PROGRESS TEST-3A MRBK-1801,1802. Test Date:

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Lenard McBride
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1 (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_05--7 [ ] PROGRESS TEST-3 MRB/MRB-80,80,803 (G & B) MRBK-80,80 NEET PTTERN Test Date: Patna-, Ph. No. : / /6/ /5 Patna-, Ph. No. : / 8, / 5

2 [ ] (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_ () rea of a parallelogram = a b where, a and b are sides of parallelogram PHYSICS Given, a p 5i 4j 3k and b q 3i j k i j k a b a b i(4 6) j( 5 9) k(0 ) a b i 4j k Thus, area, a b () (4) () 684 sq units. () For two particles to collide, the direction of the relative velocity of one with respect to other should be directed towards the relative position of the other particle r r i.e., r r direction of relative position of w.r.t.. 3. () v v and v v direction of velocity of w.r.t.. so for collision of and B r r v v r r v v For both cases t h cons tant. g Because vertical component of velocity will be zero for both the particles. 4. () u sin u sin cos R u sin g g H g H u sin g sin R g u sincos 4cos H 4cos or, R 4cot R sin H 5. () 6. () v ˆi j x = t y t (0t ) From (i) and (ii), y = x 5x On the frictionless incline plane block has acceleration a gsin its vertical component as shown in figure is a gsin Patna-, Ph. No. : / /6/ /5

3 (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_05--7 [ 3 ] Hence relative vertical acceleration of w.r.t. B is a g[sin sin ] B 7. (3) 3 g[sin 60 sin 30 ] m / s 4 4 cceleration due to gravity along inclined g' gcos(90 ) gsin Time taken, t s l g' gsin But h h sin l l sin 8. (3) h hence, t. sin sin h t sin g Normal reaction R + P sin 30 P R mg Psin30 mg Limiting friction between body and surface is given by, P F R mg 9. () For smooth d t, (gsin For rough plane d (gsin g cos )t Patna-, Ph. No. : / /6/ /5

4 [ 4 ] (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_05--7 t t d gsin d gsin gcos ccording to question, t = nt d d n gsin gsin gsin, applicable here, is kinetic friction as the block moves over the inclined plane. n k cos 45 sin45 n or k n or k k n 0. () ccording to question, mg(sin cos ) mg(sin cos ) tan 3 s, tan Som tan 3 tan. (4) u m u P S g gm m g. () ll blocks are moving with constant velocity so net force on all blocks are zero. 3. () 4. () For equilibrium of system, F F F [s 90 ] 3 In the absence of force F, cceleration = W U U q(v V ) B B B V B V W q B Net force Mass F F3 F. 5. (4) Net electric flux from a closed surface in uniform electric field is always zero. 6. () Three capacitors are in series their resultant capacity is given by m m Patna-, Ph. No. : / /6/ /5

5 (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_05--7 [ 5 ] Cs 0K 0K 0K 3 d d d 3 or d d d3 C C K K K s s d d d 3 Cs Cs 0 K K K 3 0 Cs d d d3 K K K3 7. (4) Capacitors C and C are in series with C 3 in paralle with them. Now, K 0( / ) K0 C (d / ) d 3 0 and C C 3 equivalent 3 K d CC C C C K0 K0 K30 d d d K0 K0 d d 0 K3 KK d K K K 0( / ) K 0 C (d / ) d So, none option is correct. 8. () The circuit can be redrawn as 4µF 6µF B 3µF 0 V Here 4µF and 6µF are in series. So, charge is same on both. Now equivalent capacity between and B C B 6 4.4µF 6 4 So, charge on 4 0 F capacitor Q = C B 0 =.4 0 = 4 µc Patna-, Ph. No. : / /6/ /5

6 [ 6 ] (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_ (4) rea QV 0. () 0 d energy stored in the capacitor. U (C)V CV (V ). () If ring is complete, net field at centre is zero. If small portion is cut, field opposite to this is not cancelled out.. (3) Given circuit can be redrawn as shown Capacity of each capacitor is C d 0 So, magnitude of charge on each capacitor = magnitude of charge on each plate 0 V d 0 s plate is connected with +ve terminal of battery, so charge on.v d Plante 4 comes twice and it is connected with -ve terminal of battery. So charge on plate 0V 4 d 3. (3) Given circuit can be redrawn as Potential defference between and B Patna-, Ph. No. : / /6/ /5

7 (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_05--7 [ 7 ] 5 i.e., V VB V V 500V 000 VB 500V 4. () VB B 500 V qe Here, u 0, a m s = l and v =? v = u + as qel qel v 0 v m m 5. (4) Electric field and electric potential at a general point at a distance r from the centre of the dipole is p E g (3cos ) 3 40 r 6. () pcos and Vg 4 r Flux is due to charges enclosed per 0 Total flux = ( ) nc/ C Nm / C i.e.,000 Nm C 7. (3) Initial energy, Ui C0V V Final energy, U f (KC 0 ) K U C V K or f 0 Change in energy Uf Ui C0V K 0 Patna-, Ph. No. : / /6/ /5

8 [ 8 ] (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_ (3) Common potential, Total charge V Total capacitance C V V C C V C 0 CV0 CV0 KC C C( K) V 0 V 0 V0 V V ( K K K) V V 9. () Electric field a point is equal to the negative gradient of the electrostatic potential at that point. dv Potential gradient relates with electric field according to the following relation E dr V V ˆ V ˆ V E i j kˆ r x y x ˆ ˆ ˆ 30. (3) 3 [i(xy z ) jx k3xz ] V q 4 r 0 Here, V V ve V ve q q V 4 0 L L 5 q V 4 L () The system is in equilibrium means the force experienced by each charge is zero. It is clear that charge placed at centre would be in equilibrium for any value of q, so we are considering the equilibrium of charge placed at any corner. 3. () E 33. (3) F mg V F (m m )g ( )V ( ) a r m For stable, equilibrium, the angle should be 0º Patna-, Ph. No. : / /6/ /5

9 (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_05--7 [ 9 ] 34. (3) Here, 90º 35. () tan tan or tan tan or tan tan(90º ) or tan or tan tan Resistance, I R 36. () m and Resistivity me ml R ne l l R R In series, R eq = R + R (l l ) l l eq ( l l ) l l eq (l l ) l l Patna-, Ph. No. : / /6/ /5

10 [ 0 ] (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_ () The circuit can be redrawn as follows V B 38. (3) Resistance across B RB Total resistance of circuit 8 RT 3 3 Current through ammeter 6 3 i (8 / 3) D 5 B C i 3 pplying Kirchhoff s first law at junction, B, C, D at, at B, i BC = = 6 at D, i CD = 8 5 = 3 i B = 3 at C, i CD + i = i Be or 3 + i = 6 i = (4) 40. (3) When bulbs are in series P V 3R...() When bulbs are connected in parallel V 3V P (R / 3) R 3 3P 9P [From Eq. (ii)] Patna-, Ph. No. : / /6/ /5

11 (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_05--7 [ ] 4. (3) 4. (3) R 40 > R 00. In series potential difference distributes in direction ratio of resistance. There are n rows each containing m cells Total cells = m n = 4...(i) For maximum current in the circuit, external resisrtance should be equal to net internal resistance should be equal to net internal resistance. mr R n 43. (3) 44. (3) m 6n m 3 (0.5) n From Eqs. (i) and (ii), we get m =, n = E i r R E i r R ir ir From these two equations, we get r Galvanometer current is given by S ig i S G S i G (i i ) g Shunt resistance, g i i 45. (4) 0 99 S (00 0) 3 ( 0 )(50) 0 i 3 i m Patna-, Ph. No. : / /6/ /5

12 [ ] (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_ (3) CHEMISTRY H a H b C CH H c H a H a H b C = CH H C = CH H c b H c H H a C CH b H c R.H. H a H b C = CH H c 47. (4) 48. (4) 49. () 50. (4) 5. () 5. (4) 53. (4) 54. (4) 55. () 56. () 57. (3) Carbanion in (I) is more stable than (II) Ingold effect is also called inductive effect and it arises due to difference in electronegativity and hybridisation. is conjugation. is l.p congugation 3. is vacant p-orbital conjugation so l.e. why show resonance l.p. and vacant p-orbital conjugation provide more stability Neutral C.S. are more stable CH CH contain H so i.e. why show hyperjugation 3 Patna-, Ph. No. : / /6/ /5

13 (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_05--7 [ 3 ] 58. () 59. () 60. () 6. (3) 6. (4) 63. (4) 3 K.E of n moles of N gas nrt 4 Here n moles 8 R=8.3 J/ mol/ K T=7 C =400K 3 K.E J J.493 kj.5 kj ccording to Boyle's law at constant temperature, V or PV cosntant P ccording to Graham's law of diffusion, r M where r is rate of diffusion of gas and M is its molecular weight r M r 8 9 So, r M r 00 0 r : r 9 :0 64. () Rate of diffusion depend upon molecular weight r r M r r if M M M Hence, compounds are N O and CO as both have same molar mass. Patna-, Ph. No. : / /6/ /5

14 [ 4 ] (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_ () 66. () 67. () Vandar Waal's equation is applicable for real (non-ideal) gases. Because H & CI gases may react with each other to produce HCI gas hence Dalton's law is not applicable. r M Under identical conditions, r M t M s rate of diffusion is also inversely proportional to time, we will have, t M 4 () Thus, for He, t (5s) s 0s; 3 () For O, t (5s) 0s 68. () 69. (4) 8 (3) For CO, t (5s) 5s; 44 (4) For CO, t (5s) 55s; In van der Waal's equation 'b' is for volume correction ΔH ΔE PΔV For isochoric process, ΔV (4) ΔH ΔE s ΔH ΔE Δn RT g if np n r; Δng np nr ve Hence, ΔH ΔE Patna-, Ph. No. : / /6/ /5

15 (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_05--7 [ 5 ] 7. () 7. () 73. () 74. () 75. () Δng 4, ΔH ΔE RT. For an isothermal process ΔE 0 W pδv 3(6 4) 6 L, atm J (g) B(g) C(g) 3D(g) n 5 3 H E nrt or E H nrt = =7.8 kcal M.W. of ehane (C H )=30 gm. M.W. of acetylene (C H )=6 gm. 34 heat evolved per gm of ethene.36 cal / gm Heat evolved per gm of acetylene.9 cal / gm 6 So, acetylene is better fuel. 76 (4) 77. (4) 78. (3) 79. () 80. () 8. (4) 8. () 83. (3) 84. () 85. (3) 86. (3) 87. () 88. (3) The screening effect follows the order s > p > d > f. ns p is the electronic configuration of III period. I O 3 is amphoteric oxide. While moving down in a group, effective nuclear attraction decreases due to addition of new orbitals. s a result ionisation potential decreases. Hence, the correct order Li > K > Cs. Patna-, Ph. No. : / /6/ /5

16 [ 6 ] (PT-3) MRB/MRB-80,80,803 (G & B) MRBK-80,0_NEET SOLUTION_ () Species Na + Mg + I 3+ Si 4+ Protons 3 4 Electrons Size of isoelectronic cations decreases with increase in magnitude of nuclear charge Order of decreasing size is Na + > Mg + > I 3+ > Si (3) (n )s p 6 (n )d 0 ns 0 represents the correct electronic configuration of transition elements among the given choices. BOTNY 9. () 9. (4) 93. () 94. () 95. () 96. (3) 97. (3) 98. (4) 99. () 00. (3) 0. (3) 0. (4) 03. (3) 04. () 05. () 06. () 07. () 08. () 09. () 0. (4). (). () 3. (3) 4. (4) 5. () 6. (3) 7. () 8. (4) 9. () 0. (3). (). () 3. (3) 4. (3) 5. (4) 6. (3) 7. () 8. (4) 9. () 30. () 3. (3) 3. (3) 33. (3) 34. (3) 35. (4) ZOOLOGY 36. () 37. (4) 38. (4) 39. () 40. (3) 4. () 4. (3) 43. () 44. (4) 45. (4) 46. (3) 47. (4) 48. (3) 49. () 50. () 5. (3) 5. (4) 53. () 54. () 55. (3) 56. (4) 57. (3) 58. () 59. () 60. (3) 6. () 6. () 63. () 64. (4) 65. (3) 66. (3) 67. (3) 68. (4) 69. (4) 70. () 7. (4) 7. () 73. () 74. () 75. (3) 76. (4) 77. () 78. () 79. (3) 80. () Patna-, Ph. No. : / /6/ /5

is at the origin, and charge q μc be located if the net force on q

Term: 152 Saturday, April 09, 2016 Page: 1 Q1. Three point charges are arranged along the x-axis. Charge q 3.0 0 μc 1 is at the origin, and charge q 5.0 0 μc 2 is at x = 0.200 m. Where should a third charge