XII-IC (C/F) : 11/12/ (i) m As per question, pulley to be considered as a circular disc. Angular acceleration of the pulley is

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1 CDE REVISIN NARAYANA I I T / N E E T A C A D E M Y REVISIN TEST - 0 XII-IC (C/F) : // PHYSICS CHEMISTRY MATHEMATICS The free body diagra of pulley and ass The equation of otion of ass is g T g T a or a..(i) As per question, pulley to be considered as a circular disc. Angular acceleration of the pulley is I Here, TR and I R (for circular disc) Ra g R T a (Using (ii)) a a or a g or a g R

2 V = wr V = 0 0. = /sec Sphere is rotating about a diaeter so, a = ar but, R is zero for particles on the diaeter B : <w> = w0 / I Rotational KE Traslational KE v K K R ( R ) The centre of ass of the whole carpet is originally at a high R above the floor. When the carpet unrolls itself and has a radius R/, the centre ofass is at a M ( R / ) M height R/. The ass left over unrolled is R M R 7 The decrease in PE = MgR g MgR 8 When one string is cut off, the rod will rotate about the other point A. Let a be the linear acceleration of centre of ass of the rod and be the angular acceleration of the rod about A. As is clear fro figure

3 T A g g T = a (i) g (l / ) g...(ii ) I l / l l g g a r l g g Fro (i), T = g a = g Applying principle of conservation of energy, Mgh = Mv I v v Mg v I R 6g v Mv Mv I 6 R I MR Mass of sall eleent of length d of the rod at a distance fro the centre, d d l A sin d B Perpendicular distance of this eleent fro the ais = sin Moent of inertia of this eleent about, di d ( sin ) l

4 I d ( sin ) l sin l 0. l l I sin sin l Let be the angular velocity of the rod about the centre of ass C, As angular ipulse = change in angular oentu about the centre of ass of the syste, therefore, M L/ M C L/ J=MV J.L/ = Ic ML L MV. V L As is clear fro figure Y θ θ X t T v sin / g g sin As the ball (of ass ) rolls without slipping K rot. or K trans. K total K trans. 7 Total KE of ball in position B = g (R-r) < > =.

5 R-r r. Ktrans. g ( R r ) v 7 0 g ( R r ) 7 0 g R r 7( R r ) v up R v B A upper cylinder v lower cylinder lower v=0 D B lower up v R up lower.. Let v be the velocity of the centre of ass of the sphere and be the angular velocity of the body about an ais passing through the centre of ass. J = Mv; J h R MR fro the above two equations v h R r fro the condition of pure rolling, v R R 7R h R h vcm R 0 vcm R v vcm R R K b R

6 6. 9 = R K s R R R R R Ks Kb 9 / For no slipping at B, v0 R 0 For no slipping at A, v v0 R 0 v v0, v Mv0 MR 0 Mv0 R K p Mv M v0 Mv0 K p : K c : Kc 7. Let the ass of cylinder is and its radius is R. if C has to be at a fied position, f g sin for pure rolling. 6

7 a R f.r f R / R f a or a g sin g sin 8. g R r v v v r 0 g R r 7 v 0 g R r 7 R r v 9. F h R F h R F, a R R F F h R R FR Fh FR, h 7 R 7 h R. R v v0 v0 sin v0 cos 0. 7

8 v v0 cos t T R v0 CHEMISTRY... rder of M.P or.b.p. or critical teperature: H H Te H Se H S. S S S H+H H S S H (Sulphuric acid) (Perooono-sulphuric acid) H is therally unstable and it decoposes easily. H l H l g HCl HBr HI acidic strength has etals present in traces in the glass of the vessel density on central ato I 0 HN HI 0 N H Decreasing order oftheral stability of oy acids of chlorine HCl ; HCl ; HCl; HCl in HCl, chlorine is in +7 oidation state C I C I I Na S NaI Na S 6 - Bond dissociation energy of F is les than that of Cl - Cl has higher E.A. than fluorine HF is weaker acid than HCl, dur to higher bond energy 8

9 Due to larger size of iodine ato it can accoodate upto seven sall fluorine atos around, it while due to saller sizes of chlorine and broine atosdo not accoodate seven fluorine atos, i.e., steric factor doinate in case of chlorine and broine MATHEMATICS 6. n i in Hence, least positive integral value of n =. 6. We have z z z z z z z z z z z z z Also, for z z z 6. Therefore, the greatest value of z is. a b c b c a cis cis cis cis cis cis Where cis represents cos + i sin 9

10 Equating real parts of both sides cos cos cos 6. 0 k k i cos k 0 k k i sin i cos k 0 k k i cos i sin k 0 i k 0 i k i e i e k k th i (su of roots of unit ) i 0 i. We have, 6. sin 7 cos i sin su a s Product of roots pq Equation arg z i arg y i 0, y 0, y tan 0

11 (i) y 0, 0, y arg z i arg i y y, y and tan y 0,, y..(ii) Hence, (i) and (ii) have no point of intersection. 67. z = cos + i sin = ei z z z z... z 9 z = z e e e i0 i z i ei0 e i ei cos 0 i sin 0 = i sin i cos 0 sin 0 = sin cos 0 I z sin o cos 60 = o given o sin = sin o = 68. zz zz 0 zz z z z 0 z z z 0 Let z iy so, z (i) y z y iy

12 z y iy z z y Thus equation (i) becoes y y 0 y y 7 y y y 7 Here y y 7 Let y, y 7,, y y, y 7,, y y 7, y,, y y 7, y,, y 69. Area 8 6 = 8 Let cot p, then p cos e i cot p e i e i pi. pi i cot. i cot e i. i e e i. e i e0 70. z =. and r ei arg z arg Let z rei t z rr e i z rr arg z arg

13 z r e r e r r e = 7. i.e i i i i i 8. So, =. So, =. So, = 7. a as a Circle with centre (, ) and radius. Hence C = z 7 a z in 7. z = 0; z = ; z = i ; z z z z z Note that z n = z has n + solutions. Hence, z = 0 or z = Again z = zz = z z = total nuber of roots are. 7. Let z = cos + i sin, [0, ]. Then z z z z z z z = cos isin cos isin = cos Hence, cos or cos If cos, then 7,,, If cos,then

14 , 6, 7, 8 Hence, there are eight solutions. z k cos k i sin k, k, ( + ) = ( + ) ( ) = ( + ) ( ) = ( ) ( ) = ( ) So, least value of =. a b c / ; a b c / c b c bc ac c a b a a a a a n 80. ( =. = ) n, 6 b q br qc c r 7 p p p 8 6 p Product 8 6 p p p p 8 is a coon factor, then f a b f 0 a b f c d f 0 c d 0 b d a c 0 Subtracting tan o, tan o, tan o tan o tan o tan o tan o b d c a 8.

15 We know that. But cos sin cos sin ;sin. Both can not be true siultaneously sin 8 8cos 8sin 8cos sin cos 8 9 cos 87. 8sin 8cos sin cos Let t sin cos 7 8 t or 7 t sin cos 0 or sin cos t a a a 0 a a 9 a 0a 0 a,,