Page 1 HINT – SHEET ANSWER KEY 1. Temperature of interface 1 1 22 12 KK T KK q+q = + 300 100 200 0 60C 300 200 ´ +´ = =° + 2. As the temperature of water is increased from 2°C to 3°C the density of water increases (rememeber anomalous behaviour of water), also the volume of sphere increases. Therefore buoyant force on sphere due to water shall increase. 3. Note the volume has been plotted along Y-axis. W = SPDV = (0.6 – 0.8) × 10 3 + (0.4 – 0.6) × 2 × 10 3 + (0.2–0.4) ×3×10 –3 = – 200 – 400 – 600 = –1200 J (0999DMD310319010) Test Pattern DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE NEET(UG) MINOR TEST # 09 20-10-2019 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 Test # 07 4. Rate of cooling (R) = ( ) 44 0 A TT t mc Îs- Dq = Þ 2 3 A Area r1 R m volume r r µµ µµ Þ () 1/3 11 RateR rm µµ 3 1/3 4 m = × r r m 3 éù r p Þµ êú ëû Q Þ 1/3 1/3 12 21 Rm 1 R m3 æö æö == ç÷ ç÷ èø èø 5. (i) Thermal energy Q 300 TC mc 15 J/°C T 45 25 ==== D- Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920 Ans.4 1 4 2 1 4 4 1 1 3 3 4 2 3 3 2 3 3 2 4 Que.21222324 2526 272829303132333435 3637 383940 Ans.2 4 4 3 3 3 3 3 2 4 2 1 3 4 2 1 1 4 2 1 Que.41424344 4546 474849505152535455 5657 585960 Ans.2 3 3 2 3 2 1 2 2 3 3 3 2 1 1 4 4 2 1 4 Que.61626364 6566 676869707172737475 7677 787980 Ans.4 1 3 2 1 1 3 4 3 3 2 4 2 2 4 4 1 2 2 3 Que.81828384 8586 878889909192939495 9697 9899 100 Ans.3 2 4 4 1 4 2 4 1 4 1 2 4 3 1 1 3 4 4 3 Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.2 2 3 3 1 2 1 1 2 1 4 3 3 2 1 4 3 3 3 3 Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.3 4 2 2 4 2 2 3 2 3 4 1 2 2 3 4 3 3 2 1 Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.1 1 2 1 4 2 4 3 2 3 2 4 1 2 4 2 4 3 2 2 Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.4 4 4 3 4 2 3 2 2 2 3 4 4 3 3 2 2 3 1 4 Page 2 HINT – SHEET ANSWER KEY 1. Temperature of interface 1 1 22 12 KK T KK q+q = + 300 100 200 0 60C 300 200 ´ +´ = =° + 2. As the temperature of water is increased from 2°C to 3°C the density of water increases (rememeber anomalous behaviour of water), also the volume of sphere increases. Therefore buoyant force on sphere due to water shall increase. 3. Note the volume has been plotted along Y-axis. W = SPDV = (0.6 – 0.8) × 10 3 + (0.4 – 0.6) × 2 × 10 3 + (0.2–0.4) ×3×10 –3 = – 200 – 400 – 600 = –1200 J (0999DMD310319010) Test Pattern DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE NEET(UG) MINOR TEST # 09 20-10-2019 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 Test # 07 4. Rate of cooling (R) = ( ) 44 0 A TT t mc Îs- Dq = Þ 2 3 A Area r1 R m volume r r µµ µµ Þ () 1/3 11 RateR rm µµ 3 1/3 4 m = × r r m 3 éù r p Þµ êú ëû Q Þ 1/3 1/3 12 21 Rm 1 R m3 æö æö == ç÷ ç÷ èø èø 5. (i) Thermal energy Q 300 TC mc 15 J/°C T 45 25 ==== D- Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920 Ans.4 1 4 2 1 4 4 1 1 3 3 4 2 3 3 2 3 3 2 4 Que.21222324 2526 272829303132333435 3637 383940 Ans.2 4 4 3 3 3 3 3 2 4 2 1 3 4 2 1 1 4 2 1 Que.41424344 4546 474849505152535455 5657 585960 Ans.2 3 3 2 3 2 1 2 2 3 3 3 2 1 1 4 4 2 1 4 Que.61626364 6566 676869707172737475 7677 787980 Ans.4 1 3 2 1 1 3 4 3 3 2 4 2 2 4 4 1 2 2 3 Que.81828384 8586 878889909192939495 9697 9899 100 Ans.3 2 4 4 1 4 2 4 1 4 1 2 4 3 1 1 3 4 4 3 Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.2 2 3 3 1 2 1 1 2 1 4 3 3 2 1 4 3 3 3 3 Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.3 4 2 2 4 2 2 3 2 3 4 1 2 2 3 4 3 3 2 1 Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.1 1 2 1 4 2 4 3 2 3 2 4 1 2 4 2 4 3 2 2 Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.4 4 4 3 4 2 3 2 2 2 3 4 4 3 3 2 2 3 1 4 LTS/HS-2/7 0999DMD310319010 Target : Pre-Medical 2020/NEET-UG/20-10-2019 AL LEN (ii) Specific heat is nothing but thermal capacity per unit mass 3 mc 15 C 600 J/kg-°C m 25 10 - === ´ 6. The image on the screen is real and inverted. Triangle aperture Screen 20 cm 10 cm The size of the image on the screen has aperture size given by 10 Size 1.0 0.5 cm 20 æö == ç÷ èø 7. For mixture of gases, 12 1 v 2v v 12 nC nC C nn + = + where f CR 2 = , f is degree of freedom and 12 1 p 2p p 12 nC nC C nn + = + where p f C 1R 2 æö =+ ç÷ èø For helium, n 1 = 4, f = 3 For oxygen, 2 1 n , f = 5 2 = \ p v 5R 1 7R 4 C 47 2 22 3R 1 5R C 29 4 2 22 ´ +´ == ´ +´ = 1.62 8. The efficiency of reversible engine is always greater than that of irreversible engine. In case of irreversible engine, a part of the energy may be dissipated against friction, etc. 9. t = 20 cm m = 3/2 Radius of curvature = 20 cm object Considering refraction at the curved surface, u = –20; m 2 = 1 m 1 = 3/2; R = +20 Applying 2 1 21 v uR mmm-m -= 1 3/2 1 3/2 v 10 v 20 20 - - = Þ =- - i.e. 10 cm below the curved surface or 10 cm above the actual position of flower. 10. ( ) ( ) X 125 Y 70 500 40 - - -- = ForY = 50 X = 1375.0°X 11. for lens u = wants to see = ¥; v = can see = –5 m \ From 1 11 f vu =- Þ 1 11 f5 =- -¥ Þ f = –5 m. 12. Absolute temperatures of the black body corresponding to curve P and Q are in the inverse ratio of l m (Wein's displacement law). i.e., P Q T 1987 T 2980 = Page 3 HINT – SHEET ANSWER KEY 1. Temperature of interface 1 1 22 12 KK T KK q+q = + 300 100 200 0 60C 300 200 ´ +´ = =° + 2. As the temperature of water is increased from 2°C to 3°C the density of water increases (rememeber anomalous behaviour of water), also the volume of sphere increases. Therefore buoyant force on sphere due to water shall increase. 3. Note the volume has been plotted along Y-axis. W = SPDV = (0.6 – 0.8) × 10 3 + (0.4 – 0.6) × 2 × 10 3 + (0.2–0.4) ×3×10 –3 = – 200 – 400 – 600 = –1200 J (0999DMD310319010) Test Pattern DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE NEET(UG) MINOR TEST # 09 20-10-2019 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 Test # 07 4. Rate of cooling (R) = ( ) 44 0 A TT t mc Îs- Dq = Þ 2 3 A Area r1 R m volume r r µµ µµ Þ () 1/3 11 RateR rm µµ 3 1/3 4 m = × r r m 3 éù r p Þµ êú ëû Q Þ 1/3 1/3 12 21 Rm 1 R m3 æö æö == ç÷ ç÷ èø èø 5. (i) Thermal energy Q 300 TC mc 15 J/°C T 45 25 ==== D- Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920 Ans.4 1 4 2 1 4 4 1 1 3 3 4 2 3 3 2 3 3 2 4 Que.21222324 2526 272829303132333435 3637 383940 Ans.2 4 4 3 3 3 3 3 2 4 2 1 3 4 2 1 1 4 2 1 Que.41424344 4546 474849505152535455 5657 585960 Ans.2 3 3 2 3 2 1 2 2 3 3 3 2 1 1 4 4 2 1 4 Que.61626364 6566 676869707172737475 7677 787980 Ans.4 1 3 2 1 1 3 4 3 3 2 4 2 2 4 4 1 2 2 3 Que.81828384 8586 878889909192939495 9697 9899 100 Ans.3 2 4 4 1 4 2 4 1 4 1 2 4 3 1 1 3 4 4 3 Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.2 2 3 3 1 2 1 1 2 1 4 3 3 2 1 4 3 3 3 3 Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.3 4 2 2 4 2 2 3 2 3 4 1 2 2 3 4 3 3 2 1 Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.1 1 2 1 4 2 4 3 2 3 2 4 1 2 4 2 4 3 2 2 Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.4 4 4 3 4 2 3 2 2 2 3 4 4 3 3 2 2 3 1 4 LTS/HS-2/7 0999DMD310319010 Target : Pre-Medical 2020/NEET-UG/20-10-2019 AL LEN (ii) Specific heat is nothing but thermal capacity per unit mass 3 mc 15 C 600 J/kg-°C m 25 10 - === ´ 6. The image on the screen is real and inverted. Triangle aperture Screen 20 cm 10 cm The size of the image on the screen has aperture size given by 10 Size 1.0 0.5 cm 20 æö == ç÷ èø 7. For mixture of gases, 12 1 v 2v v 12 nC nC C nn + = + where f CR 2 = , f is degree of freedom and 12 1 p 2p p 12 nC nC C nn + = + where p f C 1R 2 æö =+ ç÷ èø For helium, n 1 = 4, f = 3 For oxygen, 2 1 n , f = 5 2 = \ p v 5R 1 7R 4 C 47 2 22 3R 1 5R C 29 4 2 22 ´ +´ == ´ +´ = 1.62 8. The efficiency of reversible engine is always greater than that of irreversible engine. In case of irreversible engine, a part of the energy may be dissipated against friction, etc. 9. t = 20 cm m = 3/2 Radius of curvature = 20 cm object Considering refraction at the curved surface, u = –20; m 2 = 1 m 1 = 3/2; R = +20 Applying 2 1 21 v uR mmm-m -= 1 3/2 1 3/2 v 10 v 20 20 - - = Þ =- - i.e. 10 cm below the curved surface or 10 cm above the actual position of flower. 10. ( ) ( ) X 125 Y 70 500 40 - - -- = ForY = 50 X = 1375.0°X 11. for lens u = wants to see = ¥; v = can see = –5 m \ From 1 11 f vu =- Þ 1 11 f5 =- -¥ Þ f = –5 m. 12. Absolute temperatures of the black body corresponding to curve P and Q are in the inverse ratio of l m (Wein's displacement law). i.e., P Q T 1987 T 2980 = Leader Test Series/Joint Package Course/NEET-UG/20-10-2019 AL LEN 0999DMD310319010 LTS/HS-3/7 Area under curves represent the total power radiated by a body and is proportional to the fourth power of absolute temperature (Stefan's law) \ 4 PP QQ AT 16 A T 81 æö == ç÷ ç÷ èø 13. L H T 7002 11 T 2100 3 h=- =-= 2 % 100 66% 3 h=´= Actual efficiency is 40% which is 60% of the theoretical efficiency. 14. In the first case, the temperature difference is greater as compared to the second case. So, the rate of loss of heat quicker. 15. ( ) () rms 11 rms2 2 v T vT = Given, (V rms ) 1 = 100 m/s T 1 = 27°C = 27 + 273 = 300 K T 2 = 127°C = 127 + 273 = 400 K \ From Eq. (i) rms2 100 300 3 (v ) 400 2 == Þ( ) rms 2 2 100 200 v m/s 33 ´ == 16. ( ) ( ) 3 3 A AA AA A AA 3 BB B BB B BB 4/3 rC mC rC mC 4/3rC rC pr æö r = =´ ç÷ prr èø 3 1 2 11 2 1 3 12 æ ö æ ö æö = ´ ´= ç ÷ ç ÷ ç÷ è ø è ø èø 17. This is the statement of the second law of thermodynamics. 18. Process AB is isochoric, \ W AB = PDV = 0 Process BC is isothermal \ W BC = RT 2 · ln 2 1 V V æö ç÷ èø Process CA is isobaric \ W CA = –PDV = –RDT = –R(T 1 –T 2 ) = R(T 2 –T 1 ) (Negative sign is taken because of compression) 19. Change in L A = change in L B i.e., DL A = DL B Þa A DTL A = a B DTL B or a A L A = a B L B 20. As the ray moves toward the normal while entering medium 2 from 1, we have n 2 > n 1 For total internal reflection at interface of 2 and 3, n 2 > n 3 . Besides n 3 should also be less than n 1 or else ray would have emerged in medium 3, parallel to its path in medium 1. Hence, n 3 < n 1 < n 2 is the correct order. 21. According to Wien's law m 1 T lµ and from the figure (lm) 1 < (lm) 3 < (lm) 2 therefore T 1 > T 3 > T 2 . 22. 2N PK 3V = from the kinetic-theory account for pressure. 3 PV N 2K = AA N 3 PV n N 2 KN == 23. By symmetry I AB = I BC and I AD = I DC \ No current in BO and OD \ T B = T O = T D 24. For an adiabatic process, PV g = constant TV g – 1 = constant ; and 1 T P constant -g g = g = 5/3 (argon being a monoatomic gas). Page 4 HINT – SHEET ANSWER KEY 1. Temperature of interface 1 1 22 12 KK T KK q+q = + 300 100 200 0 60C 300 200 ´ +´ = =° + 2. As the temperature of water is increased from 2°C to 3°C the density of water increases (rememeber anomalous behaviour of water), also the volume of sphere increases. Therefore buoyant force on sphere due to water shall increase. 3. Note the volume has been plotted along Y-axis. W = SPDV = (0.6 – 0.8) × 10 3 + (0.4 – 0.6) × 2 × 10 3 + (0.2–0.4) ×3×10 –3 = – 200 – 400 – 600 = –1200 J (0999DMD310319010) Test Pattern DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE NEET(UG) MINOR TEST # 09 20-10-2019 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 Test # 07 4. Rate of cooling (R) = ( ) 44 0 A TT t mc Îs- Dq = Þ 2 3 A Area r1 R m volume r r µµ µµ Þ () 1/3 11 RateR rm µµ 3 1/3 4 m = × r r m 3 éù r p Þµ êú ëû Q Þ 1/3 1/3 12 21 Rm 1 R m3 æö æö == ç÷ ç÷ èø èø 5. (i) Thermal energy Q 300 TC mc 15 J/°C T 45 25 ==== D- Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920 Ans.4 1 4 2 1 4 4 1 1 3 3 4 2 3 3 2 3 3 2 4 Que.21222324 2526 272829303132333435 3637 383940 Ans.2 4 4 3 3 3 3 3 2 4 2 1 3 4 2 1 1 4 2 1 Que.41424344 4546 474849505152535455 5657 585960 Ans.2 3 3 2 3 2 1 2 2 3 3 3 2 1 1 4 4 2 1 4 Que.61626364 6566 676869707172737475 7677 787980 Ans.4 1 3 2 1 1 3 4 3 3 2 4 2 2 4 4 1 2 2 3 Que.81828384 8586 878889909192939495 9697 9899 100 Ans.3 2 4 4 1 4 2 4 1 4 1 2 4 3 1 1 3 4 4 3 Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.2 2 3 3 1 2 1 1 2 1 4 3 3 2 1 4 3 3 3 3 Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.3 4 2 2 4 2 2 3 2 3 4 1 2 2 3 4 3 3 2 1 Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.1 1 2 1 4 2 4 3 2 3 2 4 1 2 4 2 4 3 2 2 Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.4 4 4 3 4 2 3 2 2 2 3 4 4 3 3 2 2 3 1 4 LTS/HS-2/7 0999DMD310319010 Target : Pre-Medical 2020/NEET-UG/20-10-2019 AL LEN (ii) Specific heat is nothing but thermal capacity per unit mass 3 mc 15 C 600 J/kg-°C m 25 10 - === ´ 6. The image on the screen is real and inverted. Triangle aperture Screen 20 cm 10 cm The size of the image on the screen has aperture size given by 10 Size 1.0 0.5 cm 20 æö == ç÷ èø 7. For mixture of gases, 12 1 v 2v v 12 nC nC C nn + = + where f CR 2 = , f is degree of freedom and 12 1 p 2p p 12 nC nC C nn + = + where p f C 1R 2 æö =+ ç÷ èø For helium, n 1 = 4, f = 3 For oxygen, 2 1 n , f = 5 2 = \ p v 5R 1 7R 4 C 47 2 22 3R 1 5R C 29 4 2 22 ´ +´ == ´ +´ = 1.62 8. The efficiency of reversible engine is always greater than that of irreversible engine. In case of irreversible engine, a part of the energy may be dissipated against friction, etc. 9. t = 20 cm m = 3/2 Radius of curvature = 20 cm object Considering refraction at the curved surface, u = –20; m 2 = 1 m 1 = 3/2; R = +20 Applying 2 1 21 v uR mmm-m -= 1 3/2 1 3/2 v 10 v 20 20 - - = Þ =- - i.e. 10 cm below the curved surface or 10 cm above the actual position of flower. 10. ( ) ( ) X 125 Y 70 500 40 - - -- = ForY = 50 X = 1375.0°X 11. for lens u = wants to see = ¥; v = can see = –5 m \ From 1 11 f vu =- Þ 1 11 f5 =- -¥ Þ f = –5 m. 12. Absolute temperatures of the black body corresponding to curve P and Q are in the inverse ratio of l m (Wein's displacement law). i.e., P Q T 1987 T 2980 = Leader Test Series/Joint Package Course/NEET-UG/20-10-2019 AL LEN 0999DMD310319010 LTS/HS-3/7 Area under curves represent the total power radiated by a body and is proportional to the fourth power of absolute temperature (Stefan's law) \ 4 PP QQ AT 16 A T 81 æö == ç÷ ç÷ èø 13. L H T 7002 11 T 2100 3 h=- =-= 2 % 100 66% 3 h=´= Actual efficiency is 40% which is 60% of the theoretical efficiency. 14. In the first case, the temperature difference is greater as compared to the second case. So, the rate of loss of heat quicker. 15. ( ) () rms 11 rms2 2 v T vT = Given, (V rms ) 1 = 100 m/s T 1 = 27°C = 27 + 273 = 300 K T 2 = 127°C = 127 + 273 = 400 K \ From Eq. (i) rms2 100 300 3 (v ) 400 2 == Þ( ) rms 2 2 100 200 v m/s 33 ´ == 16. ( ) ( ) 3 3 A AA AA A AA 3 BB B BB B BB 4/3 rC mC rC mC 4/3rC rC pr æö r = =´ ç÷ prr èø 3 1 2 11 2 1 3 12 æ ö æ ö æö = ´ ´= ç ÷ ç ÷ ç÷ è ø è ø èø 17. This is the statement of the second law of thermodynamics. 18. Process AB is isochoric, \ W AB = PDV = 0 Process BC is isothermal \ W BC = RT 2 · ln 2 1 V V æö ç÷ èø Process CA is isobaric \ W CA = –PDV = –RDT = –R(T 1 –T 2 ) = R(T 2 –T 1 ) (Negative sign is taken because of compression) 19. Change in L A = change in L B i.e., DL A = DL B Þa A DTL A = a B DTL B or a A L A = a B L B 20. As the ray moves toward the normal while entering medium 2 from 1, we have n 2 > n 1 For total internal reflection at interface of 2 and 3, n 2 > n 3 . Besides n 3 should also be less than n 1 or else ray would have emerged in medium 3, parallel to its path in medium 1. Hence, n 3 < n 1 < n 2 is the correct order. 21. According to Wien's law m 1 T lµ and from the figure (lm) 1 < (lm) 3 < (lm) 2 therefore T 1 > T 3 > T 2 . 22. 2N PK 3V = from the kinetic-theory account for pressure. 3 PV N 2K = AA N 3 PV n N 2 KN == 23. By symmetry I AB = I BC and I AD = I DC \ No current in BO and OD \ T B = T O = T D 24. For an adiabatic process, PV g = constant TV g – 1 = constant ; and 1 T P constant -g g = g = 5/3 (argon being a monoatomic gas). LTS/HS-4/7 0999DMD310319010 Target : Pre-Medical 2020/NEET-UG/20-10-2019 AL LEN 25. E = e sAT 4 = es4pR 2 T 4 ( )( ) ( )( ) 24 1 24 2 e 1 4000 E E e 4 2000 s = s = ( ) ( ) 4 4 4 2 . 2000 1 16 2000 = 26. ( ) 12 1 11 1 f RR æö =m-- ç÷ èø R = 10 cm f = 20 f = 20 f' ( ) 11 f' 31 10 10 æö =-- ç÷ - èø 10 f' 4 - = eq 1 141 24 f 20 10 20 20 10 =- + =- ; f eq = –10/3 cm 27. ii WW W mix iW mL m c mm q- q= + Q m i = m w Þ i W W mix L 336 800 c 4.2 0C 22 q- +- q = = =° or 1 gm ice ® 80 cal ® 0°C 1 gm of 80°C water ® 0°C ® 80 cal so final temperature will be 0°C 28. For a given pressure, V is small for T 1 . Since V µ T, therefore, T 1 < T 2 . 29. L = f 0 + f e = 44 and 0 e f m 10 f == This gives f 0 = 40 cm 30. eq 12 2 11 K KK =+ 1 1 K A Q L Dq = and 2 2 K A Q L Dq = eq eq A 1 A QK. 1 L 2L K Dq Dq == 12 1 A . 2L 1 11 2KK Dq = æö + ç÷ èø 12 1 A Q · A Q A Q L Q · L Q · L D = DD + 12 12 12 Q Q 1 = 11 QQ QQ = + + 31. m 1 sin a 1 = m 2 sin a 2 a 1 a 2 m 1 m 2 12 12 cc sin sin vv a=a 12 12 sin sin ff aa = ll Þ 2 21 1 sin sin a l =l a Page 5 HINT – SHEET ANSWER KEY 1. Temperature of interface 1 1 22 12 KK T KK q+q = + 300 100 200 0 60C 300 200 ´ +´ = =° + 2. As the temperature of water is increased from 2°C to 3°C the density of water increases (rememeber anomalous behaviour of water), also the volume of sphere increases. Therefore buoyant force on sphere due to water shall increase. 3. Note the volume has been plotted along Y-axis. W = SPDV = (0.6 – 0.8) × 10 3 + (0.4 – 0.6) × 2 × 10 3 + (0.2–0.4) ×3×10 –3 = – 200 – 400 – 600 = –1200 J (0999DMD310319010) Test Pattern DIST ANCE LEARNING PROGRAMME (Academic Session : 2019 - 2020) PRE-MEDICAL : LEADER TEST SERIES / JOINT PACKAGE COURSE NEET(UG) MINOR TEST # 09 20-10-2019 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 Test # 07 4. Rate of cooling (R) = ( ) 44 0 A TT t mc Îs- Dq = Þ 2 3 A Area r1 R m volume r r µµ µµ Þ () 1/3 11 RateR rm µµ 3 1/3 4 m = × r r m 3 éù r p Þµ êú ëû Q Þ 1/3 1/3 12 21 Rm 1 R m3 æö æö == ç÷ ç÷ èø èø 5. (i) Thermal energy Q 300 TC mc 15 J/°C T 45 25 ==== D- Que. 1 2 3 4 5 6 7 8 9 101112131415 1617 181920 Ans.4 1 4 2 1 4 4 1 1 3 3 4 2 3 3 2 3 3 2 4 Que.21222324 2526 272829303132333435 3637 383940 Ans.2 4 4 3 3 3 3 3 2 4 2 1 3 4 2 1 1 4 2 1 Que.41424344 4546 474849505152535455 5657 585960 Ans.2 3 3 2 3 2 1 2 2 3 3 3 2 1 1 4 4 2 1 4 Que.61626364 6566 676869707172737475 7677 787980 Ans.4 1 3 2 1 1 3 4 3 3 2 4 2 2 4 4 1 2 2 3 Que.81828384 8586 878889909192939495 9697 9899 100 Ans.3 2 4 4 1 4 2 4 1 4 1 2 4 3 1 1 3 4 4 3 Que. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans.2 2 3 3 1 2 1 1 2 1 4 3 3 2 1 4 3 3 3 3 Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 Ans.3 4 2 2 4 2 2 3 2 3 4 1 2 2 3 4 3 3 2 1 Que. 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 Ans.1 1 2 1 4 2 4 3 2 3 2 4 1 2 4 2 4 3 2 2 Que. 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 Ans.4 4 4 3 4 2 3 2 2 2 3 4 4 3 3 2 2 3 1 4 LTS/HS-2/7 0999DMD310319010 Target : Pre-Medical 2020/NEET-UG/20-10-2019 AL LEN (ii) Specific heat is nothing but thermal capacity per unit mass 3 mc 15 C 600 J/kg-°C m 25 10 - === ´ 6. The image on the screen is real and inverted. Triangle aperture Screen 20 cm 10 cm The size of the image on the screen has aperture size given by 10 Size 1.0 0.5 cm 20 æö == ç÷ èø 7. For mixture of gases, 12 1 v 2v v 12 nC nC C nn + = + where f CR 2 = , f is degree of freedom and 12 1 p 2p p 12 nC nC C nn + = + where p f C 1R 2 æö =+ ç÷ èø For helium, n 1 = 4, f = 3 For oxygen, 2 1 n , f = 5 2 = \ p v 5R 1 7R 4 C 47 2 22 3R 1 5R C 29 4 2 22 ´ +´ == ´ +´ = 1.62 8. The efficiency of reversible engine is always greater than that of irreversible engine. In case of irreversible engine, a part of the energy may be dissipated against friction, etc. 9. t = 20 cm m = 3/2 Radius of curvature = 20 cm object Considering refraction at the curved surface, u = –20; m 2 = 1 m 1 = 3/2; R = +20 Applying 2 1 21 v uR mmm-m -= 1 3/2 1 3/2 v 10 v 20 20 - - = Þ =- - i.e. 10 cm below the curved surface or 10 cm above the actual position of flower. 10. ( ) ( ) X 125 Y 70 500 40 - - -- = ForY = 50 X = 1375.0°X 11. for lens u = wants to see = ¥; v = can see = –5 m \ From 1 11 f vu =- Þ 1 11 f5 =- -¥ Þ f = –5 m. 12. Absolute temperatures of the black body corresponding to curve P and Q are in the inverse ratio of l m (Wein's displacement law). i.e., P Q T 1987 T 2980 = Leader Test Series/Joint Package Course/NEET-UG/20-10-2019 AL LEN 0999DMD310319010 LTS/HS-3/7 Area under curves represent the total power radiated by a body and is proportional to the fourth power of absolute temperature (Stefan's law) \ 4 PP QQ AT 16 A T 81 æö == ç÷ ç÷ èø 13. L H T 7002 11 T 2100 3 h=- =-= 2 % 100 66% 3 h=´= Actual efficiency is 40% which is 60% of the theoretical efficiency. 14. In the first case, the temperature difference is greater as compared to the second case. So, the rate of loss of heat quicker. 15. ( ) () rms 11 rms2 2 v T vT = Given, (V rms ) 1 = 100 m/s T 1 = 27°C = 27 + 273 = 300 K T 2 = 127°C = 127 + 273 = 400 K \ From Eq. (i) rms2 100 300 3 (v ) 400 2 == Þ( ) rms 2 2 100 200 v m/s 33 ´ == 16. ( ) ( ) 3 3 A AA AA A AA 3 BB B BB B BB 4/3 rC mC rC mC 4/3rC rC pr æö r = =´ ç÷ prr èø 3 1 2 11 2 1 3 12 æ ö æ ö æö = ´ ´= ç ÷ ç ÷ ç÷ è ø è ø èø 17. This is the statement of the second law of thermodynamics. 18. Process AB is isochoric, \ W AB = PDV = 0 Process BC is isothermal \ W BC = RT 2 · ln 2 1 V V æö ç÷ èø Process CA is isobaric \ W CA = –PDV = –RDT = –R(T 1 –T 2 ) = R(T 2 –T 1 ) (Negative sign is taken because of compression) 19. Change in L A = change in L B i.e., DL A = DL B Þa A DTL A = a B DTL B or a A L A = a B L B 20. As the ray moves toward the normal while entering medium 2 from 1, we have n 2 > n 1 For total internal reflection at interface of 2 and 3, n 2 > n 3 . Besides n 3 should also be less than n 1 or else ray would have emerged in medium 3, parallel to its path in medium 1. Hence, n 3 < n 1 < n 2 is the correct order. 21. According to Wien's law m 1 T lµ and from the figure (lm) 1 < (lm) 3 < (lm) 2 therefore T 1 > T 3 > T 2 . 22. 2N PK 3V = from the kinetic-theory account for pressure. 3 PV N 2K = AA N 3 PV n N 2 KN == 23. By symmetry I AB = I BC and I AD = I DC \ No current in BO and OD \ T B = T O = T D 24. For an adiabatic process, PV g = constant TV g – 1 = constant ; and 1 T P constant -g g = g = 5/3 (argon being a monoatomic gas). LTS/HS-4/7 0999DMD310319010 Target : Pre-Medical 2020/NEET-UG/20-10-2019 AL LEN 25. E = e sAT 4 = es4pR 2 T 4 ( )( ) ( )( ) 24 1 24 2 e 1 4000 E E e 4 2000 s = s = ( ) ( ) 4 4 4 2 . 2000 1 16 2000 = 26. ( ) 12 1 11 1 f RR æö =m-- ç÷ èø R = 10 cm f = 20 f = 20 f' ( ) 11 f' 31 10 10 æö =-- ç÷ - èø 10 f' 4 - = eq 1 141 24 f 20 10 20 20 10 =- + =- ; f eq = –10/3 cm 27. ii WW W mix iW mL m c mm q- q= + Q m i = m w Þ i W W mix L 336 800 c 4.2 0C 22 q- +- q = = =° or 1 gm ice ® 80 cal ® 0°C 1 gm of 80°C water ® 0°C ® 80 cal so final temperature will be 0°C 28. For a given pressure, V is small for T 1 . Since V µ T, therefore, T 1 < T 2 . 29. L = f 0 + f e = 44 and 0 e f m 10 f == This gives f 0 = 40 cm 30. eq 12 2 11 K KK =+ 1 1 K A Q L Dq = and 2 2 K A Q L Dq = eq eq A 1 A QK. 1 L 2L K Dq Dq == 12 1 A . 2L 1 11 2KK Dq = æö + ç÷ èø 12 1 A Q · A Q A Q L Q · L Q · L D = DD + 12 12 12 Q Q 1 = 11 QQ QQ = + + 31. m 1 sin a 1 = m 2 sin a 2 a 1 a 2 m 1 m 2 12 12 cc sin sin vv a=a 12 12 sin sin ff aa = ll Þ 2 21 1 sin sin a l =l a Leader Test Series/Joint Package Course/NEET-UG/20-10-2019 AL LEN 0999DMD310319010 LTS/HS-5/7 32. app. Mass expelled Mass remained × T g= D 4 x /100 1 1.25 10 / C x 80 8000 - = = = ´° ´ 33. 3PV C M = 5 1 3 24 10 10 100 cms 20 - ´ ´ ´´ = = 6 × 10 4 cm s –1 = 600 ms –1 34. According to Newton's law 12 t q -q 12 0 k 2 q +q éù = -q êú ëû Initially, ( ) 0 80 64 80 64 K 52 - +æö = -q ç÷ èø Þ 3.2 = K[72 – q 0 ] ....(i) Finally, ( ) 0 64 52 64 52 K 102 - +éù = -q êú ëû Þ 1.2 = K[58 – q 0 ] ....(ii) On solving equation (i) and (ii) q 0 = 49°C s 35. Since 0 e f m f = Also m = Angle subtended by the image Angle subtended by the object \ 0 e f f a = b Þ 0 e f 602 a 24 f5 ´b ´ = = =° 36. The volume of matter in portional AB of the curve is almost constant and pressure is decreasing. These are the characteristics of liquid state. 37. (P = constant) PP nC TC Q5 WnRT R2 D D = == DD 38. Ideal gas equation PV = mRT = A N RT N æö ç÷ èø where N = Number of molecule, N A = Avogadro number \ 1 1 12 2 2 21 N P VT N P VT æ öæ öæ ö = ç ÷ç ÷ç ÷ è øè øè ø P V 2T4 2P V/4 T1 æ öæ öæö == ç ÷ç ÷ç÷ è øè øèø 39. Given u = (f + x 1 ) and u = (f + x 2 ), v = (f + x 2 ) the focal length ( )( ) ( )( ) 12 12 fx fx uv f u v fx fx ++ == + + ++ On solving, we get f 2 = x 1 x 2 or 12 f xx = 40. dQ KA dt Dq = l , For both rods K, A and Dq are same Þ dQ1 dt µ l So ( ) ( ) semi circular straight dQ / dt dQ / dt straight semi circular 2r2 r = == pp l l 41. 21 12 TT 1 1, T T1 h=-= -h ( ) 2 1 2 12 T 1 T T T/T1 w== -- ( ) 11 1/11 -h == h éù -h- ëû As 1 0.1 10% 0.1, = 9 0.1 - h= =w=Read More

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