Page 1 Page | 1 East Final Draft-11 March 2016 2:00 p.m. MARKING SCHEME SET 55/1/E Q. No. Expected Answer / Value Points Marks Total Marks (SECTION A) Set1,Q1 Set2,Q5 Set3,Q2 Potentiometer â€˜Qâ€™ will be preferred Reason:- ?????????????????????? ? 1 ?????????????????? ???????????????? (?? ) Since potential gradient is less, sensitivity is more. [Note: Also accept if the student just writs that potential gradient is less for potentiometer Q] ½ ½ 1 Set1,Q2 Set2,Q3 Set3,Q1 Graph of V Graph of I ½ ½ 1 Set1,Q3 Set2,Q2 Set3,Q4 [Note: If students write truth table correctly then award ½ mark.] 1 1 Set1,Q4 Set2,Q4 Set3,Q5 For a.c. source, circuit is complete due to the presence of displacement current in the capacitor. For steady dc, there is no displacement current, therefore, circuit is not complete. [Alternatively, Capacitive reactance ?? ?? = 1 2?????? = 1 ???? So, capacitor allows easy path for a.c. source. For d.c, f= 0, so X c = infinity, So capacitor blocks d.c] ½+½ ½+½ 1 Set1,Q5 Set2,Q1 Set3,Q3 Conductivity of a conductor is the current flowing per unit area per unit electric field applied. [Alternatively, conductivity ?? = ?? ?? ] ½ Page 2 Page | 1 East Final Draft-11 March 2016 2:00 p.m. MARKING SCHEME SET 55/1/E Q. No. Expected Answer / Value Points Marks Total Marks (SECTION A) Set1,Q1 Set2,Q5 Set3,Q2 Potentiometer â€˜Qâ€™ will be preferred Reason:- ?????????????????????? ? 1 ?????????????????? ???????????????? (?? ) Since potential gradient is less, sensitivity is more. [Note: Also accept if the student just writs that potential gradient is less for potentiometer Q] ½ ½ 1 Set1,Q2 Set2,Q3 Set3,Q1 Graph of V Graph of I ½ ½ 1 Set1,Q3 Set2,Q2 Set3,Q4 [Note: If students write truth table correctly then award ½ mark.] 1 1 Set1,Q4 Set2,Q4 Set3,Q5 For a.c. source, circuit is complete due to the presence of displacement current in the capacitor. For steady dc, there is no displacement current, therefore, circuit is not complete. [Alternatively, Capacitive reactance ?? ?? = 1 2?????? = 1 ???? So, capacitor allows easy path for a.c. source. For d.c, f= 0, so X c = infinity, So capacitor blocks d.c] ½+½ ½+½ 1 Set1,Q5 Set2,Q1 Set3,Q3 Conductivity of a conductor is the current flowing per unit area per unit electric field applied. [Alternatively, conductivity ?? = ?? ?? ] ½ Page | 2 East Final Draft-11 March 2016 2:00 p.m. Depends upon number density i.e. nature of material, and relaxation time i.e. temperature. ½ 1 (SECTION B) Set1,Q6 Set2,Q8 Set3,Q7 ???? =???? sin?? ???? Work done against the restoring torque ???? =t d? ?,?? = pE sin?? ???? ?? 1 ?? 0 = pE cos?? 0 - cos?? 1 ½ ½ ½ ½ 2 Set1,Q7 Set2,Q9 Set3,Q6 ? 2???? = ?? ?? ???? de Broglie wavelength, ? = ?? ???? For electron moving in the n th orbit, 2???? =???? ? mvr = ?? ?? 2?? = L (orbital angular momentum) This is Bohrâ€™s Postulate of quantization of orbital angular momentum. ½ ½ ½ ½ 2 Set1,Q8 Set2,Q10 Set3,Q9 Concept of mobile telephony is to divide the service area into a suitable number of cells centred on an office MTSO (Mobile Telephone Switching Office) / Mobile telephony means that you can talk to any person from anywhere. Explanation: 1. Entire service area is divided into smaller parts called cells. 2. Each cell has a base station to receive and send signals to mobiles. 3. Each base station is linked to MTSO. MTSO co-ordinates between ½ ½ ½ ½ 2 Derivation of expression for work done 2 de-Broglie wavelength ½ Condition of stationary orbits ½ Obtaining Bohrâ€™s Postulate of quantization of orbital angular momentum.1 Explanation of the concept of Mobile Telephony ½ Explanation of working 1½ Page 3 Page | 1 East Final Draft-11 March 2016 2:00 p.m. MARKING SCHEME SET 55/1/E Q. No. Expected Answer / Value Points Marks Total Marks (SECTION A) Set1,Q1 Set2,Q5 Set3,Q2 Potentiometer â€˜Qâ€™ will be preferred Reason:- ?????????????????????? ? 1 ?????????????????? ???????????????? (?? ) Since potential gradient is less, sensitivity is more. [Note: Also accept if the student just writs that potential gradient is less for potentiometer Q] ½ ½ 1 Set1,Q2 Set2,Q3 Set3,Q1 Graph of V Graph of I ½ ½ 1 Set1,Q3 Set2,Q2 Set3,Q4 [Note: If students write truth table correctly then award ½ mark.] 1 1 Set1,Q4 Set2,Q4 Set3,Q5 For a.c. source, circuit is complete due to the presence of displacement current in the capacitor. For steady dc, there is no displacement current, therefore, circuit is not complete. [Alternatively, Capacitive reactance ?? ?? = 1 2?????? = 1 ???? So, capacitor allows easy path for a.c. source. For d.c, f= 0, so X c = infinity, So capacitor blocks d.c] ½+½ ½+½ 1 Set1,Q5 Set2,Q1 Set3,Q3 Conductivity of a conductor is the current flowing per unit area per unit electric field applied. [Alternatively, conductivity ?? = ?? ?? ] ½ Page | 2 East Final Draft-11 March 2016 2:00 p.m. Depends upon number density i.e. nature of material, and relaxation time i.e. temperature. ½ 1 (SECTION B) Set1,Q6 Set2,Q8 Set3,Q7 ???? =???? sin?? ???? Work done against the restoring torque ???? =t d? ?,?? = pE sin?? ???? ?? 1 ?? 0 = pE cos?? 0 - cos?? 1 ½ ½ ½ ½ 2 Set1,Q7 Set2,Q9 Set3,Q6 ? 2???? = ?? ?? ???? de Broglie wavelength, ? = ?? ???? For electron moving in the n th orbit, 2???? =???? ? mvr = ?? ?? 2?? = L (orbital angular momentum) This is Bohrâ€™s Postulate of quantization of orbital angular momentum. ½ ½ ½ ½ 2 Set1,Q8 Set2,Q10 Set3,Q9 Concept of mobile telephony is to divide the service area into a suitable number of cells centred on an office MTSO (Mobile Telephone Switching Office) / Mobile telephony means that you can talk to any person from anywhere. Explanation: 1. Entire service area is divided into smaller parts called cells. 2. Each cell has a base station to receive and send signals to mobiles. 3. Each base station is linked to MTSO. MTSO co-ordinates between ½ ½ ½ ½ 2 Derivation of expression for work done 2 de-Broglie wavelength ½ Condition of stationary orbits ½ Obtaining Bohrâ€™s Postulate of quantization of orbital angular momentum.1 Explanation of the concept of Mobile Telephony ½ Explanation of working 1½ Page | 3 East Final Draft-11 March 2016 2:00 p.m. base station and TCO (Telephone Control Office) Set1,Q9 Set2,Q7 Set3,Q10 1 ?? ?????? =?? 1 ?? ?? 2 - 1 ?? ?? 2 1 ?? ?????? = 1 2² - 1 3² ?? ?? ?????? = 36 5?? The energy of the incident photon = 12.5 eV Energy of ground state = -13.6eV ?, Energy after absorption of photon can be -1.1eV This means that electron can go to the excited state ?? ?? = 3. It emits photons of maximum wavelength on going to ?? ?? = 2 i.e. = 36 5×1.1×10 7 = 6.555 × 10 -7 m = 6555 ?? ° It belongs to Balmer Series. [Note:- (1) If student just writes the formula 1 ?? ?????? =?? 1 ?? ?? 2 - 1 ?? ?? 2 for the wavelength of different levels in the Hydrogen spectrum and calculates ?? ?????? for any series, award full 3 marks. (2) Also award full 3 marks if the student writes that the energy of the excited state cannot be 12.5eV] OR ½ ½ ½ ½ Formula ½ Calculation ½ Longest Wavelength ½ Identification of Series ½ Formula 1 Calculation 1 Page 4 Page | 1 East Final Draft-11 March 2016 2:00 p.m. MARKING SCHEME SET 55/1/E Q. No. Expected Answer / Value Points Marks Total Marks (SECTION A) Set1,Q1 Set2,Q5 Set3,Q2 Potentiometer â€˜Qâ€™ will be preferred Reason:- ?????????????????????? ? 1 ?????????????????? ???????????????? (?? ) Since potential gradient is less, sensitivity is more. [Note: Also accept if the student just writs that potential gradient is less for potentiometer Q] ½ ½ 1 Set1,Q2 Set2,Q3 Set3,Q1 Graph of V Graph of I ½ ½ 1 Set1,Q3 Set2,Q2 Set3,Q4 [Note: If students write truth table correctly then award ½ mark.] 1 1 Set1,Q4 Set2,Q4 Set3,Q5 For a.c. source, circuit is complete due to the presence of displacement current in the capacitor. For steady dc, there is no displacement current, therefore, circuit is not complete. [Alternatively, Capacitive reactance ?? ?? = 1 2?????? = 1 ???? So, capacitor allows easy path for a.c. source. For d.c, f= 0, so X c = infinity, So capacitor blocks d.c] ½+½ ½+½ 1 Set1,Q5 Set2,Q1 Set3,Q3 Conductivity of a conductor is the current flowing per unit area per unit electric field applied. [Alternatively, conductivity ?? = ?? ?? ] ½ Page | 2 East Final Draft-11 March 2016 2:00 p.m. Depends upon number density i.e. nature of material, and relaxation time i.e. temperature. ½ 1 (SECTION B) Set1,Q6 Set2,Q8 Set3,Q7 ???? =???? sin?? ???? Work done against the restoring torque ???? =t d? ?,?? = pE sin?? ???? ?? 1 ?? 0 = pE cos?? 0 - cos?? 1 ½ ½ ½ ½ 2 Set1,Q7 Set2,Q9 Set3,Q6 ? 2???? = ?? ?? ???? de Broglie wavelength, ? = ?? ???? For electron moving in the n th orbit, 2???? =???? ? mvr = ?? ?? 2?? = L (orbital angular momentum) This is Bohrâ€™s Postulate of quantization of orbital angular momentum. ½ ½ ½ ½ 2 Set1,Q8 Set2,Q10 Set3,Q9 Concept of mobile telephony is to divide the service area into a suitable number of cells centred on an office MTSO (Mobile Telephone Switching Office) / Mobile telephony means that you can talk to any person from anywhere. Explanation: 1. Entire service area is divided into smaller parts called cells. 2. Each cell has a base station to receive and send signals to mobiles. 3. Each base station is linked to MTSO. MTSO co-ordinates between ½ ½ ½ ½ 2 Derivation of expression for work done 2 de-Broglie wavelength ½ Condition of stationary orbits ½ Obtaining Bohrâ€™s Postulate of quantization of orbital angular momentum.1 Explanation of the concept of Mobile Telephony ½ Explanation of working 1½ Page | 3 East Final Draft-11 March 2016 2:00 p.m. base station and TCO (Telephone Control Office) Set1,Q9 Set2,Q7 Set3,Q10 1 ?? ?????? =?? 1 ?? ?? 2 - 1 ?? ?? 2 1 ?? ?????? = 1 2² - 1 3² ?? ?? ?????? = 36 5?? The energy of the incident photon = 12.5 eV Energy of ground state = -13.6eV ?, Energy after absorption of photon can be -1.1eV This means that electron can go to the excited state ?? ?? = 3. It emits photons of maximum wavelength on going to ?? ?? = 2 i.e. = 36 5×1.1×10 7 = 6.555 × 10 -7 m = 6555 ?? ° It belongs to Balmer Series. [Note:- (1) If student just writes the formula 1 ?? ?????? =?? 1 ?? ?? 2 - 1 ?? ?? 2 for the wavelength of different levels in the Hydrogen spectrum and calculates ?? ?????? for any series, award full 3 marks. (2) Also award full 3 marks if the student writes that the energy of the excited state cannot be 12.5eV] OR ½ ½ ½ ½ Formula ½ Calculation ½ Longest Wavelength ½ Identification of Series ½ Formula 1 Calculation 1 Page | 4 East Final Draft-11 March 2016 2:00 p.m. ?? = ?? ?? 2?????? And ?? = 1 ?? ?? 2 ?? 2 4?? 2 ?? ?? 2 So, ?? =?? 2?? ?? 2 ?? ?? In first excited state n = 2 So velocity ?? 2 = 2?? ???? 2 2?? = 1.09 × 10 6 ms -1 OR Velocity of electron, ?? ?? = 1 137 ?? ?? In first excited state n =2 So velocity in first excited state (?? 2 ) = 1 137 ?? 2 = 1.09 × 10 6 ms -1 ½ ½ ½ ½ 1 ½ ½ 2 Set1,Q10 Set2,Q6 Set3,Q8 (i) Infrared waves are produced by hot bodies and molecules. Important use( Any one) To treat muscular strains/ To reveal the secret writings on the ancient walls/ For producing dehydrated fruits/ Solar heater/ Solar cooker Ozone layer protects us from harmful U-V rays ½ ½ 1 2 (SECTION C) Set1,Q11 Set2,Q15 Set3,Q12 (i) Electric flux through a Gaussian surface, ?? = ?????????? ???????????????? ?? ?? ???????? ? 0 ½ (i) How are infrared waves produced ½ One important use ½ (ii) Reason (any one) 1 (i) Electric Flux through the shell 1 (ii) Statement of Law 1 (iii) Force on charge at C 1 2 Force on charge at A 1 2 Page 5 Page | 1 East Final Draft-11 March 2016 2:00 p.m. MARKING SCHEME SET 55/1/E Q. No. Expected Answer / Value Points Marks Total Marks (SECTION A) Set1,Q1 Set2,Q5 Set3,Q2 Potentiometer â€˜Qâ€™ will be preferred Reason:- ?????????????????????? ? 1 ?????????????????? ???????????????? (?? ) Since potential gradient is less, sensitivity is more. [Note: Also accept if the student just writs that potential gradient is less for potentiometer Q] ½ ½ 1 Set1,Q2 Set2,Q3 Set3,Q1 Graph of V Graph of I ½ ½ 1 Set1,Q3 Set2,Q2 Set3,Q4 [Note: If students write truth table correctly then award ½ mark.] 1 1 Set1,Q4 Set2,Q4 Set3,Q5 For a.c. source, circuit is complete due to the presence of displacement current in the capacitor. For steady dc, there is no displacement current, therefore, circuit is not complete. [Alternatively, Capacitive reactance ?? ?? = 1 2?????? = 1 ???? So, capacitor allows easy path for a.c. source. For d.c, f= 0, so X c = infinity, So capacitor blocks d.c] ½+½ ½+½ 1 Set1,Q5 Set2,Q1 Set3,Q3 Conductivity of a conductor is the current flowing per unit area per unit electric field applied. [Alternatively, conductivity ?? = ?? ?? ] ½ Page | 2 East Final Draft-11 March 2016 2:00 p.m. Depends upon number density i.e. nature of material, and relaxation time i.e. temperature. ½ 1 (SECTION B) Set1,Q6 Set2,Q8 Set3,Q7 ???? =???? sin?? ???? Work done against the restoring torque ???? =t d? ?,?? = pE sin?? ???? ?? 1 ?? 0 = pE cos?? 0 - cos?? 1 ½ ½ ½ ½ 2 Set1,Q7 Set2,Q9 Set3,Q6 ? 2???? = ?? ?? ???? de Broglie wavelength, ? = ?? ???? For electron moving in the n th orbit, 2???? =???? ? mvr = ?? ?? 2?? = L (orbital angular momentum) This is Bohrâ€™s Postulate of quantization of orbital angular momentum. ½ ½ ½ ½ 2 Set1,Q8 Set2,Q10 Set3,Q9 Concept of mobile telephony is to divide the service area into a suitable number of cells centred on an office MTSO (Mobile Telephone Switching Office) / Mobile telephony means that you can talk to any person from anywhere. Explanation: 1. Entire service area is divided into smaller parts called cells. 2. Each cell has a base station to receive and send signals to mobiles. 3. Each base station is linked to MTSO. MTSO co-ordinates between ½ ½ ½ ½ 2 Derivation of expression for work done 2 de-Broglie wavelength ½ Condition of stationary orbits ½ Obtaining Bohrâ€™s Postulate of quantization of orbital angular momentum.1 Explanation of the concept of Mobile Telephony ½ Explanation of working 1½ Page | 3 East Final Draft-11 March 2016 2:00 p.m. base station and TCO (Telephone Control Office) Set1,Q9 Set2,Q7 Set3,Q10 1 ?? ?????? =?? 1 ?? ?? 2 - 1 ?? ?? 2 1 ?? ?????? = 1 2² - 1 3² ?? ?? ?????? = 36 5?? The energy of the incident photon = 12.5 eV Energy of ground state = -13.6eV ?, Energy after absorption of photon can be -1.1eV This means that electron can go to the excited state ?? ?? = 3. It emits photons of maximum wavelength on going to ?? ?? = 2 i.e. = 36 5×1.1×10 7 = 6.555 × 10 -7 m = 6555 ?? ° It belongs to Balmer Series. [Note:- (1) If student just writes the formula 1 ?? ?????? =?? 1 ?? ?? 2 - 1 ?? ?? 2 for the wavelength of different levels in the Hydrogen spectrum and calculates ?? ?????? for any series, award full 3 marks. (2) Also award full 3 marks if the student writes that the energy of the excited state cannot be 12.5eV] OR ½ ½ ½ ½ Formula ½ Calculation ½ Longest Wavelength ½ Identification of Series ½ Formula 1 Calculation 1 Page | 4 East Final Draft-11 March 2016 2:00 p.m. ?? = ?? ?? 2?????? And ?? = 1 ?? ?? 2 ?? 2 4?? 2 ?? ?? 2 So, ?? =?? 2?? ?? 2 ?? ?? In first excited state n = 2 So velocity ?? 2 = 2?? ???? 2 2?? = 1.09 × 10 6 ms -1 OR Velocity of electron, ?? ?? = 1 137 ?? ?? In first excited state n =2 So velocity in first excited state (?? 2 ) = 1 137 ?? 2 = 1.09 × 10 6 ms -1 ½ ½ ½ ½ 1 ½ ½ 2 Set1,Q10 Set2,Q6 Set3,Q8 (i) Infrared waves are produced by hot bodies and molecules. Important use( Any one) To treat muscular strains/ To reveal the secret writings on the ancient walls/ For producing dehydrated fruits/ Solar heater/ Solar cooker Ozone layer protects us from harmful U-V rays ½ ½ 1 2 (SECTION C) Set1,Q11 Set2,Q15 Set3,Q12 (i) Electric flux through a Gaussian surface, ?? = ?????????? ???????????????? ?? ?? ???????? ? 0 ½ (i) How are infrared waves produced ½ One important use ½ (ii) Reason (any one) 1 (i) Electric Flux through the shell 1 (ii) Statement of Law 1 (iii) Force on charge at C 1 2 Force on charge at A 1 2 Page | 5 East Final Draft-11 March 2016 2:00 p.m. Net charge enclosed inside the shell q=0 ? Electric flux through the shell ?? ? ?? =0 Award ½ mark even when the student writes - Electric flux through the shell is zero as electric field inside the shell is zero. (ii) Gauss Law- Electric flux through a Gaussian surface is 1 ? 0 times the net charge enclosed with in it. Alternatively, ?? .???? = ?? ?? 0 (iii) Force on the charge at the centre i.e. Charget ?? 2 = 0 ?? ?? = 1 4?? ?? 0 2?? × ?? + ?? 2 ?? 2 = 1 4?? ?? 0 3?? 2 ?? 2 ½ 1 ½ ½ 3 Set1,Q12 Set2,Q13 Set3,Q21 A galvanometer is converted into a voltmeter by connecting a high resistance â€˜Râ€™ in series with it. A galvanometer is converted into an ammeter by connecting a small resistance (called shunt) in parallel with it. ½ ½ ½ ½ How galvanometer is converted in to a voltmeter and an Ammeter ½ + ½ Diagram for conversion of galvanometer into a voltmeter and an ½ + ½ Ammeter. Resistance of each arrangement ½ + ½Read More

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