IIT JEE 28 JUNE 2022 EVENING SHIFT SOLVED PHYSICS PAPER

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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 28 JUNE 2022 EVENING SHIFT QUESTIONS BASED ON UNIT AND DIMENSION, LADDER PROBLEM,YDSE,RADIOACTIVITY ,DIODE FILTER CIRCUIT WAVEFORM & ZENER DIODE ARE TRICKY
we know that ✓ h=¥ 8=¥=L vi. I. =L . / %÷= ↳ → ÷,=÷ ÷=÷ . ri Eñ% =÷ 2- = -1 ÷=÷•%=(÷)(: F÷÷jm÷=÷ = ( In)=¥ m÷✗tnn=÷
✗ = 6th at att-0 ↳ Initial ✓ W=l08%c Condition =6t22t A- 4rad fdw=§t ? - t )dt w = 6t÷ 2¥ +6 10 = & att -_ 0 att -0,0=4 dad w=lora%eo W= 2-13 t ' -110 ↳ = & _dt=2t3 t ' -110 0=1--1 ¥+10T -14 do fdO=f&t3 t ' -110 )dt [ 0=2%1 ¥+10T -1C
E- Kae ✓ > Mms ' V l / 1/1/11 2=0.5 2=1-5 Mt Mt dW= F. die dW= false fdw Kae > die = E- ( x2 ) = ¥ fcis-5-co.si ] = ¥ (2) ( 1.07=1 mv tm (4) ↳ V=2M 6×2×1.0=>1×2/(112-16) at-sea V2
since ladder is under F equilibrium position < hence Fw Ff=Ñi Efa=o , Efg -0 f- = Fw , N = Mgor N L 5Mt ✓ Take torque about poentp 1mF p ¥ 4Fw (5) =wg(F¥)aˢ0 fools)=wg( ¥4)(¥) fw=%mg ✓¥÷=É÷ :%¥=¥a
9×104 kgber hour ↑ 40Mt ENERGY GENERATED = :( * ;¥oI%~ Power generated at = 1¥ watt the base = m ˢ one bulb = 100 watt = watt total bulb can be used = %¥o_- 1¥ = 50
m mf- 4md? ˢ New force f. = 4 (2-7) (4-7) m mk m+% d ? 2% 4¥ f ' f= f ' = f
Fv ↑E=omg Fr Fv= -0 & NegligibleFv= -0 ( given the✓ My = fu question) 43--11/8388=6*480 mg a- ÷ ;:ˢ g. = 4-3%1%5=41%-5×6 m5 " = 4-3%7%-431 ' × ñ ?¥ , ? # 54*0×310-3=123.4×150
AB Iso shortie DUAB = 40 DQ=0 , For BC A Upsc -11+-18<=0 UBC 50 QCA = -60 DUCA 1- WCA = 60 DU=0 HGA= 30T L for whole process

V

M molar mass

If temp gets doubled Le 2T oxygen dissociated in to atomic oxygen is %

✓ AN S
Ñ= = = v
* • ✓ 9 , 8×10-2 9<=-8×10-6 @ g% → % , -1of → Ez F- Hee = / Ei / 1-11-71 =÷ñ% +÷ña% . = ⇐•ʰd÷)× > = • (4%160)%2 6^4×104=8×9×109×01 d=smtI± d-
✓ Rt= Roca -121-7 2. = Ro (1+102) 3 = Ro (1+302) 3=4-+1%1 2+604--3+302 302=1 ✗ = ◦ = ◦ 033 6- /
✗ = I -2 ✗ I -05 No = * + a Bm = Uni Bm hour ni Bair = Moni 13m B. air = Mohini Moni Ba Moni = %%!fÉI=%÷= " 2×155-1×-11 = 1^2×10-5

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E- = hag , 2-1a.fr E- ¥ , a×É=÷??" (± 21×10-6 ✗ 0.20 12×10-7 i= 19×1%-4%-2 [ = I = 1^4 Amb
✓ E= Nd= NAdB_de- = N9ÑdB_d, P= E÷ R constant P 2 F- 22 N' 04 :÷ NII Nina -7¥ :÷ , =☒¥ˢYÉx=¥×a Pa = 4 P ,
troba gating in Advection ✓ F- → y direction B → 2 direction C. = Egg FF- Eosin ¥ ( Ot x ) Eg = 605 In 27T Bo = É°=÷g ☒,g→ (3×1081--2) = 20×10-8 = 605in -41×10 ? ( 3×108 t se ) = 2×10-7 Wb Bz = 2×1-0 > sin Ig ✗ 103 (3×1081--2)

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✓ t= a > bath difference DX = ( h 1) t ( is a) ☒ = ☒ n= a given O ' 5K = 9 a = = 2
h÷ , = Wot ki ✓ Ka = h¥ Wo k = h¥ Wo k , * =§§ ° Wo k , < ¥
✓ ✓ N = No e✗ ᵗ ⑧ loge N = loge No Xt logee loge N = loge No X t ✓ T constant ( 0 , logeNo) tana l9eNologe N ^ lose N°4 c. = * = Ye 0 ( → t " C9e% • 07

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✓ VAM = 10 ( 1+0-4 Cos ( 27×104 t) } Cos (29×107) t = 10 Cos 27×18 t + 10×04 @cos(29×104) t cos(29×10771) = 10 Cos 29 ✗ 18 t + 2 [2 Cos (29×104) t cos (29×18)-1 ] = 10 Cos (271×107)+2 / Cos (29×104-29×18) t 1- Cos (29×104+29×18) t ] = 10 Cos (29×18) t +2 { Cos 29 (104-18)+-1 Cos 29 (104+18) t ] Wm = I ✗ 27T 271 fm fm = 107 Hz = 10MHz B. AND WIDTH = 2 fm = 20MHz
150 Tau q = I 22 -11^23 + 1-19-11 20 4 = / 2 k AT = • 01 + • 02 + ' 02 + ' 0 I 4 = 0¥ = ÷ be = " 9.3¥ ✗ too = ' = ¥ , 2=150
2 1=2%0%-10 ← Zvoet → = 20mA > 10mA fled ¥ 10mA Tgv 8=4×10×153 RL= 8-↓ 10-2 = 800 -0hm Rlcmare ) -800 -0ha for Rlchrhr ) 8= Ry ✗ 20×10-3 men R¥w=%%= ≥ Ry .mn -2*1000=400 -0hm
D= ¥ , B= = ¥*=¥ ① When immersed in the medium of refractive under = 7%5 , then wave length Chapel O ' = ÷ ② 0150 % , = ¥ a o•=%= ¥ = E. as % = 0-25=1/2 2=4
vi. = 2 UR L= ✗ 153h V , = 2 VR %oo=k I 0 I/ ✗ L = 2 HIR 7 1 27 f- L = 2 R 2*1×50 ✗ = 2 ✗ 5 k¢ 100 ✗ 10-3 k= 10 k = 1¥ = Too
f- =3 R= 11=0.5 " • =¥¥÷"" Important , truthful Im tm " ≈wispier =i%¥nie=7- Lmy In nEnÉ5iwo-s-n.I-7-s-7-s.TT?--=8--=9--a=8

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too 9 , = 5× 10-6× 30=150 ME 82=0 % -19<=8 , ' -1%1 100 Uc 150 × 10-6=(4+4) V v=YY ¥ _ = lovoet 9 , '=CzV= 10×10-6×10 = 100ha
50 # § ✗% l ✗ a = 4 (e) D= f ( 4 e.) e- ¥ = %÷a◦= EmtI. N = 50 cm 25µF 25Ch " = :* ¥ D= f ✗ 34%034/0✗ 4¥25 ✗ to ≥ B-H N = 500×10-3=50 in 1×4 a-
→ = Pie -1%80,2-1/sgn=% -1%57-+581 1- Stoi ok> =P , Pm=÷¥◦ AID , -- Ails = 2ms " ( A (G) = -%D< 0<=20 , 0=24×10-3 mt% , 15 ( Vivi> =P . R IS (44--0,2)=8. Pz § -502=8 , Pa
I= ᵗ¥ = 4210¥72v7 at 20 F- 2h ① = 2×10-2=02 @ ' d) (7) = @ 2) (2) ✗ = % 1mg co a) (e) = C. 02) ( %) ʰ =RK ? { e) ( R ) # = a 5 (e) ( 10×10-2)=h a=s¥g=% FROM EQUATION ! ⑦ 20-7=22%7 20=27 F- 10N
& I & 3 / % % % d. t i tt 3 V. = 9¥ te = % , day = % ¥ +%a = ᵈ¥ ta = Fa → = g- ↳ = % , % = ¥É%T+ 9-5 = ± . + ± = 18Mt/sea

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1. Prof Kumar Sir is Masters in Electrical Engineering and had been researching scholar with IIT Delhi

2. He is also author of Many physics Education books

3. Had been in teaching Physics since he did his High School, and carries 25 years of teaching experience at IGCSE, IBDP, AP Physics, SAT Physics, NEET PHYSICS and IITJEE level

4. Tutored more than 5000 students

5. Has a strong subject grip and transfers the same to students, making students comfortable at each level

6. IGCSE tutoring is supported by complete NOTES, the first of its kind pan the world. It reduces effective teaching hours for the teacher as well as taught and reduces your expenses for IGCSE Online Classes

7. Students can avail One-o-One tutoring

8. Classes are conducted through a dedicated zoom link, and logins and passwords remain the same, with no time limit of 40 minutes

9. Each classroom session is converted as Pdf and sent to the student immediately after the class

10. 70 per cent of teaching time is spent in solving numerical, giving excellent practice to students thus making them confident for IGCSE NEET JEE Physics

11. Short cuts, tips and tricks to crack numerical are a major focus area

12. Discussion of previous year's papers are also part of the course

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