Page 1 Outside Delhi SET I MARKING SCHEME SET 55/1 Q. No Expected Answer / Value Points Marks Total Marks 1. 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 2. X – / 1 1 3. Force decreases 1 1 4. Intensity of radiation depends on the number of photons incident per unit area per unit time. [Note: Also accept the definition: ‘number of quanta of radiation per unit area per unit time’. Also accept if the student writes: All photons, of a particular frequency, have the same kinetic energy and momentum, irrespective of the intensity of incident radiation. Alternatively The amount of light energy / Photon energy, incident per metre square per second is called intensity of radiation SI Unit : W/m 2 or J/(s- m 2 ) ½ ½ 1 5. Clockwise Alternatively 1 1 Page 2 Outside Delhi SET I MARKING SCHEME SET 55/1 Q. No Expected Answer / Value Points Marks Total Marks 1. 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 2. X – / 1 1 3. Force decreases 1 1 4. Intensity of radiation depends on the number of photons incident per unit area per unit time. [Note: Also accept the definition: ‘number of quanta of radiation per unit area per unit time’. Also accept if the student writes: All photons, of a particular frequency, have the same kinetic energy and momentum, irrespective of the intensity of incident radiation. Alternatively The amount of light energy / Photon energy, incident per metre square per second is called intensity of radiation SI Unit : W/m 2 or J/(s- m 2 ) ½ ½ 1 5. Clockwise Alternatively 1 1 Outside Delhi SET I 6. Neutrinos are neutral (chargeless), (almost) massless particles that hardly interact with matter. Alternatively The neutrinos can penetrate large quantity of matter without any interaction OR Neutrinos are chargeless and (almost) massless particles. 1 1 7. 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 8. 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 9. As per Rutherford’s model = = Total energy = P.E +K.E. = - + = - . = = - 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= !" ) ½ ½ ½ ½ ½ 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 3 Outside Delhi SET I MARKING SCHEME SET 55/1 Q. No Expected Answer / Value Points Marks Total Marks 1. 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 2. X – / 1 1 3. Force decreases 1 1 4. Intensity of radiation depends on the number of photons incident per unit area per unit time. [Note: Also accept the definition: ‘number of quanta of radiation per unit area per unit time’. Also accept if the student writes: All photons, of a particular frequency, have the same kinetic energy and momentum, irrespective of the intensity of incident radiation. Alternatively The amount of light energy / Photon energy, incident per metre square per second is called intensity of radiation SI Unit : W/m 2 or J/(s- m 2 ) ½ ½ 1 5. Clockwise Alternatively 1 1 Outside Delhi SET I 6. Neutrinos are neutral (chargeless), (almost) massless particles that hardly interact with matter. Alternatively The neutrinos can penetrate large quantity of matter without any interaction OR Neutrinos are chargeless and (almost) massless particles. 1 1 7. 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 8. 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 9. As per Rutherford’s model = = Total energy = P.E +K.E. = - + = - . = = - 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= !" ) ½ ½ ½ ½ ½ 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 I = and mvr= !" ! " and m = # r = ! " Bohr’s radius (for n = 1) = $ % & / '# ½ ½ ½ 2 10. Energy stored in a capacitor = () = *) = + , (any one ) Capacitance of the (parallel) combination = C+C=2C Here, total charge, Q, remains the same initial energy = + , And final energy = + , ? ./!01 !23 /!/4/01 !23 = [Note : If the student does the correct calculations by assuming the voltage across the (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 ] ½ ½ ½ ½ 2 11. According to Ampere’s circuital Law ?6 78 9: 8 =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. ½ ½ ½ Statement of Ampere’s circuital law ½ Showing inconsistency during the process of charging 1 Displacement Current ½ Formula for energy stored ½ New value of capacitance ½ Calculation of ratio 1 Page 4 Outside Delhi SET I MARKING SCHEME SET 55/1 Q. No Expected Answer / Value Points Marks Total Marks 1. 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 2. X – / 1 1 3. Force decreases 1 1 4. Intensity of radiation depends on the number of photons incident per unit area per unit time. [Note: Also accept the definition: ‘number of quanta of radiation per unit area per unit time’. Also accept if the student writes: All photons, of a particular frequency, have the same kinetic energy and momentum, irrespective of the intensity of incident radiation. Alternatively The amount of light energy / Photon energy, incident per metre square per second is called intensity of radiation SI Unit : W/m 2 or J/(s- m 2 ) ½ ½ 1 5. Clockwise Alternatively 1 1 Outside Delhi SET I 6. Neutrinos are neutral (chargeless), (almost) massless particles that hardly interact with matter. Alternatively The neutrinos can penetrate large quantity of matter without any interaction OR Neutrinos are chargeless and (almost) massless particles. 1 1 7. 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 8. 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 9. As per Rutherford’s model = = Total energy = P.E +K.E. = - + = - . = = - 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= !" ) ½ ½ ½ ½ ½ 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 I = and mvr= !" ! " and m = # r = ! " Bohr’s radius (for n = 1) = $ % & / '# ½ ½ ½ 2 10. Energy stored in a capacitor = () = *) = + , (any one ) Capacitance of the (parallel) combination = C+C=2C Here, total charge, Q, remains the same initial energy = + , And final energy = + , ? ./!01 !23 /!/4/01 !23 = [Note : If the student does the correct calculations by assuming the voltage across the (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 ] ½ ½ ½ ½ 2 11. According to Ampere’s circuital Law ?6 78 9: 8 =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. ½ ½ ½ Statement of Ampere’s circuital law ½ Showing inconsistency during the process of charging 1 Displacement Current ½ Formula for energy stored ½ New value of capacitance ½ Calculation of ratio 1 Outside Delhi SET I 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 = ;4 ) 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 ?6 78 9: 8 = µ 0 > + µ 0 ? % 9Ø @ 9A [Note : If the student does the reasoning by using the (detailed) mathematics, relevant to displacement current, award full 2 marks ] ½ 2 12. 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 = B C 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 13. 1 Relation between V and I ½ Graph ½ Determination of emf and internal resistance ½ + ½ Circuit diagram 1 Working 1 Page 5 Outside Delhi SET I MARKING SCHEME SET 55/1 Q. No Expected Answer / Value Points Marks Total Marks 1. 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 2. X – / 1 1 3. Force decreases 1 1 4. Intensity of radiation depends on the number of photons incident per unit area per unit time. [Note: Also accept the definition: ‘number of quanta of radiation per unit area per unit time’. Also accept if the student writes: All photons, of a particular frequency, have the same kinetic energy and momentum, irrespective of the intensity of incident radiation. Alternatively The amount of light energy / Photon energy, incident per metre square per second is called intensity of radiation SI Unit : W/m 2 or J/(s- m 2 ) ½ ½ 1 5. Clockwise Alternatively 1 1 Outside Delhi SET I 6. Neutrinos are neutral (chargeless), (almost) massless particles that hardly interact with matter. Alternatively The neutrinos can penetrate large quantity of matter without any interaction OR Neutrinos are chargeless and (almost) massless particles. 1 1 7. 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 8. 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 9. As per Rutherford’s model = = Total energy = P.E +K.E. = - + = - . = = - 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= !" ) ½ ½ ½ ½ ½ 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 I = and mvr= !" ! " and m = # r = ! " Bohr’s radius (for n = 1) = $ % & / '# ½ ½ ½ 2 10. Energy stored in a capacitor = () = *) = + , (any one ) Capacitance of the (parallel) combination = C+C=2C Here, total charge, Q, remains the same initial energy = + , And final energy = + , ? ./!01 !23 /!/4/01 !23 = [Note : If the student does the correct calculations by assuming the voltage across the (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 ] ½ ½ ½ ½ 2 11. According to Ampere’s circuital Law ?6 78 9: 8 =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. ½ ½ ½ Statement of Ampere’s circuital law ½ Showing inconsistency during the process of charging 1 Displacement Current ½ Formula for energy stored ½ New value of capacitance ½ Calculation of ratio 1 Outside Delhi SET I 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 = ;4 ) 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 ?6 78 9: 8 = µ 0 > + µ 0 ? % 9Ø @ 9A [Note : If the student does the reasoning by using the (detailed) mathematics, relevant to displacement current, award full 2 marks ] ½ 2 12. 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 = B C 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 13. 1 Relation between V and I ½ Graph ½ Determination of emf and internal resistance ½ + ½ Circuit diagram 1 Working 1 Outside Delhi SET I 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”) ½ ½ 2 14. I = neA V d V d = DEF = .G H I J K I .LI J MN I .J I J MO m/s = 1.048 I10 QR m/s ( 1mm/s) ½ ½ 1 2 15. [Note : If the student just writes (without drawing any diagram) that angle of incidence for both rays ‘1’ and ‘2’ on face AC equals 45 o , and says that it is less than critical angle for ray ‘1’ (which therefore gets refracted) and more than critical angle for ray ‘2’ (which undergoes total internal reflection), award only ½ + ½ marks.] 1 1 2 16. Transducer : Any device that converts one form of energy to another. Repeater : A repeater accepts the signal from the transmitter, amplifies and retransmits it to the receiver. 1 1 2 17. Function of Transducer 1 Function of Repeater 1 Diagrams ½ + ½ Explanations ½ + ½ Tracing of Path of Ray 1 1 Tracing of Path of Ray 2 1 Formula ½ Substitution and calculation ½ + 1Read More

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