KUMAR PHYSICS CLASSES E 281 BASEMENT M BLOCK MAIN ROAD GREATER KAILASH 2 NEW DELHI 9958461445,01141032244 www.kumarphysicsclasses.com www.kumarneetphysicsclasses.com IIT JEE PHYSICS PAPER SOLUTION 26 JUNE 2022 MORNING SHIFT QUESTIONS BASED ON UNIT & DIMENSION, EMW,SCATTERING, DIODE & RL CIRCUIT WITH EMI ARE TRICKY
ANS I F- 1- loge (Ffa) ↳ Dimension less Kp÷ = 170 LOT ° ML > e< * = IT ◦ LOT ⊖ B = 1--117-2 P Dimensionless ( given ] (d) = / B) = it LF
✓ ( constant ↑R↑a ↑R↓a ↑*↑% 7ᵗʰ t.o-a.ua#C--.P4=0 t.mg cases ! -7¥ <7-t.mging ingR wg=ma my R=ma R mg=mco ) mg R=mco) R=m( gta) A- meta) R=wg R=mg1 Less than original weight
✓ 1 war height 0=0 , b- MO = 0 ↑ "
attaint @ A. = M¥YR " = mgsina-m.LI , m-Ya-IEE-y.in ma=mgaa h=§tnd mgsind mg( wgcosx 1- Mtv " -1mg C. Rana ) ✓ ± > grs.ua ] A=ʰ%R= N=mgtna -1*-6860 N= 3mg tnd ① A =% MUZ e- =M☒GgKha ) ② = 2 Mgmt
✓ _ it µ , External hn m torque hence angular momentum , , c. ◦ used & FINAL ANGULAR LM -12m) R2 Wa =L < Initial event at momentum Lp = ↳ ↳ = Aw , MAW , = ( Fl -12m) R2 w↳ LEM R2 Wa we ::÷ñ:;=¥÷m,
Inside , # 2- a < R \ ✓ A- Rat ᵈ\surface / \ F- ¥1 ga- I Inside 1- b-¥ , , ' / 8 '=Ge he \ /14=43-7835 M ' % -11A'S ¥ ' = / M•=M£}F- G¥¥É=%÷a , gaaxr ensue outside g= 9¥ for r > 0 8T ! R r
M = 1- { → 0+273=273 ↳ 100+273=373 = d- 237¥ = 0-268 2. 6- 8%
T= 27T § ✓ T ' = IT ¥+81.6 T ' = 2ft ¥.= 27T¥ T • = § (e) E- E-
KE = Total work done f- MV=n ( ¥ R ) AT f- → degree of freedom At = 17£ a = a 5R £ -5 gp To = 1^4=8
\ ✓ ↳ V C , G V → common f. botentiae 9 , = Ca V 9. I ↳ V , % " EV Total charge remains sous rail 9. = 9 , -19! 9V= @ + a) v V={¥c ↳ ei-av-c.GE#.)=(:-.-Y:)
✓ IN NONPOLAR MOLECULE CENTRE of POSITIVE CHARGE CENTRE of NEGATIVE CHARGE Hence net dipole moment ZERO when Non tolar material is placed in external field , then centre of Charge does not coincide , hence give now zero moment in field
✓ e. = d & It = dat ( 51-3+4 t +at 5.) = dat (5-13) (4-12) dat Gt3 dat (5) = 15th 8 C- z 0 at t= 2sec / / e) = / 15 (2) 2- 8 (2) so 1. = 78 volt I = 1¥ = % = 15.6 Aint
↳ = ④ / R= 1¥ ③ di =L la = @ + 9dg e) = 4004 )l But volume will remain contract&)X 100 All , = Aal Alternate method ls 1.004L= (2-00474004)×100 /A¥ = I , = # R= , LA = Constant = 0.80% Aah Quat ion ①/ equation ③ b&- ✗ 100=4%3×10 = 2×04=0.81 ¥ =%÷- = ( E.) ( ¥ ) a. ooteciooa ? % , = (1-04) ¥ -1--11^00432-1=(1.004-11) ( too 4-a.) = @ 004360043
9HB=M¥ D= M brown d- particle M 4m e de %=%÷kii¥ = ,%-×¥×=% % -5
É = 301.16 Sin C. Kz wt] AT -1452.4 Sin ( ke wt § C- I ) (j ) B- = }, sin Ckz wt > c- 5) +4524 " ᵗ ¥¥ ÷•÷ " ≥ → c- is (c) DIRECTION of PROROGATION = E E = g^ E→ = c- i.)E- ✗ B- =P I ✗ B- =L? )C- E) ✗ B- = E B = C B- =L g)
f. = 3× 1091+2 * * = :÷ . ✓ = 01 Mt ' size of the particle = ¥00 = = • ooimt here size of the particle < < a wavelength hence phenomenon scattering
electron bhoton ✓ Ee f Eph be Ppn ✗ ✗ same for both fe- f. met ME Eph mc=Eb¥ b=P=÷h✗ =L , EPh=&ph ) ( & ? -②m=¥ , %TTaÉ ② EPh=¥• Ee=E( ¥ *) ÷n=% ¥ y%=% =÷ ʰ¥ ①
✓ for ← 238✗ tartrate > Heh -203% → 8×4 q , V2 38 → g , Pb > ◦ 6 1- 8 [ Hey + 64,130 ) 92=0%-1%-6↓ Balanced equation B /
Dynamic resistance = Inverse of the slope For 2 Volt N = %j ! j ×→ = 0%1103=1007--20 -0hm for 4 Volt " =%¥÷ > = I o¥÷É = 4 F. = -5=5
Amplitude of modulated signal varies as [ per modulating or message signal
✓
Go ° 200Mt/see → 2=200 ( t) = 400 cos 0 (t ) 4 HO Mt /se Goto = % + \ ,I D= 60° I 70 & cre
10Mt Oh 0 ↑ ^ DE U2 -1 2. as 10Mt A C. 1032=0 -12g A mtseo x=5 Mt } F. 8--10-5=5 Mt1.
25 ENERGY DENSITY = { (4) ( STRAIN ) ≥ =£(÷→◦)( 5×10-92 = 1011×10-8×25 = 25× 103>43 = 25 K%t3
6 6 m 5 Y = !¥e Y ( %-) = Ya • e = , =4?; = most 24 A • eag , ;÷ = % . = %¥÷¥ % = 6,8¥ = 6m52
400 4N Volt 10 -0hm 1=20×10-317 www.x e= Ld fTet = 20 × 153*-4 = 100×10-6 so volt (0 since switch Is= ¥00 too ✗ 10-6 Closed for love time then inductor will act = ¥4 = 40×10 as a short circuit < No role of inductor ] = 400 Volt I = 2,0-0=2 AmbInitial current
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