I I T / P M T A C A D E M Y IN D IA JEE ADVANCE : 5 P PHASE TEST (.8.7) ANSWER KEY PHYSICS CHEMISTRY MATHEMATICS Q.No. Answer Key Q.No. Answer Key Q.No. Answer Key. () () (). () () (). (9) () (). () () (6) 5. (6) 5 (6) 5 () 6. () 6 () 6 () 7. (5) 7 (5) 7 () 8. (6) 8 (8) 8 () 9. (A, C, D) 9 (B, D) 9 (A, B, D). (A, B, C, D) (A, B) 5 (A, B). (A, B, C) (A, B, C, D) 5 (A, C). (A, B, C) (A, B, D) 5 (A, B). (A, B, C) (A, D) 5 (B, C, D). (B) (A,C,D) 5 (B, D) 5. (B) 5 (A,B) 55 (A, D) 6. (A, C, D) 6 (A, B, D) 56 (A, C, D) 7. (CD) 7 (A, C, D) 57 (A, B, C, D) 8. (A, C, D) 8 (B, D) 58 (A, C) 9. A p, q, r, s; B q; C p, q, r, s; D p, q, r, s. A q,r; B p; C q, r; D q. 9 A p, q, s; B p; C p, q, s; D p, q, s A s, q; B q, r; C p; D q, r 59 A p, q, t; B q, r, t; C q, r, t; D p, q, t 6 A s; B p, q, r, s; C r; D-p, r
. n = Hnt & Soluton Physcs. Applcaton of KCL n the crcut yelds = V L + V R, d R Where VL L and VR R. Then, L dt d dt d Substtutng, = 6 V, =.5 amp, R = ohm and A /s, dt we have L = H.. Let q and q be the respectve charge dstrbuted over two concentrc spheres of rad r and R such that q + q = Q As surface denstes are gven to be equal, Therefore q q q r 9 r R q R 6.e, q : q :: : So q nc and q nc 9 q q V V V 9 r R V 9V. () A C A B C B A C B RAB R 5. Crtcal angle for glass ar nterface s sn c or sn C 5 / 5 tan c
PQ R From SPQ.tan c SP 8cm r R or R 6 cm 8 6. In case of a galvanometer, I I So, G, I 5 5 Now as n case of a shunted galvanometer I IG S IGG,.e., I I IG 5 5 So, G n n = 7. Here Current for full scale deflecton. I.6A Let G be resstance of the galvanometer. g To convert the gven galvanometer nto voltmeter of range - V..e V = V, a resatance R s connected n seres wth t such that: g V I G R G 99.6 G 99 6 5 G 99 G To convert the gven galvanometer nto ammeter of range of -.5 A.e. I =.5 A a resstance of value S s connected n parallel wth t such that g I I S I G g n.5.6.6 9 n.6 9.9.6 9 n.9 N = 5.
8..6 V s balanced at meter from left end. n shuntng wth wll not affect the null pont. So t agan wll be at meter. x = m, y = m x + y = 6 meter 9. The charge stored n dfferent capactors before and after closng the swtch S are C + - + - C C + - 6C + - V V The amount of charge flown through the battery s q = C Energy suppled by the battery s 6 U qv J U.6mJ Energy stored n all the capactors before closng the swtch S s 6 U CnetV.6mJ, and after closng the swtch 6 U f CnetV.9mJ H U U U.mJ, Heat generated and charge flown through the swtch s 6C. f. (A, B, C, D) (R ) 5 (R ) 5 R 8 It s balanced wheatstone brdge hence no current flow through k. Current through upper branch = total current 5 Current through lower branch 5 PR 8 5 6 PS 5 8 PR PS 5 5
. (A, B, C) Er Z I r Z R X X 6 L 6 tan VL IX L V IR 8 R L tan ; voltage s leasng by current by. (A, B, C) v = R. (A, B, C) sn t sn t d dt d R R dt r dq d dt R dt At t =,, 5, d dt At t =,,, 6, d max max dt dq d R q R Total =, q =, nduce emf s such that t tres to oppose the cause of ts own producton W.. (B) Magnetc feld produced by a current I n a large square loop at ts center, B L Say B = K L Magnetc flux lnked wth smaller loop, = B.S. = K L (l ) 5
Therefore, the mutual nductance l M = K L or M l L 5. (B) Magnetc feld at P due to current-carryng wre AB s 7 I 6 B. T a. By rght-hand grp rule, the magnetc feld at P s drected perpendcular to the plane of the paper and n nward drecton Force on proton, 9 6 6 8 Fm qvb.6.56 N By rght-hand rule for cross product, the force on the proton acts parallel to horzontal towards rght. 6. (A, C, D) From Faraday s law, the nduced voltage d V L f rate of change of current s constant V L dt V = L V L 8 V or = V Power gven to the two cols s same,.e., V = V V or = V Energy stored W = L W = L () W L W or W = 7. Use Null pont method 9. A p, q, r, s; B q; C p, q, r, s; D p, q, r, s For A : For object between focus and pole mage s vrtual erect and magnfed For object at focus, mage s at. For object between - and focus, mage s real For B: Wherever the object les mage would be vrtual and dmnshed. For C and D: Use the concepts of lens theory. 6
. () Chemstry H S H N H N CH N, N,, N. () CH CH H H () H C CH CH CH NaH CH C C C CH H H H C C H CH aldol CH () H C CH CH C = + CH MgBr H + H CH H CH CH CH C CH H () -pentanone and methyl butanol have same molecular formula () n Wolff Kshner reducton t gves methyl butane.. () From the rate expresson, overall order of reacton s two & frst order w.r.t. [I ]. (). H + CH CH C /H HC HC 6. () 7
The aldol has -hydrogen atoms and hence gves three dehydraton products 8. (8) r.9 It wll be a body centred cubc sold and coordnaton number wll be 8. r 9. (B, D) 6 H + 6 + + 6 + H H 8 8 H H. (A, B) In esterfcaton, H+ of alcohol reacts wth H of carboxylc acd to form H and the reactvty of carboxylc acd s > >. ther statements are correct.. (A, B, C, D) R CH AgN /NHH R C H C H NHH AgN /NH H AgN /NH H N (R can be CH to C6H5) HC. (A, B, D) [R ] log [R] k t. Comparng wth y = mx + C then we get (A) and (D) t ½ does not depend on concentraton for frst order reacton. 8
8. (B, D) H H CH CHCl /H (Remer - Temann reacton) CH CH + H C CHCl Mathematcs () f x sn cos sn x cos sn cos x f ' x cos cos sn x sn sn x cos x sn sn cos x cos cos x sn x f ' x n, f sncos sn cos cossn sn sn f f sn cos There wll be one root le between,. () n n lm... lm sn n.. nn n n n lm... lm sn lm. n n n n n n n n n n n n n. () Let radus of sector = r and angle = n 9
r r r Area r r r A r r da r r dr d A maxma dr r Amax r r r 6 Maxmum area of rectangle A ' max 6. (6) 5n 5 dx ln x r r x n n ln 6 k = 6 5. () x dx Snce f x f a x Dfferentatng both sdes w.r.t.x f ' x f x. f ' x f x Integratng both sdes f x x c f x x c..() But And dx f x.() a f dx (gven) f a x then c c 5 f x x
6. () y {Sec x T an x }dx y 7. () 8. () y y {Sec x Sec x }dx {Sec x Sec x} dx y y y Sec x dx dx Sec x dx y y a b 5 I x x x x x 5 dx b Usng pro. 5 a b f x dx f a b x dx a I x x x x x 5 dx I I = b Usng pro. I I 5. (AB) a b f x dx f a b x dx dx cos x cos x I = cosec x dx x y dy xy dx a x y dy xy dx dy d x y dy y x dy xy dx y c x y c x y cy.e., dy y x y cy. 5. (AC) a =, b =, a = a + b b 6 = 8, b 6, a 8 6 7, b 8
Y X 8 7 8 X 5. (AB) Y So the three loops from = to = are alke. Now area of th loop (square) = dagonal A b b b A So, A b So the area form a G. P. seres A r So, the sum of the G. P. upto nfnte terms = f ; f t Now () f () f = () (/) = 8 () square unts. t mnmum of f(x) = m(t) = t mnmum of f(x) 5. (BCD) V AB A; V sn cos t ; m(t) t dv d or cos.e. tan ;.e. sn
5. (BD) x x, x x, x, ' x, x f x x f x, x Functon s ncreasng n each of the two ntervals (,) and (,) but not n [,] Clearly least value of f(x) s and greatest value of f(x) does not exst. 55. (AD) dx x x x tan tan dx x c x x 6 56. (ACD) y x and x y meet at (,) and P(,) x 6 S y dx dx.6 S x dx S S x dx S / 6 x S S 6 6 6 6 S. S S 6 ; S : S : S :: 57. (ABCD) Dfferentatng both sde x x e 6e Q x P 8e S x x x 9e e 9e R
58. (AC) x x ax x bx lm x x a x a b x a b lm x x a = a b = b = 59. (A-PQT; B-QRT; C-QRT; D-PQT) (A) T T 6 T : : T n... n n n T n n n lm Tn n n (B) sn x cosec x x, soluton (C) x x x x a x x,,... 6 n t
n t (D) a I x x x dx x x x dx Let Hence, f x x g y y y whch s the nverse of f x. Now, I f x dx g y dy f xdx g ydy n puttng y t Hence, I 6. (A-S; B-PQRS; C-R; D-PR) I x dx x dx (A) cos sn sn sn x dx dx 8 I 8.. (B) tan cos tan cos (C) I x dx x dx tan cos x dx tan cos x dx I. I x sn x dx Let x t dx dt 5
t sn sn t dt 8 t t dt I 8 I t cos t t cos t dt t sn t sn t dt 8I 8I. Hence, (D) sn x I x dx n x,, sn x II dx ln ln ln ln x I ln e e 6