Page 1 Page 1 of 13 Final Draft 11/03/16 03:00 p.m . MARKING SCHEME SET 55/1/S Q. No. Expected Answer / Value Points Marks Total Marks Section A Set1,Q1 Set2,Q3 Set3,Q2 (i) Manganin (ii) As ? increases A also increases Alternatively, ½ ½ 1 Set1,Q2 Set2,Q2 Set3,Q5 Phase angle = 60° [Note : If the student only writes, [ ] , give ½ mark] 1 1 Set1,Q3 Set2,Q1 Set3,Q4 Between plates of capacitor during charging / discharging Alternatively, In the region of time varying electric field 1 1 Set1,Q4 Set2,Q5 Set3,Q1 (i) P = NOT gate (ii) Q = OR gate ½ ½ 1 Set1,Q5 Set2,Q4 Set3,Q3 Def: The average time, between successive collisions of electrons, (in a conductor) is known as relaxation time 1 1 Section B Set1,Q6 Set2,Q6 Set3,Q10 The field inside a conductor is zero. Sensitive instruments are shielded from outside electrical influences by enclosing them in a hollow conductor . (any other relevant answer.) Potential inside the cavity is not zero/ potential is constant. ½ 1 ½ 2 Set1,Q7 Set2,Q7 Set3,Q8 Any two properties of electromagnetic waves Such as (a) transverse nature (b) does not get deflected by electric fields or magnetic fields (c) same speed in vacuum for all waves (d) no material medium required for propagation (e) they get refracted, diffracted and polarised / (any two properties) Electric charges present on a plane, kept normal to the direction of propagation of an e.m. wave can be set and sustained in motion by the electric and magnetic field of the electromagnetic wave. The charges thus acquire energy and momentum from the waves. ½ + ½ 1 Electrostatic Shielding ½ Using this property in actual practice 1 Potential in a cavity ½ Two properties of electromagnetic waves ½ +½ Showing e m waves have momentum 1 Page 2 Page 1 of 13 Final Draft 11/03/16 03:00 p.m . MARKING SCHEME SET 55/1/S Q. No. Expected Answer / Value Points Marks Total Marks Section A Set1,Q1 Set2,Q3 Set3,Q2 (i) Manganin (ii) As ? increases A also increases Alternatively, ½ ½ 1 Set1,Q2 Set2,Q2 Set3,Q5 Phase angle = 60° [Note : If the student only writes, [ ] , give ½ mark] 1 1 Set1,Q3 Set2,Q1 Set3,Q4 Between plates of capacitor during charging / discharging Alternatively, In the region of time varying electric field 1 1 Set1,Q4 Set2,Q5 Set3,Q1 (i) P = NOT gate (ii) Q = OR gate ½ ½ 1 Set1,Q5 Set2,Q4 Set3,Q3 Def: The average time, between successive collisions of electrons, (in a conductor) is known as relaxation time 1 1 Section B Set1,Q6 Set2,Q6 Set3,Q10 The field inside a conductor is zero. Sensitive instruments are shielded from outside electrical influences by enclosing them in a hollow conductor . (any other relevant answer.) Potential inside the cavity is not zero/ potential is constant. ½ 1 ½ 2 Set1,Q7 Set2,Q7 Set3,Q8 Any two properties of electromagnetic waves Such as (a) transverse nature (b) does not get deflected by electric fields or magnetic fields (c) same speed in vacuum for all waves (d) no material medium required for propagation (e) they get refracted, diffracted and polarised / (any two properties) Electric charges present on a plane, kept normal to the direction of propagation of an e.m. wave can be set and sustained in motion by the electric and magnetic field of the electromagnetic wave. The charges thus acquire energy and momentum from the waves. ½ + ½ 1 Electrostatic Shielding ½ Using this property in actual practice 1 Potential in a cavity ½ Two properties of electromagnetic waves ½ +½ Showing e m waves have momentum 1 Page 2 of 13 Final Draft 11/03/16 03:00 p.m . Alternatively Radiation Pressure â€“ Electromagnetic waves exert radiation pressure. Hence, they carry momentum. 2 Set1,Q8 Set2,Q8 Set3,Q9 Diffraction effects are observed for beams of electrons scattered by the crystals 1.227nm V ? ? 1.227 120 nm ? ? Value ?= 0.112nm Alternatively = =0.112nm ½ ½ ½ ½ ½ ½ ½ 2 Set1,Q9 Set2,Q10 Set3,Q7 (i) Transducer: The device which converts one form of energy into another (ii) Repeater: A repeater picks up signal, amplifies and retransmits them to receiver 1 1 2 Set1,Q10 Set2,Q9 Set3,Q6 (i) 2 0 r r n ? 21.2x10 -11 = 5.3x10 -11 n 2 implies n = 2 (ii) 2 13.6eV E n ? ? = =-3.4eV [Award ½ mark if the student just writes E= E 1 /4] OR (i) Energy of photon = = =4.5 (ii) The corresponding transition is B ½ ½ ½ ½ ½ + ½ + ½ ½ 2 2 Principle ½ Calculation of ? 1 ½ Function of Transducer 1 Function of Repeater 1 Finding the principal quantum number 1 Finding the total energy 1 Calculation of energy of photon 1½ Identification of transistion ½ Page 3 Page 1 of 13 Final Draft 11/03/16 03:00 p.m . MARKING SCHEME SET 55/1/S Q. No. Expected Answer / Value Points Marks Total Marks Section A Set1,Q1 Set2,Q3 Set3,Q2 (i) Manganin (ii) As ? increases A also increases Alternatively, ½ ½ 1 Set1,Q2 Set2,Q2 Set3,Q5 Phase angle = 60° [Note : If the student only writes, [ ] , give ½ mark] 1 1 Set1,Q3 Set2,Q1 Set3,Q4 Between plates of capacitor during charging / discharging Alternatively, In the region of time varying electric field 1 1 Set1,Q4 Set2,Q5 Set3,Q1 (i) P = NOT gate (ii) Q = OR gate ½ ½ 1 Set1,Q5 Set2,Q4 Set3,Q3 Def: The average time, between successive collisions of electrons, (in a conductor) is known as relaxation time 1 1 Section B Set1,Q6 Set2,Q6 Set3,Q10 The field inside a conductor is zero. Sensitive instruments are shielded from outside electrical influences by enclosing them in a hollow conductor . (any other relevant answer.) Potential inside the cavity is not zero/ potential is constant. ½ 1 ½ 2 Set1,Q7 Set2,Q7 Set3,Q8 Any two properties of electromagnetic waves Such as (a) transverse nature (b) does not get deflected by electric fields or magnetic fields (c) same speed in vacuum for all waves (d) no material medium required for propagation (e) they get refracted, diffracted and polarised / (any two properties) Electric charges present on a plane, kept normal to the direction of propagation of an e.m. wave can be set and sustained in motion by the electric and magnetic field of the electromagnetic wave. The charges thus acquire energy and momentum from the waves. ½ + ½ 1 Electrostatic Shielding ½ Using this property in actual practice 1 Potential in a cavity ½ Two properties of electromagnetic waves ½ +½ Showing e m waves have momentum 1 Page 2 of 13 Final Draft 11/03/16 03:00 p.m . Alternatively Radiation Pressure â€“ Electromagnetic waves exert radiation pressure. Hence, they carry momentum. 2 Set1,Q8 Set2,Q8 Set3,Q9 Diffraction effects are observed for beams of electrons scattered by the crystals 1.227nm V ? ? 1.227 120 nm ? ? Value ?= 0.112nm Alternatively = =0.112nm ½ ½ ½ ½ ½ ½ ½ 2 Set1,Q9 Set2,Q10 Set3,Q7 (i) Transducer: The device which converts one form of energy into another (ii) Repeater: A repeater picks up signal, amplifies and retransmits them to receiver 1 1 2 Set1,Q10 Set2,Q9 Set3,Q6 (i) 2 0 r r n ? 21.2x10 -11 = 5.3x10 -11 n 2 implies n = 2 (ii) 2 13.6eV E n ? ? = =-3.4eV [Award ½ mark if the student just writes E= E 1 /4] OR (i) Energy of photon = = =4.5 (ii) The corresponding transition is B ½ ½ ½ ½ ½ + ½ + ½ ½ 2 2 Principle ½ Calculation of ? 1 ½ Function of Transducer 1 Function of Repeater 1 Finding the principal quantum number 1 Finding the total energy 1 Calculation of energy of photon 1½ Identification of transistion ½ Page 3 of 13 Final Draft 11/03/16 03:00 p.m . Section C Set1,Q11 Set2,Q20 Set3,Q22 E +q = Kq / (r 2 +a 2 ) and E -q = Kq / (r 2 +a 2 ) The two Electric fields have equal magnitudes and their directions are as shown in diagram Components along dipole axis get added up while normal components cancel each other. E=-[ E -q +E +q ] cos? r so E = - = ( = Direction of electric field is opposite to that of dipole moment. 1 ½ ½ ½ ½ 3 Set1,Q12 Set2,Q15 Set3,Q16 a) Zero b) We have C series = = 1 µF Also, C parallel = (3+3+3)=9µF Energy stored = Energy in series combination = Energy in parallel combination = Ratio = 1:9 ½ ½ ½ ½ ½ ½ 3 Diagram 1 Deriving expression for E eq 1 ½ Direction of E eq ½ a) To find charge accumulated in capacitor C 2 ½ b) To find the ratio of energy stored 2 ½ Page 4 Page 1 of 13 Final Draft 11/03/16 03:00 p.m . MARKING SCHEME SET 55/1/S Q. No. Expected Answer / Value Points Marks Total Marks Section A Set1,Q1 Set2,Q3 Set3,Q2 (i) Manganin (ii) As ? increases A also increases Alternatively, ½ ½ 1 Set1,Q2 Set2,Q2 Set3,Q5 Phase angle = 60° [Note : If the student only writes, [ ] , give ½ mark] 1 1 Set1,Q3 Set2,Q1 Set3,Q4 Between plates of capacitor during charging / discharging Alternatively, In the region of time varying electric field 1 1 Set1,Q4 Set2,Q5 Set3,Q1 (i) P = NOT gate (ii) Q = OR gate ½ ½ 1 Set1,Q5 Set2,Q4 Set3,Q3 Def: The average time, between successive collisions of electrons, (in a conductor) is known as relaxation time 1 1 Section B Set1,Q6 Set2,Q6 Set3,Q10 The field inside a conductor is zero. Sensitive instruments are shielded from outside electrical influences by enclosing them in a hollow conductor . (any other relevant answer.) Potential inside the cavity is not zero/ potential is constant. ½ 1 ½ 2 Set1,Q7 Set2,Q7 Set3,Q8 Any two properties of electromagnetic waves Such as (a) transverse nature (b) does not get deflected by electric fields or magnetic fields (c) same speed in vacuum for all waves (d) no material medium required for propagation (e) they get refracted, diffracted and polarised / (any two properties) Electric charges present on a plane, kept normal to the direction of propagation of an e.m. wave can be set and sustained in motion by the electric and magnetic field of the electromagnetic wave. The charges thus acquire energy and momentum from the waves. ½ + ½ 1 Electrostatic Shielding ½ Using this property in actual practice 1 Potential in a cavity ½ Two properties of electromagnetic waves ½ +½ Showing e m waves have momentum 1 Page 2 of 13 Final Draft 11/03/16 03:00 p.m . Alternatively Radiation Pressure â€“ Electromagnetic waves exert radiation pressure. Hence, they carry momentum. 2 Set1,Q8 Set2,Q8 Set3,Q9 Diffraction effects are observed for beams of electrons scattered by the crystals 1.227nm V ? ? 1.227 120 nm ? ? Value ?= 0.112nm Alternatively = =0.112nm ½ ½ ½ ½ ½ ½ ½ 2 Set1,Q9 Set2,Q10 Set3,Q7 (i) Transducer: The device which converts one form of energy into another (ii) Repeater: A repeater picks up signal, amplifies and retransmits them to receiver 1 1 2 Set1,Q10 Set2,Q9 Set3,Q6 (i) 2 0 r r n ? 21.2x10 -11 = 5.3x10 -11 n 2 implies n = 2 (ii) 2 13.6eV E n ? ? = =-3.4eV [Award ½ mark if the student just writes E= E 1 /4] OR (i) Energy of photon = = =4.5 (ii) The corresponding transition is B ½ ½ ½ ½ ½ + ½ + ½ ½ 2 2 Principle ½ Calculation of ? 1 ½ Function of Transducer 1 Function of Repeater 1 Finding the principal quantum number 1 Finding the total energy 1 Calculation of energy of photon 1½ Identification of transistion ½ Page 3 of 13 Final Draft 11/03/16 03:00 p.m . Section C Set1,Q11 Set2,Q20 Set3,Q22 E +q = Kq / (r 2 +a 2 ) and E -q = Kq / (r 2 +a 2 ) The two Electric fields have equal magnitudes and their directions are as shown in diagram Components along dipole axis get added up while normal components cancel each other. E=-[ E -q +E +q ] cos? r so E = - = ( = Direction of electric field is opposite to that of dipole moment. 1 ½ ½ ½ ½ 3 Set1,Q12 Set2,Q15 Set3,Q16 a) Zero b) We have C series = = 1 µF Also, C parallel = (3+3+3)=9µF Energy stored = Energy in series combination = Energy in parallel combination = Ratio = 1:9 ½ ½ ½ ½ ½ ½ 3 Diagram 1 Deriving expression for E eq 1 ½ Direction of E eq ½ a) To find charge accumulated in capacitor C 2 ½ b) To find the ratio of energy stored 2 ½ Page 4 of 13 Final Draft 11/03/16 03:00 p.m . Set1,Q13 Set2,Q16 Set3,Q19 a) Intensity of radiation equals the energy of all the Photons incident normally per unit area per unit time. Alternatively, The intensity of radiation is proportional to the number of photons emitted per unit area per unit time. b) c) As per Einsteinâ€™s equation, (i) The stopping potential is same for I 1 and I 2 as they have the same frequency. (ii) The saturation currents are as shown , because I 1 > I 2 > I 3 1 1 ½ ½ 3 Set1,Q14 Set2,Q14 Set3,Q12 (i) During Forward bias of LED, electrons move from n side to p side and holes move from p side to n side. During recombination, energy is released in the form of photons having energy of the order of band gap. (ii) GaAs/ GaAsP (any one) Band gap should be 1.8 eV to 3 eV These materials have band gap which is suitable to produce desired visible light wavelengths. (iii)Low operational voltage, fast action , no warm up time required, nearly monochromatic, long life ,ruggedness, fast on and off switching capacity. (any two points) 1 ½ ½ ½ + ½ 3 Set1,Q15 Set2,Q13 Set3,Q14 Capacitance = C = = F ½ ½ a) Definition of intensity 1 b) Required graph 1 c) Explanation of nature of the curves 1 (i) To explain the process of emission 1 (ii) Material preferred to make LED and reason ½ + ½ (iii)Two advantages of using LED ½ + ½ Calculation of capacitance 1 Calculation of Impedence 1 Calculation of Power dissipitated 1 Page 5 Page 1 of 13 Final Draft 11/03/16 03:00 p.m . MARKING SCHEME SET 55/1/S Q. No. Expected Answer / Value Points Marks Total Marks Section A Set1,Q1 Set2,Q3 Set3,Q2 (i) Manganin (ii) As ? increases A also increases Alternatively, ½ ½ 1 Set1,Q2 Set2,Q2 Set3,Q5 Phase angle = 60° [Note : If the student only writes, [ ] , give ½ mark] 1 1 Set1,Q3 Set2,Q1 Set3,Q4 Between plates of capacitor during charging / discharging Alternatively, In the region of time varying electric field 1 1 Set1,Q4 Set2,Q5 Set3,Q1 (i) P = NOT gate (ii) Q = OR gate ½ ½ 1 Set1,Q5 Set2,Q4 Set3,Q3 Def: The average time, between successive collisions of electrons, (in a conductor) is known as relaxation time 1 1 Section B Set1,Q6 Set2,Q6 Set3,Q10 The field inside a conductor is zero. Sensitive instruments are shielded from outside electrical influences by enclosing them in a hollow conductor . (any other relevant answer.) Potential inside the cavity is not zero/ potential is constant. ½ 1 ½ 2 Set1,Q7 Set2,Q7 Set3,Q8 Any two properties of electromagnetic waves Such as (a) transverse nature (b) does not get deflected by electric fields or magnetic fields (c) same speed in vacuum for all waves (d) no material medium required for propagation (e) they get refracted, diffracted and polarised / (any two properties) Electric charges present on a plane, kept normal to the direction of propagation of an e.m. wave can be set and sustained in motion by the electric and magnetic field of the electromagnetic wave. The charges thus acquire energy and momentum from the waves. ½ + ½ 1 Electrostatic Shielding ½ Using this property in actual practice 1 Potential in a cavity ½ Two properties of electromagnetic waves ½ +½ Showing e m waves have momentum 1 Page 2 of 13 Final Draft 11/03/16 03:00 p.m . Alternatively Radiation Pressure â€“ Electromagnetic waves exert radiation pressure. Hence, they carry momentum. 2 Set1,Q8 Set2,Q8 Set3,Q9 Diffraction effects are observed for beams of electrons scattered by the crystals 1.227nm V ? ? 1.227 120 nm ? ? Value ?= 0.112nm Alternatively = =0.112nm ½ ½ ½ ½ ½ ½ ½ 2 Set1,Q9 Set2,Q10 Set3,Q7 (i) Transducer: The device which converts one form of energy into another (ii) Repeater: A repeater picks up signal, amplifies and retransmits them to receiver 1 1 2 Set1,Q10 Set2,Q9 Set3,Q6 (i) 2 0 r r n ? 21.2x10 -11 = 5.3x10 -11 n 2 implies n = 2 (ii) 2 13.6eV E n ? ? = =-3.4eV [Award ½ mark if the student just writes E= E 1 /4] OR (i) Energy of photon = = =4.5 (ii) The corresponding transition is B ½ ½ ½ ½ ½ + ½ + ½ ½ 2 2 Principle ½ Calculation of ? 1 ½ Function of Transducer 1 Function of Repeater 1 Finding the principal quantum number 1 Finding the total energy 1 Calculation of energy of photon 1½ Identification of transistion ½ Page 3 of 13 Final Draft 11/03/16 03:00 p.m . Section C Set1,Q11 Set2,Q20 Set3,Q22 E +q = Kq / (r 2 +a 2 ) and E -q = Kq / (r 2 +a 2 ) The two Electric fields have equal magnitudes and their directions are as shown in diagram Components along dipole axis get added up while normal components cancel each other. E=-[ E -q +E +q ] cos? r so E = - = ( = Direction of electric field is opposite to that of dipole moment. 1 ½ ½ ½ ½ 3 Set1,Q12 Set2,Q15 Set3,Q16 a) Zero b) We have C series = = 1 µF Also, C parallel = (3+3+3)=9µF Energy stored = Energy in series combination = Energy in parallel combination = Ratio = 1:9 ½ ½ ½ ½ ½ ½ 3 Diagram 1 Deriving expression for E eq 1 ½ Direction of E eq ½ a) To find charge accumulated in capacitor C 2 ½ b) To find the ratio of energy stored 2 ½ Page 4 of 13 Final Draft 11/03/16 03:00 p.m . Set1,Q13 Set2,Q16 Set3,Q19 a) Intensity of radiation equals the energy of all the Photons incident normally per unit area per unit time. Alternatively, The intensity of radiation is proportional to the number of photons emitted per unit area per unit time. b) c) As per Einsteinâ€™s equation, (i) The stopping potential is same for I 1 and I 2 as they have the same frequency. (ii) The saturation currents are as shown , because I 1 > I 2 > I 3 1 1 ½ ½ 3 Set1,Q14 Set2,Q14 Set3,Q12 (i) During Forward bias of LED, electrons move from n side to p side and holes move from p side to n side. During recombination, energy is released in the form of photons having energy of the order of band gap. (ii) GaAs/ GaAsP (any one) Band gap should be 1.8 eV to 3 eV These materials have band gap which is suitable to produce desired visible light wavelengths. (iii)Low operational voltage, fast action , no warm up time required, nearly monochromatic, long life ,ruggedness, fast on and off switching capacity. (any two points) 1 ½ ½ ½ + ½ 3 Set1,Q15 Set2,Q13 Set3,Q14 Capacitance = C = = F ½ ½ a) Definition of intensity 1 b) Required graph 1 c) Explanation of nature of the curves 1 (i) To explain the process of emission 1 (ii) Material preferred to make LED and reason ½ + ½ (iii)Two advantages of using LED ½ + ½ Calculation of capacitance 1 Calculation of Impedence 1 Calculation of Power dissipitated 1 Page 5 of 13 Final Draft 11/03/16 03:00 p.m . = 2.5 x 10 -5 F Impedence = resistance( since V and I are in phase) Impedence = 100? Power discipated = = = 400 watt ½ ½ ½ ½ 3 Set1,Q16 Set2,Q19 Set3,Q20 (i) = 2cosA/2 = A= 60° (ii) = (iii) Lies between and Hence, TIR takes place. Alternatively, which is less than angle of incidence > TIR ½ ½ ½ ½ 1 3 Set1,Q17 Set2,Q18 Set3,Q17 Nuclear force is Saturated, or short ranged [ any one] The final system is more tightly bound when heavy nucleus undergoes nuclear fission. Hence, there is a release of energy. The final system is more tightly bound when light nuclei undergoes nuclear fusion. Hence, there is a releases of energy. 1½ ½ ½ ½ (i) To calculate angle of prism 1 ½ (ii) To trace the path of incident light inside the prism 1 ½ To plot (BE/A) vs mass number graph 1½ To state the property of nuclear force ½ To explain the release of energy in fission and ½ + ½ fusion using the graphRead More

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