ANSWER OF MODEL PAPER FOR IIT-JEE

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1 NSWER OF MODEL PPER FOR IIT-JEE : NSWER OF MODEL PPER FOR IIT-JEE PHYSICS. ns. both () and () are correct. Reason: The dsplacement of the partcle s determned by the area bounded by the curve. Ths area s s π vm t 0 s π The average velocty s < v > vm t0 Such moton can not be realzed n practcal terms snce at the ntal and fnal moments the acceleraton, whch s slope of v-t graph s nfntely large. Hence both () and () are correct. sn θ. ns. g. Reason: Let a be the acceleraton sn + θ of wedge down the plane and N the normal reacton between and. The acceleraton of wll be a sn θ vertcally downwards. Then equatons of moton of and gve (N + mg) sn θ ma. () and (mg N) ma sn θ () Solvng these two equatons, we get g sn θ a acceleraton of s a a sn θ +sn θ sn θ g + sn θ. Dsplacement of n s s S sn θ a () g + sn θ.. ns. u Reason: When the strng jerks tght both partcles begn to move wth velocty components v n the drecton. Usng conservaton of momentum n the drecton mu cos 0 mv + mv. ns. or v u jerk s u. l 60. Hence the velocty of ball just after the l GM r a J υ J 0 υ u sn 0 0 Reason: Total energy of a planet n an ellptcal orbt s : E - GMm (m mass of planet) a From conservaton of mechancal energy K.E + P.E E. or GMm GMm mv or v r a GM r a π 5. ns. 6/7 Reason: t and π sn k k k or sn k s not zero. Therefore, nether of or s a node π 7π. k 6k snce π π π > > ; λ > > λ/ k k k λ. Therefore, φ π and φ k. 7 6π. φ 6. φ 7 6. ns. 6 Reason: Internal energy of n moles of an 5 deal gas at temperature T s gven by U f nrt (f degrees of freedom) U U. f n T f n T n f T ()() 6 n ft (5)() 5 7. ns. V -y + constant. Reason: dv E.dr - (y ˆ + j).(d ˆ ˆ + dyj ˆ + dzk) ˆ - (yd + dy) -d(y) Integratng, we get V -y + constant. µ 8. ns. /8. Reason: 0 M r πr or. r.e., wll become /8 when radus and current both are doubled. / 9. ns. υ [k(m+m)] / m V k. M + m Reason : PE Stored n the sprng k. mv (M +m)v or V mv (M + m) () fter collson, KE of block + bullet n t PE of the sprng. Thus (M+m) V k whch gves V k.. () (M + m) Usng eq. () and (), we have v (M + m)v m / (M + m) k [k (M+m)] / /M m M + m 0. ns. /RC R/L / LC. ns. It represents a statonary wave of frequency 60Hz. It s the result of superposton of two waves of wavelength m, frequency 60Hz each travellng wth a speed of 80 m/s n opposte drecton.

2 NSWER OF MODEL PPER FOR IIT-JEE :. ns. the ntensty of emtted radaton ncreases the mnmum wavelength of emtted radaton decreases.. ns. The nduced emf s zero n poston II The nduced emf s ant-clockwse n poston I The nduced emf s clockwse n poston III. Reason : When the loop s completely nsde the feld, the φ. constant.. ns. O. Reason: Hooke s law s obeyed n the lnear porton of the graph. 5. ns. 6. ns. Copper. Reason: Copper s a conductor. 7. ns. a current flows n the crcut for sometme and then decreases to zero. 8. ns. n alternatng current flows n the crcut. Reason: lternatng voltage mples an alternatng current flows. 9. ns.. 0. ns. 6. Reason : ccordng to conservaton of energy, hc hc + RchZ λ λ n or + RZ λ λ n λ or + λ n λλ RZ Thus /n 0.08 or n Hence the quantum number 6.. ns. 9.. ns.. Reason : s the network QCS s a balanced Wheatstone brdge, no current wll flow through C and hence the effectve resstance of the network between 6 6 QS : RQS ohm and as the resstance of the loop s ohm, the total resstance of the crcut, R + ohm. Now f the loop moves wth speed v 0, the emf nduced n the loop, e v 0l. So the current n the e v crcut 0l I R R. Substtutng the gven data, IR 0 v0 0 - ms - cm/s. In l 0. accordance wth Lenz s law the nduced current n the loop wll be clockwse.. ns. 8. Reason : s, r >> λ, So and P r + λ + rλ cos θ P r + λ rλ cos θ P P rλ cos θ rλ cos θ P P λ cos θ (P + P) [ P r, P r] For P to have mamum ntensty, λ cos θ nλ cos θ n/ where n s nteger for n 0, θ 90, 70 n ±, θ 60, 0, 0, 00 n ±, θ 0, 80 So, postons of mama are at θ 0, 60, 90, 0, 80, 0, 70 and 00.e., 8 postons wll be obtaned. d. ns. Reason : mg + (M V) d mg + m gy m g dy mg + gy but dy gy mg + mg mg. 5. ns. 0 Reason : s (t ).. () So v (d/) (t ).. () (a) v wll be zero when (t ) 0,.e., t Substtutng ths value of t n equaton (), ( ) 0.e., when velocty s zero, dsplacement s also zero. 6. ns. 6 Reason : s the rods are n seres, R eq R + R + R C wth R (L/K) L L L 7L.e., Req + + () K K 0.5K K nd hence, dq (00 ) 00K H θ θ R (7L / K) 7L Now n seres, rate of flow of heat remans same,.e., H H H H C. So for rod, (00 T )K 00K,.e. L 7L or T 00 (00/7) (600/7) 87.7 C nd for rod C, (TC 0) 0.5K 00K,.e. C L 7L or T C (00/7) 57. C Furthermore f K eq s equvalent thermal conductvty, L + L + L 7L Req [From equaton ()] Keq K.e. K eq (6/7) K. 7. ns. Reason:Let υ br be the velocty of boatman wth respect to rver or velocty of boatman n stll water and υ r the rver velocty. Velocty of raft s also υ r. ssumng rver to be at rest or raft to be at est. The boatman wll move wth υ br both durng ts downstream and upstream moton. Therefore Tme of upstream moton tme of downstream moton hr.

3 Total tme of moton of raft hrs. In ths hrs, the raft moves 6 km. Hence the raft velocty or rver velocty s km/hr. oatman hour oatman υ r + υ br hour υ br υ r Raft Raft 6 km ssumng rver to be at rest ctual moton of boatman and raft 8. ns. 5 Reason: Let be the dsplacement of straw when the frst nsect reaches the opposte end. Hence dsplacement of nsect would be a. For center of mass to reman statonary we have - m() m a or 0.a Therefore, the stuaton s as follows a a/ efore 0.75a 0.a 0.5a Let M be the mass of second nsect. υ r m 0.55a fter hour 0.a M m/ For the straw not to topple the centre of mass (of straw + two nsects) should le nsde the table or m m (0.55a) M + (0. a) or m 5 m. 9. ns. 0.0 g 0. ns. 0%. ns. Frenkel defect CHEMISTRY. ns. H. THF / H O. NaOH, Hg (Oc) / NaH. NaOH, H O / H +. ns. ll ecept I.. ns. L, Mg or l 5. ns. M 8 6. ns. orthophosphate 7. ns. (, ) 8. ns. (,, C) Reason : The reacton X + H 6 [H (X) ] + [H ] proceeds when X NH, CH NH or (CH ) NH. 9. ns. (, C, D) Reason : ll beng alkanes have only sp -hybrdzed carbons. NSWER OF MODEL PPER FOR IIT-JEE : 0. ns. (, C) Reason : Frenkel defect s also called dslocaton defect because smaller ons are dslocated from ther lattce stes nto the ntersttal stes. Hence (b) s correct. Trappng of an electron leads to the formaton of F-centre. Hence, (c) s correct.. ns. (, ) Reason: (Ph P) RhCl s Wlknson s catalyst and TCl + (C H 5) l s Zegler-Natta catalyst.. ns. Pb + Pb + Reason : Durng rechargng the col, cell reacton goes n reverse drecton.. ns. 98 Reason : Two moles of H SO are used n the net cell reacton n whch two electrons are echanged through he cell reacton.. ns. CH CH CH CN 5. ns. CH CH CH CHO 6. ns. CH CH CH COOH 7. ns. 5 Reason : H PO H + + H + + OH H O; rh on H 55.8 on H 5 kj/mol 8. ns. Reason : E 8 hc (6.6 0 J s)(.00 0 m / s) λ m J E H ( J) J; E E H (Z ) Z 8 E (6.5 0 ) E 8 H (.6 0 ) Z (helum) 9. ns. Reason : λ y λ ; 0(X) ; 0(Y) y 0(X) ln t 0(Y) y HPO ; rh? (λ y λ )t ln ; (t/)y 0 mn. 50. ns. Reason : Octahedral vod present at the centre of cube and tetrahedral vod s present at (/) th of the dstance along each body dagonal. a dstance between octahedral and tetrahedral vod. 5. ns. 5. ns. 5. ns. 5. ns. 6 Reason : Cl +6OH 5Cl + ClO 55. ns. 56. ns. + H O

4 NSWER OF MODEL PPER FOR IIT-JEE : MTHEMTICS 57. ns. W Reason: We have to fnd Z n terms of W under gven condton. Z W r say Let W re θ W re θ and arg Z π arg W π θ ( π θ) π θ θ Z re re.e re W. 58. ns. Reason: Solve the gven equatons for a and d a d /mn T mn a + (mn )d /mn (+ mn ). 59. ns. 0 Reason: Epress as quadratc and show < 0. n + Reason: n C r r + n C r + n C r 60. ns. ( ) ( n C r + n C r ) + ( n C r + n C r ) n+ C r + n+ C r n+ C r. 6. ns. /5 Reason: I f aj 0, j and aj for j. a 5λ λ /5, a 8λ 8λ 0 a λ λ 0 You may verfy other elements also satsfy the crtera. 6. ns. P (E) /, P (F) / or P (E) /, P (F) / Reason: Snce E and F are ndependent, we have P (E F) P (E) P (F) P (E) P (F) / () E and F are ndependent E c and F c are also ndependent P (E c F c ) P (E c ). P (F c ) / ( P (E)).( P(F)) ½ P (E) P (F) + P (E) P (F) / P (E) P (F) + / / P (E) + P (F) 7/ () Solvng () and (), we get P (E) /, / Then P (F) /, /. 6. ns. tan β + tan γ Reason: α + β π/ α π/ β or tan α cot β or tan α tan β () tanα tanβ tanα tanβ tan γ + tanα tanβ tan α tanβ + tan γ. 6. ns. 60º Reason: / ab / p. p.e. / b.h ab p lso a + b c 6p (a b) a + b ab 6p 8p 8p lso (a + b) a + b + ab p a b C tan cot. a + b 0 60º. π 65. ns. f f ( π) 0 Reason: π 9. Somethng [π ] 9, [ π ] 0 f() cos (9) + cos ( 0) f() cos (9) + cos (0) () () 9 0 f π cos π cos π + π cosπ + + cos( 5π) π cos as cos (nπ) ( ) n f(π) cos (9π) + cos (0π) + 0 f( π) cos ( 9π) + cos ( 0π) + 0 f(π/) cos (9π/) + cos (0π/) ns. Reason: I. If cos θ <, then + 0 II. If cos θ, then ns. dy d a + bcos Reason: m lm + cos ( π.n!) m, n d y b sn d (a + bcos ) a b y tan tan a b a + b / dy a b..sec.. d (a b ) a b a+ b +.tan a+ b dy (a + b). d (a + b) (a + b)cos + (a b)sn acos + sn + bcos sn dy d a + bcos d y b sn. d (a + bcos ) 68. ns. tan + tan + tan...to cot + sec + sec cosec Reason: Takng log on both sdes of the gven equaton, log cos log cos log cos log sn log Dfferentatng w.r.t., tan + tan + tan...to cot or tan + tan + tan...to cot +

5 gan dfferentatng w.r.t., sec + sec cosec ns. velocty s ma. at t acceleraton s mn. at t mn. dstance s at t 0, Reason: t t + t + 7 () d v t 6t 8t + () d f cc. t t + 8 () I. v s ma. dv 0, d v < 0 or f f 0 and 6t < 0 () f 0 t t / ± ( 96) ± 6 ± t 6 6 If t 6 + If t 6, then 6t > 0, (not possble), then 6t < 0, whch s true. II. cc. Is mnmum f df 0 and d f > 0 6t 0, d f 6 > 0, by () t (b) s true. III. s mn f d 0 and d > 0 or f t 6t + 8t 0 (5) and t t + 8 > 0 (6) From () t (t 6t + 8) 0 (t ) (t )t 0 t 0,,. Out of these only t 0, t satsfes (c) s true. 70. ns. + y y + 0 Reason: The centre of gven crcle s (, ) and radus s. The chord whose md-pont s M (h, k) subtends an angle of 0º at C. CM cos 60 C CM C or {(h ) + (k ) } Locus s + y y + 0. C(,) π/ π/ M(h,k) 7. ns. (/, /) Reason: Let (h, k) be any pont on the gven lne h + k and c.c. s h + ky NSWER OF MODEL PPER FOR IIT-JEE : 5 or h + ( h) y or (y ) + h ( y) 0 P + λq 0. It passes through the ntersecton of P 0 and Q 0 of (/, /) ns. Reason: y tan + tan + 5. y tan 5 tan + tan / + tan or y tan 5 + tan / etc. y 7. ns. Reason: Put sn θ and y sn φ n the gven equaton. cos θ + cos φ θ φ a or cot a snθ sn φ θ φ cot a or sn sn y cot a constant Dfferentate w.r.t., dy 0 ( ) ( y ) d 7. ns. dy y. d sn (a + y) sna sn y sn(a + y) Reason: sn y sn (a+ y), Dfferentate w.r.t.. sn(a + y) cos y sn y cos(a + y) dy. sn (a + y) d dy sn (a + y) sn (a + y). d sn(a + y y) sna z 75. ns. Reason: Here z zz z z zz z z zz (z z ) ( z z ) ( zz ) ( zz ) (z z )( z z ) ( zz ) ( zz ) zz zz zz + zz z + z + z. z ( z )( z ) 0 ut z (gven) z 0 or z. 76. ns. Reason: Rewrte the gven equaton () On replacng the last letter on the L.H.S of Eq. () by the value of epressed by Eq. () we obtan

6 NSWER OF MODEL PPER FOR IIT-JEE : (n radcal sgns) Further, let us replace the last letter by the same epresson; agan and agan yelds ( + ) (n radcal sgns) we can wrte lm N {N radal sgns} It follows that, lm f() lm f() + Hence + 0 Therefore 0, sum of roots 0 +. a + b c + b 77. ns. Reason: Let P + a b c b ac ac a + c + a + c + a + c ac ac a c a + c a + c a + c a + c c a + + a c + > a c a c + > c a a c + > c a lso, λ λ λ... λ 8 / ac b a + c λ / λ λ. 78. ns. 7 Reason: Number of ways of selectng n coupons consstng of C or n (ã the word CT can not be wrtten f at least one letters s not selected) Now number of ways of selectng n coupons bearng only n. Total number of ways n + n + n n n ( n ) Gven ( n ) 89 n 6 n 6 6 n 6 n(n + )(n + ) then n ns. Reason: pplyng C C + C + C, then (+ b ) (+ c ) f() + b (+ c ) (+ b ) + c (ã a + b + c + 0) Now, applyng R R R and R R R (+ b ) (+ c ) f() ( ) Hence, degree of f(). 80. ns. 0 Reason: ã log b c logb a a c log 5 7 log ns.. 0 ( ) 8. ns. Reason: Gven If f() + f ( ) () Replacng by, we obtan ( ) f f ( ) + f + f + () gan replacng by n Eq. () we obtan f f + f() f Now addng () and (), then we get f() f() + f(). 8. ns. Reason: tan lm 0( tan ) e.e lm e 0 () / π tan + (form ) e.e. 8. ns. 9 Reason: Number of jumps LHL RHL lm f() lm f() + lm f( h) lm f(+ h) h 0 h 0 lm 5 ( h) lm f( h) + + h 0 h 0 (6 ( ) 9.