Time : 3 Hrs. M.M. : 300 Answers & Solutions Forforfor JEE .
07/01/2020EveningRegd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005Ph.: 011-47623456 Fax : 011-47623472Time : 3 hrs.Answers & SolutionsforM.M. : 300JEE (MAIN)-2020 Phase-1(Physics, Chemistry and Mathematics)Important Instructions :1.The test is of 3 hours duration.2.The Test Booklet consists of 75 questions. The maximum marks are 300.3.There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematicshaving 25 questions in each part of equal weightage. Each part has two sections.(i) Section-I : This section contains 20 multiple choice questions which have only one correct answer.Each question carries 4 marks for correct answer and –1 mark for wrong answer.(ii) Section-II : This section contains 5 questions. The answer to each of the questions is anumerical value. Each question carries 4 marks for correct answer and there is no negative marking forwrong answer.
JEE (MAIN)-2020 Phase-1 (7E)PHYSICSSECTION - I(1) Antiparallel toMultiple Choice Questions: This section contains 20(2) Zeromultiple choice questions. Each question has 4choices (1), (2), (3) and (4), out of which ONLY ONEˆi ˆj2is correct.(3) Parallel toChoose the correct answer :(4) Parallel to k̂1.A stationary observer receives sound from twoidentical tuning forks, one of which approachesand the other one recedes with the same speed(much less than the speed of sound). Theobserver hears 2 beats/sec. The oscillationfrequency of each tuning fork is 0 1400 Hzand the velocity of sound in air 350 m/s. Thespeed of each tuning fork is close to :(1)1m/s4(3) 1 m/s(2)1m/s2(4)1m/s8Answer (1) Sol. E at t 0 at z k is given by EEE 0 ˆi ˆj cos[ ] 0 ˆi ˆj 22 FE qE Force due to electric field, FE is antiparalleltoAnswer (1)Sol. f1 f0cc vf2 f0cc vS1vO S2v BB (at t 0, z k) is 0 ˆi ˆj 2 B Fmag q v0kˆ 0 ˆi ˆj which is antiparallel22vto2 v c2 1 2 c ˆi ˆj 2ˆi ˆj Fnet FE FB is Antiparallel to22c3501 m/s v 2f0 1400 42.ˆi ˆj.2 Fmag q v B 1 1 2 f1 f2 f0 c c v c v f0 cˆi ˆj23.The electric field of a plane electromagneticwave is given by ˆi ˆjE E0cos(kz t)2At t 0, a positively charged particle is at the point (x, y, z) 0, 0, . If its instantaneousk ˆ the force acting on itvelocity at (t 0) is v0k,In a Young’s double slit experiment, theseparation between the slits is 0.15 mm. In theexperiment, a source of light of wavelength589 nm is used and the interference pattern isobserved on a screen kept 1.5 m away. Theseparation between the successive brightfringes on the screen is:(1) 3.9 mm(2) 6.9 mm(3) 5.9 mm(4) 4.9 mmAnswer (3)due to the wave is:2
JEE (MAIN)-2020 Phase-1 (7E)Sol. Fringe-width, 6.D1.5 589 10 9 md0.15 10 3 589 10–2 mm 5.89 mm 5.9 mm4.1Two ideal Carnot engines operate in cascade(all heat given up by one engine is used by theother engine to produce work) betweentemperatures, T1 and T2. The temperature ofthe hot reservoir of the first engine is T1 andthe temperature of the cold reservoir of thesecond engine is T2. T is temperature of thesink of first engine which is also the source forthe second engine. How is T related to T1 andT2, if both the engines perform equal amount ofwork?(1) T T1 T222T1T2(3) T T T121 E 2(1) c 2m (3)12c(2mE)1 E 2(2) 2m 11 2E 2(4) c m Answer (1)hhSol. e p 2mEehchc E p PE(2) T T1T2 (4) T 07.Answer (1)Sol. Let Q1 : Heat input to first engineQC : Heat rejected by first engineQ2 : Heat rejected by second engineTC : Lower temperature of first engineW Q1 – QC QC – Q2 2QC Q1 Q2 e1 EE 1 P2m c2mE cMass per unit area of a circular disc of radiusa depends on the distance r from its centre as (r) A Br. The moment of inertia of the discabout the axis, perpendicular to the plane andpassing through its centre is:4 A B (1) 2 a 4 5 aB 4 A (2) a 4 5 aB 4 A(3) 2 a 4 5 aA B (4) 2 a 4 4 5 Answer (3) 2TC T1 T2 TC 5.An electron (of mass m) and a photon have thesame energy E in the range of a few eV. Theratio of the de-Broglie wavelength associatedwith the electron and the wavelength of thephoton is (c speed of light in vacuum)T1 T2222Sol. I dmr 2 rdr ra Aa 4 Ba5 I 2 (A Br)r 3dr 2 5 40An ideal fluid flows (laminar flow) through apipe of non-uniform diameter. The maximumand minimum diameters of the pipes are 6.4 cmand 4.8 cm, respectively. The ratio of theminimum and the maximum velocities of fluid inthis pipe is:(1)81256(2)916(3)34(4)32 A Ba I 2 a 4 4 5 8.A mass of 10 kg is suspended by a rope oflength 4 m, from the ceiling. A force F is appliedhorizontally at the mid-point of the rope suchthat the top half of the rope makes an angle of45 with the vertical. Then F equals: (Takeg 10 ms–2 and the rope to be massless)(1) 75 NAnswer (2)(2) 70 NSol. A1v1 A2v2 (Equation of continuity)(3) 90 Nvmin v1 A2 (4.8)29 vmax v2 A1 (6.4)2 16(4) 100 NAnswer (4)3
JEE (MAIN)-2020 Phase-1 (7E)Sol. T2 cos45 100 N.(i)T2 sin45 F.(ii)Sol. Weight at equator mg m(g – 2R) F 100 Nmg 196 N m 19.6 kg2 2 3 mg 19.6 10 6400 10 N 24 3600 45 OF 19.6[10 – 0.034] 195.33 N11.T245 OFT1 100 NFBD of 'O'An elevator in a building can carry a maximumof 10 persons, with the average mass of eachperson being 68 kg. The mass of the elevatoritself is 920 kg and it moves with a constantspeed of 3 m/s. The frictional force opposingthe motion is 6000 N. If the elevator is movingup with its full capacity, the power delivered bythe motor to the elevator (g 10 m/s2) must beat least(1) 56300 W(2) 66000 W9.A thin lens made of glass (refractive index 1.5) of focal length f 16 cm is immersed ina liquid of refractive index 1.42. If its focallength in liquid is fl, then the ratio(3) 48000 W(4) 62360 WAnswer (2)flis closestfSol. F (10 m M)g fto the integerwhere, m 68 kg, M 920 kg, f 6000 N(1) 17(2) 1 F 22000 N(3) 5(4) 9 P FV 22000 3 66000 WAnswer (4)Sol.12.11 1 (1.5 1) Lens Maker’s formulafaRR2 11 1.51 1 1 fe 1.42 R1 R2 10.In a building there are 15 bulbs of 45 W, 15bulbs of 100 W, 15 small fans of 10 W and 2heaters of 1 kW. The voltage of electric main is220 V. The minimum fuse capacity (rated value)of the building will be(1) 15 Afe (1.5 1)1.42 1.42 142 9fa0.080.16 16(2) 10 A(3) 20 A(4) 25 AA box weighs 196 N on a spring balance at thenorth pole. Its weight recorded on the samebalance if it is shifted to the equator is close to(Take g 10 ms –2 at the north pole and theradius of the earth 6400 km)Answer (3)Sol. P VI Imain 15 (1) 194.66 N4510010103 15 15 2 220220220220(2) 195.66 N Imain (3) 195.32 N(4) 194.32 N15 155 2000 19.66 A220Answer is 20 AAnswer (3)4
JEE (MAIN)-2020 Phase-1 (7E)15. A particle of mass m and charge q has an initial velocity 0 j . If an electric field E E0 i and magnetic field B B0 i act on the particle, itsspeed will double after a time :13. Under an adiabatic process, the volume of anideal gas gets doubled. Consequently the meancollision time between the gas moleculechanges from 1 to 2. IfCP for this gasCv then a good estimate for 2 is given by 1 11 2(1) 2 1 (2) 2(3) 2(4) (1)3m 0qE0(2)3m 0qE0(3)2m 0qE0(4)2m 0qE0Answer (2)Sol. y12Answer (Bonus)Sol. 1, v Tn v 2n 1 2 1 nn T12 T1T2ax T1T2E0 qmV02 Vx2 2V0 T1V1 –1 T2(2V1) –1 2 2 2 1( 1)23 V0mE0q t 1 2 2 16. In the figure, potential difference between Aand B is :10 k AAnswer does not match with given options.14. An emf of 20 V is applied at time t 0 to acircuit containing in series 10 mH inductor and5 resistor. The ratio of the currents at time t and at t 40 s is close to30 V 10 k (3) 1.46(4) 0.84 Sol. tRL(1) Zero(2) 5 V(3) 10 V(4) 15 VAnswer (3)Answer (1)Sol. I I0 1 e10 k B(Take e2 7.389)(2) 1.152 Vx 3 V0 ax tT1 2 1T2(1) 1.06xE0B0 10 k A30 V 10 k l l (t ) l010 k BHere the diode is in forward bias. So wereplace it by a connecting wire. 40 5 20,000 l40 l t 40 s l0 1 e l0 1 e 10 10 3 Va Vb I1which is slightly greater l40 1 e 20,000 ,than 1 5l 10230 10 V 10 V15 2
JEE (MAIN)-2020 Phase-1 (7E)17. The activity of a radioactive sample falls from700 s–1 to 500 s–1 in 30 minutes. Its half life isclose to19.(1) 66 min(2) 52 minA planar loop of wire rotates in a uniformmagnetic field. Initially, at t 0, the plane of theloop is perpendicular to the magnetic field. If itrotates with a period of 10 s about an axis in itsplane then the magnitude of induced emf willbe maximum and minimum, respectively at(3) 62 min(1) 2.5 s and 7.5 s(4) 72 min(2) 5.0 s and 7.5 sAnswer (3)(3) 2.5 s and 5.0 s tln2/T1/2Sol. A A0e(4) 5.0 s and 10.0 sAnswer (3) 500 700 e tln2/T1/2 lnSol. (t) AB cos t7 30ln2ln2 T1/2 30 61.8 minute5T1/2ln1.4E T1/2 62 minutes18. d 2 AB sin t AB sin .t T dtInduced emf, is maximum when(T) B t 2.02 t 3 ,T2 2T 3Tori.e. 2.5 s or 7.5 s.44For induced emf to be minimum i.e zero1.0–150–5050 150 A/m–1.02 tT n t n ,T2H Induced emf is zero at t 5 s, 10 s–2.020.B2The dimension of, where B is magnetic2 0field and 0 is the magnetic permeability ofvaccum, isThe figure gives experimentally measured B vs.H variation in a ferromagnetic material. Theretentivity, co-ercivity and saturation,respectively, of the material are(1) ML–1 T–2(2) ML2 T–2(3) ML2 T–1(1) 1.0 T, 50 A/m and 1.5 T(4) MLT–2(2) 1.5 T, 50 A/m and 1.0 TAnswer (1)(3) 1.5 T, 50 A/m and 1.0 T(4) 150 A/m, 1.0 T and 1.5 TSol. The quantityAnswer (1)B2is the energy density of2 0magnetic field.Sol. Theoretical B2 ML2 T 2 –1 –2 ML T3 2 L 0 The values can be directly taken from thegraph.6
JEE (MAIN)-2020 Phase-1 (7E)22.SECTION - IINumerical Value Type Questions: This sectioncontains 5 questions. The answer to each of thequestions is a numerical value. Each question carries4 marks for correct answer and there is no negativemarking for wrong answer.21.A 60 pF capacitor is fully charged by a 20 Vsupply. It is then disconnected from the supplyand is connected to another uncharged 60 pFcapacitor in parallel. The electrostatic energythat is lost in this process by the time thecharge is redistributed between them is (in nJ).Answer (6)FV0 –Sol.V0/2 –C –Consider a uniform cubical box of side a on arough floor that is to be moved by applyingminimum possible force F at a point b aboveits centre of mass (see figure). If thecoefficient of friction is 0.4, the maximumpossible value of 100 Cbfor box not to toppleabefore moving is .Ui 1CV22 0Uf 1 V 2C. 0 2 2Answer (75)H Ui Uf Sol.bFmg11CV02 60 10 12 400 109 nJ44 6 nJ23.f2NWhen the block slides,The balancing length for a cell is 560 cm in apotentiometer experiment. When an externalresistance of 10 is connected in parallel tothe cell, the balancing length changes by60 cm. If the internal resistance of the cell isF f 0.4 mgN , where N is an integer then value of N is10For the block not to topple.Answer (12)a a F b mg 2 2Sol. 560 .(i) 10 500.(ii)r 10a a 0.4 mg b mg 2 2 0.2 a 0.4 b 0.5 a b 3 a 4 Maximum possible value ofr 10 56 50r 500 5601050 r 100 bis 75a6N 510 N 127
JEE (MAIN)-2020 Phase-1 (7E)24. The sum of two forces P and Q is R such that R P . The angle (in degrees) that the resultant of 2P and Q will make with Q is,.tan 90 25.Answer (90)Sol.2P Q P22P M grams of steam at 100 C is mixed with 200 gof ice at its melting point in a thermallyinsulated container. If it produces liquid waterat 40 C [heat of vaporization of water is540 cal/g and heat of fusion of ice is 80 cal/g],the value of M is .Answer (40)Q P2 Q2 2PQ. cos cos 2Psin Q 2Pcos Sol. M 540 M 1 (100 – 40) 200 80 200 1 40 600 M 24000Q.(i)2P M 40 8
JEE (MAIN)-2020 Phase-1 (7E)CHEMISTRY3.SECTION - IMultiple Choice Questions: This section contains 20Among the statements (a)-(d), the incorrectones are(a) Octahedral Co(III) complexes with strongfield ligands have very high magneticmomentsmultiple choice questions. Each question has 4choices (1), (2), (3) and (4), out of which ONLY ONEis correct.(b) When 0 P, the d-electron configuration ofChoose the correct answer :1.42Co(III) in an octahedral complex is t eg egIn the following reactions, products (A) and (B),respectively, areNaOH(hot and conc.)(c) Wavelength of light absorbed by [Co(en)3]3 is lower than that of [CoF6]3– Cl2 (A) side products(d) If the 0 for an octahedral complex ofCo(III) is 18,000, cm –1 the t for itstetrahedral complex with the same ligandwill be 16,000 cm–1Ca(OH)2 Cl2 (B) side products(dry)(1) NaOCl and Ca(OCl)2(1) (c) and (d) only(2) NaClO3 and Ca(ClO3)2(2) (a) and (d) only(3) NaOCl and Ca(ClO3)2(3) (a) and (b) only(4) NaClO3 and Ca(OCl)2(4) (b) and (c) onlyAnswer (4)Answer (2)Sol. NaOH Cl2 NaCl NaClO3 H2 OSol. (a) Co3 with strong field complex forms lowmagnetic moment complex(A)Ca(OH)2 Cl2 CaCl2 Ca(OCl)2 H2 O(B)2.For the reaction6 2(b) If 0 P configuration of Co3 will be t g e g2H2 (g) 2NO(g) N2 (g) 2H2 O(g)(c) CFSE of [Co(en)3]3 is more than [CoF6]3– absorbed of [Co(en)3]3 is less than [CoF6]3–the observed rate expression is,44 0 18000 8000 cm 199Hence, (a) and (d) are incorrectrate kf[NO]2[H2]. The rate expression for thereverse reaction is :(d) t (1) kb[N2][H2O]2/[NO] (2) kb[N2][H2O](3) kb[N2][H2O]24.(4) kb[N2][H2O]2/[H2]Answer (4)Consider the following reactions(a) Clanhyd. AlCl3(b) Cl2 (excess)anhyd. AlCl3k N H O f 22 2 2kb H NO 22Sol. keqRearranging2k f NO H2 kb N2 H2 O 2 H2 Clon comparing Rf and Rb at equilibrium,darkClClCl N H2O 2Rb kb 2 H2 Cl9Cl
JEE (MAIN)-2020 Phase-1 (7E)(c) CH2 CH — Clksanhyd.keCH CH2(d) CH2 CH — CH2Cl–Z CH3CH2O (A)anhyd.AlCl3(2) (a) and (d)(3) (a) and (b)(4) (b), (c) and (d)6.Answer (1)S o l .Vinyl halide and aryl halide do not give FriedelCraft’s reaction.The reactions which are possible are :Cldark(B)A chromatography column, packed with silicagel as stationary phase, was used to separatea mixture of compounds consisting of (A)benzanilide (B) aniline and (C) acetophenone.When the column is eluted with a mixture ofsolvents, hexane : ethyl acetate (20 : 80), thesequence of obtained compounds is(1) (A), (B) and (C)(2) (C), (A) and (B)(3) (B), (C) and (A)(4) (B), (A) and (C)ClClAnswer (2)ClClS o l .The component which is more stronglyadsorbed will be obtained later upon elutionCl(d)– A B and ke(B) ke(A)(1) (b) and (d)AlCl3OZ Which of these reactions are possible? Cl2 (excess)kskewhenCH2 — CH CH 2(b) S o l . CH3 — CH2 — CH2 — Br Z–AlCl3 CH2 CH — CH2 — ClIt also depends upon the strength of interactionbetween compound and mobile phase.anhyd. AlCl3 (A) Benzanilide(B) Aniline(C) Acetophenone5.So, (C) will be obtained first then (A) then (B).For the following reactionsCH3CH2 CH2Br ZkstionsitubuSkelim eination7.CH3CH2CH2Z BrThe number of possible optical isomers for thecomplexes MA2B2 with sp3 and dsp2 hybridizedmetal atom, respectively, isNote : A and B are unidentate neutral andunidentate monoanionic ligands, respectively.CH3CH CH2 HZ Brwhere,CH3Z CH3CH2O (A) or H3C — C — O (B),(1) 2 and 2(2) 0 and 2(3) 0 and 1(4) 0 and 0Answer (4)CH3S o l . [MA2B2]ks and ke, are, respectively, the rate constantsTetrahedralAks, thefor substitution and elimination, and kecorrect option is .Square planarBAM(1) B A and ke(A) ke(B)A(2) A B and ke(A) ke(B)BMBhas plane ofsymmetry(3) B A and ke(B) ke(A)(4) A B and ke(B) ke(A)BAplanarTotal number of optical isomer is zero in boththe cases.Answer (4)10
JEE (MAIN)-2020 Phase-1 (7E)8.1O22KClO3, Pb(NO3)2, NaNO3 on heating will releaseO2 gas.Which of the following statements is correct? H2 O S o l . H2 O2 (1) Gluconic acid is a dicarboxylic acid(2) Gluconic acid can form cyclic (acetal/hemiacetal) structure2NaBO2 2H2 O2 6H2 O (3) Gluconic acid is a partial oxidation productof glucoseSodiummetaborateNa2 OH 2 B O O 2 B OH 2 .6H2 O(4) Gluconic acid is obtained by oxidation ofglucose with HNO3SodiumPeroxoborateAnswer (3)OHHOHOHHHOHH11. Two open beakers one containing a solventand the other containing a mixture of thatsolvent with a non volatile solute are togethersealed in a container. Over time:COOHCHOSol.PartialOxidationHHOHHOHHOHOH(1) The volume of the solution and the solventdoes not changeOHgluconicacidGluconic acid is partial oxidation product ofglucose and does not form hemiacetal oracetal.OHglucose9.The bond order andcharacteristics of CN– arethe(2) The volume of the solution increases andthe volume of the solvent decreases(3) The volume of the solution does notchange and the volume of the solventdecreasesmagnetic(4) The volume of the solution decreases andthe volume of the solvent increases.1(1) 2 , paramagnetic (2) 3, diamagnetic2Answer (2)S o l .There will be lowering in vapour pressure forsolution containing non-volatile solute. So,there will be transfer of solvent moleculesfrom pure solvent to solution and hence,volume of beaker containing solvent (pure) willdecrease and volume of beaker containing1(3) 2 , diamagnetic (4) 3, paramagnetic2Answer (2)Sol. CN–Total number of electron 14solution will increase.Bond order 3It is diamagnetic in nature.12. The redox reaction among the following is:10. Among statements (a)-(d) the correct ones are:(1) Reaction of [Co(H2O)6]Cl3 with AgNO3(a) Decomposition of hydrogen peroxide givesdioxygen.(2) Formation of ozone from atmosphericoxygen in the presence of sunlight.(b) Like hydrogen peroxide, compounds, suchas KCIO3, Pb(NO3)2 and NaNO3 when heatedliberate dioxygen.(3) Combination of dinitrogen with dioxygen at2000 K(4) Reaction of H2SO4 with NaOH(c) 2-Ethylanthraquinone is useful for theindustrial preparation of hydrogen peroxide.Answer (3)S o l .The redox reaction is(d) Hydrogen peroxide is used for themanufacture of sodium perborate.(1) (a) and (c) only2000 K N2 O2 2NO(2) (a), (b) and (c) onlyNitrogen is oxidised while oxygen is reduced.(3) (a), (b), (c) and (d) (4) (a), (c) and (d) onlyReaction of [CO(H2O)6]Cl3 with AgNO3 is notredox reaction. It is a precipitation reaction.Answer (3)11
JEE (MAIN)-2020 Phase-1 (7E)13. In the following reaction sequence:Root mean square speed (Vrms); most probablespeed (Vmp); Average speed (Vav)NH2Ac2OA(1) A – Vav; B – Vrms; C – VmpBr2BAcOH(2) A – Vrms; B – Vmp; C –Vav(3) A – Vmp; B – Vrms; C – VavCH3the major product B is:(4) A – Vmp; B – Vav; C – VrmsNHCOCH3Answer 3RMS (C)(4)SpeedCH3CH2BrA – VmpAnswer Br2AcOHB – VavNHAcBrC – Vrms16. The correct order of stability for the followingalkoxides is–CH3CH3OCH314. Within each pair of elements F & CI, S & Se,and Li & Na, respectively, the elements thatrelease more energy upon an electron gain are:ONO2NO2(A)(B)––OO2N(C)(1) F, S and Li(1) (B) (A) (C)(2) (C) (B) (A)(2) F, Se and Na(3) (B) (C) (A)(4) (C) (A) (B)(3) Cl, S and LiAnswer (2)(4) Cl, Se and NaNO2Answer (3)Sol. Order of energy released upon electron gainSol.–(A)F ClS SeNO2OO2NO––(B)O(C)Stability order C B ALi Na17. The equation that is incorrect is15. Identify the correct labels of A, B and C in thefollowing graph from the options given below:no.ofmoleculesABCSpeed(1) m NaBr – m NaCl m KBr – m KCl(2) m H O m HCl m NaOH– m NaCl(3) m NaBr – m Nal m KBr – m NaBr(4) m KCl– m NaCl m KBr – m NaBr2Answer (3)12
JEE (MAIN)-2020 Phase-1 (7E) NaBr K Na ooo o S
(1) 81 256 (2) 9 16 (3) 3 4 (4) 3 2 Answer (2) Sol. A 1 v 1 A 2 v 2 (Equation of continuity) 2 min 1 2 2 max 2 1 vvA (4.8) 9 vvA 16(6.4) 6. An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c speed of .
Family Questionnaire Regarding Autism Spectrum Disorder (ASD) 41. How many hours per week is your family member with ASD engaged in each of the following activities? 0 hr 1 hr 2 hrs 3 hrs 4 hrs 5 hrs 10 hrs 15 hrs 20 hrs 25 hrs 30 hrs 35 hrs 40 hrs College/Post-Secondary School Adult Day Habilitation Employed Social Activities with
portable power systems Recharge from a Renewable Power Source Recharge Through Household AC Power Emergency Backup for Power Outages or Rolling Blackouts . AP Portable Power System 1800 800 Specifications 400 5W 8W 35W 5W 8W 40W 40W 60W 65W 5W 80W 120W 300W 700W 1000W 1200W 95 hrs 60 hrs 15.5 hrs 90 hrs 55 hrs 13 hrs 65 hrs 40 hrs 9.5 hrs 55 hrs
Science 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 30 19 Abbreviated Battery 230 212 210 189 240 203 240 201 220 187 220 187 220 186 220 185 200 160 200 160 200 160 Total Testing Time 3 hrs. 32 min. 3 hrs. 9 min. 3 hrs. 23 min. 3 hrs. 21 min. 3 hrs. 7 min. 3 hrs. 7 min. 3 hrs. 6 min. 3 .
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The Department of Basic Education (DBE) has set the question papers for Physical Sciences . Geography P1 (3 hrs) THEORY Geography P2 (1½ hrs) MAPWORK Friday 18/09/2015 Mathematics P1 (3 hrs) Mathematical Literacy P1 (3 hrs) Afrikaans HL P2 (2½ hrs) Afrikaans FAL P2 (2 hrs) ANNEXURE B (Page 1 of 2) ASSESSMENT INSTRUCTION 29 OF 2015 .
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