ANSWERS, HINTS & SOLUTIONS HALF COURSE TEST VIII PAPER-2

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1 AIITS-HCT-VIII (Paper-)-PCM(Sol)-JEE(Advanced)/7 In JEE Advanced 06, FIITJEE Students bag 6 in Top 00 AIR, 7 in Top 00 AIR, 8 in Top 00 AIR. 4 Students from Long Term Classroom/ Integrated School Program & 44 Students from All Programs have qualified in JEE Advanced, 06 FIITJEE ALL INDIA INTEGRATED TEST SERIES JEE(Advanced)-07 ANSWERS, HINTS & SOLUTIONS HALF COURSE TEST VIII PAPER- ANSWERS KEY Q. No. PHYSICS CHEMISTRY MATHEMATICS. C C A. C A C. C B A 4. B A A. C C D 6. A A A 7. B D C 8. B D A 9. A, C C, D B, C 0. A, B B, D A, C. A, C A, B, C A, B. A, B, D A, B, C A, B, C, D. A C B 4. B D B. A D D 6. B A B 7. D B C 8. A A B FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 66949, Fa 694

2 AIITS-HCT-VIII (Paper-)-PCM(Sol)-JEE(Advanced)/7 Physics PART I SECTION A mm 0 0. Equivalent mass m m 0 Mechanical energy at any instant mr v mv k r mv k 4 d 0 kv 0 dt m a k T 0 a 00 m e 0 0. y = 0 y y y = f y 0 4k f 4 y f 4 ma 4 0 4k m a 6k. Let length of wire be L T sin = mg T = mg...() strain L L L L stress Y strain T A L T Y A mg AY...() L FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 66949, Fa 694

3 AIITS-HCT-VIII (Paper-)-PCM(Sol)-JEE(Advanced)/7 4. mg mg sin( t ) k k At t = 0, = 0, mg mg cos t k k mg [ cos t] k mg k mg k Origin (negative etremity) Equilibrium position Positive etremity. Stress at stress mg mg, strain AL ALY du A d = energy density stress strain m g du YAL m g L 6AY [ d] 0 6. d T sin dm (A )Rd dm dmw R T YL R w A L L w R L Y L = R L R L R w R R Y d T sin d d T sin 4 9. Gravitational field inside the cavity is E Gr where is mass density and r is separation between centre of sphere and centre of cavity. Applying law of Conservation of energy : GMm GMm mv 0 esc R 4R Solving V esc 88GM. 4R 0. KAd d H K( rl) dr dr r dr KL d r H r dq H KL( ) 80 dt r ln r dm L 80 dt FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 66949, Fa 694

4 4 AIITS-HCT-VIII (Paper-)-PCM(Sol)-JEE(Advanced)/7 dm 80 (80 Cal/gm)= Cal/sec dt 4. dm 000 kg/sec dt 4.. Impulse = change in momentum (v v ) (i ˆ ˆj) As impulse is in the normal direction of colliding surface tan tan 90 tan. 6. A ' A 4 A tan 7 4 y A sin(wt K 7 ). A A 6A A dv dy -4. a a av av y 0 dt dt dvy ay 0 a total a ˆ ˆ ˆ i a y j av 0 i dt ay cos (ay) v a a cos t total normal acceleration 0 a v0y (ay) v 0 a a sin N total we have, av (ay) 0 v 0 a total a n -6. From conservation of momentum = V sin + sin cos V = m/s V V ( sin ) 9 90 V 0 sin 6 also e 0.7 cos 4 also I sin = I 9 I = NS 0 m/s V d sin m/s FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 66949, Fa 694

5 AIITS-HCT-VIII (Paper-)-PCM(Sol)-JEE(Advanced)/7 7. c p c v = R c c p v c v = c p = = 9 = R R c v A = c p A = R c v = B R R 7 c p = 7R B c n c pa B pb nb c n c va B vb nb n B = moles = R = R 8. Bulk Modulus B = P Compressibility K = B = P K = = P P P P Since the process is adiabatic, P V = P V V P = P V = P = P V V K = P = P P n A ng RT ( ) 8. T P = = = 4.9T V V V V K = 9 9 T = T Pa. V SECTION C. Let man catches ball at time t = T sec. For ball y ais Sy Uyt ayt 0 (0sin7)T ( g)t 40 sin7 T g u rel t = 0 v=0 m/s t = T v T? FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 66949, Fa 694

6 6 AIITS-HCT-VIII (Paper-)-PCM(Sol)-JEE(Advanced)/7 Displacement of ball w.r.t. train in horizontal direction is zero (S ball/t ) (U ball/t )t (a ball/ T )t y 0 (0cos7) ( a T ) 40 a r For train 40 VT UT art 0 VT 4 m/s ] B = 7 A. Form linear momentum conservative (for collision B & C) v0 mv0 mv v A B A C B C V 0 V 0 V 0 f = mg a A FBD A f B C g aa mg,abc g aa /B va/b UA /B aa/bsa /B a AC v0 g 0 L v 0 6 L m g For equilibrium of block B F 0 and Fy 0 Fy 0 ; N N mg mg N For equilibrium of block A N N F F 0 N ( ) or F [ ] mg ( ) = 4 N ( 0.4) ( 0.4) 0.6 F mg N N N N y 4 y FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 66949, Fa 694

7 7 AIITS-HCT-VIII (Paper-)-PCM(Sol)-JEE(Advanced)/7 4. This happens of after collision both ball and inclined plane have same horizontal velocities say V say V be initial velocity of ball V y be vertical velocity of ball after collision, m mass of ball, m mass of inclined plane Conservation of linear momentum in horizontal direction mv (m m )V...() coefficient of restitution = [V sin (V sin Vy cos )] 0 V sin y V cos V sin...() Component of velocity of ball along the inclined plane remain same V cos V cos V sin...() y Vy sin (V V )cos using() & () Vy cos V sin V V tan V V V ( tan )...(4) (m m )V m V...() Divide m m cot ; m cot 0 m m tan. V + V = (/) () = V V Solving V 4 speed = ( ) ( / 4) 7 = 8.7 m/s 6 0m/s 60 V V FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 66949, Fa 694

8 8 AIITS-HCT-VIII (Paper-)-PCM(Sol)-JEE(Advanced)/7 Chemistry PART II SECTION A. Lattice energy > Hydration Energy compound insoluble. 7. h h 0 = / mv v = 0 h = hv 0 = 4ev hc 9 4ev J = NaOH CaO R COOH R H Na CO. G = Gº + RT lnq At equilibrium G = 0, Q = K Gº =.0 RT log K 4. Gº =.0 RT log K 460 logk K K CD AB = 0.8 C 0.68 A. Cl is o-p-directing 6. CH CH CH CH conc. NHO CH NO CH NO SECTION C. nh h n =. PSO. KP ; P.P SO O on solving P 8P SO SO K PSO P 4PSO 4PO FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 66949, Fa 694

9 9 AIITS-HCT-VIII (Paper-)-PCM(Sol)-JEE(Advanced)/7 4. Ph COOH 4 HOOC Ph Ph HOOC COOH Ph Ph HOOC COOH Ph Ph HOOC COOH Ph Ph HOOC COOH Ph All Cis (All Trans). OH H OH H O FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 66949, Fa 694

10 0 AIITS-HCT-VIII (Paper-)-PCM(Sol)-JEE(Advanced)/7 Mathematics PART III SECTION A. z 6i Im(z) is parabola having focus 6i and directri as real ais (arg z) min =, (arg z)ma = 4 4 arg(z) 4 4 Arg (z) Arg z = arg (z)., sin y sin y 0 =, y = n - / ni Hence number of ordered pair solution (, y) is 4.. f '() = 4( ) ( + ) ( + ) = 0 =, - y = f() f(-) = positive - - f() = positive f(-) = K < 0 K < 9 f(-) = 7 + K > 0 K > - 7 f() = K < 0 K < 9 K (-7, 9) 4. ( + ) n = C 0 + C + C +.C n n. We have C + C + C + C 7 +.= n- () ( + i) n = C 0 + C i + C i + C i C C + C C = n n sin () 4. () () C + C = n n sin 4 Put n = 6. 6 C + 6 C C =.. cos cos y cos y cos y cos y cos cos y.cos y cos cos y = cos sin y cos ( cos y) = sin y cos sin y 4sin y cos y = cos cos y cos sin y FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 66949, Fa 694

11 AIITS-HCT-VIII (Paper-)-PCM(Sol)-JEE(Advanced)/7 cos = cos y sec cos y sec cos y. 6. y + y + y = + + (y ) + ( y ) + (y ) = 0 R D 0 (y ) 4 ( y ). (y ) 0 (y ) ((y )) 0 (y y + ) (y + y ) 0 ( y )(y ) 0 (y + ) (y ) 0 y, 7. () + (y) + (4z) () (y) (y (4z) (4z) () = 0 a + b + c ab bc ca = 0 a b b c c a 0 a = b = c = y = 4z = z =,y,z y =, z 4 4 y = z, y, z are in H.P z 8. Let the number is 4 6 We must have < < < 4 < < 6 Clearly no digit can be zero. Thus total number of such numbers = 9 C 6 = 9 C 9. Sin 4 (k + sin ) + sin (k + ) = sin Sin 4 (k+ ) sin 6 + sin (k+) = 6 4 sin sin sin k + = = sin k 4 sin sin sin sin FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 66949, Fa 694

12 AIITS-HCT-VIII (Paper-)-PCM(Sol)-JEE(Advanced)/7. Given circles touch eternally at real ais. So, the centre C of the desired circles lies on real ais, which has radius r. thus CC = CC = + r. C C = C C = CC = r ( + r) = + ( r) r = / So, the centre of desired circle is at + or 7. SO, required equation of circles are 7 z and z Y C C C (, 0) (, 0) C X. Equation of line through A, (4, ) is 4 y r cos sin A (4 + rcos, + rsin). 4 + rcos = 8 r = 4 sec. AB = 4 sec. Similarly AC = cosec 6 9 cos sin AB AC So AB + AC = 4 4sin cos cos sin sin. ae =, a = ( + ) = ) e 4. Foci are S (, ) and S (4, ) SN PS 40 S N PS. The line L meets the parabola, a a a a a 6 a 0. But and a a 6 a 6 a a, a C : y at the points for which 6. If a > 0, then a =. Coordinates of A and B are (, 6) and (, ) respectively. Equation C : is y or y coordinates of the verte O of the parabola C are (0, -). Slope of OA =, slope of OB = the angle between OA and OB is tan tan / FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 66949, Fa 694

13 AIITS-HCT-VIII (Paper-)-PCM(Sol)-JEE(Advanced)/7 7. z z and z z z, z, z form an equilateral triangle z z z arg z z, z = 8. z + z + z = z = 0, SECTION C 4y 4 y. Equation can be written as, b = 6 9 P P P P P P b. an an an a a a... a a a n n n Adding we get n r a a a a r n n an an an 4 = 9 It can be logically seen that n cannot be greater than 7 and for a to be positive integer only n = 7 is satisfied 9 n 7,a We know that in case of ellipse if PMN CQ then PM = b, PN = a hence a =9, b = 7.. Let, be the roots in 0, then the roots will be,,, and,,, Sum = (since product of roots is one) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 66949, Fa 694

ALL INDIA TEST SERIES

ALL INDIA TEST SERIES AITS-CRT-I-(Paper-)-PCM (Sol)-JEE(Advanced)/7 In JEE Advanced 06, FIITJEE Students bag 6 in Top 00 AIR, 75 in Top 00 AIR, 8 in Top 500 AIR. 54 Students from Long Term Classroom/ Integrated School Program