Page 1 Outside Delhi SET II MARKING SCHEME SET 55/2 Q. No. Expected Answer / Value Points Marks Total Marks 1. Clockwise 1 1 2. 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 3. 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 Page 2 Outside Delhi SET II MARKING SCHEME SET 55/2 Q. No. Expected Answer / Value Points Marks Total Marks 1. Clockwise 1 1 2. 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 3. 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 Outside Delhi SET II 4. They start from positive charges and end on negative charges [Alternatively: Electric field is conservative in nature.] 1 1 5. Converging lens Light rays converge, on moving from denser to rarer medium. Alternatively: Note: If explained on the basis of lens makers formulae, award 1 mark. ½ ½ 1 6. Metal A ? work functions ! and ! " # > ! " ½ ½ 1 7. Infrared radiation 1 1 8. 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 9. I = AneV d ?V d = .& .' ( ) *+ (,( ) - ( ..( ) */ = 5 X 10 -4 m/s ½ ½ 1 2 10. According to Ampereâ€™s circuital Law ?1 2 3 45 3 =µ 0 I Applying ampereâ€™s circuital law to fig (a) we see that, during charging, the right hand side in Ampereâ€™s circuital law equals µ 0 I ½ ½ Formula ½ Calculation of drift velocity 1 ½ Statement of Ampereâ€™s circuital law ½ Showing inconsistency during the process of charging 1 Displacement Current ½ Page 3 Outside Delhi SET II MARKING SCHEME SET 55/2 Q. No. Expected Answer / Value Points Marks Total Marks 1. Clockwise 1 1 2. 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 3. 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 Outside Delhi SET II 4. They start from positive charges and end on negative charges [Alternatively: Electric field is conservative in nature.] 1 1 5. Converging lens Light rays converge, on moving from denser to rarer medium. Alternatively: Note: If explained on the basis of lens makers formulae, award 1 mark. ½ ½ 1 6. Metal A ? work functions ! and ! " # > ! " ½ ½ 1 7. Infrared radiation 1 1 8. 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 9. I = AneV d ?V d = .& .' ( ) *+ (,( ) - ( ..( ) */ = 5 X 10 -4 m/s ½ ½ 1 2 10. According to Ampereâ€™s circuital Law ?1 2 3 45 3 =µ 0 I Applying ampereâ€™s circuital law to fig (a) we see that, during charging, the right hand side in Ampereâ€™s circuital law equals µ 0 I ½ ½ Formula ½ Calculation of drift velocity 1 ½ Statement of Ampereâ€™s circuital law ½ Showing inconsistency during the process of charging 1 Displacement Current ½ Outside Delhi SET II 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 ( = 6Ø 8 69 ) is added on the right hand side, Ampereâ€™s circuital law, the inconsisitency disappears. It was, therefore necessary, to generalize the Ampereâ€™s circuital law, as ?1 2 3 45 3 = µ 0 : + µ 0 ? " 4Ø < 4= [Note : If the student does the reasoning by using the (detailed) mathematics, relevant to displacement current, award full 2 marks ] ½ ½ 2 11. As per Rutherfordâ€™s model >? @ = AB CD @ ! = AB CD @ Total energy = P.E +K.E. = - AB CD @ + ! = - . AB CD @ = = - &B CD @ Negative Sign implies that Electron â€“ nucleus form a bound system. Alternatively Electron â€“ nucleus form an attractive system) OR For the electron, we have Bohrâ€™s Postulate (mvr= EF ) >? @ = AB CD @ and mvr= EF ! E F A and m! = AB G ½ ½ ½ ½ ½ ½ 2 Derivation of energy expression 1 ½ Significance of negative sign ½ Bohrâ€™s Postulate ½ Derivation of radius of nth orbit 1 Bohrâ€™s radius ½ Page 4 Outside Delhi SET II MARKING SCHEME SET 55/2 Q. No. Expected Answer / Value Points Marks Total Marks 1. Clockwise 1 1 2. 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 3. 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 Outside Delhi SET II 4. They start from positive charges and end on negative charges [Alternatively: Electric field is conservative in nature.] 1 1 5. Converging lens Light rays converge, on moving from denser to rarer medium. Alternatively: Note: If explained on the basis of lens makers formulae, award 1 mark. ½ ½ 1 6. Metal A ? work functions ! and ! " # > ! " ½ ½ 1 7. Infrared radiation 1 1 8. 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 9. I = AneV d ?V d = .& .' ( ) *+ (,( ) - ( ..( ) */ = 5 X 10 -4 m/s ½ ½ 1 2 10. According to Ampereâ€™s circuital Law ?1 2 3 45 3 =µ 0 I Applying ampereâ€™s circuital law to fig (a) we see that, during charging, the right hand side in Ampereâ€™s circuital law equals µ 0 I ½ ½ Formula ½ Calculation of drift velocity 1 ½ Statement of Ampereâ€™s circuital law ½ Showing inconsistency during the process of charging 1 Displacement Current ½ Outside Delhi SET II 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 ( = 6Ø 8 69 ) is added on the right hand side, Ampereâ€™s circuital law, the inconsisitency disappears. It was, therefore necessary, to generalize the Ampereâ€™s circuital law, as ?1 2 3 45 3 = µ 0 : + µ 0 ? " 4Ø < 4= [Note : If the student does the reasoning by using the (detailed) mathematics, relevant to displacement current, award full 2 marks ] ½ ½ 2 11. As per Rutherfordâ€™s model >? @ = AB CD @ ! = AB CD @ Total energy = P.E +K.E. = - AB CD @ + ! = - . AB CD @ = = - &B CD @ Negative Sign implies that Electron â€“ nucleus form a bound system. Alternatively Electron â€“ nucleus form an attractive system) OR For the electron, we have Bohrâ€™s Postulate (mvr= EF ) >? @ = AB CD @ and mvr= EF ! E F A and m! = AB G ½ ½ ½ ½ ½ ½ 2 Derivation of energy expression 1 ½ Significance of negative sign ½ Bohrâ€™s Postulate ½ Derivation of radius of nth orbit 1 Bohrâ€™s radius ½ Outside Delhi SET II r = B E F CD > Bohrâ€™s radius (for n = 1) = H " / IG ½ ½ 2 12. [Note : If the student does not draw the diagram but just writes that the angle of incidence for both rays â€˜2â€™ and â€˜1â€™, on face AC = 45 o and says that it is less than critical angle for ray â€˜1â€™ which therefore gets refracted and more than critical angle for ray â€˜2â€™,which undergoes total internal reflection; Award ½ + ½ marks] 1+1 2 13. Transmitter: A transmitter processes the incoming message signal so as to make it suitable for transmission through a channel and subsequent reception. Modulator: It is a device in which the amplitude/ (frequency/phase) of a high frequency carrier wave is made to change in accordance with the message signal through (appropriate) superposition. 1 1 2 Function of transmitter 1 Function of modulator 1 Tracing the path of the two rays 1 + 1 Page 5 Outside Delhi SET II MARKING SCHEME SET 55/2 Q. No. Expected Answer / Value Points Marks Total Marks 1. Clockwise 1 1 2. 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 3. 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 Outside Delhi SET II 4. They start from positive charges and end on negative charges [Alternatively: Electric field is conservative in nature.] 1 1 5. Converging lens Light rays converge, on moving from denser to rarer medium. Alternatively: Note: If explained on the basis of lens makers formulae, award 1 mark. ½ ½ 1 6. Metal A ? work functions ! and ! " # > ! " ½ ½ 1 7. Infrared radiation 1 1 8. 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 9. I = AneV d ?V d = .& .' ( ) *+ (,( ) - ( ..( ) */ = 5 X 10 -4 m/s ½ ½ 1 2 10. According to Ampereâ€™s circuital Law ?1 2 3 45 3 =µ 0 I Applying ampereâ€™s circuital law to fig (a) we see that, during charging, the right hand side in Ampereâ€™s circuital law equals µ 0 I ½ ½ Formula ½ Calculation of drift velocity 1 ½ Statement of Ampereâ€™s circuital law ½ Showing inconsistency during the process of charging 1 Displacement Current ½ Outside Delhi SET II 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 ( = 6Ø 8 69 ) is added on the right hand side, Ampereâ€™s circuital law, the inconsisitency disappears. It was, therefore necessary, to generalize the Ampereâ€™s circuital law, as ?1 2 3 45 3 = µ 0 : + µ 0 ? " 4Ø < 4= [Note : If the student does the reasoning by using the (detailed) mathematics, relevant to displacement current, award full 2 marks ] ½ ½ 2 11. As per Rutherfordâ€™s model >? @ = AB CD @ ! = AB CD @ Total energy = P.E +K.E. = - AB CD @ + ! = - . AB CD @ = = - &B CD @ Negative Sign implies that Electron â€“ nucleus form a bound system. Alternatively Electron â€“ nucleus form an attractive system) OR For the electron, we have Bohrâ€™s Postulate (mvr= EF ) >? @ = AB CD @ and mvr= EF ! E F A and m! = AB G ½ ½ ½ ½ ½ ½ 2 Derivation of energy expression 1 ½ Significance of negative sign ½ Bohrâ€™s Postulate ½ Derivation of radius of nth orbit 1 Bohrâ€™s radius ½ Outside Delhi SET II r = B E F CD > Bohrâ€™s radius (for n = 1) = H " / IG ½ ½ 2 12. [Note : If the student does not draw the diagram but just writes that the angle of incidence for both rays â€˜2â€™ and â€˜1â€™, on face AC = 45 o and says that it is less than critical angle for ray â€˜1â€™ which therefore gets refracted and more than critical angle for ray â€˜2â€™,which undergoes total internal reflection; Award ½ + ½ marks] 1+1 2 13. Transmitter: A transmitter processes the incoming message signal so as to make it suitable for transmission through a channel and subsequent reception. Modulator: It is a device in which the amplitude/ (frequency/phase) of a high frequency carrier wave is made to change in accordance with the message signal through (appropriate) superposition. 1 1 2 Function of transmitter 1 Function of modulator 1 Tracing the path of the two rays 1 + 1 Outside Delhi SET II 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. Condition : The transistor must be operated close to the centre of its active region. Alternatively The base- emitter junction of the transistor must be (suitably) forward biased and the collector â€“ emitter junction must be (suitably) reverse biased. 1 ½ ½ 2 Diagrams ½ + ½ Explanations ½ + ½ Circuit diagram 1 ½ Condition ½Read More

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