angle bisector AF : FC = DA : DC = DA : DB from (i) and (ii) we get AE : EB = AF : FC EF // BC .... (ii) The diagonal d = 25m. and area A = 168m2. Let ‘I’ be the length and ‘b’ be the width of the rectangle. Therefore, I2+b2 = d2. and Ib = A We can therefore write (I + b)2 = d2 + 2A and (I b)2 = d2 2A. Substituting and solving we get, I + b = 31 and I b = 17. Hence I = 24 and b = 7. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 18 STP1617 4. x=7+ 4 3 1 x = 2 3 5. Sn = x = 2 3 So x + 1 x =2+ 3 + 2 3 = 4 3n 2 5n 2 2 Sn–1 = 6. 3 5 (n – 1)2 + (n – 1) 2 2 an = sn – sn–1 = 3 5 (2n – 1) + 2 2 a25 = 76 In ABC, we have A+ B + C = 180° 60° + B + 50° = 180° 110° + B = 180° B = 180° – 110° B = 70° Since D and E are the mid-points of AB and AC respectively. Therefore, 1 BC 2 Now, DE || BC and the transversal AB cuts them at D and B respectively. EDB + B = 180° DE || BC and DE = Interior angles on thesameside of the transversal are sup plementary EDB + 70° = 180° EDB = 180° – 70° = 110° 7. AB is perp. bisector of PQ APQ = AQP and PBA = QBA But PQ is perpendicular bisector of AB is not always true (It is true only when the circles are of equal radius) 9. ACB = ABP = 60° ABC = ACQ = 70° BAC = 180° – (60° + 70°) = 50° BOC = 100° BTC = 180° – 100° = 80° 10. Selling price = Rs. 24 = 80% of marked price. Marked price = 30 Now selling price = 70% of marked price = 70 × 30 = 21 100 11. (a1/8 + a–1/8)(a1/8 – a–1/8)(a1/4 + a–1/4)(a1/2 + a–1/2) = (a1/4 – a–1/4)(a1/4 + a–1/4)(a1/2 + a–1/2) = (a1/2 – a–1/2)(a1/2 + a–1/2) = (a – a–1). 13. FCD = 180° – 80° = 100° [Opposite angles of a cyclic quadrilateral are supplementary] Now AB || CD ABC + DCB = 180° [Co-interior angles] FBA + 100° = 180° So. FBA = 80° Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 19 STP1617 15. We have (gesa fn;k gS ) cos – sin 1 – 3 cos sin 1 3 (cos – sin ) (cos sin ) (1 – 3 ) (1 3 ) = [Applying componendo and dividendo] [;ksxkUrjkuqikr (cos – sin ) – (cos sin ) (1 – 3 ) – (1 3 ) dk mi;ksx djus ij] 2 2 cos = – 2 sin –2 3 tan= 16. 1 cot = 3 3 tan= tan60º = 60º AOB = 2 (APB) b = AOB = 2 (85°) = 170° In AOB, b + 2a = 180° 170° + 2a = 180° a = 5° (OA = OB) 17. QRC and PRA are similar 3/5 =6:7 7 / 10 AR : AC = 6 : 13 AR : RC = 18. Let x = 0.235 10x = 2.35 35 99 990x = 233 1000x = 235..35 + x= 19. 233 990 x= – 3 , 2 y= 3 2 Sum of all natural numbers less than 400 399 400 79800 2 Sum of all natural number divisible by six. S2 = 6 + 12 + 18 +.......396 Number of terms 396 = 6 + (n–1)6 n = 66 S1 = 1 + 2 + 3 + 4 +......399 = 66 [2 6 (66 1)6] 2 S2 = 13266 S1 – S2 = 79800 – 13266 = 66534. sum Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 20 STP1617 20. Incorrect watch covers 1452 min in 1440 min. So, it will cover 1 min in 1440 × 4840 = 4800 min = 80 h. Therefore 80 h = 3 days and 8 h. 1452 4840 min in 21. 2 – a = –a – b 2 – a = –a – b 22. 1440 min. Therefore it will cover 1452 1 2 – a + 1 2 1 1 2 – + = – =0 – a ab ab ab ab Surface area of the cylinder = surface area of the square 2rh = a2 Here height ‘h’ of the cylinder = side of the square since it is rolled along its length. base radius r a 2 Ratio of base radius to side of square 52 6 52 6 24. 52 6 52 6 52 6 = 3 2 52 6 = 3 2 So, 3 2 3 2 2 2 x 2 a 1 :a 2 2 x 2 3 2 3 2 2 3 2ra = a2. x 2 x 2 3 2 25. (5a – 3b)3 + (3b – 7c)3 – (5a – 7c)3 Let 5a – 3b = x, 3b – 7c = y and 7c – 5a = z x3 + y3 + z3 = 3xyz, if x + y + z = 0. x + y + z = 5a – 3b + 3b – 7c + 7c – 5a = 0 (5a – 3b)3 + (3b – 7c)3 + (7c – 5a)3 = 3 (5a – 3b) (3b – 7c) (7c – 5a) It is divisible by 5a – 3b, 3b – 7c, 7c – 5a. 26. (2x2 + 3x + 5)1/2 + (2x2 + 3x + 20)1/2 = 15 Let 2x2 + 3x + 5 = y2 y + (y2 + 15)1/2 = 15 y2 + 15 = (15 – y)2 y2 + 15 = 225 + y2 – 30y 30y = 210 y x=3 210 7. 30 2x2 + 3x + 5 = 49 2x2 + 3x – 44 = 0 2x2 – 8x + 11x – 44 = 0 2x(x – 4) + 11 (x – 4) = 0 x = –11/2 & x = 4. Hence, two solutions x = –11/2 & x = 4 are possible. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 21 STP1617 28. a > b if ;fn c < 0 ac < bc so (A) is incorrect. xyr gSA a > b a – c > b – c so (B) is correct.lgh gS a b > so (C) is correct. lgh gSA d c c = 50, d = 2, a = 10, b = 1 a – c = – 40, b – d = – 1 If ;fn a > b > 0 , c > d > 0 If ;fn a > b , c > d =–1 1 29 to 32 Sol. 29. A,E 30. B,C 31. L,H 32. F (33-36) Sol. Let four integers be a–d, a , a + d and a + 2d where a and d are integers and d > 0. a + 2d = (a – d) 2 + a2 + (a + d)2 2d2 – 2d + 3a2 – a = 0 ...............(i) 1 1 1 2a 6a 2 2 Since d is positive integer 1 + 2a – 6a2 > 0 6a2 – 2a – 1 < 0 d= 1 7 1 7 <a< 6 6 a = 0 Put in (ii) d = 1 or 0 but d=1 The four numbers are : –1, 0, 1, 2 35. c, d, d2 1, 2, 4 they are in G.P. 36. (– 1 + 1)2 + (2 – 2)2 = 0 37. F ...............(ii) 1 r2 . If r becomes double then F reduces to |F| 38. mass nzO;eku m = = |a| a is an integer d>0 F 4 200 = 10 2 1 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 22 STP1617 Ft2 W F = FS = F 2 m 1 F 2 t , S= 2 m F 39. a = , m 40. x = a0 + a1 t + a2 t2 dx = 0 + a1 + 2a2 t dt a= d2 x dt 2 = 0 + 2a2 N A C 50º 50º 41. 40 X Y B Ray BC retraces its path when A fall on second mirror perpendicularly, so as per figure L BYC will be euqal to 50º. R 42. S Q P r f=kT;k ds o`rh; iFk ij 3/4 Hkkx r; djus ij (i) Displacement (foLFkkiu) = (ii) Distance (nwjh) = 3r 3 2r = 4 2 43. We know that v = f (i) speed of light 1 so v w < air (ii) v so w < air v 1 f = constant w > air ge tkurs gS fd v = f (i) iz dk'k dh pky v r2 r2 = r 2 f = fu;r w > air vr% v w < air (ii) v vr% w < air 44. Total power supplied dqy fn xbZ 'kfDr = 20 × 220 = 4400 Watt Already existing load tks yksM ij igys ls mifLFkr gS = 2000 + 1000 + 300 = 3300 Watt We can increase load upto 4400 Watt so A, B & C are correct options. ge yksM 4400 Watt rd c<+k ldrs gSA vr% A, B rFkk C lgh fodYi gSA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 23 STP1617 45. Based on information lwpuk ij vk/kkfjr 46. (A) Pressing force ncko cy = mg = 12 g = 120 N (B) P = F 120 = = 20 Pa A 32 F 120 = = 40 Pa A 3 1 (D) Area of surface BCFG is minimum So, pressure will be maximum. lrg BCFG dk {kS=kQy U;qure gS vr% nkc vf/kdre gksxkA (C) P = 2e 4MP For H+ e MP e For H 0 M For He2+ gy% 2e He2+ ds fy, 4M P 50. Equilibrium is affected by pressure and temperature but not by catalyst. gy- lkE; ij nkc o rki dk izHkko iMrk gS ysfdu mRizsjd dk ughaA 51. Br show oxidation state varying from –1 to +7. Br, –1 ls +7 rd ifjofrZr gksus okyh vkWDlhdj.k voLFkk n'kkZrk gSA 52. ; ; ; For He+ e 4MP 49. e H+ ds fy, M P ; e ; He+ ds fy, 4M ; P e H ds fy, 0 M c 3 108 = 3.75 × 10–8 m. 8 1015 O 53. CH3 –C–CH2 –CH3 – Ethyl methyl ketone Butanone – IUPAC name. O CH3 –C–CH2 –CH3 – ,fFkyesfFkydhVksu C;wVsukWu – IUPAC uke. 59. gy % 60. Ground state binding energy = 13.6 Z2 = 122.4 eV. Z = 3. 1st excitation energy = 10.2 Z2 = 91.8 eV. an 80 eV electron cannot excite it to a higher state. vk| voLFkk esa cU/ku ÅtkZ = 13.6 Z2 = 122.4 eV. Z = 3. 1st mÙkstu ÅtkZ = 10.2 Z2 = 91.8 eV. ,d 80 eV dk QksVkWu bls mPp voLFkk esa mÙksftr ugha dj ldrk gSA ACD are 8 carbon alkanes but B has only 6 carbons atoms. vU; lHkh esa 8 dkcZu gSa] tcfd B esa dsoy 6 dkcZu gSaA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JA/JB-PAGE # 24 STP1617 SAMPLE TEST PAPER-II (For Class-XI Appearing / Passed Students) Course : VISHWAAS (JF) Marks to be awarded S.No. Subject Section Nature of Questions No. of Questions Correct Wrong Total Section-I Straight Objective Type Questions (dsoy ,d fodYi lgh) 14 3 -1 42 Section-II Multiple Choice Questions (,d ;k ,d ls vf/kd fodYi lgh) 5 4 0 20 20 to 23 Section-III Comprehension ( vuqPNsn) (2 Comp. x 2 Q.) 4 3 -1 12 24 to 37 Section-I Straight Objective Type Questions (dsoy ,d fodYi lgh) 14 3 -1 42 Section-II Multiple Choice Questions (,d ;k ,d ls vf/kd fodYi lgh) 5 4 0 20 43 to 46 Section-III Comprehension ( vuqPNsn) (2 Comp. x 2 Q.) 4 3 -1 12 47 to 60 Section-I Straight Objective Type Questions (dsoy ,d fodYi lgh) 14 3 -1 42 PART-III 61 to 65 (Chemistry) Section-II Multiple Choice Questions (,d ;k ,d ls vf/kd fodYi lgh) 5 4 0 20 66 to 69 Section-III Comprehension ( vuqPNsn) (2 Comp. x 2 Q.) 4 3 -1 12 1 to 14 15 to 19 38 to 42 PART-I (Maths) PART-II (Physics) Total 69 222 PART - I (Hkkx - I) SECTION - I ([k.M- I) Straight Objective Type (lh/ks oLrqfu"B izdkj ) This section contains 14 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. bl [k.M esa 14 iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA 1. If |z – i| 2 and z0 = 5 + 3i then the maximum value of |iz + z0| is (A) 2 + (C) 31 31 – 2 (B) 7 (D) None of these ;fn |z – i| 2 rFkk z0 = 5 + 3i gS] rks |iz + z0| dk vf/kdre eku gS (A) 2 + (C) 2. 31 31 – 2 (B) 7 (D) bues a ls dks bZ ugha Ram sets his watch at 6 : 10 am on Sunday, which gains 12 minutes in a day. On Wednesday if this watch is showing 2 : 50 pm. What is the correct time ? jke viuh ?kMh jfookj dks 6 : 10 am ij feykrk gS] tks fd ,d fnu esa 12 feuV vkxs gks tkrh gSA cq/kokj dks ;fn ;g ?kMh 2 : 50 pm fn[kk jgh gS] rks lgh le; D;k gksxk ? (A) 1 : 50 pm (C) 2 : 30 pm (B) 2 : 10 pm (D) 3 : 30 pm Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 25 STP1617 3. 4. 5. 6. If 2576a 456b is divisible by 15 then S1 : a may take value 5 S2 : a may take value 3 S3 : a may take value 9 S4 : a may take value 8 (A) S1 and S2 are true (C) S1 , S2 and S4 are true (B) S1 and S3 are true (D) S1 and S4 are true ;fn 2576a 456b , 15 ls foHkkT; gS rc S1 : a dk eku 5 gks ldrk gS S2 : a dk eku 3 gks ldrk gS S3 : a dk eku 9 gks ldrk gS S4 : a dk eku 8 gks ldrk gS (A) S1 vkSj S2 lR; gS (C) S1, S2 vkSj S4 lR; gS (B) S1 vkSj S3 lR; gS (D) S1 vkSj S4 lR; gS 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 If ax1 = bc, by1 = ca, cz1 = ab, then xy + yz + zx = (A) xyz (B) 3 xyz (C) x+y+z ;fn ax1 = bc, by1 = ca, cz1 = ab, rc xy + yz + zx = (A) xyz (B) 3 xyz (C) x+y+z If cot = 1 ;fn cot = (A) 7. 3 , then the value of 1 3 gS] rks 1 5 2 sin 2 (B) 2 sin 2 lehdj.k 2 5 (A) 2 (D) buesa ls dksbZ ugha is (C) (B) 3 (1 x )2 + (D) none of these dk eku gS & Total number of solutions of the equation (A) 2 8. 1 cos 2 1 cos 2 (D) – 2520 (1 x )2 = 2 ds 3 5 (1 x )2 + (D) None dksbZ ugah (1 x )2 = 2 is (C) 4 (D) infinite gyksa dh la[;k gS& (B) 3 (C) 4 (D) vuUr The arithmetic mean of greatest and least integral values of so that (2 + 1, – 1) and origin lie on the same side of x – 10y – 2 = 0 is (A) 0 (B) 4 (C) 5 (D) none of these ;fn ewy fcUnq rFkk fcUnq (2 + 1, – 1) ds js[kk x – 10y – 2 = 0 ds gh vksj fLFkr gks] rks bl izdkj izkIr ds vf/kdre rFkk U;wure iw.kk±d ekuksa dk lekUrj ek/; gS & (A) 0 9. (B) 4 (C) 5 (D) buesa ls dksbZ ugha The equation of circumcircle of an equilateral triangle is x2 + y2 + 2gx + 2fy + c = 0 and one vertex of the triangle is (1, 1). The equation of incircle of the triangle is (A) 4(x2 + y2) = g2 + f 2 (B) 4(x2 + y2) + 8gx + 8fy = (1 g) (1 + 3g) + (1 f) (1 + 3f) 2 2 2 2 (C) 4(x + y ) + 8gx + 8fy = g + f (D) None of these leckgq f=kHkqt ds ifjxr o`Ùk dk lehdj.k x2 + y2 + 2gx + 2fy + c = 0 gS rFkk f=kHkqt dk ,d 'kh"kZ (1, 1) gSA rc f=kHkqt ds vUro`Ùk dk lehdj.k gS (A) 4(x2 + y2) = g2 + f 2 (B) 4(x2 + y2) + 8gx + 8fy = (1 g) (1 + 3g) + (1 f) (1 + 3f) (C) 4(x2 + y2) + 8gx + 8fy = g2 + f 2 (D) buesa ls dksbZ ugha Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 26 STP1617 10. The exhaustive range of values of ‘a’ such that the angle between the pair of tangents drawn from (a, a) to the circle x2 + y2 2x 2y 6 = 0 lies in the range , , is 3 ‘a’ ds ekuksa dk fu''ks"kh ifjlj gksxk tcfd fcUnq (a, a) ls o`Ùk x2 + y2 2x 2y 6 = 0 ij [khph xbZ Li'kZ js[kkvksa ds ;qXeksa ds e/; dks.k dk ifjlj 3 , esa fLFkr gS (A) (1, ) (B) (5, 3) (3, 5) (C) (, 2 2) (2 2, ) (D) (3, 1) (3, 5) 11. The variance of 20 observations is 5. If each observations is multiplied by 2, then the new variance of the resulting observations is 20 iz s { k.kks a dk iz l j.k 5 gS A ;fn iz R;s d iz s { k.k dks 2 ls xq . kk fd;k tkrk gS rc ifj.kkeh iz s { k.k dk u;k iz l j.k gS (A) 10 (B) 20 (C) 30 (D) 40 12. Sita has 4 different and simmi has 7 different toys. Number of ways in which they can exchange their toys so that each keeps her inital number of toys, is lhrk 4 rFkk xhrk 7 fHkUu&fHkUu f[kykSus j[krh gS] rks mu rjhdksa dh la[;k D;k gksxhA tcfd nksuksa viuh izkjfHkd f[kykSuksa dh la[;k dks leku j[krs gq, vius f[kykSuksa dks ,d&nwljs ls cny ysaA (A) 329 13. (B) 334 (C) 332 (D) 345 In a series of 5 matches in football (A team has equal chances to win, loose or draw a match). The probability that a forecast selected at random has exactly 2 correct predictions, is 5 Qq V cky es pks a dh ,d Ja [ kyk es a Bhd 2 es pks a ds iw o kZ u q eku lgh gks u s dh iz kf;drk (tcfd iz R;s d es p es a fdlh ny ds thrus] gkjus o esp Mªk gksus dh leku laHkkouk gS] gksxh& (A) 14. 163 243 (B) 40 243 (C) 60 243 (D) If , & are the roots of the equation x 3 2x 1 = 0 then, (A) zero (B) 1 (C) 4 ;fn lehdj.k x 3 2x 1 = 0 ds ewy , ,oa gks] rks (A) 'kwU; (C) 4 80 243 1 1 1 + + has the value equal to 1 1 1 (D) 1 1 1 1 + + 1 1 1 dk eku gS& (B) 1 (D) 1 SECTION - II ([k.M- II) Multiple Correct Answer Type (cgqy lgh fodYi izdkj) This section contains 5 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONE OR MORE is/are correct. bl [k.M esa 5 iz'u gSaA izR;sd iz'u ds mÙkj ds fy, 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls vf/kd lgh gSA n 15. 1 3 1 is The last term in the binomial expansion of 2 3.3 9 2 f}in O;atd (A) 10C 4 (C) 10C6 3 1 2 2 n dk vfUre in 1 3.3 9 log3 8 , then the fifth term is log3 8 gS, rks ik¡pok in gS& (B) 2. 10C4 1 (D) 2 10C 4 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 27 STP1617 16. If z1 5 12i and z2 4 then (A) maximum z1 iz 2 17 z1 (C) minimum 4 z2 z2 (B) minimum z1 1 i z 2 13 4 2 z1 13 4 4 z2 z2 (D) maximum 13 3 ;fn z1 5 12i vkSj z2 4 gks rc (A) vf/kdre z 1 iz 2 17 z1 4 (C) U;wu re z2 z2 17. z 1 1 i z 2 13 4 2 z1 13 4 4 (D) vf/kdre z2 z2 13 3 If nC4 , nC5 , nC6 are in A.P, then a value of n is ;fn nC4, nC5 , nC6 lekUrj Js<h esa gks] rks n dk eku gS& (A) 14 18. (B) U;w u re (B) 11 (C) 7 2 (D) 8 2 If (t , 2t) is one end of a focal chord of the parabola y = 4x, then which is (are) true 2 1 (A) length of the focal chord is t t 2t (B) slope of the focal chord is 2 t –1 (C) mid-point of focal chord is focus (D) focal-chord is angle bisector of tangent and the normal at (t2 , 2t) to the parabola ;fn (t2 , 2t) ijoy; y2 = 4x dh ukHkh; thok dk ,d fljk gks] rks fuEu esa ls dkSulk@ls lR; gSa& 2 1 (A) ukHkh; thok dh yEckbZ t gSA t 2t (B) ukHkh; thok dh izo.krk 2 gSA t –1 (C) ukHkh; thok dk e/; fcUnq] ukfHk gSA (D) ijoy; ds (t2 , 2t) ij Li'kZ js[kk vkSj vfHkyEc dk v)Zd ukHkh;thok gSA 19. The vertices of a ABC are A( 5, 2) ; B(7, 6) and C(5, 4) . Then : (A) measure of angle B is /4 (B) equation of the altitude drawn from the vertex C has the equation, 3x + 2y 7 = 0 (C) orthocentre of the triangle does not lie inside the ABC (D) distance between centroid and circumcentre of the ABC is 4 13 3 f=kHkqt ABC ds 'kh"kZ A( 5, 2) ; B(7, 6) vkSj C(5, 4) gks] rks (A) dks . k B dk eki /4 gS A (B) 'kh"kZ C ls [kha ps x;s 'kh"kZ yEc dk lehdj.k 3x + 2y 7 = 0 gS A (C) f=kHkq t dk yEcds Unz f=kHkq t ABC ds vUnj fLFkr ugha gS A (D) f=kHkq t ABC ds ds Unz vkS j ifjds Unz ds e/; dh nw j h 4 13 gSA 3 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 28 STP1617 SECTION - III ([k.M - III) Comprehension Type (cks/ ku çdkj) This section contains 2 paragraphs. Based upon each paragraph, there are 2 questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. bl [k.M esa 2 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 2 ç'u gSA çR;sd ç'u ds 4 fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA Paragraph for Question Nos. 20 to 21 iz'u 20 ls 21 ds fy, vuqPNsn Consider the function y = 2 |x + 2| – |3 – x| – 3x + 3. Qyu y = 2 |x + 2| – |3 – x| – 3x + 3 fn;k x;k gSA 20. 21. Number of real solutions of the equation y = 0 is (A) 1 (B) 2 (C) 3 lehdj.k y = 0 ds okLrfod gyksa dh la[ ;k gSa& (A) 1 (B) 2 (C) 3 Number of real solutions of the equation y = 4 is (A) 1 (B) 2 (C) 3 lehdj.k y = 4 ds okLrfod gyksa dh la[ ;k gSa& (A) 1 (B) 2 (C) 3 (D) infinite (D) vuUr (D) infinite (D) vuUr Paragraph for Question Nos. 22 to 23 iz'u 22 ls 23 ds fy, vuqPNsn The binomial coefficients of the 2nd, 3rd and 4th terms in the expansion of log10 (10 3 x ) 5 ( x 2 ) log10 3 2 2 log10 (10 3 x ) 5 ( x 2 ) log10 3 2 2 m represent respectively the first, third and fifth terms of an A.P.. m ds izlkj esa nwljs, rhljs rFkk pkSFks inksa ds f}in xq.kkad Øe'k% ,d lekUrj Js<+h ds izFke] r`rh; rFkk ik¡posa inksa dks iznf'kZr djrk gSA 22. 23. The value of m is equal to m dk eku gS & (A) 2 (B) 7 (C) 5 The number of values of x for which 6th term in the expansion is 21, are x ds ekuksa dh la[;k ftuds fy, izlkj esa 6ok¡ in 21 gks] gS & (A) 1 (B) 3 (C) 0 (D) 9 (D) 2 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 29 STP1617 PART - II (Hkkx - II) SECTION - I ([k.M- I) Straight Objective Type (lh/ks oLrqfu"B izdkj) This section contains 14 multiple choice questions. Each question has choices (A), (B), (C) and (D), out of which ONLY ONE is correct. bl [k.M esa 14 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA 24. 25. The current i in the circuit of figure is fp=k ds ifjiFk esa /kkjk i gS & (A) 1 amp. 45 (B) 1 amp. 15 (C) 1 amp. 10 (D) 1 amp. 5 An electron enters a magnetic field at right angles to it, as shown in figure. The direction of force acting on the electron will be : (A) to the right (nka;h rjQ) (B) to the left (cka;h rjQ) (C) out of the page (dkxt ds ry ls ckgj dh rjQ) (D) into the page (dkxt ds ry esa vUnj dh rjQ) Which of the following represents the displacement-time graph of two objects A and B moving with equal velocity? fuEu esa ls dkSulk foLFkkiu le;&oØ leku osx ls xfreku nks oLrqvksa A rFkk B dks O;Dr djrk gS ? (B) B (C) Time 27. (D) Time Displacement B A A Displacement (A) Displacement A Displacement 26. Time B Time A particle moves from position r1 3 î 2 ĵ 6k̂ to position r2 14 î 13 ĵ 9 k̂ under the action of force 4 î ĵ 3 k̂ N . The work done by this force will be ,d d.k] cy 4 î ĵ 3 k̂ N ds v/khu fLFkfr r1 3 î 2 ĵ 6k̂ ls fLFkfr r2 14 î 13 ĵ 9 k̂ rd xfr djrk gSA bl cy }kjk fd;k x;k dk;Z gksxk & (A) 100 J 28. (B) 50 J (C) 200 J (D) 75 J Two particles of combined mass M, placed in space with certain separation, are released. Interaction between the particles is only of gravitational nature and there is no external force present. Acceleration of one particle with respect to the other when separation between them is R, has a magnitude : nks d.k ftudk la;qDr nzO;eku M gS] vkdk'k esa fuf'pr nwjh ij j[ks gSa ] mudks NksM+k tkrk gSA d.kksa ds e/; dsoy xq:Rokd"kZ.k cy yxrk gS rFkk vU; dksbZ ckà; cy ugh yxrk gSA fdlh ,d d.k dk nwljs d.k ds lkis{k Roj.k dk ifjek.k] tc muds e/; R nwjh gS] gksxk & (A) GM 2R 2 (B) GM R 2 (C) 2GM (D) not possible to calculate due to lack of information R2 (lwpukvksa dh deh ds dkj.k x.kuk lEHko ugh gSA) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 30 STP1617 29. In the figure shown ABCD is a rectangular smooth tube kept fixed in a vertical plane. A particle is projected from point A to reach point C with some speed. At the corners B and D velocity changes its direction by 90º without any change of its magnitude at that corner. If time taken on paths ABC and ADC are t 1 and t 2 respectively, then: (given > b) fp=kkuqlkj ,d vk;rkdkj fpduh ufydk ABCD m/okZ/kj ry es tM+or gSA ,d d.k dks A ls C ds fy, fdlh pky ls ç{ksfir fd;k tkrk gSA dksus B o D ij dk osx fn'kk esa 90º ls cny tkrk gS] ijUrq eku leku jgrk gSA ;fn iFk ABC rFkk ADC es fy;k x;k le; Øe'k% t1 o t2 gS rks : (fn;k gS > b) (A) t 1 = t 2 30. (B) t 1 > t 2 (C) t 1 < t 2 (D) none of these (buesa ls dksbZ ugha) Two point masses of mass m1 and m2 are placed at point A and B respectively as shown in figure. Point A is the centre of hollow sphere of uniformly distributed total mass m3. Consider only gravitational interaction between all masses and neglect other gravitational forces. Select the incorrect alternative. nks fcUnq nzO;eku ftuds nzO;eku m1 rFkk m2 gS] dks Øe'k% fcUnqvksa A rFkk B ij fp=kkuqlkj j[kk x;k gSA fcUnq A ,d [kks[kys xksys ftldk nzO;eku m3 m3 m1 A a ¼le:i forfjr½ gS] dk dsUnz gSA dsoy fn;s x;s nzO;eku ds e/; gh xq:okd"kZ.k gS vU; xq:Rokd"kZ.k cyksa dks ux.; ekfu;sA vlR; dFku pqfu;sA m2 B r (A) Hollow sphere and point mass m1 moves with same acceleration. (B) m1 and m2 moves with same acceleration. (C) Net force on m1 is non-zero (D) Net force on hollow sphere and point mass m1 as a system is equal to force experienced by point mass m2 in magnitude. (A) [kks[kyk xksyk rFkk fcUnq nzO;eku m1 leku Roj.k ls xfr djsxsa (B) m1 rFkk m2 leku Roj.k ls xfreku gksxsa (C) m1 ij dqy cy v'kqU; gksxk (D) [kks[kys xksys rFkk m1 dks ,d fudk; ekurs gq;s bl ij yxus okys dqy cy dk ifjek.k m2 nzO;eku ij yxus okys dqy cy ds ifjek.k ds cjkcj gksxk& 31. A bullet of mass 10 g is fired with a rifle. The bullet takes 0.003 s to move through its barrel and leaves with a velocity of 300 ms–1. The force exerted on the bullet by the rifle : (A) 103 N (B) 104 N (C) 105 N (D) zero ,d cUnwd ls 10 xzke dh xksyh NksM+h tkrh gSA xksyh uyh ls xqtjus esa 0.003 lsd.M ysrh gS rFkk uyh ls 300 eh-/ls- ds osx ls fudyrh gSA cUnwd }kjk xksyh ij yxk;k x;k cy gS % (A) 103 U;wVu (B) 104 U;wVu (C) 105 U;wVu 32. (D) 'kwU; A machine gun fires n bullets per second and the mass of each bullet is m. If the speed of bullet is v, then the magnitude of force exerted on the machine gun is : ,d e'khuxu çfr lSd.M n xksfy;k¡ nkxrh gS vkSj çR;sd xksyh dk æO;eku m gSA ;fn xksyh dh pky v gS] rks e'khu xu ij vkjksfir cy gksxk& (A) mng 33. (B) mnv g (C) mnv The number of electrons contained in the nucleus of 92U235. (A) 92 (B) 143 (C) 235 ukfHkd 92U235 esa bySDVªkWuksa dh la[;k gS & (A) 92 (B) 143 (C) 235 (D) mnvg (D) Zero (D) 'kwU; Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 31 STP1617 34. An unnumbered clock shows time as 4 : 35 in its mirror image. The real time is : (A) 8 : 25 (B) 6 : 25 (C) 7 : 25 (D) None of these ,d fcuk vad fy[kh gqbZ ?kM+h dk niZ.k izfrfcEc 4 : 35 le; crkrk gSA rks lgh le; gS % (A) 8 : 25 (B) 6 : 25 (C) 7 : 25 (D) buesa ls dksbZ ugha 35. A transformer is used to (A) convert DC into AC (C) obtain the required DC voltage (B) convert AC into DC (D) obtain the required AC voltage VªkalQkeZj dk mi;ksx fd;k tkrk gS (A) DC dks AC es cnyus esa (C) vfHk"B DC oksYVst izkIr djus ds fy, (B) AC dks DC es cnyus esa (D) vfHk"B AC oksYVst izkIr djus ds fy, 36. A particle is projected from point 'A' with velocity u 2 at an angle of 45º with the horizontal as shown in the figure. It strikes the inclined plane BC at right angle. The velocity of the particle just before the collision with the inclined is : ,d d.k fp=kkuqlkj {kSfrt ds lkFk 45º ds dks.k ij u 2 osx ls fcUnq A ls ç{ksfir fd;k tkrk gSA ;g ry BC ij yEcor~ Vdjkrk gSA urry ls VDdj ds Bhd igys d.k dk osx gksxkA (A) 37. 3u 2 (B) u 2 2u (C) (D) u 3 A homogeneous plate PQRST is as shown in figure. The centre of mass of plate lies at midpoint A of b segment QT. Then the ratio of is (PQ = PT = b; QR = RS = ST = a) a b ,d lekax IysV PQRST fp=kkuqlkj crk;h xbZ gSA IysV dk nzO;eku dsUnz QT Hkkx ds e/; fcUnq A ij gSA rc vuqikr gS & a (PQ = PT = b; QR = RS = ST = a) (A) 13 4 (B) 13 2 (C) 13 2 (D) 13 4 SECTION - II ([k.M - II) Multiple Correct Answers Type (cgqy lgh mÙkj izdkj) This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE THAN ONE is/are correct. bl [k.M esa 5 cgq lgh mÙkj izdkj ds iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls ,d ;k ,d ls vf/kd fodYi lgh gS ¼gSa½A Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 32 STP1617 38. An ideal battery of 60 volt is connected with the resistors as shown above. A1 and A2 are two ideal ammeters and V is an ideal volt meter. Then : (A) Reading of (A1) is 2 Amp (B) Reading of (A2) is 1 Amp (C) Reading of (V) is 60 volt (D) Total power consumed by all the resistors is 120 watt 60 oks YV dh vkn'kZ cS V jh çnf'kZ r fp=kkuq l kj çfrjks /k ls tq M + h gS A ;fn A1 rFkk A2 nks vkn'kZ vehVj rFkk V vkn'kZ oksYVehVj gks rks : (A) (A1) dk ikB;ka d 2 Amp gS A (B) (A2) dk ikB;ka d 1 Amp gS A (C) (V) dk ikB;ka d 60 volt gS A (D) lHkh çfrjks /kks a es a dq y 'kfDr O;; 120 watt gS A 39. A household electric power outlet (assume 220 V constant voltage) is fused to cut at if the current equals or exceeds 20 Ampere. A 2 kW heater, 1kW Air conditioner and three 100 W bulbs are already running at rated power. If now somebody wants to run a computer then computer can run without causing fuse to burn if power requirement of computer is (neglect losses in current carrying wire) ,d ?kjsyw fo|qr 'kfDr ifjiFk (220 V vpj ekusa) esa ;fn /kkjk 20 A ;k mlls vf/kd gks tk;s rks ¶;wt VwV tkrk gSA ,d 2 kW dk ghVj, 1kW dk ,;j df.M'kuj rFkk rhu 100 W ds cYc muds vafdr 'kfDr ij dk;Zjr gSA vc ;fn dksbZ ,d dEI;wVj pykuk pkgrk gS rFkk dEI;wVj ¶;wt tyk;s fcuk gh dk;Z dj jgk gS rks dEI;wVj dks vko';d 'kfDr D;k gksxh (/kkjkokgh rkj esa gkfu ux.; ysaos) (A) 1000 W 40. (B) 1100 W (C) 100 W (D) 1200 W Choose the correct statement(s) among the following : (A) The magnetic force on a stationary charge is always zero. (B) The magnetic line of force around a straight conductor is circular. (C) The magnetic force on a moving charge is responsible for change in its KE. (D) Magnetic force is a central force. fuEu esa ls lgh dFkuksa dk p;u dhft,& (A) fLFkj vkos'k ij pqEcdh; cy lnSo 'kwU; gksxkA (B) lh/ks pkyd ds pkjksa vksj pqEcdh; cy js[kk,a o`Ùkkdkj gksxh (C) xfreku vkos'k ij pqEcd cy bldh xfrt ÅtkZ esa ifjorZu ds fy, mÙkjnk;h gSA (D) pqEcdh; cy dsfUæ; cy gSA 41. A cuboid block of mass 12 kg is lying on the ground (Assume air is absent). Take g = 10 m/sec2. (A) Pressing force applied by the block on the ground is 120 N. (B) If the surface ABCD is lying on the ground, then pressure (stress) exerted by the block on the ground will be 20 Pa. (C) If surface ABEF is lying on the ground, then the pressure (stress) exerted by the block on the ground will be 60 Pa. (D) If we place the block on the ground such that different plane surfaces lie on the ground, pressure (stress) on the ground will be maximum when surface BCFG lies on the ground. 12 kg nzO;eku dk ?kukHk ds vkdkj dk ,d Bksl CykWd tehu ij j[kk gqvk gSA g dk eku = 10 m/sec2 ysaA (A) CykWd }kjk tehu ij yxk;k x;k ncko cy 120 N gSA (B) ;fn CykWd dks lrg ABCD ds lgkjs tehu ij j[kk gS] rks CykWd }kjk tehu ij yxk;k x;k nkc (izfrcy) 20 Pa gksxkA (C) ;fn CykWd dks lrg ABEF ds lgkjs tehu ij j[kk gS] rks CykWd }kjk tehu ij yxus okyk nkc (izfrcy) 60 Pa gksxkA (D) ;fn ge CykWd dks tehu ij vyx-vyx lery lrgks ds lgkjs j[ks rks tehu ij nkc (izfrcy) vf/kdre rc gksxk tc CykWd dks lrg BCFG ds lgkjs tehu ij j[ksA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 33 STP1617 42. Two identical balls P and Q moving in the x y plane collide at the origin (x = 0, y = 0) of the coordinate system. Their velocity components just before the moment of impact were, for ball P, v x = 6 m/s, v y = 0 ; for ball Q, v x = 5 m/s, v y = 2 m/s As a result of the collision, the ball P comes to rest. The velocity components of the ball Q just after collision will be: nks le:i xsans P rFkk Q, x y ry esa xfr djrs gq, ewy fcUnq (x = 0, y = 0) ij Vdjkrh gSA VDdj ds rqjUr igys buds osx ds ?kVd Øe'k% P ds fy, v x = 6 m/s, v y = 0 ,oa Q ds fy, v x = 5 m/s, v y = 2 m/s gSA VDdj ds dkj.k xsan P :d tkrh gSA VDdj ds rqjUr i'pkr~ xsan Q ds osx ds ?kVd gksxsaA (A) v y = 1 m/s (B) v y = 2 m/s (C) v x = 2 m/s (D) v x = 1 m/s SECTION - III ([k.M - III) Comprehension Type (c) cks/ku çdkj) This section contains 2 paragraphs. Based upon each paragraph, 2 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. bl [k.M esa 2 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 2 cgq&fodYih ç'u ds mÙkj nsus gSA çR;sd ç'u ds 4 fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA Paragraph for Question Nos. 43 to 44 iz'u 43 ls 44 ds fy, vuqPNsn Impending state of motion is a critical border line between static and dynamic states of a body. A block of mass m is supported on a rough vertical wall by applying a force F as shown in figure. Coefficient of static friction between block and wall is S. The block under the influence of F sin may have a tendency to move upward or it may be assumed that F sin just prevents downward fall of the block. Read the above passage carefully and answer the following questions. xfr dh lhekUr voLFkk ,d fi.M dh LFkSfrd rFkk xR;kRed fLFkfr;ksa ds chp ,d ØkfUrd lhek js[ kk gksrh gSA m nzO;eku ds ,d CykWd dks ,d ?k"kZ. k;qä Å/okZ/kj nhokj ij fn[kk;s fp=kkuql kj ,d cy F vkjksfir dj lgkjk fn;k x;k gSA CykWd rFkk nhokj ds chp LFkSfrd ?k"kZ. k xq. kkad S gSA F sinds vUrZxr CykWd dh Åij dh vksj xfr dh izo `fÙk gks ldrh gS ;k ;g ekuk tk ldrk gS fd F sin CykWd dks uhps fxjus ls Bhd jksdrk gSA mijksä vuqPNsn dks lko/kkuhiwo Zd if<+; s rFkk fuEu iz'uksa ds mÙkj nhft,A 43. The minimum value of force F required to keep the block stationary is : CykWd dks fLFkj j[kus ds fy, vko';d cy dk U;wure eku gksxk & mg (A) cos 44. mg (B) sin cos mg (C) sin cos mg (D) tan The value of F for which friction force between the block and the wall is zero. F dk eku] ftlds fy, CykW d rFkk nhokj ds chp ?k"kZ . k cy 'kw U; gS ] gks xk & mg mg mg (A) mg (B) (C) (D) sin tan cos Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 34 STP1617 Paragraph for Question Nos. 45 to 46 iz'u 45 ls 46 ds fy, vuqPNsn An ideal gas initially at pressure p 0 undergoes a free expansion (expansion against vacuum under adiabatic conditions) until its volume is 3 times its initial volume. The gas is next adiabatically compressed back to its original volume. The pressure after compression is 3 2/3 p0. ,d vkn'kZ xSl izkjfEHkd nkc p0 ls eqä izlkj (fuokZr esa :)ks"e fLFkfr;ksa esa izlkj) djrh gS tc rd fd bldk vk;ru] izkjfEHkd vk;ru dk 3 xquk gks tkrk gSA vkxs xSl dks :)ks"e :i ls okil ewy vk;ru rd laihfM+r djrs gSaA laihM+u ds ckn nkc 32/3 p0 gSA 45. The pressure of the gas after the free expansion is : eqä izl kj ds ckn xSl dk nkc gS & (A) 46. p0 3 1/ 3 (B) p 0 (C) p0 (D) 3p0 The gas : (A) is monoatomic. (C) is polyatomic. (B) is diatomic. (D) type is not possible to decide from the given information. xSl gS & (A) ,d ijek.kq d (C) cgq ijek.kq d (B) f} ijek.kq d (D) nh xbZ lw puk ds vk/kkj ij iz dkj ugha crk;k tk ldrkA PART - III (Hkkx - III) SECTION - I ([k.M- I) Straight Objective Type (lh/ks oLrqfu"B izdkj) This section contains 14 multiple choice questions. Each question has choices (A), (B), (C) and (D), out of which ONLY ONE is correct. bl [k.M esa 14 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA 47. 48. Which of the following oxidation number is not shown by Xenon (Xe)? fuEu esas ls dkSulk vkWDlhdj.k vad ftukWu (Xe) ugha n’'kkZrk gS\ (A) +5 (B) +4 (C) +6 (D) 0 The equilibrium constant for the following two reactions are K1 and K2 respectively. fuEu nks vfHkfØ;kvksa ds fy, lkE; fu;rkad Øe'k% K1 rFkk K2 gSaA Xe F6 (g) + H2O (g) XeOF4 (g) + 2HF (g) XeO4 (g) + XeF6 (g) XeOF4 (g) + XeO3F2(g) The equilibrium constant for the given reaction is : uhps nh xbZ vfHkfØ;k dk lkE; fu;rkad gS % XeO4 (g) + 2HF (g) XeO3F4 (g) + H2O (g) (A) K1 K22 (B) K1 –K2 (C) K2 / K1 (D) K2 / K2 49. Which of the following elements does not show +4 as most stable oxidation state? fuEu esa ls dkSulk@dkSuls rRo lokZf/kd LFkk;h vkWDlhdj.k voLFkk +4 n'kkZrk gS\ (A) Si (B) Ge (C) Sn (D) Pb 50. Bonds present in CuSO4. 5H2O(s) is (A) Electrovalent and covalent (C) Electrovalent, covalent and coordinate CuSO4. 5H2O(s) esa mifLFkr ca/k gS (A) fo|qr la;kstd o lgal;kstd (C) fo|qrl;kstd ] lgl;kstd o milgl;kstd (B) Electrovalent and coordinate (D) Covalent and coordinate (B) fo|qrla;kstd o milgl;kstd (D) lgal;kstd o milgla;kstd Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 35 STP1617 51. Which one of the following statements is correct in relation to ionization enthalpy ? (A) Electron-electron interaction in the outer orbitals of the elements increases the ionization enthalpy. (B) Removal of electron from orbitals bearing higher ‘n’ value is easier than from orbitals having lower ‘n’ value. (C) The ionization enthalpies of iso-electronic species, F–, Ne and Na+ are same because their size and electron configuration are same. (D) End of valence electrons is indicated by a small jump in ionization enthalpy. vk;uu ,UFkSYih ds lUnHkZ esa fuEu esa ls dkSulk dFku lgh gS\ (A) rRoksa ds ckáre d{kksa esa bysDVªkWu&bysDVªkWu vUr% fØ;k] vk;uu ,UFkSYih c<+krh gSA (B) mPp ‘n’ eku okys d{kdksa ls bysDVªkWu dks fudkyuk fuEu ‘n’ eku okys d{kdksa dh vis{kk ljy gksrk gSA (C) lebysDVªkWfud Lih'kht F–, Ne rFkk Na+ dh vk;uu ,UFkSYih dk eku leku gksrk gS D;ksafd budk vkdkj o bysDVªkWfud vfHkfoU;kl leku gksrk gSA (D) vfUre la;ksth bysDVªkWu vk;uu ÅtkZ esa NksVs vUrj dks crkrk gSA 52. For the reaction in acidic medium between KMnO4 and H2O2 the number of electrons transferred per mol of H2O2 are : (A) one (B) two (C) three (D) four KMnO4 o H2O2 ds chp vEyh; ek/;e esa vfHkfØ;k ds fy, H2O2 ds çfr eksy LFkkukUrfjr bysDVªkWuksa dh la[;k fuEu gS% (A) ,d (B) nks (C) rhu (D) pkj 53. The correct IUPAC name of the given compound is : (A) 5-Amino-3-bromo-4-chloro-2-hydroxycyclopentane-1-carbonitrile (B) 1-Amino-3-bromo-2-chloro-4-hydroxycyclopentane-5-carbonitirle (C) 2-Amino-4-bromo-3-chloro-5-hydroxycyclopentane-1-carbonitirle (D) 2-Amino-3-chloro-3-bromo-5-hydroxycyclopentane-1-nitrile fn, x;s ;kSfxd dk lgh IUPAC uke gS % (A) 5-,ehuks-3-czkseks-4-Dyksjks-2-gkbMªkWDlhlkbDyksisUVsu-1-dkcksZukbVªkby (B) 1-,ehuks-3-czkseks-2-Dyksjks-4-gkbMªkWDlhlkbDyksisUVsu-5-dkcksZukbVªkby (C) 2-,ehuks-4-czkseks-3-Dyksjks-5-gkbMªkWDlhlkbDyksisUVsu-1-dkcksZukbVªkby (D) 2-,ehuks-3-Dyksjks-3-czkseks-5-gkbMªkWDlhlkbDyksisUVsu-1-ukbVªkby 54. Name of some compounds are given. Which one is not as per IUPAC system ? CH3 | (A) CH3 – CH2 – CH2 – CH – CH – CH2 – CH3 3–Methyl–4–ethyl heptane | CH2CH3 (B) (C) (D) 3–Methyl–2–butanol 2–Ethyl–3–methyl–but–1–ene 4–Methyl–2–pentyne Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 36 STP1617 uhps dqN ;kSfxdksa ds uke fn;s x;s gSaA dkSulk uke IUPAC i)fr ds vuqlkj ugha gS\ CH3 | CH – CH – CH – CH – CH – CH2 – CH3 3–esfFky–4–,fFky gsIVsu (A) 3 2 2 | CH2CH3 (B) 3–esfFky–2–C;wVsukWy (C) 2–,fFky–3–esfFky–C;wV–1–bZu (D) 55. 4–esfFky–2–isUVkbu In photochemical reaction h R–H + X2 R–X + HX Where X = (F, Cl, Br, I) Rate of reaction of which halide is fastest (A) F2 (B) Cl2 (C) Br2 (D) I2 (C) Br2 (D) I2 izdk'k jklk;fud vfHkfØ;k esas h R–H + X2 R–X + HX tgk¡ X = (F, Cl, Br, I) dkSuls gSykbM dh vfHkfØ;k rhoz gksxh\ (A) F2 (B) Cl2 56. Which of the following atom(s) does not have maximum oxidation state as +7? fuEu esa ls dkSulk@dkSuls ijek.kq +7 vf/kdre vkWDlhdj.k voLFkk ugha j[krs gS\ (A) Cl (B) F (C) I (D) Br 57. Which of the following equations is a balanced one- fuEu esa ls ,d larqfyr lehdj.k dkSulh gS \ (A) 5BiO3– + 22H+ + Mn2+ 5Bi3+ + 7H2O + MnO4– (B) 5BiO3– + 14H+ + 2Mn2+ 5Bi3+ + 7H2O + 2MnO4– (C) 2BiO3– + 4H+ + Mn2+ 2Bi3+ + 2H2O + MnO4– (D) 6BiO3– + 12H+ + 3Mn2+ 6Bi3+ + 6H2O + 3MnO4– 58. Which of the following are polar ? fuEu esa ls dkSu /kzqoh; v.kq gS \ (A) XeF4 59. (B) SF6 (C) XeOF4 (D) XeF5– Match Column-I with Column-II and select the correct answer using the codes given below : Column-I (Metals) Column-II (Ores) (A) Tin (p) Calamine (B) Zinc (q) Cassiterite (C) Iron (r) Cerrusite (D) Lead (s) Siderite Codes : (A) (B) (C) (D) (A) (B) (C) (D) (A) p q r s (B) q p s r (C) s r q p (D) q p r s Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 37 STP1617 dkWye-I dks dkWye-II ds lkFk lqesfyr dhft;s dkWye-I (/kkrq) (A) fVu (B) ftad (C) vk;ju (D) ysM dksM : (A) p s (A) (C) 60. (B) q r (C) r q (D) s p rFkk fn;s x;s dksM ds vuqlkj lgh mÙkj nhft;sA dkWye-II (v;Ld) (p) dSykekbu (q) dSlhVsjkbV (r) ls:lkbV (s) flMsjkbV (A) q q (B) (D) (B) p p (C) s r (D) r s C6H12 (P) has two types of alkenes that can be reduced to one type of C6H14 (Q). Q is: C6H12 (P) nks izdkj dh ,Ydhu j[krk gS tks fd ,d izdkj ds C6H14 (Q) esa vipf;r gksrk gSA vr% Q gksxk % (A) (B) (C) (D) SECTION - II ([k.M - II) Multiple Correct Answers Type (cgqy lgh mÙkj izdkj) This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE THAN ONE is/are correct. bl [k.M esa 5 cgq lgh mÙkj izdkj ds iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls ,d ;k ,d ls vf/kd fodYi lgh gS ¼gSa½A 61. Which is/are correctly matched with its common name ? OH COOH NO2 O2N (A) COOCH3 : Picric acid (B) : Aspirin NO2 COOH COOH OH (C) : Salicylic acid (D) : Phthalic acid COOH fuEu esa ls dkSu&dkSu buds lkekU; uke ds lkFk lgh lqesfyr gSa \ OH COOH NO2 O2N (A) COOCH3 : fifØd vEy (B) : ,Lizhu NO2 COOH COOH OH (C) : lsfylhfyd vEy (D) : FkSfyd vEy COOH Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 38 STP1617 62. The pair of compounds having the same general formula. (A) and (B) (C) and (D) and and H—CC—CC—H fuEu esa ls dkSulk ;qXe leku lkekU; lw=k j[krk gSaA 63. (A) rFkk (B) (C) rFkk (D) rFkk rFkk H—CC—CC—H A gas cylinder containing cooking gas can withstand a pressure of 14.9 atmosphere. The pressure guaze of cylinder indicates 12 atmosphere at 27 ºC. Due to sudden fire in the building temperature starts rising. The temperature at which cylinder will explode is : [kkuk idkus okyh xSl dk flys.Mj 14.9 ok;qe.Myh; nkc lgu dj ldrk gSA 27ºC ij flys.Mj dk nkc xSt ¼nkcekih½ 12 ok;qe.My bafxr djrk gS, fcfYMax esa vpkud vkx yxus ls rkieku c<+ tkrk gS, og rki Kkr djsa ftl ij flys.Mj QV tk;sxk& (A) 372.5 K 64. (B) 99.5 ºC (C) 199 ºC (D) 472.5 k Which of the following is/are aromatic compounds : fuEu esas ls dkSulk@dkSuls ;kSfxd ,jkseSfVd ;kSfxd gS\ (A) 65. N (B) N | H (C) O (D) A hydrogen - like atom has ground state binding energy 122.4 eV. Then : (A) its atomic number is 3 (B) a photon of 90 eV can excite it to a higher state (C) a 80 eV photon cannot excite it to a higher state (D) None ,d gkbMªkstu leku ijek.kq dh vk| voLFkk dh cU/ku ÅtkZ 122.4 eV gSa rks & (A) bldk ijek.kq Øekad 3 gSA (B) 90 eV okyk QksVkWu bls mPp voLFkk esa mÙksftr dj ldrk gSA (C) 80 eV okyk QksVkWu bls mPp voLFkk esa mÙksftr ugha dj ldrk gSA (D) dksbZ ugah SECTION - III ([k.M - III) Comprehension Type (c) cks/ ku çdkj) This section contains 2 paragraphs. Based upon each paragraph, 2 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. bl [k.M esa 2 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 2 cgq&fodYih ç'u ds mÙkj nsus gSA çR;sd ç'u ds 4 fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 39 STP1617 Paragraph for Question Nos. 66 to 67 If hydrogen atoms (in the ground state) are passed through an homogeneous magnetic field, the beam is split into two parts. This interaction with the magnetic field shows that the atoms must have magnetic moment. However, the moment cannot be due to the orbital angular momentum since = 0. Hence one must assume existence of intrinsic angular momentum, which as the experiment shows, has only two permitted orientations. Spin of the electron produces angular momentum equal to s = s(s 1) h 1 where s = + . 2 2 n n or 2 2 where n is the number of unpaired electron. The substance which contains species with unpaired electrons in their orbitals behave as paramagnetic substances. The paramagnetism is expressed in terms of magnetic moment. The magnetic moment of an atom Total spin of an atom = n n eh 1 2 2 2mc s = s(s 1) eh = 2mc s = n (n 2) B.M. s= n 2 n – number of unpaired electrons eh 4mc If magnetic moment is zero the substance is di-magnetic. 1. B.M. (Bohr magneton) = iz'u 66 ls 67 ds fy, vuqPNsn ;fn gkbMªkstu ijek.kqvksa ¼vk| voLFkk esa½ dks leku pqEcdh; {ks=k ls xqtkjk tkrk gS rks iqUt nks Hkkxksa esa foHkkftr gks tkrk gSA pqEcdh; {ks=k ds lkFk vUrj fØ;k ;g crkrh gS fd ijek.kqvksa dk pqEcdh; vk?kw.kZ gksuk pkfg,A fQj Hkh vk?kw.kZ d{kh; pqEcdh; vk?kw.kZ ds dkj.k ugha gks ldrk D;ksafd = 0 gSA bl izdkj ,d vkUrfjd dks.kh; vk?kw.kZ dh mifLFkfr ekuh tkrh gSA ;g iz;ksxkRed :i ls iznf'kZr gksrk gS fd ;g nks rjg dk pØ.k j[krk gSA bysDVªkWu ds ?kw.kZu ls mRiUu dks.kh; laosx s = s(s 1) ,d ijek.kq dk dqy pØ.k = n 2 vFkok h 2 tgka s = + 1 gksrk gSA 2 n 2 tgka n v;qfXer bysDVªkWu dh la[;k gSA rRo ftlesa d{kd esa v;qfXer bysDVªkWu ik;s tkrs gSa og vuqpqEcdh; inkFkZ dh rjg O;ogkj djrk gSA pqEcdRo dks pqEcdh; vk?kw.kZ ds inksa esa O;Dr fd;k tkrk gSA ,d ijek.kq dk pqEcdh; vk?kw.kZ n n eh 1 2 2 2mc s = s(s 1) eh = 2mc s = n (n 2) B.M. s= n 2 n – v;qfXer bysDVªkWu dh la[;k 1. B.M. (cksjesXusVku) = eh 4mc ;fn inkFkZ dk pqEcdh; vk?kw.kZ 'kwU; gS rc rRo izfrpqEcdh; gSA 66. If an ion of 25Mn has a magnetic moment of 3.873 B.M. Then Mn is in which state. ;fn vk;u 25Mn dk pqEcdh; vk?kw.kZ 3.873 B.M. gS] rks Mn fdl vkWDlhdj.k voLFkk esa gksxk (A) + 2 (B) + 3 (C) + 4 (D) + 5 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 40 STP1617 67. Which of the following is a paramagnetic substance. fuEu esa ls dkSulk ,d vuqpqEcdh; inkFkZ gS \ (A) Mg2+ (B) Cu+ (C) Mn+7 (D) Ti+2 Paragraph for Question Nos. 68 to 69 The key concepts of resonance are : Resonance occurs because of the overlaping of orbitals. Double bonds are made up of pi bonds, formed from the overlap of 2p orbitals. The electrons in these pi orbitals will be spread over more than two atoms, and hence are delocalized. Both paired and unshared electrons may be delocalized, but all the electrons must be conjugated in a pi system. If the orbitals do not overlap (such as in orthogonal orbitals) the structures are not true resonance structures and do not mix. Molecules or species with resonance structures are generally considered to be more stable than those without them. The delocalization of the electrons lower the orbital energies, imparting this stability. The resonance in benzene gives rise to the property of aromaticity. The gain in stability is called the resonance energy. All resonance structures for the same molecule must have the same sigma framework (sigma bond form from the "head on" overlap of hybridized orbitals). Furthermore, they must be correct Lewis structures with the same number of electrons (and consequent charge) as well as the same number of unpaired electrons. Resonance structures with arbitrary separation of charge are unimportant, as are those with fewer covalent bonds. These unimportant resonance structures only contribute minimum (or not at all) to the overall. From the above theory of resonance answer the followings. iz'u 68 ls 69 ds fy, vuqPNsn vuqukn dh eq[; vo/kkj.kk fuEu gSa : vuqukn d{kdksa ds vfrO;kiu ds dkj.k gksrk gSA f}cU/k] cU/kksa ls cus gksrs gaSA tks fd 2p d{kdksa ds vfrO;kiu ls curs gSaA bu d{kdksa ds bysDVªkWu nks ;k vf/kd ijek.kqvksa ds e/; foLrkfjr gks tkrs gSa] bl izdkj foLFkkuhd`r gksrs gaSA ;qfXer rFkk v;qfXer nksuksa izdkj ds bysDVªkWu foLFkkuhd`r gks ldrs gSa fdUrq lHkh bysDVªkWu iz.kkyh esa la;qfXer gksuk pkfg,A ;fn d{kd vfrO;kiu ugha djrs gSa] (tSls fd vkFkksZxksuy d{kd½ rks lajpuk;sa lgh vuquknh lajpuk;sa ugh gksrh vkSj ijLij fefJr ugha gksrhA v.kq ,oa ,slh Lih'kht (species), ftudh vuquknh lajpuk;sa gksrh gSa lkekU;r% fcuk vuquknh lajpuk okys Lih'kht dh vis{kk vf/kd LFkk;h ekuh tkrh gSaA bysDVªkWuksa dk foLFkkuhdj.k d{kdksa dh ÅtkZ dks de dj nsrk gS tks iz.kkyh dks LFkkf;Ro iznku djrh gSSA cSathu esa vuqukn bls ,jkseSfVdrk dk xq.k iznku djrh gSaA LFkkf;Ro dh miyfC/k dks vuquknh ÅtkZ (resonance energy) dgrs gSaA ,d v.kq dh lHkh vuquknh lajpuk;sa leku flXek ÝseodZ (frame work) j[krh gS vkxs Hkh ;g dg ldrs gSa fd leku bysDVªkWu la[;k rFkk leku v;qfXer bysDVªkWu ds lkFk yqbZl lajpuk dk fuekZ.k gksrk gSA vuquknh lajpuk vkSj vkos'k dk izFkDdj.k egRoghu gSa] tSlk fd cgqr FkksM+s lgla;ksth ca/k esa gksrk gSA ;s egRoghu vuquknh lajpuk;as cgqr de ;ksxnku nsrh gaS ¼;k fcYdqy ugh½ mi;qZDr vuquknh fl)kUr ds vk/kkj ij fuEu iz'uksa ds mÙkj nhft,A 68. The correct resonating structure of 1, 3-butadiene is 1, 3-C;wVkMkbbZu dh lgh vuquknh lajpuk gS – (A) C H2 — C H — CH CH2 – (C) C H2 — CH CH — C H2 69. – (B) C H2 — C H — CH CH2 (D) None of these (buesa ls dksbZ ugha) Which resonating structure is NOT CORRECT? dkSulh vuquknh lajpuk lgh ugh gS\ (A) (C) C H2 — CH CH2 (B) (D) NH — C NH 2 2 | NH2 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 41 STP1617 ANSWER KEY TO SAMPLE TEST PAPER-II 1. 8. 15. 22. 29. 36. 43. (B) (C) (AC) (B) (C) (C) (B) 2. 9. 16. 23. 30. 37. 44. (B) (B) (ABD) (D) (B) (D) (B) 3. 10. 17. 24. 31. 38. 45. (D) (D) (AC) (C) (A) (ABD) (A) 4. 11. 18. 25. 32. 39. 46. (B) (B) (AB) (D) (C) (ABC) (A) 5. 12. 19. 26. 33. 40. 47. (A) (A) (ABC) (C) (D) (AB) (A) 50. 57. (C) (B) 51. 58. (B) (C) 52. 59. (B) (B) 53. 60. (C) (B) 54. 61. 64. (AD) 65. (AC) 66. (C) 67. (D) 68. 6. 13. 20. 27. 34. 41. 48. 7. 14. 21. 28. 35. 42. 49. (D) (C) (D) (B) (D) (BD) (D) (A) 55. (ACD) 62. (A) 56. (ABD) 63. (B) (AB) (C) (D) 69. (C) (D) (A) (A) (C) (ABD) (C) 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. Incorrect watch covers 1452 min in 1440 min. So, it will cover 1 min in min in 1440 min. Therefore it will cover 4840 1452 1440 × 4840 = 4800 min = 80 h. Therefore 80 h = 3 days and 8 h. 1452 Hindi. xyr pyrh ?kMh 1440 feuV dk le; 1452 feuV esa r; djrh gSA vr% ;g 1 feuV esa vr% ;g 4840 feuV dk le; 1440 1452 feuV dk le; r; djsxh 1440 × 4840 = 4800 feuV = 80 h. Therefore 80 h = 3 days and 8 h. 1452 3. 2576a456b is divisible by 5 b = 0 or 5 (2576a456b, 5 ls HkkT; gS b = 0 ;k 5) 2576a456b is divisible by 3 35 + a + b = 3, I (2576a456b, 3 ls HkkT; gS 35 + a + b = 3, I) a = 1, 4, 7, 2, 5, 8. 4. (12 – 22) + (32 – 42) + (52 – 62) + ....+ = (1 + 2) (1 – 2) + (3 + 4)(3 – 4) + (5 + 6) = – 3 – 7 – 11..........199 a = – 3, d = – 4 and = – 199 = a + (n – 1) d – 199 = – 3 + (n – 1)(– 4) – 199 + 3 = (n – 1)(– 4) 196 n–1= = 49 4 n = 50 50 S50 = [– 3– 199] 2 = 25 [– 202] = – 5050. (992 – 1002) (5 – 6) +.....+ (99 + 100)(99 – 100) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 42 STP1617 5. ax1 = bc, by1 = ca, cz1 = ab This ax = abc = by = cz = K Taking log, x log a = log K, .........(1) log K log K log K ,y ,z log a log b log c xy + yz + zx or, x 1 1 1 (log K) 2 (log a)(l og b) (log b)(log c) (log c)(log a) = z (log K) 2 [log a log b log c] log K log K log(abc) xy (log c ) log c log c (log a)(l og b)(log c) log a log b 1 6. cot = H= 3 B P = P2 B2 So, cos log c xy. z xyz log c H= ( 3 )2 (1)2 = 4 = 2. B 1 = = H 2 P 3 sin = H 2 and 1 4 So, = 3 2 sin 2 2 4 3 1 1 cos 2 7. |1 + x| + | 1– x| = 2 x [– 1, 1] So, infinite solutions vr% vuUr gy gSA 8. Since (0, 0) and (2 + 1, – 1) lie on the same side of x – 10y – 2 = 0 –2 (2 + 1 – 10 ( – 1) – 2) > 0 2 – 10 + 9 < 0 1<<9 least integral value of is 2 greatest integral value of is 8 28 A.M. = =5 2 9. In an equilateral triangle incentre and circumcentre are the same point. Incentre = (g, f) also, 12 + 12 + 2g + 2f + c = 0 c = 2(g + f + 1) also circumradius = 2 (inradius) 5 4 = 4 3 . 5 1 2 g f2 c 2 The equation of the incircle is inradius = 1 2 1 1 g f 2 c g 2 f 2 .2 g f 1 4 4 4 4(x 2 + y2) + 8gx + 8fy = (1 g) (1 + 3g) + (1 f) (1 + 3f) 2 x g y f 10. 2 The centre of the circle is (1, 1) and radius = 2 2 . Point (a, a) musty lie outside the circle, so 2a2 4a 6 > 0 a < 1 or a > 3. Now, tan 2 2 2 (a,a) . 2 2a 4a 6 2 2 As 3 6 2 2 2 2 2 1 3 a 2 2a 3 2 3 2a 4a 6 a2 2a 15 < 0 3 < a < 5 a (3, 1) (3, 5). Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 43 STP1617 11. Let the observations be x1, x2, . . . . . . . , x 20. 20 2 x – x i (S.(D))2 = Variance = i1 20 20 2 x – x 20 i i1 2 x – x = 5 or 20 = 100 i ...(i) i 1 New observations are 2x1, 2x2, . . . . . , 2x20. Let y and x be respectively the A.M.s of new and given observations. y 2x 1 2x 2 ..... 2 x 20 x x 2 ..... x 20 2 1 2x 20 20 20 20 20 2 2 2x – y 2x – 2x 4x i New variance = i i 1 20 i 1 20 2 i – x i 1 20 20 2 x – x i =4× i1 20 12. 13. 14. 100 = 4 = 20 20 [Using (i)] Required number of ways are = Exactly one toy is exchanged + exactly two toys exchanged + exactly 3 toys exchanged + 4 toys exchanged = 4C1 7C1 + 4C2 7C2 + 4C3 7C3 + 4C4 7C4 = 329 dqy rjhdks dh la[;k = ,d f[kykSuk cnyus ds rjhdksa dh la[;k + nks f[kykSus cnyus ds rjhdksa dh la[;k + rhu f[kykSus cnyus ds rjhdksa dh la[;k + pkj f[kykSus cnyus ds rjhdksa dh la[;kA = 4C1 7C1 + 4C2 7C2 + 4C3 7C3 + 4C4 7C4 = 329 1 P = 5C2 3 2 3 2 80 = 243 3 x 3 – 2x – 1 = 0 then (rks ) 3 – 2 – 1 = 0 Let (ekukfd) 1 y 1 .............(1) y 1 y 1 3 from equation (1), we get ( lehdj.k y 1 y 1 – 1 = 0 – 2 (1) ls ) y 1 y 1 y3 + 4y2 – y = 0 is the equation whose roots are (lehdj.k y3 + 4y2 – y = 0 ds ew y then (rks ) 1 1 1 , , 1 1 1 1 1 1 , , 1 1 1 gSaA) 1 1 1 =–4 1 1 1 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 44 STP1617 15. 1 nr 3 T r+1 = Cr ( 2 ) 2 r=n r n n n T n+1 = 1 2 n 1 1 Cn (3 2 )0 2 2 1 = 3 3. 9 = 3 16. 1 2 T 4+1 n – log3 8 5 log3 8 3 2–5 = log3 8 10 1 1 = 5 = 2 2 10 1 = 2 10 1 = = 10C 1 2 3 3 8–5/3 = n 4. ( 3 2 10 4 2) n = 10 1 2 4 = 10 1 C 4 . 2 2. 2 = 2 10C 4 z1 5 12i, z2 4 z1 iz2 z1 z2 13 4 17 z1 1 i z 2 z1 1 i z 2 13 4 2 z2 4 4 z2 4 1 5 z2 z2 z2 4 4 z2 4 1 3 z2 z2 17. min z1 1 i z 2 13 4 2 max z1 4 z2 z2 13 3 and min z1 4 z2 z2 13 5 2 . nC5 = nC4 + nC6 2(n ! ) n! n! = + 5! (n – 5)! 4! (n – 4)! 6 ! (n – 6)! 2 1 1 = + 5(n – 5) (n – 4)(n – 5) 6 5 12(n – 4) = n2 – 9n + 50 n2 – 21n + 98 = 0 (n – 7) (n – 14) = 0 n = 7 or ;k n = 14 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 45 STP1617 13 2 3 = 3 =1 tan B = 13 2 1 5 . 3 3 B = 45º 5 19. (A) (B) Slope of DC = – 3 2 DC line y + 4 = – (C) 3 (x – 5) 2 3x + 2y – 7 = 0 Since D is obtuse angled triangle so orthocentre lies out side the triangle 20 to 21. Sol. For Q.No. 20, 21 at x = – 3, y at x = – 2, y at x = 3, y at x = 4, y = = = = 8 4 4 2 20. Number of real solution of equation y = 0 is one 21. Number of real solution of equation y = 4 is infinite 22. HIndi 23. m C1 ; mC2 , mC3 are in A.P. m=7 m C1 ; mC2 , mC3 lekUrj Js<+h esa gSA m=7 x ( x 2 ) log10 3 T6 = 7C5 2log(10 3 ) . 2 = 21 log10(10 – 3x) + log103x – 2 = 0 log (10 – 3x) (3x – 2) = 0 (10 – 3x) 3x – 2 = 1 24. 10 1 . 3x – 32x = 1 9 9 3x = 1, 9 32x – 10 . 3x + 9 = 0 x = 0, 2 This simplified circuit is shown in the figure. bldk ljyhÑr ifjiFk fp=k esa fufnZ"V gSA 30 2V 30 30 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 46 STP1617 = 2V 30 60 i 20 2V = i Therefore, current vr% /kkjk i = 27. 2 1 = A 20 10 W = F . (r2 – r1) = 100 J 28. a1 = GM1M2 2 / M1 R acceleation of M1 w.r.t. M2 arel. = a1 + a2 = 30. 31. 36. G(M1 M2 ) R2 = a2 = GM R2 GM1M2 R2 / M2 . a m1 = am3 0 a m1 = – am2 0 Fm1 m3 Fm2 = 0 Given (fn;k gqvk gS) : m = 10 × 10–3 kg (fdxzk-) v = 300 ms–1 (eh-/ls-), t = .003 s(ls-) F= mv t F= 10 10 3 300 F = 1000 = 103 N .003 Let the particle strikes the inclined palne BC perpendicularly with a velocity v. Then ekuk d.k urry BC ls v osx ls yEcor~ Vdjkrk gS rc 2u v cos30º = u 2 cos45º 37. v= 3 . If centre of mass is at A nzO;eku dsUnz A ij gSA a 1 1 a2 2 = ab sin b sin 3 2 or b 13 a 4 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 47 STP1617 39. 40. Total power supplied dqy fn xbZ 'kfDr = 20 × 220 = 4400 Watt Already existing load tks yksM ij igys ls mifLFkr gS = 2000 + 1000 + 300 = 3300 Watt We can increase load upto 4400 Watt so A, B & C are correct options. ge yksM 4400 Watt rd c<+k ldrs gSA vr% A, B rFkk C lgh fodYi gSA F qV B If V = 0 then FB = 0 By right hand rule, the field line is circular. Magnetic force cannot do work It cannot change kinetic energy Magnetic force is perpendicular to the line joining current elements not a central force. F qV B ;fn V = 0 rc FB = 0 nka;s gkFk ds fu;e ls {ks=k js[kk o`Ùkkdkj gksxhA pqEcdh; cy] dk;Z ugh dj ldrk gSA ;g xfrt ÅtkZ ifjofrZr ugh dj ldrkA pqEcdh; cy /kkjk vo;o dks feykus okyh js[kk ds yEcor~ gSA dsfUnz; cy ugha gSA 41. (A) Pressing force ncko cy = mg = 12 g = 120 N (B) P = F 120 = = 20 Pa A 32 F 120 = = 40 Pa A 3 1 (D) Area of surface BCFG is minimum So, pressure will be maximum. lrg BCFG dk {kS=kQy U;qure gS vr% nkc vf/kdre gksxkA (C) P = 43. F sin + f = mg and rFkk Fcos = N for minimum U;w u re ds fy, ; f = N = Fcos mg F min. = sin cos 44. As pw a fd f = 0 45. In free expansion, temperature of the gas remains constant, therefore F sin = mg mg F= sin eqDr çlkj esa xSl dk rki fLFkj jgrk gSA vr% p0 v 0 = p. 3v 0 where tgkW v 0 = initial volume. izkjfEHkd vk;ru p= 46. p0 3 For adiabatic compression, initial conditions are p0 and 3v 0. Final volume and pressure arev 0 and 32/3 3 p0. :}ks"e laihMu ds fy, çkjfEHkd fLFkfr;k¡ p0 3 rFkk 3v 0 gSaA vfUre vk;ru rFkk nkc v 0 rFkk 32/3 p0. gSA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 48 STP1617 p0 .(3v 0) = 32/3 p0(v 0) 3 or –1= 2 3 3–1 = 32/3 = 5 3 i.e. gas is monoatomic xSl ,d ijek.kqd gSA 47. +5 is not shown by Xenon. ftukWu +5 vkWDlhdj.k vad ugha n'’kkZrk gSA 48. Substracting 1st reaction from second, we will get desired reaction 3rd. So K = K2/K1 f}rh; ls 1st dks dks ?kVkus ij ge bfPNr vfHkfØ;k çkIr djsxsaA 3rd. So K = K2/K1 51. As value of ‘n’ i.e. principal quantum number increases, the size of atom increases and the distance between valence electrons and nucleus increases. As a result, the attraction between the valence electrons and nucleus decreases. This fascilitates the easier removal of valence electrons. ‘n’ dk eku vFkkZr~ eq[; DokUVe la[;k esa o`f) ds lkFk ijek.kq ds vkdkj esa o`f) gksrh gS rFkk la;ksth bysDVªkWu o ukfHkd ds e/; nwjh c<+rh gSA ifj.kkeLo:i la;ksth bysDVªkWuksa o ukfHkd ds e/; vkd"kZ.k ?kVrk gS rFkk la;ksth bysDVªkWuks dks fudkyuk ljy gksrk gSA 52. Valency factor of H2O2 in both the medium is two. nksuksa ek/;e esa H2O2 dk la;kstdrk xq.kkad nks gSA 53. 2-Amino-4-bromo-3-chloro-5-hydroxycyclopentane-1-carbonitirle 2-,ehuks-4-czkseks-3-Dyksjks-5-gkbMªkWDlhlkbDyksisUVsu-1-dkcksZukbVªkby 54. CH3 | CH3 – CH2 – CH2 – CH – CH – CH2 – CH3 | CH2CH3 4–Ethyl –3–methyl–heptane CH3 | CH3 – CH2 – CH2 – CH – CH – CH2 – CH3 | CH2CH3 4–,fFky –3–esfFkygsIVsu Reduction 57. BiO3– + Mn2+ Bi3+ + MnO4– Oxidation (i) 2e + 6H+ + BiO3– Bi3+ + 3H2O (ii) 4H2O + Mn2+ MnO4– + 8H+ + 5e ––––––––––––––––––––––––––––––––––––––––– (i) × 5 + (ii) × 2 we get 14 H+ + 5 BiO3– + 5Mn2+ 5Bi3+ + 2MnO4– + 7 H2O is the correct balanced reaction Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 49 STP1617 vi p; u Sol. BiO3– + Mn2+ Bi3+ + MnO4– vkW Dl hdj .k (i) 2e + 6H+ + BiO3– Bi3+ + 3H2O (ii) 4H2O + Mn2+ MnO4– + 8H+ + 5e ––––––––––––––––––––––––––––––––––––––––– (i) × 5 + (ii) × 2 ge izkIr djrs gS 14 H+ + 5 BiO3– + 5Mn2+ 5Bi3+ + 2MnO4– + 7 H2O lgh larqfyr vfHkfØ;k gSA 58. Polarity depend on net dipole moment. If diple moment = 0 it is nonpolar. (A) XeF4 =0 non polar (B) SF6 is regular octahedral SF6 =0 non-polar (C) XeOF4 0 polar XeF5– non polar F F (D) Xe F 0 F F gy& /kqzoh;rk dqy f}/kqzo vk?kw.kZ ij fuHkZj djrh gSA ;fn f}/kqzo vk?kw.kZ = 0 gS rks ;g v/kqzoh; gSA (A) XeF4 =0 v/kqzoh; (B) SF6 v"VQydh; lajpuk gSA SF6 =0 v/kqzoh; (C) XeOF4 0 /kqzoh; XeF5– v/kqzoh; F F (D) Xe F 0 F F 59. (A) Tin-cassiterite (SnO2) (C) Iron - siderite (FeCO3) Therefore, (B) option is correct. (B) Zinc - calamine (ZnCO3) (D) Lead - Cerrusite (PbCO3) (A) fVu&dSlhVsjkbV (SnO2) (C) vk;ju&flMsjkbV (FeCO3) blfy;s (B) lgh fodYi gSA (B) ftad&dSykekbu (ZnCO3) (D) ysM&ls:lkbV (PbCO3) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 50 STP1617 60. 63. Suppose the cylinder will burst at T2K ekuk fd flys.Mj T2K rki ij QV tkrk gSA T2 P2T1 14 .9 300 = 372.5 K P1 (V1 = V2) = 12 65. Ground state binding energy = 13.6 Z2 = 122.4 eV. Z = 3. 1st excitation energy = 10.2 Z2 = 91.8 eV. an 80 eV electron cannot excite it to a higher state. vk| voLFkk esa cU/ku ÅtkZ = 13.6 Z2 = 122.4 eV. Z = 3. st 2 1 mÙkstu ÅtkZ = 10.2 Z = 91.8 eV. ,d 80 eV dk QksVkWu bls mPp voLFkk esa mÙksftr ugha dj ldrk gSA 66. Magnetic moment = 25 Mn n (n 2) = 3.873 – [Ar] 3d5 4s2 67. 25 Mn number of unpaired electron n = 3 therefore Mn should be in + 4 pqEcdh; vk?kw.kZ = n (n 2) = 3.873 – [Ar] 3d5 4s2 v;qfXer bysDVªkWuksa dh la[;k n = 3 bl izdkj Mn, + 4 esa gksxkA There is two unpaired electron in Ti 2+ ;gk¡ Ti 2+ esa nks v;qfXer bysDVªkWu gSaA 68. 69. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JF-PAGE # 51 STP1617 SAMPLE TEST PAPER-III (For Class-XII Appearing / Passed Students) Course : VISHESH (JD) & VIJAY (JR) Marks to be awarded S.No. Subject Nature of Questions Section No. of Questions Correct Wrong Total Section-I Straight Objective Type Questions (dsoy ,d fodYi lgh) 14 3 -1 42 Section-II Multiple Choice Questions (,d ;k ,d ls vf/kd fodYi lgh) 5 4 0 20 20 to 23 Section-III Comprehension ( vuqPNsn) (2 Comp. x 2 Q.) 4 3 -1 12 24 to 37 Section-I Straight Objective Type Questions (dsoy ,d fodYi lgh) 14 3 -1 42 Section-II Multiple Choice Questions (,d ;k ,d ls vf/kd fodYi lgh) 5 4 0 20 43 to 46 Section-III Comprehension ( vuqPNsn) (2 Comp. x 2 Q.) 4 3 -1 12 47 to 60 Section-I Straight Objective Type Questions (dsoy ,d fodYi lgh) 14 3 -1 42 PART-III 61 to 65 (Chemistry) Section-II Multiple Choice Questions (,d ;k ,d ls vf/kd fodYi lgh) 5 4 0 20 66 to 69 Section-III Comprehension ( vuqPNsn) (2 Comp. x 2 Q.) 4 3 -1 12 1 to 14 15 to 19 38 to 42 PART-I (Maths) PART-II (Physics) Total 69 222 PART - I (Hkkx - I) SECTION - I ([k.M- I) Straight Objective Type (lh/ks oLrqfu"B izdkj ) This section contains 14 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. bl [k.M esa 14 iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA 1. 2. 3. dy Let x 1 y – y 1 x = 0, where –1 < x < 0, for this relation find the sum of squares of all values of at dx 1 x=– . 2 dy 1 ekukfd x 1 y – y 1 x = 0, tgk¡ –1 < x < 0, bl lEcU/k ds fy, x = – ij dx ds lHkh ekuksa ds oxksZa dk ;ksx gS& 2 (A) – 8 (B) – 4 (C) 1 (D) – 1 The area bounded by the curve y = x3 – 4x and x-axis is oØ y = x3 – 4x rFkk x-v{k ls ifjc) {ks=kQy gS& (A) 4 (B) 8 (C) 16 (D) none of these (bueas ls dksbZ ugha) In a series of 5 matches in football (A team has equal chances to win, loose or draw a match). The probability that a forecast selected at random has exactly 2 correct predictions, is 5 Qq V cky es pks a dh ,d Ja [ kyk es a Bhd 2 es pks a ds iw o kZ u q eku lgh gks u s dh iz kf;drk (tcfd iz R;s d es p es a fdlh ny ds thrus] gkjus o esp Mªk gksus dh leku laHkkouk gS] gksxh& (A) 163 243 (B) 40 243 (C) 60 243 (D) 80 243 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 52 STP1617 4. c1 a1 b1 c If a 1 b 1 c 1, a 2b 2c 2 and a 3b 3c 3 are 3-digit ev en natural numbers and = 2 a 2 b2 , then is c 3 a3 b3 c1 a1 b1 c ;fn a 1 b 1 c 1, a 2b 2c 2 ,oa a 3b 3c 3 , 3- vadh; le izkd`r la[;k,¡ gS rFkk = 2 a 2 b2 , rks c 3 a3 b3 (A) divisible by 2 but not necessarily by 4 (B) divisible by 4 but not necessarily by 8 (C) divisible by 8 (D) none of these (A)2 ls HkkT; gS fdUrq 4 ls HkkT; gksuk vko';d ugha gSA (B) 4 ls HkkT; gS fdUrq 8 ls HkkT; gksuk vko';d ugha gSA (C) 8 ls HkkT; gSA (D) buesa ls dksbZ ugha 5. Domain of definition of the function f(x) = x – 1 – x 2 is Qyu f(x) = x – 1 – x 2 dk izkUr gSA 1 , 1 (A) 2 6. 1 1 , 1 (C) (B) 1, 2 2 If , & are the roots of the equation x 32x 1= 0 then, ;fn lehdj.k x 3 2x 1 = 0 ds ewy , ,oa gks] rks (A) zero ('kwU;) 7. 8. 1 1 , 2 2 (B) 1 1 1 (D) , , 2 2 1 1 1 + + has the value equal to 1 1 1 1 1 1 + + 1 1 1 (C) 4 If x, y > 0, x + y = 60, then greatest value of x(y – 30)2 is ;fn x, y > 0, x + y = 60 gS] rks x(y – 30)2 dk vf/kdre eku gS & (A) 4000 (B) 54000 (C) does not exist (A) 4000 (B) 54000 (C) fo|eku ugha gSA dk eku gS& (D) 1 (D) none of these (D) buesa ls dksbZ ugha Sita has 4 different and simmi has 7 different toys. Number of ways in which they can exchange their toys so that each keeps her inital number of toys, is lhrk 4 rFkk xhrk 7 fHkUu&fHkUu f[kykSus j[krh gS] rks mu rjhdksa dh la[;k D;k gksxhA tcfd nksuksa viuh izkjfHkd f[kykSuksa dh la[;k dks leku j[krs gq, vius f[kykSuksa dks ,d&nwljs ls cny ysaA (A) 329 9. (B) 334 (C) 332 (D) 345 The exhaustive range of values of ‘a’ such that the angle between the pair of tangents drawn from (a, a) to the circle x2 + y2 2x 2y 6 = 0 lies in the range , , is 3 ‘a’ ds ekuksa dk fu''ks"kh ifjlj gksxk tcfd fcUnq (a, a) ls o`Ùk x2 + y2 2x 2y 6 = 0 ij [khph xbZ Li'kZ js[kkvksa ds ;qXeksa ds e/; dks.k dk ifjlj 3 , esa fLFkr gS 10. (A) (1, ) (B) (5, 3) (3, 5) (C) (, 2 2) (2 2, ) (D) (3, 1) (3, 5) The equation of circumcircle of an equilateral triangle is x2 + y2 + 2gx + 2fy + c = 0 and one vertex of the triangle is (1, 1). The equation of incircle of the triangle is (A) 4(x2 + y2) = g2 + f 2 (B) 4(x2 + y2) + 8gx + 8fy = (1 g) (1 + 3g) + (1 f) (1 + 3f) (C) 4(x2 + y2) + 8gx + 8fy = g2 + f 2 (D) None of these Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 53 STP1617 leckgq f=kHkqt ds ifjxr o`Ùk dk lehdj.k x2 + y2 + 2gx + 2fy + c = 0 gS rFkk f=kHkqt dk ,d 'kh"kZ (1, 1) gSA rc f=kHkqt ds vUro`Ùk dk lehdj.k gS (A) 4(x2 + y2) = g2 + f 2 (B) 4(x2 + y2) + 8gx + 8fy = (1 g) (1 + 3g) + (1 f) (1 + 3f) (C) 4(x2 + y2) + 8gx + 8fy = g2 + f 2 (D) buesa ls dksbZ ugha 11. 12. The tangent lines to the circle x² + y² + 6x – 4y – 12 = 0 which are perpendicular to the line 4x + 3y + 5 = 0 are given by : o`Ùk x² + y² + 6x – 4y – 12 = 0 dh Li'kZ js[ kkvksa tks js[ kk 4x + 3y + 5 = 0 ds yEcor~ gks] ds lehdj.k gS& (A) 4x + 3y 7 = 0 , 4x + 3y + 15 = 0 (B) 4x + 3y 31 = 0 , 4x + 3y + 19 = 0 (C) 3x – 4y 42 = 0 , 3x – 4y – 8 = 0 (D) 3x – 4y + 8 = 0 , 3x + 4y + 21 = 0 x4 1 x 2 (x 4 x 2 1)1/ 2 (A) x4 1 x 2 (x 4 x 2 1)1/ 2 dx is equal to : x4 x2 1 +C x (B) x2 1 1 +C x2 (C) dx dk eku gS& x4 x2 1 x2 +C (D) x 4 x2 1 x4 13. Ram sets his watch at 6 : 10 am on Sunday, which gains 12 minutes in a day. On Wednesday if this watch is showing 2 : 50 pm. What is the correct time ? jke viuh ?kMh jfookj dks 6 : 10 am ij feykrk gS] tks fd ,d fnu esa 12 feuV vkxs gks tkrh gSA cq/kokj dks ;fn ;g ?kMh 2 : 50 pm fn[kk jgh gS] rks lgh le; D;k gksxk ? (A) 1 : 50 pm (B) 2 : 10 pm (C) 2 : 30 pm (D) 3 : 30 pm 14. A pair of coins is tossed a fixed number of times. If probability of getting both head in exactly 3 trails is same as the probability of getting both heads in exactly 4 trials, then number of trails equal to flDdks dk ,d ;qXe ,d fuf'pr la[;k ckj mNkyk tkrk gSA ;fn Bhd rhu mNkyksa esa nksuksa ij fpÙk vkus dh izkf;drk] Bhd pkj mNkyksa esa nksuksa ij fpÙk vkus dh izkf;drk cjkcj gks] rks dqN mNkyksa dh la[;k gksxh& (A) 7 (B) 15 (C) 21 (D) 14 SECTION - II ([k.M- II) Multiple Correct Answer Type (cgqy lgh fodYi izdkj) This section contains 5 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONE OR MORE is/are correct. bl [k.M esa 5 iz'u gSaA izR;sd iz'u ds mÙkj ds fy, 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls vf/kd lgh gSA 15. The normal y = mx – 2am – am3 to the parabola y2 = 4ax subtends a right angle at the vertex if ijoy; y2 = 4ax dk vfHkyEc y = mx – 2am – am3, 'kh"kZ ij ledks.k vUrfjr djrk gS] ;fn (A) m = 16. (B) m = – 3 3 (C) m = – 2 (D) m = 2 If z1 5 12i and z2 4 then (A) maximum z1 iz 2 17 z1 (C) minimum 4 z2 z2 13 4 (B) minimum z1 1 i z 2 13 4 2 z1 (D) maximum 4 z2 z2 13 3 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 54 STP1617 ;fn z1 5 12i vkSj z2 4 gks rc (A) vf/kdre z 1 iz 2 17 z1 (C) U;wu re 17. z2 4 z2 (B) U;w u re z 1 1 i z 2 13 4 2 z1 13 4 (D) vf/kdre z2 4 z2 13 3 The differential equation of the curve for which the ordinate at x = 0 of any tangent is equal to the corresponding subnormal (A) is linear (B) is homogeneous of first degree (C) is of the form variable separable (D) is of second order oØ] ftlds fy, fdlh Li'kZ js[kk dh x = 0 ij dksfV laxr v/kksyEc ds cjkcj gS] dk vody lehdj.k (A) js[kh; gSA (B) izFke ?kkr dk le?kkrh; gSA (C) pj i`FkDdj.kh; :i dk gSA (D) f}rh; dksfV dk gSA 18. If a is a positive integer, then the value of lim x3 x0 ;fn a ,d /kukRed iw.kk±d gS] rks lim 6(ax – sin ax ) x3 x0 (A) 1 19. 6(ax – sin ax ) (B) 8 2 may be equal to cjkcj gks ldrk gS & (C) 27 (D) 100 2 A point on the ellipse x + 3y = 37 where the normal is parallel to the line 6x – 5y = 2 is nh?kZo`Ùk x2 + 3y2 = 37 ij fcUnq ftl ij vfHkyEc] js[kk 6x – 5y = 2 ds lekUrj gS] gksxk& (A) (5, – 2) (B) (5, 2) (C) (– 5, 2) (D) (– 5, – 2) SECTION - III ([k.M - III) Comprehension Type (cks/ ku çdkj) This section contains 2 paragraphs. Based upon each paragraph, there are 2 questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. bl [k.M esa 2 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 2 ç'u gSA çR;sd ç'u ds 4 fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA Paragraph for Question Nos. 20 to 21 iz'u 20 ls 21 ds fy, vuqPNsn Equation of a curve is defined implicitly by the equation y3 – 3y + x = 0. The curve is passing through the point P 10 2, 2 2 . Then ,d oØ dh lehdj.k y3 – 3y + x = 0 }kjk iznf'kZr dh tkrh gSA ;g oØ fcUnq P 10 2, 2 2 ls xqtjrh gS] rks 20. Number of points where the curve has vertical tangents are mu fcUnqvksa dh la[;k ftu ij oØ dh Li'kZ js[kk,a m/okZ/kj gS& (A) 0 21. d2 y dx 2 (A) (B) 1 at the point P : 4 2 3 2 7 3 (C) 2 fcUnq P ij (B) 4 2 3 2 7 3 d2 y dx 2 (D) 3 : (C) 4 2 3 7 3 (D) 4 2 7 2.3 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 55 STP1617 Paragraph for Question Nos. 22 to 23 iz'u 22 ls 23 ds fy, vuqPNsn Let us say that a real number is divisible by two non–zero real numbers a and b, if it is divisible by the least among the common integral multiples of a and b. In case there is no common integral multiple of a and b, then say that a number is divisible by both a and b if it is divisible by the product ab. ekuk fd ,d okLrfod la[;k] nksa v'kwU; okLrfod la[;kvksa a o b ls HkkT; gS] ;fn ;g nksuks a o b ds U;wure mHk;fu"V iw.kk±d xq.kt ls HkkT; gksxhA ;fn a o b dk mHk;fu"V iw.kk±d xq.kt ugha gks] rks la[;k a o b nksuksa ls HkkT; gksxh ;fn ;g xq.kuQy ab ls HkkT; gksA 22. The number of integers in {1, 2, ....... 100}, which are divisible by 2 and 3 but not by 12, is (A) 8 (B) 9 (C) 10 (D) None of these {1, 2, ....... 100} esa iw.kkZadks dh la[;k tksfd 2 o 3 ls HkkT; gS ysfdu 12 ls ugha] gS& (A) 8 (B) 9 (C) 10 (D) buesa ls dksbZ ugha 23. The number of numbers in 2, 4 , 6 , 8 , ....... 200 , which are divisible by both 2 and (A) 5 (B) 10 (C) 25 (D) None of these 2, 4 , 6 , 8 , ....... 200 (A) 5 esa la[;kvksa dh la[;k tksfd (B) 10 (C) 25 2 8 , is o 8 nksauks ls HkkT; gS] gS& (D) buesa ls dksbZ ugha PART - II (Hkkx - II) SECTION - I ([k.M- I) Straight Objective Type (lh/ks oLrqfu"B izdkj) This section contains 14 multiple choice questions. Each question has choices (A), (B), (C) and (D), out of which ONLY ONE is correct. bl [k.M esa 14 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA 24. 25. The current i in the circuit of figure is fp=k ds ifjiFk esa /kkjk i gS & (A) 1 amp. 45 (B) 1 amp. 15 (C) 1 amp. 10 (D) 1 amp. 5 An electron enters a magnetic field at right angles to it, as shown in figure. The direction of force acting on the electron will be : ,d bysDVªkWu fp=kkuqlkj pqEcdh; {ks=k eas yEcor~ :i ls izos'k djrk gS rks bysDVªkWu ij vkjksfir cy dh fn'kk gksxh A (A) to the right (nka;h rjQ) (B) to the left (cka;h rjQ) (C) out of the page (D) into the page (dkxt ds ry ls ckgj dh rjQ) (dkxt ds ry esa vUnj dh rjQ) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 56 STP1617 Which of the following represents the displacement-time graph of two objects A and B moving with equal velocity ? fuEu esa ls dkSulk foLFkkiu le;&oØ leku osx ls xfreku nks oLrqvksa A rFkk B dks O;Dr djrk gS ? A B (B) B (C) Time Time (D) Time Displacement A Displacement (A) Displacement A Displacement 26. B Time 27. Which of the following is not possible for an object of mass 5kg which is released from height 15m. (Take g = 10 m/s2) KE = Kinetic Energy of the object, PE = Potential Energy) ,d oLrq dks 15 m ÅpkbZ ls eqDr fd;k tkrk gS oLrq dk nzO;eku 5kg gS rks fuEu esa ls dkSulk vlR; gSA (;gk¡ g = 10 m/s2) KE = oLrq dh xfrt ÅtkZ, PE = fLFkfrt ÅtkZ½ (A) KE = 600 J ; PE = 150 J (B) KE = 150 J ; PE = 600 J (C) KE = 0 J ; PE = 750 J (D) KE = 800 J ; PE = – 50 J 28. Two particles of combined mass M, placed in space with certain separation, are released. Interaction between the particles is only of gravitational nature and there is no external force present. Acceleration of one particle with respect to the other when separation between them is R, has a magnitude : nks d.k ftudk la;qDr nzO;eku M gS] vkdk'k esa fuf'pr nwjh ij j[ks gSa ] mudks NksM+k tkrk gSA d.kksa ds e/; dsoy xq:Rokd"kZ.k cy yxrk gS rFkk vU; dksbZ ckà; cy ugh yxrk gSA fdlh ,d d.k dk nwljs d.k ds lkis{k Roj.k dk ifjek.k] tc muds e/; R nwjh gS] gksxk & (A) GM (B) GM 2 (C) 2GM 2R R R2 (D) not possible to calculate due to lack of information (lwpukvksa 29. 2 dh deh ds dkj.k x.kuk lEHko ugh gSA) In the figure shown ABCD is a rectangular smooth tube kept fixed in a vertical plane. A particle is projected from point A to reach point C with some speed. At the corners B and D velocity changes its direction by 90º without any change of its magnitude at that corner. If time taken on paths ABC and ADC are t 1 and t 2 respectively, then: (given > b) fp=kkuqlkj ,d vk;rkdkj fpduh ufydk ABCD m/okZ/kj ry es tM+or gSA ,d d.k dks A ls C ds fy, fdlh pky ls ç{ksfir fd;k tkrk gSA dksus B o D ij dk osx fn'kk esa 90º ls cny tkrk gS] ijUrq eku leku jgrk gSA ;fn iFk ABC rFkk ADC es fy;k x;k le; Øe'k% t1 o t2 gS rks : (fn;k gS > b) (A) t 1 = t 2 (B) t 1 > t 2 (C) t 1 < t 2 (D) none of these (mijksDr esa lss dksbZ ugha) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 57 STP1617 30. Two point masses of mass m1 and m2 are placed at point A and B respectively as shown in figure. Point A is the centre of hollow sphere of uniformly distributed total mass m3. Consider only gravitational interaction between all masses and neglect other gravitational forces. Select the incorrect alternative. (A) Hollow sphere and point mass m1 moves with same acceleration. (B) m1 and m2 moves with same acceleration. (C) Net force on m1 is non-zero (D) Net force on hollow sphere and point mass m 1 as a system is equal to force experienced by point mass m2 in magnitude. m3 m1 A a m2 B r nks fcUnq nzO;eku ftuds nzO;eku m1 rFkk m2 gS] dks Øe'k% fcUnqvksa A rFkk B ij fp=kkuqlkj j[kk x;k gSA fcUnq A ,d [kks[kys xksys ftldk nzO;eku m3 ¼le:i forfjr½ gS] dk dsUnz gSA dsoy fn;s x;s nzO;eku ds e/; gh xq:okd"kZ.k gS vU; xq:Rokd"kZ.k cyksa dks ux.; ekfu;sA vlR; dFku pqfu;sA (A) [kks[kyk xksyk rFkk fcUnq nzO;eku m1 leku Roj.k ls xfr djsxsa (B) m1 rFkk m2 leku Roj.k ls xfreku gksxsa (C) m1 ij dqy cy v'kqU; gksxk (D) [kks[kys xksys rFkk m1 dks ,d fudk; ekurs gq;s bl ij yxus okys dqy cy dk ifjek.k m2 nzO;eku ij yxus okys dqy cy ds ifjek.k ds cjkcj gksxkA 31. A bullet of mass 10 g is fired with a rifle. The bullet takes 0.003 s to move through its barrel and leaves with a velocity of 300 ms–1. The force exerted on the bullet by the rifle : (A) 103 N (B) 104 N (C) 105 N (D) zero ,d cUnwd ls 10 xzke dh xksyh NksM+h tkrh gSA xksyh uyh ls xqtjus esa 0.003 lsd.M ysrh gS rFkk uyh ls 300 eh-/ls- ds osx ls fudyrh gSA cUnwd }kjk xksyh ij yxk;k x;k cy gS % (A) 103 U;wVu (B) 104 N U;wVu (C) 105 U;wVu (D) 'kwU; 32. A machine gun fires n bullets per second and the mass of each bullet is m. If the speed of bullet is v, then the magnitude of force exerted on the machine gun is : ,d e'khuxu çfr lSd.M n xksfy;k¡ nkxrh gS vkSj çR;sd xksyh dk æO;eku m gSA ;fn xksyh dh pky v gS] rks e'khu xu ij vkjksfir cy gksxk& (A) mng 33. (B) mnv g (C) mnv The number of electrons contained in the nucleus of 92U235. (A) 92 (B) 143 (C) 235 235 ukfHkd 92U esa bySDVªkWuksa dh la[;k gS & (A) 92 (B) 143 (C) 235 (D) mnvg (D) Zero (D) 'kwU; 34. An unnumbered clock shows time as 4 : 35 in its mirror image. The real time is : (A) 8 : 25 (B) 6 : 25 (C) 7 : 25 (D) None of these ,d fcuk vad fy[kh gqbZ ?kM+h dk niZ.k izfrfcEc 4 : 35 le; crkrk gSA rks lgh le; gS % (A) 8 : 25 (B) 6 : 25 (C) 7 : 25 (D) buesa ls dksbZ ugha 35. A transformer is used to (A) convert DC into AC (C) obtain the required DC voltage (B) convert AC into DC (D) obtain the required AC voltage VªkalQkeZj dk mi;ksx fd;k tkrk gSA (A) DC dks AC es cnyus esa (C) vfHk"B DC oksYVst izkIr djus ds fy, (B) AC dks DC es cnyus esa (D) vfHk"B AC oksYVst izkIr djus ds fy, Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 58 STP1617 36. Two non-conducting hemispherical surfaces, which are having uniform charge density are placed on smooth horizontal surface as shown in figure. Assuming springs are ideal, calculate compression in each spring if both the hemispherical surface are just touching each other. nks vpkyd v/kZ&xksyh; i`"B] ftu ij le vkos'k ?kuRo gS] dks fpdus {kSfrt ry ij fp=kkuqlkj j[kk gSA fLizaxks dks vkn'kZ ekurs gq,] izR;sd fLizax es laihMu Kkr dhft, ;fn nksuksa v/kZxksyh; i`"B ijLij ek=k laifdZr gSA (A) 37. 2 R2 2 0 K (B) R (C) 2 R 2 2 0 K (D) 2 R 2 2 0K 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. The magnetic field at the centre of the coil is zero tesla. The number of turns N in the coil are : N pDdj okyh ,d dq. Myh dh f=kT;k 0.22 m gS vkS j bles a fp=kkuql kj nf{k.kkoÙkZ (clockwise) fn'kk es a /kkjk çokfgr gks jgh gSA ,d yEcs lh/ks rkj tks dq. Myh ds fdukjs ls 0.05 m nwj h ij dq. Myh ds ry esa gS] esa ck;ha rjQ 54 A /kkjk çokfgr gks jgh gSA dq. Myh ds dsUnz ij pqEcdh; {ks=k 'kwU; gSA dq. Myh esa ?ksj ksa dh la[ ;k N gS & (A) 4 (B) 6 (C) 7 (D) 8 SECTION - II ([k.M - II) Multiple Correct Answers Type (cgqy lgh mÙkj izdkj) This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE THAN ONE is/are correct. bl [k.M esa 5 cgq lgh mÙkj izdkj ds iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls ,d ;k ,d ls vf/kd fodYi lgh gS ¼gSa½A 38. An ideal battery of 60 volt is connected with the resistors as shown above. A1 and A2 are two ideal ammeters and V is an ideal volt meter. Then : (A) Reading of (A1) is 2 Amp (B) Reading of (A2) is 1 Amp (C) Reading of (V) is 60 volt (D) Total power consumed by all the resistors is 120 watt 60 oks YV dh vkn'kZ cS V jh çnf'kZ r fp=kkuq l kj çfrjks /k ls tq M + h gS A ;fn A1 rFkk A2 nks vkn'kZ vehVj rFkk V vkn'kZ oksYVehVj gks rks : (A) (A1) dk ikB;ka d 2 Amp gS A (B) (A2) dk ikB;ka d 1 Amp gS A (C) (V) dk ikB;ka d 60 volt gS A (D) lHkh çfrjks /kks a es a dq y 'kfDr O;; 120 watt gS A Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 59 STP1617 39. A household electric power outlet (assume 220 V constant voltage) is fused to cut at if the current equals or exceeds 20 Ampere. A 2 kW heater, 1kW Air conditioner and three 100 W bulbs are already running at rated power. If now somebody wants to run a computer then computer can run without causing fuse to burn if power requirement of computer is (neglect losses in current carrying wire) ,d ?kjsyw fo|qr 'kfDr ifjiFk (220 V vpj ekusa) esa ;fn /kkjk 20 A ;k mlls vf/kd gks tk;s rks ¶;wt VwV tkrk gSA ,d 2 kW dk ghVj, 1kW dk ,;j df.M'kuj rFkk rhu 100 W ds cYc muds vafdr 'kfDr ij dk;Zjr gSA vc ;fn dksbZ ,d dEI;wVj pykuk pkgrk gS rFkk dEI;wVj ¶;wt tyk;s fcuk gh dk;Z dj jgk gS rks dEI;wVj dks vko';d 'kfDr D;k gksxh (/kkjkokgh rkj esa gkfu ux.; ysaos) (A) 1000 W 40. (B) 1100 W (C) 100 W (D) 1200 W Choose the correct statement(s) among the following : (A) The magnetic force on a stationary charge is always zero. (B) The magnetic line of force around a straight conductor is circular. (C) The magnetic force on a moving charge is responsible for change in its KE. (D) Magnetic force is a central force. fuEu esa ls lgh dFkuksa dk p;u dhft,& (A) fLFkj vkos'k ij pqEcdh; cy lnSo 'kwU; gksxkA (B) lh/ks pkyd ds pkjksa vksj pqEcdh; cy js[kk,a o`Ùkkdkj gksxh (C) xfreku vkos'k ij pqEcd cy bldh xfrt ÅtkZ esa ifjorZu ds fy, mÙkjnk;h gSA (D) pqEcdh; cy dsfUæ; cy gSA 41. A cuboid block of mass 12 kg is lying on the ground (Assume air is absent). Take g = 10 m/sec2. : 12 kg nzO;eku dk ?kukHk ds vkdkj dk ,d Bksl CykWd tehu ij j[kk gqvk gSA g dk eku = 10 m/sec2. ysaA (A) Pressing force applied by the block on the ground is 120 N. (B) If the surface ABCD is lying on the ground, then pressure (stress) exerted by the block on the ground will be 20 Pa. (C) If surface ABEF is lying on the ground, then the pressure (stress) exerted by the block on the ground will be 60 Pa. (D) If we place the block on the ground such that different plane surfaces lie on the ground, pressure (stress) on the ground will be maximum when surface BCFG lies on the ground. (A) CykWd }kjk tehu ij yxk;k x;k ncko cy 120 N gSA (B) ;fn CykWd dks lrg ABCD ds lgkjs tehu ij j[kk gS] rks CykWd }kjk tehu ij yxk;k x;k nkc (izfrcy) 20 Pa gksxkA (C) ;fn CykWd dks lrg ABEF ds lgkjs tehu ij j[kk gS] rks CykWd }kjk tehu ij yxus okyk nkc (izfrcy) 60 Pa gksxkA (D) ;fn ge CykWd dks tehu ij vyx-vyx lery lrgks ds lgkjs j[ks rks tehu ij nkc (izfrcy) vf/kdre rc gksxk tc CykWd dks lrg BCFG ds lgkjs tehu ij j[ksA 42. A conducting rod of length is moved at constant velocity ‘v 0’ on two parallel, conducting, smooth, fixed rails, that are placed in a uniform constant magnetic field B perpendicular to the plane of the rails as shown in figure. A resistance R is connected between the two ends of the rail. Then which of the following is/are correct : (A) The thermal power dissipated in the resistor is equal to rate of work done by external person pulling the rod. (B) If applied external force is doubled than a part of external power increases the velocity of rod. (C) Lenz’s Law is not satisfied if the rod is accelerated by external force (D) If resistance R is doubled then power required to maintain the constant velocity v 0 becomes half. R X x V0 B X ,d pkyd NM+ ftldh yEckbZ gS nks lekUrj ?k"kZ.k fofgu NM+ksa ds chp ‘v 0’ vpj osx ls py jgh gSA fp=kuqlkj tks fd vpj pqEcdh; {ks= B esa tks fd rails ds ry ds vfHkyEcor~ gS] esa j[kk x;k gSA rails dks ,d izfrjks/k R ls tksM+k x;k gSA rc fuEu esa ls dkSu lgh gS@gSa : (A) izfrjks/k esa Å"eh; 'kfDr cká O;fDr }kjk NM+ dks [khapus esa fd;s x;s dk;Z dh nj ds cjkcj gksxhA (B) vxj cká cy dks nqxuk fd;k tk, rc] cká 'kfDr dk dqN Hkkx NM+ dh pky dks c<+krk gSA (C) vxj cká cy }kjk NM+ dks Rofjr fd;k tk, rc ySat fu;e oS| ugha gSA (D) vxj izfrjks/k R dks nqxuk fd;k tk, rc NM+ dks vpj osx v 0 ij cuk;s j[kus dh pkgh xbZ 'kfDr nqxuh gks tk;sxhA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 60 STP1617 SECTION - III ([k.M - III) Comprehension Type (c) cks/ku çdkj) This section contains 2 paragraphs. Based upon each paragraph, 2 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. bl [k.M esa 2 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 2 cgq&fodYih ç'u ds mÙkj nsus gSA çR;sd ç'u ds 4 fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA Paragraph for Question Nos. 43 to 44 iz'u 43 ls 44 ds fy, vuqPNsn In a series L-R circuit, connected with a sinusoidal ac source, the maximum potential difference across L and R are respectively 3 volts and 4 volts. L-R Js.kh ifjiFk tks fd ,d T;koØh; izR;korhZ L=kksr ls tqM+k gS L o R ds fljksa ij vf/kdre foHkokUrj Øe'k% 3 oksYV o 4 oksYV gSA 43. At an instant the potential difference across resistor is 2 volts. The potential difference in volt, across the inductor at the same instant will be : tc izfrjks/k ds fljksa ij foHkokUrj 2 oksYV gS rc mlh {k.k izsj.k dq.Myh (inductor) ds fljksa ij foHkokUrj ¼oksYV esa½ D;k gksxk\ (A) 3 cos 30° 44. (B) 3 cos 60° (C) 6 cos 45° (D) 6 At the same instant, the magnitude of the potential difference in volt, across the ac source will be mlh leku {k.k ij izR;korhZ oksYVrk L=kksr ds fljksa ij foHkokUrj dk ifjek.k oksYV esa gksxkA (A) 3 cos 67° (B) 5 cos 83° (C) 6 cos 97° (D) 0 Paragraph for Question Nos. 45 to 46 iz'u 45 ls 46 ds fy, vuqPNsn An ideal gas initially at pressure p 0 undergoes a free expansion (expansion against vacuum under adiabatic conditions) until its volume is 3 times its initial volume. The gas is next adiabatically compressed back to its original volume. The pressure after compression is 3 2/3 p0. ,d vkn'kZ xSl izkjfEHkd nkc p0 ls eqä izlkj (fuokZr esa :)ks"e fLFkfr;ksa esa izlkj) djrh gS tc rd fd bldk vk;ru] izkjfEHkd vk;ru dk 3 xquk gks tkrk gSA vkxs xSl dks :)ks"e :i ls okil ewy vk;ru rd laihfM+r djrs gSaA laihM+u ds ckn nkc 32/3 p0 gSA 45. The pressure of the gas after the free expansion is : eqä izl kj ds ckn xSl dk nkc gS & (A) 46. p0 3 1/ 3 (B) p 0 (C) p0 (D) 3p0 The gas : (A) is monoatomic. (C) is polyatomic. (B) is diatomic. (D) type is not possible to decide from the given information. xSl gS & (A) ,d ijek.kq d (C) cgq ijek.kq d (B) f} ijek.kq d (D) nh xbZ lw puk ds vk/kkj ij iz dkj ugha crk;k tk ldrkA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 61 STP1617 PART - III (Hkkx - III) SECTION - I ([k.M- I) Straight Objective Type (lh/ks oLrqfu"B izdkj) This section contains 14 multiple choice questions. Each question has choices (A), (B), (C) and (D), out of which ONLY ONE is correct. bl [k.M esa 14 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA 47. A mixture of a definite composition of ethanol and water forms : (A) Maximum boiling azeotrope. (B) Minimum boiling azeoptrope. (C) Both maximum and minimum boiling azeotrope. (D) No azeotrope. ,FksukWy rFkk ty ds fuf'pr laxBu ls cuk feJ.k gS % (A) vf/kdre DoFkukad fLFkjDokFkh feJ.kA (B) U;wure DoFkukad fLFkjDokFkh feJ.kA (C) vf/kdre DoFkukad rFkk U;wure DoFkukad fLFkjDokFkh feJ.k nksuksaA (D) fLFkjDokFkh feJ.k ughaA 48. The equilibrium constant for the following two reactions are K1 and K2 respectively. fuEu nks vfHkfØ;kvksa ds fy, lkE; fu;rkad Øe'k% K1 rFkk K2 gSaA Xe F6 (g) + H2O (g) XeOF4 (g) + 2HF (g) XeO4 (g) + XeF6 (g) XeOF4 (g) + XeO3F2(g) The equilibrium constant for the given reaction is : uhps nh xbZ vfHkfØ;k dk lkE; fu;rkad gS % XeO4 (g) + 2HF (g) XeO3F4 (g) + H2O (g) (A) K1 K22 (B) K1 –K2 49. (C) K2 / K1 (D) K2 / K2 Structures of some common polymers are given. Which is not correctly presented ? (A) Teflon ( CF2 CF2 –)n CH CH2 CH2 (B) Neoprene CH2 C | (C) Terylene n Cl (D) Nylon-66 [ NH(CH2)6NHCO(CH2)4–CO–]n uhps dqN lkekU; cgqydksa dh lajpuk,sa nh xbZ gSaA fuEu esa ls dkSulk lgh lqesfyr ugh gS\ (A) Vs¶ykWu ( CF2 CF2 –)n CH CH2 CH2 (B) fu;ksizhu CH2 C | (C) Vsjhyhu n Cl (D) uk;ykWu-66 [ NH(CH2)6NHCO(CH2)4–CO–]n 50. 106.2 g 1 molal aqueous solution of ethylene glycol is cooled to –3.72ºC. Mass of ice separated during cooling is (Kf water = 1.86, freezing point of water = 0ºC) 106.2 g ,Fkkbyhu Xykbdksy dk 1 eksyy tyh; foy;u –3.72ºC rd B.Mk fd;k tkrk gSA B.Mk djus ds nkSjku cQZ dk fdruk nzO;eku i`Fkd gksxkA (ty dk Kf = 1.86, ty dk fgekad = 0ºC) (A) 25 g (B) 50 g (C) 60 g (D) 40 g 51. On heating calcium propionate, the product formed is (A) 3-Pentanone (B) 2-Pentanone (C) 3-Methyl-2-butanone (D) Propanone dSfY'k;e izksfivksusV dks xeZ djus ij] fufeZr mRikn gS % (A) 3-isUVsukWu (B) 2-isUVsukWu (C) 3-esfFky-2-C;wVsukWu (D) izksisukWu Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 62 STP1617 52. Nitrogen and oxygen exist as diatomic but their congeners are P4 and S8 respectively because : (A) phosphorus and sulphur are solids. (B) phosphorus and sulphur catenate due to the existence of d-orbitals and form strainless structures. (C) phosphorus and sulphur polymerise as soon as they are formed. (D) catenation tendency of P and S is stronger because of the high P – P and S – S bond energies as compared to N – N and O – O bond energies. ukbVªkstu rFkk vkWDlhtu] f}ijek.kqd :i esa vfLrRo j[krs gSa] ijUrq leku oxZ esa vxys rRo ¼congeners½ Øe'k% P4 rFkk S8 gSa] D;ksafd : (A) QkLQksjl rFkk lYQj Bksl gSaA (B) QkLQksjl rFkk lYQj] d-d{kdksa ds vfLrRo ds dkj.k J`¡[kfyr ¼J`a[kykc)½ gks tkrs gSa rFkk ruko jfgr lajpuk cukrs gSA (C) QkLQksjl rFkk lYQj fufeZr gksrs gh ¼tYnh ls tYnh½ cgqyhd`r gks tkrs gSaA (D) P rFkk S dh J`a[kyu izo`fÙk izcy gksrh gS] D;ksafd P – P rFkk S – S ca/k ÅtkZ;as] N – N rFkk O – O ca/k ÅtkZvksa dh rqyuk esa mPp gksrh gSA 53. In which of the following pair of elements +1 oxidation state is most stable? rRoksa ds fuEu esa ls dkSulk@dkSuls ;qXe esa +1 vkWDlhdj.k voLFkk lokZf/kd LFkk;h gS\ (A) Na, Ag (B) Na, Tl (C) K, Sc (D) K, Ca 54. Solution having osmotic pressure nearer to that of an equimolar solution of K4[Fe(CN)6] is: og foy;u ftldk ijklj.k nkc K4[Fe(CN)6] ds leeksyj foy;u ds yxHkx cjkcj gksxk %& (A) Na2SO4 (B) BaCl2 (C) Al2(SO4)3 (D) C12 H22O11 55. What is the most stable oxidation state lead (Pb)? ysM (Pb) ds fy, lokZf/kd LFkk;h vkWDlhdj.k voLFkk D;k gS\ (A) +4 (B) 0 (C) +6 (D) +2 56. In which of the following pair of elements +1 oxidation state is most stable? rRoksa ds fuEu esa ls dkSulk@dkSuls ;qXe esa +1 vkWDlhdj.k voLFkk lokZf/kd LFkk;h gS\ (A) Na, Ag (B) Na, Tl (C) K, Sc (D) K, Ca 57. Which of the following is correct IUPAC name of Acrolein? (A) But-2-enal (B) Propenal fuEu esa ls dkSulk ,Øksyhu dk lgh IUPAC uke gS\ (A) C;wV-2-busy (B) izksihusy (C) Propanal (D) Prop-2-yn-1-al (C) izksisusy (D) izksi-2-vkbZu-1-,y 58. For the reaction in acidic medium between KMnO4 and H2O2 the number of electrons transferred per mol of H2O2 are : (A) one (B) two (C) three (D) four KMnO4 o H2O2 ds chp vEyh; ek/;e esa vfHkfØ;k ds fy, H2O2 ds çfr eksy LFkkukUrfjr bysDVªkWuksa dh la[;k fuEu gS% (A) ,d (B) nks (C) rhu (D) pkj 59. Which of the following is the least stable carbanion ? fuEu esa ls U;wure LFkk;h dkcZ_.kk;u gS ? (A) HC C– (B) (C6H5)3C– (C) (CH3)3C– 60. (D) CH3– Identify the incorrect statement / statements : (i) Alkynes are more reactive than alkenes towards electrophilic addition reaction (ii) Alkynes are less reactive than alkenes towards electrophilic addition reaction (iii) Alkynes decolourise Br2 water (iv) Addition of HBr to alkynes in presence of peroxide proceeds via Markownikoff’s rule (A) (i) & (ii) (B) (ii) & (iii) (C) (i) & (iv) (D) (ii) & (iv) xyr dFkuksa dk p;u dhft,sA (i) bysDVªkWuLusgh ;ksxkRed vfHkfØ;k ds izfr ,Ydhu dh rqyuk esa ,Ydkbu vf/kd lfØ; gksrs gSA (ii) bysDVªkWuLusgh ;ksxkRed vfHkfØ;k ds izfr ,Ydhu dh rqyuk esa ,Ydkbu de lfØ; gksrs gSA (iii) ,Ydkbu czksehu ty dks jaxghu dj nsrs gSA (iv) ijkWDlkbM dh mifLFkfr esa ,Ydkbu ij HBr dk ;ksx, ekdksZfudkWQ fu;e ls gksrk gSA (A) (i) ,oa (ii) (B) (ii) ,oa (iii) (C) (i) ,oa (iv) (D) (ii) ,oa (iv) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 63 STP1617 SECTION - II ([k.M - II) Multiple Correct Answers Type (cgqy lgh mÙkj izdkj) This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE THAN ONE is/are correct. bl [k.M esa 5 cgq lgh mÙkj izdkj ds iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls ,d ;k ,d ls vf/kd fodYi lgh gS ¼gSa½A 61. Which is/are correct among the following ? Given : the half cell emf’s are E0 2 0.34 V, E0Cu | Cu 0.52 V . Cu | Cu (A) Cu+1 disproportionates (B) Cu and Cu2+ comproportionates 0 0 (C) ECu | Cu 2 ECu | Cu is positive (D) All of these fuEu esa ls dkSulk@dkSuls lgh gS@gSa \ fn;k x;k gS % v)Z lsy ds fo|qr okgd cy E0Cu2 |Cu 0.34 V, E0Cu |Cu 0.52 V gSaA 62. (A) Cu+1 dk fo"kekuqikrhdj.k gksrk gSA (B) Cu vkSj Cu2+ lekuqikrhr ¼comproportionate½ gks tkrs gSaA 0 0 (C) ECu | Cu 2 ECu | Cu /kukRed gSA (D) mijksDr lHkh Phenol is obtained by : fQukWy fuEu esa ls fdlds }kjk izkIr fd;k tkrk gS\ 63. (A) (B) (C) (D) In a AB unit cell (Rock salt type) assumig cations forming fcc lattice point (A) The nearest neighbours of A are 6B ion. (B) The nearest neighbours of B are 6A ion. (C) The second nearest neighbours of A are 12A (D) The packing fraction of AB crystal 3 . 8 ,d AB ,dd dksf"Bdk (jkWd lkYV izdkj) esa] ekukfd /kuk;u, fcc tkyd fcUnq cukrs gSA (A) A vk;u ds fudVre lehiorhZ 6B vk;u gSA (B) B vk;u ds fudVre lehiorhZ 6A vk;u gSA (C) A vk;u ds f}rh; fudVre lehiorhZ 12A gSA (D) AB fØLVy dh ladqyu n{krk 3 8 gSA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 64 STP1617 64. Which is/are correctly matched with its common name ? OH COOH NO2 O2N COOCH3 (A) : Picric acid (B) : Aspirin NO2 COOH COOH OH (C) : Salicylic acid (D) : Phthalic acid COOH fuEu esa ls dkSulk@dkSuls buds lkekU; ukeks ds lkFk lgh lqesfyr gS@gSa \ OH COOH NO2 O2N (A) COOCH3 : fifØd vEy (B) : ,Lizhu NO2 COOH COOH OH (C) : lsfyflfyd vEy (D) : FkSfyd vEy COOH 65. Which of the following is/are aromatic compounds : fuEu esas ls dkSulk@dkSuls ;kSfxd ,jkseSfVd ;kSfxd gS\ (A) N (B) N | H (C) O (D) SECTION - III ([k.M - III) Comprehension Type (c) cks/ ku çdkj) This section contains 2 paragraphs. Based upon each paragraph, 2 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. bl [k.M esa 2 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 2 cgq&fodYih ç'u ds mÙkj nsus gSA çR;sd ç'u ds 4 fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA Paragraph for Question Nos. 66 to 67 iz'u 66 ls 67 ds fy, vuqPNsn Alcohols undergo acid catalysed elimination reactions to produce alkenes. Because water is lost in the elimination , this reaction is called dehydration reaction. Secondary and tertiary alcohols always give E1 reaction in dehydration. Primary alcohols whose -carbon is branched also give E1 reaction. The reactivity of alcohol for elimination reaction is tertiary alcohol > Secondary alcohol > Primary alcohol. ,YdksgkWy vEy mRizsfjr foyksiu vfHkfØ;k }kjk ,Ydhu cukrs gSa] D;ksafd foyksiu vfHkfØ;k esa H2O ckgj fudyrk gS vr% ;g vfHkfØ;k futZyhdj.k dgykrh gSA f}rh;d vkSj r`rh;d ,YdksgkWy futZyhdj.k esa lnSo E1 vfHkfØ;k nsrs gSaA izkFkfed ,YdksgkWy ftldk -dkcZu 'kkf[kr gks oks Hkh E1 vfHkfØ;k nsrk gSA ,YdksgkWyksa dh foyksiu dh fØ;k'khyrk dk Øe r`rh;d ,YdksgkWy > f}rh;d ,YdksgkWy > izkFkfed ,YdksgkWy gSA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 65 STP1617 66. conc .H2SO 4 Alkenes In the given reaction : Total number of alkenes (Including stereo isomers) formed will be (A) Two (B) Six (C) Four l kUnzH2SO 4 nh xbZ vfHkfØ;k esa] (D) Five ,Ydhu cuus okyh dqy ,Ydhuksa ¼f=kfoe leko;oh dks lEefyr djrs gq,½ dh la[;k gksxh & (A) nks (B) N% (C) pkj (D) ik¡p 67. Which of the following dehydration product (major) is incorrect ? (A) (B) conc .H2SO 4 conc .H SO 2 4 conc .H SO 2 4 (C) CH3 CH2 CH2CH2OH CH3–CH=CH–CH3 (D) conc .H2SO 4 fuEu esa ls dkSulk futZyhdj.k mRikn ¼eq[;½ xyr gSa \ (A) (B) l kUnzH2SO 4 l kUnzH2SO 4 l kUnzH2SO 4 (C) CH3 CH2 CH2CH2OH CH3–CH=CH–CH3 (D) l kUnzH2SO 4 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 66 STP1617 Paragraph for Question Nos. 68 to 69 For any polyprotic acid, we always consider successive dissociation. The value of equilibrium constant of successive dissociation decreases due to common ion effect. For example : H2A is a dibasic acid. H2A H + HA [H ][HA ] K1 = [H2 A ] HA– H+ + A– – K2 = + – [H ][ A ] [HA ] K1 is greater than K2. iz'u 68 ls 69 ds fy, vuqPNsn fdlh cgqizksfVd vEy esa ge ges'kk Øekxr fo;kstu dk voyksdu djrs gSaA levk;u izHkko ds dkj.k Øekxr fo;kstu ds lkE; fu;rkad dk eku ?kVrk tkrk gSA mnkgj.k ds fy, : H2A ,d f}{kkjh; vEy gSA H2A H + HA [H ][HA ] K1 = [H2 A ] HA– H+ + A– – K2 = + – [H ][ A ] [HA ] K1dk eku K2 ls vf/kd gksrk gSA 68. Find the pH of 0.1 M Na2HAsO4. Use data : (For H3AsO4, pK1 = 5.6, pK2 = 9.6, pK3 = 13.6) (A) 7.6 (B) 9.6 (C) 11.6 (D) None of these 0.1 M Na2HAsO4 dk pH Kkr dhft;sA vk¡dM+s iz; qDr dhft;s % (H3AsO4 ds fy,, pK1 = 5.6, pK2 = 9.6, pK3 = 13.6) : (A) 7.6 69. (B) 9.6 (C) 11.6 (D) buesa ls dksbZ ugha Find the concentration of H+ ions in an aqueous solution which is saturated with H2S (0.1 M) as well as H2CO3 (0.2 M). Use data : [K1 = 10–7, K2 = 10–14 for H2S, K1 = 4 × 10–7, K2 = 4 × 10–11 for H2CO3] (A) 3 × 10–4 M (B) 3.83 × 10–4 M (C) 2.83 × 10–4 M (D) None of these H2S (0.1 M) rFkk (0.2 M) H2CO3 ds lkFk ,d lar`Ir tyh; foy;u esa H+ dh lkUnzrk Kkr dhft;sA vk¡dM+s iz; qDr dhft;s % [H2S ds fy;s K1 = 10–7, K2 = 10–14, H2CO3 ds fy;s K1 = 4 × 10–7, K2 = 4 × 10–11 ] : (A) 3 × 10–4 M (B) 3.83 × 10–4 M (C) 2.83 × 10–4 M (D) buesa ls dksbZ ugh Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 67 STP1617 ANSWER KEY TO SAMPLE TEST PAPER-III 1. (C) 2. (B) 3. (D) 4. (A) 5. (A) 6. (C) 7. (C) 8. (A) 9. (D) 10. (B) 11. (C) 12. (C) 13. (B) 14. (B) 15. (CD) 16. (ABD) 17. (AB) 18. (ABC) 19. (BD) 20. (C) 21. (B) 22. (A) 23. (A) 24. (C) 25. (D) 26. (C) 27. (D) 28. (B) 29. (C) 30. (B) 31. (A) 32. (C) 33. (D) 34. (C) 35. (D) 36. (C) 37. (A) 38. (ABD) 39. (ABC) 40. (AB) 41. (ABD) 42. (ABD) 43. (A) 44. (B) 45. (A) 46. (A) 47. (A) 48. (C) 49. (B) 50. (B) 51. (A) 52. (D) 53. (B) 54. (C) 55. (D) 56. (B) 57. (B) 58. (B) 59. (C) 60. (C) 61. (AC) 62. (ACD) 63. 68. (C) 69. (A) (ABC) HINTS & SOLUTION TO SAMPLE TEST PAPER-III 1. x 1 y – y 1 x = 0 x2 (1 + y) = y2(1 + x) and xy > 0 2 x – y2 + xy (x – y) = 0 and xy > 0 (x – y) (x + y + xy) = 0 and xy > 0 either x x = y or y = x 1 x x and 1 x 0 y 0 = y y 1 x xy 0 dy = 1, for all x (–1, 0) dx Hindi. x 1 y – y 1 x = 0 x2 (1 + y) = y2(1 + x) rFkk xy > 0 x 2 – y2 + xy (x – y) = 0 rFkk xy > 0 (x – y) (x + y + xy) = 0 rFkk xy > 0 ;k x 2. rks x = y ;k y = x 1 x x and 1 x 0 y 0 = y y 1 x dy = 1, lHkh x (–1, 0) ds dx xy 0 fy, The required area vHkh"V {ks=kQy Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 68 STP1617 0 = x 3 2 – 4 x dx – –2 x 3 – 4x dx 0 0 2 x4 x4 2 2 = 4 – 2x – 4 – 2x – 2 0 = – (4 – 8) – (4 – 8 ) = 8. 5C 1 3 2 3 2 80 = 3 243 3. P= 4. As a 1b 1c 1, a 2b 2c 2 and a 3b 3c 3 are even natural numbers, each of c 1,c 2,c 3 is divisible by 2. Let C i = 2k i for i = 1,2, 3 thus 2 k1 a1 b1 k = 2 2 a2 b2 = 2m, where m is some natural number.. k 3 a3 b3 Thus, is divisible by 2. m may not be divisible by 2 that we can see by taking the three number as 112, 122, 134 = 2 1 1 2 4 1 1 2 3 = 2 Hindi pwafd a 1b 1c 1, a 2b 2c 2 ,oa a 3b 3c 3 le izkd`r la[;k,¡ gSA c 1,c 2,c 3 esa ls izR;sd 2 ls HkkT; gSA ekukfd i = 1,2, 3 ds fy, C i = 2k i k1 a1 b1 k = 2 2 a2 b2 = 2m, tgk¡ m dksbZ izkd`r la[;k gSA k 3 a3 b3 vr% 5. = , 2 ls HkkT; gSA m, 2 ls HkkT; ugha gks ldrk gSA ge rhu la[;k,¡ tSls 112, 122, 134 ysdj ns[k ldrs gSA 2 1 1 2 4 1 1 2 3 = 2 2x 2 1 0 –1 x 1) squaring both the sides ( nks u ks a rjQ oxZ djus ij ) x 1 x2 x2 1 – x2 6. (1 – x 2 x 1 x 2 0 (x x2 1 2 x 0) 1 or x 2 1 2 1 x , 1 2 x 3 – 2x – 1 = 0 then (rks ) 3 – 2 – 1 = 0 Let (ekukfd) 1 y 1 .............(1) y 1 y 1 3 from equation (1), we get ( lehdj.k y 1 y 1 – 1 = 0 – 2 (1) ls ) y 1 y 1 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 69 STP1617 y3 + 4y2 – y = 0 is the equation whose roots are (lehdj.k y3 + 4y2 – y = 0 ds ew y 1 1 1 , , 1 1 1 gSaA) 1 1 1 =–4 1 1 1 then (rks ) 7. 1 1 1 , , 1 1 1 y = 60 – x x(y – 30)2 = x(60 – x – 30)2 = x(x – 30)2 Let f(x) = x(x – 30)2 f (x) = (x – 30)2 + 2x(x – 30) = (x – 30) (3x – 30) f (x) = 0 x = 10, 30 f(10) = 4000, f(30) = 0 lim f(x) = 0 x 0 lim f(x) = 54000 x 60 greatest value of x(y – 30)2 does not exist Hindi. y = 60 – x x(y – 30)2 = x(60 – x – 30)2 = x(x – 30)2 ekukfd f(x) = x(x – 30)2 f (x) = (x – 30)2 + 2x(x – 30) = (x – 30) (3x – 30) f (x) = 0 x = 10, 30 f(10) = 4000, f(30) = 0 lim f(x) = 0 x 0 lim f(x) = 54000 x 60 x(y – 30)2 dk vf/kdre eku fo|eku ugha gSA 8. Required number of ways are = Exactly one toy is exchanged + exactly two toys exchanged + exactly 3 toys exchanged + 4 toys exchanged = 4C1 7C1 + 4C2 7C2 + 4C3 7C3 + 4C4 7C4 = 329 Hindi. dqy rjhdks dh la[;k = ,d f[kykSuk cnyus ds rjhdksa dh la[;k + nks f[kykSus cnyus ds rjhdksa dh la[;k + rhu f[kykSus cnyus ds rjhdksa dh la[;k + pkj f[kykSus cnyus ds rjhdksa dh la[;kA = 4C1 7C1 + 4C2 7C2 + 4C3 7C3 + 4C4 7C4 = 329 9. The centre of the circle is (1, 1) and radius = 2 2 . Point (a, a) musty lie outside the circle, so 2a2 4a 6 > 0 a < 1 or a > 3. Now, tan 2 2 2 (a,a) . 2 2a 4a 6 2 2 As 3 6 2 2 2 2 2 2a 4a 6 10. 1 3 a 2 2a 3 2 3 a2 2a 15 < 0 3 < a < 5 a (3, 1) (3, 5). In an equilateral triangle incentre and circumcentre are the same point. Incentre = (g, f) also, 12 + 12 + 2g + 2f + c = 0 c = 2(g + f + 1) also circumradius = 2 (inradius) inradius = 1 2 g f2 c 2 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 70 STP1617 The equation of the incircle is 1 2 1 1 g f 2 c g 2 f 2 .2 g f 1 4 4 4 4(x 2 + y2) + 8gx + 8fy = (1 g) (1 + 3g) + (1 f) (1 + 3f) 2 x g y f x 12. 13. 2 1 x3 dx 1 2 subtitute 1 2 x 1 2 x x2 + 1 + 1 x2 = t2 Incorrect watch covers 1452 min in 1440 min. So, it will cover 1 min in 1440 min. Therefore it will cover 4840 1452 1440 × 4840 = 4800 min = 80 h. Therefore 80 h = 3 days and 8 h. 1452 min in Hindi. xyr pyrh ?kMh 1440 feuV dk le; 1452 feuV esa r; djrh gSA vr% ;g 1 feuV esa 1440 1452 feuV dk le; r; djsxh 1440 vr% ;g 4840 feuV dk le; 1452 × 4840 = 4800 feuV = 80 h. Therefore 80 h = 3 days and 8 h. 1 P (success in one trial ) 4 P(HH) 14. P(3 success ) P ( 4 success ) 3 n 1 3 C3 . 4 4 n3 4 1 3 n C 4 . 4 4 n 4 n = 15 1 P (,d iz;kl esa lQyrk) 4 P(3 lQyrk) = P(4 lQyrk) Hindi. P(HH) 3 n 1 3 C3 . 4 4 n3 4 1 3 n C 4 . 4 4 n 4 n = 15 15. Making y2 = 4ax homogeneous with the help of y = mx – 2am – am3 ........(i) y = mx – 2am – am3 ........(i) dh lgk;rk ls y2 = 4ax dk le?kkrh; cukrs gSA mx – y or 2am am 3 =1 mx – y then y2 = 4ax 3 2am am y2 = 4 x mx – y 2m m3 (2m + m3) y2 – 4mx2 + 4xy = 0 Angle between the lines represented by equation (i) is /2. Coefficient of x2 + coefficient of y2 = 0 lehdj.k (i) js[kkvks ds e?; dksa.k /2 iznf'kZr djrh gSA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 71 STP1617 x2 dk xq.kkad + y2 dk xq.kkad = 0 2m + m 3 – 4m = 0 m 3 – 2m = 0 m2 = 2, m 0 16. m=± 2 z1 5 12i, z2 4 z1 iz2 z1 z2 13 4 17 z1 1 i z 2 z1 1 i z 2 min z1 1 i z 2 13 4 2 17. z2 4 4 z2 4 1 5 z2 z2 z2 4 4 z2 4 1 3 z2 z2 13 4 2 z1 13 13 4 max and min 5 z2 4 3 z2 z2 z2 z1 If y = f(x) is the curve, Y – y = f (x) (X – x) is the equation of the tangent at (x, y) with f (x) = = 0, the ordinate of the tangent is y – x f (x). The subnormal at this point is given by y dy . Putting X dx dy , so we have y dx y dy dy dy =y–x = xy dx dx dx This is a homogeneous equation and, by rewriting it as dx xy x = = +1 dy y y dx x – =1 dy y we see that it is also a linear equation. Hindi dy ;fn y = f(x) oØ gSA Y – y = f (x) (X – x) fcUnq (x, y) ij izo.krk f (x) = dx okyh Li'kZ js[kk dk lehdj.k gSA X = 0 j[kus ij Li'kZ js[kk dh dksfV y – x f (x) gSA bl fcUnq ij v/kksyEc y vr% y dy dy =y–x dx dx dy dx }kjk fn;k tkrk gSA y dy = xy dx ;g ,d le?kkrh; lehdj.k gS rFkk bls fuEu rjg Hkh fy[k ldrs gS& dx xy x = = +1 dy y y dx x – =1 dy y ge ns[krs gS fd ;g ,d js[kkh; lehdj.k Hkh gSA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 72 STP1617 18. lim 6(a – a cos ax ) 3 x2 x0 lim x0 6a 2 sin ax = a3 3.2 x a3 = 1 i.e. a3 = 8 i.e. a3 = 27 i.e. a3 = 100 Now Hindi : lim 6(a – a cos ax ) 3 x2 x0 lim x0 a=1 a=2 a=3 i.e. a = (100)1/3 (not a positive integer) by L'Hospital's Rule. 6a 2 sin ax = a3 3.2 x vc 19. by L'Hospital's Rule. a3 = 1 a3 = 8 a3 = 27 a3 = 100 vFkkZr~ vFkkZr~ vFkkZr~ vFkkZr~ At (x1, y1) slope of tangent a=1 a=2 a=3 a = (100)1/3 (,d /kukRed iw.kk±d ugha gSA) dy x1 = ((x1, y1) ij dx 3 y1 Slope of normal (vfHkyEc dk >qdko ) = x1 = dy x 1 Li'kZ js[kk dk >qdko dx = ) 3 y1 3 y1 6 = x1 5 5y 1 2 25 y12 + 3y12 = 37 4 y12 = 4 y= 2 x= 5 (5,2), (–5, –2) dx 0 3 3y 2 0 y 1 dy 20. x = 3y – y3 21. dy 1 d2 y 2 y dy 2y . 2 2 2 2 2 dx 3(1 y ) 3 (1 y ) dx 3 (1 y 2 )3 dx 22. Hindi 23. LCM of 2 and 3 6 number of numbers divisible by 6 is 16 {6, 12, 18, ........96} out of which 8 are not divisible by 12 2 o 3 dk LCM 6 6 ls HkkT; la[;kvksa dh la[;k 16 gSA {6, 12, 18, ........96} esa ls 8 la[;k,sa 12 ls HkkT; ugha gSA LCM of i.e. 2 & 8 is 8 number of numbers divisible by both 2 & 8 is 5. 8, 2 8, 3 8, 4 8, 5 8 8 , 32 , 72 , 128 , 200 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 73 STP1617 Hindi: 2 o 8 dk LCM 8 gS 2 & 8 nksauks ls HkkT; la[;kvksa dh la[;k 5 gSA i.e. 8, 2 8, 3 8, 4 8, 5 8 8 , 32 , 72 , 128 , 200 24. This simplified circuit is shown in the figure. bldk ljyhÑr ifjiFk fp=k esa fufnZ"V gSA 30 30 2V 30 = 2V 30 60 i 20 2V = i Therefore, current vr% /kkjk i = 2 1 = A 20 10 28. a1 = GM1M2 / M1 R2 acceleation of M1 w.r.t. M2 arel. = a1 + a2 = 30. 31. 35. G(M1 M2 ) R 2 = GM R2 a2 = GM1M2 R2 / M2 . a m1 = am3 0 a m1 = – am2 0 Fm1 m3 Fm2 = 0 Given (fn;k gqvk gS) : m = 10 × 10–3 kg (fdxzk-) v = 300 ms–1 (eh-/ls-), t = .003 s(ls-) F= mv t F= 10 10 3 300 F = 1000 = 103 N .003 Obtain the required AC voltage vfHk"B AC oksYVst izkIr djus ds fy, Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 74 STP1617 36. 2 . R2 = kx 2 0 37. 0 .i1 N 0 i2 = 2(d r ) 2r N= i1 r . i2 (d r ) = 0.22 7 54 . =4 22(0.27) 3. 5 38. Reading of (A1) is 2 Amp Reading of (A2) is 1 Amp Total power consumed by all the resistors is 120 watt (A1) dk ikB;ka d 2 Amp gS A (A2) dk ikB;ka d 1 Amp gS A lHkh çfrjks/kksa esa dqy 'kfDr O;; 120 watt gSA 39. Total power supplied dqy fn xbZ 'kfDr = 20 × 220 = 4400 Watt Already existing load tks yksM ij igys ls mifLFkr gS = 2000 + 1000 + 300 = 3300 Watt We can increase load upto 4400 Watt so A, B & C are correct options. ge yksM 4400 Watt rd c<+k ldrs gSA vr% A, B rFkk C lgh fodYi gSA 40. F qV B If V = 0 then FB = 0 By right hand rule, the field line is circular. Magnetic force cannot do work It cannot change kinetic energy Magnetic force is perpendicular to the line joining current elements not a central force. F qV B ;fn V = 0 rc FB = 0 nka;s gkFk ds fu;e ls {ks=k js[kk o`Ùkkdkj gksxhA pqEcdh; cy] dk;Z ugh dj ldrk gSA ;g xfrt ÅtkZ ifjofrZr ugh dj ldrkA pqEcdh; cy /kkjk vo;o dks feykus okyh js[kk ds yEcor~ gSA dsfUnz; cy ugha gSA 41. (A) Pressing force ncko cy = mg = 12 g = 120 N (B) P = F 120 = = 20 Pa A 32 F 120 = = 40 Pa A 3 1 (D) Area of surface BCFG is minimum So, pressure will be maximum. lrg BCFG dk {kS=kQy U;qure gS vr% nkc vf/kdre gksxkA (C) P = 42. Rate of work done by external agent is : dw BL.dx = = BLv & thermal power dissipated in the resistor = e = (BvL) dt dt clearly both are equal, hence (A) . Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 75 STP1617 If applied external force is doubled, the rod will experience a net force and hence acceleration. As a result velocity increases, hence (B) e Since ; = R On doubling ‘R’, current and hence required power becomes half. Since, P = BLv Hence (D). cká dkjd }kjk dk;Z fd;s tkus dh nj gS % dw BL.dx = = BLv & Å"eh; dt dt ÅtkZ tks izfrjks/k esa {k; gqbZ gS = e = (BvL) nksuksa cjkcj gS blfy, (A) vxj yxk;k x;k cy nqxuk fd;k tk, rc NM+ ij ,d ifj.kkeh cy o Roj.k yxsxkA ifj.kkekuqlkj osx c<+rk gS] blfy, (B) pw¡fd ; = e R ‘R’, dks nqxuk djus ij] /kkjk vkSj pkgh xbZ 'kfDr vk/kh jgh tk,xhA blfy, , P = BLv blfy, (D). 43. Let at an instant v R = (VR)mcost ekuk fdlh {k.k v R = (VR)m cost 2 = 4 cost cost = ½ cost = 60°. Since D;ksafd VL is 90° ahead of VR v L = (VL)m cos (t + 90 ) = – 3 sint = – 3 sin 60° = – 3 cos 30° | (VL)m| = 3 cos 30° 44. From phasor diagram (VS)m = 2 ( VR )m ( VL )m 2 = 5 volt. dyk&dks.k fp=k ls(VS)m = ( VR )m 2 ( VL )m 2 = 5 volt. VL )m 3 tan = V ) 4 R m = 37° | v S | = | (VS)m cos (t + 37°)| = 5 | cos (60° + 37°)| = 5 | cos 97°| = 5 cos 83° 45. In free expansion, temperature of the gas remains constant, therefore eqDr çlkj esa xSl dk rki fLFkj jgrk gSA vr% p0 v 0 = p. 3v 0 where tgkW v 0 = initial volume. izkjfEHkd vk;ru p= 46. p0 3 For adiabatic compression, initial conditions are p0 and 3v 0 . Final volume and pressure are 3 v 0 and 32/3 p0. :}ks"e laihMu ds fy, çkjfEHkd fLFkfr;k¡ p0 .(3v 0) = 32/3 p0(v 0) 3 or –1= i.e. gas is monoatomic 2 3 p0 3 rFkk 3v 0 gSaA vfUre vk;ru rFkk nkc v 0 rFkk 32/3 p0. gSA 3–1 = 32/3 = 5 3 xSl ,d ijek.kqd gSA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 76 STP1617 47. Maximum boiling azeotrope is formed when Hmix is negative and attractive force molecules of liquids getting mixed is more. vf/kdre DoFkukad fLFkj DokFkh curk gS tc HfeJ.k _.kkRed gks rFkk nzo dk vkd"kZ.k cy v.kq vf/kd fefJr gksrk gSaA 48. Substracting 1st reaction from second, we will get desired reaction 3rd. So K = K2/K1 f}rh; ls 1st dks dks ?kVkus ij ge bfPNr vfHkfØ;k çkIr djsxsaA 3rd. So K = K2/K1 50. Mass of solute in 1 kg water = 62 g Mass of solute in 106.2 g solution = 3.72 = 1.86 x g y- 62 x 106.2 = 6.2 g 1062 6.2 x 1000 62 x w w = 50 g Mass of ice separated = 50 g 1 kg ty esa foys; dk nzO;eku = 62 g 106.2 g foy;u esa foys; dk nzO;eku = 3.72 = 1.86 x 62 x 106.2 = 6.2 g 1062 6.2 x 1000 62 x w w = 50 g izFk`d cQZ dk nzO;eku = 50 g 51. O O || || CH CH – C – O 3 2 2 Ca CH3 – CH2 – C – CH2 – CH3 + CaCO3 52. N and O have ability to form p-p multiple bonds with it self on account of smaller size of atoms. N – N and O – O bond energies are less on account of repulsion between non-bonded pairs of electrons due to smaller size of atoms. S – S bond energy (265 kJ mol–1) is next to C – C. N rFkk O esa] buds NksVs vkdkj ds dkj.k] Lo;a ds ijek.kq ds lkFk p-p ca/k fuekZ.k dh {kerk gksrh gSA NksVs ijek.kq vkdkj ds dkj.k] vcaf/kr bysDVªksu ds ;qXeksa ds e/; izfrd"kZ.k ds dkj.k] N – N rFkk O – O ca/k ÅtkZ;sa de gksrh gSA S–S ca/k ÅtkZ (265 kJ mol–1) C – C ds ckn vxyh ca/k ÅtkZ gSA 53. Most stable oxidation state of Na is +1 and most stable oxidation state of Tl is +1 because of inert pair effect. Na dh lokZf/kd LFkk;h vkWDlhdj.k voLFkk +1 gS rFkk Tl dh lokZf/kd LFkk;h vkWDlhdj.k voLFkk vfØ; ;qXe izHkko ds dkj.k +1 gksrh gSA 54. Osmotic pressure will be same for equimolar solutions if Van't Hoff factor is same. ;fn okWUVgkWQ xq.kkad leku gS rks leeksyj foy;uksa dh fy;s ijklj.k nkc Hkh leku gksxk K4[Fe(CN)6] Al2(SO4)3 56. i = 1 + (n–1) = 1 + 4 = 5 i = 1 + (n–1) = 1 + 4 = 5 Most stable oxidation state of Na is +1 and most stable oxidation state of Tl is +1 because of inert pair effect. vfØ; ;qXe izHkko ds dkj.k Na dh lokZf/kd LFkk;h vkWDlhdj.k voLFkk +1 rFkk Tl lokZf/kd LFkk;h vkWDlhdj.k voLFkk +1 gksrh gSA 58. Valency factor of H2O2 in both the medium is two. nksuksa ek/;e esa H2O2 dk la;kstdrk xq.kkad nks gSA 60. Conceptual (lS)kfUrd) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 77 STP1617 61. 2Cu+1 Cu+2 Cu + 2Cu+1 + 2e 2Cu Cu – 2e Cu+2 –––––––––––––––––––––– 2Cu+1 Cu+2 + Cu –––––––––––––––––––––– 62. E° 2 0.521 2( 0.337 ) = 0.184 2 (A) + Na2SO3 (C) 63. NaOH CaO NaOH CaO ( ) (B) + Mg (OH)Br fodkcksZ fDl y hdj .k decarboxyl ation NaNO HCl (D) HO 2 2 Based on FCC crystal, like NaCl lattice 4 4 4 r 3 4 r 3– A B 3 3 Packing fraction = = a3 4 4 a3 (0.414)3 3 (2 2)3 a3 4 4 a3 3 (2 2)3 = 0.79 FCC tkyd ij vk/kkfjr] NaCl tkyd dh rjg ladqyu n{krk 4 4 4 r 3 4 r 3– A B 3 3 = = a3 4 4 a3 (0.414)3 3 (2 2)3 a3 66. H 4 4 a3 3 (2 2)3 = 0.79 –H2O + + + 67. Most stable alkene (saytzeff alkene) is major product. vf/kd LFkkbZ ,Ydhu (lSRtSQ ,Ydhu) eq[; mRikn gksxkA 68. pH = 69. [H+] = pK 2 pK 3 9.6 13.6 = = 11.6 2 2 K 1C1 K 2C 2 = 10 7 0.1 4 10 7 0.2 = 3 × 10–4 M Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PLC024029 JD/JR-PAGE # 78 cim Sp e en Co p y Corporate Office: Resonance Edventures Ltd., A-46 & A-52, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Tel. No.: 0744-3192222, 3012222, 3022222, 2437144 | Fax : 022-39167222, 0744-2427144 | e-mail : [email protected]
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