Page 1 Outside Delhi SET III MARKING SCHEME SET 55/3 Q. No. Expected Answer / Value Points Marks Total Marks 1. Anticlockwise 1 1 2. Metal A The minimum frequency, at which photoemission starts, is more for metal A Alternatively: Work function of A is more. ½ ½ 1 3. Definition : One ampere is the value of steady current which when maintained in each of the two very long, straight, parallel conductors of negligible cross section and placed one metre apart in vaccum, would produce on each of these conductors a force equal of 2 x 10 -7 N/m of its length. Alternatively If the student writes F = L and says that when 1 R= 1 meter and L = 1 meter, then F= 2 x 10 -7 N Award full 1 mark Alternatively If the student draws any one of the two diagram, as shown , Award full 1 mark 1 1 4. As a diverging lens Light rays diverge on going from a rarer to a denser medium. [Alternatively Also accept the reason given on the basis of lens marker’s formula.] ½ ½ 1 5. At the point of intersection of the two field lines, there will be two directions for the electric field. This is not acceptable. 1 1 6. Short radio waves (or) microwaves 1 1 7. Neutrinos are neutral (chargeless), (almost) massless particles that hardly 1 Page 2 Outside Delhi SET III MARKING SCHEME SET 55/3 Q. No. Expected Answer / Value Points Marks Total Marks 1. Anticlockwise 1 1 2. Metal A The minimum frequency, at which photoemission starts, is more for metal A Alternatively: Work function of A is more. ½ ½ 1 3. Definition : One ampere is the value of steady current which when maintained in each of the two very long, straight, parallel conductors of negligible cross section and placed one metre apart in vaccum, would produce on each of these conductors a force equal of 2 x 10 -7 N/m of its length. Alternatively If the student writes F = L and says that when 1 R= 1 meter and L = 1 meter, then F= 2 x 10 -7 N Award full 1 mark Alternatively If the student draws any one of the two diagram, as shown , Award full 1 mark 1 1 4. As a diverging lens Light rays diverge on going from a rarer to a denser medium. [Alternatively Also accept the reason given on the basis of lens marker’s formula.] ½ ½ 1 5. At the point of intersection of the two field lines, there will be two directions for the electric field. This is not acceptable. 1 1 6. Short radio waves (or) microwaves 1 1 7. Neutrinos are neutral (chargeless), (almost) massless particles that hardly 1 Outside Delhi SET III interact with matter. Alternatively The neutrinos can penetrate large quantity of matter without any interaction OR Neutrinos are chargeless and (almost) massless particles. 1 8. Any two of the following (or any other correct) reasons : i. AC can be transmitted with much lower energy losses as compared to DC ii. AC voltage can be adjusted (stepped up or stepped down) as per requirement. iii. AC current in a circuit can be controlled using (almost) wattless devices like the choke coil. iv. AC is easier to generate. ½ + ½ 1 9. According to Ampere’s circuital Law ? =µ 0 I Applying ampere’s circuital law to fig (a) we see that, during charging, the right hand side in Ampere’s circuital law equals µ 0 I However on applying it to the surfaces of the fig (b) or fig (c), the right hand side is zero. Hence, there is a contradiction. We can remove the contradiction by assuming that there exists a current (associated with the changing electric field during charging), known as the displacement current. When this current ( = Ø ) is added on the right hand side, Ampere’s circuital law, the inconsisitency disappears. It was, therefore necessary, to generalize the Ampere’s circuital law, as ? = µ 0 + µ 0 ? Ø ! [Note : If the student does the reasoning by using the (detailed) mathematics, relevant to displacement current, award full 2 marks ] ½ ½ ½ ½ 2 Statement of Ampere’s circuital law ½ Showing inconsistency during the process of charging 1 Displacement Current ½ Page 3 Outside Delhi SET III MARKING SCHEME SET 55/3 Q. No. Expected Answer / Value Points Marks Total Marks 1. Anticlockwise 1 1 2. Metal A The minimum frequency, at which photoemission starts, is more for metal A Alternatively: Work function of A is more. ½ ½ 1 3. Definition : One ampere is the value of steady current which when maintained in each of the two very long, straight, parallel conductors of negligible cross section and placed one metre apart in vaccum, would produce on each of these conductors a force equal of 2 x 10 -7 N/m of its length. Alternatively If the student writes F = L and says that when 1 R= 1 meter and L = 1 meter, then F= 2 x 10 -7 N Award full 1 mark Alternatively If the student draws any one of the two diagram, as shown , Award full 1 mark 1 1 4. As a diverging lens Light rays diverge on going from a rarer to a denser medium. [Alternatively Also accept the reason given on the basis of lens marker’s formula.] ½ ½ 1 5. At the point of intersection of the two field lines, there will be two directions for the electric field. This is not acceptable. 1 1 6. Short radio waves (or) microwaves 1 1 7. Neutrinos are neutral (chargeless), (almost) massless particles that hardly 1 Outside Delhi SET III interact with matter. Alternatively The neutrinos can penetrate large quantity of matter without any interaction OR Neutrinos are chargeless and (almost) massless particles. 1 8. Any two of the following (or any other correct) reasons : i. AC can be transmitted with much lower energy losses as compared to DC ii. AC voltage can be adjusted (stepped up or stepped down) as per requirement. iii. AC current in a circuit can be controlled using (almost) wattless devices like the choke coil. iv. AC is easier to generate. ½ + ½ 1 9. According to Ampere’s circuital Law ? =µ 0 I Applying ampere’s circuital law to fig (a) we see that, during charging, the right hand side in Ampere’s circuital law equals µ 0 I However on applying it to the surfaces of the fig (b) or fig (c), the right hand side is zero. Hence, there is a contradiction. We can remove the contradiction by assuming that there exists a current (associated with the changing electric field during charging), known as the displacement current. When this current ( = Ø ) is added on the right hand side, Ampere’s circuital law, the inconsisitency disappears. It was, therefore necessary, to generalize the Ampere’s circuital law, as ? = µ 0 + µ 0 ? Ø ! [Note : If the student does the reasoning by using the (detailed) mathematics, relevant to displacement current, award full 2 marks ] ½ ½ ½ ½ 2 Statement of Ampere’s circuital law ½ Showing inconsistency during the process of charging 1 Displacement Current ½ Outside Delhi SET III 10. I = AneV d V d = .# .$ % & '( % .)% & '* %+% & , = 7.5 X 10 -4 m/s ½ ½ 1 2 11. The relation between V and I is V = E – Ir Hence, the graph, between V and I, has the form shown below. For point A, I=0, Hence, V A = E For point B, V=0, Hence, E=I B r Therefore, r = - . Alternatively: emf (E) equals the intercept on the vertical axis. Internal resistance ( r) equals the negative of the slope of the graph. ½ ½ ½ ½ 2 12. Energy stored in a capacitor = /0 = 10 = 2 3 (any one ) Capacitance of the (parallel) combination = C+C=2C Here, total charge, Q, remains the same initial energy = 2 3 And final energy = 2 3 ? 56789 :7:;<= 676689 :7:;<= = [Note : If the student does the correct calculations by assuming the voltage across the ½ ½ ½ ½ Formula ½ Calculation of drift velocity 1 ½ Relation between V and I ½ Graph ½ Determination of emf and internal resistance ½ + ½ Formula for energy stored ½ New value of capacitance ½ Calculation of ratio 1 Page 4 Outside Delhi SET III MARKING SCHEME SET 55/3 Q. No. Expected Answer / Value Points Marks Total Marks 1. Anticlockwise 1 1 2. Metal A The minimum frequency, at which photoemission starts, is more for metal A Alternatively: Work function of A is more. ½ ½ 1 3. Definition : One ampere is the value of steady current which when maintained in each of the two very long, straight, parallel conductors of negligible cross section and placed one metre apart in vaccum, would produce on each of these conductors a force equal of 2 x 10 -7 N/m of its length. Alternatively If the student writes F = L and says that when 1 R= 1 meter and L = 1 meter, then F= 2 x 10 -7 N Award full 1 mark Alternatively If the student draws any one of the two diagram, as shown , Award full 1 mark 1 1 4. As a diverging lens Light rays diverge on going from a rarer to a denser medium. [Alternatively Also accept the reason given on the basis of lens marker’s formula.] ½ ½ 1 5. At the point of intersection of the two field lines, there will be two directions for the electric field. This is not acceptable. 1 1 6. Short radio waves (or) microwaves 1 1 7. Neutrinos are neutral (chargeless), (almost) massless particles that hardly 1 Outside Delhi SET III interact with matter. Alternatively The neutrinos can penetrate large quantity of matter without any interaction OR Neutrinos are chargeless and (almost) massless particles. 1 8. Any two of the following (or any other correct) reasons : i. AC can be transmitted with much lower energy losses as compared to DC ii. AC voltage can be adjusted (stepped up or stepped down) as per requirement. iii. AC current in a circuit can be controlled using (almost) wattless devices like the choke coil. iv. AC is easier to generate. ½ + ½ 1 9. According to Ampere’s circuital Law ? =µ 0 I Applying ampere’s circuital law to fig (a) we see that, during charging, the right hand side in Ampere’s circuital law equals µ 0 I However on applying it to the surfaces of the fig (b) or fig (c), the right hand side is zero. Hence, there is a contradiction. We can remove the contradiction by assuming that there exists a current (associated with the changing electric field during charging), known as the displacement current. When this current ( = Ø ) is added on the right hand side, Ampere’s circuital law, the inconsisitency disappears. It was, therefore necessary, to generalize the Ampere’s circuital law, as ? = µ 0 + µ 0 ? Ø ! [Note : If the student does the reasoning by using the (detailed) mathematics, relevant to displacement current, award full 2 marks ] ½ ½ ½ ½ 2 Statement of Ampere’s circuital law ½ Showing inconsistency during the process of charging 1 Displacement Current ½ Outside Delhi SET III 10. I = AneV d V d = .# .$ % & '( % .)% & '* %+% & , = 7.5 X 10 -4 m/s ½ ½ 1 2 11. The relation between V and I is V = E – Ir Hence, the graph, between V and I, has the form shown below. For point A, I=0, Hence, V A = E For point B, V=0, Hence, E=I B r Therefore, r = - . Alternatively: emf (E) equals the intercept on the vertical axis. Internal resistance ( r) equals the negative of the slope of the graph. ½ ½ ½ ½ 2 12. Energy stored in a capacitor = /0 = 10 = 2 3 (any one ) Capacitance of the (parallel) combination = C+C=2C Here, total charge, Q, remains the same initial energy = 2 3 And final energy = 2 3 ? 56789 :7:;<= 676689 :7:;<= = [Note : If the student does the correct calculations by assuming the voltage across the ½ ½ ½ ½ Formula ½ Calculation of drift velocity 1 ½ Relation between V and I ½ Graph ½ Determination of emf and internal resistance ½ + ½ Formula for energy stored ½ New value of capacitance ½ Calculation of ratio 1 Outside Delhi SET III (i) Parallel or (ii) Series combination to remain constant (=V) and obtain the answers as (i) 2:1 or (ii) 1:2 , award full marks ] 2 13. As per Rutherford’s model >? ; = @A B: ; C = @A B: ; Total energy = P.E +K.E. = - @A B: ; + C = - . @A B: ; = = - DA B: ; Negative Sign implies that Electron – nucleus form a bound system. Alternatively Electron – nucleus form an attractive system) OR For the electron, we have Bohr’s Postulate (mvr= 7E ) >? ; = @A B: ; and mvr= 7E C 7 E @ and mC = @A F r = A 7 E B: > Bohr’s radius (for n = 1) = G H / IF ½ ½ ½ ½ ½ ½ ½ ½ 2 2 Derivation of energy expression 1 ½ Significance of negative sign ½ Bohr’s Postulate ½ Derivation of radius of nth orbit 1 Bohr’s radius ½ Page 5 Outside Delhi SET III MARKING SCHEME SET 55/3 Q. No. Expected Answer / Value Points Marks Total Marks 1. Anticlockwise 1 1 2. Metal A The minimum frequency, at which photoemission starts, is more for metal A Alternatively: Work function of A is more. ½ ½ 1 3. Definition : One ampere is the value of steady current which when maintained in each of the two very long, straight, parallel conductors of negligible cross section and placed one metre apart in vaccum, would produce on each of these conductors a force equal of 2 x 10 -7 N/m of its length. Alternatively If the student writes F = L and says that when 1 R= 1 meter and L = 1 meter, then F= 2 x 10 -7 N Award full 1 mark Alternatively If the student draws any one of the two diagram, as shown , Award full 1 mark 1 1 4. As a diverging lens Light rays diverge on going from a rarer to a denser medium. [Alternatively Also accept the reason given on the basis of lens marker’s formula.] ½ ½ 1 5. At the point of intersection of the two field lines, there will be two directions for the electric field. This is not acceptable. 1 1 6. Short radio waves (or) microwaves 1 1 7. Neutrinos are neutral (chargeless), (almost) massless particles that hardly 1 Outside Delhi SET III interact with matter. Alternatively The neutrinos can penetrate large quantity of matter without any interaction OR Neutrinos are chargeless and (almost) massless particles. 1 8. Any two of the following (or any other correct) reasons : i. AC can be transmitted with much lower energy losses as compared to DC ii. AC voltage can be adjusted (stepped up or stepped down) as per requirement. iii. AC current in a circuit can be controlled using (almost) wattless devices like the choke coil. iv. AC is easier to generate. ½ + ½ 1 9. According to Ampere’s circuital Law ? =µ 0 I Applying ampere’s circuital law to fig (a) we see that, during charging, the right hand side in Ampere’s circuital law equals µ 0 I However on applying it to the surfaces of the fig (b) or fig (c), the right hand side is zero. Hence, there is a contradiction. We can remove the contradiction by assuming that there exists a current (associated with the changing electric field during charging), known as the displacement current. When this current ( = Ø ) is added on the right hand side, Ampere’s circuital law, the inconsisitency disappears. It was, therefore necessary, to generalize the Ampere’s circuital law, as ? = µ 0 + µ 0 ? Ø ! [Note : If the student does the reasoning by using the (detailed) mathematics, relevant to displacement current, award full 2 marks ] ½ ½ ½ ½ 2 Statement of Ampere’s circuital law ½ Showing inconsistency during the process of charging 1 Displacement Current ½ Outside Delhi SET III 10. I = AneV d V d = .# .$ % & '( % .)% & '* %+% & , = 7.5 X 10 -4 m/s ½ ½ 1 2 11. The relation between V and I is V = E – Ir Hence, the graph, between V and I, has the form shown below. For point A, I=0, Hence, V A = E For point B, V=0, Hence, E=I B r Therefore, r = - . Alternatively: emf (E) equals the intercept on the vertical axis. Internal resistance ( r) equals the negative of the slope of the graph. ½ ½ ½ ½ 2 12. Energy stored in a capacitor = /0 = 10 = 2 3 (any one ) Capacitance of the (parallel) combination = C+C=2C Here, total charge, Q, remains the same initial energy = 2 3 And final energy = 2 3 ? 56789 :7:;<= 676689 :7:;<= = [Note : If the student does the correct calculations by assuming the voltage across the ½ ½ ½ ½ Formula ½ Calculation of drift velocity 1 ½ Relation between V and I ½ Graph ½ Determination of emf and internal resistance ½ + ½ Formula for energy stored ½ New value of capacitance ½ Calculation of ratio 1 Outside Delhi SET III (i) Parallel or (ii) Series combination to remain constant (=V) and obtain the answers as (i) 2:1 or (ii) 1:2 , award full marks ] 2 13. As per Rutherford’s model >? ; = @A B: ; C = @A B: ; Total energy = P.E +K.E. = - @A B: ; + C = - . @A B: ; = = - DA B: ; Negative Sign implies that Electron – nucleus form a bound system. Alternatively Electron – nucleus form an attractive system) OR For the electron, we have Bohr’s Postulate (mvr= 7E ) >? ; = @A B: ; and mvr= 7E C 7 E @ and mC = @A F r = A 7 E B: > Bohr’s radius (for n = 1) = G H / IF ½ ½ ½ ½ ½ ½ ½ ½ 2 2 Derivation of energy expression 1 ½ Significance of negative sign ½ Bohr’s Postulate ½ Derivation of radius of nth orbit 1 Bohr’s radius ½ Outside Delhi SET III 14. A paramagnetic material tends to move from weaker to stronger regions of the magnetic field and hence increases the number of lines of magnetic field passing through it. [Alternatively: A paramagnetic material, dipole moments are induced in the direction of the field.] A diamagnetic material tends to move from stronger to weaker regions of the magnetic field and hence, decreases the number of lines of magnetic field passing through it. [Alternatively: A diamagnetic material, dipole moments are induced in the opposite direction of the field.] [Note: If the student just writes that a paramagnetic material has a small positive susceptibility (0< X < J ) and a diamagnetic material has a negative susceptibility (-1 X < ), award the ½ mark for the second part of the question.] ½ ½ ½ ½ 2 15. Working: During one half of the input AC, the diode is forward biased and a current flows through R L . During the other half of the input AC, the diode is reverse biased and no current flows through the load R L. Hence, the given AC input is rectified [Note : If the student just draws the waveforms, for the input AC voltage and output voltage (without giving any explanation) (award ½ mark only for “working”) 1 ½ ½ 2 Diagrams ½ + ½ Explanations ½ + ½ Circuit diagram 1 Working 1Read More

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