Page 1 HINT – SHEET ANSWER KEY (0999DMD310319007) Test Pattern DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE NEET(UG) MINOR TEST # 07 29-09-2019 LTS/HS-1/7 Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005 +91-744-2757575 dlp@allen.ac.in www.allen.ac.in Test Type : Unit - 06 1. The given AC circuit is the combination of two pure parallel circuits with the applied voltage. In which I 2 is in phase with V and I 1 leads V by 90º. 2. Work done by the retarding torque = initial KE i.e., tq = K Since t is same, hence q or the number of revolutions will also be same. 3. X C > X L . Hence, current will lead the voltage. Z = 22 CL R (X –X) + = 102W \ I rms = rms V Z = 400 / 2 102 = 20 A cos f = R1 Z 2 = V R = I rms R = (20) (10) = 200 V 4. Mw 2 L = T 0 Þ 0 T ML w= 0 1T f 2 2 ML w == pp 5. tan f = L 2 50 0.21 R 12 w p´´ = = 5.5 as tan 45º = 1 and tan 90º = ¥ so best answer is only in between 45º and 90º which is f = 80º Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Ans.1 2 4 2 3 2 3 3 4 3 1 1 4 2 2 3 2 3 3 2 Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Ans.1 1 4 4 1 4 1 2 3 3 3 1 3 4 2 1 1 3 2 4 Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans.3 2 2 3 4 4 4 2 3 3 3 3 4 2 2 2 4 2 3 3 Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Ans.1 1 3 4 2 3 4 2 2 4 2 2 2 4 2 4 1 4 4 1 Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 Ans.3 2 4 1 2 4 3 2 1 3 1 3 3 4 3 3 2 4 2 4 Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.4 4 1 2 2 2 2 2 3 4 4 2 4 3 4 3 2 2 4 3 Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.3 2 4 4 2 2 2 3 3 2 4 3 3 1 3 4 2 4 3 3 Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.2 3 2 3 4 1 1 4 3 4 1 3 4 2 2 2 3 2 1 1 Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.4 1 3 3 4 1 1 1 4 3 1 2 2 3 2 2 4 4 2 3 Page 2 HINT – SHEET ANSWER KEY (0999DMD310319007) Test Pattern DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE NEET(UG) MINOR TEST # 07 29-09-2019 LTS/HS-1/7 Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005 +91-744-2757575 dlp@allen.ac.in www.allen.ac.in Test Type : Unit - 06 1. The given AC circuit is the combination of two pure parallel circuits with the applied voltage. In which I 2 is in phase with V and I 1 leads V by 90º. 2. Work done by the retarding torque = initial KE i.e., tq = K Since t is same, hence q or the number of revolutions will also be same. 3. X C > X L . Hence, current will lead the voltage. Z = 22 CL R (X –X) + = 102W \ I rms = rms V Z = 400 / 2 102 = 20 A cos f = R1 Z 2 = V R = I rms R = (20) (10) = 200 V 4. Mw 2 L = T 0 Þ 0 T ML w= 0 1T f 2 2 ML w == pp 5. tan f = L 2 50 0.21 R 12 w p´´ = = 5.5 as tan 45º = 1 and tan 90º = ¥ so best answer is only in between 45º and 90º which is f = 80º Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Ans.1 2 4 2 3 2 3 3 4 3 1 1 4 2 2 3 2 3 3 2 Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Ans.1 1 4 4 1 4 1 2 3 3 3 1 3 4 2 1 1 3 2 4 Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans.3 2 2 3 4 4 4 2 3 3 3 3 4 2 2 2 4 2 3 3 Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Ans.1 1 3 4 2 3 4 2 2 4 2 2 2 4 2 4 1 4 4 1 Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 Ans.3 2 4 1 2 4 3 2 1 3 1 3 3 4 3 3 2 4 2 4 Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.4 4 1 2 2 2 2 2 3 4 4 2 4 3 4 3 2 2 4 3 Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.3 2 4 4 2 2 2 3 3 2 4 3 3 1 3 4 2 4 3 3 Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.2 3 2 3 4 1 1 4 3 4 1 3 4 2 2 2 3 2 1 1 Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.4 1 3 3 4 1 1 1 4 3 1 2 2 3 2 2 4 4 2 3 LTS/HS-2/7 0999DMD310319007 Target : Pre-Medical 2020/NEET-UG/29-09-2019 AL LEN 6. The average value of energy density (energy/volume) is given by u av = 2 00 1 E 2 e Total volume of the cylinder V = A·1 \ Total energy contained in the cylinder, U = (u av ) (V) = 2 00 1 E (Al) 2 æö e ç÷ èø Substituting the values, we have U = 1 2 × (8.86×10 –12 ) (50) 2 (10×10 –4 ) (50×10 –2 ) J = 5.5 × 10 –12 J 7. KE in both cases should be same and equal to the loss in potential energy (according to law of conservation of energy). Hence, 2 2 22 2 1 1 v 1 5v Mv MkM 2 2 R 24 æö += ç÷ èø Þ k = 3R 4 8. I 0 = 0 V 100 R 20 = = 5 A \ I rms = 0 I5 A 22 = 9. E 0 = B 0 c = 2 × 10 –7 × 3 × 10 8 = 60 N/C EB ^ rr and as direction of propogation is along x so E is along z. 10. The cylinder will just begin to slip and will not topple, if Moment of force Mg sin q about A = moment of N about A 45º G h/2 Mg sin q Mg cos q Mg r N A 45º Mg sin q × h 2 = Mg cos q × r Þ r = h tan 2 q for q = 45º, r = h 2 11. e = 22 12 12 2 2 22 1 2 12 esin t ecost ee e e ee æö ww ç÷ ++ ç÷ ++ èø Let cos f = 1 22 12 e ee + sin f = 2 22 12 e ee + then e = 22 12 ee + sin(wt + f) \ e rms = 22 22 12 12 ee ee 2 2 + + = 12. Intensity I = 2 00 1 EC 2 Î Þ E 0 = 0 2I C Î = 128 24 8.8 10 3 10 - ´ ´ ´´ = 54.87 N/C and B 0 = 0 E c = 8 54.87 3 10 ´ = 1.83 × 10 –7 T 13. According to law of conservation of linear momentum mv = MV .......(1) According to law of conservation of angular momentum, mv· 2 L ML 2 12 =w .......(2) Since the collision is elastic, therefore 2 22 1 11 mv MVI 2 22 = +w Putting the values of w and V in above equation, m = M 4 Page 3 HINT – SHEET ANSWER KEY (0999DMD310319007) Test Pattern DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE NEET(UG) MINOR TEST # 07 29-09-2019 LTS/HS-1/7 Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005 +91-744-2757575 dlp@allen.ac.in www.allen.ac.in Test Type : Unit - 06 1. The given AC circuit is the combination of two pure parallel circuits with the applied voltage. In which I 2 is in phase with V and I 1 leads V by 90º. 2. Work done by the retarding torque = initial KE i.e., tq = K Since t is same, hence q or the number of revolutions will also be same. 3. X C > X L . Hence, current will lead the voltage. Z = 22 CL R (X –X) + = 102W \ I rms = rms V Z = 400 / 2 102 = 20 A cos f = R1 Z 2 = V R = I rms R = (20) (10) = 200 V 4. Mw 2 L = T 0 Þ 0 T ML w= 0 1T f 2 2 ML w == pp 5. tan f = L 2 50 0.21 R 12 w p´´ = = 5.5 as tan 45º = 1 and tan 90º = ¥ so best answer is only in between 45º and 90º which is f = 80º Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Ans.1 2 4 2 3 2 3 3 4 3 1 1 4 2 2 3 2 3 3 2 Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Ans.1 1 4 4 1 4 1 2 3 3 3 1 3 4 2 1 1 3 2 4 Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans.3 2 2 3 4 4 4 2 3 3 3 3 4 2 2 2 4 2 3 3 Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Ans.1 1 3 4 2 3 4 2 2 4 2 2 2 4 2 4 1 4 4 1 Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 Ans.3 2 4 1 2 4 3 2 1 3 1 3 3 4 3 3 2 4 2 4 Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.4 4 1 2 2 2 2 2 3 4 4 2 4 3 4 3 2 2 4 3 Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.3 2 4 4 2 2 2 3 3 2 4 3 3 1 3 4 2 4 3 3 Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.2 3 2 3 4 1 1 4 3 4 1 3 4 2 2 2 3 2 1 1 Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.4 1 3 3 4 1 1 1 4 3 1 2 2 3 2 2 4 4 2 3 LTS/HS-2/7 0999DMD310319007 Target : Pre-Medical 2020/NEET-UG/29-09-2019 AL LEN 6. The average value of energy density (energy/volume) is given by u av = 2 00 1 E 2 e Total volume of the cylinder V = A·1 \ Total energy contained in the cylinder, U = (u av ) (V) = 2 00 1 E (Al) 2 æö e ç÷ èø Substituting the values, we have U = 1 2 × (8.86×10 –12 ) (50) 2 (10×10 –4 ) (50×10 –2 ) J = 5.5 × 10 –12 J 7. KE in both cases should be same and equal to the loss in potential energy (according to law of conservation of energy). Hence, 2 2 22 2 1 1 v 1 5v Mv MkM 2 2 R 24 æö += ç÷ èø Þ k = 3R 4 8. I 0 = 0 V 100 R 20 = = 5 A \ I rms = 0 I5 A 22 = 9. E 0 = B 0 c = 2 × 10 –7 × 3 × 10 8 = 60 N/C EB ^ rr and as direction of propogation is along x so E is along z. 10. The cylinder will just begin to slip and will not topple, if Moment of force Mg sin q about A = moment of N about A 45º G h/2 Mg sin q Mg cos q Mg r N A 45º Mg sin q × h 2 = Mg cos q × r Þ r = h tan 2 q for q = 45º, r = h 2 11. e = 22 12 12 2 2 22 1 2 12 esin t ecost ee e e ee æö ww ç÷ ++ ç÷ ++ èø Let cos f = 1 22 12 e ee + sin f = 2 22 12 e ee + then e = 22 12 ee + sin(wt + f) \ e rms = 22 22 12 12 ee ee 2 2 + + = 12. Intensity I = 2 00 1 EC 2 Î Þ E 0 = 0 2I C Î = 128 24 8.8 10 3 10 - ´ ´ ´´ = 54.87 N/C and B 0 = 0 E c = 8 54.87 3 10 ´ = 1.83 × 10 –7 T 13. According to law of conservation of linear momentum mv = MV .......(1) According to law of conservation of angular momentum, mv· 2 L ML 2 12 =w .......(2) Since the collision is elastic, therefore 2 22 1 11 mv MVI 2 22 = +w Putting the values of w and V in above equation, m = M 4 Leader Test Series/Joint Package Course/NEET-UG/29-09-2019 AL LEN 0999DMD310319007 LTS/HS-3/7 14. In the given graph current is leading the voltage by 45º. Therefore circuit is C–R tan f = C X R or X C = R 16. The cylinder rotating under gravity has both translational and rotational motions. Let v be the linear velocity of its centre of mass and w its angular velocity about the axis of rotation. We can consider as it is case of inclined plane with angle of inclination a = 90º a = 2 2 gsin K 1 R a + T T T T M = gsin90º 1 1 2 + = 2 g 3 17. For capacitive circuits X C = 1 C w \ i = C V X = VwC Þ i µ w 18. I = MR 2 8 × (0.2) 2 = 0.32 kg × m 2 t = Ia 0.32 × 3 = 0.96 Nm But F = 0.96 4.8N R 0.2 t == 19. Inductive reactance X L = Lw = 2pfL also, E I = Lw = X L E = IX L E = 5 1256 1000 ´ E = 6.28 volt (rms) 20. As v = 2 2gh I 1 MR + hence velocity is independent of the inclination of the plane and depends only on height h through which body descends But because t = 2 1 2hI 1 sin g MR æö + ç÷ q èø depends on the inclination also, hence greater the inclination lesser will be the time of descend. Hence in present case, the speeds will be same (because h is same) but time of descend will be different (because of different inclinations). 21. tan f = L X R æö ç÷ èø and X L = wL = (2pfL) = (2p) (50) (0.01) = pW also R = 1 W f = tan –1 (p) 22. t = Ia Þ a = I t = 500 100 = 5 rad/s 2 w = w 0 + at (w – w 0 ) = at = 5 × 2 = 10 rad/s 23. I = 2 sin [wt + 42 pp + ] Hence phase difference between V and I = 2 p P = 0 24. By defination, angular momentum x yz ˆ ˆˆ i jk L r p mx yz v vv = ´= r rr Now for motion in xy-plant z = 0, v z = 0 So L r = yx xy ˆ ˆˆ i jk ˆ m x y 0 mk(xv yv) vv0 =- 25. I rms = rms L E X where X L = wL = 2 pfL is the reactance of the inductor I rms = 3 220 22 2 50 44 10 7 - ´ ´ ´´ = 10 1000 100 6 ´ ´ = 15.9 A Page 4 HINT – SHEET ANSWER KEY (0999DMD310319007) Test Pattern DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE NEET(UG) MINOR TEST # 07 29-09-2019 LTS/HS-1/7 Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005 +91-744-2757575 dlp@allen.ac.in www.allen.ac.in Test Type : Unit - 06 1. The given AC circuit is the combination of two pure parallel circuits with the applied voltage. In which I 2 is in phase with V and I 1 leads V by 90º. 2. Work done by the retarding torque = initial KE i.e., tq = K Since t is same, hence q or the number of revolutions will also be same. 3. X C > X L . Hence, current will lead the voltage. Z = 22 CL R (X –X) + = 102W \ I rms = rms V Z = 400 / 2 102 = 20 A cos f = R1 Z 2 = V R = I rms R = (20) (10) = 200 V 4. Mw 2 L = T 0 Þ 0 T ML w= 0 1T f 2 2 ML w == pp 5. tan f = L 2 50 0.21 R 12 w p´´ = = 5.5 as tan 45º = 1 and tan 90º = ¥ so best answer is only in between 45º and 90º which is f = 80º Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Ans.1 2 4 2 3 2 3 3 4 3 1 1 4 2 2 3 2 3 3 2 Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Ans.1 1 4 4 1 4 1 2 3 3 3 1 3 4 2 1 1 3 2 4 Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans.3 2 2 3 4 4 4 2 3 3 3 3 4 2 2 2 4 2 3 3 Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Ans.1 1 3 4 2 3 4 2 2 4 2 2 2 4 2 4 1 4 4 1 Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 Ans.3 2 4 1 2 4 3 2 1 3 1 3 3 4 3 3 2 4 2 4 Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.4 4 1 2 2 2 2 2 3 4 4 2 4 3 4 3 2 2 4 3 Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.3 2 4 4 2 2 2 3 3 2 4 3 3 1 3 4 2 4 3 3 Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.2 3 2 3 4 1 1 4 3 4 1 3 4 2 2 2 3 2 1 1 Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.4 1 3 3 4 1 1 1 4 3 1 2 2 3 2 2 4 4 2 3 LTS/HS-2/7 0999DMD310319007 Target : Pre-Medical 2020/NEET-UG/29-09-2019 AL LEN 6. The average value of energy density (energy/volume) is given by u av = 2 00 1 E 2 e Total volume of the cylinder V = A·1 \ Total energy contained in the cylinder, U = (u av ) (V) = 2 00 1 E (Al) 2 æö e ç÷ èø Substituting the values, we have U = 1 2 × (8.86×10 –12 ) (50) 2 (10×10 –4 ) (50×10 –2 ) J = 5.5 × 10 –12 J 7. KE in both cases should be same and equal to the loss in potential energy (according to law of conservation of energy). Hence, 2 2 22 2 1 1 v 1 5v Mv MkM 2 2 R 24 æö += ç÷ èø Þ k = 3R 4 8. I 0 = 0 V 100 R 20 = = 5 A \ I rms = 0 I5 A 22 = 9. E 0 = B 0 c = 2 × 10 –7 × 3 × 10 8 = 60 N/C EB ^ rr and as direction of propogation is along x so E is along z. 10. The cylinder will just begin to slip and will not topple, if Moment of force Mg sin q about A = moment of N about A 45º G h/2 Mg sin q Mg cos q Mg r N A 45º Mg sin q × h 2 = Mg cos q × r Þ r = h tan 2 q for q = 45º, r = h 2 11. e = 22 12 12 2 2 22 1 2 12 esin t ecost ee e e ee æö ww ç÷ ++ ç÷ ++ èø Let cos f = 1 22 12 e ee + sin f = 2 22 12 e ee + then e = 22 12 ee + sin(wt + f) \ e rms = 22 22 12 12 ee ee 2 2 + + = 12. Intensity I = 2 00 1 EC 2 Î Þ E 0 = 0 2I C Î = 128 24 8.8 10 3 10 - ´ ´ ´´ = 54.87 N/C and B 0 = 0 E c = 8 54.87 3 10 ´ = 1.83 × 10 –7 T 13. According to law of conservation of linear momentum mv = MV .......(1) According to law of conservation of angular momentum, mv· 2 L ML 2 12 =w .......(2) Since the collision is elastic, therefore 2 22 1 11 mv MVI 2 22 = +w Putting the values of w and V in above equation, m = M 4 Leader Test Series/Joint Package Course/NEET-UG/29-09-2019 AL LEN 0999DMD310319007 LTS/HS-3/7 14. In the given graph current is leading the voltage by 45º. Therefore circuit is C–R tan f = C X R or X C = R 16. The cylinder rotating under gravity has both translational and rotational motions. Let v be the linear velocity of its centre of mass and w its angular velocity about the axis of rotation. We can consider as it is case of inclined plane with angle of inclination a = 90º a = 2 2 gsin K 1 R a + T T T T M = gsin90º 1 1 2 + = 2 g 3 17. For capacitive circuits X C = 1 C w \ i = C V X = VwC Þ i µ w 18. I = MR 2 8 × (0.2) 2 = 0.32 kg × m 2 t = Ia 0.32 × 3 = 0.96 Nm But F = 0.96 4.8N R 0.2 t == 19. Inductive reactance X L = Lw = 2pfL also, E I = Lw = X L E = IX L E = 5 1256 1000 ´ E = 6.28 volt (rms) 20. As v = 2 2gh I 1 MR + hence velocity is independent of the inclination of the plane and depends only on height h through which body descends But because t = 2 1 2hI 1 sin g MR æö + ç÷ q èø depends on the inclination also, hence greater the inclination lesser will be the time of descend. Hence in present case, the speeds will be same (because h is same) but time of descend will be different (because of different inclinations). 21. tan f = L X R æö ç÷ èø and X L = wL = (2pfL) = (2p) (50) (0.01) = pW also R = 1 W f = tan –1 (p) 22. t = Ia Þ a = I t = 500 100 = 5 rad/s 2 w = w 0 + at (w – w 0 ) = at = 5 × 2 = 10 rad/s 23. I = 2 sin [wt + 42 pp + ] Hence phase difference between V and I = 2 p P = 0 24. By defination, angular momentum x yz ˆ ˆˆ i jk L r p mx yz v vv = ´= r rr Now for motion in xy-plant z = 0, v z = 0 So L r = yx xy ˆ ˆˆ i jk ˆ m x y 0 mk(xv yv) vv0 =- 25. I rms = rms L E X where X L = wL = 2 pfL is the reactance of the inductor I rms = 3 220 22 2 50 44 10 7 - ´ ´ ´´ = 10 1000 100 6 ´ ´ = 15.9 A LTS/HS-4/7 0999DMD310319007 Target : Pre-Medical 2020/NEET-UG/29-09-2019 AL LEN 26. In electromagnetic waves, electric field E r and magnetic field B r are perpendicular to the direction of wave. So direction of wave propagation is given by EB ´ rr . 27. r 3L = 60º r O L The desired moment of inertia about O is I = 6 × I one side = 2 2 m(2L) 6 mr 12 éù + êú ëû = 2 2 mL 6 3mL 3 éù + êú ëû = 20mL 2 28. T = 11 s f 60 = w = 2pf = 377 rad/s X L = wL = (377) (0.040) = 15.08 W Z = 22 L xR + = 25.05 W Amplitudes (maximum value) are, 0 0 V 150 i 6A Z 25.05 = =» (V 0 ) R = i 0 R = 120 V (V 0 ) L = i 0 X L = 90.5 V 29. I = I c + mR 2 = mR 2 + mR 2 = 2mR 2 = 2 × 3 × (1) 2 = 6 gm cm 2 30. 3 /2 m /2 I sin t dt i 3 22 pw pw w < >= pp - ww ò = 3 /2 m /2 cost I 0 pw pw w æö - ç÷ w èø = p w 31. k 2 11 2c p = =- w l ´ pn ln 32. I' = 2I 1 = 2kg m 2 I'w 2 = Iw 1 Þ 2Iw 2 = Iw 1 Þw 2 = 1 100 50rad/s 22 w == DKE = 22 1 1 22 11 II 22 w-w = 22 1 1 (100) 2 (50) 2 éù ´ -´ ëû = 2 1 (50) (4 2) 2 ´- = 2500 J = 2.5 kJ 33. cos f = R Z = R V IR IZV = 22 8 84 105 8 (12 6) == +- 35. 2kg 10 N T T a a For block : 10 – T = 2a ...(i) For pulley : TR = Ia T × 0.25 = 2a Þ T = 8a ...(ii) aRa 4 a = aÞ= ...(iii) 1082 42 aa -aÞ ´= 17 10 2 a Þ = Þa= 20 17 rad/s 2 37. CL XX tan R - f= 1 2 fL 2 fC tan45 R -p p °= ( ) 1 C 2f2 fL R éù = êú p p+ êú ëû Page 5 HINT – SHEET ANSWER KEY (0999DMD310319007) Test Pattern DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE NEET(UG) MINOR TEST # 07 29-09-2019 LTS/HS-1/7 Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA 324005 +91-744-2757575 dlp@allen.ac.in www.allen.ac.in Test Type : Unit - 06 1. The given AC circuit is the combination of two pure parallel circuits with the applied voltage. In which I 2 is in phase with V and I 1 leads V by 90º. 2. Work done by the retarding torque = initial KE i.e., tq = K Since t is same, hence q or the number of revolutions will also be same. 3. X C > X L . Hence, current will lead the voltage. Z = 22 CL R (X –X) + = 102W \ I rms = rms V Z = 400 / 2 102 = 20 A cos f = R1 Z 2 = V R = I rms R = (20) (10) = 200 V 4. Mw 2 L = T 0 Þ 0 T ML w= 0 1T f 2 2 ML w == pp 5. tan f = L 2 50 0.21 R 12 w p´´ = = 5.5 as tan 45º = 1 and tan 90º = ¥ so best answer is only in between 45º and 90º which is f = 80º Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Ans.1 2 4 2 3 2 3 3 4 3 1 1 4 2 2 3 2 3 3 2 Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Ans.1 1 4 4 1 4 1 2 3 3 3 1 3 4 2 1 1 3 2 4 Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans.3 2 2 3 4 4 4 2 3 3 3 3 4 2 2 2 4 2 3 3 Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Ans.1 1 3 4 2 3 4 2 2 4 2 2 2 4 2 4 1 4 4 1 Que. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 Ans.3 2 4 1 2 4 3 2 1 3 1 3 3 4 3 3 2 4 2 4 Que.101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.4 4 1 2 2 2 2 2 3 4 4 2 4 3 4 3 2 2 4 3 Que.121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.3 2 4 4 2 2 2 3 3 2 4 3 3 1 3 4 2 4 3 3 Que.141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.2 3 2 3 4 1 1 4 3 4 1 3 4 2 2 2 3 2 1 1 Que.161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.4 1 3 3 4 1 1 1 4 3 1 2 2 3 2 2 4 4 2 3 LTS/HS-2/7 0999DMD310319007 Target : Pre-Medical 2020/NEET-UG/29-09-2019 AL LEN 6. The average value of energy density (energy/volume) is given by u av = 2 00 1 E 2 e Total volume of the cylinder V = A·1 \ Total energy contained in the cylinder, U = (u av ) (V) = 2 00 1 E (Al) 2 æö e ç÷ èø Substituting the values, we have U = 1 2 × (8.86×10 –12 ) (50) 2 (10×10 –4 ) (50×10 –2 ) J = 5.5 × 10 –12 J 7. KE in both cases should be same and equal to the loss in potential energy (according to law of conservation of energy). Hence, 2 2 22 2 1 1 v 1 5v Mv MkM 2 2 R 24 æö += ç÷ èø Þ k = 3R 4 8. I 0 = 0 V 100 R 20 = = 5 A \ I rms = 0 I5 A 22 = 9. E 0 = B 0 c = 2 × 10 –7 × 3 × 10 8 = 60 N/C EB ^ rr and as direction of propogation is along x so E is along z. 10. The cylinder will just begin to slip and will not topple, if Moment of force Mg sin q about A = moment of N about A 45º G h/2 Mg sin q Mg cos q Mg r N A 45º Mg sin q × h 2 = Mg cos q × r Þ r = h tan 2 q for q = 45º, r = h 2 11. e = 22 12 12 2 2 22 1 2 12 esin t ecost ee e e ee æö ww ç÷ ++ ç÷ ++ èø Let cos f = 1 22 12 e ee + sin f = 2 22 12 e ee + then e = 22 12 ee + sin(wt + f) \ e rms = 22 22 12 12 ee ee 2 2 + + = 12. Intensity I = 2 00 1 EC 2 Î Þ E 0 = 0 2I C Î = 128 24 8.8 10 3 10 - ´ ´ ´´ = 54.87 N/C and B 0 = 0 E c = 8 54.87 3 10 ´ = 1.83 × 10 –7 T 13. According to law of conservation of linear momentum mv = MV .......(1) According to law of conservation of angular momentum, mv· 2 L ML 2 12 =w .......(2) Since the collision is elastic, therefore 2 22 1 11 mv MVI 2 22 = +w Putting the values of w and V in above equation, m = M 4 Leader Test Series/Joint Package Course/NEET-UG/29-09-2019 AL LEN 0999DMD310319007 LTS/HS-3/7 14. In the given graph current is leading the voltage by 45º. Therefore circuit is C–R tan f = C X R or X C = R 16. The cylinder rotating under gravity has both translational and rotational motions. Let v be the linear velocity of its centre of mass and w its angular velocity about the axis of rotation. We can consider as it is case of inclined plane with angle of inclination a = 90º a = 2 2 gsin K 1 R a + T T T T M = gsin90º 1 1 2 + = 2 g 3 17. For capacitive circuits X C = 1 C w \ i = C V X = VwC Þ i µ w 18. I = MR 2 8 × (0.2) 2 = 0.32 kg × m 2 t = Ia 0.32 × 3 = 0.96 Nm But F = 0.96 4.8N R 0.2 t == 19. Inductive reactance X L = Lw = 2pfL also, E I = Lw = X L E = IX L E = 5 1256 1000 ´ E = 6.28 volt (rms) 20. As v = 2 2gh I 1 MR + hence velocity is independent of the inclination of the plane and depends only on height h through which body descends But because t = 2 1 2hI 1 sin g MR æö + ç÷ q èø depends on the inclination also, hence greater the inclination lesser will be the time of descend. Hence in present case, the speeds will be same (because h is same) but time of descend will be different (because of different inclinations). 21. tan f = L X R æö ç÷ èø and X L = wL = (2pfL) = (2p) (50) (0.01) = pW also R = 1 W f = tan –1 (p) 22. t = Ia Þ a = I t = 500 100 = 5 rad/s 2 w = w 0 + at (w – w 0 ) = at = 5 × 2 = 10 rad/s 23. I = 2 sin [wt + 42 pp + ] Hence phase difference between V and I = 2 p P = 0 24. By defination, angular momentum x yz ˆ ˆˆ i jk L r p mx yz v vv = ´= r rr Now for motion in xy-plant z = 0, v z = 0 So L r = yx xy ˆ ˆˆ i jk ˆ m x y 0 mk(xv yv) vv0 =- 25. I rms = rms L E X where X L = wL = 2 pfL is the reactance of the inductor I rms = 3 220 22 2 50 44 10 7 - ´ ´ ´´ = 10 1000 100 6 ´ ´ = 15.9 A LTS/HS-4/7 0999DMD310319007 Target : Pre-Medical 2020/NEET-UG/29-09-2019 AL LEN 26. In electromagnetic waves, electric field E r and magnetic field B r are perpendicular to the direction of wave. So direction of wave propagation is given by EB ´ rr . 27. r 3L = 60º r O L The desired moment of inertia about O is I = 6 × I one side = 2 2 m(2L) 6 mr 12 éù + êú ëû = 2 2 mL 6 3mL 3 éù + êú ëû = 20mL 2 28. T = 11 s f 60 = w = 2pf = 377 rad/s X L = wL = (377) (0.040) = 15.08 W Z = 22 L xR + = 25.05 W Amplitudes (maximum value) are, 0 0 V 150 i 6A Z 25.05 = =» (V 0 ) R = i 0 R = 120 V (V 0 ) L = i 0 X L = 90.5 V 29. I = I c + mR 2 = mR 2 + mR 2 = 2mR 2 = 2 × 3 × (1) 2 = 6 gm cm 2 30. 3 /2 m /2 I sin t dt i 3 22 pw pw w < >= pp - ww ò = 3 /2 m /2 cost I 0 pw pw w æö - ç÷ w èø = p w 31. k 2 11 2c p = =- w l ´ pn ln 32. I' = 2I 1 = 2kg m 2 I'w 2 = Iw 1 Þ 2Iw 2 = Iw 1 Þw 2 = 1 100 50rad/s 22 w == DKE = 22 1 1 22 11 II 22 w-w = 22 1 1 (100) 2 (50) 2 éù ´ -´ ëû = 2 1 (50) (4 2) 2 ´- = 2500 J = 2.5 kJ 33. cos f = R Z = R V IR IZV = 22 8 84 105 8 (12 6) == +- 35. 2kg 10 N T T a a For block : 10 – T = 2a ...(i) For pulley : TR = Ia T × 0.25 = 2a Þ T = 8a ...(ii) aRa 4 a = aÞ= ...(iii) 1082 42 aa -aÞ ´= 17 10 2 a Þ = Þa= 20 17 rad/s 2 37. CL XX tan R - f= 1 2 fL 2 fC tan45 R -p p °= ( ) 1 C 2f2 fL R éù = êú p p+ êú ëû Leader Test Series/Joint Package Course/NEET-UG/29-09-2019 AL LEN 0999DMD310319007 LTS/HS-5/7 38. w M y O x R Angular momentum of disc about origin L = mvR + Iw and v = wR, So L = MwR 2 + 2 MR 2 w = 2 3 MR 2 w 39. At the time of maximum speed of point P, the rod should be vertical DK + DU = 0 Þ 2 2 1m 0 mg0 232 éù æö éù w-+-= êú ç÷ êú ëû èø ëû ll Þ 3g w= l Velocity of point P V P = w l = 3g l . 42. I = I larger + I smaller I = 4M(4R) 2 + [MR 2 + M(5R) 2 ] = 90 MR 2 44. Net current : I 68 2sint =+w Now find ( ) T 2 0 v 6 8 2 sin t dt I 10A T +w == ò 45. 2 2 mr I mr 2 =+ L = 2pr m = r×L 2 3 mr 2 = r = L 2p 23 22 3 L 3L L 2 48 r =´r´= pp 46. 2NO 2 + H 2 O ¾¾® HNO 2 + HNO 3 (X) (Y) (X) & (Y) both are acidic 47. Reactivity order. White P > Red P > Black P P P P > P P P P – > P P P P P P P P P P P P P – P due to angle strain 48. R–SiCl 3 HOH ¾¾¾® R–Si(OH) 3 Polymerisation Cross-linked silicone 49. Al 4 C 3 + 12 HOH ¾¾® 4Al(OH) 3 + 3CH 4 Be 2 C + 4HOH ¾¾® 2 Be (OH) 2 + CH 4 50. I 2 Cl 6 ® I Cl Cl I Cl Cl Cl Cl Hybridisation – sp 3 d 2 (4s + 2LP) – planar 51. Pyrosulphuric acid – H 2 S 2 O 7 S O O O H O S O O O H all Bond length are not equal 52. Acidic strength µ +ive oxidation state 53. BF 3 ® F B F F 54. In pseudo halide 'N' must be present. 55. H 2 S + KMnO 4 ® MnO + S + H O 22 ¯ Colloidal sulphurRead More

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