SOLUTIONS JEE Entrance Examination - Advanced/Paper-2 Code-7. Vidyamandir Classes. VMC/Paper-2 1 JEE Entrance Exam-2017/Advanced. p h 1.

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1 SOLUTIONS 7-JEE Entrance Examination - Advanced/Paper- Code-7 PAT-I PHYSICS p h.(a) K m md.(c) hc h md hc d h d d md From the figure, we can write P b S (From ΔOPS) and Q P (Given) P b( Q P) S S b P bq d d. (D) Magnetic field due to one section of star can be calculated as follows i B sin 6 sin 4a i B 4a Total field B B i a 4.(D) 5.(B) For any n-sided polygon h h cos x x n cos n h x h h cos n h cos n Total mass = constant 4 constant C C Differentiating with respect to time d d dt dt VMC/Paper- JEE Entrance Exam-7/Advanced

2 6.(C) Now Since d v dt d d v dt dt d is given to be constant dt v Time taken for the stone to reach the bottom of the well. t L g time taken by the sound to reach from the bottom of the well to the observer t L V Total time T t t s 7.(B) L L T g V Percentage error, For earth, For Sun + Earth, dt dl g L V s s. 5 dl. dl 6 6 dl l 6 % Ve GM 5 GM GM Potential energy = 4 m.5 GMm GMm mv GM v. v e v 4.4 km / s 4 km / s. e GMm L di di 8.(BCD) Let the currents through inductors be i and i. Then, VL L dt dt t t Li Li i.e., d L i d L i i L i.e. is fixed at all times. i L Hence (D) is correct Also, since inductor acts as an open circuit at t =, current through all branches is zero at t =. After a long time current through battery V i and L i L i and i i i VMC/Paper- JEE Entrance Exam-7/Advanced v

3 i L i L i and i L L L L 9.(AD) Since the charge lies outside the sphere, net flux passing through the sphere is zero. curved surface disc Option (C) is incorrect cos disc E d A curved surface disc Q E r 4 r Q rdr cos Q rdr 4 r 4 r r Q rdr r / Q curved surface Option (A) is correct r / Q / Potential at any point on the circumference of the flat surface is Hence it is equipotential Option (D) is correct Q E cos 4 ( / cos ) 4 E normal Which is not constant Option (B) is in correct.(ab) For (A) For Q E cos 4 8 p B Q p pq r QB Q8p 8 Clearly x cos r x Hence particle enters region through point P (r = radius of circle in which charge moves) For (B) The particle will reenter region if p B p QB Q / Q Q r Q Q 4 4 VMC/Paper- JEE Entrance Exam-7/Advanced

4 For (C) Distance between point p and point of re-entry into region = r mv y m qb For (D) When particle re-enters region through longest possible path, it will re-enter horizontally only. Hence at farthest point from y axis, if must be vertical with magnitude of momentum being p. hence.(ac) At P, x change in its linear momentum must be p. P, x d so, between P and P 999 bright fringes will be formed. So, A is correct. At P, there will be bright fringe. So, B is incorrect At P, there will be th bright fringe At angle θ, x d sin (assuming the circle to be of large radius) d d ate of change of path difference is x d cos Which decreases with increasing θ. so for small values of θ, path difference increases sharply with increases in θ resulting in closer fringes. So, angular separation between consecutive bright sports increases as we move from P to P. So, D is incorrect.(bd) If connected across X and Y : VXY VX VY V sin t V sin t V sin t sin t cos cos t sin V sin t sin t cos t V sin t cos t rms V VXY V If connected across X and Z. VXZ VX VZ V sin t V sin t If connected across Y and Z: 4 V sin 6 YYZ VY V 4 Z V sin t V sin t V cos t V is same if the voltmeter is connected across any two terminals. rms.(bcd) (A) L x sin y L cos x Hence L y L Hence NOT parabola (B) Torque about (C) B L mg sin Hence correct V sin t As there is no external horizontal force and zero initial velocity Hence it will move downwards VMC/Paper- 4 JEE Entrance Exam-7/Advanced

5 L L (D) Displacement of mid point cos 4.(BD) (A) Distance of F from point Q increases Torque of F increases (clockwise) Distance of force Mg from Q decreases. torque of Mg decreases (anti clockwise) Net torque is anti-clockwise direction increases Incorrect (B) Net torque = (C) As the wheel begins to climb, the distance of the force F from point Q increases initially Torque of F through X increases clockwise. Torque of mg decreases anticlockwise Net torque increases clockwise initially Incorrect (D) When force is applied vertically, then there will be no normal reaction at point Q and hence friction will also be zero. This will cause slipping of the wheel. Since, it is given that there is no slipping at point Q, therefore we can write F4 cos Mg cos (Clockwise) F Mg cos 4 As increases from to, cos decreases from to. decreases as the wheel climbs. V 5. (D) When voltage is set to, Correct CV charge supplied by battery V CV CV CV When voltage is raised to, additional charge supplied. VMC/Paper- 5 JEE Entrance Exam-7/Advanced

6 When voltage is raised to V, additional charge supplied CV CV CV V CV V CV CV Total energy supplied by cell V CV Final charge on capacitor CV Energy stored in capacitor CV Energy dissipated across resistor 6. (C) Energy supplied by cell V CV CV CV Energy stored in capacitor Energy dissipated across resistor E D E E C D E C ED CV CV CV 6 CV CV CV 7.(B) The motion can be visualized as a larger ring spinning around a smaller rotating disk, without slipping. Let angular speeds of smaller and center larger ring be &. In addition to rotation about its own axis of larger ring it is moving say with a speed v. Center of larger ring is moving in a circle of radius ( - r). So, V r...(i) Also, no slipping at the point of contact gives r V...(ii) (i) and (ii) give Now draw the FBD of ring N ma s m r...(iii) c Balancing forces in vertical direction for translational equilibrium, mg N mg m r i.e. 8. No answer matches Kinetic energy of ring can be calculated by two methods (i) Basic method KE(Total) mvcm Icm m r m g r m r (ii) Using ICO : Centre of smaller disk is the Instantaneous Centre of otation for ring so, KE I ICO m m r m r VMC/Paper- 6 JEE Entrance Exam-7/Advanced

7 PAT-II CHEMISTY 9. (A) Zn(s) CuSO 4(aq) ZnSO 4(aq) Cu(s).T Zn E E log F Cu.T.T. log. F F.T G FE F. F.F.T =. T. F. (C) IV > I > II > III Gaunidine is most basic due to more number of equivalent resonating structures of conjugate acid hence more stability of conjugate acid..(a).(c) III is least basic because of electron withdrawing effect of CH = CH group. G P V.9 kjmol.9 Jmol 9.45 Nm 6 6 m mol m mol 9.45 P bar bar 45bar VMC/Paper- 7 JEE Entrance Exam-7/Advanced

8 . (D) Acid oxidation state Acid oxidation state HPO + HPO 4 +5 HPO + H4P O (C) Mole of CH5OH m Tf 46 5 f f f T T 7 T Tf 7 7K 5. (D) 4Au+8NaCN + H O + O 4Na Au(CN) +4NaOH Cu+4HNO Cu NO NO H O conc. (Fe become passive on reaction with concentrated HNO ) Fe 6HNO Fe O (s) 6NO H O conc Zn NaOH(aq) Na ZnO H 6.(AC) BeO, AlO, SnO, PbO, CrO, SnO, PbO are amphoteric oxides. CrO is basic oxide, BO is acidic oxide, NO is Neutral oxide 7.(BC) Cloud is not emulsion type colloid. It is an aerosol type colloid. For adsorption rh and r S. Higher is critical temperature greater is extent of adsorption. Brownian motion of colloidal particles depends on the size of the particles and viscosity of the solution. 8.(ABC) VMC/Paper- 8 JEE Entrance Exam-7/Advanced

9 eactivity order for nucleophilic substitution reaction is IV > III > I > II 9.(ABD) Dimeric structure of BH : Dimeric structure of Al(CH ) : Dimeric structure of AlCl : Lewis acidity decreases down the group. Ea T.(AC) K P. Z e (According to Arrhenius equation).(ab) Ea T K A e A P Z According to collision theory, P value is generally less than unity but for some reactions P is greater than one and for such reactions, observed rate is greater than rate predicted from Arrhenius equation. For a reaction with P value greater than, implies that the experimentally determined value of frequency factor (A) is higher than that predicted by Arrhenius equations and such reactions proceeds rapidly without the use of a catalyst. (A) and (C). VMC/Paper- 9 JEE Entrance Exam-7/Advanced

10 .(BD) G T n K, G H TS H S T n K H TS, nk T H, S K decreases with increasein T ; H, S K increases with increase in T H, S K decreases with increase in T ; H, S K decreases with increase in T qsys Ssurr Tsurr S surr favourable means Ssurr is positive while Ssurr unfavourable means Ssurr is negative. If value of K increases with increase in temperature for endothermic reaction it means reaction shift toward forward direction because of unfavourable change in entropy of the surrounding decreases. Similarly if value of K for exothermic reaction decreases with increase in temperature it means reaction shift toward backward direction due to decrease in favourable change in entropy of surrounding. -4..(A) 4.(A) MnO KClO (s) KCl(s) O (g) 4 4 (W) (C) P (s) O (g) P O (s) HNO P4O 6NO5 4HPO4 or HPO VMC/Paper- JEE Entrance Exam-7/Advanced

11 (C) 6.(C) PAT-III MATHEMATICS 7.(C) x y z Total number of non-negative integers satisfying this equation If z is even i. e, z m; m x y m,,,, 4,5 m m No of. Solutions C C m 5 m 66 6 Therefore number of favourable cases Probability C 66 m C C 8.(A) f " x Since f, f Let g x f x x g g Therefore, g '( x) e (, ) f '( e) f '' x f '' x dx dx e ' f ' f e ' f ' f e f ' e 9.(C) The normal to the plane is : iˆ ˆj kˆ iˆ ˆj 4 6 kˆ 4iˆ ˆj 5kˆ 6 Therefore, the equation of the plane is x y z 4 5. VMC/Paper- JEE Entrance Exam-7/Advanced

12 4.(B) Taking st two OP OQ O OS OQ O OP OS OQ O PS Q Similarly QS P 4.(B) S PQ S is orthocentre a a a M b b b c c c a b c T M a b c a b c a b c a a a T M M a b c b b b a b c c c c T Tr M M a b c a b c a b c 5 5 5, s s, 4,,, 7 s C 5 9 C7 C 4.(D) 4.(C) 5 4 N C C4 5 4 N C C 5 4 N C C 5 4 N4 C4 C 5 4 N5 C5 C N... N5 C5 6 4 dy dx, 8 x 9 x 4 9 x y dx dt 4 9 x 9 x x y 4 9 x C y 7 C y x 4 9 x t 44.(BC) x( x) lim cos x x x VMC/Paper- JEE Entrance Exam-7/Advanced

13 x x cos ( x) x ( x ) lim cos x ( x) x x ( x ) lim cos x ( x ) x does not exist 45.(CD) cos cos cos cos cos cos cos cos Assume x tan and y tan y x ( x ) ( y ) ( x )( y ) y x ( x )( y ) Apply C and D x y x y ( x y x y ) x y x y ( x y x y ) ( x )( y ) ( x ) ( y ) x y ( x y ) x y. 46.(BC) cos x () cos x ( cos x sin x) sin x( cos x.sin x cos x.sin x) f ( x) cos x cos4x f '( x) sin x 4sin 4x sin x[ 4cos x] sin x cos x 4,,. 47.(AD) x y x x ( x x ) dx ( x x ) dx (CD) k,,,98 k x k k x x x x k k k dx dx dx x x x k k k k k k n dx n k x x k k VMC/Paper- JEE Entrance Exam-7/Advanced

14 n... I n n5 I n99 49.(CD) f x f x f x f x dy y dx x dy x e e y dx d e x y dx and x x e f x is increasing x f x e f x x e f x f 5. g x sin sin x sin x t dt g x sin sin x cos x sin sin x cos x sin sin cos sin g sin cos g 5.(D) OZ OY OX OY OX OY sin ( ). sin sin ( P Q) 5.(A) cos ( P Q) cos ( Q ) cos( P) In a ABC OX cos A cos B cos C [cos P cosq cos ] has minimum value /. 5.(D) 54.(A) n n an an p q n n p q VMC/Paper- 4 JEE Entrance Exam-7/Advanced

15 n n an an p q an an an a 4 a a a a a a a a a a p q p q 5 5, 5 p 5 q p q 7 p 7q 8, p q 8 7 p p 4 VMC/Paper- 5 JEE Entrance Exam-7/Advanced