Physics Past Paper Solutions PDF
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2024
ALLEN
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This document includes solutions to a physics exam paper, dated 15-09-2024, from the ALLEN board. The paper covers various concepts in physics, including magnetism and electromagnetism.
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15-09-2024 9610WMD801018240009 MD PHYSICS SECTION-A 1) A cylindrical bar magnet is kept along the axis of a circular coil. If the...
15-09-2024 9610WMD801018240009 MD PHYSICS SECTION-A 1) A cylindrical bar magnet is kept along the axis of a circular coil. If the magnet is rotated about its axis, then (1) A current will be induced in a coil (2) No current will be induced in a coil (3) Only an e.m.f. will be induced in the coil (4) An e.m.f. and a current both will be induced in the coil 2) An aluminium ring B faces an electromagnet A. The current I through A can be altered (1) Whether I increases or decreases, B will not experience any force (2) If I decrease, A will repel B (3) If I increases, A will attract B (4) If I increases, A will repel B 3) The flux linked with a circuit is given by ϕ = t3 + 3t – 7. The graph between time (x-axis) and induced emf (y-axis) will be (1) Straight line through the origin (2) Straight line with positive intercept (3) Straight line with negative intercept (4) Parabola not through the origin 4) The magnetic flux linked with a coil of N turns of area of cross section A held with its plane parallel to the field B is (1) (2) NAB (3) (4) 0 5) An electric potential difference will be induced between the ends of the conductor shown in the diagram, when the conductor moves in the direction (1) P (2) Q (3) L (4) M 6) A 50 mH coil carries a current of 2 ampere. The energy stored in joules is (1) 1 (2) 0.1 (3) 0.05 (4) 0.5 7) A wheel with ten metallic spokes each 0.50 m long is rotated with a speed of 120 rev/min in a plane normal to the magnetic field at the place. If the magnitude of the field is 0.4 Gauss, the induced e.m.f. between the axle and the rim of the wheel is equal to (1) (2) (3) (4) 8) The momentum in mechanics is expressed as m × v. The analogous expression in electricity is (1) I × Q (2) I × V (3) L × I (4) L × Q 9) A coil of N = 100 turns carries a current I = 5 A and creates a magnetic flux ϕ = 10–5 Tm–2 per turn. The value of its inductance L will be (1) 0.05 mH (2) 0.10 mH (3) 0.15 mH (4) 0.20 mH 10) What is the coefficient of mutual inductance when the magnetic flux changes by and change in current is 0.01A (1) 2 henry (2) 3 henry (3) (4) Zero 11) Find out the e.m.f. produced when the current changes from 0 to 1 A in 10 second, given L = 10 µH (1) 1 V (2) 1 µV (3) 1 mV (4) 0.1 V 12) There are two long co-axial solenoids of same length. The inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2 respectively. The ratio of mutual inductance to the self-inductance of the inner-coil is (1) (2) (3) (4) 13) Two conducting circular loops of radii R1 and R2 are placed in the same plane with their centres coinciding. If R1 >> R2, the mutual inductance M between them will be directly proportional to (1) R1/R2 (2) R2/R1 (3) (4) 14) Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be (1) Maximum in situation (A) (2) Maximum in situation (B) (3) Maximum in situation (C) (4) The same in all situations 15) A proton, deutron and an α - particle are accelerated by same potential, enter in uniform magnetic field perpendicularly. Ratio of radii of circular path respectively :- (1) (2) 2 : 2 : 1 (3) 1 : 2 : 1 (4) 1 : 1 : 1 16) A wire PQRS carrying a current I runs along three edges of a cube of side ℓ as shown. There exists a uniform magnetic field of magnitude B along one of the sides of the cube. The magnitude of the force acting on the wire is (1) 0 (2) IℓB (3) IℓB (4) 2IℓB 17) A uniform conducting wire ABC has a mass of 10gm. A current of 2A flows through it. The wire is kept in a uniform magnetic field B = 2T. The acceleration of the wire will be (1) Zero (2) 12 ms–2 along y-axis (3) 1.2 × 10–3 ms–2 along y-axis (4) 0.6 × 10–3 along y-axis 18) The net magnetic force on the loop ABCD due to the wire is (1) (2) Zero (3) (4) 19) Points A and B are situated perpendicular to the axis of a 2cm long bar magnet at large distances x and 3x from its centre on opposite sides. The ratio of the magnetic fields at A and B will be approximately equal to (1) 1 : 9 (2) 2 : 9 (3) 27 : 1 (4) 9 : 1 20) A square loop of side is kept in a uniform magnetic field B such that its plane makes an angle α with. The loop carries a current I. The torque experienced by the loop in this position is: (1) BI 2 (2) BI 2sinα (3) BI 2cosα (4) zero 21) A current carrying square loop is placed near an infinite long current carrying wire as shown in figure. The torque acting on loop is :- (1) (2) (3) (4) Zero 22) The network shown in the figure is a part of a complete circuit. If at a certain instant the current is 5 A and is decreasing at the rate of 103 A/s then VA–VB is:- (1) 5 V (2) 10 V (3) – 15 V (4) 20 V 23) A square of side ℓ meters lies in the x-y plane in a region, where the magnetic field is given by T. Where a,b,c are constant. The magnitude of flux in Wb passing through the square is. (1) ℓ2b (2) ℓ2a (3) ℓa2 (4) ℓ2c 24) The magnetic energy stored in a solenoid is terms of magnetic field B, area A and length ℓ of the solenoids (1) (2) (3) (4) 25) A conducting wire frame is placed in a magnetic field, which is directed into the paper. The magnetic field is increasing at a constant rate. The directions of induced currents in wires AB and CD are :- (1) A to B and C to D (2) B to A and C to D (3) A to B and D to C (4) B to A and D to C 26) A varying current in a coil changes from 10 amp to zero in 0.5 sec. If average EMF is induced in the coil is 220 volts, the self-inductance of coil is :- (1) 5 H (2) 10 H (3) 11 H (4) 12 H 27) When the current changes from +2 amp to –2 amp in 0.05 sec an emf of 8 volt is induced in a coil. The coefficient of self-inductance of the coil is:- (1) 0.1 H (2) 0.2 H (3) 0.4 H (4) 0.8 H 28) A semi circular loop of radius R is placed in a uniform magnetic field as shown. It is pulled with a constant velocity. The induced emf in the loop is : (1) (2) (3) (4) 29) Two coils of self inductances 9 mH and 4 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is :- (1) 10 mH (2) 6 mH (3) 4 mH (4) 16 mH 30) A metal disc rotates freely, between the poles of a magnet in the direction indicated. Brushes P and Q make contact with the edge of the disc and the metal axle. What current, if any, flows through R? (1) a current from P to Q (2) a current from Q to P (3) no current, because the emf in the disc is opposed by the back emf no current, because the emf induced in one side of the disc is opposed by the emf induced in the (4) other side. 31) A uniform but time-varying magnetic field B(t) exists in a circular region of radius a and is directed into the plane of the paper, as shown. The magnitude of the induced electric field at point P at a distance r from the centre of the circular region : (1) Is zero (2) Decreases as (3) Increases as r (4) Decreases as 32) In the inductive circuit given in the figure, the current rises after the switch is ON. At instant when the current is 0.25 A, then potential difference across the inductor will be : (1) Zero (2) 240 V (3) 100 V (4) 60 V 33) If a current is passed through a spring then the spring will (1) Expand (2) Compress (3) Remains same (4) None of these 34) For given arrangement (in horizontal plane) the possible direction of magnetic field: (1) Towards right (2) Towards left (3) Inside the plane (4) Outside the plane 35) A solenoid having 500 turns and length 2 m has radius of 2 cm, then self inductance of solenoid ? (1) 4 × 10–4 H (2) 2 × 10–4 H (3) 8 × 10–4 H (4) 16 × 10–4 H SECTION-B 1) A circular coil of radius R = 10cm having 500 turns and total resistance 2Ω is placed initially perpendicular to the magnetic field of 3 × 10–5 T. The coil is rotated about its vertical diameter by an angle 2π in 0.5 seconds. The induced current in the coil is (1) 0.5 mA (2) 1.0 mA (3) 1.5 mA (4) 3.0 mA 2) Figure shows a circuit that contains three identical resistors with resistance R = 9.0Ω each, two identical inductors with inductance L = 2.0 mH each, and an ideal battery with emf ε = 18 V. The current 'i' through the battery just after the switch closed is,...... :- (1) 0.2 A (2) 2 A (3) 0 ampere (4) 2 mA 3) A metallic rod with negligible resistance and length approximately equal to 0.5 m completes its circuit as shown in the figure. The circuit is normal to a magnetic field of B = 0.15 tesla. If the resistance is 3Ω the force required to move the rod with a constant velocity of 2 m/sec is- (1) 3.75 × 10–3 N (2) 3.75 × 10–2 N (3) 3.75 × 102 N (4) 3.75 × 10–4 N 4) A conducting rod AC of length 4l is rotated about a point O in a uniform magnetic field directed into the paper. AO = l and OC = 3l. Then (1) (2) (3) (4) 5) A circular coil having N turns and radius r carries a current I. It is held in the XZ plane in a magnetic field. The torque on the coil due to the magnetic field is : (1) (2) (3) Zero (4) 6) The figure shows three circuits with identical batteries, inductors, and resistors. Rank the circuits, in the decreasing order, according to the current through the battery (i) just after the switch is closed and (ii) a long time later (1) (i) i2 > i3 > i1 (i1 = 0) (ii) i2 > i3 > i1 (2) (i) i2 < i3 < i1 (i1 ≠ 0) (ii) i2 > i3 > i1 (3) (i) i2 = i3 = i1 (i1 = 0) (ii) i2 < i3 < i1 (4) (i) i2 = i3 > i1 (i1 ≠ 0) (ii) i2 > i3 > i1 7) A uniform magnetic field B and a uniform electric field E act in a common region. An electron is entering this region of space. The correct arrangement for it to escape undeviated is :- (1) (2) (3) (4) 8) A magnetised wire of moment M is bent into an arc of a circle subtending an angle of 60° at the centre; then the new magnetic moment is :- (1) (2) (3) (4) 9) A magnet is parallel to a uniform magnetic field. If it is rotated by 60°, the work done is 0.8 J. How much work is done in moving it 30° further (1) 0.8 × 107 ergs (2) 0.4 J (3) 8 J (4) 0.8 ergs 10) Two parallel, long wires carry currents i1 and i2 with i1 > i2. When the current are in the same direction, the magnetic field at a point midway between the wires is 10 mT. If the direction of i2 is reversed, the field becomes 30mT. The ratio i1 /i2 is (1) 4 (2) 3 (3) 2 (4) 1 11) A magnetic field of 2 × 10–2 T acts at right angles to a coil of area 100 cm2 with 50 turns. The average emf induced in the coil is 0.1V, when it is removed from the field in time t. The value of t is :- (1) 0.1 sec (2) 0.01 sec (3) 1 sec (4) 20 sec 12) A toroidal solenoid with an air core has an average radius of 15 cm, area of cross-section 12 cm2 and 1200 turns. Ignoring the field variation across the cross-section of the toroid, the self-inductance of the toroid is :- (1) 4.6 mH (2) 6.9 mH (3) 2.3 mH (4) 9.2 mH 13) Two coils have a mutual inductance M = 5×10–3H. The current changes in first coil according to equation I = I cos ωt where I = 2A and ω = 100 π 0 0. The maximum value of induced emf. in the second coil is:- (1) 4π volt (2) 3π volt (3) 2π volt (4) π volt 14) A metallic rod of length is tied to a string of length and made to rotate with angular speed ω on a horizontal table with one end of the string fixed. If there is a vertical magnetic field B in the region, the emf induced across the ends of the rod is :- (1) (2) (3) (4) 15) An equilaterial triangular loop having a resistance R and length of each side 'ℓ' is placed in a magnetic field which is varying at T/s. The induced current in the loop will be :- (1) (2) (3) (4) CHEMISTRY SECTION-A 1) In the following compounds the decreasing order of B.P. is (1) (I) > (II) > (III) (2) (I) > (III) > (II) (3) (II) > (III) > (I) (4) (III) > (II) > (I) 2) Addition of HBr to 2-methyl-1-propene in the presence of hydrogen-peroxide produces (1) 1-Bromobutane (2) 2-Bromopropane (3) 2-Bromo-2-methylpropane (4) 1-Bromo-2-methylpropane 3) C6H6 + Z Toluene, Z is (1) Acetic acid (2) Acetic anhydride (3) Acetone (4) Chloromethane 4) In Kharasch effect, reaction follows (1) Free radical substitution (2) Electrophilic addition (3) Free radical addition (4) Nucleophilic addition 5) The catalyst used to reduce an alkyne to alkene is (1) Raney Nickel (2) Palladium (3) Lindlar’s catalyst (4) Iron 6) Benzene undergoes substitution reaction more easily than addition because (1) It has a cyclic structure (2) It has three double bonds (3) Aromatic nature of ring will loose (4) Delocalization of π bonds 7) The electrophile which attacks in Friedel-Craft acylation is (1) (2) (3) (4) Both (1) & (3) 8) Which of the following shows geometrical isomerism ? (1) But-1-ene (2) But-2-ene (3) Prop-1-ene (4) Pent-1-ene 9) ; The compound “A” is (1) (2) (3) (4) 10) In Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorbenzene (I) (II) (III) The decreasing order of reactivity towards electrophilic substitution reaction is (1) (I) > (II) > (III) (2) (I) > (III) > (II) (3) (II) > (I) > (III) (4) (III) > (I) > (II) 11) The configuration of the compound is (1) E (2) R (3) S (4) Z 12) The number of stereoisomers possible for 2–bromo–3–chloro butane is :- (1) 2 (2) 1 (3) 4 (4) 3 13) Which of the following compound can exist in meso form– (1) 1, 2–Dichlorobutane (2) 2, 3–Dichloropentane (3) 2, 3–Dichloro butane (4) 1, 2–Dichloropentane 14) Total stereoisomers of compound 'x' is : (1) 2 (2) 8 (3) 6 (4) 4 15) Following structure of tartaric acid represents : (1) dextrorotatory form (2) Laevorotatory form (3) meso form (4) trans form 16) Which of the following compound has ‘S’ configuration ? (1) (2) (3) (4) 17) Number of chiral carbon present in the following compound : (1) 2 (2) 3 (3) 4 (4) 5 18) Compound will show (1) Geometrical isomerism (2) Optical isomerism (3) Both 1 and 2 (4) Double bond equivalent = 2 19) Which of the following is meso compound : - (1) (2) (3) (4) 20) Rate of SN1 reaction is :- (1) S > Q > R > P (2) S > R > P > Q (3) P > Q > R > S (4) S > R > Q > P 21) Which of the following species is most reactive in an SN2 reaction ? (1) CH3CH2–Cl (2) CH3CH2–Br (3) CH3CH2–I (4) CH3CH2–F 22) Which of the following is minimum reactive towards SN2 reaction? (1) CH3—CH2—Cl (2) (3) (4) 23) Order of nucleophilicity in polar protic solvent :- (1) IΘ > BrΘ > ClΘ > FΘ (2) FΘ > ClΘ > BrΘ > IΘ (3) IΘ > ClΘ > BrΘ > FΘ (4) FΘ > IΘ > BrΘ > ClΘ 24) Which one is most reactive towards SN1 reaction ? (1) (2) (3) (4) 25) Which is most reactive for SN1 reaction :- (1) (2) (3) (4) 26) This reaction would follow which of the following pathway predominantly ? (1) SN1 (2) SN2 (3) E1 (4) E2 27) Which of the following is polar aprotic solvent? (1) N, N–dimethyl foramide (DMF) (2) Water (3) Ethanol (4) Methanol 28) Which is Finkelstein reaction ? (1) R–X + NaI (2) R–X + AgF (3) R–X + NaF → (4) R–F + AgCl → 29) Correct order of leaving group tendency is :- (1) I⊝ > Br⊝ > Cl⊝ > F⊝ (2) F⊝ > Cl⊝ > Br⊝ > I⊝ (3) Cl⊝ > F⊝ > Br⊝ > I⊝ (4) I⊝ > Cl⊝ > Br⊝ > F⊝ 30) (P) ; Major product (P) is : (1) (2) (3) (4) 31) (1) (2) (3) (4) 32) Select the correct statement about SN2 reaction (1) Molecularity of reaction is 2 (2) Order of reaction is 2 (3) On increasing concentration of nucleophile rate of reaction increases (4) All of these 33) Chloroform on reaction with acetone gives :– (1) Acetylene (2) Chloretone (3) Nitrochloroform (4) Chloroacetone 34) Which of the following undergoes nucleophilic substitution by SN mechanism at fastest rate : 1 (1) CH3–CH2–Cl (2) (3) (4) 35) Which gives SN2 product as a major product ? CH3—CH2—CH2—Cl + NaI (1) (2) (3) (4) SECTION-B 1) Addition of HBr/Peroxide to propene (1) Follows Markovnikov’s rule (2) Does not follow Markovnikov’s rule (3) Follows Markovnikov’s rule but the product rearranges to give anti-Markovnikov’s product (4) Follows free radical substitution mechanism. 2) Compound “A” is (1) Propyl benzene (2) Isopropyl benzene (3) Biphenyl (4) None of above 3) In sulphonation of benzene; electrophilic reagent used is (1) Cl (2) SO3 (3) NO2 (4) 4) Which one is o, p-directing group for electrophilic substitution reaction ? (1) (2) (3) (4) –NO2 5) Toluene Y. Here Y is (1) Benzaldehyde (2) Toluene (3) Benzoic acid (4) Ethylbenzene 6) (1) (2) (3) (4) 7) Consider the following two structure Choose the correct statement (1) Both (i) & (ii) have non-conjugated system (2) (i) & (ii) both are non aromatic (3) (i) & (ii) both are aromatic (4) (ii) is less stable than (i) 8) In above reaction product (B) is (1) (2) (3) (4) 9) Number of chiral center present in above compounds are :- (1) 0, 1 (2) 0, 2 (3) 0, 3 (4) 1, 2 10) Which are not diastereomers :- (1) (2) (3) (4) All of these 11) Which of the following does not contain the plane of symmetry : (1) (2) (3) (4) All of these 12) Relation between given compound is (1) Homomer (2) Enantiomer (3) Functional isomer (4) Chain isomer 13) Stereoisomers which are not mirror image of each other are (1) Enantiomers (2) Diastereomers (3) Meso compounds (4) None of these 14) Which have centre of symmetry and plane of symmetry both :- (1) (2) (3) (4) All of these 15) Correct match is Column-I Column-II (P) Compound of which mirror image is super imposable (a) Enantiomers (Q) Stereo isomers which are non-super imposable mirror images (b) Meso (R) Stereo isomers which are not mirror images (c) Diastereomers (1) P → a ; Q → b ; R → c (2) P → b ; Q → a ; R → c (3) P → b ; Q → c ; R → a (4) P → c ; Q → a ; R → b BIOLOGY-I SECTION-A 1) A and B genes are linked. What shall be genotype of progeny in a cross between AB/ab and ab/ab ? (1) AAbb and aabb (2) AaBb and aabb (3) AABB and aabb (4) None 2) What proportion of offsprings of the cross AABBcc × AaBbCc will be completely homozygous for all gene :- (1) (2) (3) (4) 3) "Segregation of one pair of characters is independent of the other pair of characters". This explains :- (1) Independent assortment (2) Law of segregation (3) Dominance phenomenon (4) Postulate of paired factors 4) Male heterogamy found in case of :- (1) XO type male in Grasshopper (2) XY type male in human (3) ZW type male in birds (4) Both (1) or (2) 5) Genetic maps by Sturtevant was based on the frequency of :- (1) Non-disjunction (2) Crossing-over (3) Mutations (4) Translocation 6) Percentage of recombination between A and B is 9%, A anc C is 17%, B and C is 26 %, then the arrangement of genes is (1) ABC (2) ACB (3) BCA (4) BAC 7) If two person with 'AB' blood group marry and have sufficiently large number of children, these children could be classified as 'A', 'AB' and 'B' blood group in 1 : 2 : 1 ratio. Modern technique of protein electrophoresis reveals presence of both 'A' and 'B' type proteins in 'AB' blood group individuals. This is an example of : (1) Complete dominance (2) Co-dominance (3) Incomplete dominance (4) Partial dominance 8) The genotype of a plant showing the dominant phenotype, can be determined by : (1) Pedigree analysis (2) Back cross (3) Test cross (4) Dihybrid cross 9) In garden pea, seeds with genotype Bb, have _______ shape and ______ size starch grain. It is an example of _______ :- (1) Round, Large, Pleiotropy (2) Round, Medium, Multiple allele (3) Wrinkled, Small, Pleiotropy (4) Round, Medium, Pleiotropy 10) Above representative figure showing an individual affected with chromosomal disorder, The affected individual have all the given features except :- (1) Palm crease. (2) Normal heart condition (3) Mental retardation. (4) Furrowed tongue 11) In the cross PpQq × Ppqq, the genotype PPQq;PpQQ; ppqq and ppQq will be obtained in the ratio : (1) 1:0:2:1 (2) 2:1:4:0 (3) 1:0:1:1 (4) 1:2:1:2 12) Choose correct option based on following statement :- (i) Emasculation is done in male parent. (ii) Emasculation is done in female parent. (iii) Emasculation prevent underisable cross-pollination. (iv) Emasculation is done to prevent self-pollination. (1) i, ii (2) ii, iii (3) i, iii (4) ii, iv 13) In Drosophila crossing over occurs in female but not in male. Gene A and B are 10 map unit apart on chromosome. A female Drosophila with genotype and male Drosophila with genotype. How many type of gametes are produced by female and male Drosophila respectively (1) 4 types : 2 types (2) 2 types : 2 types (3) 4 types : 4 types (4) 4 types : one types 14) How many types of phenotypes would be produced in F1 progeny in the following cross that follows Mendel's law of independent assortment AABBCC × aabbcc ? (1) 1 (2) 64 (3) 27 (4) 8 15) If the minimum and maximum height of sugarcane plant is 50 cm and 200 cm respectively and this character is regulated by 3 polygenes. Then find out the height of plants produced in the progeny of cross aabbcc × AABBcc :- (1) 75 cm (2) 100 cm (3) 50 cm (4) 125 cm 16) In an example of incomplete dominance, pure red flowered plant are crossed with pure white flowered plant and the F1 individuals have pink flowers. Which one of the following is not correct? (1) Pink flowered plant will produce only pink flowered offsprings if self pollinated (2) Pink flowered plant will produce offsprings having three kinds of flowers, if self pollinated (3) The alleles of the hybrid pink flowered plant will segregate, if self pollinated (4) Half of the offsprings of the pink flowered plant will be homozygous, if it was self-pollinated 17) Observe the cross given here :- What is the genotype of crossed plants ? (1) YyRr × yyRr (2) YYRR × yyrr (3) YyRR × YYRr (4) YyRr × YYRr 18) In a cross CcDdEe × ccddEe, what will be the probability of organisms, which are homozygous for first character and heterozygous for second and third characters? (1) (2) (3) (4) 19) If Tt pea plant is crossed with Tt pea plant, then the offsprings will be - A- 75% plants are Tall B- 25% plants are Dwarf C- 50% plants are homozygous D- 50% plants are heterozygous Select the correct option - (1) A, C, D (2) A, B, C, D (3) B, C, D (4) A, B, C 20) In which of the following condition phenotype may not be affected ? (1) When modified allele produce a normal enzyme. (2) When modified allele produces a non-functional enzyme. (3) When the unmodified allele produces no enzyme. (4) All the above 21) Which is not always true for genes in diploid organisms ? (1) They occur in pairs (2) They segregate at the time of gamete formation (3) Only one of each pair is transmitted to a gamete (4) One pair segregates independently of another pair 22) A true - breeding line is one that :- (1) Have undergone continuous self pollination (2) Have undergone continuous cross pollination (3) Shows stable trait inheritance & expression for several generations (4) Both (1) & (3) 23) Which one of the following trait of garden pea is not a recessive trait :- (1) (2) (3) (4) 24) Which one of the following trait of garden pea is not a Dominant trait :- (1) (2) (3) (4) 25) Statement I : Mendel crossed pure tall and pure dwarf pea plants to study the inheritance of one gene. Statement II : Mendel observed that 1/4 of the F1 plants were dwarf, while 3/4 of the F1 plants were tall. (1) Both statement-I and statement-II are correct (2) Statement-I is incorrect and statement-II is correct (3) Both statement-I and statement-II are incorrect (4) Statement-I is correct and statement-II is incorrect 26) Assertion : Genes are the units of inheritance. Reason : Genes contain the information that is required to express a particular trait in an organism. (1) Both Assertion and Reason are true and Reason is the correct explanation of Assertion. (2) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (3) Assertion is true but Reason is false. (4) Both Assertion and Reason are false. 27) Select the incorrect match regarding pea plant :- Characters Recessive trait Dominant trait (A) Seed colour Green Yellow (B) Flower colour White Violet (C) Pod colour Green Yellow (D) Stem height Dwarf Tall (1) D (2) C (3) B (4) A 28) Identify the set of correct statements:- (A) ABO blood groups are controlled by the gene O. (B) Both alleles IA and IB produce a similar form of the sugar while allele i does not produce any sugar. (C) When IA and i are present only IA expresses and when IB and i are present, IB expresses. (D) There are six different combinations of genotype these three alleles of gene I. (E) There are four different phenotype of ABO blood groups. Choose the correct answer from the options given below: (1) A, C, D and E only (2) B, C, D and E only (3) C, D and E only (4) A, B, C and D only 29) Which statement is wrong about Law of dominance:- In a dissimilar pair of factors one member of the pair dominates (dominant) the other (1) (recessive). homozygous parent produces two kinds of gametes each having one allele with equal (2) proportion. (3) Characters are controlled by discrete units called factors. (4) Factors occur in pairs. 30) Which of the following genotype of pea seed produces intermediate size starch grains. (1) BB (2) Bb (3) bb (4) Both (1) and (2) 31) which of the following genotype of F2 generation in dihybrid cross have four progeny :- (1) RrYY (2) rrYY (3) RrYy (4) Rryy 32) Assertion : Sickle-cell anaemia transmitted from parents to the offspring when both the partners are carrier for the gene. Reason : Sickle-cell anaemia is an sex linked recessive trait. (1) Assertion is true but Reason is false. (2) Assertion is false but Reason true. (3) Both Assertion and Reason are true and Reason is the correct explanation of Assertion. (4) Both Assertion and Reason true but Reason is not the correct explanation of Assertion. 33) Which of the following conclusions of Mendel can not be explained by monohybrid cross ? (1) Dominance (2) Unit factor and segregation (3) Independent assortment (4) All the above 34) Absence of SBE in Pisum leads to the development of :- (1) Round seeds with large sized starch grains (2) Round seeds with small sized starch grains (3) Wrinkled seed with intermediate sized starch grains (4) Wrinkled seed with small sized starch grains 35) Heterozygous tall plants of garden pea were self crossed. In the progeny 520 plants were produced. How many of these plants were homozygous? (1) 390 (2) 260 (3) 130 (4) 412 SECTION-B 1) An example of polygenic inheritance in human beings is :- (1) skin colour (2) sickle cell anaemia (3) colour blindness (4) phenylketonuria 2) Children in a family have blood types O, A, AB and B respectively. What are the genotypes of their parents? (1) IA i and IB i (2) IA IB and ii (3) IB IB and IA IA (4) IA IA and IB i 3) The Punnett square shown below represents the pattern of inheritance in dihybrid cross when yellow (Y) is dominant over white (y) and round (R) is dominant over wrinkled (r) seeds:- YR Yr yR ry YR F J N R Yr G K O S yR H L P T ry I M Q U A plant of 'H' type will produce seeds with the genotype identical to seeds produced by the plants of: (1) Type M (2) Type J (3) Type P (4) Type N 4) Assertion (A) :- Emasculation is required in Pea plant for cross hybridisation. Reason (R) :- Pea plant has bisexual flowers. (1) Both (A) and (R) are correct but (R) is not the correct explanation of (A) (2) (A) is correct but (R) is not correct (3) (A) is incorrect but (R) is correct (4) Both (A) and (R) are correct and (R) is the correct explanation of (A) 5) The figure below shows three types of sex determination. Select the option giving correct identification. (1) C- ZZ males, ZW females (2) A - XO Males, XX females (3) B - XY females, XX males (4) C - ZZ females, XX males 6) Identify the set of correct statements about sex determination in honey bee:- (a) Based on the number of sets of chromosomes an individual receives. (b) Unfertilised egg develops as a queen by means of parthenogenesis. (c) Males produce sperms by mitosis. (d) Grandfather can have grandsons. (e) Males do not have father and thus cannot have sons. Choose the correct answer from the options given below: (1) A, C, D and E only (2) B, C, D and E only (3) C, D and E only (4) A, B, C and D only 7) A cross is being given here :- If B stands for pink flower then (a) Identify the plant with this characteristic inheritance pattern. (b) What percentage of F2 progeny will produce A? (1) (a) Garden pea, (b) 25% (2) (a) Snapdragon, (b) 25% (3) (a) Garden pea, (b) 50% (4) (a) Snapdragon, (b) 50% 8) Select the incorrect statement about Punnett square :- (1) It was developed by a British geneticist, Thomas Hunt Morgan. (2) It calculate the probability of all possible genotypes of offspring in a genetic cross. (3) The possible gametes are written on two sides, usually the top row and left columns. All possible combinations are represented in boxes below in the squares, which generates a (4) square output form. 9) In maize coloured endosperm (C) is dominant over colourless (c) and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1-generation was test crossed it produced four phenotypes in the following percentage Coloured and Full = 45% Coloured – Shrunken = 5% Colourless – Full = 4% Colourless – Shrunken = 46% From these data what would be distance between the two non allelic genes :- (1) 48 unit (2) 9 unit (3) 4 unit (4) 12 unit 10) Which one is wrong statement about genetic material? (1) It should be able to express itself in the form of Mendelian characters. (2) DNA is chemically less reactive and structurally more stable than RNA. (3) RNA can directly code for the synthesis of proteins. (4) Only DNA not RNA is able to mutate. 11) Hershey and Chase used which organism to prove that DNA is genetic material :- (1) Streptococcus (2) TMV (3) T2 - bacteriophage (4) λ - bacteriophage 12) The unequivocal proof that DNA is the genetic material came from the experiment of:- (1) Avery, MacLeod and McCarty, they worked with E.coli (2) Frederick Griffith, he worked with Streptococcus pneumoniae (3) Avery, MacLeod and McCarty, they worked with Bacteriophage (4) Hershey and Chase, they worked with bacteriophage and bacteria 13) The Hershey-Chase experiments show the following steps chronologically :- (1) Blending, Infection, Centrifugation (2) Centrifugation, Infection, Blending (3) Infection, Centrifugation, Blending (4) Infection, Blending, Centrifugation 14) In the following representation A and B is - Pneumococcus S Strain → injected into Mice → A Pneumococcus R Strain → injected into Mice → B A → Mice die (1) B → Mice live A → Mice die (2) B → Mice die A → Mice live (3) B → Mice live A → Mice live (4) B → Mice die 15) Assertion :- DNA is perferred over RNA for storage of genetic information. Reason :- DNA is more stable than RNA. (1) Both Assertion & Reason are True & the Reason is a correct explanation of the Assertion. (2) Both Assertion & Reason are True but Reason is not a correct explanation of the Assertion. (3) Assertion is True but the Reason is False. (4) Both Assertion & Reason are False. BIOLOGY-II SECTION-A 1) In a dihybrid cross, how many plants are impure for one character in F2 - generation out of 2000 plants ? (1) 1000 (2) 500 (3) 250 (4) 2000 2) Law, based on fact that the characters don't show any blending and both the traits are recovered as such in F2-generation although one trait was absent in F1-progeny, is :- (1) law of purity of gametes (2) law of independent assortment (3) law of incomplete dominance (4) law of dominance 3) In Antirrhinum majus when a pink flowered plant is crossed with white flowered plant then total 120 plants are obtained. What will be percentage of phenotype in offspring ? (1) 100% red flowered (2) 50% white flowered (3) 75% pink flowered (4) 100 pink flowered 4) Starch synthesis in the pea seed is controlled by one gene. It has two alleles B and b. If starch grain size is considered as the phenotype, then from this angle allele show (1) Multiple allele (2) Incomplete Dominance (3) Co-dominance (4) Polygenic inheritance 5) The figure given below is of experimental technique of Mendel. Select the option giving correct identification of A, B & C : A-Artificial cross pollination; (1) B-Hybrid offsprings; C-Emasculation. A-Emasculation; B-Bagging & tagging; (2) C-Hybrid offsprings. A-Emasculation; B-Artificial cross (3) pollination; C-Hybrid offsprings. A-Bagging & tagging; B-Emasculation; (4) C-Hybrid offsprings. 6) In an example of incomplete dominance, pure red flowered plant are crossed with pure white flowered plant and the F1 individuals have pink flowers. Which one of the following is not correct ? (1) Pink flowered plant will produce only pink flowered offsprings (2) Pink flowered plant will produce offsprings having three kinds of flowers, if self pollinated (3) The genes of the hybrid pink flowered plant will segregate, if self pollinated (4) Half of the offsprings of the pink flowered plant will be homozygous, if it was self-pollinated 7) When a Pea plant with intermediate sized starch grain in seed is crossed with other plant having small sized starch grain in seed and 700 seeds are produced. Then find out correct option? (1) 350 (large size), 350 (small size) (2) 330 (small size), 630 (medium size) (3) 350 (intermediate size), 350 (small size) (4) 158 (large size), 158 (small size), 314 (intermediate size) 8) The number of types of phenotypes in the progeny of a cross between dihybrids involving two genes with incomplete dominance would be - (1) Four (2) Six (3) Nine (4) Sixteen 9) Sickle cell anemia is :- (1) Characterized by elongated sickle like RBCs with a nucleus (2) An autosomal linked dominant trait (3) Caused by substitution of valine by glutamic acid in the beta globin chain of haemoglobin (4) Caused by a change in a single base pair of DNA 10) Phenotypic and genotypic ratio is not similar in case of :- (1) Complete dominance (2) Incomplete dominance (3) Co-dominance (4) Test cross 11) The year 1900 was important for geneticist because of :- (1) Discovery of gene (2) Rediscovery of Mendel's work (3) Mendel's results were published. (4) Historical experiments of heredity on pea plant 12) In case of qualitative inheritance, phenotype of homozygous dominant and heterozygous is - (1) Same (2) Different (3) Intermediate (4) Uncertain 13) 81 genotypes in a hybrid selfing will produce how many phenotypes ? (1) 81 (2) 16 (3) 8 (4) 32 14) Given below structure represent :- (1) Nucleotide of DNA (2) Nucleotide of RNA (3) Nucleoside of DNA (4) Nucleoside of RNA 15) In a polynucleotide chain a free phosphate moiety is present at which place :- (1) 5' carbon of ribose sugar (2) 3' carbon of ribose sugar (3) 1' carbon of ribose sugar (4) 2' carbon of ribose sugar 16) Beads on string structure in chromatin is :- (1) Known as nucleosome and is made up of acidic histone protein. (2) Made up of 200 bp of DNA wrapped around positively charge histone. (3) Common in prokaryotes and is known as nucleoid (4) Made up of RNA and non histone protein rich in lysine and arginine. 17) DNA was first discovered by- (1) Meischer (2) Robert Brown (3) Flemming (4) Watson & Crick 18) Which of the following enzyme will inhibit transformation when we add into above test tube :- (1) Protease (2) RNase (3) DNase (4) Lipase 19) The number of linkage groups in a Klinefelter syndrome person having 47 chromosomes in somatic cell :- (1) 24 (2) 46 (3) 45 (4) 23 20) A male human is heterozygous for autosomal genes X and Y and is also hemizygous for hemophilic gene h. What proportion of his sperms will be xyh (1) (2) (3) (4) 21) A colour blind man marries a woman with normal sight who has no history of colour blindness in her family. What is the probability of their grandson (son's son) being colour blind ? (1) 0.25 (2) 0.5 (3) 1 (4) Nil 22) In which genetic condition, each cell in the affected person has one sex chromosomes X ? (1) Thalassemia (2) Klinefelter Syndrome (3) Phenylketonuria (4) Turner's Syndrome 23) The number of contrasting characters studied by Mendel in his experiments were : (1) 14 (2) 4 (3) 2 (4) 7 24) In a cross between a male and female in which, both are heterozygous for sickle cell anaemia, what percentage of the progeny will be diseased ? (1) 50% (2) 75% (3) 25% (4) 100% 25) If a DNA molecule having 2000 bp how many helical turns are present in it ? (1) 100 (2) 200 (3) 400 (4) 1000 26) If the length of a DNA molecule is 1.56 mm, what will be the approximate number of base pairs ? (1) 6.6 × 109 bp (2) 4.6 × 106 bp (3) 4.6 × 109 bp (4) 3.3 × 109 bp 27) A polygenic trait is controlled by 3 genes A, B and C. In a cross AaBbCc × AaBbCc, the phenotypic ratio of the offsprings was observed as :- 1 : 6 : 15 : x : 15 : y : 1 what is the possible value of x and y respectively? (1) 20 and 6 (2) 20 and 20 (3) 10 and 20 (4) 10 and 6 28) What is the expected percentage of F2 progeny with yellow and inflated pod in dihybrid cross experiment involving pea plants with green coloured, inflated pod and yellow coloured constricted pod ? (1) 100% (2) 56.25% (3) 6.25% (4) 18.75 % 29) A normal girl, whose mother is haemophilic marries a male with no ancestral history of haemophilia. What will be the possible phenotypes of the offsprings ? (a) Haemophilic son and haemophilic daughter. (b) Haemophilic son and carrier daughter. (c) Normal daughter and normal son. (d) Normal son and haemophilic daughter. Choose the most appropriate answer from the options given below : (1) (a) and (b) only (2) (b) and (c) only (3) (a) and (d) only (4) (b) and (d) only 30) Given below are two statements :- Statement I : Genetic material should be chemically and structurally be stable. Statement II : DNA being more stable is preferred for storage of genetic information. Choose correct answer from options given below :- (1) Both statement I and statement II are true. (2) Both statement I and statement II are false. (3) Statement I is correct but statement II is false. (4) Statement I is incorrect but statement II is true. 31) Read the following statements and choose the set of correct statements- (a) Loosely packed chromatin region are referred to as Euchromatin. (b) Heterochromatin is transcriptionally inactive. (c) Histone octamer contains eight molecules of histone proteins. (d) Packaging of chromatin at higher level requires additional set of proteins that referred to as Histone chromosomal protein. (e) In human diploid cell 6.6 × 109 nucleosomes are present. Choose the correct answer from the options given below- (1) (a), (c), (d) only (2) (a), (b), (c) only (3) (a), (c), (e) only (4) (b), (e) only 32) (A) Lambda bacteriophage have 48502 base pairs in it's DNA. (B) Escherichia coli have 4.6 × 106 base pairs in it's DNA. (C) Bacteriophage ϕ × 174 have 5386 base pairs in it's DNA. (D) Haploid DNA of human have 3.3 × 109 base pairs How many of these statements are correct ? (1) One (2) Two (3) Three (4) Four 33) Match the following terms with their concepts: Column A Column B A Phenotype 1 Physical expression of a trait B Genotype 2 Genetic make up C Allele 3 Alternate form of a gene D Homozygous 4 Identical alleles (1) A-1, B-2, C-3, D-4 (2) A-4, B-3, C-1, D-2 (3) A-3, B-4, C-2, D-1 (4) A-2, B-1, C-4, D-3 34) Percentage of recombination between A and B is 9% and between A and C is 17%, B and C is 26%, then the arrangement of genes is (1) BAC (2) ACB (3) BCA (4) ABC 35) Given pedigree represents inheritance of myotonic dystrophy which is an autosomal dominant disorder. What will be genotype of parents ? (1) Mother - aa Father - AA (2) Mother - AA Father - aa (3) Mother - Aa Father - aa (4) Mother - aa Father - aa (1) 1 (2) 2 (3) 3 (4) 4 SECTION-B 1) In a cross between individuals homozygous for and 600 out of 1000 individuals were of parental type. Then the distance between a and b is :- (1) 90 map unit (2) 5 map unit (3) 40 map unit (4) 15 map unit 2) Which of the following pairs is wrongly matched ? (1) Starch synthesis in pea : Multiple alleles (2) AB blood grouping : Co-dominance (3) XO type sex determination : Grasshopper (4) T.H. Morgan : Drosophila 3) Which of the following conditions of the zygotic cell would lead to the birth of a normal human female child? (1) Two X-chromosomes (2) Only one Y-chromosomes (3) Only one X-chromosomes (4) One X and Y-chromosomes 4) Which of the following occurs due to the presence of autosome linked dominant trait ? (1) Myotonic dystrophy (2) Haemophilia (3) Thalassemia (4) Sickle cell anaemia 5) The unequivocal proof that DNA is the genetic material came from the experiments of :- (1) Griffith (2) Avery, MacLeod & McCarty (3) Hershey and Chase (4) Watson and Crick 6) Down's syndrome found in - (1) Only male (2) Only female (3) Either male or female (4) Female with XO 7) In this pedigree the genotype of all affected children will be :- (1) AA (2) Aa (3) aa (4) AA or Aa 8) Which symbol of pedigree is correctly matched ? (1) – Female (2) – affected offspring (3) – Affected male of autosomal recessive disease (4) – Marriage between relatives 9) How many boys are normal for a Haemophillic carrier female and affected man : (1) 50% (2) 25% (3) 100% (4) 0% 10) Read the following statement for autosomal recessive trait and select the correct option - (a) Trait express equally in both male and female (b) Trait can skip generation (c) Affected offspring can have unaffected parents (1) All are correct (2) only a and b are correct (3) only a is correct (4) b and c are incorrect 11) Above given pedigree chart show inheritance of (1) SCA (2) Haemophillia (3) Pseudorickets (4) Myotonic dystrophy 12) Match the following - Column-I Column-II Board plam with characteristic A Haemophilia (i) palm crease B Down's Syndrome (ii) Delayed clotting of blood Klinefelter's C (iii) Few feminine characters syndrome D Turner's Syndrome (iv) Rudimentary ovaries (1) (A)-(i), B-(ii), C-(iii), D-(iv) (2) (A)-(iii), B-(ii), C-(i), D-(iv) (3) (A)-(ii), B-(i), C-(iv), D-(iii) (4) (A)-(ii), B-(i), C-(iii), D-(iv) 13) Study the pedigree chart given below : What does it show ? (1) Inheritance of a condition like phenylketonuria as an autosomal recessive trait (2) The pedigree chart is wrong as this is not possible (3) Inheritance of a recessive sex - linked disease like haemophilia (4) Inheritance of a sex - linked inborn error of metabolism like phenylketonuria 14) A couple has six daughters. What is the possibility of their having a girl next time ? (1) 10 % (2) 50 % (3) 90 % (4) 100 % 15) Given pedigree chart represent which type of inheritance:- (1) Autosomal (2) X-linked recessive (3) Y-linked (4) Cytoplasmic inheritance ANSWER KEYS PHYSICS SECTION-A Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 A. 2 4 4 4 4 3 4 3 4 1 2 4 4 1 1 3 2 1 3 3 Q. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 A. 4 3 4 1 4 3 1 4 2 1 2 3 2 3 2 SECTION-B Q. 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 A. 2 2 1 3 1 1 3 4 1 3 1 3 4 1 1 CHEMISTRY SECTION-A Q. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 A. 1 4 4 3 3 3 1 2 2 2 2 3 3 4 3 3 2 3 1 1 Q. 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 A. 3 3 1 4 2 2 1 1 1 1 3 4 2 3 1 SECTION-B Q. 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 A. 2 2 2 3 3 2 4 4 3 1 3 2 2 3 2 BIOLOGY-I SECTION-A Q. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 A. 2 1 1 4 2 4 2 3 4 2 3 4 1 1 2 1 4 2 2 1 Q. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 A. 4 4 4 3 4 1 2 3 2 2 3 1 3 4 2 SECTION-B Q. 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 A. 1 1 4 4 1 1 2 1 2 4 3 4 4 1 1 BIOLOGY-II SECTION-A Q. 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 A. 1 1 2 2 3 1 3 3 4 1 2 1 2 2 1 2 1 3 1 2 Q. 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 A. 4 4 4 3 2 2 1 3 2 1 2 3 1 1 3 SECTION-B Q. 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 A. 3 1 1 1 3 3 3 4 1 1 1 4 1 2 2 SOLUTIONS PHYSICS 1) Because there is no change in flux linked with coil. 2) If current through A increases, crosses (X) linked with coil B increases, hence anticlockwise current induces in coil B. As shown in figure both the current produces repulsive effect. 3) e = 3t2 + 3 It is not in the form y = 4ax2. 4) Magnetic flux linked with a coil, ϕ = NBA cosθ Since the magnetic field B is parallel to the area A, i.e., θ = 90° ∴ ϕ = 0. 5) Conductor cuts the flux only when it moves in the direction of M. 6) Energy stored 7) 8) Magnetic flux ϕ = LI By analogy, since physical quantities mass (m) and linear velocity (v) are equivalent to electrical quantities inductance (L) and current (I) respectively. Thus magnetic flux ϕ = LI is equivalent to momentum p = m × v. 9) Nϕ = Li ⇒ 100 × 10–5 = L × 5 ⇒ L = 0.2 mH. 10) ⇒ 11) 12) ⇒. 13) Mutual inductance between two coil in the same plane with their centers coinciding is given by. 14) The mutual inductance between two coils depends on their degree of flux linkage, i.e., the fraction of flux linked with one coil which is also linked to the other coil. Here, the two coils in arrangement (a) are placed with their planes parallel. This will allow maximum flux linkage. 15) 16) Force on the part of wire which is parallel to will be zero. The forces on the other two parts will each be IlB in perpendicular directions. 17) = 2 × 3 × 10–2 × 2 Use right hand thumb rule 18) Magnetic force 19) 20) θ = (90 –α) τ = NIABsinθ = 1×I × 2 Bsin(90–α) τ = BI 2cosα 21) 22) By using Kirchhoff's voltage law VA – i R + E + L 23) 24) ∵ E= 25) Due to increasing field, the induced current in both loops individually must be anticlockwise which actually oppose each other. However as the emf induced in larger loop is more, the net current in larger loop is anticlockwise and in smaller loop is clockwise. Hence, its direction is B to A and D to C. 26) ⇒ L = 11 H 27) L = 0.1H 28) vsinθ ⊥ 2R ⊥ B thus e = 29) 30) 31) Eout = ⇒ Eout ∝ 32) VR = IR = 0.25 × 400 = 100V remaining 100V drop across the inductor 33) 34) conceptual 35) ⇒ ⇒ L = 2 × 10–4H 36) Here, Radius, R = 10 cm = 10 × 10–2 m Number of turns, N = 500 Magnetic field, B = 3 × 10–5 T and Resistance = 2Ω Initial flux through the coil, = 3 × 10–5 × π × (10 × 10–2)2 cos0° = 3π × 10–7 Wb Final flux after rotation = 3 × 10–5 × π × (10 × 10–2)2 × cos180° = –3π × 10–7 Wb Emf induced in the coil, Induced current in the coil,. 37) Just after switch closed then Here E = 18V, R = 9Ω I = E/R = = 2A 38) fext = 39) By using For part AO ; eOA = eO – eA = Bl2ω For part OC; eOC = eO – eC = B(3l)2ω ∴ eA – eC = 4 Bl2ω 40) Magnetic moment of coil = NIA Torque on loop (coil) = 41) Just before closing the switch. , so After a long time closing the switch Hence Req = R 42) act in opposite direction 43) M' = m' ℓ' =mR = m (ℓ/π/3) M' = 44) W = MB(cosθ1 – cosθ2) 0.8 = MB(cos0° – cos60°) W = MB(cos60° – cos90°) ⇒ W = 0.8J 45) The magnetic field due to the current-carrying long, straight wire at point a is given by. When both the wires carry currents and in the same direction, they produce magnetic fields in opposite directions at any point in between the wires............... (1) Here, a is the distance of the midpoint from both the wires. When both the wires carry currents in opposite directions, they produce fields in the same direction at the midpoint of the two wires............ (2) On solving eqs. (1) and (2), we get 46) | e | = ⇒e= ⇒e= ⇒ 0.1 = = t = 0.1 sec 47) For Toroid Nϕ = LI L= = =N = = = 2.3 mH 48) |e2| = –M |e2| = +5 × 10–3 [I0ω sinωt] e2 = 5 × 10–3 × 100π × 2 × (1) = π volt ∴ (sinωt)max = 1 49) (e.m.f)OB = Bω(5ℓ)2 = Bω25ℓ2 and (e.m.f)OA= Bω(2ℓ)2 Now (e.m.f)AB = Bωℓ2 – Bωℓ2 = 50) CHEMISTRY 51) With increase in branching boiling point decreases. This is due to decrease in the magnitude of van der Waals forces with increase in branching. Therefore boiling point of n-hexane is higher than 2- methyl pentane and 2, 2-dimethylbutane. 52) Mechanism 53) 54) Kharasch effect is the anti-Markovnikov’s addition of HBr to an unsymmetrical alkene in the presence of peroxide. This reaction follows free radical addition. 55) Lindlar’s catalyst is palladium deposited on CaCO3/BaSO4 which is poisoned with lead or sulpher compounds. It is used for the hydrogenation of alkyne to alkene in suitable solvent. 56) Benzene consist of sp2 hybridized carbon atoms whose π-charge cloud is delocalized over the ring. Because of the high stability of benzene due to aromaticity, it does not show addition reaction but since it contain electron charge cloud, it shows electrophilic substitution reaction. 57) 58) 59) Major Product in Bromination follow 3° > 2° > 1° 60) Reactivity 61) ⇒ R-Configuration 62) ∴ Total no. of stereoisomerism ⇒ 2n ⇒ 22 ⇒ 4 63) 2,3-dichlorobutane can exist as meso compound because it has two chiral centers and symmetry is possible. 64) Unsymmetrical compound So No. of Stereoisomer = 2n where n → no. of stereogenic center = 22 = 4 65) This represents meso compound 66) "S" configuration 67) No. of chiral carbon ⇒ 3 68) Different group attached to each sp2 carbon, so it will show G.I. and chiral carbon is present so it is optically active. 69) Meso compound :- Compounds having two or more chiral carbon having POS/COS is called meso compounds. 70) Rate of SN1 ∝ Stability of carbocation 71) Due to presence of best leaving group in CH3–CH2–I, It is most reactive in SN2 reaction. 72) 3° Alkyl halide is minimum reactive towards SN2 reaction. 73) Order of nucleophilicity in polar protic solvent : IΘ > BrΘ > ClΘ > FΘ Reason :- Due to solvation, the solvated FΘ is bulkiest hence, nucleophilicity is minimum, but for IΘ, due to large size and low electron density, solvation is minimum and nucleophilicity is maximum. 74) For SN1 reaction check stability of carbocation. 75) is most reactive towards SN1 reaction because it give the most stable carbocation among the given chlorides. 76) The reaction would follow SN2 mechanism. 77) N,N-dimethyl formamide (DMF) is polar aprotic solvent, as in it H-atom is not directly bonded to an E.N. atom. 78) Finkelstein reaction Halide exchange reaction in polar aprotic solvent like acetone. R–X + NaI R–I + NaX 79) Leaving tendency ∝ stability of anion of leaving group weaker the conjugate base leaving tendency will be more. I⊝ > Br⊝ > Cl⊝ > F⊝ 80) Configuration in product is inversion of reactant. 81) Electrophilic aromatic substitution reaction of chlorobenzene give para as a major product. 82) For SN2 reaction. r = K[Substrate] [Nu–] · Molecularity = 2 · Order = 2 · [Nu–]↑ rate↑ 83) 84) 85) CH3—CH2—CH2—Cl + NaI CH3—CH2—CH2—I Finkelstein reaction is a type of SN2 reaction that involves the exchange of halide ion. 86) When HBr is adds to propene, Markovnikov’s rule is obeyed. However in the presence of peroxide, free radical addition of HBr to propene takes place which does not follow Markovnikov’s rule. 87) Cumene is also known as isopropyl benzene which is an alkyl benzene. 88) 89) –N(CH3)2 is an electron donating group. It is an ortho and para directing group for electrophilic substitution reaction. and –NO2 are electron withdrawing group. They are meta directing group for electrophilic substitution reaction. 90) Acidified K2Cr2O7 is an strong oxidizing agent and its oxidizes –CH3 group of toluene to –COOH 91) Carbocation (I) is more stable than carbocation II because I has more number of hyperconjugating structures. 92) Benzene is an aromatic compound having conjugated double bonds and it is highly stable due to resonance. Cyclooctatetraene (structure ii) on the other hand is a non-aromatic compound which is not planar and hence it does not exhibit resonance despite of having conjugated double bonds. 93) H 2 /Pd-BaSO 4 (Lindlar’s catalyst) reduces/hydrogenates alkyne to give cis-alkene whereas Na/NH3(liquid) hydrogenates alkyne to give trans alkene. 94) 95) So they are not diastereomers. They are enantiomers. 96) POS absent 97) So the compounds are enantiomers. 98) Diastereomers 99) POS present ; COS present 100) P→b;Q→a;R→c BIOLOGY-I 105) NCERT XII Pg.# 83 112) NCERT Pg. # 71 116) NCERT XII Pg # 60 119) NCERT-XII, Pg. # 57,58 120) NCERT-XII, Pg. # 60,61 121) NCERT-XII, Pg. # 59 122) NCERT-XII, Pg. # 54 133) NCERT-XII, Pg. # 59 134) NCERT-XII, Pg. # 62 135) NCERT-XII, Pg. # 57 137) NCERT XII Page No. # 77, para 2 139) NCERT-XII, Pg # 33 142) NCERT XII Pg. # 76 145) NCERT-XII, Pg. # 87 146) NCERT-XII, Pg. # 85 147) NCERT-XII, Pg. # 86 148) NCERT-XII, Pg. # 86 149) NCERT-XII, Pg. # 84 150) BIOLOGY-II 151) NCERT-XII, Pg. # 63 152) NCERT-XII, Pg. # 59 153) NCERT-XII, Pg. # 60 154) NCERT-XII, Pg. # 62 155) NCERT XII Pg. # 55 156) NCERT-XII, Pg. # 60 157) NCERT-XII, Pg. # 62 165) NCERT-XII, Pg. # 81 166) NCERT-XII, Pg. # 83 167) 168) NCERT-XII, Pg. # 85 180) NCERT Pg. No. # 87 181) NCERT Pg. # 83, 84 182) NCERT, Pg. # 80 183) NCERT-XII, Pg. # 56, 57 193) NCERT XII Pg. # 88