CLASSROOM CONTACT PROGRAMME

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1 CLASSROOM CONTACT PROGRAMME (Academic Session : 05-06) ENTHUSIAST,LEADER & ACHIEVER COURSE PHASE : ENTHUSIAST, MLM, MLN, MLP, MLQ, MAZE, MAZF, MAZG, MAZH & MAZI TARGET : PRE-MEDICAL 06 Test Type : MAJOR Test Pattern : AIPMT TEST DATE : TEST SYLLABUS : SYLLABUS - 0 ANSWER KEY Que Ans. Que Ans. Que Ans. Que Ans. Que Ans. Que Ans. Que Ans. Que Ans. Que Ans. HINT SHEET. Since, LHS is displacement, so RHS should hae dimensions of displacement. Also argument of a trigonometric function should be dimensionless. In expression () argument is not dimensionless and in expression () a/t has not the dimensions of displacement.. Aerage elocity Displacement Time Here. ag R H T / H maximum height and sin R range g y R H T time of flight sin g x 00CM sin g ag cos HS - /8

2 . K Emax. E T V min. HS - /8 Gien: U(x) x x du x x x dx When U is minimum, (du/dx) 0 x 0, ± U min. (x ± ) J (KE) max. (KE) max. x 9 J mmax. (KE) 9 9 m max max. max. m / s. 5. Moment of inertia of the rod about a perpendicular axis PQ passing through the centre of the mass C, A I CM ML P C Q L Let N be the point which diides the length of the rod AB in ratio :. This point will be at a distance L from C. Thus, the moment of inertia 6 I' about an axis parallel to PQ and passing through the point N. N L L ML ML ML I' ICM M If K be the radius of gyration, then K B I' L L. M 9 Maj Test F Target : Pre-Medical 06/ Here, Mass of a body, M 5.00 ± 0.05 kg Volume of a body, V.00 ± 0.05 m Density, M V Relatie err in density is, M V M V Percentage err in density is, M V M V % + 5% 6 % (m m m )F 7. T m m m m T F (m m )F Similarly, T m m m m T F (m )F T m m m m T F 8. If the mot pumps water (density ) continuously through a pipe of area of cross-section A with elocity, then mass flowing out per second. m A...(i) Rate of increase of kinetic energy m (A )...(ii) Mass m, flowing out per, sec, can be increased to m' by increasing be ', then power increases from P to P'. A ' P' P' ' P P A m' A' ' Now, m A As, m' nm; ' n P' P n P' n P. 00CM056508

3 9. Difference in kinetic energy 00CM / m[( 5gr ) ( gr ) ] mgr 0 0J. 0. The moment of inertia of the unifm rod about an axis through one end and perpendicular to its length is, m I where m is the mass of the rod and is the length. Tque ( I) acting on the centre of graity of rod is gien by mg I mg m mg a g.. The ball is thrown ertically upwards then accding to equation of motion, (0) u gh...(i) and (0) u gt...(ii) From eqns. (i) and (ii) gt h When the ball is falling downwards after reaching the maximum height. s ut' + g(t ') h (0)t ' g(t ') h t t ',t ' g Hence, the total time from the time of projection to reach a point at half of its maximum height while returning t + t' t + t. Phase/ENTHUSIAST, MLM,N,P,Q MAZE,F,G,H & MAZI/ p(t) A(icos kt jsin kt) F x F F y p y p x p d F [p(t)] Ak( isin kt jcos kt) dt F.p A k( coskt sin kt sin kt cos kt) 0. In first case, let at point P, its kinetic and potential energies are equal, i.e., m mgh h g...(i) V P h PEKE O Ground In second case, when body's elocity is then at the same point P, () mg P g KE m. Velocity of A befe collision gh from conseration of momentum A m h B m m gh mvv gh From conseration of mechanical energy gh m h h mgh HS - /8

4 Maj Test F Target : Pre-Medical 06/ I I HS - /8 R MR M 5 5 n n 9. From Newton's second law, F dp dt If F 0, then dp 0 dt p constant. Thus, if total external fce acting on the system is zero, then linear momentum of the system remains consered. MR V I r R E K MV MV E 0. R :. Net displacement of the cyclist is zero, since the initial position coincides with the final position. Aerage speed of the cyclist total dis tance traelled total time taken OP PQ QO km/min 0 km/min km/hr. km/hr. 0. Let T is the tension of the rope. When monkey climbs up with acceleration a, then T m g m a...(i) When another monkey goes, down with same acceleration a, then m g T m a...(ii) Hence, (m m )g (m + m )a a g 5 m m g a m m g a. 5. m/s m/s Befe collision m/s After collision Let be the elocity of ball after collision. Collision is elastic. e relatie elocity of separationrelatie elocity of approach + 6 m/s (away from the wall) ˆi ˆj kˆ L r p L r p ˆi( ) ˆj( ) k( ˆ 6) 0iˆ j ˆ kˆ L has components along y-axis and -z-axis but it has no components in the x-axis. The angular momentum is in Y Z plane, i.e. perpendicular to x-axis. drel t 0 s rel Fce, F i ˆ 5ˆj 6kN ˆ Displacement, s 0i ˆ 0j ˆ 0kˆ m wk done W F s ( i ˆ 5ˆj 6k) ˆ (0i ˆ 0ˆj 0l) ˆ 50 J. 9. It is clear from figure that co-dinates of the centre of mass C, X CM (0, a) (m ) kg (m ) 6 kg (a,0) O (0, 0) (m ) kg m x m x m x m m m 6 0 a 0 a CM056508

5 Phase/ENTHUSIAST, MLM,N,P,Q MAZE,F,G,H & MAZI/ Y CM (X CM, Y CM ) 00CM m y m y m y m m m a a 6 5 a a, 5 5 a a Hence, OC ˆi ˆj 5 5 Angle made by OC with x-axis y a / 5 x a / 5 tan tan 5 0. Accding to theem of parallel axes, I R R M M(R) M 5 5 MR +. Aerage elocity. a C. MR MR 5 5 displacement total time taken Displacement 0, as boat comes back at the starting point Aerage elocity 0. r r r r Momentum p m m r tan rg tan 90 0 tan 5 rg 0m / s. Here, the centre of mass of the system remains unchanged when the mass m moed a distance L cos, let the mass (m + M) moes a distance x in the backward direction. (M + m) x ml cos 0 x (ml cos)/(m+m) 5. n u n u [kg][m][sec] n [0gm][0cm][0.sec] 000gm 00cm sec n 0gm 0 0.sec 000 n 0.N 0 6. F ertical motion, time of flight t g f hizontal motion R t g 7. Wk done area between F x cure and x-axis area of trapezium (6 + ).5 J 9. No external hizontal fce is applied, a CM 0 Since, CM 0, hence x CM Let magnitude of P is x. then P Q xj ˆ Q xj ˆ xi ˆ Q (x) ( x) 8. ai ˆ bj ˆ R H max x 8 5 u cos a and u sin b tan b a also u sin u sin g g tan b a b a. It is not possible that after collision one ball moes along the iginal line of motion while the other ball moes along some angle (a) with iginal line.the momuntum perpendicular to iginal line of motion cannot be consered in this situation. Initial momentum along perpendicular direction zero. Final momentum along perpendicualr direction m sina. Hence momentum is not consered. Hence the situation is physically impossible. HS - 5/8

6 . Rg ms.. Moment of inertia of the square plate about ma XY is moment of inertia about ZZ' can be 6 computed using parallel axis theem, Z Y HS - 6/8 Z X a I ZZ I XY + m a ma ma ma 6 I0 ( ) 5. Iag 6. (A) NH OH + NH Cl Basic buffer. (C) NH OH + HCl NH Cl + H O 0 0 mol 0 mole 7. (In BaI ) AgI (S) Ag (aq.) I (aq.) S S C K SP [S] [S + C] K SP S + SC {S negligible K SP SC K SP M In pure water s K SP 0 8 M 8. PCl 5 (g)pcl (g) + Cl (g) 5 mol 0 0 ( ) 9. N N 50. Cl fms fcc and Na + occupy all o.. per unit cell Na + and Cl. 5. n r 0.59 A Z () r r 0.59 (0.59 () ) 0.59 (6 9).855 A Maj Test F Target : Pre-Medical 06/ H e ng e Addition of inert gas at constant olume does not affect equilibrium bacause actie mass is not changed. 5. CH (g) + O (g) CO (g) + H O() L L 00L 00L 00L Vo 0 00 V air V air Vo. L B ion fms ccp No. of B ion is fcc 56. h No. of A + ions T.V 8 Fmula A B AB Kw K K a b, Independent from concentration. 57. T r n h / m ze ze / nh 58. K p T n /z Hg mmhg P Fe (s) + H O() Fe O (s) H 56g 8 g gm assume ml water g L.R is Fe 56 g Fe gies g Fe O 77. g 60. a (r + +r ) r + +r a pm 00CM056508

7 Phase/ENTHUSIAST, MLM,N,P,Q MAZE,F,G,H & MAZI/ eq () eq() eq () ( 0) ( 0) H H +800 KJ 6. m + then n 6. H e ng e 6. mg N 0 mol mol No. of N atoms mol.5 0 N A atoms. 65. in bcc a r in scc a r. / 5 KSP F M(OH), S / KSP 7 0 F M(OH), S s s p s p d n f d sub shell s / / 0 5 M 0 6 M. 7. CH (g) + O (g) CO (g) + H O() mol mol? (L.R) mol O gie mol H O mol O gie mol H O 75. T f ( + ). K f m 0.68 (+) %. 76. ph 7 + PKa log C log R H.Z n n n n R. (5) 9 9 5R 78. w nrt n P P mol 69. stoichiometric coefficient min 70. Fact 7. H e ng e 7. G T.S LR G 00.0 n R log V V n J 79. mol Fe (SO ) mol Fe mol FeSO mol Fe 80. i c c c log 8 59 J/ mol 05 c 6 c c c 5 Fie times dilution 00CM HS - 7/8

8 Maj Test F Target : Pre-Medical 06/ F water [H + ] [OH ] K w [H + ] [OH ] K w 0 is applicable at 90 C 8. In melting process entropy increases 8. In one unit cell No of selinide ion No of Li ion 8 Fmula Li 8 Se Li Se. p n 85. o p n N 0.5 n N m n n N n m N It is a salt of strong acid and weak base therefe it's nature will be acidic. In case of basic solution solubility increases due to odd ion effect. 87. RT RT M M w O w H To On compresion T as V 89. O fms molecular solid (molecule) 90. [Cu(NH ) Cl]Cl (i > ) HS - 8/8 00CM056508

CIRCULAR MOTION EXERCISE 1 1. d = rate of change of angle

CIRCULAR MOTION EXERCISE 1 1. d = rate of change of angle CICULA MOTION EXECISE. d = rate of change of angle as they both complete angle in same time.. c m mg N r m N mg r Since r A r B N A N B. a Force is always perpendicular to displacement work done = 0 4.