Narayana IIT Academy

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1 INDIA Sec: Sr. IIT_IZ JEE-MAIN Date: 7--8 Time: 7:3 AM to :3 AM GTM-4 Ma.Marks: 36 KEY SHEET PHYSICS CHEMISTRY MATHS

VSD=9MSD=9mm. 3. v u lv T l T v v u Least count =MSD-VSD=.mm. Zero error=. 8. mm. Thickness= Main scale reading +L.C V.C+error. i.e, Thickness=.3..5mm 4.

2 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s SOLUTIONS PHYSICS. At any instant let be the angle made by v with u and let they meet after line T T v cos dt ut and v u cos dt l T T T v dt ucos dt l T vt u cos dt l ut vt u l v. VSD=9MSD=9mm. 3. v u lv T l T v v u Least count =MSD-VSD=.mm. Zero error=. 8. mm. Thickness= Main scale reading +L.C V.C+error. i.e, Thickness=.3..5mm 4. The initial etension of spring is mg k. Just after collision of B and A the speed of combined mass is v ( mv m v ). For spring to just attain natural length the combined mass must rise up mg k by (see figure) and come to rest. Now applying conservation of energy v m k m g and mg. k Sec: Sr.IIT_IZ Page

6m So get v g. k 5. 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s 6.

3 6m So get v g. k _Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s 6. ) As mutual repulsive force between the particles is internal for the system and as there is no other eternal force on the system,linear momentum of the system is conserved in any direction ) As the forces on the particles due to one on the other are equal in magnitude, opposite in direction and act along the line joining them always, net torque on the system due to these forces about any point in space is zero. therefore angular momentum of the system remains constant about any point in space 3) As center of mass of the system lies on the line joining the particles always and force on any of them is passing through C.M always, torque due to this force on any particle about C.M is zero. Hence angular momentum of any particle about C.M is conserved individually. 4) About any other point ecept C.M, torque on any individual particle is not zero. Hence angular momenta of individual particles change but total angular momentum of the system remains constant. L L and sin v v v tan 7. cos v Sec: Sr.IIT_IZ Page 3

4 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s 8. From conservation of angular momentum of the mass about the centre of the black hole, mvr mva and from conservation of energy GMm mv mv. a 9.. F mg sin mg tan ( is small). i.e, F dy mg mg d 4 a g P 5g g g geff 5g. Therefore pressure at centre= P 5Rg. If is temperature of the body at any instant and temperature of surroundings, from Newton s law of Cooling e, A,, s are same for both. 3 d ea 4 dt ms. i,e At any temperature d dt as As area is same for both mass of cube is less than that of sphere. Slope of the curve at any for cube is greater than that of sphere. At triple point substance eits is three states in equilibrium m Sec: Sr.IIT_IZ Page 4

5 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s Keeping pressure constant if temperature is increased substance will be in vapour state and if temperature is decreased substance will be solid state. Keeping temperature constant if pressure increased substance will be in liquid state and pressure decreased substance will be vapour state. 3. Internal energy 5 U nrt and PV nrt same for both). As final pressures are equal U f U f 4. Let the wave equation is y Asin t k Ui U PV U P U ( V is At the given instant particle at.9 m is moving towards mean position from positive etreme. i.e, 3 4 sin t k t k sin rad. its phase is i P P.589rad. At the given instant particle at is at mean position and is moving towards positive etreme. its phase is or rad or 4 rad etc. As wave is moving along positive -direction its phase is more than that at.9m and as these are within one wave length range phase of particle should be rad. Phase difference between the particles rad 3.79 T k k. Wave velocity v 3.79ms.9 k k dv E d 6. Potential at any point inside the shell=potential at any point on the surface potential at A=potential at C due to q and induced charges = Sec: Sr.IIT_IZ Page 5

6 q q q q 4 r 4 R 4 R 4 r 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s 7. When the switches open charges on differtent plates are as shown When all the switches are closed the two capacitors will be in parallel and net charge on the isolated plates B and C together (-+4=-6 c )is shared by them in the the ratio of their capacities final charges on different plates are as shown 8C flows through s fromplate A to ground, 8C C to B and 8C charge flows through s 3 from ground to plate D. 8. V V (as they are earthed ) E D V V, V V, V 6 6V B C A V A B Current through C D Current through 3 4 V V V 3 A A charge flows through s from plate From Kirchhoff s junction rule we can show that current through wire DE= 9. Electric field is induced and is the form of concentric circles about centre of cylindrical region. Sec: Sr.IIT_IZ Page 6

7 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s d db r db r dt dt dt E r r r E path AB at distance r from the centre. Consider a small path length dl along tan 3 AB L r L L L L L L de E. dl Edl cos dl eab r de dl AB 3 3. When the right wire starts moving, due to motional emf developed in it anti clock wise current flows in the frame, due to this there is mutual attraction between the wires and due to this the wires attain common velocity after some times. When they attain common velocity there will be no net emf hence no current flow, hence wires continue to move with the common velocity mv mv v v = common velocity. Loss in v ke mv m mv 4 Ni L. B ni. For permanent magnet we prefer a material with high retentivity (so as to make a stronger magnet) and high coercivity (so that magnetization any not be wiped out easily). For electromagnet we prefer high saturated magnetism low coercivity and least possible area of hysteresis loop so that electromagnet develops high magnetization, is easily demagnetized and energy loss in a magnetization cycle is least. Therefore, P is suitable for making permanent magnets and Q for making electromagnets. sin d E and id dt 3. J E t k c J E t k d J cos J de A dt AE t k c d cos Sec: Sr.IIT_IZ Page 7

8 4. Replacing it with string block system 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s Let at initial position F force is a applied then W.E.T. from A to B W S W F 3F X K Net elongation in spring 3F Qma 3CE K v D 5. m u f e v u 5 v 5 5 u 6 5 For eye lens 6 u 5 v 4 u e e cm v u f 5 u 5 u e e e e e and u 6v 59 cm cm v u f f 3 For objective f. 6. Due to total internal reflection light comes out through a cone of semi verte angle which is equal to critical angle. Total light energy is distributed over 4 solid angle. Sec: Sr.IIT_IZ Page 8

9 Solid angle made by cone at source cos T P cos 53 P For st minimum bsin bsin 3 () For st secondary maimum b sin 3 Sec: Sr.IIT_IZ Page _Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s..() sin 3 sin 3 4 sin 3 sin sin Charge is conserved. In order to fully convert an electron into energy, a positron (the electron s antiparticle must be involved). That is, electron+positron energy, Not electron positron+energy 9. For ecitation with least KE of colliding particle the collision should be perfectly in elastic and all the loss in KE is to be used for ecitation. mu mu M ( M m) v v M m maimum loss in KE is where v=common velocity after collision. m u M M K mu ( M m) mu K K M m M m M m 4 i. e.. 4 K K 63.75eV 4 3. y A B. B CHEMISTRY 3. NCERT-XI, VOL-II, PAGE NO : 43, NCERT-XII, VOL-II, PAGE NO : 45, NCERT-XII, VOL-I, PAGE NO : Cis - Co NH 35. Cl 3 4 H 3 H 488 N 4 - violet and trans- Co NH Cl 3 4 Ionization energy (in Kj/mol) - green

10 N _Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s O 34 O 75 F 68 F 55 If energy of electrons in HOMO of molecule is higher than energy of last election in its atom then ionization energy of molecule will be less than its atom 36. NCERT-XII, VOL-I, PAGE NO : NCERT-XII, VOL-I, PAGE NO : 5,5,5,54 In a group pairing energy of 3d elements greater than that of 4d and 5d series elements because 3d-oribital is more compact ( small size) than 4d-and 5d 38. NCERT-XII, VOL-I, PAGE NO : NCERT-XI, VOL-II, PAGE NO : 3 4. From Si to P to S to Cl, P d bond strength with oygen increases due to decrease in size(or energy) difference of 3d of central atom and p of oygen. As a result polymerization tendency decreases. 4. NCERT-XII, VOL-I, PAGE NO : Weak base and strong nuclophiles like I and RS give best yield for substitution than elimination 43. -I,+M groups OH, OR, OCOCH, NH, NR, NHCOCH, CH CH, C H 44. activates O and p positions due to +M effect but deactivates m-position due to I effect, But +I groups like alkyl groups activates all O, m and p positions. Sec: Sr.IIT_IZ Page

11 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s 45. Sucrose is non-reducing sugar, polysaccharides are non-reducing non-sugar 46. ConcHSO4, E e H H Less stable alkenes rearrange to more stable alkene by acid calatyst but not by base. 47. Molecule which is non superinposable on its mirror image is chiral molecule. Structures I, II and IV are nonsuperinposable on their mirror images. But compound one is optically inactive because its enantiomers are inter convertable ( conformers) compound two is optically inactive due to pyramidal inversion. 48. NCERT-XII, VOL-II, PAGE NO : 9,3 49. NCERT-XII, PRACTICAL MANUAL, PAGE NO : 9 Compounds Phthalein dye test OH CH 3 Red OH Bluish purple CH 3 OH No colour CH 3 OH OH Blue but takes longer time to appear OH Green fluorescent colour of fluorescein OH Sec: Sr.IIT_IZ Page

12 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s Take.g of organic compound and.g of phthalic anhydride in a clean dry test tube and add - drops of Canc. H SO 4 heat the test tube for about min in oil bath. Cool and pour the reaction miture carefully into a beaker containing 5 ml of dil. NaOH. Appearance of pink, blue, green, red etc. Colour indicates the presence of phenol OH group in the compound E 5. G VdP, area of ABCA and CDEC are equal AB A 5. In case of acetones at any concentration of it slow step is nucleophilic attack of carbanion at carbonyl carbon because ketones are less reactive in nucleophilic addition 5. Number of milli moles of AgNO Number of milli moles of NaCl 5. AgNO3 NaCl AgCl NaNO Cl 5 / 5 AgCl Ag Cl s Ag 5 Ag 9 M 53. NCERT-XI, VOL-I, PAGE NO : NCERT-XI, VOL-I, PAGE NO : I, II, III, IV 56. X X, Y Y, Z Z 57. M Na K CO4 NaKCO X Y Z NaKC 4 O4 X Y Z, / Number of atoms per unit cell= 4 4 X Sec: Sr.IIT_IZ Page

13 58. For inilial solution Tb w 6 6 w 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s wurea 6g For final solution left Tb g Weight of water vaporized =-5=75g 59. NCERT-XII, VOL-I, PAGE NO : NCERT-XII, VOL-I, PAGE NO :,, 4 MATHS 6. (4) Since, are the roots of the equation p q p and a q. /4 /4 4 /4 /4 Now, ( ) [( ) ] / / /4 [ ( ) ] ( ) p q (q) /4 /4 p 6 q 4q p q /4 6. (3) p 6 q 4q p q /4 /4 /4 /4 The total numbers of factors of n yz. is equal to the number of ways of selecting one or more out of n identical quantities of one type and remaining m distinct quantities. Hence, the required numbers of factors = (n + ) m 63. () F() is defined for [] [] 6 ([] 3)([] ) [] or [] 3 Sec: Sr.IIT_IZ Page 3

14 But [] [] 3, 4, 5, _Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s Domain of f (, ) [4, ) 64. () Let the roots of the given equation be + ip and ip, where p R product of roots ( ip)( ip) p, p R (, ) 65. () Given condition A B (A B)(A B) A B A AB BA B AB BA. 66. () Since abc, a bc and a 3b3c3are divisible by k, therefore a b c n k a b c n k a b c n k where n, n, n 3 are integers. Now a b c a b c a b c a b a b c a b a b c a b a b c (Applying C3 C3 C C ) a b n k a b n k a b n k = k a b n a b n k a b n is divisible by k (Since elements of are integers, is an integer) 67. (3) The four digits 3, 3, 5, 5 can be arranged at four even places in Sec: Sr.IIT_IZ Page 4 4! 6ways and the!! remaining digits viz.,,, 8, 8, 8 can be arranged at the five odd places in ways. Thus, the number of possible arrangements is (6) () = 6. 5!!3!

15 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s 68. (3) term independent of will be C (sin ) 5 (cos ) 5 C 5 sin 5 ( ) 5 5 So maimum value will be 69. (4) Given: C (i ) 3 5 i () A.M. > G.M. 7. () y z y z / 3.. y z y z y z 3 y z y is an integer is an integer A is Refleive /3 y is an integer y is an integer A is symmetric y, y z are integers As sum of two integers is an integer. ( y) (y z) z is an integer A is transitive. Hence statement is true. Also is a rational number B y is rational is refleive need not be rational i.e., is rational is not rational Sec: Sr.IIT_IZ Page 5

16 Hence B is not symmetric B is not an equivalence relation. 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s 7. () Let Ei denote the event that Ai dies in a year, then P(E i ),..n P (none of A, A,. A 3 dies in a year) P(E E E ) P(E )P(E )...P(E ) ( p) ' ' ' ' ' ' n n n Because E, E, E n are independent. ' p and P(E ) p i for I =, Let E denote the event that at least one of A, A,...A n dies in a year, then P(E) P(E E... E ) ( p) ' ' ' n n Let F denote the event that A is the first to die, then P(F / E).Also, P(F) = P(E).P(F/E) n = ( p) n n 73 () Since, P(A B) 3 P(A B) P(AB) 3 3 P(A B) P(A) P(B) P(AB) 3 3 p p 3 74 () lim 3 z (z ) 8z 4 8z lim z z 4 8z 4/3 3 3 z z lim 4 4 4/ /3.z 8z 4 8z 8z 75. (3) In the neighbourhood of =, f() = log sin g() = f(f()) = log sin (f()) log sin(log sin ) Since g() is differentiable at =, Sec: Sr.IIT_IZ Page 6 4

17 g ' () = cos(log sin) ( cos) 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s g '() cos(log ) (4) f () f '() f "() f "'(3) f '() 3 f '() f "() f "() 6 f '() f "'() 6 Now, f '() 3 f '() f "() f "() f '() 3 () Again f "() f '() f "() f '() () Again f "'(3) 6 (3) From () and () f "() f "() 6 f "() (4) () gives f '() 3 f '() 5 3 f () 5 6 (5) f () 6 f () f () f (3) f () f () 6 f () (a) is false f () f (3) 6 6 f () f (3) ( 5 6) 6 f () f () f (3) f () [ f () 6] (4) is false (4) Let, f () 3 a f '() 3 3 3( )( ) Now, f () a,f ( ) a Sec: Sr.IIT_IZ Page 7

18 The roots would be real and distinct if, f ()f ( ) (a ) (a ) or a 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s Thus the given equation would have real and distinct roots if a (, ) 78. () Let 79. () n n n f () a a a... a n a n n n n Then f() is continuous and differentiable in [, ], as it is a polynomial function of. Also, f() = a a a a f ()... a n n n n n and. (Given) Hence, by Rolle s theorem, there eists atleast one real number c (,) f '(c) Put n n i.e., c is a root of the equation a a... a a. d I ( ) tan t n n such that 8. () Let I f (r )d r I f ()d f ( )d f ( )d... f (99 )d 3 I f ()d f ()d f ()d... f ()d 99 I f ()d a(given) 8. () We have, sin for sin(cos ) cos for < < / / / sin(cos ) d cos d I I () 3 Sec: Sr.IIT_IZ Page 8

19 and,, Now, cos cos sin cos cos(sin ) 7--8_Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s / / cos d cos(sin )d I I () 3 from () and () I I 3 I 8. (4) a and ah R abc ()()()(3) 3 4 3(4) cos cos cos 83. (3) 4 cos ec 4 sin sin 4 6 n (3) Given equation of lines 3y 4y 3 3y 3y y 3 ( 3y )(y 3) y, y 3 3 o APO 75 o AM In AMP, sin 75 o AM 3sin 75 3 Now, length of chord of contact AB = AM o o (3sin 75 ) 6sin 75 Sec: Sr.IIT_IZ Page 9

20 3 ( 3 ) _Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s 86. () y 4b( (a b)) or y 4bX where (a b) X 4a(y (a b)) or 4aY where y (a b) Y For y 4bX, etremities of latus rectum (b, b) and (b, b) w.r.t. X-Y ais. i.e., (a, b) and (a, b)w.r.t. -y ais. For 4aY, etremities of latus rectum (a, a) and ( a, a) w.r.t. X-Y ais. i.e., (a, b) and ( a, b) hence the common end of latus rectum (a, b) Now for st parabola dy y 4b d dy b d y at (a, b) Also for nd parabola dy dy 4a or d d a Hence parabolas intersect orthogonally at (a, b) 87. () 9( 3) 9(y 4) y 9( 3) 8y 7y 44 9( 3) 8(y 9y) ( 3) 8 y ( 3) 8 y Sec: Sr.IIT_IZ Page

21 9 8 y 9( 3) y ( 3) e 9 9 e _Sr.IIT_IZ_Ph-I_JEE-MAIN_GTM-4_Key&Sol s y 88. (4) 6 9 Locus will be the auiliary circle y (3) Range of if and only if range of g() 3 b c is f () tan (3 b c) is, [, ). This is possible only when discriminate of the equation 3 b c is equal to zero. i.e. b = c 9. () Vector ((3i ˆ j ˆ 4k) ˆ (4i ˆ 3j ˆ 4k)) ˆ is perpendicular to i ˆ ˆ j mk ˆ m Sec: Sr.IIT_IZ Page

Narayana IIT Academy

Narayana IIT Academy INDIA Sec: Jr. IIT_N10(CHAINA) JEE-MAIN Date: 7-08-18 Time: 08:00 AM to 11:00 AM CTM- Ma.Marks: 60 KEY SHEET MATHS 1 4 5 6 7 8 9 10 4 4 1 11 1 1 14 15 16 17 18 19 0 4 1 1 4 1 4 5 6 7 8 9 0 4 4 4 1 PHYSICS