Page 1 MARKING SCHEME SET 55/1/3 (DELHI) Q.No. Expected Answer/Value Points Marks Total Marks 1. Drift velocity , is the directed velocity acquired by the electrons for a unit value of the applied Electric field. Current, ½ ½ 1 2. Modulation index = = = 0.5 ½ ½ 1 3. Perpendicular to the plane formed by / [Note: Give full credit for writing the expression only.] ½ ½ 1 4. 1 1 5. 20cm 1 1 6. X: Channel It connects Transmitter and Receiver ½ ½ 1 7. Glass. In glass there is no electromagnetic induction, due to presence of Earthâ€™s magnetic field, unlike in the case of metallic ball. ½ ½ 1 8. If Electric field is not normal, it will have non-zero component along the surface. In that case, work would be done in moving a charge on an equipotential surface. 1 1 9. = => = 12 Potential energy ½ ½ Calculation of Potential energy of the dipole 2 Page 2 MARKING SCHEME SET 55/1/3 (DELHI) Q.No. Expected Answer/Value Points Marks Total Marks 1. Drift velocity , is the directed velocity acquired by the electrons for a unit value of the applied Electric field. Current, ½ ½ 1 2. Modulation index = = = 0.5 ½ ½ 1 3. Perpendicular to the plane formed by / [Note: Give full credit for writing the expression only.] ½ ½ 1 4. 1 1 5. 20cm 1 1 6. X: Channel It connects Transmitter and Receiver ½ ½ 1 7. Glass. In glass there is no electromagnetic induction, due to presence of Earthâ€™s magnetic field, unlike in the case of metallic ball. ½ ½ 1 8. If Electric field is not normal, it will have non-zero component along the surface. In that case, work would be done in moving a charge on an equipotential surface. 1 1 9. = => = 12 Potential energy ½ ½ Calculation of Potential energy of the dipole 2 Delhi Set III U = -pE cos = -12 x cos 60 0 = -6J ½ ½ 2 10. It makes use of the principle that the energy of the charged particles / ions can be made to increase in presence of crossed Electric and magnetic fields. A normal Magnetic field acts on the charged particle and makes them move in a circular path .While moving from one dee to another; particle is acted upon by the alternating electric field, and is accelerated by this field, which increases the energy of the particle. 1 1 2 11. (i) Reactance of the capacitor will decrease, resulting in increase of the current in the circuit. Therefore the bulb will glow brighter. (ii) Increased resistance will decrease the current in the circuit, which will decrease glow of the bulb. [Note : Do not deduct any mark for not giving the reasons] 1 1 2 12. (a) deutron ?= = = => ? d > ? a (b) deuteron KE= qV , for the same accelerating potential, we have ½ ½ ½ ½ 2 Part (a) and its reason ½ + ½ Part (b) and its reason ½ + ½ Underlying principle 1 Brief working 1 Effect on glow of bulb in Part (i) 1 Part (ii) 1 Page 3 MARKING SCHEME SET 55/1/3 (DELHI) Q.No. Expected Answer/Value Points Marks Total Marks 1. Drift velocity , is the directed velocity acquired by the electrons for a unit value of the applied Electric field. Current, ½ ½ 1 2. Modulation index = = = 0.5 ½ ½ 1 3. Perpendicular to the plane formed by / [Note: Give full credit for writing the expression only.] ½ ½ 1 4. 1 1 5. 20cm 1 1 6. X: Channel It connects Transmitter and Receiver ½ ½ 1 7. Glass. In glass there is no electromagnetic induction, due to presence of Earthâ€™s magnetic field, unlike in the case of metallic ball. ½ ½ 1 8. If Electric field is not normal, it will have non-zero component along the surface. In that case, work would be done in moving a charge on an equipotential surface. 1 1 9. = => = 12 Potential energy ½ ½ Calculation of Potential energy of the dipole 2 Delhi Set III U = -pE cos = -12 x cos 60 0 = -6J ½ ½ 2 10. It makes use of the principle that the energy of the charged particles / ions can be made to increase in presence of crossed Electric and magnetic fields. A normal Magnetic field acts on the charged particle and makes them move in a circular path .While moving from one dee to another; particle is acted upon by the alternating electric field, and is accelerated by this field, which increases the energy of the particle. 1 1 2 11. (i) Reactance of the capacitor will decrease, resulting in increase of the current in the circuit. Therefore the bulb will glow brighter. (ii) Increased resistance will decrease the current in the circuit, which will decrease glow of the bulb. [Note : Do not deduct any mark for not giving the reasons] 1 1 2 12. (a) deutron ?= = = => ? d > ? a (b) deuteron KE= qV , for the same accelerating potential, we have ½ ½ ½ ½ 2 Part (a) and its reason ½ + ½ Part (b) and its reason ½ + ½ Underlying principle 1 Brief working 1 Effect on glow of bulb in Part (i) 1 Part (ii) 1 Delhi Set III 13. A: Paramagnetic B: Diamagnetic Susceptibility For A: positive For B: negative ½ ½ ½ ½ 2 14. Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction. Alternatively, ? i =0 Justification : Conservation of charge Loop rule: The Algebraic sum of changes in the potential around any closed loop involving resistors and cells in the loop is zero. Alternatively, ? =0 , where is the changes in potential Justification : Conservation of energy ½ ½ ½ ½ 2 15. ½ ½ ½ ½ 2 16. (a) power = , where n = no. of photons per second 0909 photons/second =9.1 photons/second ½ ½ (a) Estimation of number of photons per second 1 (b) plot showing the variation 1 Finding flux in two cases 1+1 Identification of magnetic material ½ + ½ Susceptibility ½ + ½ Statements of two Laws ½ + ½ Justification ½ + ½ Page 4 MARKING SCHEME SET 55/1/3 (DELHI) Q.No. Expected Answer/Value Points Marks Total Marks 1. Drift velocity , is the directed velocity acquired by the electrons for a unit value of the applied Electric field. Current, ½ ½ 1 2. Modulation index = = = 0.5 ½ ½ 1 3. Perpendicular to the plane formed by / [Note: Give full credit for writing the expression only.] ½ ½ 1 4. 1 1 5. 20cm 1 1 6. X: Channel It connects Transmitter and Receiver ½ ½ 1 7. Glass. In glass there is no electromagnetic induction, due to presence of Earthâ€™s magnetic field, unlike in the case of metallic ball. ½ ½ 1 8. If Electric field is not normal, it will have non-zero component along the surface. In that case, work would be done in moving a charge on an equipotential surface. 1 1 9. = => = 12 Potential energy ½ ½ Calculation of Potential energy of the dipole 2 Delhi Set III U = -pE cos = -12 x cos 60 0 = -6J ½ ½ 2 10. It makes use of the principle that the energy of the charged particles / ions can be made to increase in presence of crossed Electric and magnetic fields. A normal Magnetic field acts on the charged particle and makes them move in a circular path .While moving from one dee to another; particle is acted upon by the alternating electric field, and is accelerated by this field, which increases the energy of the particle. 1 1 2 11. (i) Reactance of the capacitor will decrease, resulting in increase of the current in the circuit. Therefore the bulb will glow brighter. (ii) Increased resistance will decrease the current in the circuit, which will decrease glow of the bulb. [Note : Do not deduct any mark for not giving the reasons] 1 1 2 12. (a) deutron ?= = = => ? d > ? a (b) deuteron KE= qV , for the same accelerating potential, we have ½ ½ ½ ½ 2 Part (a) and its reason ½ + ½ Part (b) and its reason ½ + ½ Underlying principle 1 Brief working 1 Effect on glow of bulb in Part (i) 1 Part (ii) 1 Delhi Set III 13. A: Paramagnetic B: Diamagnetic Susceptibility For A: positive For B: negative ½ ½ ½ ½ 2 14. Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction. Alternatively, ? i =0 Justification : Conservation of charge Loop rule: The Algebraic sum of changes in the potential around any closed loop involving resistors and cells in the loop is zero. Alternatively, ? =0 , where is the changes in potential Justification : Conservation of energy ½ ½ ½ ½ 2 15. ½ ½ ½ ½ 2 16. (a) power = , where n = no. of photons per second 0909 photons/second =9.1 photons/second ½ ½ (a) Estimation of number of photons per second 1 (b) plot showing the variation 1 Finding flux in two cases 1+1 Identification of magnetic material ½ + ½ Susceptibility ½ + ½ Statements of two Laws ½ + ½ Justification ½ + ½ Delhi Set III [Note: Even if the student doesnâ€™t write the formula and calculates correctly, give full credit to this part] (b) 1 2 17. Truth Table Input Output A B Yâ€™ Y 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 Gate R: OR Gate S: AND Gate OR P: NAND Gate Q: OR Gate Truth Table Input Output A B X 0 0 1 1 0 1 0 1 1 1 1 1 1 ½ ½ ½ ½ 1 2 2 Truth Table 1 Names of gates used ½ + ½ Identification 1 Truth Table 1 Page 5 MARKING SCHEME SET 55/1/3 (DELHI) Q.No. Expected Answer/Value Points Marks Total Marks 1. Drift velocity , is the directed velocity acquired by the electrons for a unit value of the applied Electric field. Current, ½ ½ 1 2. Modulation index = = = 0.5 ½ ½ 1 3. Perpendicular to the plane formed by / [Note: Give full credit for writing the expression only.] ½ ½ 1 4. 1 1 5. 20cm 1 1 6. X: Channel It connects Transmitter and Receiver ½ ½ 1 7. Glass. In glass there is no electromagnetic induction, due to presence of Earthâ€™s magnetic field, unlike in the case of metallic ball. ½ ½ 1 8. If Electric field is not normal, it will have non-zero component along the surface. In that case, work would be done in moving a charge on an equipotential surface. 1 1 9. = => = 12 Potential energy ½ ½ Calculation of Potential energy of the dipole 2 Delhi Set III U = -pE cos = -12 x cos 60 0 = -6J ½ ½ 2 10. It makes use of the principle that the energy of the charged particles / ions can be made to increase in presence of crossed Electric and magnetic fields. A normal Magnetic field acts on the charged particle and makes them move in a circular path .While moving from one dee to another; particle is acted upon by the alternating electric field, and is accelerated by this field, which increases the energy of the particle. 1 1 2 11. (i) Reactance of the capacitor will decrease, resulting in increase of the current in the circuit. Therefore the bulb will glow brighter. (ii) Increased resistance will decrease the current in the circuit, which will decrease glow of the bulb. [Note : Do not deduct any mark for not giving the reasons] 1 1 2 12. (a) deutron ?= = = => ? d > ? a (b) deuteron KE= qV , for the same accelerating potential, we have ½ ½ ½ ½ 2 Part (a) and its reason ½ + ½ Part (b) and its reason ½ + ½ Underlying principle 1 Brief working 1 Effect on glow of bulb in Part (i) 1 Part (ii) 1 Delhi Set III 13. A: Paramagnetic B: Diamagnetic Susceptibility For A: positive For B: negative ½ ½ ½ ½ 2 14. Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction. Alternatively, ? i =0 Justification : Conservation of charge Loop rule: The Algebraic sum of changes in the potential around any closed loop involving resistors and cells in the loop is zero. Alternatively, ? =0 , where is the changes in potential Justification : Conservation of energy ½ ½ ½ ½ 2 15. ½ ½ ½ ½ 2 16. (a) power = , where n = no. of photons per second 0909 photons/second =9.1 photons/second ½ ½ (a) Estimation of number of photons per second 1 (b) plot showing the variation 1 Finding flux in two cases 1+1 Identification of magnetic material ½ + ½ Susceptibility ½ + ½ Statements of two Laws ½ + ½ Justification ½ + ½ Delhi Set III [Note: Even if the student doesnâ€™t write the formula and calculates correctly, give full credit to this part] (b) 1 2 17. Truth Table Input Output A B Yâ€™ Y 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 Gate R: OR Gate S: AND Gate OR P: NAND Gate Q: OR Gate Truth Table Input Output A B X 0 0 1 1 0 1 0 1 1 1 1 1 1 ½ ½ ½ ½ 1 2 2 Truth Table 1 Names of gates used ½ + ½ Identification 1 Truth Table 1 Delhi Set III 18. In the first case, the overlapping of the contributions of the wavelets from two halves of a single slit produces a minimum because corresponding wavelets from two halves have a path difference of 2 ? . In the second case, the overlapping of the wavefronts from the two slits produces first maximum because these wavefronts have the path difference of . (Alternatively, if a student writes the conditions given below, give full credit.) Condition for first minimum in single slit diffraction is , ? / a, Whereas in case of two narrow slits separated by distance a, first maximum occurs at angle ? / a [Note: Award 1 mark even if the candidate attempts this question partly.] 1 1 2 19. (a) Statement of law Expression of the law in integral form: = (Award 1 mark if the student just writes the integral form of Ampereâ€™s circuital law) (b) B = I Magnitude of net magnetic field inside the combined system on the axis , B =B 1 - B 2 I Also accept if the student writes B = I (iii)Outside the combined system, the net magnetic field is zero. 1 ½ ½ ½ ½ 3 20. Energy required to excite hydrogen atoms from ground state to the second state = E final â€“ E initial = -1.51 â€“ (-13.6)eV = 12.09 eV ½ Finding maximum energy level of hydrogen atoms ½ Calculation of wavelengths 2 ½ Explanation of the given statement 1 + 1 (a) Statement of Ampereâ€™s circuital Law 1 ½ (b) Calculation of net magnetic field (i) inside and (ii) outside 1 ½Read More

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