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Page 1 HINT – SHEET LTS/HS-1/8 0999DMD310319001 DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) LEADER TEST SERIES / JOINT PACKAGE COURSE TARGET : PRE-MEDICAL 2020 Test Type : Unit Test Test # 01 Test Pattern : NEET-UG TEST DATE : 07 - 07 - 2019 ANSWER KEY 1. Electric field E = – dV dr æö ç÷ èø –30V +80V (1) 20V –10V +70V +90V (2) (3) For the set of plates as in figure (I) E 1 = – 1 300 ( 80) 110 Vm dr dr - - -++ éù = êú ëû Similarly E II = – 1 20 ( 70) 50 Vm dr dr - +-++ éù = êú ëû And E III = – 1 10 ( 90) 100 Vm dr dr - - -+ éù = êú ëû \ E 1 > E 2 > E 3 The order of the pair of plates (for E r is descending order) is I, III and II. 2. dq i dt = Q /2 00 dq idt p = òò Q = () ( ) /2 /2 0 0 4 4sin 2t dt cos2t 2 p p =- ò = [ ] 4 cos cos0 2 - p- = [ ] 4 1 1 4coulomb 2 - --= Que. 1 2 3 4 5 6 7 8 9 101112 13 14151617 181920 Ans.4 1 2 1 3 1 4 2 3 1 2 3 2 3 4 3 3 1 3 3 Que.21222324 25262728 29 303132 33 34353637 383940 Ans.1 1 4 3 3 4 4 3 3 3 3 2 4 1 3 3 1 2 3 3 Que.41424344 45464748 49 505152 53 54555657 585960 Ans.1 4 1 1 1 2 1 2 1 4 3 2 3 3 2 2 3 4 3 1 Que.61626364 65666768 69 707172 73 74757677 787980 Ans.2 4 1 3 4 1 2 1 2 2 4 1 3 1 3 3 1 4 2 2 Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 Ans.4 1 2 2 2 4 2 4 3 2 4 3 1 3 2 1 2 4 3 4 Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.1 1 4 1 4 2 3 4 2 2 2 3 1 3 2 2 2 4 4 4 Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.1 2 1 2 1 1 2 2 4 2 2 3 1 4 3 1 2 3 2 2 Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.2 4 4 2 3 4 1 2 3 1 3 3 4 1 3 3 1 2 2 3 Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.2 4 1 4 3 1 2 3 3 4 1 4 3 4 4 3 1 3 4 2 Page 2 HINT – SHEET LTS/HS-1/8 0999DMD310319001 DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) LEADER TEST SERIES / JOINT PACKAGE COURSE TARGET : PRE-MEDICAL 2020 Test Type : Unit Test Test # 01 Test Pattern : NEET-UG TEST DATE : 07 - 07 - 2019 ANSWER KEY 1. Electric field E = – dV dr æö ç÷ èø –30V +80V (1) 20V –10V +70V +90V (2) (3) For the set of plates as in figure (I) E 1 = – 1 300 ( 80) 110 Vm dr dr - - -++ éù = êú ëû Similarly E II = – 1 20 ( 70) 50 Vm dr dr - +-++ éù = êú ëû And E III = – 1 10 ( 90) 100 Vm dr dr - - -+ éù = êú ëû \ E 1 > E 2 > E 3 The order of the pair of plates (for E r is descending order) is I, III and II. 2. dq i dt = Q /2 00 dq idt p = òò Q = () ( ) /2 /2 0 0 4 4sin 2t dt cos2t 2 p p =- ò = [ ] 4 cos cos0 2 - p- = [ ] 4 1 1 4coulomb 2 - --= Que. 1 2 3 4 5 6 7 8 9 101112 13 14151617 181920 Ans.4 1 2 1 3 1 4 2 3 1 2 3 2 3 4 3 3 1 3 3 Que.21222324 25262728 29 303132 33 34353637 383940 Ans.1 1 4 3 3 4 4 3 3 3 3 2 4 1 3 3 1 2 3 3 Que.41424344 45464748 49 505152 53 54555657 585960 Ans.1 4 1 1 1 2 1 2 1 4 3 2 3 3 2 2 3 4 3 1 Que.61626364 65666768 69 707172 73 74757677 787980 Ans.2 4 1 3 4 1 2 1 2 2 4 1 3 1 3 3 1 4 2 2 Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 Ans.4 1 2 2 2 4 2 4 3 2 4 3 1 3 2 1 2 4 3 4 Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.1 1 4 1 4 2 3 4 2 2 2 3 1 3 2 2 2 4 4 4 Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.1 2 1 2 1 1 2 2 4 2 2 3 1 4 3 1 2 3 2 2 Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.2 4 4 2 3 4 1 2 3 1 3 3 4 1 3 3 1 2 2 3 Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.2 4 1 4 3 1 2 3 3 4 1 4 3 4 4 3 1 3 4 2 LTS/HS-2/8 0999DMD310319001 Target : Pre-Medical 2020/NEET-UG/07-07-2019 3. For balance mg = eE Þ E = mg e Also m = 3 73 4 4 22 r d (10 ) 1000kg 3 37 - p=´´´ Þ E = 73 19 4 22 (10 ) 1000 10 37 260N / C 1.6 10 - - ´ ´ ´´ = ´ (g = 10 newton/kg, e = 1.6 × 10 –19 coulomb) 4. Let A and B be two forces Greatest resultant = A + B = 10 .....(1) Least resultant = A – B = 6 .... (2) Solving (1) & (2) we get A = 8N and B = 2N when each force is increased by 3N A' = A + 3 = 8 + 3 = 11 N B' = B + 3 = 2 + 3 = 5 N As the forces are acting at an angle of 90º So R' = ( ) () 22 22 A' B' 115 + =+ = 146 N 5. Flux through curved surface will be equal to flux through flat surface. 6. In case of spherical metal conductor the charge quickly spreads uniformly overthe entire surface because of which charges stay for longer time on the spherical surface. While in case of non-spherical surface, the charge concentration is different at different points due to which the charges do not stay on the surface or longer time. 7. Trailing zeroes will be counted if unit placed. 8. q y 80V 120V a O –q x 1cm 1cm It may be that charges of same magnitude but opposite sign may have placed on y-axis in such a manner that potential due to them on x-axis is zero, but they will produce electric field in addition to 120 80 20 V / cm 2 - = 9. The dimensions of ( )( ) ( ) 2 22 21 2 522 5 132 MLT MLT EJ 1 MG M M LT -- -- == So, it represents dimensions quantity like angle. 10. The net flux through each closed surface is determined by the net charge inside 1 S 00 ( Q 3Q) 2Q +-- f== ee 2 S 0 ( Q 2Q 3Q) 0 ++- f== e 3 S 00 (2Q3Q)Q +- f = =- ee 4 S 0 Q f =+ e 11. The cork floats motionless if the weight of the cork is equal to electrostatic force due to uniformly charged plane of earth's surface mg = F e = QE Since earth is considered as infinite that E = 0 2 s e mg = 0 2 s e Q 123 0 8 2 mg 2 8.85 10 10 9.8 Q 1 10 -- - e ´ ´ ´´ s== ´ = 17.3 × 10 –8 C/m 2 12. cos – 1 2 æö - =q ç÷ èø cos q = 1 2 - q > 90° q = 120° as cos (120) = –sin(90+30) = –sin30 = 1 2 - 13. The potential energy of the system of charges relative to infinite separation is U = B C AC AB 0 q q qq qq 1 4 AB BC AC éù ++ êú pe ëû =8.55×10 –4 J A B C 1×10 C –8 2×10 C –8 3×10 C –8 x=0 x=1cm x=2cm x=3cm U = 9 16 2 9 10 10 1 2 2 3 13 1 12 10 - - ´ ´ ´ ´´ æö ++ ç÷ èø = 9 × 10 –5 (9.5) = 85.5 × 10 –5 = 8.55 × 10 –4 J Page 3 HINT – SHEET LTS/HS-1/8 0999DMD310319001 DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) LEADER TEST SERIES / JOINT PACKAGE COURSE TARGET : PRE-MEDICAL 2020 Test Type : Unit Test Test # 01 Test Pattern : NEET-UG TEST DATE : 07 - 07 - 2019 ANSWER KEY 1. Electric field E = – dV dr æö ç÷ èø –30V +80V (1) 20V –10V +70V +90V (2) (3) For the set of plates as in figure (I) E 1 = – 1 300 ( 80) 110 Vm dr dr - - -++ éù = êú ëû Similarly E II = – 1 20 ( 70) 50 Vm dr dr - +-++ éù = êú ëû And E III = – 1 10 ( 90) 100 Vm dr dr - - -+ éù = êú ëû \ E 1 > E 2 > E 3 The order of the pair of plates (for E r is descending order) is I, III and II. 2. dq i dt = Q /2 00 dq idt p = òò Q = () ( ) /2 /2 0 0 4 4sin 2t dt cos2t 2 p p =- ò = [ ] 4 cos cos0 2 - p- = [ ] 4 1 1 4coulomb 2 - --= Que. 1 2 3 4 5 6 7 8 9 101112 13 14151617 181920 Ans.4 1 2 1 3 1 4 2 3 1 2 3 2 3 4 3 3 1 3 3 Que.21222324 25262728 29 303132 33 34353637 383940 Ans.1 1 4 3 3 4 4 3 3 3 3 2 4 1 3 3 1 2 3 3 Que.41424344 45464748 49 505152 53 54555657 585960 Ans.1 4 1 1 1 2 1 2 1 4 3 2 3 3 2 2 3 4 3 1 Que.61626364 65666768 69 707172 73 74757677 787980 Ans.2 4 1 3 4 1 2 1 2 2 4 1 3 1 3 3 1 4 2 2 Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 Ans.4 1 2 2 2 4 2 4 3 2 4 3 1 3 2 1 2 4 3 4 Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.1 1 4 1 4 2 3 4 2 2 2 3 1 3 2 2 2 4 4 4 Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.1 2 1 2 1 1 2 2 4 2 2 3 1 4 3 1 2 3 2 2 Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.2 4 4 2 3 4 1 2 3 1 3 3 4 1 3 3 1 2 2 3 Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.2 4 1 4 3 1 2 3 3 4 1 4 3 4 4 3 1 3 4 2 LTS/HS-2/8 0999DMD310319001 Target : Pre-Medical 2020/NEET-UG/07-07-2019 3. For balance mg = eE Þ E = mg e Also m = 3 73 4 4 22 r d (10 ) 1000kg 3 37 - p=´´´ Þ E = 73 19 4 22 (10 ) 1000 10 37 260N / C 1.6 10 - - ´ ´ ´´ = ´ (g = 10 newton/kg, e = 1.6 × 10 –19 coulomb) 4. Let A and B be two forces Greatest resultant = A + B = 10 .....(1) Least resultant = A – B = 6 .... (2) Solving (1) & (2) we get A = 8N and B = 2N when each force is increased by 3N A' = A + 3 = 8 + 3 = 11 N B' = B + 3 = 2 + 3 = 5 N As the forces are acting at an angle of 90º So R' = ( ) () 22 22 A' B' 115 + =+ = 146 N 5. Flux through curved surface will be equal to flux through flat surface. 6. In case of spherical metal conductor the charge quickly spreads uniformly overthe entire surface because of which charges stay for longer time on the spherical surface. While in case of non-spherical surface, the charge concentration is different at different points due to which the charges do not stay on the surface or longer time. 7. Trailing zeroes will be counted if unit placed. 8. q y 80V 120V a O –q x 1cm 1cm It may be that charges of same magnitude but opposite sign may have placed on y-axis in such a manner that potential due to them on x-axis is zero, but they will produce electric field in addition to 120 80 20 V / cm 2 - = 9. The dimensions of ( )( ) ( ) 2 22 21 2 522 5 132 MLT MLT EJ 1 MG M M LT -- -- == So, it represents dimensions quantity like angle. 10. The net flux through each closed surface is determined by the net charge inside 1 S 00 ( Q 3Q) 2Q +-- f== ee 2 S 0 ( Q 2Q 3Q) 0 ++- f== e 3 S 00 (2Q3Q)Q +- f = =- ee 4 S 0 Q f =+ e 11. The cork floats motionless if the weight of the cork is equal to electrostatic force due to uniformly charged plane of earth's surface mg = F e = QE Since earth is considered as infinite that E = 0 2 s e mg = 0 2 s e Q 123 0 8 2 mg 2 8.85 10 10 9.8 Q 1 10 -- - e ´ ´ ´´ s== ´ = 17.3 × 10 –8 C/m 2 12. cos – 1 2 æö - =q ç÷ èø cos q = 1 2 - q > 90° q = 120° as cos (120) = –sin(90+30) = –sin30 = 1 2 - 13. The potential energy of the system of charges relative to infinite separation is U = B C AC AB 0 q q qq qq 1 4 AB BC AC éù ++ êú pe ëû =8.55×10 –4 J A B C 1×10 C –8 2×10 C –8 3×10 C –8 x=0 x=1cm x=2cm x=3cm U = 9 16 2 9 10 10 1 2 2 3 13 1 12 10 - - ´ ´ ´ ´´ æö ++ ç÷ èø = 9 × 10 –5 (9.5) = 85.5 × 10 –5 = 8.55 × 10 –4 J LTS/HS-3/8 0999DMD310319001 Leader Test Series/Joint Package Course/NEET-UG/07-07-2019 14. T =cos30° T T =sin30° A 0 A 0 30° W A 0 0 A /2 T 30 = 0 A T cos30 W °= 0 A T 60N = 3 60W 2 = W = 30 3N 15. a x = 67 qE 10 2 10 m2 - ´´ = = 10 m/s 2 u=10m/s 45º Y X Time of flight T = 1 2 10 2u sin 2 2s g 10 ´´ q == Hence, Range R = u x T + 1 2 a x T 2 R = 10 cos 45º × T + 1 2 a x T 2 101 2 10 2 20m 2 2 ´ + ´ ´= (use g = 10 m/s 2 ) 16. Initially, force between A and C F = 2 2 Q k r C B A +Q +Q –Q r/2 r/2 r FF A C When a similar sphere B having charge +Q is kept at the mid point of line joining A and C, then Net force on B is F net = F A + F C = k 22 22 Q kQ (r / 2) (r / 2) + = 8 2 2 kQ 8F r = (direction is shown in figure) 17. because dimension of resistance is [ML 2 T –3 A –2 ] 18. Force of repulsion between the charges, F = 9 12 2 9 10 40 40 10 160 N 9 (0.9) - ´ ´´´ = Since F > mg, so string always remains tight at any position even if velocity of bob is zero.So minimum speed at lowest point will be such that it is sufficient to take particle to highest point, i.e., velocity becomes zero at highest point. For this u min = 4gl 4 10 0.9 6m / s = ´ ´= 20. Pont P lies at equatorial positions of dipole 1 and 2 and axial osition of dipole 3. Hence field at P due to dipole 1 E 1 = 3 k.p x (towards left) due to dipole 2 E 2 = 2 k.p x (towards left) due to dipole 3 E 3 = 3 k(2p) x (toward righ) So net field at P will be zero. +Q –Q +Q –Q +Q –Q 3 2 1 E 3 E 1 E 2 P 21. If a r is perpendicular to b r a.b0 = r r 3 + l + 6 = 0 l = –9 22. 2 00 mv q 2kq qv r 2r 2mm l ll = Þ== pe pe T = 2rm 2r v 2kq p =p l 23. The electric field E at any point on an equipotential surface acts perpendicular to it. Therefore, it cannot have any component parallel (tangential) to the surface. Page 4 HINT – SHEET LTS/HS-1/8 0999DMD310319001 DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) LEADER TEST SERIES / JOINT PACKAGE COURSE TARGET : PRE-MEDICAL 2020 Test Type : Unit Test Test # 01 Test Pattern : NEET-UG TEST DATE : 07 - 07 - 2019 ANSWER KEY 1. Electric field E = – dV dr æö ç÷ èø –30V +80V (1) 20V –10V +70V +90V (2) (3) For the set of plates as in figure (I) E 1 = – 1 300 ( 80) 110 Vm dr dr - - -++ éù = êú ëû Similarly E II = – 1 20 ( 70) 50 Vm dr dr - +-++ éù = êú ëû And E III = – 1 10 ( 90) 100 Vm dr dr - - -+ éù = êú ëû \ E 1 > E 2 > E 3 The order of the pair of plates (for E r is descending order) is I, III and II. 2. dq i dt = Q /2 00 dq idt p = òò Q = () ( ) /2 /2 0 0 4 4sin 2t dt cos2t 2 p p =- ò = [ ] 4 cos cos0 2 - p- = [ ] 4 1 1 4coulomb 2 - --= Que. 1 2 3 4 5 6 7 8 9 101112 13 14151617 181920 Ans.4 1 2 1 3 1 4 2 3 1 2 3 2 3 4 3 3 1 3 3 Que.21222324 25262728 29 303132 33 34353637 383940 Ans.1 1 4 3 3 4 4 3 3 3 3 2 4 1 3 3 1 2 3 3 Que.41424344 45464748 49 505152 53 54555657 585960 Ans.1 4 1 1 1 2 1 2 1 4 3 2 3 3 2 2 3 4 3 1 Que.61626364 65666768 69 707172 73 74757677 787980 Ans.2 4 1 3 4 1 2 1 2 2 4 1 3 1 3 3 1 4 2 2 Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 Ans.4 1 2 2 2 4 2 4 3 2 4 3 1 3 2 1 2 4 3 4 Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.1 1 4 1 4 2 3 4 2 2 2 3 1 3 2 2 2 4 4 4 Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.1 2 1 2 1 1 2 2 4 2 2 3 1 4 3 1 2 3 2 2 Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.2 4 4 2 3 4 1 2 3 1 3 3 4 1 3 3 1 2 2 3 Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.2 4 1 4 3 1 2 3 3 4 1 4 3 4 4 3 1 3 4 2 LTS/HS-2/8 0999DMD310319001 Target : Pre-Medical 2020/NEET-UG/07-07-2019 3. For balance mg = eE Þ E = mg e Also m = 3 73 4 4 22 r d (10 ) 1000kg 3 37 - p=´´´ Þ E = 73 19 4 22 (10 ) 1000 10 37 260N / C 1.6 10 - - ´ ´ ´´ = ´ (g = 10 newton/kg, e = 1.6 × 10 –19 coulomb) 4. Let A and B be two forces Greatest resultant = A + B = 10 .....(1) Least resultant = A – B = 6 .... (2) Solving (1) & (2) we get A = 8N and B = 2N when each force is increased by 3N A' = A + 3 = 8 + 3 = 11 N B' = B + 3 = 2 + 3 = 5 N As the forces are acting at an angle of 90º So R' = ( ) () 22 22 A' B' 115 + =+ = 146 N 5. Flux through curved surface will be equal to flux through flat surface. 6. In case of spherical metal conductor the charge quickly spreads uniformly overthe entire surface because of which charges stay for longer time on the spherical surface. While in case of non-spherical surface, the charge concentration is different at different points due to which the charges do not stay on the surface or longer time. 7. Trailing zeroes will be counted if unit placed. 8. q y 80V 120V a O –q x 1cm 1cm It may be that charges of same magnitude but opposite sign may have placed on y-axis in such a manner that potential due to them on x-axis is zero, but they will produce electric field in addition to 120 80 20 V / cm 2 - = 9. The dimensions of ( )( ) ( ) 2 22 21 2 522 5 132 MLT MLT EJ 1 MG M M LT -- -- == So, it represents dimensions quantity like angle. 10. The net flux through each closed surface is determined by the net charge inside 1 S 00 ( Q 3Q) 2Q +-- f== ee 2 S 0 ( Q 2Q 3Q) 0 ++- f== e 3 S 00 (2Q3Q)Q +- f = =- ee 4 S 0 Q f =+ e 11. The cork floats motionless if the weight of the cork is equal to electrostatic force due to uniformly charged plane of earth's surface mg = F e = QE Since earth is considered as infinite that E = 0 2 s e mg = 0 2 s e Q 123 0 8 2 mg 2 8.85 10 10 9.8 Q 1 10 -- - e ´ ´ ´´ s== ´ = 17.3 × 10 –8 C/m 2 12. cos – 1 2 æö - =q ç÷ èø cos q = 1 2 - q > 90° q = 120° as cos (120) = –sin(90+30) = –sin30 = 1 2 - 13. The potential energy of the system of charges relative to infinite separation is U = B C AC AB 0 q q qq qq 1 4 AB BC AC éù ++ êú pe ëû =8.55×10 –4 J A B C 1×10 C –8 2×10 C –8 3×10 C –8 x=0 x=1cm x=2cm x=3cm U = 9 16 2 9 10 10 1 2 2 3 13 1 12 10 - - ´ ´ ´ ´´ æö ++ ç÷ èø = 9 × 10 –5 (9.5) = 85.5 × 10 –5 = 8.55 × 10 –4 J LTS/HS-3/8 0999DMD310319001 Leader Test Series/Joint Package Course/NEET-UG/07-07-2019 14. T =cos30° T T =sin30° A 0 A 0 30° W A 0 0 A /2 T 30 = 0 A T cos30 W °= 0 A T 60N = 3 60W 2 = W = 30 3N 15. a x = 67 qE 10 2 10 m2 - ´´ = = 10 m/s 2 u=10m/s 45º Y X Time of flight T = 1 2 10 2u sin 2 2s g 10 ´´ q == Hence, Range R = u x T + 1 2 a x T 2 R = 10 cos 45º × T + 1 2 a x T 2 101 2 10 2 20m 2 2 ´ + ´ ´= (use g = 10 m/s 2 ) 16. Initially, force between A and C F = 2 2 Q k r C B A +Q +Q –Q r/2 r/2 r FF A C When a similar sphere B having charge +Q is kept at the mid point of line joining A and C, then Net force on B is F net = F A + F C = k 22 22 Q kQ (r / 2) (r / 2) + = 8 2 2 kQ 8F r = (direction is shown in figure) 17. because dimension of resistance is [ML 2 T –3 A –2 ] 18. Force of repulsion between the charges, F = 9 12 2 9 10 40 40 10 160 N 9 (0.9) - ´ ´´´ = Since F > mg, so string always remains tight at any position even if velocity of bob is zero.So minimum speed at lowest point will be such that it is sufficient to take particle to highest point, i.e., velocity becomes zero at highest point. For this u min = 4gl 4 10 0.9 6m / s = ´ ´= 20. Pont P lies at equatorial positions of dipole 1 and 2 and axial osition of dipole 3. Hence field at P due to dipole 1 E 1 = 3 k.p x (towards left) due to dipole 2 E 2 = 2 k.p x (towards left) due to dipole 3 E 3 = 3 k(2p) x (toward righ) So net field at P will be zero. +Q –Q +Q –Q +Q –Q 3 2 1 E 3 E 1 E 2 P 21. If a r is perpendicular to b r a.b0 = r r 3 + l + 6 = 0 l = –9 22. 2 00 mv q 2kq qv r 2r 2mm l ll = Þ== pe pe T = 2rm 2r v 2kq p =p l 23. The electric field E at any point on an equipotential surface acts perpendicular to it. Therefore, it cannot have any component parallel (tangential) to the surface. LTS/HS-4/8 0999DMD310319001 Target : Pre-Medical 2020/NEET-UG/07-07-2019 24. ˆˆ b ij =+ r component of a along b r r a cos q ˆ b = a.b ˆ b b æö ç÷ èø r r 22 22 ˆ ˆ ˆˆ ˆˆ (2i 3j).(i j) (i j) . 1 1 11 + ++ = ++ = 23 ˆˆ (i j) 2 + + = 5 ˆˆ (i j) 2 + 25. mgsinq mgcosq qEcosq q qEsinq qE mg N = mg cos q + qE sin q mg sin q = qE cos q + m (mg cos q + qE sin q) Þ q = mg(1) E(1) -m +m (for q = 45º, sin q = cos q) = 2 1 10(1 0.5) 10 0.5 3.3 10 C 100(1 0.5) 100 1.5 - ´-´ = =´ +´ 26. Potential at large distances (¥) is V¥ = 0. Potential at the surface of the sphere V s = 0 1Q 4R pe . The potential at the centre of the spherical shell V centre = 0 3 1Q 24R ´ pe Kinetic energy at the surface = –q (V s – V ¥ ) i.e., (KE) s = –q × 0 1Q 4R ´ pe Kinetic energy at the centre, (KE) C = –q(V C –V ¥ ) = –q 0 3 1Q 24R ´ pe 2 centre c 2 surface (KE) (1/ 2)mv 33 (KE) 22 (1/ 2)mv == Þ v c = 1.5v 27. MI I = MR 2 I M 2R 100 100 100 I MR D DD æö ´ =± ´ +´ ç÷ èø = ± (1 + 2 × 0.5) = ± 2% 28. No field inside the hollow conducting sphere. 29. Zero error : Least count of circular scale 1 0.02 mm 50 == Reading of main scale = 0.0 cm Number of division coincied = 4 reading of circular scale = 4 × 0.02 Zero error = 0.08 mm = +0.008 cm For diameter sphere : Reading of main scale = 1.1 cm Number of division coincied = 14 reading of circular scale = 14 × 0.02 = 0.28 mm = 0.028 cm Reading = 1.128 cm Diameter = Reading – zero error = 1.128 – 0.008 = 1.120 cm 30. A = 5t 2 + 4t + 8 dA 10t4 dt =+ t = 3 sec t dA dt = 10 × 3 + 4 = 34 31. Q Q/2 –Q 4cm x 2 22 kQ kQ kQ 0 2 4 x 2(x 4) - -= ´- Þ 111 0 8 x 2(x 4) --= - Þ x(x – 4) – 8x(x – 4) – 4x = 0 Þ x 2 – 16x + 32 = 0 Þ x ; 13 cm Page 5 HINT – SHEET LTS/HS-1/8 0999DMD310319001 DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) LEADER TEST SERIES / JOINT PACKAGE COURSE TARGET : PRE-MEDICAL 2020 Test Type : Unit Test Test # 01 Test Pattern : NEET-UG TEST DATE : 07 - 07 - 2019 ANSWER KEY 1. Electric field E = – dV dr æö ç÷ èø –30V +80V (1) 20V –10V +70V +90V (2) (3) For the set of plates as in figure (I) E 1 = – 1 300 ( 80) 110 Vm dr dr - - -++ éù = êú ëû Similarly E II = – 1 20 ( 70) 50 Vm dr dr - +-++ éù = êú ëû And E III = – 1 10 ( 90) 100 Vm dr dr - - -+ éù = êú ëû \ E 1 > E 2 > E 3 The order of the pair of plates (for E r is descending order) is I, III and II. 2. dq i dt = Q /2 00 dq idt p = òò Q = () ( ) /2 /2 0 0 4 4sin 2t dt cos2t 2 p p =- ò = [ ] 4 cos cos0 2 - p- = [ ] 4 1 1 4coulomb 2 - --= Que. 1 2 3 4 5 6 7 8 9 101112 13 14151617 181920 Ans.4 1 2 1 3 1 4 2 3 1 2 3 2 3 4 3 3 1 3 3 Que.21222324 25262728 29 303132 33 34353637 383940 Ans.1 1 4 3 3 4 4 3 3 3 3 2 4 1 3 3 1 2 3 3 Que.41424344 45464748 49 505152 53 54555657 585960 Ans.1 4 1 1 1 2 1 2 1 4 3 2 3 3 2 2 3 4 3 1 Que.61626364 65666768 69 707172 73 74757677 787980 Ans.2 4 1 3 4 1 2 1 2 2 4 1 3 1 3 3 1 4 2 2 Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 Ans.4 1 2 2 2 4 2 4 3 2 4 3 1 3 2 1 2 4 3 4 Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.1 1 4 1 4 2 3 4 2 2 2 3 1 3 2 2 2 4 4 4 Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.1 2 1 2 1 1 2 2 4 2 2 3 1 4 3 1 2 3 2 2 Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.2 4 4 2 3 4 1 2 3 1 3 3 4 1 3 3 1 2 2 3 Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.2 4 1 4 3 1 2 3 3 4 1 4 3 4 4 3 1 3 4 2 LTS/HS-2/8 0999DMD310319001 Target : Pre-Medical 2020/NEET-UG/07-07-2019 3. For balance mg = eE Þ E = mg e Also m = 3 73 4 4 22 r d (10 ) 1000kg 3 37 - p=´´´ Þ E = 73 19 4 22 (10 ) 1000 10 37 260N / C 1.6 10 - - ´ ´ ´´ = ´ (g = 10 newton/kg, e = 1.6 × 10 –19 coulomb) 4. Let A and B be two forces Greatest resultant = A + B = 10 .....(1) Least resultant = A – B = 6 .... (2) Solving (1) & (2) we get A = 8N and B = 2N when each force is increased by 3N A' = A + 3 = 8 + 3 = 11 N B' = B + 3 = 2 + 3 = 5 N As the forces are acting at an angle of 90º So R' = ( ) () 22 22 A' B' 115 + =+ = 146 N 5. Flux through curved surface will be equal to flux through flat surface. 6. In case of spherical metal conductor the charge quickly spreads uniformly overthe entire surface because of which charges stay for longer time on the spherical surface. While in case of non-spherical surface, the charge concentration is different at different points due to which the charges do not stay on the surface or longer time. 7. Trailing zeroes will be counted if unit placed. 8. q y 80V 120V a O –q x 1cm 1cm It may be that charges of same magnitude but opposite sign may have placed on y-axis in such a manner that potential due to them on x-axis is zero, but they will produce electric field in addition to 120 80 20 V / cm 2 - = 9. The dimensions of ( )( ) ( ) 2 22 21 2 522 5 132 MLT MLT EJ 1 MG M M LT -- -- == So, it represents dimensions quantity like angle. 10. The net flux through each closed surface is determined by the net charge inside 1 S 00 ( Q 3Q) 2Q +-- f== ee 2 S 0 ( Q 2Q 3Q) 0 ++- f== e 3 S 00 (2Q3Q)Q +- f = =- ee 4 S 0 Q f =+ e 11. The cork floats motionless if the weight of the cork is equal to electrostatic force due to uniformly charged plane of earth's surface mg = F e = QE Since earth is considered as infinite that E = 0 2 s e mg = 0 2 s e Q 123 0 8 2 mg 2 8.85 10 10 9.8 Q 1 10 -- - e ´ ´ ´´ s== ´ = 17.3 × 10 –8 C/m 2 12. cos – 1 2 æö - =q ç÷ èø cos q = 1 2 - q > 90° q = 120° as cos (120) = –sin(90+30) = –sin30 = 1 2 - 13. The potential energy of the system of charges relative to infinite separation is U = B C AC AB 0 q q qq qq 1 4 AB BC AC éù ++ êú pe ëû =8.55×10 –4 J A B C 1×10 C –8 2×10 C –8 3×10 C –8 x=0 x=1cm x=2cm x=3cm U = 9 16 2 9 10 10 1 2 2 3 13 1 12 10 - - ´ ´ ´ ´´ æö ++ ç÷ èø = 9 × 10 –5 (9.5) = 85.5 × 10 –5 = 8.55 × 10 –4 J LTS/HS-3/8 0999DMD310319001 Leader Test Series/Joint Package Course/NEET-UG/07-07-2019 14. T =cos30° T T =sin30° A 0 A 0 30° W A 0 0 A /2 T 30 = 0 A T cos30 W °= 0 A T 60N = 3 60W 2 = W = 30 3N 15. a x = 67 qE 10 2 10 m2 - ´´ = = 10 m/s 2 u=10m/s 45º Y X Time of flight T = 1 2 10 2u sin 2 2s g 10 ´´ q == Hence, Range R = u x T + 1 2 a x T 2 R = 10 cos 45º × T + 1 2 a x T 2 101 2 10 2 20m 2 2 ´ + ´ ´= (use g = 10 m/s 2 ) 16. Initially, force between A and C F = 2 2 Q k r C B A +Q +Q –Q r/2 r/2 r FF A C When a similar sphere B having charge +Q is kept at the mid point of line joining A and C, then Net force on B is F net = F A + F C = k 22 22 Q kQ (r / 2) (r / 2) + = 8 2 2 kQ 8F r = (direction is shown in figure) 17. because dimension of resistance is [ML 2 T –3 A –2 ] 18. Force of repulsion between the charges, F = 9 12 2 9 10 40 40 10 160 N 9 (0.9) - ´ ´´´ = Since F > mg, so string always remains tight at any position even if velocity of bob is zero.So minimum speed at lowest point will be such that it is sufficient to take particle to highest point, i.e., velocity becomes zero at highest point. For this u min = 4gl 4 10 0.9 6m / s = ´ ´= 20. Pont P lies at equatorial positions of dipole 1 and 2 and axial osition of dipole 3. Hence field at P due to dipole 1 E 1 = 3 k.p x (towards left) due to dipole 2 E 2 = 2 k.p x (towards left) due to dipole 3 E 3 = 3 k(2p) x (toward righ) So net field at P will be zero. +Q –Q +Q –Q +Q –Q 3 2 1 E 3 E 1 E 2 P 21. If a r is perpendicular to b r a.b0 = r r 3 + l + 6 = 0 l = –9 22. 2 00 mv q 2kq qv r 2r 2mm l ll = Þ== pe pe T = 2rm 2r v 2kq p =p l 23. The electric field E at any point on an equipotential surface acts perpendicular to it. Therefore, it cannot have any component parallel (tangential) to the surface. LTS/HS-4/8 0999DMD310319001 Target : Pre-Medical 2020/NEET-UG/07-07-2019 24. ˆˆ b ij =+ r component of a along b r r a cos q ˆ b = a.b ˆ b b æö ç÷ èø r r 22 22 ˆ ˆ ˆˆ ˆˆ (2i 3j).(i j) (i j) . 1 1 11 + ++ = ++ = 23 ˆˆ (i j) 2 + + = 5 ˆˆ (i j) 2 + 25. mgsinq mgcosq qEcosq q qEsinq qE mg N = mg cos q + qE sin q mg sin q = qE cos q + m (mg cos q + qE sin q) Þ q = mg(1) E(1) -m +m (for q = 45º, sin q = cos q) = 2 1 10(1 0.5) 10 0.5 3.3 10 C 100(1 0.5) 100 1.5 - ´-´ = =´ +´ 26. Potential at large distances (¥) is V¥ = 0. Potential at the surface of the sphere V s = 0 1Q 4R pe . The potential at the centre of the spherical shell V centre = 0 3 1Q 24R ´ pe Kinetic energy at the surface = –q (V s – V ¥ ) i.e., (KE) s = –q × 0 1Q 4R ´ pe Kinetic energy at the centre, (KE) C = –q(V C –V ¥ ) = –q 0 3 1Q 24R ´ pe 2 centre c 2 surface (KE) (1/ 2)mv 33 (KE) 22 (1/ 2)mv == Þ v c = 1.5v 27. MI I = MR 2 I M 2R 100 100 100 I MR D DD æö ´ =± ´ +´ ç÷ èø = ± (1 + 2 × 0.5) = ± 2% 28. No field inside the hollow conducting sphere. 29. Zero error : Least count of circular scale 1 0.02 mm 50 == Reading of main scale = 0.0 cm Number of division coincied = 4 reading of circular scale = 4 × 0.02 Zero error = 0.08 mm = +0.008 cm For diameter sphere : Reading of main scale = 1.1 cm Number of division coincied = 14 reading of circular scale = 14 × 0.02 = 0.28 mm = 0.028 cm Reading = 1.128 cm Diameter = Reading – zero error = 1.128 – 0.008 = 1.120 cm 30. A = 5t 2 + 4t + 8 dA 10t4 dt =+ t = 3 sec t dA dt = 10 × 3 + 4 = 34 31. Q Q/2 –Q 4cm x 2 22 kQ kQ kQ 0 2 4 x 2(x 4) - -= ´- Þ 111 0 8 x 2(x 4) --= - Þ x(x – 4) – 8x(x – 4) – 4x = 0 Þ x 2 – 16x + 32 = 0 Þ x ; 13 cm LTS/HS-5/8 0999DMD310319001 Leader Test Series/Joint Package Course/NEET-UG/07-07-2019 32. ˆˆ ˆ i jk r F 3 23 2 34 t= ´= - r r r = ˆˆ [(2 4) – (3 3)]i [(2 3)– (3 4)]j ´ ´- +´´ ˆˆ ˆˆ [(3 3)– (2 2)]k 17i 6 j 13k + ´- ´ = -- 33. The facing surfaces have equal and opposite charges. Q O A B C Q O A B C –Q +Q –Q +Q 34. 1.26 2.3 3.56 + Rounding off to 1 decimal place we get 3.6. In addition or subtraction the number of decimal places in the result should be equal to the number of decimal places of that term in the operation which contain lesser number of decimal places. e.g. 12.587 – 12.5 = 0.087 = 0.1 (Q second term contain lesser i.e. one decimal place) 35. Let T µ P a d b E c Writing dimensions on both sides. [M 0 L 0 T] = [ML –1 T –2 ] a [ML –3 ] b [ML 2 T –2 ] c [M 0 L 0 T] = M a+b+c L –a–3b+2c T –2a–2c Thus, a + b + c = 0, –a – 3b + 2c = 0, –2a – 2c = 1 On solving these equation, we get a = 5 6 - , b = 1 2 and c = 1 3 36. because electric field applies the force on electron in the direction opposite to its motion. 37. 38.7 If the digit to be rounded off is more than 5, then the preceding digit is increased by one. e.g. 6.87» 6.9. If the digit to be rounded off is less than 5, then the preceding digit is left unchanged. e.g. 3.94 » 3.9 38. Total enclosed charge q = 100 Q coulomb f E = 00 q 100Q = ee 39. The electric field at a point x is E(x) = 2 dV(x)d [4(1 x )] 8x dx dx - =- + =- The electric field and hence force on a positive charge of 1C, is in negative x-axis direction. (Q F = qE) 40. AB ´ r r is ^ to plane of AandB r r 41. Metal plate acts as an equipotential surface, therefore the field lines should enter normal to the surface of the metal plate. 42. g = 2 2 4 T p l g 100 g D ´ = T 100 2 100 T DD ´ +´ l l Dl and DT are minimum for option (4) and also maximum number of observations are taken in option (4) only. 43. Distance between A(1, 2, 4) and B(3, 2, 1) 2 22 r (3 1) (2 2) (1 4) 13 = - + - +-= V = 98 0 1 Q 9 10 2 10 50volt 4r 13 - ´ ´´ == pÎ E = 98 2 0 1 Q 9 10 2 10 180 N/C 4 13 13 r - ´ ´´ == pÎ Unit vector along AB, ˆ ˆ 2i 3k ˆ r 13 - = 180 ˆ ˆ ˆ E Er (2i 3k) 13 13 ==- r 44. At t = 1 x A = 4, x B = 7 y A = 3, y B = 7 distance = 22 (7–4) (7–3) + = 5 45. V = 4 + 5x 2 (i) x = 1, V 1 = 9 x = –2, V 2 = 24 V 2 – V 1 = 15 V, (i) is O.K. (ii) E x = – dV dx = –10x at x = –1m, E = 10 NC F = qE = 1 × 10 = 10N, (ii) is O.K. (iii) ˆˆ F qE 10xi 10 iN = =-= rr , (iii) is O.K.Read More

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