NEET PHYSICS PAPER SOLUTION 2020

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KUMAR PHYSICS CLASSES E 281 BASEMENT M BLOCK MAIN ROAD GREATER KAILASH 2 NEW www.kumarneetphysicsclasses.com9958461445,01141032244DELHIwww.kumarphysicsclasses.com NEET MAGNETISMELECTRICITY,HEATFOCUSSOLUTIONPAPERPHYSICS2020ONCURRENT&THEORY

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Tutors For Physics in Ghaziabad HO ANS 1 At earth G- 72N g=a÷ ' 8. = "¥ñi ✓ gl g- =¥¥a¥=÷ñ¥ 8 '=g( §) , ing ' mg ( E.) Mg '= 72×4/9 ✓ = 32N ANSI FA fB= 6-10 f=-m08 f- B- f- A = 6-② f- ✗ F. Tension B IS②gg☆Ffgff÷@ then Quat'm ① a slightly decreased linenIs valid 530 fB=6 ✗ Querrey of B well☆☆☆fgg@☆ÉfgJÉÉ g- • = 530-6 * " " " = 524 Hz

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ANS 3 c. = Eo , c. ' = k A ÷ = K K = 3% = 5 C- = K to = 5×8.85×10-12 = 044 × 10-10 < 2N ' m2 ✓ ANS 4 pitch = Least count ✗ Number of division on circular scale = • 01×50 = 0.5mm ✓ BEBB

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ANS 5 FOR → small UP _-gq÷-ᵗ¥g dipole = 9×109×16×159 ✗ Cos 60° (0-6)-2 = 200 Volt ANS 6 A Atblf snail 's { 90° Law at AB i , i ' h Sini µ sink N+8< = A÷: :)B 8 , = A IS In i=µshA4 [ =µA

ANS -7 outside the • P spherical conductor 15cm F- = ¥T→%2 10cm = 9×109×3,;} ¥ ÷z = 1.28 ✗ 105% ✓ % ᵗᵗÉasecollector length of collector > length of emitter D > length of base ✓ E > c. > B. Less ← 1 doped hlghlf moderatelydoped doped

ANS 9 STRESS = fLa = = kgmt Mt Eec" ✓ = ME ' -1-2 ANS 10 If potential is constant throughout the region it means it should be conducting Region In Which Charge should always lie at the surface F- = %r V= covenant F- = 0

ANS 11 89 A 2350 + on '→ Kj , -13 (on ' ) + ✗ z 92 ✓ compare both the side 92+0=36-101-2 2=56 235-11=891-3 -1A A = 144 ANS -12 1 DIODE AS P ! A region ✓ Reverse ⇐ ✓%!:*: BIASED ANS 13 # 20m51 MOTION BETWEEN 0 &P cord ¥ V§=Ug " -12dg Sy sg= h Ug = 80m55 Ug= -20ms]ag= -10§ ME C- 605=(-205+26-10) C- b) p # 80m51 6400=400-1 > ◦ a h=¥o0 = 300Mt

AN S 14 DO = MS AT = 43-1-1838 S AT A @ ✗ 83 ✓ 7,87 (E) ˢ=Cis→=¥ AN S 15 PM =P Rt f=%÷ ✓ = (249×103) (2×10-3) (8-3) × 300 = 0-2 k8/mt3

ANS 16only valid for single electron ✓ → more than one electron in orbit hence not valid A¥% Al =L , L 7- "¥¥ñ ANS 18 , Rt = Ro ( • + ✗ AT ) ✗ Is De for insulator & semi conductor ✓

AN S 19 DISPLACEMENT f. = hfnwt Acceleration ( A = whasinwt A = W a sin Cwt -19)[ comparebotheration to hate difference = IT N S 20 ENERGY RECEIVED = Intensity ✗ AREA ✗ time = 20×20×60 = 24×1035

AHS -21 Varg { C- otfo ≥ ENERGY contribution Vava / B=£Bµ÷ ✓ In electric field and magnetic field F◦ = CBO are equal Therefore tneoatiowUAVG-z-C-C.RO ) ≥ of contribution be E 2B is 181 = £ B µ = Varg ANS -22 ps = Did Be = ① I 132=2%1 ③ egn①/eeh② :÷=%¥*%=÷ 132=4 By

ANS -23 ✗ = l ¥ A° E- 12¥ _ ¥ñ=?¥¥¥◦→ ✓ V = 104 Volt A- MS -24 Conductor fAvg relaxation time It with ↑ in temp → Resulting A- in resistivity ✓ Variation of resistivity of copper truth temp is parabolic in nature

AN S 25 MONOATOMIC GAS → DEGREE of FREEDOM =3Hence Avg Thermal energy per ✓ molecule = 3- KBT ANS -26 B- Mon f D= % B--471×187×200×2.5 = too ✓ 50×10-2 = 6.28×10-4 Tesla = 200 tummy

ANS -27 ✗ m= 599 Ur=1tXm= 600 µ =MrMo ✓ = 600×49×10-7 = 24001-1×10-7 =3 -41-1×10-4 -1m £ ' 1-15-28 9^99 IN DELHI 9.9801 Mt PHYSICS TUTOR IN BANGALORE T.EEEU.EB.EE " PHYSICS TUTOR IN CHENNAI 9-91.→mt

ANS 29 µ = Dd E= = 2.5×106 Mtv 's " ANS 30 ← 1Mt a-=mIm ? =M#t (%) 5kg ( to 10kg✓ = = =%mt= 3×100=25-0 = 67cm ANs- → According to ✗ = ¥nFd2✓ please mugwb # this formula

ANS 32 F- = MLK O' 5×10-3 (3×108) ≥ ✓ =4 -5× 10135 AHS-33 MF tuff 11 10 12 In a % = } 8--302=15 -0hm ,R= Ri * F- = b. =oimt ✓ =a × imt ANS 34 → µ = tanep 16h20 ✓ p < tarlp < N tan 'CD< ↳ < railcar ) 45° < ↳ 290°

ANS -35 fB= 2178T Cos @ = Mg Mar - m¥=2¥ ma=iog AHS -36 z=J✗f→=&É×3É ) = 6612×5 )= Gi E×j= i ✓ ANS -37 MhmÑÉ I %%÷ moved case -2-6 removed nutty mirror c•%=÷É ↳1-0-1 ' • % ;¥→ , É JÑ¥a=¥ É✗c=✗a✓ Hence cos @ =/ ← Resonance

ANS 38 > 1eV= 1.6× 10-195 10-20-5=1%9--5,9 ✓ ✓ = Ofer AHS -39 for 6kg block Tvnevt• f- T= 6h ① AT for 4kg block f. T 4g __4a -02 4g fa. ADD , equation ① & ② ↳ 6g 2g = 10h a=%-=%m5 "

ANS -40 Vrms = Arms ✗ c. Forms = Vrms_=Vrms (29 f- c.) ( Bafa) -13ms = @ 20)(2×3-14×50×40×156) = 2.5A ANS -41 tolerance I[ ↓ multiplier 2ND significantFIRST (7)47×18-+5 % ⇐ sag Number number ) ✓ 4701=5 % (4) / 1- sj

AN S 42 1. 22 ✗ OR = D✓ = 1^22×600×189 2 = 3.66×10-7 Radian AN 5- 43 ↳ system is thermallyinsulated , so no heat exchange takes place , hence ADIABATIC

AN 5-44 hi = Not @ E) man hi No = # E) man If 8 Is halved , no emission of electron , even if Intensity is ✓ increased Ii ¥ Ñ=Y ✓ BA B AB Y=AÑ O O O =ÉB 0 I 0 = A- B. I 0 0 ( ( AND GATE] I 1 I * Remember Dem organ 's Law A-B _A- 1- B- ← AtB_- A- B-

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