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Q61: Identify the pair of physical quantities that have same dimensions (A) velocity gradient and decay constant (B) wien’s constant and Stefan constant

angular frequency and angular momentum

wave number and Avogadro number

A- H S 61

= mtm_ec_seE1-EFE@ggBj.e "÷÷=÷=

Locality

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AN S 62 As best the Kepler 's 3RD Law 72283 ¥ ( %) " :( % •#Bg ¥ ☒ ÷ = ✗ " " " =

✓ H At buy instant e- wgooso=m¥ To " 7mg - oso -1m¥ ht lowest point ◦ L %gGa•• 0=00 , E- mg -1m¥ at middle point TH< Ty < % 0=900 T = wgcoseo + MIIKE at highest poiiit 0=1800 7- my -6188 -1M¥= ng -1m¥ It

✓ TR TR f< > F f. I> 4 MR UR ↓ mgmg Under equilibrium condition f- MR ÷j{÷=MCmg ? " =÷ñ;÷mg] L=¥¥ = 0.12MtL=J a ?o=kcm

AN S 65 Ta = 727 1- 273 = I 000 K % = 227 1- 273 = 4 00 K 7=4 = a Q a w = • . ( • E.) = 5 ooo ( a 400am ) ✗ 10 ≥ 1 cal = 4-2 tow = 5 ooo ✗ § ✗ 103 cal = 3 ✗ I 06 Gal =3 ✗ I 0£ ✗ 4 2 = I 2 6 ✗ I 0° Joule

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ANS 68 → Image = hw haw , = dawn f=¥¥r-BE-a-BE-fggg.GE:÷÷:÷÷÷÷:*=F÷.IE

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AN S 67 mm ÷ : ÷ " -B⑤g§ u¥ñÉ Reg = 3- +6 = 2 ¥ = 2% n

Q68: The soft-iron is a suitable material for making an electromagnet. This is because soft- iron has (A) low coercivity and high retentivity (B) low coercivity and low permeability (C) high permeability and low retentivity (D) high permeability and high retantivity ANS 68 Always high permeability ✓ and low retentivity ( Because electromagnet is always magnets e. & demagnetrse quickly ]

VERY Important kE= 1- mÑ question Asked many times in a- ¥7T✓ All competitive exams qB=m¥ r=m¥*Remember the result to save time D= hf-=ÑE →vi. vi. ra : : ←¥ •* : do % I % % I

ANS -70 ✓ STATEMENT Ites Reactance of ac circuit is ZERO when ✗ < = ✗ & ✗ = ✗ L ✗ • = @ GCORRECE ) STATEMENT 2. → In Inductor or capacitor 1 Pay = VI cos 90=0 FALSE

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✓ -EEEoEÉhgg_g= day = o Addy 8' ◦ + B dy 8- 5 = 0→-Bs ↳•osE•g§÷÷÷÷ : : 10 A Ñ " 5B ¥ ÷ 0 • = ( ¥ )%-

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1-15-72 Time of ← TA TB → " m&descent ascent 4=1 ( gta) TAZ -10 / 4=1 @ a) Tpf ② % ( gta) tab = £ ( g- a) Tps " ¥•=T÷,=É÷ ,→B&&É8BEE⑧g⑧gggg g%§ ¥ -1771¥

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+ at * Ns 73 5- = ✗ c ' ≥✗=5ra% , D= wt + % at 0 , → FIRST SECOND @eat second @ a + Or = % ✗ ( 25 →EEBEogo÷:÷÷÷÷÷÷÷ . 02 =3 ( 0,2 = 3%

ANS 74 student _ wage me unit % th of the enemy of the hammerShould ✓ ☒ Slight = for heating the nail ✓ G"l hammering 2) = MNAIL ( s ) ( AO ) Remember unit game D. • = % HChammer M NAIL ( S ) = % ✗ ( I ' 5) 56032 - × aTo = 16 07°C

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✓ AHS -75 u = % @ " = NEW charge = Q -12 U' = ( Q -1 2) Z-EFFI-fggg.f.LU ' = New energy = Ut U ✗ 4,470 BEBER ÷ -_BBag§÷÷÷ :

Volume of the cylinder = TIR≥ (l ) ✓ 6=94 9=6 V T =TR2( e.) S T Tf• = Tato # =%Tq%r e e e 4E P =÷i•¥ ' to gmé=%% ears " ee = Fetor → for c. KEVLAR MOTION 9E=¥- 1- R → I[ 9(¥¥¥)=m¥ -1 @fˢ÷•D=@r ×±

ANS TT ✓ Power = 200 watt Intensity =P ;e÷= ⇐?¥ Ta(a → (w ᵈm ¥ ) TooI= BOI Remember / 2h0 %"¥mt, this formula Bo = =F÷ii÷÷::• = 1.2×10-87

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* N S T q BBBgg←" " " ± :÷=÷÷ = ¥÷÷ = ÷

ANS 80 ✓ PE= -2¥ " TE = -1k¥ ' when 8 ↑ ↑ TE = KE + PE then KE = TE PE = ¥ § -12k¥ kE=±kr ¥ KE = * 1- Kaze ' P= 2k¥ ' ↑ TE = =£¥- ↑-

20 Mt Vx= Use -1 Art vltna V ' Lost = Ubos 45° ① F- ↳ + art " "" V " and =Ufn45 g. (2) -20 SQUARING AND ADDING EQUATION ◦ 74 $ } ① I② UCos450 N' = ,ÉsÉnus- ! MAX HEIGHT d) " = ¥-1%+400-4 , ¥ H=u%g→=(¥÷¥É* = U2 -4,0=4-+40.01 = ( 2032 -20 = 20Mt u=¥- mtsea

6×109 Ha µ = ¥ 6 1=144 ✗ = 4 ( l ) = = = 2-5 A- Fem → m=%=%÷¥ Um = f ( X ) f- = 3%-1%-475×153 = 32%0,7%103×1060 6×1091-1-2 = 3000¥09 6GHz ¥74201

25 25W im -0hm current in primary /circuit I = I. VI 30+20 ← , , ma -1¥ {to "mʰT Circuit = = -1 but g2.5Mt } SecondaResistance of lout = 20 -0hm × ⑥ circuit ✗ a 2 5Mt = # £ 2.5 = +05ohm E E = 1- (5) = § volt = 2,5-0 volt = % 2=25

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as " " ← • N¥N°µ¥¥ji , for A → 16 = Na (4) nu = 4 16 = Nz (8) n , = 2•BBgzB@÷÷÷÷ ÷ & = 7-oo * = 25

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an " " ° "" ÷?µ•aw =(00034073 × 1 × 2 × 3.14 = 440 Volt -3333735455733rad of

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