Tiêu đề 08.01.24_SR.STAR CO-SC(MODEL-A,B_C)_Jee_Main_GTM-10(N)_KEY _ SOL.pdf ✅

Sec: SR.IIT_*CO-SC(MODEL-A,B&C) GTM-10(N) Date: 10-01-24 Time: 3 HRS JEE-MAIN Max. Marks: 300 KEY SHEET PHYSICS 1 D 2 D 3 B 4 B 5 C 6 A 7 A 8 D 9 A 10 B 11 C 12 B 13 A 14 D 15 B 16 A 17 C 18 D 19 B 20 D 21 2 22 5 23 3 24 8 25 246 26 5 27 27 28 4 29 3 30 50 CHEMISTRY 31 A 32 B 33 A 34 D 35 D 36 A 37 D 38 A 39 C 40 D 41 B 42 A 43 D 44 B 45 B 46 B 47 B 48 A 49 A 50 B 51 1758 52 3 53 25 54 6 55 1 56 2 57 9 58 6 59 2 60 8 MATHEMATICS 61 C 62 D 63 A 64 A 65 D 66 A 67 B 68 A 69 B 70 A 71 D 72 C 73 D 74 D 75 B 76 C 77 C 78 A 79 B 80 B 81 3 82 30 83 7 84 2 85 7 86 1 87 2 88 8 89 4 90 4
Narayana IIT Academy 08-01-24_SR.IIT_*CO-SC(MODEL-A,B&C)_JEE-MAIN_GTM-10(N)_KEY&SOL SR.IIT_*CO-SC Page NO: 2 SOLUTIONS PHYSICS 1. 1 2 mgd T   2. Angular momentum is defined by the equation L=MVR 3. Let height of liquid above second hole (2) be ‘H’    v g H h 1 2   Thus the force experienced by the tank at point ‘A’ is 2 1 1 d F A V dt     Similarly 2 F A V 2 2   V2 is the velocity of liquid coming out at second hole (2). At point B V gH 2  2 The net force on tank is, F F F  2 1        2 2 2 1     A V V A gh A g h  2 2 F h net 4. ' YA K L  ' ' eq KYA KK KYL L K K K KL YA YA K L          2 2 eq m T K m KL YA KYA       5. The given graph represent isothermal process and for isothermal process internal energy is constant. 6.
Narayana IIT Academy 08-01-24_SR.IIT_*CO-SC(MODEL-A,B&C)_JEE-MAIN_GTM-10(N)_KEY&SOL SR.IIT_*CO-SC Page NO: 3 A B C D +q -q 0 3 3 4 Kq KQ Kq a a a    4 q Q   4 2 3 4 6 A q Kq Kq Kq K V a a a a     6 A C Kq V V a   7. 1 8 sin ,cos 3 3     According to conservation of momentum mu 2mvcos   8 u 2v 3  v 9 e u cos 16    8. As the branch of the circuit containing 3 resistor is open so no current flows through it. 9. In equilibrium electrostatic attraction between the plates = spring force 2 2 0 q kx  A       2 0 0.8 2 CE k d d  A    2 0 2 0 0.8 0.2 2 A E d dk A           2 2 0 0 3 3 4 0.256 AE AE k d d      10. As magnetic flux is same through inductors, L i L i 1 1 2 2  1 2   i i 2 .....(i)
Narayana IIT Academy 08-01-24_SR.IIT_*CO-SC(MODEL-A,B&C)_JEE-MAIN_GTM-10(N)_KEY&SOL SR.IIT_*CO-SC Page NO: 4 Also, 1 2 10V 1A 10Ω i i     R  .....(ii) (i) and (ii) 1 2 A 3   i 11. Time constants of both the circuits 1 2 1 2    R R C R R but that does not ensure same charge at any time on capacitors or same current in branches of the capacitor. 12. At the bottom, 2 mv r N mg   13. 2 1 ; c r A r c      1 c r A   Sin r A 1   sin  c    sin sin c i A     sin sin cos cos sin i A A       c c  7 3 3 3 1 1 1 1 1 3 2 7 7 2 2 2            1 0 sin 30 2 i or i   14. According to Lenz’s law e.m.fs of the same magnitude in the clockwise direction are induced in the two loops into which the figure is divided. So, current is induced in the clockwise direction in the outer boundary but no current in wire AB. 15. 1 4.84 D mm d   Let required wavelength is 2 the according to given information 2 4.84 2 D mm d   1 2 2 1 1200 / 2 mm        16.