KUMAR PHYSICS CLASSES E 281 BASEMENT M BLOCK MAIN ROAD GREATER KAILASH 2 NEW www.kumarneetphysicsclasses.com9958461445,01141032244DELHIwww.kumarphysicsclasses.com NEETSOLUTIONPAPERPHYSICS2019 IN THIS PAPER FOCUS ON RIVER MAN PROBLEM MAGNETIC FIELD DUE TO SOLID CYLINDER ELECTRIC FIELD DUE TO SPHERICAL SHELL AND GATE-LOGIC GATE
AN S I \ U= £ ( WORK DONE BY GRAVITY £ mg ( l ) ① ✓ ANS 2 7- mgcosO=M¥ position ① ② 0=1000 T , mgcosioi % =D=m¥ 1 €Ta = MUZ g- mg -10 ③ mgwpcoso position ② Tz mgcosso=M¥ FROM eanQ ② , ③⑨%iHm③ position Is is makimum ⑨ I = MRI? ② Hence brokeatlowpqtmq-mgcoso-M-%ea-mgcosso.mu " T T3= MRI-1mg ③ Tae = m¥ -1MHz ⑥
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MalviyaDwarkaPatelBURARIHauzRohiniKhasNagarNagar AN S 3 90 B = M¥ 8 = M = :#÷¥÷÷÷ ÷ = E. = : BEBBqBBq§☆B§ HEEtPH4S1CstUToRlNÉt÷ffrTLI
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Mr Mi Te , mi " Mt v , Tr Milly 1- Mz Uz = Mind , + Mr VL e=%u, Va Va = Un Un % ENERGY Lost of A = 1- Mi Ui { m , v12 " 2=9 ↳ = U , V1 = ? , V. = % Mi U = M , V , + Mila * 2- I , = Up Iz = U a +V1 Ui = U 2. MI UI 2 Ma U = Mio , + Much -1013 = Jmu = moi + man -1min UCM , Ma) =D , Cmi 1- M I v. = uem-m ˢ ÉÉg!=%cm , -1mn ) ↑= • .cm#i.-+m:-;Ir-ormn=em+*-YnEm?;m-I .±gmm;m÷m ="%m%÷÷ ¥
A5 Angular miotn D= 1% a- ¥ *=÷ When immersed in water ✓ O ' = % f. = ¥ te , O' =%=% ¥ jo ANS -7 ELECTRIC HEATER ↳ Joules heating effect}→ eddy Curren ✓
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NEET Coaching in Hyderabad Best NEET Coaching in Chennai Best NEET Coaching in Kolkata Best NEET Coaching in Lucknow Best NEET Coaching in Patna Best NEET Coaching in Mumbai AHS -7 Under equilibrium position Po -14£ Ptfgzo☒ ¥¥tSgzo 70=41 Reg = 4×25×10-2 10-3×103×10=1 < M ANS-8 ✓ -Effgg V1 BGYOR increasing AR > ✗ a > ✗ is > ✗ v.
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AN S 9 WORK REQUIRED = CH AG C- IN KE = kf Ki = o ( 1- mist aw :) = o ( 1- mr + ± ( m_ ¥ ) = o ⇐ mv + tgmv :) = 2- MOI § ✗ 100×(20×154) " * NS 10 = 3.5 Y = Aot A sin wt 1- B cos wt ✓ 8 = Ao + jars {Asnwt-y.at?-.-.coswt)JA2-BF=Ao-j-A-B-fcosasnwt-tnaeos wt ]¥ÉÉ→ , = Ao -1*+5 sncwt -14 Aof I t Amplitude / fnlwtta3=1
ANS 11 glass 1 CASE -1 / µ=I -5 F. = -1-+17=27 CASE -2 glass / t- 1. 5 ↳ glycerin (U=1 's:) ✓ ¥=¥+¥¥=¥ ¥=± AHS-12 U=§n RT✓ Of T ↑↑ , KE FF
ANS 13 ✗ = A°→ eaccelerated through P D. V ✗ = 12--27×-10-10=12.27 ✗ 1512Mt I 10 , ooo Hs -14 → It = to ( Attn AT ) lt lo = lot AT Al = lot AT ↳ Increase In length Lou Lou = ✗ Ae t.ae 17×10-5×88=2-2×10-5 ✗ LAI Lae = " ¥90 = -68cm
AHS 15 Rainbow can not be observed when observer faces towards the sun ✓ 08 8=8 ( a- %) sRmj.mg ( • %) ✓ = > • ( • %) = too N
ANS 17 CASE I ✗ ¥ All are glowing ummm hmm ÉEWut 41-1↳ 1-1 urge 3% %¥ˢwEwÉ 41-1 -54T¥-11-1 E 2%3 MT H•= ᵗ¥g, -101+2=8%4 , -20 4 ¥ : INS-10 Intrinsic semiconductor ¥÷%FÉÉ¥ L + DOPING ( TRIVALENT ? =%=P type that creates deficiency of Valency @ called holes emagnritg carrier
AHS -19 Varg = ¥ wcoswtdwt = -0 1OR IN OTHER WORDS TOTAL AREA OF THE GRAPH -0 five + file} ← ( Utt down? = O ANS -20 do a= KA date k=d% ˢ A ( dear ? k= ÷:c:* .ie#e.-m-.--wwEi--.--l--1m-'kI
AN S 21 e. = I (d) ✓ WE W -12 (d) ( O ) ( O) 1- (29×365)<-22 ( 2T) (297 %÷T◦=a a÷ e=M¥×÷•= ⇐%✗ Too 81 = 2×1%-15 ↳✗¥qg= 2×10-6 N Mt *Hs fdw = ffdf = f&o -110 f) D8✓ 0 = / 208+108=238--120 -1¥ o ] = 25 I
ANS 23 Us c- WIRE Less melting faint Due to heat it melts AN S 24 Ideal Ammeter ↳ Res = -0 Ideal voltmeter Res =D CIRCUIT I CIRCUIT-2 Mr ↳ =i] Try] ✗ 4--1%0=1 Amp ↓ / hmÉ% ^ Be -1%0=1 AND by"-1✓ 01 10 Volt V1 = V2 , I , = to
ANS -25 → e8=h< R Abbey gauss theorem e §Eds=%o e 1 ¥ ✓ E. 41%2=0,9--0 eharee ' IIF- =0 / inside resisted e e e attire At surface surface8=R §Eds=%o Chale enclosed → F- Mhk E. 49Rh=% = @ F- =IaTo%iEmae e=◦É outside the surface 8=b > R , §E.ds=%0 E. 4tb2=% E=£q f- 2%2 1-15-26 s 028<900 positive in NEGATIVE NORTHERN IN SOUTHERN HEMISPHERE HEMISPHERE
A-1-15-27 As fer Bohr 's model TE= -3-4 er kE= TE = 3. GeV PE = -26kE) = 6. Bev AHS 28 RARER ✗ > 900 MEDIUM y P ☒ ¥ DENSER MEDIUM Refracted axle -700 bud axle of Incidence will be critical axle
A- MS -20 WORK DONE = Chafe In § eh potential enereg✓ 9- = -41¥ , er } . Ui = amr.MN D= Uf Uc = a :k-Ek = 9MM * + amr = 9m¥ 4- %) = 41¥ g=q÷ Renee: = -8¥ 9+-1=8 R2
ANS 30 u 0%600 ¥ 3 }gsinÉ-g 8h30 g VK.tl ' -12ha , v2 _U2 -12daL ( O )= Uh Zgfn 6021 (o ) = U2 2gtn3ok↳ Uh =2g 5m60sec Ñ=☒fn30)RL fn6ok , -2*30*0 ¥zxi=E %÷=¥✓ ANS 31Heft 268010N 2 Neutron}- only
AN S 32 NORTH ✓ U RIVER , GROUND FFF.FM#ground * r $3B ^ US WIMMER , GROUND = U swimmer w.at WEST EAST river ← 0ggf-Tfgnggggggq.tw .org swimmer,&worDO AB sin @ = EE ° 0A SOUTH = " :÷:÷:÷→ = 1%0--11 , 0=300
ANS 33 → All forces are forming close loop hence Fnet = 0 = MDV Tet ✓ = constant *NS-34w = ZI T FOR particle ① , W , = 21¥ FOR PARTICLE Wz= 2¥ Since T , =Tz=T = :
→ w ↑ MR Under equilibrium motion → R MR = mg ① t.mg✓ R = MW 25 ③ EQUATION ①/EQUATION ③ u¥=¥→r ← 8 → h = w&-p WE % , w2= , w Foo = 108001sec
A- 1-15-36 keep unit taint charge at point PF- NET _- Et Es f ✓ =Éo±r+±To¥ E→ Ei =4¥z¥ ) ← R →← r → F- = _ ¥◦R + ✗ 4m ✗ 4m
CASE I < r > Q Q ✓ # ÷i••¥ ① CASE -2 EQUATION OVATION ② 9 , -= O Qq 9<=-8+04 = 3¥ %= -3¥ ' ÷i¥¥; ÷ Ht=¥ ② Y¥=¥÷°=¥,E=%f
✓ Rate of flow of water }e( Mt % e.) Velocity of water a , = 2×156 which is coming out from → it with the orifice = # Rate of flow of liquid = (a) (D) = C. me) ( mtsec)=¥e? = a × # = 2×10-6×12=10×-2 = 2×10-6 ✗ 2X 3- 3. = 12-6 ✗ 15° Mt%ea
ANS 39 When , A,B=1 ( switch Is closed ) Wher A / B → 0 ( switch Is open ? BULB A- B. Y EFFECT ON GLOWS BULB MEANS ⑨ 0 0 I → BULB GLOWS BULB 0 I / → BULB DOES GLOWS NOT I 0 I → BULB GLOWS GLOW ✓ BULB DOGS Jtence MEANS 1 I ① → Not Glad HAND GATE ⑨ ANS -40 NO C- ✗ CHANGE OF HEAT ✓ STUDENT ADIABATIC SHOULD REMEMBER EACH YEAR of NEET This type ofONE QUESTION IS ASKED IN NEET PAPER , ie ON HEAT
AN S 4 I e. = H d= di = BA cos 0 = B A & f- = BA cos 90 = 0 e"¥÷^→=¥% = 800 × 5 × 10-5 × • 05 ( 0 1.) = 8*5,15*55×10 = 80 × 25 × 10-5 = 2000 ✗ 10-5 = 02 Volt
ANS 42 att -0F- maximum Hence equation IS J = a coswt w=¥=¥y= a c. oswt = '%•% = 3 Cos IT = 1y =3 Cos t ✓
ANS 43 9= CV D8 It = added V g=3 Volt ✓ Tea 6=20 × 10-6 f D8 g= -20 × 10-6× 3 = 60×10-6 A = 60 UA Id = Is = 60 µ A ( As per maxwell modified Law § Edt =Mo ( f- + Id ) continuity of current
A N S 4 4 ¥÷: % ? r r o r 1- ✗ too % = 2 AIA ✗ I 00 % + ≤ ¥7 Iwi ✓ + § Af ✗ too % + 3 ✗ too % = 2 ( I %) + 1- G %) +§ ( 3 % ) + 3 ( 4 % ) = @ + I + • + I 2) % = I 6 %
↳ ¥¥¥É☆§§F * •s Not ≤ c. this quest " m is area §F§ £ É I ✓ so please write Ms p.pk??zderivakm twice Imp for 6B£ Cas well as state board exam
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