Page 1 SET 55/1/1/N Page 1 of 22 Final Draft 11/03/16 2:00 p.m. MARKING SCHEME Q. No. Expected Answer / Value Points Marks Total Marks Set1,Q1 Set2,Q2 Set3,Q4 SECTION A Zero / No work done / None 1 1 Set1,Q2 Set2,Q5 Set3,Q3 Drift velocity per unit field (directly proportional to relaxation time) ½ ½ 1 Set1,Q3 Set2,Q4 Set3,Q2 Charged particle moves inclined to the magnetic field (angle between and is neither nor 0) (component of , parallel to , is not zero.) 1 1 Set1,Q4 Set2,Q1 Set3,Q5 (some) light gets deviated / scattered / absorbed Scattering of light ½ ½ 1 Set1,Q5 Set2,Q3 Set3,Q1 = 2005 kHz ; 1995 kHz (Give full 1 mark if the student straightaway writes the answer as 2005 kHz and 1995 kHz) ½ ½ 1 Set1,Q6 Set2,Q8 Set3,Q7 SECTION B ; Alternatively, (Award this 1 mark even if the student writes the formula for directly as such) ½ ½ ½ ½ 2 Formulae: 1 Substitution and calculation: 1 Page 2 SET 55/1/1/N Page 1 of 22 Final Draft 11/03/16 2:00 p.m. MARKING SCHEME Q. No. Expected Answer / Value Points Marks Total Marks Set1,Q1 Set2,Q2 Set3,Q4 SECTION A Zero / No work done / None 1 1 Set1,Q2 Set2,Q5 Set3,Q3 Drift velocity per unit field (directly proportional to relaxation time) ½ ½ 1 Set1,Q3 Set2,Q4 Set3,Q2 Charged particle moves inclined to the magnetic field (angle between and is neither nor 0) (component of , parallel to , is not zero.) 1 1 Set1,Q4 Set2,Q1 Set3,Q5 (some) light gets deviated / scattered / absorbed Scattering of light ½ ½ 1 Set1,Q5 Set2,Q3 Set3,Q1 = 2005 kHz ; 1995 kHz (Give full 1 mark if the student straightaway writes the answer as 2005 kHz and 1995 kHz) ½ ½ 1 Set1,Q6 Set2,Q8 Set3,Q7 SECTION B ; Alternatively, (Award this 1 mark even if the student writes the formula for directly as such) ½ ½ ½ ½ 2 Formulae: 1 Substitution and calculation: 1 SET 55/1/1/N Page 2 of 22 Final Draft 11/03/16 2:00 p.m. Set1,Q7 Set2,Q10 Set3,Q8 (i) is more for particle, we have (Also, accept if the student writes ) (ii) K.E. = q V q is less for proton, we have =2) ½ ½ ½ ½ 2 Set1,Q8 Set2,Q9 Set3,Q6 When the electron jumps from the orbit with n=3 to n=2 (Longest wavelength of the Balmer series / First line of the Balmer series) = = = Hz Hz. (If the student just writes , award ½ mark) (Alternatively, = = ) 1 ½ ½ ½ ½ 2 Formulae ½ + ½ Conclusions in the two cases ½ + ½ Indicating the transition 1 Calculation of frequency 1 Page 3 SET 55/1/1/N Page 1 of 22 Final Draft 11/03/16 2:00 p.m. MARKING SCHEME Q. No. Expected Answer / Value Points Marks Total Marks Set1,Q1 Set2,Q2 Set3,Q4 SECTION A Zero / No work done / None 1 1 Set1,Q2 Set2,Q5 Set3,Q3 Drift velocity per unit field (directly proportional to relaxation time) ½ ½ 1 Set1,Q3 Set2,Q4 Set3,Q2 Charged particle moves inclined to the magnetic field (angle between and is neither nor 0) (component of , parallel to , is not zero.) 1 1 Set1,Q4 Set2,Q1 Set3,Q5 (some) light gets deviated / scattered / absorbed Scattering of light ½ ½ 1 Set1,Q5 Set2,Q3 Set3,Q1 = 2005 kHz ; 1995 kHz (Give full 1 mark if the student straightaway writes the answer as 2005 kHz and 1995 kHz) ½ ½ 1 Set1,Q6 Set2,Q8 Set3,Q7 SECTION B ; Alternatively, (Award this 1 mark even if the student writes the formula for directly as such) ½ ½ ½ ½ 2 Formulae: 1 Substitution and calculation: 1 SET 55/1/1/N Page 2 of 22 Final Draft 11/03/16 2:00 p.m. Set1,Q7 Set2,Q10 Set3,Q8 (i) is more for particle, we have (Also, accept if the student writes ) (ii) K.E. = q V q is less for proton, we have =2) ½ ½ ½ ½ 2 Set1,Q8 Set2,Q9 Set3,Q6 When the electron jumps from the orbit with n=3 to n=2 (Longest wavelength of the Balmer series / First line of the Balmer series) = = = Hz Hz. (If the student just writes , award ½ mark) (Alternatively, = = ) 1 ½ ½ ½ ½ 2 Formulae ½ + ½ Conclusions in the two cases ½ + ½ Indicating the transition 1 Calculation of frequency 1 SET 55/1/1/N Page 3 of 22 Final Draft 11/03/16 2:00 p.m. OR m ½ ½ 1 2 Set1,Q9 Set2,Q6 Set3,Q10 If base band signal were to be transmitted directly 1. The height of the antennae needed will be impractically large. 2. The effective power radiated would be too low. 3. There would be a high probability of different signals getting mixed up with one another. (Any two) 1+1 2 Set1,Q10 Set2,Q7 Set3,Q9 Since AQ = AR, we have QR|| BC is the angle of minimum deviation. (Alternatively: Since AQ=AR, we get is the angle of minimum deviation.) = = 60 or =60 0 ½ ½ ½ ½ 2 Formula 1 Calculation of 1 Identifying that is the angle of minimum deviation ½ Formula ½ Calculation of 1 Two Reasons 1+ 1 Page 4 SET 55/1/1/N Page 1 of 22 Final Draft 11/03/16 2:00 p.m. MARKING SCHEME Q. No. Expected Answer / Value Points Marks Total Marks Set1,Q1 Set2,Q2 Set3,Q4 SECTION A Zero / No work done / None 1 1 Set1,Q2 Set2,Q5 Set3,Q3 Drift velocity per unit field (directly proportional to relaxation time) ½ ½ 1 Set1,Q3 Set2,Q4 Set3,Q2 Charged particle moves inclined to the magnetic field (angle between and is neither nor 0) (component of , parallel to , is not zero.) 1 1 Set1,Q4 Set2,Q1 Set3,Q5 (some) light gets deviated / scattered / absorbed Scattering of light ½ ½ 1 Set1,Q5 Set2,Q3 Set3,Q1 = 2005 kHz ; 1995 kHz (Give full 1 mark if the student straightaway writes the answer as 2005 kHz and 1995 kHz) ½ ½ 1 Set1,Q6 Set2,Q8 Set3,Q7 SECTION B ; Alternatively, (Award this 1 mark even if the student writes the formula for directly as such) ½ ½ ½ ½ 2 Formulae: 1 Substitution and calculation: 1 SET 55/1/1/N Page 2 of 22 Final Draft 11/03/16 2:00 p.m. Set1,Q7 Set2,Q10 Set3,Q8 (i) is more for particle, we have (Also, accept if the student writes ) (ii) K.E. = q V q is less for proton, we have =2) ½ ½ ½ ½ 2 Set1,Q8 Set2,Q9 Set3,Q6 When the electron jumps from the orbit with n=3 to n=2 (Longest wavelength of the Balmer series / First line of the Balmer series) = = = Hz Hz. (If the student just writes , award ½ mark) (Alternatively, = = ) 1 ½ ½ ½ ½ 2 Formulae ½ + ½ Conclusions in the two cases ½ + ½ Indicating the transition 1 Calculation of frequency 1 SET 55/1/1/N Page 3 of 22 Final Draft 11/03/16 2:00 p.m. OR m ½ ½ 1 2 Set1,Q9 Set2,Q6 Set3,Q10 If base band signal were to be transmitted directly 1. The height of the antennae needed will be impractically large. 2. The effective power radiated would be too low. 3. There would be a high probability of different signals getting mixed up with one another. (Any two) 1+1 2 Set1,Q10 Set2,Q7 Set3,Q9 Since AQ = AR, we have QR|| BC is the angle of minimum deviation. (Alternatively: Since AQ=AR, we get is the angle of minimum deviation.) = = 60 or =60 0 ½ ½ ½ ½ 2 Formula 1 Calculation of 1 Identifying that is the angle of minimum deviation ½ Formula ½ Calculation of 1 Two Reasons 1+ 1 SET 55/1/1/N Page 4 of 22 Final Draft 11/03/16 2:00 p.m. SECTION C Set1,Q11 Set2,Q17 Set3,Q22 We have by Gaussâ€™s law Let Q be the total charge on the shell (i) For the point M outside the shell, we have (ii) For the point N inside the shell, as charge enclosed inside the shell is zero. The graph is as shown ½ ½ ½ ½ ½ ½ 3 Statement of Gaussâ€™s Law ½ Calculation of field (i) Outside the shell 1 (ii) Inside the shell 1 Graph ½ Page 5 SET 55/1/1/N Page 1 of 22 Final Draft 11/03/16 2:00 p.m. MARKING SCHEME Q. No. Expected Answer / Value Points Marks Total Marks Set1,Q1 Set2,Q2 Set3,Q4 SECTION A Zero / No work done / None 1 1 Set1,Q2 Set2,Q5 Set3,Q3 Drift velocity per unit field (directly proportional to relaxation time) ½ ½ 1 Set1,Q3 Set2,Q4 Set3,Q2 Charged particle moves inclined to the magnetic field (angle between and is neither nor 0) (component of , parallel to , is not zero.) 1 1 Set1,Q4 Set2,Q1 Set3,Q5 (some) light gets deviated / scattered / absorbed Scattering of light ½ ½ 1 Set1,Q5 Set2,Q3 Set3,Q1 = 2005 kHz ; 1995 kHz (Give full 1 mark if the student straightaway writes the answer as 2005 kHz and 1995 kHz) ½ ½ 1 Set1,Q6 Set2,Q8 Set3,Q7 SECTION B ; Alternatively, (Award this 1 mark even if the student writes the formula for directly as such) ½ ½ ½ ½ 2 Formulae: 1 Substitution and calculation: 1 SET 55/1/1/N Page 2 of 22 Final Draft 11/03/16 2:00 p.m. Set1,Q7 Set2,Q10 Set3,Q8 (i) is more for particle, we have (Also, accept if the student writes ) (ii) K.E. = q V q is less for proton, we have =2) ½ ½ ½ ½ 2 Set1,Q8 Set2,Q9 Set3,Q6 When the electron jumps from the orbit with n=3 to n=2 (Longest wavelength of the Balmer series / First line of the Balmer series) = = = Hz Hz. (If the student just writes , award ½ mark) (Alternatively, = = ) 1 ½ ½ ½ ½ 2 Formulae ½ + ½ Conclusions in the two cases ½ + ½ Indicating the transition 1 Calculation of frequency 1 SET 55/1/1/N Page 3 of 22 Final Draft 11/03/16 2:00 p.m. OR m ½ ½ 1 2 Set1,Q9 Set2,Q6 Set3,Q10 If base band signal were to be transmitted directly 1. The height of the antennae needed will be impractically large. 2. The effective power radiated would be too low. 3. There would be a high probability of different signals getting mixed up with one another. (Any two) 1+1 2 Set1,Q10 Set2,Q7 Set3,Q9 Since AQ = AR, we have QR|| BC is the angle of minimum deviation. (Alternatively: Since AQ=AR, we get is the angle of minimum deviation.) = = 60 or =60 0 ½ ½ ½ ½ 2 Formula 1 Calculation of 1 Identifying that is the angle of minimum deviation ½ Formula ½ Calculation of 1 Two Reasons 1+ 1 SET 55/1/1/N Page 4 of 22 Final Draft 11/03/16 2:00 p.m. SECTION C Set1,Q11 Set2,Q17 Set3,Q22 We have by Gaussâ€™s law Let Q be the total charge on the shell (i) For the point M outside the shell, we have (ii) For the point N inside the shell, as charge enclosed inside the shell is zero. The graph is as shown ½ ½ ½ ½ ½ ½ 3 Statement of Gaussâ€™s Law ½ Calculation of field (i) Outside the shell 1 (ii) Inside the shell 1 Graph ½ SET 55/1/1/N Page 5 of 22 Final Draft 11/03/16 2:00 p.m. Set1,Q12 Set2,Q19 Set3,Q21 We have, for a single cell, =0.5 (Alternatively, And This gives A = 0.4A = 0.25 ) ( Note: If the student just draws the circuit diagram of the setup but does not do any calculations, award 1 mark only.) 1 ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ 3 Formulae 1 Calculation of r 2Read More

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