NARAYANA I I T / P M T A C A D E M Y. C o m m o n P r a c t i c e T e s t 1 6 XII STD BATCHES [CF] Date: PHYSIS HEMISTRY MTHEMTIS

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. (D). (A). (D). (D) 5. (B) 6. (A) 7. (A) 8. (A) 9. (B). (A). (D). (B). (B). (C) 5. (D) NARAYANA I I T / P M T A C A D E M Y C o m m o n P r a c t i c T s t 6 XII STD BATCHES [CF] Dat: 8.8.6 ANSWER PHYSIS HEMISTRY MTHEMTIS 6.

1 . (D). (A). (D). (D) 5. (B) 6. (A) 7. (A) 8. (A) 9. (B). (A). (D). (B). (B). (C) 5. (D) NARAYANA I I T / P M T A C A D E M Y C o m m o n P r a c t i c T s t 6 XII STD BATCHES [CF] Dat: ANSWER PHYSIS HEMISTRY MTHEMTIS 6. (A). (C) 6. (D) 6. (C) 7. (C). (C) 7. (A) 6. (B) 8. (B). (B) 8. (A) 6. (C) 9. (B). (C) 9. (A) 6. (B). (C) 5. (B) 5. (B) 65. (B). (C) 6. (C) 5. (C) 66. (C). (C) 7. (A) 5. (A) 67. (A). (A) 8. (B) 5. (A) 68. (A). (A) 9. (C) 5. (A) 69. (A) 5. (C). (D) 55. (C) 7. (A) 6. (C). (C) 56. (B) 7. (C) 7. (C). (B) 57. (B) 7. (A) 8. (B). (C) 58. (C) 7. (D) 9. (A). (A) 59. (A) 7. (D). (B) 5. (B) 6. (A) 75. (D) (Hint & Solution) PART A : PHYSICS 76. (A) 77. (A) 78. (D) 79. (A) 8. (A) 8. (A) 8. (C) 8. (B) 8. (D) 85. (C) 86. (B) 87. (A) 88. (A) 89. (D) 9. (C). Both V and I ar in th sam phas. So, lt us calculat th tim takn by th voltag to chang from pak valu to rms valu. Now, = sin t or t or t s Again, sint or sint or t or t s Rquird tim t t s.5 s 7. Sinc, imag formd by mirror will b at distanc OM on right sid, so imag will b formd for conv mirror at OM MP. So, = OM MP and objct is at OP. u = OP 8. Sinc, imag is virtual, v is +v. f = 5, u =? m or, u u NARAYANA IIT/PMT ACADEMY ()

R + L X L= L Powr factor cos = R R L R. Phas diffrnc for R-L circuit lis btwn,.

2 Applying mirror formula, w gt; u f or u u 5 u 5 9. Th impdanc triangl for rsistanc (R) and inductor (L) connctd in sris is shown is th fig. R + L X L= L Powr factor cos = R R L R. Phas diffrnc for R-L circuit lis btwn,. According to th following ray diagram HI AB d and d DS CD d AH AD GH CD d Similarly IJ d so GJ GH HI IJ d d d d. From th following ray diagram. l d. tan d. /. If nd A of rod acts an objct for mirror thn it s imag will b A and if f 5 f u f so by using f v u NARAYANA IIT/PMT ACADEMY ()

3 5 v f f v 5 f 5 f Lngth of imag f f. From fig. it is clar that rlativ vlocity btwn objct and it s imag = v cos 5. I f I I mm O f u 6. Sinc f v u v u f Putting th sign convntion proprly v u f v u f Comparing this quation with y m c Slop m tan 5 or 5 and intrcpt C f 7. i = 9t 9t dt i dt i rms A NARAYANA IIT/PMT ACADEMY ()

4 8. X C 6 C AC ammtr givs rms valu of currnt. Erms Irms A ma X C 9. As VL VC V, and V V V V R L C VR V V V Also I.A R. I sin t Amp 6 Initial valu of currnt It sin 6 sin Amp 6. Powr loss = watt Output powr = VI = = 88watt Powr gnratd by dynamo = 9 watt 9 9 El 9 E 5volt I. Compar with i i sin t i sin t cos i cos t sin p p p Thus i p cos, i p sin 8. Hnc tan 5. A solnoid consists of inductanc and rsistanc. Whn V dc is applid, = Z = R Vrms Z R Irms Whn V, 5 Hz ac is applid Vrms Z I.5 rms Z R X X L C XL fl L.55 H 5 NARAYANA IIT/PMT ACADEMY ()

5 . Sinc, V L = V C = V R X L = X C = R and V = V R = V whn capacitor is short circuitd i as X L R R X R L VL ixl R V R 6. I rms rms T / I dt T sin t cos t sin t cos.dt dt / I 5 / Ampr. 7. Z R L C 8. Ev 6 V 9. Hr, V L = V C. Thy ar in opposit phas. Hnc,. Thy will cannot ach othr. Now th rsultant potntial diffrnc is qual to th applid potntial diffrnc = V Z = R ( X L = X C ) rms rms I V V rms A Z R 5. Givn that E = V, t s 6 E E cos ft cos 5 6 cos / 6 / 5 V. A B C da db dc dt dt dt. A yb C da db dc.5 dt dt dt da db dc dt dt dt PART B : CHEMISTRY. NO Cl NOCl Rat kno Cl NARAYANA IIT/PMT ACADEMY (5)

6 K.6. Fraction of B = 76.8% 5 K K (.6 ) (.8 ) Fraction of C = K K K (.6 ) (.8 ).7% E a / RT 5. showing fraction of molculs having nrgy gratr than E a E a / RT Ea ln RT 8. r = K[O ] [O] O O Also Kc O O r = K [O ]. [O ] - O O Kc 9. r = k[x]....(i) whr [X] =. M. X K = log t X..... (ii) whr t = min, [X] =., [X] t =.5. k = from (i) and (ii).. r log (.).5 =.7 - M min - Ea / RT A t [Arrhnius Equation] log K = log A - Ea RT or log k = Ea log A R T y = m + c mans ngativ slop and non-zro intrcpt 5. E is activation nrgy. Thus ordr with rspct to NO is. If E a + E R = E Thrshold nrgy thn only molcul may ract. 6. Rat quation is drivd from slowst stp, thus first stp in mchanism A is rat dtrmining stp. K Cl H S Rat 8. G.75 J mol 6 G ne F 6.75 E =.96 V Cathod: H H Anod: OH HO O 5. E cll E E... V OPM / M RPX / X For M X M X. Thus raction is non-spontanous. Th spontanous raction in M X M X ; E.V NARAYANA IIT/PMT ACADEMY (6)

7 5. Th salt bridg posssss th lctrolyt having narly sam ionic mobilitis of its cation and anion. 5. Equivalnt conductivity of F (SO ) is givn by q q F q SO 5. Mor ngativ potntial will constitut ngativ trminal (i.. anod) of th galvanic cll. 5. A positiv potntial implis that th givn ion has a larg tndncy to rduc as compard to H +. Thus, Cu + ion can b rducd by H (g). 55. For th raction H aq H g, th standard potntial is givn by / RT ph / bar E ln nf H / mol dm / c / E will b ractiv if th numrical valu of p / H H. 56. For th raction RT, Nrnst quation is E E ln F Cu / c Cu Cu Cu RT RT.59V E E ln ln..59v F. Cu F 6. Th S.R.P. of E.8V is highr B / B It acts as cathod. PART C : MATHEMATICS 6. F f t dt Diffrntiating both th sids, w gt f. f or f or 6. I log d log log d d / log log 5 / NARAYANA IIT/PMT ACADEMY (7)

8 6. cos f d sin d d 6. g Now f t dt f t dt f dt f dt and f t dt f t dt f t dt g g 65. lim n r n r n r lim n r / n n r r / n d From th adjacnt of y = sin, w find that sin attains intgral valu for 5 7,,,, 6 6 / / sin d 5 /6 7 /6 / d d d d / 5 /6 7 / / d I / cos / d / cos / d / cos NARAYANA IIT/PMT ACADEMY (8)

9 / I d / cos cos / d cot / / / sin I 68. f t dt tf t dt Diffrntiating both th sids with rspct to, w hav f f 69. If f f f thn thn If I f ' f d d d 7. g cos t dt cos t dt cos t dt cos t dt cos t dt ( cos t is priodic with priod ) g g k 7. k I f d k k f d k k b a b f d f a b d a f d I I I I I I 7. For < <, w hav NARAYANA IIT/PMT ACADEMY (9)

10 I 7 sin, 6 7 sin, 6 6 sin, 6 sin d 7 /6 /6 d d d 7 /6 / f A B A f ' cos f ' A cos A A Now f d A Asin B d 7. sin A A cos B A A B B So, A and B 7. / d I tan / cos d cos sin cos / d cos sin / sin d sin cos / I d I NARAYANA IIT/PMT ACADEMY ()

11 75. Lt h f f g g h f f g g h h is odd function. / / h d 76. Givn 5 y =, and y + 9 = Eliminating y, w gt 5 ( + 9) = = 9 =, Thrfor, rquird ara 9 5 d 9 d 9 9 sq. units 77. Curv tracing: y = log Clarly, > For < <, a log < and for >, log > Also, log = = dy Furthr, log /, which is a point of minima. d Rquird ara d log d NARAYANA IIT/PMT ACADEMY ()

12 log lim log y 79. Ara tan d sq. units Intgrating along th -ais, w gt A cosc sc d Intgrating along th y-ais, w gt / A sc y dy / log sc y tan y y log log sq. units NARAYANA IIT/PMT ACADEMY ()

13 8. A d d 6 sq. units 8. A ln d log log Rquird ara = ( ln ) 6 ln sq. units. 8. First considr y = - NARAYANA IIT/PMT ACADEMY ()

14 For < ; y = ( ) = For ; y = ( ) = 6 6 Considr y 6 For ; y y 6 6 For ; y 5 Rquird ara d d log 5 6log 6 ln sq. units 8. Givn curvs ar y = log and y = (log ) Solving log = (log ) log =, = and = Also, for < <, < log < log > (log ) For >, log < (log ) Y = (log ) > for all > and whn, (log ). From ths information, w can plot th graph of th functions. Thn th rquird ara log log log log d d d log log log d log sq. units NARAYANA IIT/PMT ACADEMY ()

15 8. Curv tracing: y = + sin dy cos d d y Also sin whn = n, nz d Hnc, = n ar points of inflction, whr curv changs its concavity Also for (, ), sin > + sin >. And for (, ), sin < > sin < From ths information, w can plot th graph of y = f() and its invrs. Rquird ara = A, whr 85. f A sin d d cos cos sq. units d sin cos / Diffrntiating both sids w.r.t., w gt f cos sin sin f ' sin cos cos cos f ' d d d d.5 d d. 87. Rquird ara is / a a d a a a 88. Rquird ara is / ( ) d sq. units NARAYANA IIT/PMT ACADEMY (5)

16 89. Rquird ara d sin. 9. y = y = 8 = ( ) = =,, y = min at, A ( )d / TOPIC/SUB TOPIC Q. NOS. PHYSICS Rflction of light,,, 5, 6, 7, 8,,,, 5 LR Circuit 9,, 8 Rflction, 5 Mirror quation 6 AC Circuit, 8, 9,,,, 6, LCR Circuit, 7, 9 CHEMISTRY Chmical Kintics,,,, 5, 6, 7, 8, 9,,,,,, 5, 6, 7 Elctrochmistry 8, 9, 5, 5, 5, 5, 5, 55, 56, 57, 58, 59, 6 MATHEMATICS Ara 76, 77, 78, 79, 8, 8, 8, 8, 8, 85, 87, 88, 89, 9 Dfinit Intgration 6, 6, 6, 6, 65, 66, 67, 68, 69, 7, 7, 7, 7, 7, 75, 86 NARAYANA IIT/PMT ACADEMY (6)

SAFE HANDS & IIT-ian's PACE EDT-15 (JEE) SOLUTIONS

SAFE HANDS & IIT-ian's PACE EDT-15 (JEE) SOLUTIONS It is not possibl to find flu through biggr loop dirctly So w will find cofficint of mutual inductanc btwn two loops and thn find th flu through biggr loop Also rmmbr M = M ( ) ( ) EDT- (JEE) SOLUTIONS